Solutions Manual for Engineering Economy 17th Global Edition By William Sullivan Elin Wicks Patrick

Solutions to Chapter 1 Problems A Note To Instructors: Because of volatile energy prices in today's world, the instructor is encouraged to vary energy prices in affected problems (e.g. the price of a gallon of gasoline) plus and minus 50 percent and ask students to determine whether this range of prices changes the recommendation in the problem. This should make for stimulating inclass discussion of the results.

1-1

(a)

66 million tons per year × (0.05) = 3.3 million tons per year of reduced greenhouse gas $2.1 billion ÷ 3.3 million tons per year = $636.36 per ton

(b)

3 billion tons per year × (0.02) = 60 million tons per year of reduced greenhouse gas $2.1 billion/3.3 million tons = $X billion/60 million tons X = $40.65 billion

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1-2

The primary disadvantage is the installed cost of $8,000 (rather expensive!). A key advantage of synthetic turf is the annual savings due to lawn maintenance. This may be worth up to $1,000 per year when periodic lawn mowing, watering, and feeding are avoided.

See what other monetized advantages and disadvantages members of your class can identify. Be creative!

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1-3

Cost per Watt-hour = $0.90/1.5 Watt-hours = $0.60 per Watt-hour At a cost of $0.60 per Watt-hour, it would cost (1,000)($0.60 per Watt-hour) = $600 per kilo Watt-hour for power from a single AAA battery. This is 3,000 times more costly than energy from your local utility. No wonder we turn off our battery operated devices when we're not using them!

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1-4

At first glance, Tyler’s options seem to be: (1) immediately pay $803 to the owner of the other person’s car or (2) submit a claim to the insurance company. If Tyler keeps his Nissan for five more years (an assumption), the cost of option 2 is $500 + ($60 × 2 payments/year) × 5 years = $1,100. This amount is more than paying $803 out-of-pocket, so Tyler appears to have made the most economical choice.

What we don’t know in this problem is the age and condition of the other person’s car. If we assume it’s a clunker, another option for Tyler is to offer to buy the other person’s car and fix it himself and then sell it over the internet. Or Tyler could donate the unrepaired (or repaired) car to his favorite charity.

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1-5

(a)

18,000 miles per year/25 mpg = 720 gallons per year of E20 Savings = 720 gallons per year ($4.40 ‒ $3.73) = $482.4 per year

(b)

Gasoline saved = 0.20 (720 gal/yr)(1,000,000 people) = 144 million gallons per year

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1-6

An economic tradeoff is trading (exchanging) one economic outcome for another. Examples include (a) whether to ride the bus or drive your car to campus today, (b) whether to pack your lunch or eat at a local fast food restaurant and (c) whether to carry an umbrella (an inconvenience) or hope for no rain if you don’t take your umbrella.

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1-7

The cost of a wood-framed home is 2,500 ft2 × $140/ft2 = $350,000. The ICF home will cost $350,000 (1.10) = $385,000. So, the cost difference to construct an ICF home is $35,000.

The energy savings per month is 5% × $200 = $100 per month. Therefore, it will take 350 months to pay back the extra ICF construction cost through monthly energy savings.

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1-8

Increased earnings with new technology = $2,000,000(0.05) = $100,000

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1-9

Strategy 1: Change oil every 4,000 miles. Cost = (16,000/4,000)($35) = $140/year Strategy 2: Change oil every 8,000 miles. Cost = (16,000/8,000)($35) = $70/year Savings = $70 per year

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1-10

The halogen bulb will use 60 × 30 = 1,800 kWh over 30,000 hours. Electricity costs with a halogen bulb will be (1,800 − 300)($0.2) = $300 higher than with an LED bulb. Additionally, the halogen bulb will have to be replaced every 1,000 hours, which will cost (30)($5) = $150, which is $120 more than the cost of a single LED bulb. Therefore, over the span of 30,000 usage hours, the LED bulb saves $420.

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1-11

100 gallons × $4.50 per gallon = $450 saved over 45,000 miles of driving. This comes down to $450 ÷ 45,000 = $0.01 per mile driven. So, Brand A saves a penny for each mile driven.

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1-12

(a)

Problem:

To find the least expensive method for setting up capacity to produce drill bits.

(b)

Assumptions: The revenue per unit will be the same for either machine; startup costs are negligible; breakdowns are not frequent; previous employee’s data are correct; drill bits are manufactured the same way regardless of the alternative chosen; in-house technicians can modify the old machine so its life span will match that of the new machine; neither machine has any resale value; there is no union to lobby for inhouse work; etc.

(c)

Alternatives: (1) Modify the old machine for producing the new drill bit (using in-house technicians); (2) Buy a new machine for $450,000; (3) Get McDonald Inc. to modify the machine; (4) Outsource the work to another company.

(d)

Criterion:

Least cost in dollars for the anticipated production runs, given that quality and delivery time are essentially unaffected (i.e., not compromised).

(e)

Risks:

The old machine could be less reliable than a new one; the old machine could cause environmental hazards; fixing the old machine in-house could prove to be unsatisfactory; the old machine could be less safe than a new one; etc.

(f)

Non-monetary Considerations: Safety; environmental concerns; quality/reliability differences; “flexibility” of a new machine; job security for in-house work; image to outside companies by having a new technology (machine); etc.

(g)

Post Audit:

Did either machine (or outsourcing) fail to deliver high quality product on time?

Were maintenance costs of the machines acceptable? Did the total production costs allow an acceptable profit to be made?

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1-13

(a)

Problem A: Subject to energy that Farah is willing/able to exert, her ability to do well in the Masters program, and expectations about the future state of the job market, Problem A might be "How can Farah motivate herself to complete the Masters program and get a job after two years, which is at least as good as the one in hand?”

Problem B: Subject to knowledge of the job market, mobility, and professional ambition, as well as the probability of winning another scholarship a few years from now, Problem B could be "How can Farah use the new job as a springboard to better pay and career advancement, such that she can either pay for a Masters degree, increase her chances of winning a scholarship, or dispense with the degree altogether?”

(b) Problem A – Some feasible solutions could include 1) Spend some time finding out about future job opportunities after the Masters program. Speak to the employer who has offered her a job and see if they would be willing to take her on after two years. Use university resources to establish employment prospects after the Masters.

2) Take a holiday or some time off to motivate herself to perform well on a second degree and in future exams.

Problem B – Some feasible solutions could include 1) Explore the possibility of working part-time with the company and finishing the Masters degree simultaneously; or of working for the company for a few years and then being given a sabbatical to finish her Master’s degree.

2) Find out about scholarship chances for working professionals who are re-entering school after taking some time off to work.

3) Project future earnings to see when she would be able to save enough to either pay for the degree herself or get access to a student loan.

4) Develop a career path that can give her the experience and the skills that she would have got on the Masters program by working on different assignments or getting on-the-job training, for example.

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1-14

A Typical Discussion/Solution: (a)

One way to phrase this problem is: How do the three students satisfy their hunger in the most economical way possible, both in terms of time and money?

(b)

Principle 1 - Develop the Alternatives i) Alternative A is to sign up for college meals ii) Alternative B is to cook on their own Other options probably exist but we’ll stick to these two alternatives Principle 2 - Focus on the Differences The biggest difference is likely to be in explicit money costs. Another difference is in the time taken to satisfy hunger in either case. A third difference could be quality of the food and of the ingredients. We’ll concentrate our attention on cost differences in part (c) to follow.

Principle 3 - Use a Consistent Viewpoint Consider your problem from the perspective of three students as customers wanting to get a good deal. Use the customer’s point of view in this situation rather than that of the principal of the college, or the owners of the local supermarket, or the neighbors of the students.

Principle 4 - Use a Common Unit of Measure Most people use “dollars” as one of the most important measures for examining differences between alternatives. In deciding which meal system to select, we’ll use a cost-based metric in part (c).

Principle 5 - Consider All Relevant Criteria Factors other than cost may affect the decision about which meal system to commit to. For example, students may not end up eating every meal at college even if they have signed up for it.

Another factor is how well prepared these students are for the final exams. Dynamics of group decision-making may also introduce various “political” considerations into the final selection (can you name a couple?)

Principle 6 - Make Uncertainty Explicit Students should check the timings of college meals and whether there is a holiday period when the college kitchen would be closed. Another area of uncertainty is about the shopping and cooking time associated with preparing your own meals. Are all students equally efficient cooks? Since the time taken by each student will vary, how would the three friends allocate cooking time between themselves?

Principle 7 - Revisit Your Decision After you’ve completed the semester, how did you do on the final exam? Did your decision have an impact on study times? How good was the food that you consumed? How much money did you save? You will keep these sorts of things in mind (good and bad) when you plan for your next semester!

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1-14

continued c)

Finally some numbers to crunch – don’t forget to list any key assumptions that underpin your analysis to minimize the costs of time and money (Principles 1, 2, 3, 4 and 6 are integral to this comparison).

Assumptions: (i) the costs given in the example are all accurate, (ii) cooking and meal preparation time will be equally divided between all three, (iii) students will not miss a single college meal if they have signed up for the system, (iv) students never eat out at a restaurant or a friend’s place.

Analysis: Alternative A Cost of meals on weekdays: $500 per term per student, or $50 per week or $5 per meal.

Cost of meals on weekends: (1/2)($5) × 4 = $2.50 per meal × 4 = $10 Cost of meals in the week = $50 + $10 = $60.

Time taken to prepare home meals: 1 student-hour per meal or 20 minutes a student for all three.

Value of time per meal = $15/hr × 20 mins = $5 per meal.

Value of time spent cooking per week = $5 × 4 = $20 Total cost per week per student: $60 + $20 = $80.

Alternative B Cost of meals for the week: $2.50 per meal × 7 × 2 = $35 per student Time taken to prepare home meals: 20 minutes × 14 = 280 mins per student.

Value of time spent cooking per week = $15/hr × 280 mins = $70.

Total cost per week per student: $35 + $70 = $105.

Therefore, sign up for the college meal system to minimize total cost.

Typical other criteria you and your friends could consider are: (i) minimize explicit money costs paid out, ignoring the value of time (pick cooking yourselves), (ii) minimize total time spent, ignoring money costs (pick college meal system), and (iii) minimize wasted funds (pick cooking yourself because you will invariably miss a few college meals because of your social calendar).

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1-15

Definition of Need Some homeowners need to determine (confirm) whether a storm door could fix their problem. If yes, install a storm door. If it will not basically solve the problem, proceed with the problem formulation activity.

Problem Formulation The homeowner’s problem seems to be one of heat loss and/or aesthetic appearance of their house.

Hence, one problem formulation could be: “To find different alternatives to prevent heat loss from the house.”

Alternatives  Caulking of windows  Weather stripping  Better heating equipment  Install a storm door  More insulation in the walls, ceiling, etc. of the house  Various combinations of the above

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1-16

STEP 1—Define the Problem: Your basic problem is that you need transportation.

STEP 2—Develop Your Alternatives (Principle 1 is used here.): Your problem has been reduced to buying a new car or a used one. The alternatives would appear to be 1. Buy a new car for $20,000. Spend $10,000 of your savings and borrow $10,000. You will have to repay $10,000 after two years, with 10% in interest over two years, generating an interest cost of $1,000. The car will be brand new with 0 miles of prior use. Your savings account will be run down to 0. You can resell the car after two years for $10,000 (50% less than the original cost).

2. Buy a used car for $9,000. Spend the money out of your savings. You will have $1,000 left in your account. Your car will have 25,000 miles of prior use. You can resell the car for $6,000.

3. Buy a new car for $20,000 but finance more of it on loan from the company. For example, if the entire amount is borrowed from the company, then the interest cost due will be $2,000; however, you will have more savings through the year.

4. Buy a used car for $9,000 but finance more of it on loan from the company. For example, if the entire amount is borrowed from the company, then the interest cost due will be $900 but you will have more savings through the year.

ASSUMPTIONS: 1. Each car will perform at a satisfactory operating condition (as it was originally intended) and will provide the same total mileage before being sold.

2. Each car will have similar operating efficiency in terms of gallons per mile (alternatively, assume fixed values of gallons per mile for each car).

3. Interest earned on money remaining in savings is negligible (or assume a given rate of interest).

STEP 3—Estimate the Cash Flows for Each Alternative (Principle 2 should be adhered to in this step.) a. Your net outflow of cash over two years will be ($10,000 + $10,000 + $1,000 – $10,000) = $11,000. b. Your net cash outflow will be ($9,000 – $6000) = $3,000. c. The net cash flow at the end of the period will be ($20,000 + $2,000 – $10,000) = $12,000. d. Your net cash outflow will be ($9,000 + $900 – $6000) = $3,900.

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1-16

continued STEP 4—Select a Criterion: It is very important to use a consistent viewpoint (Principle 3) and a common unit of measure (Principle 4) in performing this step. The viewpoint in this situation is yours (the prospective owner of a car).

The value of the car to the owner is its market value (i.e., $20,000 for the new car and $9,000 for the used car). Hence, the dollar is used as the consistent value against which everything is measured. This reduces all decisions to a quantitative level, which can then be reviewed later with qualitative factors that may carry their own dollar value (e.g., how much is low mileage or reliability worth?).

STEP 5—Analyze and Compare the Alternatives: Make sure you consider all relevant criteria (Principle 5).

If interest on savings is negligible, there is no benefit to holding savings. In this case, options 3 and 4 are eliminated because they involve higher cash flows and the same end result as options 1 and 2.

Alternative 2 is a good alternative to consider, because it spends the least amount of cash. The choice between 1 and 2 depends on how highly you value having a new car – i.e. is it worth $8,000 over two years?

STEP 6—Select the Best Alternative: When performing this step of the procedure, you should make uncertainty explicit (Principle 6). Among the uncertainties that can be found in this problem, the following are the most relevant to the decision. How reliable will the used car be? How different would projected maintenance and repair costs on both cars be? How does mileage on both cars vary – is the new car much better than the old car? Do you have a complete history of the used car – such as whether it’s been in an accident before? How reliable are the projected end-period values of both cars?

Based on the information in all previous steps, you decide that though the new car may be very good to drive, the sharp depreciation in its value makes it a poor choice to own for only two years. Alternative 2 was actually chosen.

STEP 7—Monitor the Performance of Your Choice: This step goes hand-in-hand with Principle 7 (revisit your decisions). The used car turned out to be great. Mileage was great, and no repairs were needed. The systematic process of identifying and analyzing alternative solutions to this problem really paid off!

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1-17

Employees might be able to come to work earlier and stay later if they are not restricted by public transport timings or by the need to come and go by carpool. If they were previously parking their cars farther away, then they might have been reluctant to work late; especially if the area in which they were parking was unsafe or had poor security. Finally, poor security for employees and their cars could have invited lawsuits and high legal costs if the company was found negligent in providing for its employees.

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1-18

(a)

Value of metal in collection = (5,400/135 lb)(0.95)($3.80/lb) + (5,400/135 lb)(0.05) ($1.20/lb) = $146.8 Each penny is worth about 2.7 cents for its metal content. The numismatic value of each coin is likely to be much greater. Note: It is illegal to melt down coins.

(b)

This answer is left to the individual student. In general, the cost of purchases would go up slightly. The inflation rate would be adversely affected if all purchases were rounded up to the nearest nickel. Additional note: The cost of producing a nickel is almost 10 cents. Maybe the U.S. government should get out of the business of minting coins and turn over the minting operation to privately-owned subcontractors.

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1-19

Left to student.

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1-20

Left to student.

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1-21

One assumption is that water at 60 degrees and 40 degrees has the same efficiency in washing and removing stains. If clothes washed at 40 degrees require longer or more frequent washing, or replacement, then the savings would not be as large. Another assumption would be related to the time for which the wash is run – for example, was there a rinse cycle, or a pre-wash cycle, and what temperatures were these run at? Another factor is whether the same detergent would be effective at both temperatures or whether different temperatures require different detergents. Finally, the material of clothing affects the temperature at which it can be most effectively washed.

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Solutions to Chapter 2 Problems A Note To Instructors: Because of volatile energy prices in today's world, the instructor is encouraged to vary energy prices in affected problems (e.g. the price of a gallon of gasoline) plus and minus 50 percent and ask students to determine whether this range of prices changes the recommendation in the problem. This should make for stimulating inclass discussion of the results.

2-1

The total mileage driven would have to be specified (assumed) in addition to the variable cost of fuel per unit (e.g. $ per gallon). Also, the fixed cost of both engine blocks would need to be assumed. The efficiency of the traditional engine and the composite engine would also need to be specified

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2-2

(a) (b) (c) (d) (e) (f) (g) (h)

4 – sunk 5 – opportunity 3 – fixed 2 – variable 6 – incremental 1 – recurring 7 – direct 8 – nonrecurring

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2-3

(a)

# cows =

1,500,000 miles/year = 273.97 or 274 cows (365 days/year)(15 miles/day)

Annual cost = (1,500,000 miles/year)($10/60 miles) = $250,000 per year (b)

Annual cost of gasoline =

1,500,000 miles/year  $4/gallon = $200,000 per year 30 miles/gallon

It would cost $50,000 less per year to fuel the fleet of cars with gasoline.

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2-4

Cost Annual rental fee Hauling cost Total cost

Site A = $100,000 (3)(150,000)($2) = $900,000 $1,000,000

Site B = $60,000 (4)(150,000)($2) = $1,200,000 $1,260,000

Note that the revenue of $8.00/yd3 is independent of the site selected. Thus, we can maximize profit by minimizing total cost. The solid waste site should be located in Site A.

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2-5

Present cost to company = $1,300,000 + (44 repairs × 20 hrs/repair × $2,500/hr) = $3,500,000 With Ajax, the cost to company = $2,180,000 + 44 repairs × R hrs/repair × $2,500/hr Ajax is preferred if 2,180,000 + 110,000R ≤ $3,500,000 or R ≤ 12 hours/breakdown.

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2-6

This reasoning is incorrect. The time and money spent on the ten hours so far cannot be recovered and it is a sunk cost, which should be ignored. Any further decision on whether to attend the classes or not should not be affected by the size of the sunk cost.

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2-7

If the probability that you are able to find a tenant for the furnished flat is quite high then you should furnish the apartment. Your investment of $2,400 will be recovered in 8 months (the rent differential of $300 per month over 8 months). Even if you cannot find a tenant who will pay $800, but who is willing to pay, say $700, then the investment will still be recovered, though now in 12 months.

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2-8

The opportunity cost to your friend of quitting her current job is losing a certain income of $80,000 per year. She expects to earn an additional (25 – 5) × 12 × $500 = $120,000 per year baking cookies, which yields a net increase in income of $120,000 – $80,000 – $10,000 = $30,000. Provided her projections about future events per month (25) are relatively certain, quitting her job and switching to catering fulltime would be a good decision. If she is able to cater at least 20 events a month, she will be able to turn a profit from switching careers.

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2-9

(a) If you invest in risky property, you could end up losing all your money or not making any return at all. Your opportunity cost would be a guaranteed return of 5% per annum by investing funds in a bank deposit. (b) If you keep your money in a safe, you will not earn any financial return. Your opportunity cost will be at least 5% per annum, earned on a bank deposit. For a risk-loving person, the opportunity cost will be even higher because they might have considered the property market’s potential 25% over the next year as an alternative return.

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2-10

Let X equal dollars per item delivered, and set total revenue equal to total cost: $X (15 items/hr) = $42.00/hr + ($0.50/item)(15 items/hr) X = ($49.50/hr) / (15 items/hr) X = $3.30 per item At least $3.30 per item delivered on Sunday will be needed to break even.

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2-11

(a)

Students can use Equation (2-10) to determine D*.

D* = Then

.

= 225 units/month

p = 75 – 0.1(225) = $52.50 per unit TR = 225 units × $52.50/unit = $11,812.50 and

CT = $1,000 + $30/unit × 225 units = $7,750 Maximum profit = TR – CT = $11,812.50 − $7,750 = $4,062.50 per month

(b)

Using the quadratic equation to solve for the breakeven points: 𝐷

45

45

4 0.1 1,000 0.2

𝐷

45

𝐷

24 (rounded up from 23.5)

𝐷

426 (rounded down from 426.5)

40.3 0.2

Thus, the profitable range of demand is from 24 to 426 units per month.

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2-12

Re-write the price-demand equation as follows: p = 3,200 – 0.1D. Then, TR = p D = 3,200D – 0.1D2 The first derivative of TR with respect to D is d(TR)/dD = 3,200 – 0.2D This, set equal to zero, yields the D̂ that maximizes TR. Thus, 3,200  0.2 Dˆ  0 D̂ = 16,000 units per month

What is needed to determine maximum monthly profit is the fixed cost per month and the variable cost per lash adjuster.

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2-13

P = 500 − 0.2D; CF = $100,000; Cv = $150 Revenue = PD = (500 – 0.2D)D = 500D – 0.2D2 d(Revenue) / dD = 500 – 0.4D = 0 D = 500 / 0.4 = 1,250 units per year. d2R / dD = −0.4 which yields a maximum for revenue.

Profit = PD − CF − Cv = (500 – 0.2D)D – 100000 – 150D = 500D – 0.2D2 – 100000 – 150D = 350D – 0.2D2 − 100000 d(Profit) / dD = 350 − 0.4D = 0 D = 875 units per year d2Profit / dD = −0.4 which yields a maximum for profit.

Unit price at this level of demand is P = 500 – 0.2 (875) = $325.

Profits at this price = $(325 – 150) × 875 – 100000 = $53,125

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2-14

(a)

P = 450 – 0.1D; Cv = $25/racquet; CF = $25,000 + lease value Optimal D* = (a – Cv)/2b = (450 – 25)/2(0.1) = 2,125 units (Eqn 2-10).

(b)

Profit = [450 – 0.1(2,125)] × 2,125 – 25,000 – Lease value – 25(2,125) For profits to be at least 0, Lease value can be no greater than L = $426,562,50

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2-15

(a)

Profit = PD − 2,000 − 35D = (5 + 4800/D – 3000/D2)D – 2,000 – 35D = −30D + 4800 – 3000/D − 2000 dProfit/dD = −30 + 3000/D2 = 0 D* = 10

(b)

d2Profit/dD = −6000/D3 < 0 at D* which implies that D* maximizes profit.

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2-16

Profit = Total revenue – Total cost = (20X – 0.5X2) – (20 + 0.8X + 0.25X2) = 19.2X – 0.75X2 – 20 dProfit = 0 = 19.2  1.5X dX

X = 12.8 megawatts Note:

d 2 Profit =  1.5; thus, X = 12.8 megawatts maximizes profit dX 2

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2-17

Annual sales (units) = $480,000/$40 = 12,000 units/yr.

D’ = CF/(P – CV) = 150000/(40 – 25) = 10,000 units/yr. (from Eqn 2-13)

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2-18

20,000 tons/year (2,000 pounds/ton) = 40,000,000 pounds of zinc per year are produced.

The variable cost per pound is $30,000,000/40,000,000 pounds = $0.75 per pound. (a) Profit/year = (40,000,000 pounds/year)($1.00 – $0.75) – $10,000,000 = $10,000,000 – $10,000,000 = $0 per year Because Profit = 0, 20,000 tons per year is the breakeven point production level for this mine. A loss would occur for production levels < 20,000 tons/year and a profit for levels > 20,000 tons per year. (b) The mine will be profitable. If 21,000 tons (= 42,000,000 pounds) are produced, then Profit/yr = (42,000,000 pounds/year)($1.00 – $0.75) – $10,000,000 = $50,000

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2-19

(a)

BE = $500,000 / ($50 − $10) = 12,500visitors per month.

(b)

New BE point = $500,000 / ($60 − $15) = 11,112visitors per month.

(c)

For 25,000 visitors per month, profit equals 25,000 ($60 − $15) − $500,000 = $625,000 per month, so the park makes a profit.

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2-20

(a)

BE point = $70,000/($10 – $3) =10,000 tubs or 66.67% of capacity.

Profits = 15,000 ($10 – $3) – $70,000 = $35,000.

(b)

Profits = 15,000 ($7 – $3) – $70,000 = –$10,000.

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2-21

Annual savings are at least equal to ($120/lb)(500 lb) = $60,000. So the company can spend no more than $60,000 (conservative) and still be economical. Other factors include ease of maintenance and cleaning, passenger comfort, and aesthetic appeal of the improvements. Yes, this proposal appears to have merit, so it should be supported.

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2-22

Afua is willing to pay $400 to eliminate the risk of data loss, in case she is unable to back up all her data to the portable hard drive, suggesting she is relatively risk-averse. Perhaps, she expects there could be further complications with an unstable hard drive that could make the computer unusable even with a portable hard drive. In such an eventuality, investing in a new laptop would eliminate the costs of any future repairs to the old computer.

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2-23

Over 81,000 miles, the gasoline-only car will consume 2,700 gallons of fuel. The flex-fueled car will use 3,000 gallons of E85. So we have (3,000 gallons)(X) + $1,000 = (2,700 gallons)($3.89/gal) and X = $3.17 per gallon This is 18.5% less expensive than gasoline. Can our farmers pull it off – maybe with government subsidies?

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2-24

(a)

𝐶 =𝑐 𝑋

Eqn. (2-8)

$2,000,000 = 𝑐 2000 𝑡𝑜𝑛𝑠 𝑐 = $1,000/ton = $0.50/ lb.

Profit = [𝑠 - 𝑐 ]X - 𝐶 = [ $0.80/lb - $0.50/ lb] 4,000,000 lb - $700,000 = $500,000 (b)

X’

=𝐶 / 𝑠 -𝑐 )

Eqn. (2-13)

= $700,000 / ($0.80/lb - $0.50/ lb) = 2,333,333 lbs 𝑐 = 𝐶 / X’ = $700,000 / 2,333,333 lbs = $0.30 / lb OR X’ 𝑠 - 𝑐 ) = 𝑐 X’ at breakeven point 𝑐 (c)

𝑠 - 𝑐 ) = ($0.80/lb - $0.50/ lb) = $0.30/ lb

𝑐 = (X𝑐 + 𝐶 ) / X = [ 4,000,000 lb ($0.50/lb) + $700,000] / 4,000,000 lb = $0.675/ lb

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2-25

Annual profits = Annual revenues (R) – Total fixed costs – Total variable costs $15,000 = R – $10,000 – .5R .5R = $25,000 R = $50,000 The correct choice is (b).

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2-26 CT = Co + Cc = knv2 +

$1,500n v

dCT 1,500 = 0 = 2kv- 2 = kv3 - 750 dv v 750 k

v=3

To find k, we know that Co = $100/mile at v = 12 miles/hr n

Co = kv 2 = k(12) 2 = 100 n and k = 100 / 144 = 0.6944

so,

v = 3

750 = 10.25 miles / hr .

0.6944

The ship should be operated at an average velocity of 10.25 mph to minimize the total cost of operation and perishable cargo.

Note: The second derivative of the cost model with respect to velocity is: d 2CT n = 1.388n + 3,000 3 2 dv v

The value of the second derivative will be greater than 0 for n > 0 and v > 0. Thus we have found a minimum cost velocity.

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2-27

Solve for k: CG / n = kv

v in miles/hr

1/18 mils/gal = k (70 miles/hr) ; k = 1 hr – gal / 1,260 mi2 = 0.000794 hr-gal/mi2 CG = (0.000794 hr-gal/mi2)(3000)(v)($3.60/gal) = 8.5752 hr-gal/mi2 CFSS = ($15,000/hr)(1/v)

Find CT CT = ($8.5752 hr/mi2)(v mi/hr) + ($15,000/hr)(v-1hr/mi)($/mi)

d CT / dv = 8.5752 −15,000/v2 = 0 v2 = 15,000/8.5752 v* = 41.82 mi/hr

Check d2 CT / dv2 = 30,000v-3 which is positive for v > 0, therefore we have minimized the total cost

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2-28 (293 kWh/106 Btu)($0.15/kWh) = $43.95/106 Btu

R11 R19 R30 R38 A. Investment cost $2,400 $3,600 $5,200 $6,400 6 B. Annual Heating Load (10 Btu/yr) 74 69.8 67.2 66.2 C. Cost of heat loss/yr $3,252 $3,068 $2,953 $2,909 D. Cost of heat loss over 25 years $81,308 $76,693 $73,836 $72,737 E. Total Life Cycle Cost = A + D $83,708 $80,293 $79,036 $79,137 Select R30 to minimize total life cycle cost.

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2-29

(a)

C dC   I  CR t = 0 d 2

or, 2 = CI/CRt and, * = (CI/CRt)1/2; we are only interested in the positive root. (b)

d 2C d 2



2C I

3

 0 for  > 0

Therefore, * results in a minimum life-cycle cost value. (c)

Investment cost versus total repair cost

C CRꞏꞏt $

CI



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2-30

180 rpm

1 cycle = (12 jacks/refurb.)(1 +4 refurb.) / (7 jacks/hr) = 60/7 = 8.57 hrs Cost per cycle = 1 brush ---- $90 4 refurb. --- 4($30) Oper. Cost – 8.57 hr. ($70/hr) = $810/cycle Cost per jack = $810/60 = $13.50/jack

240 rpm

1 cycle = (8 x 3) / 10 = 2.4 hrs Cost per cycle = $90 + 2($30) + 2.4hr ($70/hr) = $318/cycle Cost per jack = $318/24 = $13.25/jack

300 rpm

1 cycle = (6 x 2) / 12 = 1hr Cost per cycle = $90 + 1($30) + 1hr ($70/hr) = $190/cycle Cost per jack = $190/12 = $15.83/jack

Select 240 rpm

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2-31

(a)

With Dynolube you will average (20 mpg)(1.01) = 20.2 miles per gallon (a 1% improvement).

Over 50,000 miles of driving, you will save 50,000 miles 50,000 miles   24.75 gallons of gasoline.

20 mpg 20.2 mpg

This will save (24.75 gallons)($4.00 per gallon) = $99. (b)

Yes, the Dynolube is an economically sound choice.

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2-32

The cost of tires containing compressed air is ($200 / 50,000 miles) = $0.004 per mile. Similarly, the cost of tires filled with 100% nitrogen is ($220 / 62,500 miles) = $0.00352 per mile. On the face of it, this appears to be a good deal if the claims are all true (a big assumption). But recall that air is 78% nitrogen, so this whole thing may be a gimmick to take advantage of a gullible public. At 200,000 miles of driving, one original set of tires and three replacements would be needed for compressed-air tires.

One original set and two replacements (close enough) would be required for the 100% nitrogen-filled tires. What other assumptions are being made?

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2-33

(a)

Speed A B C

Drilling Rate (ft/min) 2 3 4

Bit life at this Speed (min) 10 6 3

Given: rock drill with 3 operating speeds Cost= bit cost + operator cost (+ blasting penalty) 1 cycle = 96 ft. of drilling Operator cost = $30/hr = $0.50/min Bit cost = $10.00/each Speed A – Cycle = 96ft / 2fpm = 48 min Bit cost- 48 min/ 10 min/bit = 4.8 bits x $10 each = $48.00 Oper. Cost- 48 min x $0.50/min

$24.00 $72.00/cycle

Cost/ft. = $72.00/cycle / 96 ft/cycle = $0.75/ft Speed B – Cycle = 96ft / 3fpm = 32 min Bit cost- 32 min/ 6 min/bit = 5.33 bits x $10 each = $53.33 Oper. Cost- 32 min x $0.50/min

$16.00 $69.33/cycle

Cost/ft. = $69.33/cycle / 96 ft/cycle = $0.72/ft Speed C – Cycle = 96ft / 4fpm = 24 min Bit cost- 24 min/ 3 min/bit = 8 bits x $10 each =

$80.00

Oper. Cost- 24 min x $0.50/min

$12.00 $92.00/cycle

Cost/ft. = $92.00/cycle / 96 ft/cycle = $0.96/ft

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Choose Speed B

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2-33

(b)

Penalty of $60.00/hour for cycle time greater than 30 minutes $60/hr = $1.00/min Speed A- Cycle Time = 48 min Total cost/cycle = $72.00 + (48-30)$1.00 = $90.00 (includes penalty) ($0.9375/ft) Speed B- Cycle Time = 32 min Total cost/cycle = $69.33 + (32-30)$1.00 = $71.33 (includes penalty) ($0.743/ft) Speed C- Cycle Time = 24 min (no penalty Total cost/cycle = $92.00 ($0.958/ft) Speed B still has the lowest cost per cycle, now the cost per foot is $0.743/ft

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2-34

(a)

Manufacturing Option A Labor = (40 x 5)($10) = $2000/wk.

Rental = $20,000 Σ = $22,000 Material = $15/unit Purchase Option B = $20/unit $22,000 + $15(x) = $20(x) $22,000/5 = 𝑥 𝑥 = 4,400 units/wk = Breakeven amount

(b)

$22,000 + $15(3500) [< or > ] $20 (3500) $74,500 > $70,000 Purchase this item

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2-35

Strategy: Select the design which minimizes total cost for 125,000 units/year (Rule 2). Ignore the sunk costs because they do not affect the analysis of future costs. (a)

Design A Total cost/125,000 units = (12 hrs/1,000 units)($18.60/hr)(125,000) + (5 hrs/1,000 units)($16.90/hr)(125,000) = $38,463, or $0.3077/unit Design B Total cost/125,000 units = (7 hrs/1,000 units)($18.60/hr)(125,000) + (7 hrs/1,000 units)($16.90/hr)(125,000) = $33,175, or $0.2654/unit Select Design B

(b)

Savings of Design B over Design A are: Annual savings (125,000 units) = $38,463 − $33,175 = $5,288 Or, savings/unit = $0.3077 − $0.2654 = $0.0423/unit.

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2-36

Profit per day = Revenue per day – Cost per day = (Production rate)(Production time)($30/part)[1-(% rejected+% tested)/100] – (Production rate)(Production time)($4/part) – (Production time)($40/hr) Process 1: Profit per day = (35 parts/hr)(4 hrs/day)($30/part)(1-0.2) – (35 parts/hr)(4 hrs/day)($4/part) – (4 hrs/day)($40/hr) = $2640/day Process 2: Profit per day = (15 parts/hr)(7 hrs/day)($30/part) (1-0.09) – (15 parts/hr)(7 hrs/day)($4/part) – (7 hrs/day)($40/hr) = $2155.60/day Process 1 should be chosen to maximize profit per day.

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2-37

At 70 mph your car gets 0.8 (30 mpg) = 24 mpg and at 80 mph it gets 0.6(30 mpg) = 18 mpg. The extra cost of fuel at 80 mph is: (400 miles/18mpg – 400 miles/24 mpg)($4.00 per gallon) = $22.22 The reduced time to make the trip at 80 mph is about 45 minutes. Is this a good tradeoff in your opinion? What other factors are involved?

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2-38

(a)

Operation 1

cycle time = 1 hr + 0.333 hr = 1.333 hr/cycle Cycles/day = (8hr/day)(1 cycle/ 1.333 hr) = 6 cycle/day Value added = (2000 parts/cycle)(6 cycles/day)($0.40/part) = $4,800/day

Cost1 = 8 hr/day ($20/hr) = $160/day Value – cost = $4,800 - $160 = $4,640/day Operation 2

cycle time = 2 hr + 0.5 hr = 2.5 hr/cycle

Cycles/day = (8hr/day)(1 cycle/ 2.5 hr) = 3.2 cycle/day Value added = (3500 parts/cycle)(3.2 cycles/day)($0.40/part) = $4,480/day Cost2 = 8 hr/day ($11/hr) = $88/day Value – cost = $4,480 - $88 = $4,392/day Select Operation 1 to maximize profit (b)

Output/day for Operation 1 = 12,000 parts and output/.day for Operation 2 = 11,200 parts.

Downtime for Operation 1 = 6 x 20 min = 120 minutes/day and downtime for Operation 2 = 3.2 x 30 = 96 minutes/day. So increased production for Operation 1 is being traded off for increased tool changing time (downtime), and the balance is favorable for Operation 1 compared to Operation 2.

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2-39

Apache: (24 hr/day)(7 days/wk) – 4 = 164 hrs/wk uptime (90 hits/min) (60 min/hour) = 5,400 hits/hr (5,400 hits/hr) (164 hrs/wk) = 885,600 hits/wk @ $0.015/hit = $13,284/wk Profit/yr. = ($13,284/wk)(52 wk/yr) = $690,768 Windows IIS: (24 hr/day)(7 days/wk) – 0.75 = 167.25 hrs/wk uptime (5,400 hits/hr) (167.25 hrs/wk) = 903,150 hits/wk @ $0.015/hit = $13,547.25/wk Profit/yr. = ($13,547.25/wk)(52 wk/yr) - $5,000 = $699,457 Go with Windows software.

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2-40

Option A (Purchase): CT = (10,000 items)($8.50/item) = $85,000 Option B (Manufacture): Direct Materials = $5.00/item Direct Labor = $1.50/item Overhead = $3.00/item $9.50/item CT = (10,000 items)($9.50/item) = $95,000 Choose Option A (Purchase Item).

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2-41

This statement makes sense, up to a point. Airlines do have very high fixed costs relative to variable costs, particularly no-frills airlines, which do not offer meals or other services that would vary on a perpassenger basis. However, the fuel carried by a plane does depend on the weight of the plane: with fewer passengers, planes can carry less fuel. As long as this variable cost is covered, any additional revenue from discounted ticket sales will help offset the fixed costs of operating the flight.

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2-42

Assumptions: You can sell all the metal that is recovered Method 1:

Recovered ore = (0.62)(100,000 tons) = 62,000 tons Removal cost = (62,000 tons)($23/ton) = $1,426,000 Processing cost = (62,000 tons)($40/ton) = $2,480,000 Recovered metal = (300 lbs/ton)(62,000 tons) = 18,600,000 lbs Revenues = (18,600,000 lbs)($0.8 / lb) = $14,880,000 Profit = Revenues - Cost = $14,880,000 - ($1,426,000 + $2,480,000) = $10,974,000

Method 2:

Recovered ore = (0.5)(100,000 tons) = 50,000 tons Removal cost = (50,000 tons)($15/ton) = $750,000 Processing cost = (50,000 tons)($40/ton) = $2,000,000 Recovered metal = (300 lbs/ton)(50,000 tons) = 15,000,000 lbs Revenues = (15,000,000 lbs)($0.8 / lb) = $12,000,000

Profit = Revenues - Cost = $12,000,000 - ($750,000 + $2,000,000) = $9,250,000 Select Method 1 (62% recovered) to maximize total profit from the mine.

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2-43

Profit per ounce (Method A) = $1,750 - $550 / [(0.90 oz. per ton)(0.90)] = $1,750 - $679 = $1,071 per ounce Profit per ounce (Method B) = $1,750 - $400 / [(0.9 oz. per ton)(0.60) =$1,750 - $741 = $1,009 per ounce Therefore, by a slim margin we should recommend Method A.

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2-44

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r) (s)

True False False False True True True True False True False True False True True True False True False

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1,750,000 Btu  2-45

(a) Loss 

lb coal    12,000 Btu   486 lbs of coal 0.30

(b) 486 pounds of coal produces (486)(1.83) = 889 pounds of CO2 in a year.

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2-46

(a) Let X = breakeven point in miles Fuel cost (car dealer option) = ($2.00/gal)(1 gal/20 miles) = $0.10/mile Motor Pool Cost = Car Dealer Cost ($0.36/mi)X = (6 days) ($30/day) + $0.20/mi + $0.10/miX $0.36X = 180 + $0.30X and X = 3,000 miles

(b)

6 days (100 miles/day) = 600 free miles If the total driving distance is less than 600 miles, then the breakeven point equation is given by: ($0.36/mi)X = (6 days)($30 /day) + ($0.10/mi)X X = 692.3 miles > 600 miles This is outside of the range [0, 600], thus renting from State Tech Motor Pool is best for distances less than 600 miles.

If driving more than 600 miles, then the breakeven point can be determined using the following equation: ($0.36/mi)X = (6 days)($30 /day) + ($0.20/mi)(X - 600 mi) + ($0.10/mi)X X = 1,000 miles

(c)

The true breakeven point is 1000 miles.

The car dealer was correct in stating that there is a breakeven point at 750 miles. If driving less than 900 miles, the breakeven point is: ($0.34/mi)X = (6 days)($30 /day) + ($0.10/mi)X X = 750 miles < 900 miles However, if driving more than 900 miles, there is another breakeven point. ($0.34/mi)X = (6 days)($30/day) + ($0.28/mi)(X-900 mi) + ($0.10/mi)X X = 1800 miles > 900 miles The car dealer is correct, but only if the group travels in the range between 750 miles and 1,800 miles. Since the group is traveling more than 1,800 miles, it is better for them to rent from State Tech Motor Pool.

This problem is unique in that there are two breakeven points. The following graph shows the two points.

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2-46 continued

Total Cost

Car Dealer v. State Tech Motor Pool $1,000 $900 $800 $700 $600 $500 $400 $300 $200 $100 $0

X2' = 1,800 miles

X1' = 750 miles Car Dealer State Tech

0

500

1000

1500

2000

2500

Trip Mileage

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2-47

This problem is location specific. We’ll assume the problem setting is in Tennessee. The eight years ($2,400 / $300) to recover the initial investment in the stove is expensive (i.e. excessive) by traditional measures. But the annual cost savings could increase due to inflation. Taking pride in being “green” is one factor that may affect the homeowner’s decision to purchase a corn-burning stove.

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2-48 F

G

H

I

J

K

0

L

M

N

O

P

Net Income

250

$20,000 $1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

$(20,000) $(40,000) $(60,000) $(80,000) $(100,000) Volume (Demand)

$600,000

$500,000

$400,000 Total Revenue

$300,000

Total Expense

$200,000

$100,000

10 00 15 00 20 00 25 00 30 00 35 00 40 00 45 00 50 00

0

$50 0

Monthly Price per Total Demand Unit Revenue Total Expense Net income 6 7 0 $ 180 $ $ 73,000 $ (73,000) 8 250 $ 175 $ 43,750 $ 93,750 $ (50,000) 9 500 $ 170 $ 85,000 $ 114,500 $ (29,500) 10 750 $ 165 $ 123,750 $ 135,250 $ (11,500) 11 1000 $ 160 $ 160,000 $ 156,000 $ 4,000 12 1250 $ 155 $ 193,750 $ 176,750 $ 17,000 13 1500 $ 150 $ 225,000 $ 197,500 $ 27,500 14 1750 $ 145 $ 253,750 $ 218,250 $ 35,500 15 2000 $ 140 $ 280,000 $ 239,000 $ 41,000 16 2250 $ 135 $ 303,750 $ 259,750 $ 44,000 17 2500 $ 130 $ 325,000 $ 280,500 $ 44,500 18 2750 $ 125 $ 343,750 $ 301,250 $ 42,500 19 3000 $ 120 $ 360,000 $ 322,000 $ 38,000 20 3250 $ 115 $ 373,750 $ 342,750 $ 31,000 21 3500 $ 110 $ 385,000 $ 363,500 $ 21,500 3750 $ 105 $ 393,750 $ 384,250 $ 9,500 22 4000 $ 100 $ 400,000 $ 405,000 $ (5,000) 23 4250 $ 95 $ 403,750 $ 425,750 $ (22,000) 24 4500 $ 90 $ 405,000 $ 446,500 $ (41,500) 25 4750 $ 85 $ 403,750 $ 467,250 $ (63,500) 26 5000 $ 80 $ 400,000 $ 488,000 $ (88,000) 27 5250 $ 75 $ 393,750 $ 508,750 $ (115,000) 28 5500 $ 70 $ 385,000 $ 529,500 $ (144,500) 29 30 31 32 Summary of impact of changes in cost components on optimum 33 demand and profitable range of demand.

34 35 Percent Change D 1' cv D2' CF D* 36 -10% -10% 2,633 724 4541 37 0% -10% 2,633 824 4443 38 10% -10% 2,633 928 4339 39 -10% 0% 2,425 816 4036 40 0% 0% 2,425 932 3918 41 10% 0% 2,425 1060 3790 42 -10% 10% 2,218 940 3495 43 0% 10% 2,218 1092 3343 44 10% 10% 2,218 1268 3167 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

$40,000

Volume (Demand)

$600,000 $500,000 $400,000 $300,000 Total Revenue $200,000

Total Expense Net income

$100,000 $$(100,000) $(200,000) 0

83 180 0.02

E

50 0 10 00 15 00 20 00 25 00 30 00 35 00 40 00 45 00 50 00 55 00

73,000

$ $ $

D Demand Start point (D) = Demand Increment =

Net Income

$

C

Cash Flow

2 3 4 5

B

Cash Flow

1

A Fixed cost/ mo. = Variable cost/unit = a= b=

Volume (Demand)

Reducing fixed costs has no impact on the optimum demand value, but does broaden the profitable range of demand. Reducing variable costs increase the optimum demand value as well as the range of profitable demand.

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2-49

New annual heating load = (230 days)(72 °F  46 °F) = 5,980 degree days. Now, 136.7  106 Btu are lost with no insulation. The following U-factors were used in determining the new heating load for the various insulation thicknesses.

R11 R19 R30 R38

Energy Cost

Investment Cost Annual Heating Load (106 Btu) Cost of Heat Loss/yr Cost of Heat Loss over 25 years Total Life Cycle Cost

$

U-factor 0.2940 0.2773 0.2670 0.2630

Heating Load 101.3  106 Btu 95.5  106 Btu 92  106 Btu 90.6  106 Btu

$/kWhr $0.086

$/106 Btu $25.20

R11 900 $

R19 1,350 $

R30 1,950 $

R38 2,400

101.3 $2,553 $63,814 $64,714

95.5 $2,406 $60,160 $61,510

92 $2,318 $57,955 $59,905

90.6 $2,283 $57,073 $59,473

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2-50

In this problem we observe that "an ounce of prevention is worth a pound of cure." The ounce of prevention is the total annual cost of daylight use of headlights, and the pound of cure is postponement of an auto accident because of continuous use of headlights. Clearly, we desire to postpone an accident forever for a very small cost.

The key factors in the case study are the cost of an auto accident and the frequency of an auto accident.

By avoiding an accident, a driver "saves" its cost. In postponing an accident for as long as possible, the "annual cost" of an accident is reduced, which is a good thing. So as the cost of an accident increases, for example, a driver can afford to spend more money each year to prevent it from happening through continuous use of headlights. Similarly, as the acceptable frequency of an accident is lowered, the total annual cost of prevention (daytime use of headlights) can also decrease, perhaps by purchasing less expensive headlights or driving less mileage each year.

Based on the assumptions given in the case study, the cost of fuel has a modest impact on the cost of continuous use of headlights. The same can be said for fuel efficiency. If a vehicle gets only 15 miles to the gallon of fuel, the total annual cost would increase by about 65%. This would then reduce the acceptable value of an accident to "at least one accident being avoided during the next 16 years." To increase this value to a more acceptable level, we would need to reduce the cost of fuel, for instance.

Many other scenarios can be developed.

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2-51

Suppose my local car dealer tells me that it costs no more than $0.03 per gallon of fuel to drive with my headlights on all the time. For the case study, this amounts to (500 gallons of fuel per year) x $0.03 per gallon = $15 per year. So the cost effectiveness of continuous use of headlights is roughly six times better than for the situation in the case study.

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2-52

p = 400 – D2 TR = p  D = (400 – D2) D = 400D – D3 TC = $1125 + $100  D Total Profit / month = TR – TC = 400D - D3 - $1125 - $100D = - D3 + 300D – 1125 dTP = -3D2 + 300 = 0  D2 = 100  D* = 10 units dD d 2 TP = -6D; at D = D*, dD 2

d 2 TP = - 60 dD 2

Negative, therefore maximizes profit.

Select (a)

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2-53

- D3 + 300D – 1125 = 0 for breakeven At D = 15 units; -153 + 300(15) – 1125 = 0 Select (b)

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2-54

CF = $100,000 + $20,000 = $120,000 per year CV = $15 + $10 = $25 per unit p = $40 per unit D =

CF $120,000 = = 8,000 units/yr p - c v ($40 - $25)

Select (c)

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2-55

Profit = pD – (CF + CVD) At D = 10,000 units/yr, Profit/yr = (40)(10,000) – [120,000 + (25)(10,000)] = $30,000 Select (e)

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2-56 Profit = pD – (CF + CVD) 60,000 = 35D – (120,000 + 25D) 180,000 = 10D; D = 18,000 units/yr

Select (d)

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2-57

Annual profit/loss = Revenue - (Fixed costs + Variable costs) = $300,000  [$200,000 + (0.60)($300,000)] = $300,000  $380,000 = $80,000 Select (d)

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2-58

Savings in first year = (7,900,000 chips) (0.01 min/chip) (1 hr/60 min) ($8/hr + 5.50/hr) = $17,775 Select (d)

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Solutions To Chapter 3 Problems 3-1

Power generation = 100 kW  1450 hours  .9 = 130,500 kWh

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3-2

A representative cost and revenue structure for construction, 10-years of ownership and use, and the sale of a farm is: Cost or revenue category Captial investment

Typical cost or revenue elements Real estate costs, construction costs, working capital (tools, initial operating supplies, etc.), initial planting of strawberry and blackberry bushes, storage facilities Annual operating and maintenance costs Fertilizer usage, seeds used for replanting, labour costs, electricity and irrigation, vehicles used for transportation Major repair or replacement costs Repairing the farmhouse, driveways, testing the plants for disease and damage Real estate fees Acquisition and selling Asset sales Sale of farm (year 10)

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3-3

The estimated cost is $20,000 + ($12/ft2)(9 ft)(16 ft)(1,000)(1.8) = $3,130,400

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3-4

(a)

Natasha’s reduction in energy consumption = 0.2  5,000 kWh = 1,000 kWh Cost savings = $250 Cost of 1kWh = $250/1,000kWh = $0.25/kWh

(b)

Further reduction in energy consumption = .1  (5000 – 1000) = 400 kWh per year Cost of energy consumption = $0.25/kWh Total saving = $0.25  400 kWh = $100 per year

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3-5

(24,000 ft2)(72,000 Btu/ft2) = 1,728 million Btu during the heating season. This is 1,728 thousand cubic feet of natural gas, and the cost would be (1,728,000 ft3)($12/1000 ft3) = $20,736 for the heating season.

Side note: The building uses 0.3 million kWhr of electricity × $0.10 per kWhr = $30,000 to cool the area.

The total bill will be about $45,000. The owner must take this into account when she decides on a price to charge per square foot of leased space.

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3-6

(a)

Standard electric bill = (480 kWhr)(12 months/year)($0.15/kWhr) = $864 per year.

Green power bill = (12 months/year)($5/month) = $60 per year.

Total electric bill = $924 per year.

(b)

$924/5,760 kWhr = $0.16 per kWhr (a 6.67% increase due to green power usage).

(c)

The technology used to capture energy from solar power, wind power, and methane is more expensive than traditional power generation methods (coal, natural gas, and so on).

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3-7

(a)

Relatively easy to estimate. Contact a real estate firm specializing in commercial property.

(b)

Marketing costs can be easily estimated across the medium term by designing an appropriate strategy involving billboards, advertisements in magazines, or special discounts offered at certain times of the year.

(c)

Total capital investment can be reasonably estimated from (a) and (b) plus the cost of other fixed assets such as such as a cash machine, air-conditioning, a vending machine for employees, etc.

(d)

Reasonably accurate per unit cost estimates can be obtained by contacting suppliers and signing contracts over the long-term for the supply of accessories. However, the demand for accessories is not easy to estimate.

(e)

It is not easy to estimate total annual labor costs due to the initial uncertainty of sales demand for the product. However, these costs can be controlled since they occur over time and the expenditure can be matched to the demand.

(f)

First-year revenues for a new product are difficult to estimate. A marketing consultant with experience in this field could be helpful.

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3-8

C2019 = C2015 (Ī2019/Ī2015) = $440,000 (298/238) = $550,924

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3-9

 64   51   65  0.60   0.10   0.30   32   34   40   100  183.75 I 2018  0.60  0.10  0.30

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3-10

(CA / CB) = (SA / SB)x CA = $1,200,000 (32,000/24,000)0.88 = 1,200,000 (1.288) CA = $1,545,708 for the larger warehouse

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3-11

Let

CA = cost of new boiler, CB = cost of old boiler, today

SA = 1.24X SB = X

 245  CB = $160,000   = $220,225  178  0.82

 1.24 X  CA = $220,225   = $262,707  X  Total cost with options = $262,707 + $34,000 = $2,96,707

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3-12

The estimated capital investment for the seven MW solar farm in four years is: $16 million (F/P, 6%, 6) = $16 million (1.4185) = $22.696 million Next, the capital investment (C) for the seven MW solar farm in six years can be estimated by using Equation 3-4: C = $22.696 million (7/8)0.8 = $20.397 million

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3-13

(a)

 194  Cnow(80-kW) = $160,000   = $165,989  187   120  Cnow(120-kW) = $165,989    80 

0.6

= $211,707

Total Cost = $211,707 + $18,000 = $229,707 (b)

 40  Cnow(40-kW) = $165,989    80 

0.6

= $109,512

Total Cost = $109,512 + $18,000 = $127,512

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3-14

Let

CA = cost of new plant

SA = 450,000 gal/yr

CB = cost of similar plant

SB = 250,000 gal/yr

= $6,000,000  450,000  CA = $6,000,000    250,000 

X = 0.59 0.59

= $8,487,153

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3-15

$600,000 = $300,000 (100,000 /40,000)x 2 = 2.5x log 2 = x log 2.5 x = 0.756 This is the cost-capacity factor for this technology.

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3-16

(a)

(500-425) / 500 = 0.15 85% learning curve

(b)

n = log 0.85 / log2 = -0.234 Z4 = 500(4)-0.234 = 361.5 hours

(c)

Z1 = 500hrs Z2 = 425hrs Z3 = 500(3)-0.234 = 387 hrs Z4 = 361.5hrs ΣZi = 1,673.5 Average $ = (1673.5/4)($15) = $6,275.63

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3-17

n = log (0.9) / log 2 = -0.152 Z6 = 10 (6)n = 10 [(6)-0.152] = 7.6 hours

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3-18

S0 = 200 MW; S30 = 1000 MW (a)

C0 = $300 mn; C30 = $1000 mn (1000/300) = (1000/200)x X = ln(10/3)/ln(5) = 0.75

(b)

C0 = 300; X = 0.8 C30 = $300 (1000MW/200MW).8 mn = $1,087.17 mn

(c)

C30/C0 = 4.5 4.5 = (1000 MW/200 MW)X X = ln(4.5)/ln(5) = 0.93.

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3-19

(a)

∑x = 687 ∑y = 2,559 ∑xy = 442,844 ∑x2 = 118,831 𝑥̅ = 687/4 = 171.75 𝑦 = 2,559/4 = 639.75 𝑏 = [4(442,844) – 687(2,559)] / [4(118,831) – 6872] = 3.977 𝑎 = [2,559 – 3.977(687)] / 4 = -43.308 ∴ 𝑦 = -43.308 + 3.977(x)

(b)

𝑦 = -43.308 + 3.977(170) 𝑦 = $632.78

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3-20

∑x = 1,732 ∑y = 3,532 ∑xy = 644,176 ∑x2 = 325,586 b = [644,176 – 173.2(3,532)] / [325,586 – 173.2(1,732)] = 1.2668 a = 353.2 – 1.2668(173.2) = 133.79 So y = 133.79 + 1.2668x When x = 198, y = 133.79 + 1.2668(198) = 384.6 (call it 385 units per quarter)

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3-21

The following table facilitates the intermediate calculations needed to compute the values of b0 and b1 using Equations (3-8) and (3-9).

I 1 2 3 4 5 6 7 8 9 10 Totals

xi 12,500 13,000 13,500 14,200 15,000 17,000 18,400 21,000 22,500 25,000 172,100

yi 700,000 725,000 750,000 800,000 840,000 950,000 1,000,000 1,125,000 1,225,000 1,400,000 9,515,000

xi2 156,250,000 169,000,000 182,250,000 201,640,000 225,000,000 289,000,000 338,560,000 441,000,000 506,250,000 625,000,000 3,133,950,000

x iy i 8,750,000,000 9,425,000,000 10,125,000,000 11,360,000,000 12,600,000,000 16,150,000,000 18,400,000,000 23,625,000,000 27,562,500,000 35,000,000,000 172,997,500,000

b1 

(10)(172,997,500,000)  (172,100)(9,515,000)  53.7 (using the excel values) (10)(3,133,950,000)  (172,100) 2

b0 

9,515,000  (53.7)(172,100)  27,113 10

(a) The resulting CER relating supermarket building cost to building area (x) is: Cost = 27,113 + 53.7x So the estimated cost for the 23,000 ft2 store is: Cost = $27,113 + ($53.7/ft2)(23,000 ft2) = $1,262,213. (b) The CER developed in part (a) relates the cost of building a supermarket to its planned area using the following equation: Cost = 27,113 + 53.7x Using this equation, we can predict the cost of the ten buildings given their areas.

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3-21

continued

x

i

xi

yi

Costi

(yi - Costi)2

( x i  x )( y i  y ) ( xi  x ) 2

( yi  y ) 2

1

12,500

700,000

698,516

2,203,381

1,184,565,000

22,184,100

63,252,250,000

2

13,000

725,000

725,372

138,170

953,565,000

17,724,100

51,302,250,000

3

13,500

750,000

752,228

4,963,108

747,565,000

13,764,100

40,602,250,000

4

14,200

800,000

789,826

103,503,541

456,015,000

9,060,100

22,952,250,000

5

15,000

840,000

832,796

51,896,508

246,415,000

4,884,100

12,432,250,000

6

17,000

950,000

940,220

95,639,761

315,000

44,100

2,250,000

7

18,400

1,000,000 1,015,417

237,699,215

57,715,000

1,416,100

2,352,250,000

8

21,000

1,125,000 1,155,069

904,155,062

657,565,000

14,364,100

30,102,250,000

9

22,500

1,225,000 1,235,637

113,155,233

1,446,815,000

27,984,100

74,802,250,000

10

25,000

1,400,000 1,369,918

904,932,688

3,493,815,000

60,684,100

201,152,250,000

Totals 172,100

9,515,000 9,515,000

2,418,286,667

9,244,350,000

172,109,000 498,952,500,000

172,100  17,210 10

y

9515000  951,500 10

Using Equations (3-10) and (3-11), we can compute the standard error and correlation coefficient for the CER.

SE  R=

2,418,286,667 = 17,386.4 10  2 9,244,350,000 = 0.9976 (172,109,000)(498,952,500,000)

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3-22

xi = weight of order (lbs) yi = packaging and processing costs ($) (a)

y = b0 + b1x

 x = 1,033;  x = 112,777;

x = 103.3

 y = 754;  y = 59,768;

y = 75.4

2 i

i

2 i

i

 x y = 81,716; i

b1 =

i

10(81,716)  (754)(1,033) = 0.631.

10(112,777)  (1,033) 2

754  (0.631)(1,033) = 10.22.

10 y = 10.22 + 0.631x b0 =

(b)

R=

S xy S xx S yy

Sxy = 81,716 – (754)(1,033)/10 = 3,827.8 Sxx = 112,777 – (1,033)2/10 = 6,068.1 Syy = 59,768 – (754)2/10 = 2,916.4 R= (c)

3,827.8 = 0.9099 ( 2,916.4 )(6,068.1)

y = 10.22 + 0.631x = 10.22 + 0.631(250) = $167.97.; ; ;; ; ;; = 1. = 26.99; = 0.9951

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3-23

 150  Cost150 ft = $15,250    250 

0.6

 1029    = $13,915  830 

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3-24

K = 3 hours; u = 4; s = 0.95; n = log s/log 2 = log 0.95 / log 2 = –0.074 Z4 = 3(4)–0.074 = 2.71 hours.

Z8 = 3(8)–0.074 = 2.57 hours.

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3-25

The amount of the FICO score affected is (0.35)(720) = 252. If this drops by 10%, the payment history score will be (0.90)(252) = 227 and the overall FICO score will be 695. This lower value could adversely affect the interest rate you’ll be quoted on your next loan.

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3-26

mW 0.8 Boiler Cost = $300,000 106 mW  = $451,440 9 mW 0.6 Generator Cost = $400,000  6 mW  = $510,170

Tank Cost = $106,000





91,500 gal 0.66 80, 000 gal

= $115,826

Total Cost = (2)($451,440) + (2)($510,170) + $115,826 + $200,000 = $2,239,046

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3-27

The following spreadsheet was used to calculate a 2019 estimate of $320,274,240 for the plant.

Element Code

Units/Factors

1.1.1-2

600

Price/Unit

Subtotal

$2,000

1.1.3

$

1,200,000

$

3,000,000

1.1

Totals

$

1.2-3

2.3969 a

1.4

4.4727 b $

$ 110,000,000

4,200,000

$ 263,659,000

$ 263,659,000

5,000,000

$

22,363,500

$

22,363,500

60

$

4,823,400

$

15,000,000

$

900,000 $

20,723,400

$

9,328,340

1.5.1-3 labor

80,390 c

$

materials 1.5.4

600

$

1,500

1.5 1.9

3%

$ 310,944,672

$ 320,274,240

TOTAL ESTIMATED COST IN 1996

a

Factor value for boiler and support system (WBS elements 1.2 and 1.3):  492   1      110   2 

b

0.9

 2.3969

Factor value for the coal storage facility (WBS element 1.4):

 492     4.4727  110  c

Labor time estimate for the 3rd facility (WBS elements 1.5.1, 1.5.2, and 1.5.3): K = 95,000 hours, s = 0.9, n = log(0.9)/log(2) = -0.152 Z3 = 95,000(3)-0.152 = 80,390 hours

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3-28

K u

100 5 s 0.75 0.77 0.79 0.81 0.83 0.85 0.87 0.89 0.91 0.93 0.95

Zu 51.27 54.51 57.85 61.31 64.88 68.57 72.37 76.29 80.33 84.49 88.77

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3-29

(a)

Based on the constant reduction rate of 8% each time the number of homes constructed doubles, a 92% leanring curve applies to the situation. The cumulative average material cost per square foot for the first five homes is $24.12.

(b)

The estimated material cost per square foot for the 16th home is $19.34.

S K

0.92 $27

Home 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Material Cumulative Cumulative Cost per ft2 Sum Average $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

27.00 24.84 23.66 22.85 22.25 21.76 21.37 21.02 20.73 20.47 20.23 20.02 19.83 19.66 19.49 19.34

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

27.00 51.84 75.50 98.35 120.60 142.36 163.73 184.75 205.48 225.95 246.18 266.21 286.04 305.69 325.19 344.53

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

27.00 25.92 25.17 24.59 24.12 23.73 23.39 23.09 22.83 22.59 22.38 22.18 22.00 21.84 21.68 21.53

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3-30 A B C D E F G H I J K Spacecraft Weight Cost (millions) 1 0 100 $ 600 2 1 400 $ 278 3 2 530 $ 414 4 3 750 $ 557 5 4 900 $ 689 6 5 1,130 $ 740 7 6 1,200 $ 851 8 9 10 SUMMARY OUTPUT 11 Regression Statistics 12 0.705850276 13 Multiple R 0.498224612 14 R Square 0.397869535 15 Adjusted R Square 151.9036833 16 Standard Error 7 17 Observations 18 19 ANOVA df SS MS F Significance F 20 1 114557.2121 114557.212 4.9646179 0.076342997 21 Regression 5 115373.645 23074.729 22 Residual 6 229930.8571 23 Total 24 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% 25 341.9170907 125.2152843 2.73063383 0.0412491 20.04148132 663.7927001 20.04148132 663.7927001 26 Intercept 0.346423227 0.155476261 2.22814225 0.076343 -0.053240574 0.746087027 -0.053240574 0.746087027 27 Weight 28 29 30 31 RESIDUAL OUTPUT 32 Observation Predicted Cost Residuals 33 1 376.5594133 223.4405867 34 2 480.4863813 -202.4863813 35 3 525.5214008 -111.5214008 36 4 601.7345106 -44.73451063 37 5 653.6979946 35.30200539 38 6 733.3753367 6.624663274 39 7 757.6249626 93.37503741 40 41 42 43 44 45 46 47 48 49 50 0 0 0 0 0 0 0 0 0 0 0 0 0 51 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 , 0 0 , 1 0 ,2 0 , 3 0 1 1 1 1 52 53 54 55 56

Cost (m illions)

Observed vs. Predicted Values

$900 $800 $700 $600 $500 $400 $300 $200 $100 $-

Weight

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3-31

Other cost factors include maintenance, packaging, supervision, materials, among others. Also, the case solution presents a before-tax economic analysis.

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3-32

Left as an exercise for the student. However, by observation, it appears that the factory overhead and factory labor are good candidates since they comprise the largest percentage contributions to the per unit demanufacuring cost.

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3-33

A 50% increase in labor costs equates to a factor of 15%; a 90% increase in Transportation equates to a factor of 38%. The corresponding demnaufacturing cost per unit is $5.19. The per unit cost of using the outside contractor (i.e., the target cost) is $11.70. Should the proposed demanufacturing method be adopted, the revised per unit cost savings is $6.51 for a 55.6% reduction over the per unit cost for the outside contractor.

DE-MANUFACTURING COST ELEMENTS

Unit Elements Factor Estimates Units Cost/Unit Factor of Row

A: Factory Labor 24.5 hrs $ 12.00/hr B: Quality Costs - Training 15% C: TOTAL LABOR D: Factory Overhead - Set up Costs 150% E: Transportation Cost 38% F: TOTAL DIRECT CHARGE G: Facitily Rental H: TOTAL DE-MANUFACTURING COST I: Quantity - Lot Size J: De-manufacturing Cost/Unit Outside Cost/Unit - Target Cost

A C C

Row Total $ $ $ $ $ $

294.00 44.10 338.10 507.15 192.72 699.87 $ 1,037.97 200 $ 5.19

$ 11.70

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3-34

The estimate of direct labor hours is based on the time to produce the 50th unit.

K = 1.76 hours s = 0.8 (80% learning curve) n = (log 0.80)/(log 2) = -0.322 Z50 = 1.76(50)-0.322 = 0.5 hours Factory Labor Production Material Factory Overhead Packing Costs Total Manufacturing Cost Desired Profit Unit Selling Price

= ($15/hr)(0.5 hr/widget) = $375 / 100 widgets = (1.25)($7.50 / widget) = (0.75)($7.50 / widget) = (0.20)($26.25 / widget)

= $7.50 / widget = $3.75 / widget = $9.375 / widget = $5.625 / widget = $26.25 / widget = $5.25 / widget = $31.50 / widget

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3-35

Profit = Revenue – Cost $25,000 = ($20.00/unit)(x) – [($21.00/unit)(.2 hours/unit)(x) + ($4.00/unit)(x) + (1.2)($21.00/unit)(.2 hours/unit)(x) + ($1.20/unit)(x)] $25,000 = 5.56x;

x = 4,497 units

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3-36

K = 460 hours; s = 0.92 (91% learning curve); n = (log 0.92)/(log 2) = -0.120 30

C30 = T30/30; T30 = 460  u -0.120 = 10,419.63 hrs; u =1

C30 = 10,419.63 / 30 = 347.3211 Select (d)

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3-37

−1,500 + 800 + (0.07 - 0.05)(4.00)(10)x = 0 −700 + 0.80x = 0 x = 700/0.80 = 875 miles/year Select (a)

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3-38

ACcurrent = $4,000 Proposed: N = 13 years, SV = 11% of first cost $4,000 = I (A/P,12%,13) – (0.11)I (A/F,12%,13) $4,000 = I(0.1557) – (0.003927)I $4,000 = I (0.1517) I = $26,358 Select (c)

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3-39

Let X = average time spent supervising the average employee. Then the time spent supervising employee A = 2X and the time spent supervising employee B = 0.5X. The total time units spent by the supervisor is then 2X + 0.5X + (8)X = 10.5X. The monthly cost of the supervisor is $3,800 and can be allocated among the employees in the following manner: $3,800/10.5X = $361.90 / X time units.

Employee A (when compared to employee B) costs (2X – 0.5X)($361.90/X) = $542.85 more for the same units of production. If employee B is compensated accordingly, the monthly salary for employee B should be $3,000 + $542.85 = $3,542.85.

Select (a)

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3-40

Type X filter: cost = $5, changed every 7,000 miles along with 5 quarts oil between each oil change 1 quart of oil must be added after each 1,000 miles Type Y filter: cost = ?, changed every 5,000 miles along with 5 quarts of oil no additional oil between filter changes oil = $1.08 / quart Common multiple = 35,000 miles For filter X = 5 oil changes: 5($5 + 5($1.08) + 6($1.08)) = (5)$16.88 = $84.40 For filter Y = 7 oil changes: 7CY + 7(5)($1.08) = 7X + $37.8 $84.40 = 7CY + $37.8 $46.60 = 7CY CY = $6.66 Select (d)

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3-41

 200  C2008(new design) = $900,000   150 

0.92

 450   $1,125,000   200 

0.87

 175   $750,000   100 

0.79

= $4,617,660

C2018 = $4,617,660(1.12)10 = $14,341,751 Select (c)

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Solutions to Chapter 4 Problems

4-1

Interest = ($50,000)(.02/month)(12 months) = $12,000.

Total owed = $50,000 + $12,000 = $52,000.

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4-2

Interest = $500 (2.5) (0.08) = $100.

F = P + I = $500 + $100 = $600.

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4-3

F

=10,000 euros (F/P, −1%, 5)

=10,000 euros 1

0.01

=10,000 euros (0.95099)

= 9,509.90 euros

In essence, the customer is paying 490.10 euros for the preservation of his/her money. In US dollars, this is approximately 490.10 euros ($1.15/euro) = $563.62

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4-4

Interest from the first investment = (3%)(P)(2) = 0.06P Interest from second investment = (1.02)(1.02)P – P = 0.0404P Therefore, the return on the first investment (simple interest at 3%) will be higher.

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Year 1

Amount owed at beginning of year 1000

Interest accrued for year 90

Total amount owed at end of year 1090

Principal payment 0

Total end of year payment 90

2

1000

90

1090

0

90

3

1000

90

1090

1000

1090

4

500

45

545

0

45

5

500

45

545

0

45

6

500

45

545

500

545

4-5

Total interest= $405. $90 in interest is payable each year for the first three years. $45 in interest is payable each year for the second three years. The total interest paid over the six-year period is $405.

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4-6

(a)

A = $15,000 (A/P, 12%, 4) = $15,000 (0.3292) = $4,938 per year

Total interest paid = ($4,938)(4) – 15,000 = $4,752 (b) Year 1 2 3 4

Amt. at BOY $15,000 $15,000 $7,500 $7,500

Interest Total owed during yr. at EOY $1,800 $16,800 $1,800 $16,800 $900 $8,400 $900 $8,400 $5,400 = total interest

Principal Repaid 0 $7,500 0 $7,500

Total EOY Payment $1,800 $9,300 $900 $8,400

Principal is repaid faster in Part (a).

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4-7

(a) (The monthly payment is $24,000 (A/P, 1%, 60) = $24,000 (0.0222) = $532.80.

(b) Payment

Interest

Principal

Remaining Loan

Number

Payment

Repayment

Balance

1

$240.00

$292.80

$23,707.20

2

$237.07

$295.73

$23,411.47

3

$234.11

$298.69

$23,112.78

Interest in the third month is $234.11 and principal repayment is $298.69.

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4-8

A = $8,000 (A/P, 5%, 5) = $1,847.80

Payment

Owed at beginning of year

Principal payment

Interest payment

Payment at end of year

1

8,000

£1,447.80 £400.00

£1,847.80

2

£6,152.20

£1,520.19 £327.61

£1,847.80

3

£4,304.40

£1,596.20 £251.60

£1,847.80

4

£2,456.60

£1,676.01 £171.79

£1,847.80

5

£608.81

£1,759.81 £87.99

£1,847.80

Payment

Owed at beginning of year

Principal payment

Interest payment

Payment at end of year

1

8,000

£1,447.80 £400.00

£1,847.80

2

£6,152.20

£1,520.19 £327.61

£1,847.80

3

£5,032.01

£1,596.20 £251.60

£1,847.80

4

£3,435.81

£1,676.01 £171.79

£1,847.80

5

£1759.80

£1,759.81 £88.00

£1,847.80

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4-9

Here we are given A = $624 per year, i = 7% and N = 10 years. We want to determine F at the end of 10 years: F = $624 (F/A, 7%,10) = $8,621.43

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4-10

Consumer Loan: F = $15,000 (F/P, 10%, 5) = $24,157.50.

Interest = $24,157.50− $15,000 = $9,157.50.

PLUS Loan: F = $15,000 (F/P, 7%, 5) = $21,039.

Interest = $6,039.

Difference = $9,157.50− $6,039 = $3,018.50

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4-11

The compounded future amount, given P = $6,000, i = 8% per year and N = 23 years, will be F = $6,000 (F/P, 8%, 23) = $35,229.

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4-12

The affordable amount, P, is the following.

P = $100 (P/A, 12%, 15) = $100 (6.8109) = $681.09

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4-13

P = $20,000 (P/F, 3%, 20) = $11,074.

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4-14

Here we are given P = $32,000, A = $618.65 per month and i = 0.5% per month. The equivalence relationship is: $32,000 = $618.65 (P/A, 0.5% per month, N) We can determine N by inspecting Table C-2. From Table C-2 when N = 60 months, we see the following: $618.65(51.7256) = $32,000. Thus, the loan will be completely repaid in 60 months.

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4-15

F33 = $1,200(F/A, 5%, 25)(F/P, 5%,8) − $4,500(F/A, 5%,5)(F/P, 5%,3) − $3,000(F/A, 5%,2)(F/P, 5%,1) F33 = $1,200(47.7271)(1.4775) − $4,500(5.5256)(1.1576) − $3,000(2.0500)(1.0500) F33 = $49,378.69.

F33 = $1,200(F/A, 5%, 25)(F/P, 5%,8) − $4,500(F/A, 5%,5)(F/P, 5%,3) − $3,000(F/A, 5%,2)(F/P, 5%,1) F33 = $1,200(47.7271)(1.4775) − $4,500(5.5256)( 1.1576) − $3,000(2.0500)(1.0500) F33 = 84620.15 – 28783.96 – 6457.50 = $49,378.69

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4-16

Solution by Rule of 72: N 

72  7.2 years.

10

Compare to the exact solution: $10,000 = $5,000 (F/P, 10%, N) = $5,000 (1 + 0.10)N 2 = 1.1 N ; ln(2) = N ln(1.1); and N = 7.2725 years

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4-17

This is a “given P, find F” type of problem. Specifically, we use the following relationship to find F in 2024: F = $74.50 (F/P, 7%, 7) = $74.50 (1.6058) = $119.63 per month

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4-18

The simplest way to determine the future value of the 15 payments is: F = $2,000 (F/A, 4%, 15)(F/P, 4%, 1)= $2,000(20.0236)(1.04) = $41,649.09.

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4-19

In this problem we know P ($32,300), A ($300 per month) and N (10 years). We can calculate the value of the unknown i' % as follows: $32,300 = $300 (P/A, i' % per month, 120 months) Using Goal Seek in Excel, we determine i' = 0.1827% per month.

Goal Seek Setup: Cell A1 = unknown interest rate = 0.01 (starting guess) Cell B1 = 300*((1+a1)^120-1)/(a1*(1+a1)^120) In Goal Seek, set value of cell B1 = 32300 by changing cell A1

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4-20

N = 2014 – 1955 = 59 years 60 = 0.10 (1 + i)59 = .1145 i  59 60 .1  1 = 11.45% per year.

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4-21

(a)

N = 2012 – 1985 = 27 years $100,000 = $10,000 (1 + i)27 i = 27 10 − 1 = 0.089 = 8.9%.

Barney and Lynne had a stock return at just below the standard for a good mutual fund.

(b)

$100,000 = $800 (F/A, i′, 27) (F/A, i′, 27) = 125. i′ = 10.2% per year.

In this situation they did marginally better on their mutual fund investment.

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4-22

We can solve for the compound annual increase as follows: $42,700 = $1,900 (F/P, i'% per year, 57 years) or i = 57 427 19 − 1 = 0.056 = 5.6%.

This is more than double the average consumer price index over this 57-year interval of time.

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4-23

At time zero, an amount of $330,000 is equivalent to $35,000 received at the end of years one through ten. This cash flow pattern may be solved for the unknown value of i'% in the following equation: $330,000 = $35,000 (P/A, i'%, 10) Using Goal Seek in Excel we find i'% = 1.08% Goal Seek Setup: Cell A1 = unknown interest rate = 0.05 (starting guess) Cell B1 = 35000*((1+a1)^10-1)/(a1*(1+a1)^10) In Goal Seek, set value of cell B1 = $330,000 by changing cell A1

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4-24

Let's compute the equivalent value of the cash flows at the present time (year 0): P = −$70,000 + $8,500 (P/A, 9%, 10) = −$70,000 + $8,500 (6.4177) = −$15,450 So the investment is not worthwhile.

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4-25

The F-equivalent amount of this series of deposits can be determined as follows: F = $1,000 (F/A, 4%, 25) = $1,000 (41.6459) = $41,645.90.

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4-26

Coach Blooper's monthly buyout amount is determined as follows: A = $1,200,000 (A/P, 2%, 24) = $1,200,000 (0.0529) = $63,480 per month.

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4-27

0.01 ($20,000) = $200 per year will be placed in the savings account. This will accumulate to F = $200 (F/A, 5%,15) = $200 (21.5786) = $4,315.72

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4-28

The monthly amount of the mortgage cannot exceed $900 – $200 = $700. This is a “given A to find P” type of problem. The monthly interest rate is 6%/12 = 0.5% (0.005) on the loan. The value of the affordable amount is: P = $700 (P/A, 0.5%/month, 300 months) P = $700 [(1 + 0.005)300 – 1] / [0.005(1 + 0.005)300] P = $700 (155.2069) P = $108,645

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4-29

(a)

F = $763 (F/A, 8%, 35) = $763 (172.3168) = $131,477.72. Liam’s friend has exaggerated the truth because the claim is much too high.

(b)

Liam is trading off the answer in part (a) against the likelihood that he will survive to age 60.

IfLiam’s health is good, he may choose to gamble and go ahead with the mutual fund.

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4-30

(a)

The monthly payment will be: A/month = $100,000 (A/P, ½%, 360 months) = $100,000

.

. .

= $100,000 (0.0060) = $600 per month (b)

If Will doubles the payment to $1,200 per month, he will repay the loan in: $100,000 = $1,200 (P/A, 0.5%, N) N ≈ 108 months

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4-31

(a)

P = $13,000 (P/A, 3%, 12) = $13,000 (9.9540) = $129,402.

(b)

P = $13,000 + $13,000 (P/A, 3%, 11) = $133,284.

(c)

The present equivalent in part (b) is higher because the cash flows are not as far into the future, so less discounting occurs.

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4-32

Since payments are made monthly, first find the monthly interest rate in effect. $30,000 = $1,177 (P/A, i′, 36) (P/A, i′, 36) = 25.4885 Using Appendix C(Table C-5) we find that (P/A, 2%, 36) = 25.4888, so i′ = 2% per month.

The effective annual interest rate is (1.02)12 – 1 = 0.2682 or 26.82% per year.

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4-33

The carbon fiber automobile would average around 40 miles per gallon of gasoline. The dollar savings in fuel per year is

135,000 / 5 135,000 / 5  A =   ($4.20) = $945 30 40 

The extra sticker price that can be afforded is $945 (P/A, 15%, 5) = $3,168. This is not a whole lot of wiggle room compared to the sticker price of the conventional car. So, the cost of manufacturing carbon fiber automobiles must be made much more competitive against conventional ferrous−aluminum car bodies and parts.

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4-34

First investment: Net present value = −$70,000 + $15,000(P/A, 5%, 5) = −$70,000 + $15,000(4.3295) = −$70,000 − $64,942.5 = −$5,057.5.

Second investment: Net present value = −$40,000 + $5,000 (P/A, 5%, 5) = −$40,000 + $5,000(4.3295) = −$40,000 + $21,647.5 = −$18,352.5.

Since both projects cost more than they deliver in savings, the company should not invest in either.

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4-35

P = $400,000 + $400,000 (P/A, 2%, 24) = $7,965,560.

If Spivey is offered a lump-sum payment now that exceeds $7,965,560, he should take it.

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4-36

Set up an equation for the unknown P-equivalent amount: P – 0.35P = 0.8 ($3,000)(P/A, 8%, 16), or 0.65P = $2,400 (P/A, 8%, 16) or, 0.65P = $2,400 (8.8514) The value of P (affordable amount) is $32,682. This type of heat pump is also a cleaner choice for the environment.

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4-37

Assuming $3 million per year can be saved, the equivalent present value of savings will be: PW (savings) = $3,000,000 (P/A, 5%, 18) = $3,000,000 (11.6896) = $35,068,800 Since the present equivalent of the savings is greater than the cost of the improvements, Green Corporation will not have to pay back any money.

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4-38

(a) In 40 years you would have $40,000 (F/P, 6%, 40) = $40,000 (10.2857) = $411,428 in real purchasing power. (b) In 20 years you would have $4,000 (F/A, 6%, 20) = $4,000 (36.7856) = $147,142. At the end of 40 years you would have $147,142 (F/P, 6%, 20) = $147,142 (F/P, 6%, 20) = $471,899.

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4-39

N = 12 × 40 = 480 months A = $900,000 (A/F, 0.75%, 480) = 900 ,000

0.0075 = $192.25 per month.

1.0075 480  1

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4-40

(a)

Annual reduction in electricity consumption = $100,000/$0.25 = 400,000 kWh.

(b)

Annual reduction in electricity consumption = $300,000(A/F, 10%, 3)/$0.25 = 362,520 kWh.

(c)

The compounding of interest in part (b) compared to no interest in (c).

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4-41

The annual equivalent of F = $750,000 in 35 years at an annual interest rate of 5% is: A = $750,000 (A/F, 5%, 35) = $750,000 (0.0111) = $8,325 per year The engineering graduate earns an average of $8,325 per year more than the Liberal Arts BA holder.

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4-42

A = $10,000,000 (A/P, 15%, 4) = $10,000,000 (0.3503) = $3,503,000

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4-43

Fuel savings over eight years =

100,000 miles 100,000 miles  321 gallons, which is about 40.1  24 miles/gallon 26 miles/gallon

gallons per year. Let X represent the cost of gasoline ($/gal).

$1,000 = (40.1 gal/yr)(X/gal)(P/A, 12%, 8)

or, X =

$1,000 or X = $5.02/gal (40.1 gal/yr)(4.9676)

If the actual cost of gasoline is less than $5.02 per gallon, the CVT is not a good deal.

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4-44

A = $14,000,000 (A/P, 9%, 7) = $14,000,000 (0.1987) = $2,781,800 per year

Total interest paid = $2,781,800 (7) − $14,000,000 = $5,472,600

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4-45

A = $200,000 (A/P, 10%, 8) = $200,000 (0.1874) = $37,480.

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4-46

With traditional financing, the monthly payment will be A = $90,000 (A/P, 0.5%, 48) = $2,115. The monthly payment with the 0% financing plan would be $90,000/48 = $1,875 per month. Total savings = ($2,115 − $1,875)(48) = $11,520. Thus the claim is untrue.

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4-47

$35 million + $39 million = $35 million (F/P, i, 40)

(1 + i)40 = 74/35

i = 1.89% per year

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4-48

0

F=?

1

2

3

4

5

6

7

8

9

10

11 12

A = $250 / year

r = 22%, M = 4 i/yr = (1 + 0.22/4)4 – 1 = 23.88% P0 = $250 + $250 (P/A, 23.88%, 11) = $250 + $250 (3.7905) = $250 + $947.61 = $1,197.61 F12= P0 (F/P, 23.88%, 12) = $1,197.61(1.2388)12 = $1,197.61(13.0621) = $15,643.35

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4-49

(a)

P = A ( P/A, i%, N); $1,000 = $200(P/A, 12%,N); (P/A, 12%, N) = 5.0000 By looking at the 12% interest table in Appendix C under the P/A column, N  8 years.

(b)

P = A ( P/A, i%, N); $1,000 = $200 (P/A, i%,10); (P/A, i%, 10) = 5.0000 (P/A, 15%, 10) = 5.0188 and (P/A, 18%, 10) = 4.4941, thus 15% < i < 18% By using linear interpolation -

18% - 15% i% - 15%  5.0188 - 4.4941 5.0188 - 5.00

 i = 15.11%

(c)

P = A (P/A, i%, N) = $200 (P/A, 12%, 5) = 200 (3.6048) = $720.96

(d)

A = P (A/P, i%, N) = $1,000 (A/P, 12%, 5) = $1,000 (0.2774) = $277.40

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4-50

$250,000 = $5,000,000 (A/P, 4%, N), so (A/P, 4%, N) = 0.05. From Table C-7, 40 ≤ N ≤ 45. If i = 2%, Table C-5 gives us 25 ≤ N ≤ 30.

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4-51

$3,800 = $100 (P/A, 2%, N), so (P/A, 2%, N), = 38. From table C-5, we see that 72 < N < 84 months to repay the credit card value.

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4-52

$300,000 = $25,000 (P/A, 4%, N), so (P/A, 4%, N) = 12. From Table C-7, 16 ≤ N ≤ 17 years.

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4-53

First we determine how much gasoline is consumed each year: At 24 mpg, 12,000/24 mpg = 500 gallons of gasoline will be consumed in one year. At 28 mpg, 12,000/ 28 mpg = 428.6 gallons of gasoline will be consumed each year.

The more fuel-efficient car will save 71.4 gallons of gasoline each year, which translates into 71.4  $5 = $357 annual savings. Over a six-year period, the present equivalent amount of savings is $357 (P/A, 5%, 6) = $1,812.02. Based on fuel savings, this is the extra amount that you can afford to spend on the new car relative to your present car.

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4-54

F12 = $25,000(F/P, 0.75%, 12) = $25,000(1.0938) = $27,345.

A = F12(A/P, 0.75%, 60) = $27,345(0.0208) = $568.78 per month

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4-55

F = $2,000 (F/A, 8%, 8)(F/P, 8%, 30) = $2,000(10.6366)(10.0627) = $214,066.

This sounds too good to be true, but it really is true! The stock market is one place where a return of 8% might be earned over the span of time in this problem.

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4-56

The instructor is encouraged to draw a cash flow diagram for this problem.

P10 (of deposits) = A (F/A, 5%, 10), and P11 = P10 (F/P, 5%, 1) P11' (of withdrawals) = $400 (P/A, 5%, 6) By letting P11 = P11', we have [A (F/A, 5%,10)](F/P, 5%, 1) = $400 (P/A, 5%, 6) [A(12.5779)] (1.05) = $400 (5.0757) A = $153.73.

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4-57

This is a deferred annuity, the time periods are months, and i = 1/2 % per month: P59 = $600 (P/A, 1/2%, 60) = $600 (51.7256) = $31,035.36.

P0 = $31,035.36 (P/F, 1/2%, 59) = $31,035.36(0.74508) = $23,123.80.

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4-58

The instructor is encouraged to draw a cash flow diagram for this problem. Lisa will make 35 annual payments of $A each into her retirement account. So at age 65 she will have F = $A (F/A, 5%, 35) = $ 90.3203A in her account. The amount A that will fund Lisa's withdrawals of $15,000 for 15 years will be: 90.3203A = $15,000 (P/A, 5%, 15), or 90.3203A = $15,000 (10.3797) A = $1,724.

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4-59

(a)

Sum of expenses = $3,000 + $2,000 + 3($1,000) = $8,000.

(b)

P0 = $3,000(P/F, 10%, 1) + $2,000(P/F,10%,2) + $1,000 (P/A,10%,3) (P/F,10%,3) P0 = $3,000(0.9091) + $2,000(0.8264) + $1,000 (2.4869) (0.7513) P0 = $6,248.51.

(c)

A = $6,248.51(A/P,10%,5) = $1,648.36

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4-60

The present equivalent amount at time zero (now) is: P = −$30,000(P/A, 7%, 10) − $80,000(P/F, 7%, 15) − $80,000(P/F, 7%, 20) P = −$30,000(7.0236) − $80,000(0.3624) − $80,000(0.2584) P = −$210,708 − $28,992 − $20,672 = −$260,372.

Then the equivalent annual amount over 30 years is: A = −$260,372(A/P, 7%, 30) = −$260,372(0.0806) = −$20,986.

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4-61. (a) (b)

F65 = $2,500 (F/A, 5%, 40) = $2,500 (120.7998) = $302,000.

A66−90 = $302,000 (A/P, 5%, 25) = $302,000 (0.0710) = $21,442 per year or about $1,787 per month.

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4-62

(a)

The future compound value of cash inflows (deposits) to the account is = X (F/A, 4%, 4)(1.04) = X(4.2465)(1.04) = 4.41636X = $25,000.

We find X = $5,660.77.

(b)

The future compound value of cash inflows (deposits) to the account is = X(F/A, 4%, 16)(1.04)4 = X(21.8245)(1.1699) = 25.53248X = $25,000.

We find X = $979.15.

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4-63

(a)

Cost in 12 years = $80,000 (F/P, 8%, 12) = $80,000 (2.5182) = $201,456

A = $201,456 (A/F, 5%, 12) = $201,456 (0.0628) = $12,651

(b)

Because $220,000 is more than $201,456, this is not a good deal (the college’s cost increases are less than 10% per year).

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4-64

The future equivalent value of the “investment” in an engineering bachelor’s degree is:

F = $20,000 (F/A, 18%, 4) = $20,000 (5.2154) = $104,308 Next the annual equivalent “return” on this investment can be determined: A = $104,308 (A/P, 18%, 15) = $104,308 (0.1964) = $20,486 This is the extra annual amount that an engineering degree will be worth.

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4-65

Let the reference point in time be EOY 0. The left-hand diagram’s discounted value at EOY 0 is: P0 = 5M (P/F, 10%, 1) + 6M (P/F, 10%, 2) + 4M (P/F, 10%, 3) + 3M (P/F, 10%, 4) = 5M (0.9091) + 6M (0.8264) + 4M (0.7513) + 3M (0.6830) = 14.558M The right-hand diagram’s discounted value at EOY 0 is: P0 = 5M (P/A, 10%, 4) = 5M (3.1699) = 15.850M Since the values at the EOY 0 are not equal, the cash flow diagrams are not equivalent.

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4-66

CFD from viewpoint of investor in this fund.

R

$6,050 $2,000

0

1

2

3

4

5

6

7

8

9

10

11

12

$5,000 Reference point $10,000

F5 = $5,000(1.1)2 = $6,050

Use EOY 7 as the reference point. $10,000(F/P, 12%, 7) + $5,000(F/P, 12%, 4) = $6,050(F/P, 12%, 2) + $2,000(P/A, 12%, 5) + R(P/F, 12%, 5)

Solving for R, R = $26,745

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4-67

The cash flow diagram is shown below: $2.25 million 2017

A

2018

A

2019

A

2020

A

2021

A

2022

A

End of year, i = 15%

To “convert” this diagram to a situation that can be solved using Table C-15, we first need to discount $2.25 million to its equivalent value at the end of year 2016 (one-time period before the start of the annual payments).

P2016 = $2.25 million (P/F, 15%, 1) = $2.25 million (0.8696) = $1,956,600 Next, the value of A in years 2017 through 2022 can be determined: A = $1,956,600 (A/P, 15%, 6) = $1,956,600 (0.2642) = $516,934

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4-68

Using time = 0 as the reference point, set P0(LHS) = P0(RHS) $1,000(P/F,12%,1) + $1,000(P/F,12%,3) − $1,000(P/F,12%,5) = W + W(P/F,12%,7) $1,000(0.8929) + $1,000(0.7118) − $1,000(0.5674) = W + W(0.4523) $1,037.30 = 1.4523 W W = $714.25

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4-69

One method- solve at year 6 X

= 120,000 (F/P, 9%, 6) – 20,000 (F/A, 9%, 6) + 20,000 + 10,000 (F/P, 9%, 3) = 120,000 (1.6771) – 20,000 (7.5233) + 20,000 + 10,000 (1.295)

X

= 201,252 – 150,466 + 20,000 + 12,950

X

= $83,736

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4-70

First draw a cash flow diagram: EOY F

0

1,000

1

2

3

1,000

4

5

1,000

6

1,000

7

8

9

10

11

12

1,000

Next set up an equivalence equation: F = 1,000 (F/P, 6%, 12) + 1,000 (F/P, 6%, 10) + 1,000 (F/P, 6%, 8) + 1,000 (F/P, 6%, 6) + 1,000 (F/P, 6%, 4) Finally, crunch the numbers to determine the value of F F = 1,000 (2.0122) + 1,000 (1.7908) + 1,000 (1.5938) + 1,000 (1.4185) + 1,000 (1.2625) F = $8,077.80

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4-71

Profit = ($0.10/gal)(60,000 gal/mo) = $6,000 per month. $67,500 = $6,000 (P/A, 1%, N), so (P/A, 1%, N) = 11.25 From Table C-4, N = 12 months.

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4-72

P0 = $1400 (P/A. 20%, 4) - $300 (P/G, 20%, 4) + 500 (P/A, 20%, 4) (P/F, 20%, 4) + 2,000 (P/F, 20%, 10) P0 = $3,581.87

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4-73

Left side: PL = −10 (P/F, 15%, 1) + H (P/A, 15%, 16−4) (P/F, 15%, 4) + 0.7H (P/A, 15%, 6) (P/F, 15%, 7) = −10 (0.8696) + H(5.4206) (0.5718) + 0.7H (3.7845) (0.3759) = 4.0953H – 8.696

Right side: PR = −P0 + 2P0 (P/F, 15%, 10) = −P0 + 2P0 (0.2472) = −0.5056 P0

set PR = PL: −0.5056 P0 = 4.0953H – 8.696 P0 = 17.2 – 8.1H

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4-74 $10,000 $10,000 $10,000

0

1

2

3

4

5

6

7

8

9

End of Year

X=?

(r/M) = (12%/2) = 6% (1.062) – 1 = 0.1236 X + X (P/A, 6%, 5) = $10,000 (P/A, 12.36%, 3)(P/F, 12.36%, 6) 5.2124X = $11,862.91 X = $2,275.90

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4-75

P0 (rental income) = $1,500 (P/A, 8%, 10) − $50 (P/G, 8%,10) = $1,300 (6.7101) − $50 (25.977) = $8,723.13 − $1,298.85 = $7,424.28.

The present equivalent of the rental income is greater than the present equivalent of the $7,000 investment, so the rental property appears to be a good investment.

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4-76

P0 (loan amount) = P0 (loan repayment) $12,000 = Z(P/G,12%,6)(P/F,12%,4) $12,000 = Z(8.93)(0.6355) $12,000 = 5.675015Z Z = $2114.53.

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4-77

(a)

F = $600 (P/G, i%, 6)(F/P i%, 6) = $10,000 If i = 7%, F = $9,884.81 < $10,000,  i > 7% If i = 8%, F= $10,019.37 > $10,000,  i < 8% Thus, 7% < i < 8%. Using linear interpolation, i% - 7% 8% - 7% = $10,000 - $9,884.81 $10,019.37 - $9,884.81

i = 7.86% (b)

F = $600 (P/G, 5%, N) (F/P, 5%, N) = $10,000 If N = 6, F = $9,622.99 < $10,000,  N > 6 If N = 7, F= $13,704.03 > $10,000,  N < 7 Thus, 6 < N < 7. Using linear interpolation, N-6 7-6 = $10,000 - $9,622.99 $13,704.03 - $9,622.99

N = 6.1 periods. If an integer value of N is desired, choose N = 7 periods.

(c)

F = $1,000(P/G,10%,12)(F/P,10%,12) = $1,000(29.901)(3.1384) = $93,841.30

(d)

G = F(P / F,10%,6)

1 1  $8,000( 0.5645)  $466.34 (P / G,10%,6) 9.684

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4-78

(a)

$3,000 = $100 (P/A, 1%, 24) + G (P/G, 1%, 24) = $100 (21.2434) + G (234.18) G = $3.74 per month beginning at the end of month 2

(b)

A = $3,000 (A/P, 1%, 24) = $3,000 (0.0471) = $141.30

(c)

G=

$3,000  $120(21.2434) = $1.92 per month beginning at end of month 2.

234.18

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4-79

Using time 1 as the reference point, set P1(LHS) = P1(RHS) K(P/A,12%,2) (P/F,12%,2) = $100(P/A,12%,6) + $110(P/G,12%,6) K(1.6901) (0.7972) = $100(4.1114) + $110(8.93) 1.3473 K = $1,393.44 K = $1,034.25

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4-80

920 900 880 860

i = 6% per year

840 820 800

0

1

2

3

4

5

6

7

300

300

300

8

9

10

500 EOY

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4-81

A = [$2,000 (P/A,8%,4) + $400 (P/G,8%,4)] (P/F,8%,2) (A/F,8%,11) = [$2,000 (3.3121) + $400 (4.650)] (0.8573) (0.0601) = $437.14

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4-82

We can set up an equivalence equation with time 0 as the reference point: 2,000 = 100 + 500 (P/A, 6%, 5)(P/F, 6%, 1) – G(P/G, 6%, 5)(P/F, 6%, 1) Or 2,000 = 100 + 500 (4.2124)(0.9434) – G(7.935)(0.9434) 2,000 = 2,087 – 7.4859G G = $11.62

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4-83

Using time 1 as the equivalence point: − R (P/G,15%,N − 1) = −27R (P/F,15%,3) + 27R (P/F,15%,9) − R (P/G,15%,N − 1) = −27R (0.6575) + 27R (0.2843) (P/G,15%,N − 1) = 10.0764 From Table C-15, (P/G,15%,7) = 10.192, thus N − 1 = 7 and N = 8 years.

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4-84

The present equivalent of energy savings can be determined with Equation (4-30):

 P  P  1   F , 18%, 15  F , 15%, 15     P = $24,000   0 .

18  0 .

15    



$24,000(0.3206) 0.03

= $256,480.

Therefore, the investment in the insulation is justified by a wide margin.

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4-85

Note N = 6  P  P  1   F , 20%, 6  F , 9%, 6     P0 = $200,000   0 .

20  0 .

09    

1  (0.3349)(1.6771)   $200,000   0.11 

= $796,980 You can afford to spend as much as $796,980 for a higher quality heat exchanger.

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4-86

N = 40 years

A1 = $500

𝑓 ̅ = 2%

i = 4%

The P-equivalent of Jay’s account when he is 20 years old (from Equation 4-30) is:

P=

$

⁄ , %, .

⁄ , %, .

=

$

–

.

. .

=

$

. .

= $13,502.50

The future worth of the account (just after Jay’s 40th deposit) will be:

F = $13,502.50(F/P, 4%, 40) = $13,502.50(4.8010) = $64,825.50

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4-87

P=

$500[1  P / F ,12%,15)( F / P,6%,15)] 0.12  0.06

= $4,684.51

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4-88

 P  P  1   F , 20%, 10  F , 12%, 10      = $69,778 PGas = $11,200   0.20  0.12    

 P  P  1   F , 20%, 10  F , 18%, 10      = $5,158 PMaint. = $565(1.18)   0.20  0.18     A = ($69,778 + $5,158)(A/P, 20%, 10) = $17,872

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4-89

̅ One year prior to your start of Refer to Equation 4-30 when i is not equal to 𝑓 ̅ (in this problem i < 𝑓 ). college is considered "year 0" in our calculations below:

P0 = $12,000 [1- (P/F, 5%, 4)(F/P 8%, 4)] / (0.05 - 0.08)

= $12,000 [1 - (0.8227)(1.3605)] / -0.03

=$12,000 (-0.11928) / -0.03

= $47,712

You and your parents need to deposit $47,712 in the mutual fund at the start of your senior year in high school.

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4-90

A = 0.1  $40,000 = $4,000.

F=

$4,000[1  ( P/F, 10%, 40)(F/P, 2%, 40)] (F/P, 10%, 40) 0.1  0.02

F=

$4,000[1  (0.0221)(2.2080)] (45.2593) = $2,152,539.55.

0.1  0.02

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4-91

(a) P0 = =

$8,000[1  ( P / F ,10%,10)( F / P,5%,10)] 0.10  0.05 $8,000[1  (0.3855)(1.6289)] 0.05

= $59,529

We can justify spending up to $59,529 for the device.

(b) Using the result of part (a):

A = $59,529 (A/P, 10%, 10) = $59,529 (0.1627) = $9,685

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4-92

P = $30,000(P/A, 6%, 4) + P = $30,000(3.4651) +

$12,000[1  ( P/F, 6%, 4)(F/P, 5%, 4)] 0.06  0.05

$12,000[1  (0.7921)(1.2155)] = $148,595.94.

0.06  0.05

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4-93

$3,000[1  ( P/F, 6%, 8)(F/P, 5%, 8)] $3,000[1  (0.6274)(1.4775)] = = $21,904.95.

0.06  0.05 0.01

(a)

P0 =

(b)

P0′ = $1,000 (P/A, 6%, 8) + G (P/G, 6%, 8)

(c)

Set P0 = P0′ and solve for G: $21,904.95 = $1,000 (6.2098) + G (19.842) G = $791.

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4-94

P = 100  $ 10(P/F, 4%, 1) + 100  $ 10(P/F, 4%, 2) + 100  $10(P/F, 4%, 3) + 100  $10(P/F, 4%, 3) (P/F, 6%, 1) + 100  $10(P/F, 4%, 3)(P/F, 6%, 2)[For the first 3 years $1000 dividend is an annuity] P = $1,000(P/F, 4%, 1) + $1,000(P/F, 4%, 2) + $1,000(P/F, 4%, 3)[1 + (P/F, 6%, 1) + (P/F, 6%, 2)] P = $1,000(.9615) + $1,000(.9246) + $1,000(0.8890)[1 + 0.9434 + 0.89] = $961.50 + $924.60 + $2,518.89 = $4,404.89.

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4-95

F = $1,500 (F/P, 8%, 1) (F/P, 9%, 1) (F/P, 10%, 3) F = $1,500 (1.08) (1.09) (1.331) = $2,350.28

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4-96

First we discount the September payment of $10,000 for the three months to bring it back to June 15 (and add it to $13,000): 𝑃

= $10,000 (P/F, ¾%, 3 months) + $13,000 = $10,000 (0.9778) + $13,000 = $22,778

Then we discount $22,778 for two months and add it to $12,000: 𝑃

= $22,778 (P/F, ½%, 2) + $12,000 = $22,778 (0.9901) + $12,000 = $34,552.50

Finally we see that 𝑃

= $34,552.50 (P/F, ¼%, 3) + $10,000= $44,293.35

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4-97

P = $1,000 (P/F, 8%, 1) + $2,000 (P/F, 10%, 1)(P/F, 8%, 1) + $1,000 (P/F, 6%, 1)(P/F, 8%, 1)(P/F, 10%, 1)(P/F, 8%, 1) + $2,000 (P/F, 10%, 1)(P/F, 6%, 2)(P/F, 8%, 1)(P/F, 10%, 1)(P/F, 8%, 1) = $1,000 (0.9259) + $2,000 (0.9091)(0.9259) + $1,000 (0.9434)(0.9259)(0.9091)(0.9259) + $2,000 (0.9091)(0.8900)(0.9259)(0.9091)(0.9259) = $4,606

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4-98

This solution assumes a 10-year repayment period (common practice) to compute the minimum monthly payments.

ACard 1 = $4,500 (A/P, 21/12 %, 120) = $89.97 ACard 2 = $5,700 (A/P, 2%, 120) = $125.67 ACard 3 = $3,200 (A/P, 1.5%, 120) = $57.66 AConsolidated = $13,400 (A/P, 16.5/12 %, 120) = $228.66

Mary’s current monthly payments total $273.30. The consolidated payment represents a ($273.30 − $228.66)/$273.30 × 100% = 16.3%. So while consolidating the credit cards will benefit Mary, the company has overstated the amount of this savings.

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4-99

(a)

Using Equation 4-32, 0.1 = [1 + (i/4)]4 – 1 Solving for i, we get 9.6%.

(b)

Using Equation 4-32, i = [1 + (.07/12)]12 – 1 = 7.23%.

(c)

Since compounding is annual, the effective annual interest rate is equal to the nominal interest rate. 4 = 1 (F/P, 5% per year, N years), or 4 = (1.05)N, so we can compute that N = log (4)/log (1.05) = 28.4 years.

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4-100 Using Equation 4-32, i = [1 + (.04/365)]365 – 1 = 4.08%.

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4-101 Using Equation 4-32, i = [1 + (.35/52)]52 – 1 = 41.74%.

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4-102 (a) (b)

(1 + 0.2 / M)M = 1.219, or M = 12 (i.e. monthly compounding) (1 + 0.2 / M)M = 1.221, or M = 365 (i.e. daily compounding)

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4-103 The number of installments is 5 × 12 = 60 and the interest rate per month is 24/12 = 2%.

A = P (A/P, 2%, 60) = $20,000(0.0288) = $576 per month for 60 months.

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4-104 A = F (A/F, 4%, 40) = $10,000(.0105) = $105 per quarter.

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4-105 Present value of second option = $200 + A(P/A, .5%, 12) = $200 + $120(11.6189) = $1,594.27.

The difference in price is $194.27.

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4-106 P = A(P/A, 3%, 6) = $200(5.4172) = $1,083.44.

The value of the washing machine is only $1,083, rather than $1,200.

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4-107 (a) (b)

A = P(A/P, .5%, 48) = $5,000(0.0235) = $117.50.

A = P(A/P, .75%, 60) = $5,000(0.0208) = $104.

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4-108 (a)

(b)

The total interest rate payable depends on the flexible portion of the interest rate. If this is linked to the same inflation index that increases by 3% every year, then the total interest rate will be the sum of the fixed interest rate and the inflation-linked interest rate, which is 6%.

F = A(F/A, .5%, 36) = $200(39.3361) = $7,867.22.

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4-109 (a)

F1 = $1,000 (F/P, 1%, 60) = $1,000(1.8167) = $1,816.70.

F2 = $1,000 (F/P, 0.75%, 48) = $1,000(1.4314) = $1,431.40.

(b)

The five-year bond will yield a higher interest income; however, you need to be willing to invest your principal for an additional year. If you expect alternative investment opportunities to emerge after the fourth year that pay a higher return than the five-year bond, then you may choose the four-year bond.

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4-110 Under scheme 1: the deposit is equal to $250,000(.25) = $62,500. A mortgage is taken out on the remainder, which is $250,000 − $62,500 = $187,500.

The monthly payments required to pay off this loan are = A = $187,500(A/P, 1%, 100) = $187,500(0.0159) = $2,981.25. The total interest cost of the loan is given by $2,981.25(100) − $187,500 = $110,625.

Under scheme 2: the deposit is equal to $250,000(0.05) = $12,500. A mortgage is taken out on the remainder, which is $250,000 − $12,500 = $237,500.

The monthly payments required to pay off this loan are = A = $237,500(A/P, 2%, 100) = $237,500(0.0232) = $5,510. The total interest cost of the loan is given by $5,510(100) − $187,500 = $363,500.

Therefore, the difference in monthly payments is $2,528.75 and the difference in interest costs is $252,875.

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4-111 (a)

$1,100 = $19.80 (P/A, 1.5%, N), so (P/A, 1.5%, N) = 55.5556

This can be solved by trial and error or using Excel.

N = NPER (0.015, 19.8, −1100) = 120

(b)

N = NPER (0.015, 29.8, −1100) = 54

(c)

Difference = $19.80 (120) − $29.80 (54) = $766.80

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4-112 The weekly interest rate is 5.2% /52, or 0.1% per week. The present equivalent value is P= $5,000 (P/A, 0.1%/ wk., 208 wks.)

P= $5,000 (187.7085) = $938,543 The future equivalent value at the end of 4 years is F= P(F/P, 0.1%/wk., 208 wks.)

F= 938,543 (1.2311) = $1,155,440

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4-113 10% = ei − 1 ln (1.1 ) = i or i = 9.53%.

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4-114 (a)

A = $2,500(A/P, 3%, 24) = $2,500(0.0590) = $147.50.

(b)

A = $2,500(A/P, 3%, 24) = $2,500(0.0593) = $148.25.

(c)

The increased frequency of compounding leads to a higher interest cost with b, compared to a, and, hence, a higher monthly payment to repay the debt.

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4-115 (a)

A = $8,000(A/F,8%,10) =

$8,000 $8,000 = $543.67  (F / A,8%,10) 14.7147

(b)

P = $1,000(P/A,8%,12) = $1,000(7.4094) = $7,409.40

(c)

r = 8%/2% = 4%

 e (0.04)(12)  1 F = $243(F/A,4%,12) = $243  0.04  = $243(15.0959) = $3,668.30 1   e (d)

F = $1,000(F/P,8%,9) = $1,000(2.0544) = $2,054.40

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4-116 Set Equivalent cash outflows = Equivalent cash inflows Using time 9 as the equivalence point, Z (F/P,20%,9) = $500(F/A,20%,5) + Z (F/P,20%,6) 6.0496 Z = $500(7.7609) + 3.3201 Z 2.7295 Z = $3,880.45 Z = $1,421.67

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4-117 P = $1,000(e−0.073) = $1,000(0.8106) = $810.60.

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4-118 F = $200 (F/A, 0.5%, 120) = $32,802.61.

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4-119 A = $20,000(A/P, .75%, 120) = $20,000(0.0127) = $253.72.

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4-120 F = $400 (F/A, 3%, 15) = $400 (18.661) = $7,464.40.

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4-121 F = P(F/P, i, 6) 12000 = 6000(e6i) i = ln 2/6 = 11.55%

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4-122 (a) (b)

F = $4,000 (F/P, 12%, 5) = $4,000(1.8221) = $7,288.40.

Now, Will earns accumulated interest on the $4,000 invested when he was 21 for 25 years, as well as the accumulated savings from investing $100 every month for ten years and $200 every month for the next ten years.

F = $4,000(F/P, 12%, 25) + $100(F/A, 1%, 120)(F/P, 12%, 10) + $200(F/A, 1%, 120) F = $80,342.15 + $76,646.10 + $46,170.7 = $203,159

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4-123 (a) False; (b) False; (c) False; (d) True; (e) False; (f) True; (g) False; (h) False; (i) False

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4-124 (a) (b)

True; i/yr = er −1 = e0.1 −1 = 0.1052 or 10.52% > r = 10% True; In fact, more than half of the principal is still owed after the tenth monthly payment is made. r = 10% / 12 = 0.833% per month compounded continuously. 

    e 0.00833(24) ( e 0.00833  1)  e 0.00833(24)  1

P0 = $185(P/A,0.833%,24) = $185  = $185(21.6627) = $4,008

Amount still owed immediately following = $185(P/A,0.833%,10) = $185(13.1595) = $2,435 tenth payment $2,435 * 100 = 61% of principal is still owed after tenth payment. $4,008 2

(c)

 0.08   1 = 0.0816 = 8.16% False; r = 8%, M = 2, i/yr = 1  2   F = P(F/P, i%, N) ?

$1,791  $900 (F/P,8.16%,10) ?

$1,791  $900 (1.0816)10 $1,791  $1,972 (d)

True

(e)

Part i)

(P/A,i%,N) (F/P,i%,N) = (F/A,i%,N)

Note:

 (1  i) N  1  i (1  i) N  ( F / A , i%, N ) = (F/A,i%,N)(A/P,i%,N) =    N i ( P / A , i%, N )    (1  i)  1 = (1+i)N = (F/P,i%,N)

Part ii) (A/G,i%,N)(P/A,i%,N) = (P/G,i%,N) Note:

1   (1  i) N  1 N (A/G,i%,N)(P/A,i%,N) =    N N   i (1  i)  1  i(1  i)  =

1  (1  i) N  1 N     N i  i(1  i) (1  i) N 

= (P/G,i%,N)

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4-125 If an investment of $100 was made in gold 40 years ago, that investment would be worth $700 now.

700 = 100 (e40i) i = ln 7/40 = 4.86%.

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4-126 (a)

A = $22,000(A/P, 0.75%, 40) = $22,000(0.029) = $638.

(b)

This is an open-ended exercise. Monthly expenses include the $638 repayments on the loan, the cost of gas and other maintenance bills.

(c)

A = $22,000(A/P, 0.75%, 60) = $22,000(0.0208) = $457.05.

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4-127 F = P(e.035n) 600 = 160(e.035n) n = ln 3.75/.035 = 37.8 years.

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4-128 P = A (P/A, 5.87/12%, 18) = $212(17.1881) = $3,643.90.

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4-129 Credit Card Balance: Credit Card APR:

$17,000 12%

Plan 1: Pay interest due at end of each month and principal at the end of fourth month.

1 $17,000 $170 $17,170 $0 $170 2 $17,000 $170 $17,170 $0 $170 3 $17,000 $170 $17,170 $0 $170 4 $17,000 $170 $17,170 $17,000 $17,170 $‐mo. = $68,000 $680 = total interest

Plan 2: Pay off the debt in four equal end‐of‐month installments (principal and interest).

1 $17,000 $170 $17,170 $4,187 $4,357 2 $12,813 $128 $12,941 $4,229 $4,357 3 $8,585 $86 $8,670 $4,271 $4,357 4 $4,314 $43 $4,357 $4,314 $4,357 $‐mo. = $42,711 $427 = total interest

Plan 3: Pay principal and interest in one payment at end of fourth month.

1 $17,000 $170 $17,170 $0 $0 2 $17,170 $172 $17,342 $0 $0 3 $17,342 $173 $17,515 $0 $0 4 $17,515 $175 $17,690 $0 $17,690 $‐mo. = $69,027 $690 = total interest

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4-130 EOY

Cash Flow

0

$

(15,000)

1

$

2,000

2

$

2,500

3

$

3,000

4

$

3,500

5

$

4,000

6

$

4,000

7

$

4,000

8

$

4,000

P

$

2,189

A

$

410

F

$

4,692

Notice that a cell is not designated for the interest rate. Therefore, the interest rate must be entered specifically in the NPV, PMT, and FV financial functions. The following entries in the designated cells will yield the results for P, A, and F:

B11 = NPV (0.1, B3:B10) + B2

P = $2,189.02

B12 = PMT (0.1, A10, −B11) or = PMT (0.1, 8, −B11)

A = $410.32

B13 = FV (0.1, A10, −B12) or = FV (0.1, 9, −B12)

F = $4,692.38

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4-131 i / yr =

25%

f=

20%

A1 =

$ 1,000

N=

10

EOY 1

$ 1,000

2

$ 1,200

3

$ 1,440

4

$ 1,728

5

$ 2,074

6

$ 2,488

7

$ 2,986

8

$ 3,583

9

$ 4,300

10

$ 5,160

P=

$ 6,703

A=

$ 1,877

F=

$ 62,430

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4-132 Note: This spreadsheet assumes monthly compounding.

Loan Amount # of Payments APR 0% 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12%

$25,000 60

Payment $416.67 $427.34 $438.19 $449.22 $460.41 $471.78 $483.32 $495.03 $506.91 $518.96 $531.18 $543.56 $556.11

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4-133 Amount set aside each month = $311.40 Interest rate = 3%/12 = ¼% per month First payment occurs at the end of the first month of year 6 Amount available at end of 10−year time frame:

F = $311.40 (F/A, ¼%, 60) = $311.40 (64.6467) = $20,130.98

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4-134 Assume the utility cost remains at $150 per month.

Assume a property tax and insurance rate equal to 25% of the principal and interest (P&I) payment.

Total Monthly Payment = $800 = 1.25 (P&I) P&I = $640

Assume a 30−year mortgage at 6% compounded monthly.

Mortgage amount = $640 (P/A, ½%, 360) = $640 (166.7916) = $106,747

Maximum purchase price = $106,747 + $40,000 = $146,747

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4-135 A = $320; i = 10%/12 = 0.833% per month; N = 120 months

F = $320 (F/A, 0.833%, 120) = $320 (204.7981) = $65,535.39

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4-136

I = (P)(N)(i) = $3,000 (7) (0.06) = $1,260 F = P + I = $3,000 + $1,260 = $4,260

Select (e)

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4-137 P = $100,000 (P/F, 10%, 25) = $100,000 (0.0923) = $9,230

Select (b)

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4-138 F = $2,000 (F/A,2%,30) = $2,000 (40.5681) = $81,136

Select (d)

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4-139

i / mo. = 0.5%, N = 30 x 12 = 360 months P = $1,500 (P/A, 0.5%, 360)

 (1.005) 360  1  = $1,500  360   (0.005)(1.005)  = $1,500 (166.7916) = $250,187

Select (c)

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4-140

A = $8,000; G = $7,000 Atotal = $8,000 + $7,000 (A/G, 12%, 5) = $8,000 + $7,000 (1.7746) = $20,422

Select (a)

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4-141

$9,982 = $2,500 (P/A, i', 5) 3.9928 = (P/A, i', 5); i' = 8% / yr $9,982 = G (P/G. 8%, 5) $9,982 = G (7.372) G = $1,354

Select (d)

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4-142 A = $10,000 (A/P, 3%, 20) = $672

Select (c)

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4-143 imonthly =

6.00% / year = 1/2% / month 12 months / year

A = $100,000 (A/F, 1/2%, 60) = 100,000 (0.0143) = $1,430 Select (d)

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4-144 A = $20,000 (A/P,1%,60) = $444 P (of remaining 40 payments) = $444 (P/A,1%,40) or P = $14,579 Select (c)

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4-145

i / mo. = 12%/12 = 1% per month; N = 4 x 12 = 48 months A = $5,000 (A/P, 1% per month, 48 months) = $5,000 (0.0263) = $131.50

Select (a)

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4-146 0.192 = er – 1

r = ln (1.192) = 17.56%

Select (c)

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4-147

r = 12%

i = e0.12 – 1 = 0.1275 or 12.75% compounded annually

Select (c)

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4-148

F

= $7,000 (F/P, 12%, 3) = $7,000 e(0.12)(3) = $7,000e0.36 = $7,000 (1.4333) = $10,033

Select (c)

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4-149 A = $200,000 (A/P, 4.2%/12, 360) = $200,000 (0.0049) = $980

Select (b).

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4-150 (P/A, 5%/12, N) = $100,000 / $550 per month, so (P/A, 5%/12, N) = 181.8182

Using Goal Seek in Excel, N = 341, so the correct answer is (b) = 360 – 341 = 19 months.

Select (b).

Goal Seek Setup:

Cell A1 = 360 (initial guess)

Cell A2 = ((1 + 0.05/12)^A1 - 1)/((0.05/12)*(1 + 0.05/12)^A1)

Set Cell A2 = 181.8182 by changing Cell A1

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4-151 F = $200 (F/A, 0.25%, 120) = $200 (139.7414) = $27,948.28.

Select (c).

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4-152 F = $6,000 (F/A, 8%, 15) F = $6,000(27.8569) = $167,141 Select answer (d).

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4-153 i = 7%/12 = 0.583% per month F = $100 (F/A, 0.583%, 12 × 40) = $100 [(1.00583)480 – 1]/0.00583 = $262,187 Select answer (c).

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Solutions to Chapter 5 Problems

5-1

All five projects are more profitable than the 10% being earned on municipal bonds and are therefore feasible investments. We need to consider rank-ordered projects (based on the annual rate of profit) versus a cumulative capital investment: Project ranking

C

E

A

B

D

Profit rate per year (%)

28

25

18

15

12

Cumulative investment ($millions)

40

65

95

105

120

Because of limited funds, only projects C, E, and A can be undertaken. The best rejected project is B, so Sydney Tool's MARR is 15% per year. Only whole projects are possible, so B and D cannot be funded.

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5-2

Let’s calculate the present worth (PW) of the venture: PW (10%) = −30 – 20 (P/F, 10%, 1) – 10 (P/F, 10%, 2) + [10 (P/A, 10%, 6) + 10 (P/G, 10%, 6)](P/F, 10%, 3) PW (10%) = $49.03 million, so it’s a good investment.

Note: IRR = 21.3% Yes, PW = $49.03 million

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5-3

The annual equivalent amount of the bi-annual payment is: $25,000 (A/F, 2%, 3) = $25,000 (0.3268) = $8,170 The capitalized worth is CW 

A $8,170   $408,500 i 0.02

So, $408,500 must be deposited now into an account earning 2% per annum such that $25,000 can be paid out every three years indefinitely (starting three years from now).

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5-4

Let X = annual savings required.

AW (15%) = 0 = X – [$8,000,000 (A/P, 15%, 8) − $500,000 (A/F, 15%, 8)] X = $1,746,750/year Net savings per pallet = ($1,746,750/year) / (2,000,000 pallets/year) = $0.87 per pallet

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5-5

First, we determine the equivalent uniform annual cost (EUAC) of the initial cash flow diagram.

EUAC = $500 + $500 (A/G, 10%, 5) = $500 + $500 (1.7746) = $1,387.30 Next, the capitalized worth can be computed: CW (10%) = EUAC/0.10 = $1,387.30/0.10 = $13,873

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5-6

Demand charge / year =

Energy charge / year =

($92 / kW/yr)(100hp)(0.746 kW/hp) = $7,460 / year 0.92

($0.08 / kWh)(100 hp)(0.746 kW/hp)(8,760 hr/yr) = $56,826 / year 0.92

Total annual cost of operation = $64,286

PW(15%) = −$4,000 − $64,286 (P/A, 15%, 10) = −$326,639.

Notice that the annual cost of operating the motor is 16 times the initial investment cost of the motor!

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5-7

PW (9%) = −$20,000 − $35,000 (P/F, 9%, 1) + $48,000 (P/F, 9%, 2) + $67,800 (P/F, 9%, 3) − $5,000 (P/F, 9%, 4) PW (9%) = −$20,000 − $35,000 (0.9174) + $48,000 (0.8417) + $67,800 (0.7722) − $5,000 (0.7084) PW (9%) = $37,105.76

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5-8

The PW of the incremental investment is: mi

PW 8%

-$400

15,000 yr 19 mpg

mi

-

15,000 yr

$3.50

24 mph

gal

(P/A, 8%, 10)

= −$400 + (164.5 gal/yr) ($3.50/gal) (6.7101) = $3,463 This is a very attractive investment in the Ford truck ($3,463 > 0).

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5-9

Assume miles driven each year are roughly constant at (120,000/12) = 10,000. A four-speed transmission car will consume (10,000/40) = 250 gallons of gasoline each year, and the fuel cost is (250 × $5) = $1,250 per year. The six-speed transmission car will attain (1.04)(40 mpg) = 41.6 mpg for an annual consumption of (10,000/41.6) = 240.38 gallons and a fuel cost of (240.38 × $5) = $1,201.90. The annual savings due to a six-speed transmission car is $1,250 – $1,201.90 = $48.10. Over a 12-year life, the present worth of savings will be $48.10 (P/A, 8%, 12) = $48.10 (7.5361) = $362.49. This is how much extra the motorist should be willing to pay for a six-speed transmission car.

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5-10

The bond pays (0.05)($5,000) = $250 once per year. The yield, or effective annual interest rate, can be found as follows: 0 = −$5,500 + $250 (P/A, i’%, 10) + $5,000 (P/F, i’%, 10) i’% = 3.78%

$5,000

A = $250 / year

Buyer’s Viewpoint:

0

1

2

3

4

5

6

7

8

9

10

End of year

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5-11

P = $500(P/A, 7%, 10) + $10,000(P/F, 7%, 10) + $200(P/A, 7%, 3) + $10,000(P/F, 7%, 3) P = $500(7.0236) + $10,000(.5083) + $200(2.6243) + $10,000(.8163) = $17,282.66.

The following modifications aresuggested: P = $500(7.0236) + $10,000(.5083) + $200(2.6243) + $10,000(.8163) = $17,282.66.

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5-12

Price1 = $40(P/A, 3%, 5) + $1,000(P/F, 3%, 5) = $40(4.5797) + $1,000(.8626) = $1,045.79.

Price2 = $40(P/A, 4%, 5) + $1,000(P/F, 4%, 5) = $40(4.4518) + $1,000(.8219) = $999.97.

Price3 = $40(P/A, 5%, 5) + $1,000(P/F, 5%, 5) = $40(4.3295) + $1,000(.7835) = $956.68.

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5-13

Using now as our reference point and letting X = selling price of the bond, X = 1,000 (F/P, 8%, 6) – 100 (F/A, 8%, 6) = 1,000 (1.5869) – 100 (7.3359) = $853.31

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5-14

$1,000,000 + $500,000 (P/F, 25%, 1) − $500,000 (P/F, 25%, 2) = ($Y/0.25)(P/F, 25%, 2) $1,000,000 + $500,000 (0.80) − $500,000 (0.64) = 0.64Y/0.25 1,080,000 = 2.56Y Y = $421,875 Factors: (P/F, 25%, 1) = 0.80; (P/F, 25%, 2) = 0.64

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5-15

From the viewpoint of WRC, they can determine the annuity equivalent of their services: A = $375,000 + $25,000 (A/G, 18%, 80) = $513,885.

Then PW (18%) = $513,885 (P/A, 18%, 80) = $2,854,888.

WRC will receive more upfront money than the value of its services over an 80-year period, so this is a profitable agreement for WRC. They will earn more than 18% on this deal.

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5-16

 A  $20,000 , 15%, 4    $6,000 F   (P/F, 15%, 1) = $63,224  (a) CW(15%) = 0.15  0.15    (b) Find the value for N for which (A/P, 15%, N) = 0.15 From Table C-15, N = 60 years

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5-17

Assuming the team spends $4 million on the plane, the annual worth will be AW = –$4,000,000(A/P, 6%, 20) – $750,000 = –$4,000,000(0.0872) – $750,000 = –$1,098,800.

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5-18 P = $300 (P/A, 3%/qtr, 60 qtr) + $20,000 (P/F, 3%/qtr, 60 qtr) P = $300 (27.6756) + $20,000 (0.1697) P = $8,302.68 + $3,394 P = $11,696.68

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5-19

$2,900 $2,900 $2,800 $2,700 $2,500 $2,400 $2,300 $2,200 $2,100

$500

1

2

3

4

5

6

7

8

9

10

(EOY)

Let A = $2,900, G = −$100 (delayed 1 year) F6 = −$2,000

P0 = $2,900 (P/A,6$,10) – 100(P/G,6%,9)(P/F,6%,1) − $2,000(P/F,6%,6) = $2,900 (7.3601) − $100 (24.5768) (0.9434) − $2,000 (0.7050) = $17,615.71

FW10 = $17,615.71(F/P, 6%, 10) = $17,615.71(1.7908) = $31,546.21

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5-20

The cash flow diagram looks like this: 58k

45k 40k 35k 30k 25k 20k 0

1

2

3

4

5

6

7

EOY 200k The equivalence relationship is the following: AW (10%) = −$200,000(A/P, 10%, 7) + $20,000 + $5,000 (A/G, 10%, 7) + $8,000 (A/F, 10%, 7) AW (10%) = −$200,000 (0.2054) + $20,000 + $5,000 (2.6216) + $8,000 (0.1054) = −$7,130 This is not a good investment

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5-21

FW (15%) = −$15,000 (F/P, 10%, 8) + ($8,800 − $5,800)(F/A, 10%, 8) − $2,500 = −$15,000 (2.1436) + $3,000 (11.4359) − $2,500 = –$346.30 Since FW (10%) < 0, the project is not acceptable.

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5-22

PW = $120,000[1 – (P/F, 5%, 15)(F/P, 1%, 15)]/(.05 – .01) PW = $120,000[1 – (.481)(1.161)]/0.04 = $1,324,677.

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5-23

Opportunity Cost = Investment at BOY x (P/F, 15%, 1) = BOY (0.15) Capital Recovery Amount = Opportunity Cost + Loss in Value During Year

Investment Opportunity at Beginning Cost of Interest Year of Year (i = 15%) 1 $10,000 $10,000 (0.15) = 1,500 2 10,000−3000 = 7,000 7,000 (0.15) = 1,050 3 7,000 − 2,000 = 5,000 5,000 (0.15) = 750 4 5,000 − 2,000 = 3,000 3,000 (0.15) = 450

Loss in Capital Value Recovery During Year Amount $3,000 1,500 + 3,000 = 4,500 2,000 1,050 + 2,000 = 3,050 2,000 750 + 2,000 = 2,750 450 + 1,000 = 1,450 1,000

P0 = $4,500 (P/F,15%,1) + $3,050 (P/F,15%,2) + $2,750 (P/F,15%,3) +$1,450 (P/F, 15%, 4) = $4,500 (0.8696) + $3,050 (0.7561) + $2,750 (0.6575) + $1,450 (0.5718) = $ 8,856.54 A = $8,856.54 (A/P,15%,4) = $8,856.49 (0.3503) = $3,102.45 This same value can be obtained and confirmed with Equation (5-5): CR(i%) = I (A/P, i%, N) − S (A/F, i%, N) = $10,000 (A/P,15%,4) − $2,000 (A/F,15%,4) = $10,000 (0.3503) − $2,000 (0.2003) = $3,102.12 Note: The Annual Worth from the table and the CR amount from Equation (5-5) are the same.

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5-24

The capital recovery amount based on Equation (5-5) is: CR (18%) = $10 million (A/P, 18%, 12) – [–$4 million (A/F, 18%, 12)] CR (18%) = $10 million (0.2086) + $4 million (0.0286) = $2.2004 million

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5-25

We are asked whether the trade-off between extra cash outflow on the front end of a proposed heating system investment ($200,000) and annual downstream energy savings is a smart investment to make.

Let's calculate the PW of the project when the interest rate is 12%:

PW = -$200,000 + 500,000 kWhr ($0.10/ kWhr)(P/A, 12%, 8) + $20,000(P/F, 12%, 8)

PW = -$200,000 + $50,000(P/A, 12%, 8) + $20,000(P/F, 12%, 8), so

PW = $56,458.

The PW is greater than zero, so this is an attractive investment.

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5-26

The capital recovery amount is 0.55 ($9,713) = $5,342 per year.

Fuel-related operating expense is 16,000 miles/year × $0.25/mile = $4,000 per year.

So, the other operating expenses (insurance, tax, license and registration) are: $9,713 − $5,342 − $4,000 = $371 per year.

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5-27

Capital Investment = −$300,000 − $600,000 − $250,000 − $100,000 = −$1,250,000 Revenue = $750,000 per year Market Value = $400,000 + $350,000 + $50,000 = $800,000 Expenses = −$475,000 per year AW(15%) = $750,000 − $475,000 − $1,250,000(A/P, 15%, 10) + $900,000(A/F, 15%, 10) = $275,000 −$1,250,000(0.1993) + $900,000(0.0493) = $70,245 > 0 Therefore, they should invest in the new product line (Note: all of the $100,000 working capital recovered at the end of year 10).

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5-28

(a)

Initial investment = $54,000 (one-time cost) Operating expenses = $450 + $7.50(3,000) = $22,950 per year (recurring expense) Disposal = $8,000 (one-time cost)

AW(15%) = −$54,000(A/P, 15%, 8) − $22,950 − $8,000(A/F, 15%, 8) = −$35,570

(b)

Fuel related cost = ($22,500 / $35,570) × 100% = 63.3% of the total.

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5-29

(a)

AW(12%) = -$450 million (A/P,12%, 30) + $50 million = -$55.845 million + $50 million = -$5.845 million per year.

This is not a profitable investment.

(b)

Theta = $450 million / $50 million per year = 9 years.

A payback period of 9 years is usually not acceptable. For energy projects, a 4-5 year payback period is more the norm.

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5-30

The capital recovery amount (CR) at 3% interest per year is: CR(3%) = $1,000,000 (A/P, 3%, 15) = $83,800 per year The CR at 6% annual interest is: CR(6%) = $1,000,000 (A/P, 6%, 15) = $103,000 per year The CR at 12% annual interest is: CR(12%) = $1,000,000 (A/P, 12%, 15) = $146,800 per year The CR at 6% interest is 22.9% higher than 3% interest. The CR at 12% interest is 75.2% higher than 3% interest.

0.16 0.16 0.16 CR ($M/year) 0.14 0.16 0.12 0.14

0.14 0.14

0.1 0.12 0.12

0.1

CR CR($M/year) ($M/year) 0 0

0.12 0.12 0.1

0.1 CR Interest ($M/year) Rate/year 0.08 0.08 0.08 0.1 Series1 CR ($M/year) CR ($M/year) 0.08 Interest CR ($M/year) Rate / year 0.08 CRCR ($M/year) ($M/year) 0 0.06 0.06 0.06 0.06 0.06 0.04 0.04 0.04 0.04 0.04 0.02 0.02 0.02 0.02 0.02 0 0 0 0 0 11 22 21233201344 54045 6246572 2 7686 489 67 91010 8 12 1193810 12 12 10 14 16 58310 611 13 810413 10 15 10 1212 15 14

Interest Rate/year

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5-31

The incremental investment is (0.08)($250,000) = $20,000. The equivalent annual savings (A) is then $20,000(A/P, 10%, 30) = $2,122 to justify the extra investment. If this savings represents 15% of the total annual heating and cooling expense, the total annual expenditure would have to be $14,147. This is high in most parts of the U.S., so green homes are difficult to justify on economic grounds alone. Can you list some non-economic considerations that may bear on the decision to build a green home?

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5-32

Let X = units produced per year.

Then the breakeven equation becomes: AW (12%) = −$650,000 (A/P, 12%, 6) − $54,000 + X ($85 − $12.80) = 0 Solving yields X = 2,937 units per year.

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5-33

Monthly gasoline savings = 10 gallons x $4.00 per gallon = $40 per month.

$1,200 = $40 per month × (P/A, 0.5% per month, N months) (P/A, 0.5%, N) = 30

From Table C-2, 30 < N < 36

Interpolating yields N = 32.6 or 33 months.

Using Excel, NPER(0.005, 40, −1200) = 32.585

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5-34

Let A equal the annual revenue that permits the shopping center to break even in its operation: 0 = −$50,000,000 + (A − $15,000,000)(P/A, 20%, 15) + $20,000,000(P/F, 20%, 15) 0 = −$50,000,000 + 4.6755A − $70,132,500 + $1,298,000 $118,834,500 = 4.6755A A = $25,416,426 per year If annual revenue exceeds A, the project is worthwhile.

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5-35

EUAC of the equipment = $1,000,000 (A/P, 15%, 8) = $222,900.

Letting C = the output (sq. yards) of re-manufactured carpet, we have

$1C + $222,900 = $3C,

so C = 111,450 square yards per year.

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5-36

The time 0 capital investment equivalent is $2.1 million + $1.2 million (P/F, 25%, 1) = $2.1 million + $0.96 million = $3.06 million By using Equation (5-5), the capital recovery cost can be determined: CR = $3.06 million (A/P, 25%, 25) + $2 million (A/F, 25%, 25) = $3.06 million (0.2509) + $2 million (0.0009) = $0.7696 million per year

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5-37

The capital investment in the trail is $4,000 and the annual maintenance is $1,200. The EUAC for this project is AW(7%) = −$4,000 (A/P, 7%, 20) − $1,200 = −$4,000 (0.0944) − $1,200 = − $377.60 − $1,200 = −$1,577.60 This is the annual amount that should be budgeted for the construction and maintenance of the greenway walking trail.

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$255 5-38

0 1

2

3

14

15

$3,000 $3,000 = $255 (P/A, i′, 15) i′ = 3.2% per month r = 12 × 3.2% = 38.4% compounded monthly (APR) ieff = (1.032)12 – 1 = 0.459 or 45.9% per year

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5-39

We can equate PW of the “pay cash” plan to the PW of the “monthly payment” plan to determine the monthly interest rate. −$8,000 = −$1,000 −$800 (P/A, i’%, 12) −$7,000 = −$800 (P/A, i’%, 12) By trial and error, i’ = 5.23% per month. The effective annual interest rate is: i’ = (1 + 0.0523)12 – 1 = 84.36% per year

Alternatively, we can use Goal Seek in Excel to determine the monthly interest rate. Goal Seek Setup: Cell A1 = unknown interest rate = 0.01 (starting guess) Cell B1 = 1000+800*((1+a1)^12−1)/(a1*(1+a1)^12) In Goal Seek, set value of cell B1 = 8000 by changing cell A1. Doing so, we determine i' = 5.229% per month.

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5-40 F= $3500 (F/P, 8%, 4) = $4,761.75

N=8

P = 1500 + 138 (P/A, 8%, 8) = $2,293.03

$2,293.03 (F/P, i’, 8) = $4,761.75 (F/P, i’, 8) = 2.0766

(1 + i )8 = 2.0766

i’ ≈ 9.6% The ERR is < 10% MARR, not an attractive investment.

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5-41

$3,000 = $30(P/A, i'% per month, 120 months)

By trial and error or with Excel we can determine that i' equals 0.31% per month. Assuming monthly compounding, this equates to i' = 3.78% per year. This is not a particularly good investment, but it is favorable to the environment. (The payback period is 100 months, or 8.33 years.)

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$375 5-42 0

1

2

3

11

12

$350 The CFD is from the lender’s viewpoint. Equating outflows to inflows:

$350 + $350(P/A, i′, 11) = $375(P/A, i′, 12) i′ = 7.14% per month

The effective annual interest rate is (1.0714)12 – 1 = 1.288 (128.8%). Jess’s wife is correct, and in fact, the U.S. military is moving to correct predatory lending practices near its installations.

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5-43

The average annual fuel savings of a hybrid vehicle is $3,720/ 10= $372 per year. For a life of 15 years, the IRR on the difference is calculated as follows. This assumes that $372 per year will continue for 15 years. $3,720 = $372 (P/A, i’, 15) 10= (P/A, i’, 15) From Excel we find that i’ = 5.56% per year. What assumptions are necessary to rework this problem when the price of gasoline goes to $3.00 per gallon?

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5-44

PW(i'%) of outflows = PW(i'%) of inflows

$85.71 (P/A,i'%,35) = $9,800

(P/A,i'%,35) = 25.4077

(P/A,1%,35) = 29.4085 and (P/A,2%,35) = 24.9986, Therefore, 1% < i'% < 2%.

Using Excel, i'% = 1.9% per month

A.P.R. = (1.9% per month)(12 months/year) = 22.8% compounded monthly

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5-45

General Equation:

PW(i'%) = 0 = −$450,000 − $42,500(P/F,i'%,1) + $92,800(P/F,i'%,2) + $386,000(P/F, i'%, 3) + $614,600(P/F, i'%, 4) − $202,200(P/F, i'%, 5) PW(20%) = $17,561 > 0, i'% > 20% PW(25%) = −$41,497 < 0, i'% < 25%

Linear interpolation between 20% and 25% yields: i'% = 21.5% > 10%, so the new product line appears to be profitable.

However, due to the multiple sign changes in the cash flow pattern, the possibility of multiple IRRs exists. The following graph of PW versus i indicates that multiple IRRs do not exist for this problem.

PW(i%) $400,000 $300,000

IRR = 21.5%

$200,000 $100,000 $0 -$100,000

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

-$200,000 -$300,000 -$400,000

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5-46

Assume that the lump-sum equivalent amount of the cost of a college education is $100,000. Then the extra annual income must be the equivalent of $100,000 (A/P, 12%, 40) = $12,130.

A = $12,130

0

1

2

3 EOY

38

39

40

i = 12%

$100,000

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5-47

IRR method: PW(i'%) = 0 = $500,000(P/F,i'%,1) + $300,000(P/F,i'%,2) + [$100,000 + $100,000(P/A,i'%,7) + $50,000(P/G,i'%,7)](P/F,i'%,3) − $2,500,000 (P/F,i'%,4)

i'% 1 2 3 4 5

Present Worth $103,331.55 63,694.68 30,228.14 2,175.18 −21,130.28

i'% 30 31 32

Present Worth −$12,186.78 −5,479.09 1,182.76

There are two internal rates of return: 4.09% and 31.8% per year.

ERR Method: −$2,400,000(P/F,8%,4)(F/P,i'%,10) = $500,000(F/P,8%,9) +$300,000(F/P,8%,8) + $100,000(F/P,8%,7) + $150,000(P/A,8%,6)(F/P,8%,6) + $50,000(P/G,8%,6)(F/P,8%,6) After solving, the external rate of return is 7.6% per year.

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5-48

The general equation to find the internal rate of return is: PW(i'%)= 0 = −$520,000 + $200,000 (P/A, i'%, 10) − $1,500,000 (P/F, i'%, 10) (a)

i'%

PW(i'%)

i'%

PW(i'%)

0 0.5 1 4 10

− $ 20,000 0 10,250 89,000 131,000

15 20 25 30 40

$113,000 76,000 33,000 − 10,350 − 89,100

i'% = 1/2% and 28.8% per year.

$150,000 $100,000 PW(i%)

IRR = 28.8%

$50,000 $0 5%

10%

15%

20%

25%

30%

35%

-$50,000 -$100,000

IRR = 0.5%

i%

(b) Assume  = 20% per year. −$520,000−$1,500,000(P/F,20%,10)(F/P,i'%,10)=$200,000(F/A,20%,10)

[$520,000 + $1,500,000(0.1615)](F/P,i'%,10) = $200,000(25.9587) $762,250 (1+i')10 = $5,191,740

i' = 0.2115 or 21.15%

ERR > 20%, therefore the project is economically acceptable.

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5-49

(a) The simple payback period is $1,400 / $24 per month = 58 months, or about five years. This is a marginally good investment.

(b) The IRR can be determined from this equation:

0 = $1,400 + $24 (P/A, i'% per month, 120 months)

From Excel we can determine i' = 1.39% per month (approximately), or about 18% compounded annually. This is a pretty good investment, assuming the salvage value is negligible. If the solar panels can last for 10 years, they are a good hedge against any increase in the cost of electricity (hedges are investments to protect against price increases, i.e. they reduce risk).

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5-50

(a)

Average electrical output per year = 285 MW (0.75) = 213.75 MW. This translates into 8,760 hours per year × 213,750 kW = 1,872.45 × 106 kWh per year of output. Thus, the annual profit will be $56,173,500 when the profit of a kiloWatt hour is $0.03. The simple payback period (once the plant is operating) is $570,000,000 / $56,173,500 = 10.15 years (round it up to 11 years since we are using the end of year cash flow convention). This is not a liquid investment and should be considered “high risk”.

(b)

The IRR is determined from the following equation (all figures in $ millions): 0 = −$285 − $285(P/F, i′, 1) + $56.1735(P/A, i′, 20)(P/F, i′, 1)

Solving yields i′ = 7.1% per year. Given a MARR = 6% per year, the plant would be considered profitable.

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5-51

(a)

0 = −$4,900 + $1,875(P/A, i′, 5);

i′ = 26.4%

(b)

θ = $4,900 / $1,875 = 3 years (to the integer year)

(c)

The IRR will signal an acceptable (profitable) project if the MARR is less than 26.4% and the value of θ may indicate a poor project in terms of liquidity.

(d)

1/ θ = 33.3%. This is the payback rate of return, and it over-estimates the actual IRR.

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5-52

(a)

The simple payback period, θ, is as follows. θ = $239,000 / $40,300 per year = 5.93 = 6 years

(b)

The IRR is determined from this equation: 0 = −$239,000 + $40,300 (P/A, i’%, 20) 5.93 = (P/A, i’%, 20) Using Excel, i’% = 16%

(c)

If MARR ≤ 16% this is a profitable investment (even though the simple payback period is only marginally attractive).

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5-53

PW(i'%) = 0 = −$100,000 + $20,000 (P/A,i'%,5) + $10,000 (P/G,i'%,5) + $10,000 (P/F,i'%,5)

PW(20%) = $12,891 > 0,  i'% > 20% PW(25%) = −$897

< 0,  i'% < 25%

By linear interpolation, i'% = IRR = 24.7%

EOY 1 2 3 4

Cumulative Cash Flow −$100,000 + $20,000 = −$80,000 < 0 −80,000 + 30,000 = −50,000 < 0 −50,000 + 40,000 = −10,000 < 0 −10,000 + 50,000 = 40,000 > 0  = 4 years

Although this project is profitable (IRR > MARR), it is not acceptable since  = 4 years is greater than the maximum allowable simple payback period of 3 years.

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5-54

Year 0 1 2 3 4

Cash Flow For Year $275,000  35,000 55,000 175,000 250,000

PW(8%) of Cash Flow $275,000  32,407 47,154 138,921 183,757

Cumulative PW thru Year $275,000  307,407  260,254  121,333 62,424

>0

The project meets the benchmark.

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5-55

Annual fuel savings are ($3/gallon)(45,000 gallons/year) = $135,000 per year. The payback period (θ) is $1,000,000/$135,000 per year = 7.4 years.

The discounted payback period (θ’) is calculated as follows: $135,000 (P/A, 20%, θ’) − $1,000,000 = 0 (P/A, 20%, θ’) = 7.407 This investment can never meet this condition since (P/A, 20%, ∞) = 5, so θ’ = never.

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5-56

(a)

Annual Savings = $2.5 billion

Affordable amount: PW = $2.5 billion (P/A, 7%, 40) = $33.33 billion

(b)

θ = $25 billion / $2.5 billion = 10 years

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5-57

(a) The cumulative positive cash flows overtake the initial investment in seven years. Therefore, the simple payback period is seven years.

(b) The discounted payback period, θ’, is 8 years. This is when the discounted positive cash flows exceed the initial investment.

PW(6%) of first 7 years = -$4,179

PW(6%) of first 8 years = $8,369

(c) The present worth of this investment is positive, so undertaking this project appears sound, even though theta and theta prime are a bit high for this speculative venture. As an aside, the internal rate of return is 16.4% per year, which clearly exceeds the MARR.

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5-58

(a)

The affordable amount that can be spent on “smart home” apps is $2,000 / year × 5 years = $10,000. Any amount less than $10,000 will lower the simple payback period.

(b)

$10,000 = $2,000 (P/A, i%, 5) i = 0%

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5-59

(a)

The simple payback period for the PV system is: 𝜃

(b)

$28,000 $200/month

140 months or 11

2 years .

3

The IRR of the PV system is: $28,000 = $200 (P/A, i’/month, 240 months) Using Excel we determine i’ = 0.50% per month ( or 6% nominal per year).

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5-60

(a)

(b)

In all three cases, IRR = 15.3%. This is true for EOY 0 as a reference point in time, and also for EOY 4 as a reference point in time.

PW1(10%) = $137.24

at EOY 0

PW2(10%) = $137.24 at EOY 4 PW2(10%) = $93.73

at EOY 0

PW3(10%) = $686.18 at EOY 4 PW3(10%) = $468.67 at EOY 0

Select (3) to maximize PW(10%). However, the PW(IRR=15.3%) would be zero for all three situations.

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5-61

(a)

Loan repayment amount = $30 million (A/P, 6%, 40) = $1.995 million per year.

Recurring expenses = $4,000 (300) = $1.2 million per year Total annual expenses = $1.995 million + $1.2 million = $3.195 million per year Revenue = $12,000 (300)(0.80) = $2.88 million per year Annual Profit (loss) = $2.88 million − $3.195 million = −$0.315 million per year

(b)

Revenue = $12,000 (300)(0.95) = $3.42 million per year Annual Profit (loss) = $3.42 million − $3.195 million = +$0.225 million per year

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5-62

Assume the hospital’s average occupancy rate is 85%. The revenue per year is (300 beds)(0.85)(365 days/year)($1.00 per day) = $93,075 per year The simple payback period, Θ, is Θ = ($104,000 / $93,075/yr) = 1.12 years (round it up to 2 years) Maurice appears to be a good investment.

Display of Θ versus occupancy rate: 1.27 1.19 Θ (years)

1.12 1.05 1.00 0.95 75

80

85

90

95

100

Occupancy Rate (%)

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5-63

(a)

AW(15%)

= −$710,000(A/P, 15%, 5) + $198,800 + $100,000(A/F, 15%, 5) = $1,828 per year.

Yes, it is a good investment opportunity

(b)

IRR:

0

= −$710,000(A/P, i′, 5) + $198,800 + $100,000(A/F, i′, 5)

i′

= 15.3%

 = 4 years

' = 5 years

(c)

Other factors include sales price of reworked units, life of the machine, the company’s reputation, and demand for the product.

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5-64

Desired Ending Balance Current Age Age at Retirement

$ 250,000 25 60

4% N

5 10 15 20

$ $ $ $

4,458 6,003 8,395 12,485

Interest Rate per Year 8% $ 2,207 $ 3,420 $ 5,463 $ 9,207

$ $ $ $

12% 1,036 1,875 3,470 6,706

The formula used in cell C7 is:

=−PMT(C$6,$D$3−$D$2−$B7,,$D$1)

Note that this uses the fv parameter of the PMT function instead of the pv parameter. The formula in cell C7 was copied over the range C7:E10.

The trend in the table shows that as the interest rate increases, less has to be saved each year. Also, the longer Jane delays the start of her annual savings, the larger the annual deposit will have to be.

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5-65

0.1 P = P (A/P,8%,N) so N ≈ 21 years

A payout duration table can be constructed for selected payout percentages and compound interest rates as follows:

Interest Rate/Year Payout/Yr

4%

6%

8%

10%

10%

13.0*

15.7

20.9

never

20%

5.7

6.1

6.6

7.3

30%

3.7

3.8

4.0

4.3

* Note: Table entries are years.

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5-66 MARR = Reinvestment rate = Capital Investment = Market Value = Useful Life = Annual Savings = Annual Expense = EOY 0 1 2 3 4 5 5

$ $ $ $ $

20% 20% 25,000 5,000 5 8,000 −

$ $ $ $ $ $ $

Cash Flow (25,000) 8,000 8,000 8,000 8,000 8,000 5,000

EOY 0 1 2 3 4 5

Present Value = Annual Worth = Future Worth = Internal Rate of Return = External Rate of Return =

Cash Flow $ (25,000) $ 8,000 $ 8,000 $ 8,000 $ 8,000 $ 13,000

$ $ $

934.28 312.41 2,324.80 21.58% 20.88%

The following table displays the results of different MARRs on the profitability measures.

Present Worth Annual Worth Future Worth IRR ERR

MARR = 18% MARR = 22% $ 2,202.91 $(240.89) $ 704.44 $ (84.12) $ 5,039.73 $(651.04) 21.58% 21.58% 20.88% 20.88%

The original recommendation is unchanged for a MARR = 18%. However, the recommendation does change for MARR = 22% (which is greater than the IRR of the project's cash flows).

Note that the ERR is unaffected by changes in the MARR. This is because 1) the reinvestment rate was assumed to remain at 20%, and 2) there is only a single net cash outflow occurring at t=0.

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5-67

EOY 0 1 2 3 4 5 6 7 8 9 10

Labor savings

Operating expenses

Net savings

$ 15,000.00 $ 16,050.00 $ 17,173.50 $ 18,375.65 $ 19,661.94 $ 21,038.28 $ 22,510.96 $ 24,086.72 $ 25,772.79 $ 27,576.89

$ (3,500.00) $ (3,750.00) $ (4,000.00) $ (4,250.00) $ (4,500.00) $ (4,750.00) $ (5,000.00) $ (5,250.00) $ (5,500.00) $ (5,750.00)

$ 11,500.00 $ 12,300.00 $ 13,173.50 $ 14,125.65 $ 15,161.94 $ 16,288.28 $ 17,510.96 $ 18,836.72 $ 20,272.79 $ 21,826.89

PW

$ 93,560.18

The hospital can afford to pay $93,560 for this device.

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5-68

Given:

Inputs:

Loan Amount Interest Rate Repayment Periods Annual Expenses (per apartment) Number of units

$30,000,000 6% 40 $4,000

Annual Rental Fee Occupancy Rate

$12,000 89%

Intermediate Calculations: Annual Loan Payment Results:

Annual Profit (Loss)

300

$1,993,846 $0

The Goal Seek function was used to find the breakeven occupancy rate of 89%.

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5-69

Average number of wafers per week:

(10 wafers/hr)(168 hr/wk)(0.90) = 1,512

Added profit per month:

(1,512 wafers/wk)(4.333 wk/month)($150/wafer) = $982,724

PW(1%) = −$250,000 − $25,000(P/A, 1%, 60) + $982,724(P/A, 1%, 60) = $42,804,482

The increase in CVD utilization serves to make the project even more attractive.

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5-70

Average number of wafers per week:

(15 wafers/hr)(168 hr/wk)(0.80) = 2,016

New breakeven point: X

$1,373,875  $3.50 / wafer (2,016)(4.333)(44.955)

$3.50/$100 = 0.035 extra microprocessors per wafer.

The retrofitted CVD tool would significantly reduce the breakeven point.

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5-71

Average number of wafers per week:

(10 wafers/hr)(150 hr/wk)(0.90) = 1,350

Added profit per month:

(1,350 wafers/wk)(4.333 wk/month)($150/wafer) = $877,433

PW(1%) = −$250,000 − $25,000(P/A, 1%, 60) + $877,433(P/A, 1%, 60) = $38,071,126

X

$1,373,875  $5.225 / wafer (1,350)(4.333)(44.955)

New breakeven point:

$5.225/$100 = 0.05225 extra microprocessors per wafer.

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5-72

The remaining loan principal immediately after the 240th payment of $200 is:

$200(P/A, 7%/12, 120) = $200(86.1264) = $17,225

The IRR on this investment can be dtermined by solving for i′ in this equation:

0 = −$20,000 − $200(P/A, i′, 240) + [$100,000 − $17,225](P/F, i′, 240)

By trial and error we find that i′ is about 0.128% per month. This corresponds to an effective annual interest rate of (1.00128)12 – 1 = 0.015 (1.5%)

Select (d)

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5-73

i / mo. = 9%/12 = ¾% per month; N = 4 x 12 = 48 months

A = $8,000 (A/P, ¾% per month, 48 months) = $8,000 (.0249) = $199.20

Select (d)

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5-74

EOY

Cumulative PW (i = 12%)

0

−$300,000

1

−$300,000 + $111,837.50(P/F,12%,1) = −$200,140.30

2

−200,140.30 + $111,837.50(P/F,12%,2) = −$110,983.45

3

−$110,983.45 + $111,837.50(P/F,12%,3) = −$31,377.52

4

−$31,377.52 + $111,837.50(P/F,12%,4) = $39,695.21 > 0

 Ø' = 4

Select (a)

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5-75

Capitalized Cost (10%) = $120,000 + $6,500 (P/A, 10%, 20) + $8,500 (P/F, 10%, 20) / 0.1 = $187,969.

Select (b)

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5-76

Find i% such that PW(i%) = 0

0 = −$3,345 + $1,100(P/A,i'%,4) PW(10%) = $141.89 tells us that i'% > 10% PW (12%) = −$3.97 tells us that i'% < 12% (but close!)

 IRR = 11.95%

Select (b)

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5-77

VN = C(P/F,i%,N) + rZ(P/A,i%,N) = $981 N = 8 periods r = 10% per period

($1,000/$100 = 10%)

C = Z = $1,000

$981= $1,000(P/F,i%,8) + (.10)($1,000)(P/A,i%,8) $981 = $1,000(P/F,i%,8) + $100(P/A,i%,8)

Try i = 10%; $466.50 + $533.49 = $999.99 Try i = 12%; $403.90 + $496.76 = $900.66

By observation, i% is > 10% but very close to 10%

 rate of return = 10.35%

Select (c)

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5-78

CW(10%) = -$5,123 + $1,110 / 0.10 = $5,977.

Select (b)

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5-79 AW = ($5,000 + $500 (A/G.1%24))*(F/P,1%,12)

= ($5,000 + $500*11.02337)*1.0100*12.6825 = $134,649

Select (e)

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f  5% 5-80

i = 12%

$7000[1  (P/F, 12%, 6)(F/P, 5%, 6)] 0.12  0.05 $7000[1  (0.5066)(1.3401)]  $9,000  0.07  $41,110.53 AW(12%)  $41,110.53 (A/P, 12%, 6)  $41,110.53(0.2432)  $9,998.08 PW (12%)  $9,000 

Select (c)

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5-81

0 = $500,000 + $50,000 (P/A, i'%, 25), or 10 = (P/A, i', 25)

By trial and error (or Excel) we can determine that i' = 8.8% per year.

Select (c)

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5-82

P = $386(P/A, 1%, 38)(P/F, 1%, 1) = $12,033

Select (b)

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5-83

The simple payback period is $5,000 / $725 per year = 6.9 years (call it seven years). So the discounted payback period will be equal to or greater than the simple payback period. We want the discounted savings to be marginally greater than $5,000 in the following table:

N

PW(6%) of savings

7

$4,047

8

$4,502

9

$4,931

10

$5,336

Thus, the discounted payback period is ten years.

Select (d).

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5-84

Set the PW of the cash flows equal to zero at the unknown interest rate: PW (i%) = −$48 million + $45 million (P/F, i%, 1) + $45 million (P/F, i%, 2) − $21 million (P/F, i%, 3) By trial and error (or Excel), we find the IRR to be 38.6%, so choice (b) is correct.

Note: If the MARR = 20%, PW(20%) = $8.6 million

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5-85

Set the net PW equal to zero at the unknown interest rate i’% (the IRR).

0 = −$120,000 + $20,000 (P/F, i’%, 1) + $30,000 (P/F, i’%, 2) + $25,000 (P/F, i’%, 3) + $35,000 (P/F, i’%, 4) + $35,000 (P/F, i’%, 5) By trial and error we find that i’% = 6.1%, so answer (b) is correct.

Net PW(6%) = $435.60

Net PW(7%) = −$3,042

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5-A-1 [$50 + $360(P/F, 8%, 2)](F/P, ERR, 3) = $235(F/P, 8%, 2) + $180

$358.63 (1 + ERR)3 = $454.10

ERR = 8.19% per year

A maximum of three IRR values are suggested by Descartes’ rule of signs.

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5-A-2 [$5,000 + $1,000(P/F, 12%, 2)](F/P, ERR, 3) = $6,000(F/P, 12%, 2) + $4,000

$5,797.20 (1 + ERR)3 = $11,526.40

ERR = 25.75% per year.

Nordstrom’s criterion indicates that a unique IRR exists since there is only one sign change in the cumulative cash flow. A plot of PW v. i confirms a single IRR value of 45%.

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5-A-3 Descartes’ rule allows for the possiblity of two IRRs. A plot of PW v. i confirms this (26.73% and 37.12%). The ERR metric is a more useful measure in this situation.

[$1,810(P/F, 10%, 4)](F/P, ERR, 10) = $120(F/P, 10%, 10) + $90(F/P, 10%, 9) + … + $100

$1,236.23 (1 + ERR)10 = $3,624.30

ERR = 11.36% per year

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5-A-4 Descartes' Rule and Nordstrom's Criterion each suggest the possibility of a maximum of three IRRs. If we plot PW (i%) versus i%, we discover there is a unique IRR of 33.1%:

i

0%

10%

20%

30%

40%

50%

60%

70%

80%

100%

PW (i) 400

243

123

26

-52

-119

-175

-223

-266

-338

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Solutions to Chapter 6 Problems

6-1

A: In-house treatment

0

1

2

3

4

5

6

7

8

9

10 G = $3,000

A = $30,000

B: Outside treatment

0

1

2

3

4

5

6

7

8

9

10

$15,000

AW(A) = -$30,000 + 3,000 (A/G, 10%, 10) = −$18,824 AW(B) = −$15,000 (F/P, 10%, 1) = −$16,800

Outside processing is the most economical alternative.

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6-2

Present Worth Method, MARR = 10% per year PWD1 (10%) = −$600,000 − $780,000(P/A,10%,8) = −$4,761,222 PWD2 (10%) = −$760,000 − $728,000(P/A,10%,8) = −$4,643,807 PWD3 (10%) = −$1,240,000 − $630,000(P/A,10%,8) = −$4,600,987 PWD4 (10%) = −$1,600,000 − $574,000(P/A,10%,8) = −$4,662,233 Select Design D3 to minimize the present worth of costs.

Future Worth Method, MARR = 10% per year FWD1 (10%) = −$600,000(F/P,10%,8) − $780,000(F/A,10%,8) = −$10,206,162 FWD2 (10%) = −$760,000(F/P,10%,8) − $728,000(F/A,10%,8) = −$9,954,471 FWD3 (10%) = −$1,240,000(F/P,10%,8) − $630,000(F/A,10%,8) = −$9,862,681 FWD4 (10%) = −$1,600,000 (F/P,10%,8) – $574,000(F/A,10%,8) = −$9,993,967 Select Design D3 to minimize the future worth of costs.

Annual Worth Method, MARR = 1% per year AWD1 (10%) = −$600,000 (A/P,10%,8) – $780,000 = −$892,440 AWD2 (10%) = −$760,000 (A/P,10%,8) – $728,000 = −$870,424 AWD3 (10%) = −$1,240,000 (A/P,10%,8) – $630,000 = −$862,376 AWD4 (10%) = −$1,600,000 (A/P,10%,8) – $574,000 = −$873,840 Select Design D3 to minimize the annual worth of costs.

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6-3

Consider the incremental cash flow, ∆ (Sauer – Glock): P P/2 0

5

P

EOY

10

The IRR on ∆ (Sauer – Glock) is: 0 = −P + P (P/F, i%, 5) + P/2 (P/F, i%, 10) 0 = −1 + (P/F, i%, 5) + ½ (P/F, i%, 10) By trial and error, i% = 6.44%, so select Glock.

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6-4

Assume all units are produced and sold each year.

AWA(20%) = −$30,000(A/P,20%,10) + 15,000($3.50 − $1.00) − $15,000 = $15,345 AWB(20%) = −$60,000(A/P,20%,10) + 20,000($4.40 − $1.40) − $30,000 + $20,000(A/F,20%,10) = $16,460 AWC(20%) = −$50,000(A/P,20%,10) + 18,000($4.10 − $1.15) − $25,000 + $15,000(A/F,20%,10) = $15,753 Select Design B to maximize the annual worth.

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6-5

With a 20-year study period, we can compare the present worth of the two alternatives: Gas:

PW(8%)

= -$225,000 - $2,000(P/A, 8%, 20) – $5,000 [1- (P/F, 8%, 20)(F/P, 5%, 20)] / (0.08-0.05)

= -$225,000 - $2,000(9.8181) – $5,000 [1- (0.2145)(2.6533)] / (0.03) = -$316,453

Geothermal:

PW(8%)

= -$200,000 - $10,000 (P/A, 8%, 20)

= -$200,000 – $98,181 = -$298,181 The lower cost (higher PW) alternative is the geothermal heating option.

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6-6

Wet Tower, Mechanical Draft AW(12%) = −$3,000,000 (A/P,12%,30)  200hp   0.746kw  − 40    (8,760 hr/yr) ($0.022/kWh)  0.9   hp   150hp   0.746kw  − 20    (8,760 hr/yr) ($0.022/kWh) − $150,000  0.9   hp  = −$372,300 − $1,277,948 − $479,230 − $150,000 = −$2,279,478/yr.

Wet Tower, Natural Draft AW(12%) = −$8,700,00(A/P,12%,30)  150hp   0.746kw  − 20    (8,760 hr/yr) ($0.022/kWh) − $100,000  0.9   hp  − $1,076,670 − $479,230 − $100,000 = −$1,658,900/yr.

Dry Tower, Mechanical Draft AW(12%) = −$5,100,000(A/P,12%,30)  200hp   0.746kw  − 20    (8,760 hr/yr) ($0.022/kWh)  0.9   hp   100hp   0.746kw  − 40    (8,760 hr/yr) ($0.022/kWh) − $170,000  0.9   hp  = −$632,910 − $638,974 − $638,974 − $170,000 −$2,080,858/yr.

Dry Tower, Natural Draft AW(12%) = −$9,000,000(A/P,12%,30)  100hp   0.746kw  − 40    (8,760 hr/yr) ($0.022/kWh) − $120,000  0.9   hp  = −$1,116,900 − $638,974 − $120,000 = −$1,875,874/yr.

The wet cooling tower with natural draft heat removal from the condenser water is the most economical (i.e., least costly) alternative.

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6-7

Rank order MEAs:

DN → C → B → A

First check Δ(C – DN) 0 = -$300,000 + $57,000 (P/A, i’, 15) + $35,000 (P/F, i’, 15) Plug in MARR = 12% 0 < -$300,000 + $57,000(6.8109) + $35,000(0.1827) 0 < $94,615.8 so i´> 12% Select C, reject DN. Next check Δ(B – C) 0 = -$100,000 + $5,000 (P/A, i’, 15) + $5,000 (P/F, i’, 15) By inspection, ∑ neg. cash flows > ∑ pos. cash flows so reject B, keep C. Finally check Δ (A – C) 0 = -$150,000 + $13,000 (P/A, i’, 15) + $15,000 (P/F, i’, 15) 0 < -$150,000 + $13,000(6.8109) + $15,000(0.1827) 0 > -$58,717.80 so i’ < 12% Reject A, keep C. Therefore we would select Furnace C.

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6-8

ILB: Installation cost is $2,000 plus the cost of the bulbs ($500) for an investment cost of $2,500. This is assumed to occur at the beginning of each year in the 8-year study period. The cost of electricity for 60,000 kWh at $0.12 per kWh is $7,200 per year. Let's assume that the electricity expense is incurred at the end of each year for eight years:

PW(12%) = $2,500  $2,500(P/A, 12%, 7)  $7,200(P/A, 12%, 8) = $49,677

CFL: Installation cost is $3,000 plus the cost of the 1,000 CFLs for a total investment cost of $5,000.

This is assumed to be incurred at the beginning of the 8-year study period. The cost of electricity for 13,000 kWh at $0.12 per kWh is $1,560 at the end of each year for eight years:

PW(12%) = $5,000  $1,560(P/A, 12%, 8) = $12,749

The boss will be happy to learn that CFLs offer tremendous cost savings over the ILBs. CFLs cost about 26% of the PW of cost of the ILBs over the 8-year study period. A side note: In Europe ILBs will not be sold in stores beginning in September of 2009. ILBs are simply too energy inefficient and create too much of a carbon footprint! Let's become "enlightened" and make the change.

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6-9

(a)

Assume fuel expense is the only expense to be considered (ignore tire wear, oil changes, etc.)

Composite auto annual fuel expense: (20,000 mi./yr.)(1 gal. / 50 mi.)($3.50/gal.) = $1,400 Conventional auto annual fuel expense: (20,000 mi./yr.)(1 gal. / 25 mi.)($3.50/gal.) = $2,800 Spending an additional P = ($2,800 - $1,400)(P/A, 10%, 6) = $6,097 will be justified by the annual fuel savings. If the composite auto costs more than $6,097 relative to the conventional auto, the purchase should not be made. This ignores intangibles such as “prestige value” of being first on the block with a composite car.

(b)

Reduced fuel expense per year is being traded off against extra initial investment for the composite car.

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6-10

Approach 1: Determine PW(10%) on ∆(1-2). If PW∆(10%) is greater than or equal zero, select Alt. 1.

Otherwise select Alt. 2.

EOY k 0 1 2 3 4 5

Cash Flow ∆(1-2) -$40,000 $10,000 $50,000 -$6,500 -$4,500 -$3,500

PW∆ (10%) = $283 so select Alt. 1 Approach 2: Compare FWs at EOY 5 FW1 (10%) = -$40,000 (F/P, 10%, 5) + $10,000 (F/A, 10%, 5) + $2,000 (F/P, 10%, 1) + $3,000 = $1,830.6 FW2 (10%) = -$40,000 (F/P, 10%, 3) + $16,500 (F/A, 10%, 3) = $1,375 Select Alt. 1 to maximize FW (10%)

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6-11

From the table given, we know that A is preferred to DN since the IRR of Δ(A−DN) = 10.6% > the MARR = 10%. The same logic tells us that alternative B is preferred to alternative A. The next step is to decide if the incremental investment needed by alternative C over B is justified.

Use linear interpolation to find the IRR of Δ(C−B).

IRR on Δ(C−B):

0 = −$1,000 + $170(P/A, iΔ’%, 10) 5.882 = (P/A, iΔ’%, 10) iΔ’% (P/A, iΔ’%, 10) 10% 6.1446 12% 5.6528

So iΔ’% falls between 10% and 12%.

6.1446 5.882 6.1446 5.6528 0.2626 0.4918 𝑖 Since 𝑖

2%

𝑖 , % 10% 12% 10% 10%

𝑖

11.07%

10%, alternative C is preferred to alternative B.

Now to see if alternative D is better than C. A quick inspection of the incremental cash flows for Δ(D−C) reveals that the sum of the cash inflow ($150 × 10 = $1,500) = sum of the cash outflow ($1,500). Thus, 𝑖 = 0%. So the incremental investment required for alternative D is not justified and alternative C should be selected.

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6-12

Design A: All components have a 20 year life.

Capital Investment Concrete pavement: ($90/ft)(5,280 ft/mi) = $475,200 /mile Paved ditches: 2($3/ft)(5,280 ft/mi) = $31,680 /mile Box culverts: (3 culverts/mile)($9,000/culvert) = $27,000 /mile Total Capital Investment = $533,880 /mile Maintenance Annual maintenance: $1,800 /mile Periodic cleaning of culverts*: (3 culverts/mile)($450/culvert) = $1,350 /mile every 5 years AWA(6%) = −$533,880(A/P,6%,20) − $1,800 − $1,350(A/F,6%,5)* = −$48,594 /mile PWA(6%) = −$533,880 − [$1,800 + $1,350(A/F,6%,5)](P/A,6%,20) = −$557,273 /mile * assumes a cleaning also occurs at the end of year 20.

Design B: All components have a 10 year life.

Capital Investment (Year 0) Bituminous pavement: ($45/ft)(5,280 ft/mi) Sodded ditches: 2($1.50/ft)(5,280 ft/mi) Pipe culverts: (3 culverts/mile)($2,250/culvert) Total

= $237,600 /mile = $15,840 /mile = $6,750 /mile = $260,190 /mile

Capital Investment (EOY 10) Bituminous pavement: ($45/ft)(5,280 ft/mi) = $237,600 /mile Sodded ditches: 2($1.50/ft)(5,280 ft/mi) = $15,840 /mile Replacement culverts: (3 culverts/mile)($2,400/culvert) = $7,200 /mile Total = $260,640 /mile Maintenance Annual pavement maintenance: = $2,700 /mile Annual cleaning of culverts: (3 culverts/mile)($225/culvert) = $675 /mile Annual ditch maintenance: 2($1.50/ft)(5280 ft/mi) = $15,840 /mile Total = $19,215 /mile AWB(6%) = −[$260,190 + $260,640(P/F,6%,10)](A/P,6%,20) − $19,215 = −$54,595 /mile PWB(6%) = −$260,190 − $260,640(P/F,6%,10) − $19,215(P/A,6%,20) = −$626,126 /mile Select Design A (concrete pavement) to minimize costs. © 2019 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6-13

Method: Incremental PW Order alternatives by increasing capital investment: ER3, ER1, ER2.

Is ER3 an acceptable base alternative?

PWER3(12%) = −$81,200 + $19,750(P/A,12%,6) = $0.15  0.

Since PW(MARR=12%)  0, ER3 is an acceptable base alternative.

Analyze  (ER1 − ER3)

$25,800[1  ( P / F ,12%,6)( F / P,6%,6)] − $19,750(P/A,12%,6) 0.12  0.06 $25,800(0.2814) = −$17,400 + − $19,750(4.1114) 0.06 = $22,402 > 0

PW(12%)= −($98,600 − $81,200) +

The additional capital investment earns more than the MARR. Therefore, design ER1 is preferred to design ER3.

Analyze  (ER2 − ER1) PW(12%) = −($115,000 − $98,600) + $29,000(P/A,12%,6) + $150(P/G,12%,6) $25,800[1  ( P / F ,12%,6)( F / P,6%,6)] − 0.12  0.06 $25,800(0.2814) = −$16,400 + $29,000(4.1114) + $150(8.93) − 0.06 = −$16,832 < 0 The additional capital investment required by design ER2 has a negative PW (earns less than the MARR). Therefore, design ER1 is preferred to design ER2.

Decision: Recommend Design ER1

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6-14

Rank order alternatives from low capital investment to high: B → C → A Examine Δ(C – B) Δ Investment = -$8,600 Δ Savings = $2,050/yr. for 7 years IRR = 14.72% Short cut: 0 = -$8,600 + $2,050 (P/A, 12%, 7) The right hand side is positive so IRR on Δ (C – B) > 12%, and we should select alternative C (delete B). Actual IRRΔ = 14.72%.

Examine Δ(A - C) Δ Investment = -$13,800 Δ Savings = $2,650/yr. for 7 years IRR = 8% Short cut: 0 = -$13,800 + $2,650 (P/A, 12%, 7) The right hand side is negative so IRR on Δ (A - C) < 12%, and we should keep alternative C as best (delete A). IRRΔ = 8.9%.

Therefore we conclude that alternative C should be chosen.

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6-15

Look at incremental cash flows. Compare PW (MARR = 15%) to 0.

DN → I (∆I-DN):

-$22,000 + $4,500 (P/A, i’, 10) = 0;

i’ > 15% (it’s equal to 15.7%) I → II (∆II-I): -$7,000 + $2,250 (P/G, i’, 10) - $250 (P/G, i’, 10) = 0; i’ > 15% (it’s 15.3%).

Therefore, select alternative II.

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6-16

(a)

Cost of electricity for the 90% efficient motor: [30 hp/0.90](0.746 kW/hp)($0.10/kWh)(4,000 hr/yr) = $9,946.67 per year

Cost of electricity of the 93% efficient motor: [30 hp/0.93](0.746 kW/hp)($0.10/kWh)(4,000 hr/yr) = $9,625.81 per year

PW90%(15%) = −$2,200 − $9,946.67(P/A, 15%, 8) = −$46,834

PW93%(15%) = −$3,200 − $9,625.81(P/A, 15%, 8) = −$46,394

The 93% efficient motor is the better choice. Notice that energy expense dominates the analysis.

For instance, the 93% efficient motor has a PW of energy expense that is $43,194 / $3,200, or over 13 times its purchase price. Generally speaking, the potential for savings is enourmous for high efficiency motors and pumps.

(b)

$9,946.67 − $9,625.81 = $320.86 per year savings

PW(15%) of savings = $320.86(P/A, 15%, 8) = $1,440, which exceeds the extra investment cost of $1,000, so the more efficienct motor is preferred. Thus, the PW of the incremental investment is $440.

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6-17

Cost of electricity of the 90% efficient motor = $9,946.67 (0.60) = $5,968 per year Cost of electricity of the 93% efficient motor = $9,625.81 (0.60) = $5,775 per year

PW90%(15%) = −$2,200 − $5,968(P/A, 15%, 8) = −$28,980 PW93%(15%) = −$3,200 − $5,775(P/A, 15%, 8) = −$29,114

In this case, the 90% efficient motor should be selected by a slim margin.

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6-18

A:

Incandescent lighting system: 10kW (8,760 h/yr) = 87,600 kWh/yr.

EOY 0

1

2

3

4

N=5

$3,942/yr = A

B:

Fluorescent lighting system: 4.5kW (8,760 h/yr) = 39,420 kWh/yr.

EOY 0

1

2

3

4

N=5

$1,773/yr = A $11,000 Net PWA (15%) = -$3,942 (P/A, 15%, 5) = -$13,214.20 Net PWB (15%) = -$11,000 - $1,773 (P/A, 15%, 5) = -$16,943.45

ACA = 3,942 ACB = 5,054

∴ Keep the existing incandescent lighting system Assume zero salvage value for both alternatives

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6-19

Drop Alternative A because 15% < 18%. Alternative B is acceptable (25% > 18%). The IRR on ∆(C-B) is 12.5%, so keep Alternative B and drop C. The IRR on ∆(D-B) is 22% so take Alternative D and drop B. Next the IRR on ∆(E-D) = 20%, so accept E and drop D. The last comparison is ∆(F-E) which has an IRR = 15% which is less than 18%. Thus, Alternative E is the best choice.

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6-20

AWA (8%)

= -$1,200 (A/P, 8%, 5) – $60 – [50hp(0.746 kw/hp)(1,000 hr./yr.)($0.07/kwh)] / 0.82 = -$1,200 (0.2505) – $60 – [50hp(0.746 kw/hp)(1,000 hr./yr.)($0.07/kwh)] / 0.82 = -$300.6 – $60 – $3,184.15 = -$3,544.75

AWB (8%)

= -$1,000 (A/P, 8%, 5) – $100 – [50hp(0.746 kw/hp)(1,000 hr./yr.)($0.07/kwh)] / 0.77 = -$250.50 - $100 - $3,390.91 = -$3,741.41

Select Motor A

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6-21

PW1 (8%) = -18,000 + 6,100 (P/A, 8%, 12) – 16,000[(P/F, 8%, 4) + (P/F, 8%,8)] + 2,000 (P/F, 8%, 12) = $8,071.61 PW2 (8%) = -22,000 + 7,000 (P/A, 8%, 12) + 500 (P/F, 8%, 12) = $30,951.25 Choose 2 to maximize profitability

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6-22

Initial cost Net Revenue Salvage value

Dynasty -13,000 2,000 1,300

Might-E -13,500 2,300 1,600

Z -14,000 1,500 2,000

Drop Zenn because its IRR is less than 8% per year.

PWD

= -13,000 + 2,000 (P/A, 8%, 10) + 1,300 (P/F, 8%, 10) = -13,000 + 2,000 (6.7101) + 1,300 (0.4632) = $1,022.36

PWM = -13,500 + 2,300 (6.7101) + 1,600 (0.4632) = $2,624.35 Select Might-E

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6-23

PW (15%) of ER1 = -$110,000 + $25,800 [1- (P/F, 15%,6)(F/P,6%,6} / 0.09 = -$110,000 + $110,878 = $878

PW(15%) of ER2 = -$115,000 + $29,000 (P/A,15%, 6) + $250 (P/G, 15%, 6) = -$115,000 + $109,751 + $1,984 = -$3,265

PW(15%) of ER3 = -$81,200 + $19,750 (P/A, 15%,6) = -$6456

Therefore, ER1 should be selected.

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6-24 1.1 M

0

1

2

3

4

N=18

4.3M

1. Δ (B – A) 2. A is base (one will be constructed) 3. Analyze Δ (B – A) Δ investment = -34.5 mil – (-30.2mil) = -4,300,000 Δ revenues = 35.2 mil – 31.6 mil = 3,600,000 Δ costs = -25.2 mil – (-22.7 mil) = -2,500,000 0 = -4,300,000 + (3,600,000 – 2,500,000)(P/A, i’, 18) PW(25%) = positive (barely) i’ > 25%

(i’ = 25.1%)

Select B. Extra investment earns more than MARR It is interesting to note that i’A = 29.2% and i’B = 28.7% (but as one might suspect, A is not the correct choice).

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6-25

X = operating hours per year Energy CostABC= (100 hp/ 0.8)(0.746 KW/ hp)($0.10/ kW-hr) X = 9.325 X Energy CostXYZ = (100 hp/ 0.9)(0.746 KW/ hp)($0.10/ kW-hr) X = 8.289 X ACABC = ACXYZ 1900 (A/P, 12%, 10) + 170 + 9.352X = 6200 (A/P, 12%, 10) + 310 + 8.289X 1900 (0.1770) + 170 + 9.352X = 6200 (0.1770) + 310 + 8.289X 506.3 + 9.352X = 1407.4 + 8.289X 1.036X = 901.1 X = 869.79 ≈ 870 hrs./yr.

Select Motor XYZ for 2000 hrs./yr. > 870 hrs./yr. to minimize annual costs.

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6-26

Rank alternatives by increasing capital investment: A,B,C,D and E. Since the ERR on Equipment A > MARR, it is an acceptable base alternative. Therefore, ERR on (B−A) must be examined.

(B−A) ($50,000 − $38,000)(F/P, i′, 10) = ($14,100 − $11,000)(F/A, 18%, 10) i′ = 19.8% > MARR, therefore select B.

(C−B) $5,000 (F/P, i′, 10) = $2,200(F/A, 18%, 10) i′ = 26.3% > MARR, therefore select C.

(D−C) $5,000 (F/P, i′, 10) = $500(F/A, 18%, 10) i′ = 8.9% < MARR, therefore keep C.

(E−C) $15,000 (F/P, i′, 10) = $2,900(F/A, 18%, 10) i′ = 16.4% < MARR, therefore keep C as best.

Answer: Select C.

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6-27

Rank order: D1  D2  D3  D4

D1  D2: $160,000(1 + i′)8 = $52,000(F/A, 12%, 8)

i′ = 18.9%, select D2

D2  D3: $480,000(1 + i′)8 = $98,000(F/A, 12%, 8)

i′ = 12.2%, select D3

D3  D4: $360,000(1 + i′)8 = $56,000(F/A, 12%, 8)

i′ = 8.4%, keep D3

Recommend D3.

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6-28

(a)

Use AW to deal with different useful lives

AWlead acid(5%) = −$6,000(A/P,5%,12) − $2,500 = −$3,176.80

AWlithium ion(5%) = −$14,000(A/P,5%,18) + $2,800(A/F,5%,18) − $2,400 = −$3,497.60

Select Alternative X (could also calculate PW over 36 years and compare)

(b)

PWlead acid(5%) = −$6,000 − $2,500(P/A,5%,12) − $8,000(P/A,5%,6)(P/F,5%,12) = −$50,767.45

PWlithium ion(5%) = −$3,497.60(P/A,5%,18) = −$40,885.54

Select lithium ion battery system (could also calculate AW over 18 years and compare).

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6-29

(a)

Assume repeatability so that AWs can be directly compared (over a 15−year study period).

AWA(8%) = −$2,400(A/P,8%,3) − $320 −

60hp (0.746 kW/hp)(800 hrs/yr.)($0.07/kWh) 0.92

= −$3,976

AWB(8%) = −$2,000(A/P,8%,6) − $200 −

60hp (0.746 kW/hp)(800 hrs/yr.)($0.07/kWh) 0.80

= −$3,766

By a slim margin, select Motor B (b)

Decreased capital investment of Motor B (relative to Motor A) is being traded off for lower electrical efficiency and higher annual energy expenses.

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6-30

(a)

X = $/kWh Motor A: Motor B:

(b)

.

2000

$

.

2000

$

.

.

,

.

,

.

Assume repeatability and MVs = $0. Set AWA = AWB and solve for X. −$1,200(A/P, 8%, 5) – $660 – 90,975.61X = −$500(A/P, 8%, 16) − $100 – 120,322.58X −90,975.61X −$960.60 = −120,322.58X −$156.50 X = $0.0274/kWh

(c)

− AW (8$) 6000

5000

4000 Motor A 3000

Motor B

2000

1000

0

$/kWh 0.0274

(d)

If X < $0.0274/kWh, select B. If X > $0.0274/kWh, select A. If X = $0.0274/kWh, we are indifferent.

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6-31 Number of machines needed:

D1:

3,450 = 2.40 (3 machines) 2,000(0.8)(0.9)

D2:

2,350 = 1.96 (2 machines) 2,000(0.75)(0.8)

Annual equivalent cost of ownership:

D1: $16,000(3)(A/P,15%,6) − $3,000(3)(A/F,15%,6) = $11,653.80 D2: $24,000(2)(A/P,15%,8) − $4,000(2)(A/F,15%,8) = $10,116.00

Annual operating expenses (assume repeatability):

D1: $5,000(3) = $15,000 D2: $7,500(2) = $15,000

Total equivalent annual cost:

D1: $11,653.60 + $15,000 = $26,653.80 D2: $10,116.00 + $15,000 = $25,116.00

Select Machine D2 to minimize total equivalent annual cost.

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6-32

(a) PWA(15%) = −$272,000 − $28,800(P/A,15%,9) + $25,000(P/F,15%,6) − $66,000(P/A,15%,3)(P/F,15%,6) = −$463,758 PWB(15%) = −$346,000 − $19,300(P/A,15%,9) + $40,000(P/F,15%,9) = −$426,720 Select B to minimize equivalent cost. (b) Alternative A is the base alternative because it requires the least capital investment.

Year 0 1−5 6 7−8 9

 (B − A) cash flow −$346,000 − (−$272,000) = −$74,000 − 19,300 − (−$28,800) = 9,500 − 19,300 − (−$28,800 + $25,000) = − 15,500 − 19,300 − (−$94,800) = 75,500 − 19,300 + $40,000 − (−$94,800) = 115,500

PW(i′%) = 0 = −$74,000 + $9,500(P/A, i′%,5) − $15,500(P/F, i′%,6) + $75,500(P/A, i′%,2)(P/F, i′%,6) + $115,500(P/F, i′%,9) By trial and error, i′% = 22.5% > MARR. Therefore the incremental investment is justified and Alternative B should be selected. Note that there are multiple sign changes. It is possible that there are multiple IRRs. (c) Again, Alternative A would be the base alternative. Using the incremental cash flows computed in part (b), the ERR is found by solving the following equation: −$74,000 − $15,500(P/F,15%,6) (F/P, i′%,9) = $9,500(F/A,15%,5)(F/P,15%,4) + $75,500(F/A,15%,3) + $40,000 $80,701(1 + i′)9 = $404,202 ERR = 19.9% > MARR. Therefore, select Alternative B (d) PWL(15%) = −$94,800 (P/A,15%,9) = −$452,348 Thus, leasing crane A is not preferred to the selected alternative (B), but would be preferred to the purchase of crane A.

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6-33

A:

-$22,000 (A/P, 15%, 10) - $15,000 – 0.15X + $2,000 (A/F, 15%, 10) -$22,000 (0.1993) - $15,000 – 0.15X + $2,000 (0.0493) Where X = annual production value

B:

-$15,000 (A/P, 15%, 8) - $13,000 – 0.20X + $2,000 (A/F, 15%, 8) -$15,000 (.2229) - $13,000 – 0.20X + $2,000 (0.0729)

Equate EUACs and solve for x -$19,286 – 0.15x = -$16,197.7 – 0.20x X = 61,766 tees/year Select A for x ≥ 61,766 tees/year

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6-34 (a)

PW1(8%) = −$100,000 − $80,000(P/F, 8%, 5) + $20,000(P/F, 8%, 10) + $28,000(P/A, 8%, 10) = $42,699

PW2(8%) = −$150,000 + $30,000(P/A, 8%, 10) = $51,303

Therefore, select alternative 2 to maximize profitability.

(b)

PW1(15%) = $5,694

PW2(15%) = $564

If the MARR is changed to 15% per year, alternative 1 becomes the better choice. The principal assumption in parts (a) and (b) is the repeatability of cash flows for alternative 1.

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6-35

(a)

$30,000/$13,000 = 2.3

Simple payback period is three years.

(b)

AW2 = −$60,000(A/P, 20%, 5) + $23,500 + $10,000(A/F, 20%, 5) = $4,780

(c)

Assuming repeatability and reinvestment of positive cash balances at 20% per year, IRRΔ(2−1) can be found by equating AWs: -$60,000 (A/P, i’, 5) + $10,000 (A/F, i’, 5) + $23,500 = -$30,000 (A/P, i’, 5) + $10,000 (A/F, i’, 5) + $13,000; i’ ≈ 22.1%

(d)

AW1 (20%) = $4,312 (over 5 years) AW2 (20%) = $4,780 (over 5 years) AW3 (20%) = $1,979 (over 6 years) Select alternative 2 to maximize AW.

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6-36

(a)

AW is the easiest method to use.

(b)

The lowest common denominator is 60 years.

(c)

AWA(20%) = -13,000 (A/P, 20%, 5) – 6,800 + 11,000 + 2,000 (A/F, 20%, 5) = 121.6 AWB(20%) = -11,000 (A/P, 20%, 4) – 5,800 + 10,000 + 1,000 (A/F, 20%, 4) = 137 AWC(20%) = -15,000 (A/P, 20%, 6) – 7,500 + 12,200 + 2,000 (A/F, 20%, 6) = 391 Therefore select System C.

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6-37

(a)

AW1(10%)

AW2(10%)

= -$70,000 (A/P, 10%, 10) +$10,000 + $30,000 (A/F, 10%, 10) = -$70,000 (0.1627) +$10,000 + $30,000 (0.0627) = $492 = -$68,000 (A/P, 10%, 5) +$15,300 + $29,000 (A/F, 10%, 5) = -$68,000 (0.2638) +$15,300 + $29,000 (0.1638) = $2,111.80

Select Alt. 2 (b)

AW1(10%)

= -$70,000 (A/P, 10%, 5) +$10,000 + $45,000 (A/F, 10%, 5) = -$70,000 (0.2638) +$10,000 + $45,000 (0.1638) = -$1,095

AW2(10%)

= $2,111.80 (as in part a)

Select Alt. 2

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6-38

Assume repeatability over an 18-year study period, and compare equivalent uniform annual costs (EUAC) over the useful lives of the two systems.

EUAC (15%) of AC = $60,000 (A/P, 15%, 6) - $10,000 (A/F, 15%, 6) +$40,000 = $54,710

EUAC (15%) of CC = $80,000 (A/P, 15%, 9) + $32,000 = $48,768

Recommend the compression chilling system.

An alternative solution involves the comparison of PWs over the common 18-year study period:

PW(15%) of AC = -$54,710 (P/A, 15%, 18) = -$335,263

PW(15%) of CC = -$48,768 (P/A, 15%,18) = -$298,850

As expected, select the compression chilling system.

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6-39

Assume repeatability. (a) Machine A: CR cost/yr = $35,000(A/P,10%,10) − $3,500(A/F,10%,10) = $5,475.05 Maint. Cost/yr = $1,000 $5,475.05  $1,000 = $0.6475 CR and maintenance cost/part = 10,000 parts $16 / hr = $5.3333 3 parts / hr Total cost/part (to nearest cent) = $5.98 Labor cost/part =

Machine B: CR cost/yr = $150,000(A/P,10%,8) − $15,000(A/F,10%,8) = $26,799 Maint. Cost/yr = $3,000 $26,799  $3,000 CR and maintenance cost/part = = $2.98 10,000 parts $20 / hr = $3.33 6 parts / hr Total cost/part (to nearest cent) = $6.31 Labor cost/part =

Select Machine A to minimize total cost per part. (b) Machine A: $8 / hr = $2.67 3 parts / hr. (other costs remain unchanged) Total cost/part (to nearest cent) = $3.31

Labor cost/part =

Machine B: $10 / hr = $1.67 6 parts / hr. (other costs remain unchanged) Total cost/part (to nearest cent) = $4.65

Labor cost/part =

Select Machine A to minimize total cost per part.

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6-40

Assume repeatability and compare equivalent uniform annual costs over the life of the roadway.

Asphalt:

EUAC = $800,000 (A/P, 5%, 10) + $2,000 = $105,600

Concrete:

EUAC = $1,100,000 (A/P, 5%, 18) + $1,000 = $95,050

Select the concrete option to minimize EUAC.

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6-41

(a) Assume repeatability.

AWA = −$20,000(A/P,20%,5) − $5,500 + $1,000(A/F,20%,5) = −$12,053.60 AWB = −$38,000(A/P,20%,10) − $4,000 + $4,200(A/F,20%,10) = −$12,901.30

Select A

(b) AWA(20%) = −$12,053.60 AWB(20%) = −$38,000(A/P,20%,5) − $4,000 +$15,000(A/F,20%,5) = −$14,691.20 Select A

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6-42

Trail on Flat Terrain:

Paving cost = ($3/m2)(2 m)(14,000 m) = $84,000 Site preparation = (0.2)($84,000) = $16,800 Annual Maintenance = (0.05)($84,000) = $4,200

CW = −$16,800 −$84,000 − $4,200/0.06 = −$170,800

Trail on Hilly Terrain:

Paving cost = ($3/m2)(2 m)(12,000 m) = $72,000 Site preparation = (0.2)($72,000) = $14,400 Annual Maintenance = (0.08)($72,000) = $5,760

CW = −$14,400 −$72,000 − $5,760/0.06 = −$182,400

Recommend the trail on flat terrain.

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6-43

We can analyze the annual worth of the three mutually exclusive facility concepts:

AW (10%) of A = -$85,000,000 (A/P, 10%, 60) + $85,000,000 (A/F, 10%, 60) + $11,300,000 = $2,800,000

AW (10%) of B = -$100,000,000 (0.10) + $12,000,000 = $2,000,000

AW (10%) of C = -$50,000,000 (0.10) + $7,500,000 = $2,500,000

Therefore, Concept A should be selected by a slim margin.

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6-44

Assume repeatability (repeating cash flow patterns, etc.)

AW(15%) of Caterpillar = $22,000 (A/P, 15%, 4) + $4,000 (A/F, 15%, 4) + $7,000 = $94.60

AW(15%) of Deere = $26,200 (A/P, 15%, 3) + $5,000 (A/F, 15%,3) + $9,500 = $535.60

AW(15%) of Case = $17,000 (A/P,15%,5) + $3,500(A/F, 15%, 5) + $5,200 = $647.95

To maximize AW(15%), select the Case equipment.

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6-45

(a)

Assume repeatability and compute AW over useful life.

AWE1(15%) = −$14,000 (A/P, 15%, 5) − $14,000 + $8,000 (A/F, 15%, 5) = −$16,990

AWE2(15%) = −$65,000 (A/P, 15%, 20) − $9,000 + $13,000 (A/F, 15%, 20) = −$19,620

Select E1 to minimize costs.

(b) Compute the implied market value of E2 at the end of year 5.

CRE2(15%) = $65,000 (A/P, 15%, 20) − $13,000 (A/F, 15%, 20) = $10,260

Implied MV5 = $10,260 (P/A, 15%, 15) = $59,994

AWE2(15%) = −$65,000 (A/P, 15%, 5) − $9,000 + $59,994 (A/F, 15%, 5) = −$19,492

The small difference in the value of AWE2(15%) in part (a) and part (b) is due to rounding.

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6-46

Wet storage: EUAC (10%) = $2,000M (A/P, 10%, 10) + $300M = $325.4M + $300M = $625.4M Dry storage:

EUAC (10%) = $2,500M (A/P, 10%, 7) + $150M = $513.5 + $150 = $663.5 million

Wet storage should be recommended.

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6-47

Equate the AW of each alternative over its own useful life and solve for i: −$400,000(A/P, i’%, 8) − $30,000 = −$250,000(A/P, i’%, 6) − $60,000 Solving yields i’ = 20.75% > MARR. Extra investment in the hydrogen fuel-cell fork lift is justified and this alternative should be selected.

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6-48

Back Scatter (B) PWB(15%) = −$210,000 − $31,000 (P/A,15%,6) − $2,000 (P/G,15%,6) + $21,000 (P/F,15%,6) = − $334,118 Millimeter Wave (W) Calculate an imputed MV at EOY 6: Capital Recovery Amount = −$264,000 (A/P,15%,10) + $38,000 (A/F,15%,10) = −$50,742 MV6 = | −$50,742 | (P/A,15%,4) + $38,000 (P/F,15%,4) = $166,597

iCR =

015 .  0.057 (1088 . )6 1 = 0.088 or, 8.8%; (P/A,8.8%,6) = = 4.5128 1057 .

0.088(1088 . )6

PWW(15%) = −$264,000 −

$19,000 (P/A,8.8%,6) + $166,597 (P/F,15%,6) 1057 .

= −$273,100 Select Millimeter Wave.

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6-49

(a)

IRRA = 29.9%

IRRB = 14.8%

IRRC = 16.8%

Drop B because its IRR < 15%. Consider the incremental IRR of C minus A. IRRΔ(C – A) < 15% so select alternative A. (b)

AWA = −$20,000(A/P, 15%, 3) + $10,000 + $4,000(A/F, 15%, 3) = $2,392 AWB = −$33,000(A/P, 15%, 4) + $9,500 + $10,000(A/F, 15%, 4) = −$57 AWC = −$45,000(A/P, 15%, 6) + $10,500 + $18,000(A/F, 15%, 6) = $667 Recommend alternative A.

Check the Δ (B-A) for System B versus A:

0 = -$6,000 - $6,000 (P/F, i'%, 5) + $1,600 (P/A, i'%, 10) i"% = 10.4% < 15%, so keep A as better than B.

Check the Δ (C-A) for System C versus A:

0 = -$18,000 + $2,000 (P/F i'%,5) + $3,000 (P/A, i'%, 10) i"% = 12.1% < 15%, so keep A as better than C.

Final recommendation: select System A as best (and as expected).

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6-50

(a)

PWA(8%) = [−$250 + $40.69(P/A, 8%, 10)][1 + (P/F, 8%, 10)] = $33.70 PWB(8%) = −$375 + $44.05(P/A, 8%, 20) = $57.49 PWC(8%) = [−$500(A/P, 8%, 5) + $131.90](P/A, 8%, 20)] = $65.29

Select C to maximize PW over a common 20-year study period.

(b)

PWA(8%) = −$250 + $40.69(P/A, 8%, 10) = $23.03 PWB(8%) = −$375 + $44.05(P/A, 8%, 20) = $57.49 PWC(8%) = −$500 + $131.90(P/A, 8%, 5) = $26.64

Select B to maximize PW over a common 20-year study period with reinvestment in alternatives A and C occurring at 8% per year (instead of at 10% per year as in Part a). Notice that we continue to analyze MEAs over a common 20-year period.

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6-51

Calculate the imputed market value of Alternative 2 at the end of year five.

MV5 = [ X (A/P, 12%, 7) − $4,000 (A/F, 12%, 7) ] (P/A, 12%, 2) + $4,000 (P/F, 12%, 2) = 0.3703X + $2,518.84 PWAlt 1 = −$3,000 + $800(P/A, 12%, 5) + $2,000(P/F, 12%, 5) = $1,018.64 PWAlt 2 = −X + $1,300(P/A, 12%, 5) + (0.3703X + $2,518.84)(P/F, 12%, 5) The breakeven value of X will result in PWAlt 1 = PWAlt 2. $1,018.64 = X + $1,300(P/A, 12%, 5) + (0.3703X + $2,518.84)(P/F, 12%, 5) $1,018.64 = −X + $4,686.24 + 0.2101X + $1,429.19 0.7899X = $5,096.79 X = $6,452 Note that this corresponds to an imputed market value at the end of year five of $4,908 for Alternative 2.

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6-52 Assume repeatability and a study period of 20 years.

AWA(15%)

= -$50,000 (A/P, 15%, 20) + $10,000 (A/F, 15%, 20) - $5,000 (A/F, 15%, 5) - $5,000 - $500 (A/G, 15%, 20) = -$50,000 (0.1598) + $10,000 (0.0098) - $5,000 (0.1483) - $5,000 - $500 (5.3651) = -$16,316.05

Then PWA(15%) = AW(15%)(P/A, 15%, 20) = AW(15%)(6.2593) = -$102,127.57

AWB(15%)

= -$20,000 (A/P, 15%, 10) - $10,000 - $1,000 (A/G, 15%, 10) = -$20,000 (0.1993) - $10,000 - $1,000 (3.3832) = -$17,369.20

Then PWB(15%) = AW(15%)(P/A, 15%, 20) = AW(15%)(6.2593) = -$108,719.60

Choose alternative A.

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6-53

Life of 100yr. bulb = 100H hrs. / H hrs./yr. = 100 yrs.

Life of standard bulb = H hrs. / H hrs./yr. = 1 yrs.

AC100 = ($24.95/4)(A/P, 5%, 100) + (60W/0.8)(kW/1000W)($0.0574/kW-hr)(H hrs./yr.) = ($24.95/4)(0.0504) + (60W/0.8)(kW/1000W)($0.0574/kW-hr)(H hrs./yr.) = 0.31437 + 0.004305H ACstandard= ($2.60/4)(A/P, 5%, 1) + (60W/0.95)(kW/1000W)($0.0574/kW-hr)(H hrs./yr.) = ($2.60/4)(1.05) + (60W/0.8)(kW/1000W)($0.0574/kW-hr)(H hrs./yr.) = 0.6825 + 0.0036253H AC100 = ACstandard 0.31437 + 0.004305H = 0.6825 + 0.0036253H 0.0006797 H = 0.36813 H = 541.58 hrs./yr.

Note: 100-yr preferred for H < 541.58 hrs./yr.

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6-54

The present worth of Alt. J is: PWJ(12%)

= -45 – 12(P/A, 12%, 6) = -45 – 12(4.114) = -94.34

The imputed market value of Alt. K is: MV6

= [60 (A/P, 12%, 8)](P/A, 12%, 2) = 60(0.2013)(1.6901) = 20.413

Now the present worth of Alt. K over 6 years is: PWK(12%)

= -60 – 14(P/A, 12%, 6) + 20.413(P/F, 12%, 6) = -60 – 14(4.1114) + 20.413(0.5066) = -107.22

Select Alt. J

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6-55

Both alternatives will be compared with the present worth technique over a 10-year study period. Only the imputed market value for Alternative A needs to be computed.

For Alt. A the annual capital recovery cost is CRA = $50,000 (A/P, 15%, 20) + $5,000 (A/F, 15%, 5) − $10,000 (A/F, 15%, 20) = $8,633.50 The imputed market (salvage) value at the end of year 10 is MV10 = $8,633.50 (P/A, 15%, 10) = $43,330.

The PW of Alt. A is PWA = −$50,000 − $5,000 (P/F, 15%, 5) − $5,000 (P/F, 15%, 10) − $5,000 (P/A, 15%, 10) − $500 (P/G, 15%, 10) + $43,330 (P/F, 15%, 10) = −$76,595 The life of Alternative B is 10 years, so the PW can be computed directly.

PWB = −$20,000 − $10,000 (P/A, 15%, 10) − $1,000 (P/G, 15%, 10) = −$87,168

Therefore, Alternative A is the best choice.

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6-56

1.

Rank: DN→B→A

2.

B is better than DN, since IRRB = 16.6% > MARR

3.

Analyze ∆(A – B)

Assumptions: Repeatability ∆(A−B) cash flows:

$6000

$300 $300 $300 0

1

2

$300

$300

$300

$300

3

4

5

6

$3000 PW(i’(A-B)) = 0 = -$3,000 + $300 (P/A, i’B→A, 6) + $6,000 (P/F, i’B→A, 3) + $300 (P/F, i’B→A, 6) By interpolation i’(A-B) ≈ 38% Or PW (15%) = $237.54 > 0

∴ i’(A-B) > 15% Select A

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6-57

Assume repeatability. Let: NA = life of blower costing $42,000 NB = life of the more efficient blower H = hours of operation per year C = cost of electricity per kilowatt−hour X = cost of the more efficient blower

Operation Cost of A = OCA =

90 hp (0.746 kW / hp)(H hrs / yr)(C $ / kW - hr) 0.72

AWA(20%) = −$42,000 (A/P,20%,NA) − (OCA)

Operation Cost of B = OCB =

90 hp (0.746 kW / hp)(H hrs / yr)(C $ / kW - hr) 0.81

AWB(20%) = − X (A/P,20%,NB) − (OCB)

Set AWA(20%) = AWB(20%) and solve for X. $42,000 (A/P,20%,NA) − X (A/P,20%, NB) = OCB − OCA

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6-58

Assume repeatability over the study period of 60 years, and use the Annual Worth criterion to make the comparison:

AW (10%) of A = -$100,000 (A/P, 10%, 15) - $9,500 = -$22,650

AW (10%) of B = -$130,000 (A/P, 10%, 20) - $8,000 = -$23,275

By a slim margin, select Generator A. A side question is "how sensitive is this choice to changes in the annual operating and maintenance (O&M) expenses?" There is only $625 to play with, so if A's O&M increases to greater than $10,125, then B would be selected. Or if B's O&M decreases to less than $7,375, then B becomes the preferred choice.

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6-59

PW of home filtered water = $75  182.5 gal/yr ($0.01 per gal.)(P/A, 10%,3) = $75  $1.825(2.4869) = $79.54

PW of bottled water = $1.29/bottle(8 bottles/gal)(182.5 gal/yr.)(P/A, 10%, 3) = $4,684

The high cost of bottled water vs. filtered tap water is borderline unbelievable! Let's go with the tap water from home to save money and keep the environment free of used plastic water bottles that take thousands of years to degrade in a landfill (or they can be re-cycled).

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6-60

For Edward, the future worth of his account in 36 months will be

$3,000 (F/P, 1.5%, 36) = $3,000 (1.7091) = $5,127.

The future worth of Jorge's account in 36 months will be

$3,000 (F/P, 0.60833%, 36) = $3,000 (1.2440) = $3,732.

So over a period of three years, Edward will repay $1,395 more on his credit card than Jorge will have to repay. Clearly, the idea is to keep a "clean" credit score so you can save big bucks on loans.

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6-61

Net annual receipts equal $760,000 - $500,000 = $260,000 for Alt. A, and $580,000 - $360,000 = $220,000 for Alt. B.

AW (8%) of A = -$1,000,000 (A/P, 8%, 6) + $260,000 + $100,000 (A/F, 8%, 6) = $57,330

AW (8%) of B = -$1,250,000 (A/P, 8%, 8) + $220,000 + $ 150,000 (A/F, 8%, 8) = $16,600

Select Alternative A to maximize annual worth.

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6-62

(a)

F = $10,000 with interest-only being repaid each month.

(b)

F = $5,000(F/P, 0.75%, 48) + $5,000(F/P, 0.75%, 24) = $13,139 with no interest repaid while in school.

(c)

Monthly payment (interest repaid each month) = $10,000(A/P, 0.75%, 60) = $208 Monthly payment with no interest repaid = $13,139(A/P, 0.75%, 60) = $273.29

(d)

Total interest (interest paid while in school) = $37.50(48) + $37.50(24) + $208(60) − $10,000 = $5,180

Total interest (with no interest paid while in school) = $273.29(60) − $10,000 = $6,397

Therefore the difference is $1,217 more interest if Sara does not repay any interest while she is in school. This is 23.5% higher than the $5,180 Sara repays when interest-only is paid while she is attending college.

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6-63

Loan-balance calculation: A=$500,000 (A/P, ½%, 72), or A= $8,300 per month. The loan balance is then $8,300 (P/A, ½%, 48) = $353,416.50. For the new loan we have: A= $353,416.50 (A/P, ¼%, 48) = $7,810.50. The monthly savings equals $8,300 - $7,810.50 = $489.50. Finally, the PW of the difference between the original financing plan and the new (proposed) financing plan is PW(1%) = -$10,000 + $489.50 (P/A, 1%, 48) = $8,588.27 Yes, financing is recommended.

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6-64

With 2.9% financing, the monthly payment will be

$30,000 (A/P, 2.9%/12, 48 months) = $30,000 ( 0.0221) = $663.00 per month.

With 8.9% financing, the monthly payment will be

$28,000 (A/P, 8.9%/12, 48 months) = $28,000 ( 0.0248) = $694.40 per month.

The 2.9% financing plan is the better choice because it saves $31.40 per month.

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6-65

The incremental amount that you have available to invest now on the certificate of deposit is $8,500 (if you pay $9,500 now and invest the remainder). It can grow to $8,500 (1.03) = $8,755 by January 1. It will not cover the $9,500 due at that time, so the discounted plan should be chosen.

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6-66

Electrical demand / year (incandescent bulbs) = 1,000 Watts × 3,000 hours = 3,000 kWh Annual energy cost (incandescent bulbs) = 3,000 kWh × $0.10/kWh = $300 Annual purchase cost of incandescent bulbs = 10 × $0.75 = $7.50 Annual CO2 penalty for incandescent bulbs = (150 lb/bulb)(10 bulbs)($0.02/lb) = $30

Total annual expenses = $300 + $7.5 + 30 = $337.50

PW(costs of incandescent) = $500 + $337.50(P/A, 8%, 10) = $2,765

Now the maximum affordable cost (X) for the CFB fixtures and bulbs is as follows:

Annual energy cost (CFB) = $300 (1 – 0.7) = $90

10X + $90(P/A, 8%, 10) = $2,765

So, X = $216 (which includes the CFB fixture and a CFB bulb that will last 10 years). If a CFB bulb costs $5, this leaves $211 for the purchase of each CFB fixture. Because CFB fixtures generally cost less than $211, it is economical for Bob and Sally to install fluorescent lighting in their new home. Ask your students to list non-monetary advantages and disadvantages of fluorescent and incandescent lighting. Is their choice the same based on economics alone?

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6-67

(a)

How many machines will be needed?

6,150/(3,000*0.85*0.95) = 2.54 (3 lathes of type L1 Or 4,400/(3,000*0.90*0.90) = 1.81 (2 lathes of type L2)

(b)

The annual cost of ownership, based on capital investment, is

$18,000*3*(A/P,18%,7) = $14,169.60 for lathe L1 and $25,000*2*(A/P,18%,11) = $10,740.00 for L2.

Repeatability is assumed.

(c)

Since the workers are paid for idle time we cannot make any reductions in annual expenses. Thus, annual expenses for 3 L1is 3*$5,000 = $15,000/yr. For 2 L2s, the annual expense equals 2*$9,500 = $19,000/yr.

(d)

The total equivalent annual cost for the two options equals: $14,169.60 + $15,000 = $29,169.60 for lathe L1 and $10,740.00 + $19,000 = $29,740.00 for lathe L2. Hence, lathe L1 is the preferred choice to minimize equivalent annual choice.

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6-68

(a)

MARR =

15%

Useful Life = Expenses Capital Investment Power Labor Maintenance Tax & Insurance

Annual Output Capacity Selling price =

5

120,000 $

0.375

$

P1 24,000

$

P2 30,400

$

P3 49,600

$

P4 52,000

$ $ $ $

2,720 26,400 1,600 480

$ $ $ $

2,720 24,000 1,800 608

$ $ $ $

4,800 16,800 2,600 992

$ $ $ $

5,040 14,800 2,000 1,040

Reject Rate Revenue

$

8.4% 41,220

$

0.3% 44,865

$

2.6% 43,830

$

5.6% 42,480

EOY 0 1 2 3 4 5

$ $ $ $ $ $

P1 (24,000) 10,020 10,020 10,020 10,020 10,020

$ $ $ $ $ $

P2 (30,400) $ 15,737 $ 15,737 $ 15,737 $ 15,737 $ 15,737 $

P3 (49,600) $ 18,638 $ 18,638 $ 18,638 $ 18,638 $ 18,638 $

P4 (52,000) 19,600 19,600 19,600 19,600 19,600

PW = AW = FW =

$ $ $

9,589 2,860 19,286

$ $ $

22,353 6,668 44,960

12,877 3,842 25,901

13,702 4,088 27,560

$ $ $

$ $ $

For a MARR = 15% per year, P2 is still the recommended alternative.

The overall ranking of the alternatives remains P2 > P4 > P3 > P1.

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6-68

(b)

MARR = Useful Life = Expenses Capital Investment Power Labor Maintenance Tax & Insurance

Annual Output Capacity Selling price =

10% 5 P1

P2

$

P3

120,000 0.500 P4

$ $ $ $

24,000 2,720 26,400 1,600

$ $ $ $

30,400 2,720 24,000 1,800

$ $ $ $

49,600 4,800 16,800 2,600

$ $ $ $

52,000 5,040 14,800 2,000

$

480

$

608

$

992

$

1,040

Reject Rate Revenue

$

8.4% 54,960 $

2.6% 58,440 $

5.6% 56,640

EOY 0 1 2 3 4 5

$ $ $ $ $ $

P1 (24,000) 23,760 23,760 23,760 23,760 23,760

$ $ $ $ $ $

P2 (30,400) 30,692 30,692 30,692 30,692 30,692

$ $ $ $ $ $

P3 (49,600) 33,248 33,248 33,248 33,248 33,248

$ $ $ $ $ $

P4 (52,000) 33,760 33,760 33,760 33,760 33,760

PW = AW = FW =

$ $ $

66,069 17,429 106,405

$ $ $

85,947 22,673 138,418

$ $ $

76,436 20,164 123,101

$ $ $

75,977 20,043 122,362

0.3% 59,820 $

For a selling price of $0.50, P2 is still the recommended alternative.

However, the overall ranking of the alternatives is now P2 > P3 > P4 > P1.

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6-68

(c)

MARR =

10%

Useful Life =

Expenses Capital Investment Power Labor Maintenance Tax & Insurance

Annual Output Capacity Selling price = Scrap Price =

5

120,000 $ $

0.375 0.100

$

P1 24,000

$

P2 30,400

$

P3 49,600

$

P4 52,000

$ $ $ $

2,720 26,400 1,600 480

$ $ $ $

2,720 24,000 1,800 608

$ $ $ $

4,800 16,800 2,600 992

$ $ $ $

5,040 14,800 2,000 1,040

Reject Rate Revenue

$

8.4% 42,228

$

0.3% 44,901

$

2.6% 44,142

$

5.6% 43,152

EOY 0 1 2 3 4 5

$ $ $ $ $ $

P1 (24,000) 11,028 11,028 11,028 11,028 11,028

$ $ $ $ $ $

P2 (30,400) $ 15,773 $ 15,773 $ 15,773 $ 15,773 $ 15,773 $

P3 (49,600) $ 18,950 $ 18,950 $ 18,950 $ 18,950 $ 18,950 $

P4 (52,000) 20,272 20,272 20,272 20,272 20,272

PW = AW = FW =

$ $ $

17,805 4,697 28,675

$ $ $

29,392 7,754 47,336

22,235 5,866 35,810

24,847 6,555 40,016

$ $ $

$ $ $

Including a scrap price of $0.10, P2 is still the recommended alternative. The overall ranking of the alternatives remains P2 > P4 > P3 > P1.

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6-68

(d)

MARR = Useful Life =

Expenses Capital Investment Power Labor Maintenance Tax & Insurance

Annual Output Capacity Selling price = Scrap Price =

15% 5

P1

P2

$ $

P3

120,000 0.375 0.100 P4

$ $ $ $

24,000 2,720 26,400 1,600

$ $ $ $

30,400 2,720 24,000 1,800

$ $ $ $

49,600 4,800 16,800 2,600

$ $ $ $

52,000 5,040 14,800 2,000

$

480

$

608

$

992

$

1,040

Reject Rate Revenue

$

8.4% 42,228 $

2.6% 44,142 $

5.6% 43,152

EOY 0 1 2 3 4 5

$ $ $ $ $ $

P1 (24,000) 11,028 11,028 11,028 11,028 11,028

$ $ $ $ $ $

P2 (30,400) 15,773 15,773 15,773 15,773 15,773

$ $ $ $ $ $

P3 (49,600) 18,950 18,950 18,950 18,950 18,950

$ $ $ $ $ $

P4 (52,000) 20,272 20,272 20,272 20,272 20,272

PW = AW = FW =

$ $ $

12,968 3,868 26,082

$ $ $

22,474 6,704 45,202

$ $ $

13,923 4,154 28,005

$ $ $

15,955 4,760 32,091

0.3% 44,901 $

When all changes occur simultaneously, P2 is still the recommended alternative.

The overall ranking of the alternatives remains P2 > P4 > P3 > P1.

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6-69

Investment Amount Interest Rate

$4,000 5%

Class A:

0 1 2 3 4 5 6 7 8 9 10

Class B: Fee 5% 0.61% 0.61% 0.61% 0.61% 0.61% 0.61% 0.61% 0.61% 0.61% 0.61%

Account Value at EOY $ 3,800.00 $ 3,966.82 $ 4,140.96 $ 4,322.75 $ 4,512.52 $ 4,710.62 $ 4,917.42 $ 5,133.29 $ 5,358.64 $ 5,593.89 $ 5,839.46

0 1 2 3 4 5 6 7 8 9 10

Fee 0 2.35% 0.34% 1.37% 1.37% 1.37% 1.37% 1.37% 1.37% 1.37% 1.37%

Account Value at EOY $ 4,000.00 $ 4,106.00 $ 4,297.34 $ 4,453.33 $ 4,614.99 $ 4,782.51 $ 4,956.12 $ 5,136.03 $ 5,322.46 $ 5,515.67 $ 5,715.89

It takes eight years for the Class A account to be preferred to the Class B account.

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6-70

(a)

MARR

20%

Capital investment Annual energy expense Annual maintenance: Starting year First year Increase per year Useful life (years) Market Value

EOY 4 5 6 7 8 9 AW(20%)

SP240 $ 33,200 $ 2,165

$ $ $

1 1,100 500 5 −

HEPS9 $ 47,600 $ 1,720

$ $ $

4 500 100 9 5,000

$ $

SP240 (4,765) (5,265)

HEPS9 $ (2,220) $ (2,320) $ (2,420) $ (2,520) $ (2,620) $ 2,280

$

(15,187)

$ (13,621)

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6-70

(b)

MARR

20%

Capital investment Annual energy expense Annual maintenance: Starting year First year Increase per year Useful life (years) Market Value EOY 0 1 2 3 4 5 6 7 8 9 AW(20%)

SP240 $ 28,519 $ 2,165

$ $ $

1 1,100 500 5 −

HEPS9 $ 47,600 $ 1,720

$ $ $

4 500 100 9 5,000

$ $ $ $ $ $

SP240 (28,519) (3,265) (3,765) (4,265) (4,765) (5,265)

HEPS9 $ (47,600) $ (1,720) $ (1,720) $ (1,720) $ (2,220) $ (2,320) $ (2,420) $ (2,520) $ (2,620) $ 2,280

$

(13,621)

$ (13,621)

Difference $ −

If the capital investment for SP240 was $28,519, we would be indifferent as to which pump was implemented.

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6-71

A B C $28,000 $55,000 $40,000 $15,000 $13,000 $22,000 $23,000 $28,000 $32,000 $6,000 $8,000 $10,000 10 years 10 years 10 years

Investment cost Annual expenses Annual revenues Market value Useful life MARR

0 1 2 3 4 5 6 7 8 9 10 IRR Good?

20% A ‐ DN ($28,000) $8,000 $8,000 $8,000 $8,000 $8,000 $8,000 $8,000 $8,000 $8,000 $14,000 26.4% Yes

C‐A ($12,000) $2,000 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000 $2,000 $6,000 13.1% No

B‐A ($27,000) $7,000 $7,000 $7,000 $7,000 $7,000 $7,000 $7,000 $7,000 $7,000 $9,000 22.8% Yes

Invest in alternative B.

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6-72

Other factors that could be included are distribution costs to grocers, storage costs at grocery stores and projected revenues in the likely event that half gallons are sold for less than twice the price of a quart, and quarts are sold for less than twice the price of a pint. Such non−proportional pricing is common for consumer food products.

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6-73

If landfill costs double, the recommendation for the assumptions given in the case study will be to "produce ice cream and yogurt in half−gallon containers."

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6-74

At your grocery store you will probably discover that the profit associated with a smaller container of ice cream is greater than that of a larger container. For a fixed amount of consumption (e.g. 20,000,000 gallons per year), it is likely that packaging in pint containers is the way to go. Ned and Larry's Ice Cream Company apparently already knows this to be true.

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6-75

The up-front savings in commission is 0.0575 ($23,000) = $1,322.50. Try i= 12% to determine whether this equation is met: $1,322.50 = $4,000 (P/F, 12%, 10). So $1,322.50 is quite close to $1,288 and the correct choice is (b).

Select (b)

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6-76

Savings per plan = $40,000 − $30,000 = $10,000 / year Let X = number of planes operated per year.

$500,000 = $10,000 (X) (P/A,10%,15) + $100,000 (P/F,10%,15)

X=

$500,000  $100,000(0.2394) $10,000(7.6061)

X = 6.26 or 7 planes per year

Select (a)

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6-77

PWB(15%) = −$16,000 + $5,500 (P/A,15%,3) + $6,150 (P/F,15%,3) = $601

Select (b) − Alternative A to maximize PW

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6-78

FWC(12%) = −$13,000 (F/P,12%,10) − $500 (F/A,12%,10) + $1,750 = −$47,400

Select (c) – Alternative C to minimize costs.

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6-79

AWA(12%) = −$12,000(A/P, 12%, 4) + $4,000 + $3,000(A/F, 12%, 4) = $677 AWB(12%) = −$15,800(A/P, 12%, 4) + $5,200 + $3,500(A/F, 12%, 4) = $730 AWC(12%) = −$8,000(A/P, 12%, 4) + $3,000 + $1,500(A/F, 12%, 4) = $680

Select (b)

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6-80

IRR on Δ(B – C): 0 = −$3,000 + ($460 − $100) (P/A,i′,6) + $3,350 (P/F,i′,6) i′ = 13.4% > 10%

Select (c) – Alternative B

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6-81

ΔPWW→X(15%) = −$550 + $15 (P/A,15%,8) + $200 (P/F,15%,8) = −$417.31 < 0

Select (a) – Alternative W

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6-82

Rank Order: DN→D→C→A→E→B Assuming the MARR ≤ 42.5%, Alternative D is the base alternative. The first Comparison to be made based on the tank ordering would be D→C.

Select (e)

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6-83

PWA→B(15%) = [−$90,000 − (−$60,000)] + ($12,000 − $20,000) (P/A,15%,10) + ($15,000 − $10,000) (P/F,15%,10) = −$68,914

Select (a)

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6-84

PWA(i′) = 0 = −$60,000 + $20,000 (P/A,i′,10) i′ = 31.5%

Select (a)

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6-85

Eliminate Alt. B and Alt. E (IRR < 15%) PWA(15%) = −$60,000 + $20,000 (P/A,15%,10) + $10,000 (P/F,15%,10) = $42,848 PWC(15%) = −$40,000 + $13,000 (P/A,15%,10) + $10,000 (P/F,15%,10) = $27,716 PWD(15%) = −$30,000 + $13,000 (P/A,15%,10) + $10,000 (P/F,15%,10) = $37,716

Select (b) – Alternative A to maximize PW.

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6-86

AWX(10%) = −$500,000(A/P, 10%, 5) + $131,900 = 0 AWY(10%) = −$250,000(A/P, 10%, 10) + $40,690 = $15 AWZ(10%) = −$400,000(A/P, 10%, 20) + $44,050 = −$2,950

Select (b)

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6-87

By the time Mary quits making payments into her savings plan at age 55, her investment will be worth:

F55 = $100 (F/A, 8%/12, 240) = $100 (589.0233) = $58,902.

At age 65 Mary's plan (compounded with interest) will be worth

F65 = F55 (F/P, 8%/12, 120) = $58,902 ( 2.2196) = $130,740

At age 65 Bill's plan will be worth:

F65 = $100 (F/A, 8%/12, 240) = $100 (589.0233) = $58,902.

Mary's plan will be worth ($130,740 - $58,902) / $58,902, or 122% more than Bill's plan.

Select (a)

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Solutions to Chapter 7 Problems 7-1

The actual magnitude of depreciation cannot be determined until the asset is retired from service (it is always paid or committed in advance). Also, throughout the life of the asset we can only estimate what the annual or periodic depreciation cost is. Another difference is that relatively little can be done to control depreciation cost once an asset has been acquired, except through maintenance expenditures.

Usually much can be done to control the ordinary out-of−pocket expenses such as labor and material.

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7-2

To be considered depreciable, a property must be: 1) used in a business to produce income; 2) have a determinable life of greater than one year; and 3) lose value through wearing out, becoming obsolete, etc.

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7-3

Personal property is generally any property that can be moved from one location to another, such as equipment or furniture. Real property is land and anything erected or growing on it.

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7-4

The cost basis is usually the purchase price of the property, plus any sales taxes, transportation costs, and the cost of installation or improving the property to make it fit for intended use. Salvage value is not considered, nor is the cost of the land the property is on.

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7-5

B = $28,000; R = 2/N = 2/8 = 1/4 (a)

d8 = B (1–R)7(R) = $28,000 (1 – 1/4)7(1/4) = $934.38

(b)

BV7 = B (1 – R)7 = $28,000 (1–1/4)7 = $3,737.54

(c)

Gain = MV7 – BV7 = $1,000 – $3,737.54 = –$2,737.4 (a loss).

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7-6

Basis = $100,000 + $12,000 = $112,000 (a)

d2 = ($112,000 – $8,000)/10 = $10,400

Depreciation during second year = $10,400 (b)

BV2 = $112,000 – $10,400 = $101,600

(c)

BV8 = $112,000 – $10,400  8 = $28,800

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7-7 Year, k 0 1 2 3 4 5

DB = BVk-1(2/5)

SL = BVk-1/[5 – (k – 1)]

“Winner”

32,000 19,200 11,520 6,912 4,147.20

16,000 12,000 9,600 8,640 8,640

32,000 19,200 11,520 8,640 8,640

BVk 80,000 48,000 28,800 17,280 8,640 0

Switchover occurs in year 4.

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7-8

Basis = $60,000 and SVN = $12,000. Find d3 and BV5. d3 = dk =

(a)

B  SVN $60,000  $12,000  = $3,428.57 N 14

BV5 = $60,000 − (5)($3,428.57) = $42,857.15 (b) (1)

Year, k 1 2 3 4 5

(2) (3) (4) Depreciation Method Beginning of 200% Declining Straight−Line Depreciation Year BVa Balance Methodb Method c Amount Selectedd $ 60,000.00 $8,571.43 $3,428.57 $8,571.43 51,428.57 7,346.94 3,032.97 7,346.94 44,081.63 6,297.38 2,673.47 6,297.38 37,784.25 5,397.75 2,344.02 5,397.75 32,386.50 4,626.64 2,038.65 4,626.64

a

Column 1 for year k − column 4 for year k = the entry in column 1 for year k+1 Column 1 x (2 / 14) c Column 1 minus estimated SVN divided by remaining years from the beginning of the current year through the fourteenth year. d Select the larger amount of column 2 or column 3. b

From the above table, d3 = $6,297.38 and BV5 = $32,386.50 − $4,626.64 = $27,759.86 (c) From Table 7-2, the GDS recovery period is 7 years. d3

= $60,000 (0.1749) = $10,494

BV5 = $60,000 − $60,000 (0.1429 + 0.2449 + 0.1749 + 0.1249 + 0.0893) = $13,386 (d)

From Table 7-2, the ADS recovery period is 14 years.  $60,000  d1  d15  (0.5)   $2,142.86  14   $60,000  d 2  d 3   d 14    = $4,285.71  14  BV5 = $60,000 − [$2,142.86 + 4($4,285.71)] = $40,714.30

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7-9

B = $5,000; SV5 = $500 (a)

dk = (B – SVN)/N= ($ 5,000 – $500)/5 = $900 = d2

Useful life = 5 years (b)

dk* = B [1 – (1 – R)k] ;

d3* = $5,000 [1 – (1 – 2/5)3] = $5,000 [1 – (3/5)3 ] = $3,920

R = 2/5 (c)

d4* = $5,000 [0.2 + 0.32 + 0.192 + 0.1152] = $ 4,136

Property class life is taken as 5 years. (d)

BV4 = B – 4d = $5,000 – 4 × $900 = $1,400

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7-10

Referring to Equation 7-9, the depreciation rate is: ($240,000 – $24,000) / (12 years × 180 days/year) = $216,000 / 2,160 rental days = $100 per day

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7-11

From Table 7-2, the GDS recovery period is seven years. (a)

Basis = $420,000 d*3 = $420,000 (0.1429 + 0.2449 + 0.1749) = $236,334

(b)

d4 = 0.1249 ($420,000) = $52,458

(c)

BV2 = $240,000 (1 – 0.1429 – 0.2449) = $146.928

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7-12

(a) EOY, k

(1) Straight-Line Methoda $48,000 48,000 48,000 48,000 48,000 48,000 48,000 48,000 48,000 48,000

1 2 3 4 5 6 7 8 9 10

(2) Declining Balance Methodb $100,000 80,000 64,000 51,200 40,960 32,768 26,214 20,972 16,777 13,422

(3) MACRS Methodc $71,450 122,450 87,450 62,450 44,650 44,600 44,650 22,300 0 0

a

Column 1 = ($500,000 - $20,000) / 10 = $48,000 Column 2 = $500,000 (1 – R)k-1(R) where R = 2/10 = 0.2 c Column 3 = $500,000 rk , where rk is taken from the 7-year column of Table 7-3 b

Note: From Table 7-2, a 10-year class life corresponds to a 7-year GDS recovery period (b)

PWSL (10%) = $48,000 (P/A, 10%, 10) = $294,941 PWDB (10%) = ∑

𝑑k (P/F, 10%, k) =$319,534

PWMACRS (10%) = ∑ (c)

𝑑k (P/F, 10%, k) = $360,720

The MACRS (GDS) depreciation method has the maximum PW, thus it is the most desirable from an income tax point of view.

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7-13

(a)

From Table 7-2, the GDS recovery period is 5 years and the ADS recovery period is 6 years.

GDS depreciation deductions: d1 = 0.2000 ($300,000) = $60,000 d2 = 0.3200 ($300,000) = $96,000 d3 = 0.1920 ($300,000) = $57,600

d4 = 0.1152 ($300,000) = $34,560 d5 = 0.1152 ($300,000) = $34,560 d6 = 0.0576 ($300,000) = $17,280

ADS depreciation deductions:  $300,000  d 1  d 7  (0.5)   $25,000   6 $300,000 d2 = d3 =  = d6   $50,000 6 (b)

Assume calculation of PW at time of purchase and the depreciation deduction is taken at the end of the year.

PWGDS = $60,000 (P/F,12%,1) + $96,000(P/F,12%,2) + $57,600(P/F,12%,3) + $34,560(P/F, 12%,4) + $34,560(P/F,12%,5) + $17,280(P/F,12%,6) = $221,431.15 PWADS = $25,000 (P/F,12%,1) + $50,000 (P/A,12%,5)(P/F,12%,1) + $25,000(P/F,12%,7) = $194,566.30 Difference = PW = $221,431.15 − $194,566.30 = $26,864.85

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7-14

From Table 7-4, the GDS recovery period is five years.

B = $120,000 + $20,000 = $140,000 (a)

d4 = 0.1152 ($140,000) = $16,128

(b)

BV5 = $140,000 – $140,000 (0.1 + 0.18 + 0.144 + 0.1152 + 0.0922) = $51,604

(c)

R = 2/10 = 1/5 d5* = $140,000 [1 – (1 – 1/5)5] = $45,875.2

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7-15

(a)

Cost basis = $1,500,000 + $35,000 + $50,000 = $1,585,000

(b)

From Table 7-2, the class life is 10 years.

(c)

The GDS recovery period is seven years. Thus, the MACRS depreciation in year five is $1,585,000(0.0893) = $141,540.50.

(d)

Remember that only half the normal depreciation amount can be claimed in the year an asset is disposed of. BV6 = (1 – 0.8215)($1,585,000) = $282,922.50. The depreciation recaptured is MV6 – BV6 = $360,000 − $282,922.50 = $77,077.50.

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7-16

Depreciation per unit of production =

$90,000  $8,000 = $0.328/unit 250,000 units

d4 = (80,000 units) × ($0.328/unit) = $26,240 BV4 = $90,000 – (80,000 units + 20,000 units)($0.328/unit) = $57,200

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7-17

Total units of production over the six-year life = 133,000 m3 d4 

24, 000  ($80, 000 – $8, 000)  $12,992.48 133, 000

 20, 000 28, 000  BV2 = $80,000 – (d1 + d2) = $80,000 –    × ($80,000 – $8,000)  133, 000 133, 000  = $54,015.037

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7-18

Taxable income = gross income – all expenses (excluding capital investment) – depreciation deductions In our case we have Taxable income = $800,000 – $500,000 – $70,000 = $230,000 From Table 7-5, we see that the federal income tax rate is t = $22,250 + 39% of the amount over $100,000 for this tax year.

Thus, the federal income tax = $22,250 + 39/100 ($230,000 – $100,000) = $72,950

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7-19

(a) Year

0 1 2 3 4a 4b 4 (b)

BTCF

Depr.

-$150,000 --80,000 $49,995 80,000 66,675 80,000 22,215 80,000 11,115 50,000 --130,000 11,115

Taxable In co m e --$30,005 13,325 57,785 68,885 50,000 118,885

Income Tax

---$6,001 -2,665 -11,557 -13,777 -10,000 23,777

ATCF

-$150,000 73,999 77,335 68,443 66,223 40,000 106,223

BV2 = $150,000 – $49,995 – $66,675(0.5) = $66,667.5 Gain = $100,000 – $66,667.5 = $33,332.50

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7-20

Basis = $12,000 + $600 = $12,600 A EOY BTCF 0 –$12,600 1 5,000 2 5,000 3 5,000 4 5,000 5a

5,000

5b

3,000 (= MV)

B Depreciation – 0.2  $12,600 0.32  $12,600 0.192  $12,600 0.1152  $12,600 0.1152  $12, 600 2 –

D C=A–B Income Tax ATCF Taxable Income 40% (A – D) – – –$12,600 $2.480 –$992 4008 0.968 –387.2 4612.8 2,580.8 –1032.82 3967.68 3,548.48 –1419.392 3580.60 4274.24

–1709.696

3290.30

1,548.48 (= MV – BV)

–619,392

2380.608

BV = $12,600 – (0.2 + 0.32 + 0.192 + 0.1152 + 0.0576) × $12,600 = $1,451.52 PW = –$12,600 + $4,008 (P/F, 10%, 1) + $4,612.8 (P/F, 10%, 2) + $3,967.68 (P/F, 10%, 3) + $3,580.6 (P/F, 10%, 4) + $(3,290.30 + 2,380.60) (P/F, 10%, 5) + $3,003.27 The sample payback period is three years.

IRR: 0 = –$12,600 + $4,008 (P/F, i’, 1) + $4,612.8 (P/F, 2%, 2) + $3,967.68 (P/F, 2%, 3) + $3,580.6 (P/F, 2%, 4) + $5,670.9 (P/F, 2%, 5) At IRR = 10%, PW = $3603.22 At IRR = 12%, PW = $2973.40 $3803.22  $2973.40 $3803.22  2 x  10

IRR = x = 19.166%

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7-21 t 0 1 2 3 4 5 6 6 PW = ∑

BTCF

Dt

-20,000 -2000 + R -3000 + R -4000 + R -5000 + R -6000 + R -7000 + R +3000

-----4000 6400 3840 2304 2304 1152 ------

Taxable income

Tax = .4(TI)

ATCFt

R – 6000 R – 9400 R – 7840 R – 7304 R – 8304 R – 8152 +3000

.4R – 2400 .4R – 3760 .4R – 3136 .4R – 2921.60 .4R – 3321.60 .4R – 3260.80 1200

-20000 .6R + 400 .6R + 760 .6R – 864 .6R – 2078.40 .6R – 2678.40 .6R – 3739.20 1800

𝐴𝑇𝐶𝐹 t (P/F, 15, t) = -23003.94 + .6R(P/A, 15, 6) = 0 = -23003.94 + .6R(3.7845) = 0 R = $10130.77

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7-22

(a) EoY 0 1 2 3 4 5 6 6 (b)

(A) Before-tax Cash Flow

(B) Depreciation Deduction

–$160,000 32,000 30,800 29,600 28,400 27,200 26,000 30,000 (= MV)

– $25,000 25,000 25,000 25,000 25,000 25,000 –

(C) = (A) – (B) Taxable Income – $7,000 5,800 4,600 3,400 2,200 1,000 20,000 (= MV – BV)

– $2,800 2,320 1,840 1,360 880 400

(E) = (A) – (D) After-tax Flow –$160,000 29,200 28,480 27,760 27,040 26,320 25,600

8,000

22,000

(D) = 0.4(C) Income Taxes

PW = –$160,000 + $29,200 (P/A, 12%, 6) – $720 (P/G, 12%, 6) + $22,000 (P/F, 12%, 6) = –$35,155.6

(c)

MARRBT = (0.12/0.6) = 20% Let X = increase in annual revenues $60,000 = X (P/A, 20%, 6) X

$60, 000  $18, 042.39 3.3255

≈ 45.10% Total before-tax annual revenues = $58.042.39

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7-23

(a)

Before−tax MARR =

After - tax MARR 0.12   0.2 = 20% 1- t 1  0.4

(b) Year 1 Depreciation $14,290 (c)

2 24,490

3 17,490

4 12,490

5 8,930

6 8,920

7 8,930

8 4,460

BV8 = 0, therefore TI8 = $12,000 (Property having a seven-year recovery period is fully depreciated after N + 1 = 8 years).

(d) (A) (B) (C) = (A – B) (D) = t(c) (E) = (A) – (D) EOY BTCF Depreciation Taxable Income Taxes (40%) ATCF PW (12%) 0 –$100,000 – – – –$100,000 –$100,000 1 20,000 $14,290 $5,710 $2,284 $17,716 15,818 2 20,000 24,490 –4,490 -1,796 21,796 17,375.77 3 20,000 17,490 2,510 1,004 18,996 13,521.35 4 20,000 12,490 7,510 3,004 16,996 10,800.95 5 20,000 8,930 11,070 4,428 15,572 8,835.55 6 20,000 8,920 11,080 4,432 15,568 7,886.74 7 20,000 8,930 11,070 4,428 15,572 7,043.21 8 20,000 4,460 15,540 6,216 13,784 5,567.35 8 12,000 – 12,000 4,800 8,000 3,231.2 PW (12%) = –$22,088.88 (e)

The machine should not be recommended for purchase. Since PW (12%) of after-tax cash flow is < 0.

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7-24

Machine 1: Year 0 1 2 3 4 5a 5b

BTCF -$60,000 -$16,000 -$16,000 -$16,000 -$16,000 -$16,000 $0

d

TI

T (40%)

$12,000 $19,200 $11,520 $6,912 $3,456

-$28,000 -$35,200 -$27,520 -$22,912 -$19,456 -$6,912

$11,200 $14,080 $11,008 $9,165 $7,782 $2,765

d

TI

T (40%)

$13,200 $21,120 $12,672 $7,603 $7,603 $3,802

-$31,200 -$39,120 -$30,672 -$25,603 -$25,603 -$21,802 -$18,000 -$18,000 $0

$12,480 $15,648 $12,269 $10,241 $10,241 $8,721 $7,200 $7,200 $0

ATCF -$60,000 -$4,800 -$1,920 -$4,992 -$6,835 -$8,218 $2,765

AWM1(12%) = $21,307 Machine 2: Year 0 1 2 3 4 5 6 7 8a 8b

BTCF -$66,000 -$18,000 -$18,000 -$18,000 -$18,000 -$18,000 -$18,000 -$18,000 -$18,000 $0

ATCF -$66,000 -$5,520 -$2,352 -$5,731 -$7,759 -$7,759 -$9,279 -$10,800 -$10,800 $0

AWM2(12%) = $20,163 Therefore, select M2.

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7-25

Printing M/C X Year 0 1 2

Inv. $25,000   6,000

Exp.  $1,000 1,000 1,000

Dep.

BV $25,000  $3,572.5 21,427.5 6,122.5 15,305 4,372.5 10,932.5

Taxable Income  $4,572.5 7,122.5 5,372.5

Income Tax  $1,829 $2,849 $2,149

ATCF $25,000 829 1,849 1,149

PW (15%) = $25000 + $829 × 0.8696 + 1849 × 0.7561 + 1149 × 0.6575 = $22,125.60 Printing M/C Y PW = $11,200 (1  0.4)(P/A, 15%, 3) = 11,200 × 0.6 × 2.2832 = $15,343.10 PW (15%) of X is less than the PW of Y. Therefore, go with printing machine Y, on lease.

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7-26

(a) EOY BTCF d TI T (40%) ATCF 0 $280,000 $280,000 1-7 $80,000 $40,000 $40,000 16,000 64,000 IRR - ?

O = $280,000 + $64000 (P/A, ?, 7)

At 10%,

PW = $280,000 + $64000 (4.8684) = $31,577.6

At 12%

PW = $280,000 + $64000 (4.5638) = $12,083.2 19494.4 12.083.2  2 ( x  12) x = 13.23%

IRR = 13.23%, after-tax IRR is more than 10%, the robot should be acquired (justified).

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(b) EOY 0 1 2 3 4 5 6 7

BTCF 280,000 80,000 80,000 80,000 80,000 80,000 80,000 80,000

d  56,000 89,600 53,760 32,256 32,256 16,128

TI  $24,000 9,600 26,240 47,744 47,744 63,872 80,000

T (40%)  9,600 3,840 10,496 19,097.6 19,097.6 25,548.8 32,000

ATCF 280,000 70,400 83,840 69,504 60,9024 60,902.4 54,451.2 48,000

PW (10%) after-tax cash flow = $280,000 + 70,400 (P/F, 10%, 1) + 83,840 (P/F, 10%, 2) + 69,504 (P/F, 10%, 3) + 60,902.4 (P/F, 10%, 4) + 60,902.4 (P/F, 10%, 5) + 54,451.2 (P/F, 10%, 6) + 48,000 (P/F, 10%, 7) = $40,969.31 PW (12%) after-tax cash flow = $21,725.23 18, 744.07 21, 725.23  2 ( x  12) IRR = x = 14.31% IRR in case (a) = 13.23% The after-tax IRR is lower.

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7-27

The sales revenue is $40 per ticket x 60,000 tickets per year = $2,400,000 per year and the investment in working capital is (1/12)($2,400,000) = $200,000. The the total investment is $800,000 = $200,000 = $1,000,000 and depreciation is $800,000 / 4 = $200,000 per year. Total annual operating costs are $24 (60,000 tickets) + $400,000 = $1,840,000.

(A)

(B)

(C) = (A) − (B)

(D) = −t (C)

(E) = (A) + (D)

EOY 0

BTCF −$1,000,000

Depr.

---

TI ---

T (50%) ---

ATCF

1 2 3 4a 4b

$560,000 $560,000 $560,000 $560,000 $200,000

$200,000 $200,000 $200,000 $200,000 −−

$360,000 $360,000 $360,000 $360,000 −−

−$122,400 −$122,400 −$122,400 −$122,400 −−

− $1,000,000 $437,600 $437,600 $437,600 $437,600 $200,000

PW(15%) = −$1,000,000 + $437,600 (P/A,15%,4) + $200,000 (P/F,15%,4) = $363,690 >> 0; the investment should be made.

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7-28

B = 3  ($480,000 + $20,000) = $1,500,000 EOY 0 1 2 3 4a 4b

BTCF $1,500,000 960,000 960,000 960,000 960,000 75,000 (= MV)

Depr. $300,000 480,000 288,000 172,800 -

TI $660,000 480,000 672,000 787,200 454,200 (= MV  BV)

T (40%) $264,000 192,000 268,800 314,880

ATCF $1500,000 696,000 768,000 691,200 645,120

181,680

256,680

BV4 = $1500,000 – $300,000 – 480,000 – $288,00 – $172,800 = $529,200 PW (18%) = –$ 1500,000 + 696,000 (P/F, 18%, 1) + 768,000(P/F, 18%, 2) + 691,200 (P/F, 18%, 3) + 645,120(P/F, 18%, 4) + 256,680 (P/F, 18%, 4) = $527,250.36 PW (20%) = $448,211.58 IRR = 31.06 % >18% Recommend the investment.

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7-29 EOY 0 1 2 3a 3b

BTCF −$200,000 −65,000 −65,000 −65,000 70,000

d

TI

$40,000 −$105,000 64,000 −129,000 19,200 −84,200 −6,800

T $42,000 51,600 33,680 2,720

ATCF −$200,000 −23,000 −13,400 −31,320 72,720

EUAC(12%) = [$200,000 + $23,000(P/F, 12%, 1) + $13,400(P/F, 12%, 2) + ($31,320 − $72,720)(P/F, 12%, 3)](A/P, 12%, 3) = $83,989

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7-30 EOY 0 1 2 3 4 5 5

(A)

(B)

(C) = (A) − (B)

(D) = −t (C)

(E) = (A) + (D)

BTCF −$160,000 $35,000 $35,000 $35,000 $35,000 $35,000 $30,000**

Depr* --$26,000 $41,600 $24,960 $14,976 $7,488 ---

TI --$9,000 −$6,600 $10,040 $20,024 $27,512 0−$14,976** *

T (50%) --−$4,500 $3,300 −$5,020 −$10,012 −$13,756 $7,488

ATCF −$160,000 $30,500 $38,300 $29,980 $24,988 $21,244 $37,488

*d = $130,000 ꞏ rk(p) (see Table 7-3) **Assume land recovered at original cost of $30,000 ***MV – BV; Market Value of equipment (purchased) is negligible at the end of year 5.

Assume 1/2 year on year 5 depreciation (recapture = $14,976) PW(5%) = −$160,000 + $30,500(P/F,5%,1) + $38,300(P/F,5%,2) + $29,980(P/F,5%,3) + $24,988(P/F5%,4) + ($21,244+$37,488)(P/F,5%,5) = −$160,000 + $30,500(0.9524) + $38,300(0.9070) + $29,980(0.8638) + $24,988(0.8227) + $58,732(0.7835) = −$3,742.83 PW(MARRAT) < 0, not a profitable investment

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7-31

(a) TI = 20,000 – BV = 20,000 – 42,048 = -22,048

(b)

EOY

BTCF

0 1 2 3 4 5 5

-$365,000 $90,000 $90,000 $90,000 $90,000 $90,000 $20,000

Depreciation --$73,000 $116,800 $70,080 $42,048 $21,024 $0

Taxable Income --$17,000 -$26,800 $19,920 $47,952 $68,976 -$22,048

Income Tax (37%) ---$6,290 $9,916 -$7,370 -$17,742 -$25,521 $8,158

ATCF -$365,000 $83,710 $99,916 $82,630 $72,258 $64,479 $28,158

PW = -365,000 + 83,710 (P/F, 10%, 1) + 99,916 (P/F, 10%, 2) + 82,630 (P/F, 10%, 3) + 72,258 (P/F, 10%, 4) +92,637 (P/F, 10, 5) PW = -$37,376 Not a good investment

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7-32

X = annual production level EOY 0 1–5 6–7

BTCF $800,000 $60,000 + 70x $60,000 + 70x

d $160,000 -

Taxable Income $220,000 + 70x $60,000 + 70x

Income Tax $88,000 – 28x $24000 – 28x

ATCF $800,000 28000 + 42X −36,000 + 42X

PW = 0 = $ 800,000 + (28,000 + 50x)(P/A, 12%, 5) + (36,000 + 42x)(P/A, 12%, 2)(P/F, 12%, 5) O = $800,000 + (2,8000 + 50x)(3.6048) + (3,6000 + 42x) (1.6901)(0.5674) O =  695,613.33 + 220.516 x. x=

695, 613.33 = 3154.47 ≃ 3,155 units must be produced for this project to be economically viable.

220.516

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(A)

7-33 EOY 0 1 2 3 4 5 6 . .

N

BTCF −$50,000 14,000 14,000 14,000 14,000 14,000 14,000 14,000 14,000 14,000

(B)

Depr --$10,000 10,000 10,000 10,000 10,000 0 0 0 0

(C) = (A) − (B)

(D) = −t (C)

(E) = (A) + (D)

TI --$4,000 4,000 4,000 4,000 4,000 14,000 14,000 14,000 14,000

T (20%) --−$800 −800 −800 −800 −800 −2,800 −2,800 −2,800 −2,800

ATCF −$50,000 13,200 13,200 13,200 13,200 13,200 11,200 11,200 11,200 11,200

Cumulative

PW(10%) −$50,000 −38,000 −27,091 −17,174 −8,158 38

The equipment must be operated for a minimum of five years in order to earn 10% on the investment.

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7-34

After-tax MARR = 7.2% per year.

EOY 0 1 2 3 4 4

BTCF -$12,000 4,000 4,000 4,000 4,000 3,000

d

TI

T

4,000 5,334 1,777 889 ---

$0 -1,334 2,223 3,111 3,000

$0 534 -889 -1,244 -1,200 A: PW

EOY 0 1 2 3 4 4

BTCF -$15,800 5,200 5,200 5,200 5,200 3,500

d

TI

T

$5,266 7,023 2,340 1,171 ---

-$66 -1,823 2,860 4,029 3,500

$26 729 -1,144 -1,612 -1,400 B: PW

EOY 0 1 2 3 4 4

BTCF -$8,000 3,000 3,000 3,000 3,000 1,500

d

TI

T

$2,666 3,556 1,185 593 ---

$334 -556 1,815 2,407 1,500

-$133 222 -726 -963 -600 C: PW

ATCF -$12,000 4,000 4,534 3,111 2,756 1,800 $

1,651 ATCF -$15,800 5,226 5,929 4,056 3,588 2,100

$

1,835 ATCF -$8,000 2,867 3,222 2,274 2,037 900

$

1,548

Choose alternative B to maximize after-tax PW. Alternative B was also chosen when a before-tax analysis was done in Problem 6-79.

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7-35

t = s + f (1 − s) = 0.04 + 0.34(1 − 0.04) = 0.3664, or 36.64% (A)

EOY 0 1 2 3 3 3

(B)

(C)

Investment Revenue Expenses s −$1,000,000 ----X $636,000 X 674,160 X 714,610 c 280,000 ----580,000d -----

(D)=(A)+(B)+(C)

(E)

BTCF

Depra

− $1,000,000 X − 636,000 X − 674,160 X − 714,610 280,000 580,000

--$139,986 186,690 31,101b -----

a

Cost basis for depreciation calculations = $420,000 Only a half-year of depreciation is allowable c Market value of depreciable investment d Assumed value of non−depreciable investment (land and working capital) b

EOY 0 1 2 3 3 3 e

(F)=(D)−(E)

(G)= −t (F)

(H)=(D)+(G)

TI --X − 775,986 X − 860,850 X − 745,711 217,777e ---

T(36.64%)

ATCF −$1,000,000 0.6336X − 351,679 0.6336X − 358,745 0.6336X − 441,381 200,207 580,000

0 −0.3664X + 284,321 −0.3664X + 315,415 −0.3664X + 273,229 −79,793 ---

MV − BV3 = $280,000 − $62,223 = $217,777

PW(10%) = 0 = −$1,000,000 + (0.6336 X)(P/A,10%,3) − $351,679(P/F,10%,1) − $358,745(P/F,10%,2) − $441,381(P/F,10%,3) + $780,207(P/F,10%,3) X=

$1,361,618 = $864,135 / year (2.4869)(0.6336)

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7-36

(a)

The equivalent uniform annual cost (EUAC) of Alternative A is $10,000. The EUAC of Alternative B is $50,000 (A/F, 15%, 5) = $50,000 (0.1483) = $7,415. Alternative B should be chosen.

(b)

The after-tax MARR is 15%(1-0.40) = 9% (use this for discounting purposes).

Alt. A: EOY BTCF T.I. Income Tax ATCF 1-5 - $10,000 -$10,000 $4,000 (credit) -$6,000 EUAC of Alt. A is -$6,000 Alt. B: EOY BTCF T.I. Income Tax ATCF 1-4 0 0 0 0 5 -$50,000 -$50,000 $20,000 (credit) -$30,000 EUAC of Alt. B equals $30,000 (A/F, 9%, 5) = $30,000 (0.1671) = $5,013. Again, Alt.B should be recommended.

(c)

There is no difference in recommendation for this problem.

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7-37 EOY 0 1 2 3 4 5 6 7 8 9 10

EOY 0 1 2 3 4 5 6 7 8 9 10

BTCF d -$1,140,000 -115,500 $162,906 -115,500 279,186 -115,500 199,386 -115,500 142,386 -115,500 101,802 -115,500 101,688 -115,500 101,802 -115,500 50,844 -115,500 -115,500

BTCF d -$992,500 -73,200 $141,828 -73,200 243,063 -73,200 173,588 -73,200 123,963 -73,200 88,630 -73,200 88,531 -73,200 88,630 -73,200 44,266 -73,200 -73,200

TI

T

-$278,406 $111,362.4 -394,686 157,874.4 -314,886 125,954.4 -257,886 103,154.4 -217,302 86,920.8 -217,188 86,875.2 -217,302 86,920.8 -166,344 66,537.6 -115,500 46,200.0 -115,500 46,200.0

ATCF -$1,140,000.0 -4,137.6 42,374.4 10,454.4 -12,345.6 -28,579.2 -28,624.8 -28,579.2 -48,962.4 -69,300.0 -69,300.0

PW(9%)

-$1,255,661

TI

T

-$215,028 -316,263 -246,788 -197,163 -161,830 -161,731 -161,830 -117,466 -73,200 -73,200

$86,011.3 126,505.3 98,715.3 78,865.3 64,732.1 64,692.4 64,732.1 46,986.2 29,280.0 29,280.0

ATCF -$992,500.0 12,811.3 53,305.3 25,515.3 5,665.3 -8,467.9 -8,507.6 -8,467.9 -26,213.8 -43,920.0 -43,920.0

PW(9%)

-$983,060

Select new ESP to maximize after-tax present worth.

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7-38

Assume repeatability.

Purchase Option: From Table 7-2, the ADS recovery period is 6 years (asset class 36.0). Applying the half year convention, depreciation deductions can be claimed over a 7-year period. d1 = d7 = (0.5)($30,000/6) = $2,500 d2 = d3 = .... = d6 = ($30,000/6) = $5,000

EOY 0 1 2 3 4 5 6 7 8

(A)

(B)

BTCF −$30,000 0 0 0 − 10,000 0 0 0 0

Depr.

--$2,500 5,000 5,000 5,000 5,000 5,000 2,500 ---

(C) = (A) − (B)

TI --−$2,500 − 5,000 − 5,000 −15,000 − 5,000 − 5,000 − 2,500 0

(D) = −t (C)

(E) = (A) + (D)

T (25%) --$625 1,250 2,000 3,750 1,250 1,250 625 0

ATCF − $30,000 625 1,250 1,250 − 6,250 = − 10,000 + 3,750 1,250 1,250 625 0

PW(12%)

= − $30,000 + $625(P/A,12%,7) + $625(P/A,12%,5)(P/F,12%,1) − 7,500(P/F,12%,4) = − $29,902

AW(12%)

= − $26,030.47 (A/P,12%,8) = − $6,019

Leasing Option ATCF = − (1 − 0.25)(Leasing Cost) = AW(12%) For the leasing option to be more economical than the purchase option, − (0.75) (Leasing Cost) < − $6,019 Leasing Cost < $8,025 If Leasing Cost < $8,025 per year, lease the tanks; otherwise, purchase the tanks.

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7-39

Assume repeatability.

Fixture X (A)

EOY BTCF 0 −$30,000 1−5 −$3,000 6 −$3,000 6* $6,000 *Market Value

(B)

Depr.

--$6,000 0 ---

(C) = (A) − (B)

(D) = −t (C)

(E) = (A) + (D)

TI T (25%) ATCF ----− $30,000 −$9,000 $2,250 −$750 −$3,000 $750 −$2,250 $6,000** −$1,500 $4,500 **Depreciation Recapture

AWX(8%) = −$6,830 Fixture Y (A)

EOY 0 1 2 3 4 5 6 7 8 8

(B)

BTCF Depr. −$40,000 --−$2,500 $8,000 −$2,500 $12,800 −$2,500 $7,680 −$2,500 $4,608 −$2,500 $4,608 −$2,500 $2,304 −$2,500 0 −$2,500 0 $4,000* --* Market Value

(C) = (A) − (B)

(D) = −t (C)

(E) = (A) + (D)

TI --−$10,500 −$15,300 −$10,180 −$7,108 −$7,108 −$4,804 −$2,500 −$2,500 $4,000

T (25%) --$2,625 $3,825 $2,545 $1,777 $1,777 $1,201 $625 $625 −$1,000

ATCF − $40,000 $125 $1,325 $45 −$723 −$723 −$1,299 −$1,875 −$1,875 $3,000

AWB(8%) = −$7,141 Select Fixture X.

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7-40

Pump A: EOY 0 1 2 3 4 5 5

BTCF -$2,000 -400 -400 -400 -400 -400 400

d

TI

T(40%) $320 320 320 320 320 -160

ATCF -$2,000 -80 -80 -80 -80 -80 240

$400 400 400 400 400

-$800 -800 -800 -800 -800 400

PWA(10%)

-$2,154

TI

T(40%)

-$1,133.35 -1,244.55 -948.15 -874.15 -800.05 200

$453.34 497.82 379.26 349.66 320.02 -80

ATCF -$1,000.00 -346.71 -302.23 -420.79 -450.39 -480.03 120.00

PWB(10%)

-$2,412

Pump B: EOY 0 1 2 3 4 5 5

BTCF d -$1,000 -800 $333.30 -800 444.50 -800 148.10 -800 74.10 -800 --200

Incremental Analysis (A – B): EOY 0 1 2 3 4 5

ATCF -$1,000.00 266.71 222.23 340.79 370.39 520.03

IRR

18.4%

The incremental investment is justified, select Pump A.

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7-41

M/C A: EOY

(A) (B) (C) = (A) – (B) TI D = – 0.4 (C) ATCF BTCF Depr 0 –$425,000 –$25,000 1–10 $20,000 $2,300 $17,700 –$7,080 $12,920 10 $2,000 0 0 $2,000 (MV = BV)

AW = –$25,000 (A/P, 8%, 10) + $12,920 + $2,000 (A/F, 8%, 10) = $9156.6 per year M/C B: EOY 0 16

(A) (B) (C) = A – B D = -0.4 (C) (E) = (A) + (D) BTCF Depr TI (IT) ATCF $40,000 $40,000 $20,000 $6,666.67 $13,333.33 $5,333.33 $14,666.67

AW = –$40,000 (A/P, 8%, 6) + $14,666.67 = $30,681.33 per year Machine B is the better option.

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7-42

Assume repeatability and compare AW over useful life.

Design A: EOY 0 1 2 3 4 5 6 7a 7b

BTCF d -$1,000,000 200,000 $200,000 200,000 320,000 200,000 192,000 200,000 115,200 200,000 115,200 200,000 57,600 200,000 1,000,000

TI

T(20%)

$0 -120,000 8,000 84,800 84,800 142,400 200,000 1,000,000

$0 24,000 -1,600 -16,960 -16,960 -28,480 -40,000 -200,000

ATCF -$1,000,000 200,000 224,000 198,400 183,040 183,040 171,520 160,000 800,000

PW(10%) AW(10%)

$ $

T(20%) $0 48,000 -3,200 -33,920 -33,920 -56,960 -220,000

ATCF -$2,000,000 400,000 448,000 396,800 366,080 366,080 343,040 880,000

PW(10%) AW(10%)

$ $

344,126 70,685

Design B: EOY 0 1 2 3 4 5 6 6

BTCF d TI -$2,000,000 400,000 $400,000 $0 400,000 640,000 -240,000 400,000 384,000 16,000 400,000 230,400 169,600 400,000 230,400 169,600 400,000 115,200 284,800 1,100,000 1,100,000

199,725 45,858

Select Design A to maximize after-tax AW.

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7-43

(a)

Method I

EOY 0 1−5 5

(A) (B) (C) = (A) − (B) (D) = −t (C) (E) = (A) + (D) BTCF Depr TI T (40%) ATCF −$10,000 ------−$10,000 $14,150 $1,800 −$15,950 $6,380 −$7,770 $1,000 --0 0 $1,000

PW0(15%) = −$10,000 − $7,770(P/A,15%,5) +$1,000(P/F,15%,5) = −$35,549.39 To have a basis for computation, assume that Method I is duplicated for years 6−10. Transform the additional PW5(15%) = −$35,549.39 to the present and get: PW(15%) = PW0(15%) + (P/F,15%,5)PW5(15%) = −$53,224.55 Method II EOY 0 1−10 10

(A) (B) (C) = (A) − (B) (D) = −t (C) (E) = (A) + (D) BTCF Depr TI T (40%) ATCF −$40,000 ------−$40,000 −$7,000 $3,500 −$10,500 $4,200 −$2,800 $5,000 --0 0 $5,000

PW(15%) = −$40,000 − $2,800(P/A,15%,10) +$5,000(P/F,15%,10) = −$52,816.64 Thus, Method II is the better alternative. (b)

Method I: Assume that MV5 is $1,000. The MACRS property class is 5 years. This means that the tax−life is 6 years which is greater than the useful life of 5 years.

EOY 0 1 2 3 4 5 5 6

(A) (B) (C) = (A) − (B) (D) = −t (C) (E) = (A) + (D) BTCF Depr TI T (40%) ATCF −$10,000 ------−$10,000 −$14,150 $2,000 −$16,150 $6,460 −$7,690 −$14,150 $3,200 −$17,350 $6,940 −$7,210 −$14,150 $1,920 −$16,070 $6,428 −$7,722 −$14,150 $1,152 −$15,302 $6,120.80 −$8,029.20 −$14,150 $1,152 −$15,302 $6,120.80 −$8,029.20 $1,000 --$1,000 −$400 $600 0 $576 −$576 $230.40 $230.40

PW (15%) = −$35,400.73 PW(15%) over 10 years = −$35,400.73[1+(P/F,15%,5)] = −$44,155.33

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7-43

continued To get a figure for comparison, convert PW(12%) to AW over the useful life of 5 years: AW(15%) = −$35,400.73(A/P,15%,5) = −$10,560.04 With straight line depreciation, AWSL(15%) = −$35,549.39(A/P,15%,5) = $10,604.38. We note that the annual worths are basically the same. This is due to the fact that, whatever depreciation method we use, the depreciation deductions are small relative to the annual expenses.

Method II: EOY 0 1 2 3 4 5 6 7 8 9 10 10

The MACRS class life is 7 years. Assume that MV5 is $5,000

(A) BTCF −$40,000 − $7,000 − $7,000 − $7,000 − $7,000 − $7,000 $7,000 $7,000 $7,000 $7,000 $7,000 $5,000

(B) (C) = (A) − (B) (D) = −t (C) (E) = (A) + (D) Depr TI T (40%) ATCF ------−$40,000 $5,716 −$12,716 $5,086.40 −$1,913.60 $9,796 −$16,796 $6,718.40 −$281.60 $6,996 −$13,996 $5,598.40 −$1,401.60 $4,996 −$11,996 $4,798.40 −$2,201.60 $3,572 −$10,572 $4,228.80 −$2,771.20 $3,568 −$10,568 $4,227.20 −$2,772.80 $3,572 −$10,572 $4,228.80 −$2,771.20 $1,784 −$8,784 $3,513.60 −$3,486.40 0 −$7,000 $2,800 −$4,200 0 −$7,000 $2,800 −$4,200 --$5,000 −$2,000 $3,000

PW(15%) = −$50,305.84 AW over the useful life of 10 years: AW(15%) −$50,305.84(A/P,15%,10) = −$10,025.95 Thus, Method II is also chosen in this case.

AWSL(15%) = −$52,816.64 (A/P,15%,10) = −$10,526.36 We notice a more significant difference in this case. Here, the timing of the depreciation deductions is of greater importance.

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7-44 EOY 0 1 2 3 4 5a

BTCF $12,000 4,000 4,000 4,000 4,000 4,000

d 3,000 3,000 3,000 -

5b

3,000

-

TI 1,000 1,000 1,000 4,000 4,000 0 (MV = SV)

T (40%) 400 400 1,600 1,600 1,600

ATCF $12,000 3,600 3,600 2,400 2,400 2,400

0

3,000

T (40%) 799.52 2400.8 1867.36 2932.96 4000 1200

ATCF $36000 10799.52 12400.8 8132.64 7067.04 6000 1800

PW (8%) = $12,000 + 3,600 {(P/F, 8%, 1) + (P/F, 8%, 2) + (P/F, 8%, 3)} + 2,400{(P/F, 8%, 4) + (P/F, 8%, 5)} + 3000 (P/F, 8%, 5). = $2,716.44 M/C 2 EOY 0 1 2 3 4 5a 5b

BTCF $36,000 10,000 10,000 10,000 10,000 10,000 3,000

d 11,998.8 16,002 5331.6 2667.6 -

TI 1998.8 6002 4668.4 7332.4 10,000 3,000

PW (8%) = $36,000 + 10,799.52 (P/F, 8%, 1) + 12,400.8 (P/F, 8%, 2) + 8,132.64 (P/F, 8%, 3) + 7,067.04 (P/F, 8%, 4) + 7800 (P/F, 8%, 5) = $1,589.12 M/C 1 should be selected.

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7-45 Manufacturing designed for varying degrees of automation: (A) (A)

EOY 0 1−5

BTCF −$10,000 −$9,500

(B)

(C) = (A) − (B)

Depr --$2,000

TI --−$11,500

(D) = −t (C)

T (40%) --$4,600

(E) = (A) + (D)

ATCF −$10,000 −$4,900

Straight Line Depreciation: ($10,000−0)/5 = $2,000 (B) (A)

EOY 0 1−5

BTCF −$14,000 −$8,300

(B)

(C) = (A) − (B)

Depr --$2,800

(D) = −t (C)

TI --−$11,100

T (40%) --$4,440

(C) = (A) − (B)

(D) = −t (C)

(E) = (A) + (D)

ATCF −$14,000 −$3,860

Depreciation: ($14,000−0)/5 = $2,800 (C) (A)

EOY 0 1−5

BTCF −$20,000 −$6,000

(B)

Depr --$4,000

TI --−$10,000

T (40%) --$4,000

(E) = (A) + (D)

ATCF −$20,000 −$2,000

Depreciation: ($20,000−0)/5 = $4,000 (D) (A)

EOY 0 1−5

BTCF −$30,000 −$4,500

(B)

Depr --$6,000

(C) = (A) − (B)

TI --−$10,500

(D) = −t (C)

T (40%) --$4,200

(E) = (A) + (D)

ATCF −$30,000 −$300

Depreciation: ($30,000−0)/5 = $6,000 AWA = −$10,000(A/P,15%,5) − $4,900 = −$7,883 AWB = −$14,000(A/P,15%,5) − $3,860 = −$8,036.20 AWC = −$20,000(A/P,15%,5) − $2,000 = −$7,966 AWD= −$30,000(A/P,15%,5) − $300 = −$9,249 Select to automate to Degree A.

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7-46 Use a study period of 3 years Quotation I EOY BTCF Capital

BTCF Operating

Depr.

Fact.

Depr.

Book Value

Gain (Loss) On Disp.

0 $(180,000) $180,000 1 --$(28,000) 0.2000 $(36,000) $144,000 2 --$(28,000) 0.3200 $(57,600) $ 86,400 3 $50,000 $(28,000) 0.0960 $(17,280) $ 69,120 $ (19,120) PW of ATCF, Quotation I: $(143,174) IT = Income Taxes Quotation II EOY BTCF Capital

BTCF Operating

Depr.

Fact.

Depr.

Book Value

Gain (Loss) On Disp.

0 $(200,000) $200,000 1 --$(17,000) 0.2000 $(40,000) $160,000 2 --$(17,000) 0.3200 $(64,000) $ 96,000 3 $60,000 $(17,000) 0.0960 $(19,200) $ 76,800 $ (16,800) PW of ATCF, Quotation II: $(136,848) IT = Income Taxes

 in Ord. Inc

 Cash Flow  Cash Flow for IT (Cap) for IT (Oper)

$(64,000) $(85,600) $(45,280) $ 7,648

 in Ord. Inc

$ 25,600 $ 34,240 $ 18,112

 Cash Flow  Cash Flow for IT (Cap) for IT (Oper)

$(57,000) $(81,000) $(36,200) $ 6,720

$ 22,800 $ 32,400 $ 14,480

ATCF $(180,000) $ (2,400) $ 6,240 $ 47,760

ATCF $(200,000) $ 5,800 $ 15,400 $ 64,200

Accept Quotation II.

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7-47

Depreciation schedule (MACRS 5-year property class). The cost basis (B) is assumed to be $345,000.

Year 1 2 3 4 5 6

BVk−1 $345,000 276,000 165,600 99,360 59,616 19,872

rk 0.2000 0.3200 0.1920 0.1152 0.1152 0.0576

dk $ 69,000 110,400 66,240 39,744 39,744 19,872

BVk $276,000 165,600 99,360 59,616 19,872 0

(a) Economic Value Added (EVA): −t = −0.5 T(50%) − $21,500 − 800 − 22,880 − 36,128 − 36,128 − 46,064 − 60,000

EOY BTCF Depr TI NOPAT 1 $112,000a $ 69,000 $ 43,000 $21,500 2 110,400 1,600 800 3 66,240 45,760 22,880 4 39,744 72,256 36,128 5 39,744 72,256 36,128 6 112,000 19,872 92,128 46,064 6 120,000b 120,000c 60,000 a BTCFk = $120,000 − $8,000 = $112,000 b MV6 = $120,000 c Gain on disposal = MV6 − BV6 = $120,000 − 0 = $120,000 EOY 1 2 3 4 5 6 6

EVA $ 21,500 − 0.10 ($345,000) = − $13,000 800 − 0.10 ( 276,000) = − 26,800 22,880 − 0.10 ( 165,600) = 6,320 36,128 − 0.10 ( 99,360) = 26,192 36,128 − 0.10 ( 59,616) = 30,166 46,064 − 0.10 ( 19,872) = 44,077 60,000

PW(10%) − $11,818 − 22,148 4,748 17,888 18,730 24,880 33,870 Total: $66,150

The present equivalent of the EVA = $66,150.

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7-47

(b)

After−tax cash flow (ATCF): EOY 0 1 2 3 4 5 6 6

BTCF −$345,000 112,000

112,000 120,000

T(50%) 0 − $21,500 − 800 − 22,880 − 36,128 − 36,128 − 46,064 − 60,000

ATCF −$345,000 90,500 111,200 89,120 75,872 75,872 65,936 60,000

PW(10%) −$345,000 82,274 91,896 66,957 51,822 47,109 37,222 33,870 Total: $66,150

Yes; the PW(10%) of the ATCF ($66,150) is the same as the present equivalent of the annual EVA amounts.

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7-48

Beginning-of-year book values: Year 1 2 3 4 5 6

BVk−1 $ 180,000 144,000 86,400 51,840 31,104 10,368

dk $ 36,000 57,600 34,560 20,736 20,736 10,368

BVk $ 144,000 86,400 51,840 31,104 10,368 0

EOY NOPATa (0.10) BVk−1 EVAb 1 0 $ 18,000 − $ 18,000 2 − $ 13,392 14,400 − 27,720 3 893 8,640 − 7,747 4 9,464 5,184 4,280 5 9,464 3,110 6,354 6 15,892 1,037 14,855 7 22,320 0 22,320 8 22,320 0 22,320 9 22,320 0 22,320 10 22,320 0 22,320 10 18,600 --18,600 a From Table 7-6: Column (C) algebraically added to Column (D). b Equation 7-22: EVAk = NOPATk − i  BVk−1 PW(10%) = − $18,000 (P/F,10%,1) −  + $14,855 (P/F,10%,6) + $22,320 (P/A,10%,4) (P/F,10%,6) + $18,600 (P/F,10%,10) = $17,208

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7-49

(a) EOY 0 1 2 3 4 4

(b)

BTCF -75,000 25,000 22,500 20,000 17,500 30,000

D --15,000 24,000 14,400 4,320 ---

TI --10,000 -1,500 5,600 13,180 12,720

T ---3,500 525 -1,960 -4,613 -4,452

ATCF -75,000 21,500 23,025 18,040 12,887 25,548

PW (15%) = -75,000 + 21,500 (P/F, 15%, 1) + 23,025 (P/F, 15%, 2) + 18,040 (P/F, 15%, 3) + (12,887 + 25,548)(P/F, 15%, 4) PW (15%) = -$5,056 Do not purchase this machine since the PWAT < 0 it is not profitable.

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7-50

Leasing Option: PW(10%) = −$166,036; AW(10%) = −$43,800 EOY

BTCF (A)

Depr

Interest

Principal Payment (B)

0 0 1 2 3 4 5

−$75,000 −$55,000 −$55,000 −$55,000 −$55,000 −$55,000 +$75,000

−− 0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

TI −− −$55,000 −$55,000 −$55,000 −$55,000 −$55,000

Taxes Payable (40%) (C) −− $22,000 $22,000 $22,000 $22,000 $22,000

ATCF (A) + (B) + (C) −$75,000 −$33,000 −$33,000 −$33,000 −$33,000 −$33,000 $75,000

Purchasing Option: PW(10%) = −$191,197; AW(10%) = −$50,438 EOY

BTCF (A)

Depr

Interest

0 1 2 3 4 5a 5b

−$350,000 −$20,000 −$20,000 −$20,000 −$20,000 −$20,000 +$150,000

−− $116,655 $155,575 $51,835 $25,935 0 −−

−$28,000 −$19,376 −$10,062 0 0 −−

Principal Payment (B)

TI

$350,000 −$107,800 −$116,424 −$125,738 0 0 −−

−− −$164,655 −$194,951 −$81,897 −$45,935 −$20,000 +$150,000

Taxes Payable (40%) (C) −− $65,862 $77,980 $32,759 $18,374 $8,000 −$60,000

ATCF (A) + (B) + (C) $0 −$89,938 −$77,820 −$123,041 −$1,626 −$12,000 +$90,000

Payment = −$350,000 (A/P,8%,3) = −$135,800 Select the lease.

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7-51

In 30 years your tax-exempt traditional IRA will be worth $1,000 (F/A, 10%, 30) = $164,494. After paying the 24% tax, this becomes $120,015. The Roth IRA is better, and it offers additional flexibility in terms of penalty-free early withdrawals. Such flexibility leads most investors to choose the Roth IRA. We’re assuming the income tax rates remain unchanged for the time periods in this problem.

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7-52

F = $5,000(1 – 0.28)(F/P, 8%, 30) = $36,226

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7-53

PW(i) = 0 = −$9,000 + ($10,000)(0.10/2)(1−0.28)(P/A,i%,15) + [$10,000 – ($10,000 − $9,000)(0.28)](P/F,i%,15) = −$9,000 + $360(P/A,i%,15) + $9,720(P/F,i%,15) i% 4% i% 5%

PW(i) $400.14 0 −$587.99

i% = 4.4%/6 months, r = 8.8%/year, ieff. = 8.99%/year

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7-54

529 Plan: Roth Plan:

F = $10,000(F/A, 8%, 10) = $144,866 F = $10,000(1 – 0.28)(F/A, 8%, 10) = $104,304 (not quite enough)

Thus, the 529 plan accumulates 38.9% more in future worth compared to the Roth IRA. It’s a much better deal if the parents can meet the withdrawal restrictions placed on the 529 fund.

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7-55

(a)

Roth IRA: F = $1,440(F/A, 8%, 30) = $163,128 Tax-deductible IRA: F = $2,000(F/A, 8%, 30)(1 – 0.28) = $163,128 Both plans are equivalent when the income tax rate is constant at 28% (a big assumption).

(b)

Roth IRA: F = 163, 128 Tax-deductible IRA: F = $2,000(F/A, 8%, 30)(1 – 0.30) = $158,596 For this assumption, the Roth IRA is better. In reality, the income tax rate will vary year-by-year, so it’s virtually impossible to get an “actual” comparison between the two plans. The ROTH IRA has more flexibility regarding how it can be cashed out over multiple years after the retiree reaches age 70.5.

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7-56

Alt 1 EOY 0 1 2 3 4 5 5

BTCF -11,000 3,000 3,000 3,000 3,000 3,000 2,000

DEP --3,000 3,000 3,000 ----2,000*

*Book value = 2000

T1 --0 0 0 3,000 3,000 0

I+ ---------1,200 -1,200 0

ATCF -11,000 3,000 3,000 3,000 1,800 1,800 2,000

I+ --799.56 2,267.4 -1,634.52 -2,621.88 -3,600 -800

ATCF -33,000 9,799.56 11,267.4 7,365.5 6,378.1 5,400 1,200

PW (1) = -$494.35

Alt 2 EOY 0 1 2 3 4 5 5 *Book value = 0

BTCF -33,000 9,000 9,000 9,000 9,000 9,000 2,000

DEP --10,998.9 14,668.5 4,913.7 2,445.3 0 0*

T1 ---1,998.9 -5,668.5 4,086.3 6,554.7 9,000 2,000 PW (2) = -$2,227

Select Alt 1.

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7-57 MARR Cost Basis Useful Life Market Value DB Rate MACRS Recovery Period EOY 1 2 3 4 5 6 7 8 9 10 PW(10%)

10% $500,000 10 $ 20,000 200% 7

SL Method DB Method $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $294,939

$100,000 $ 80,000 $ 64,000 $ 51,200 $ 40,960 $ 32,768 $ 26,214 $ 20,972 $ 16,777 $ 13,422 $319,534

MACRS Method $ 71,429 $122,449 $ 87,464 $ 62,474 $ 44,624 $ 44,624 $ 44,624 $ 22,312

$360,721

The MACRS method results in the largest PW of the depreciation deductions.

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7-58

Goal seek was used to find the solution below.

M1 0 1 2 3 4 5a 5b

d

TI

-$60,000 -$16,000 -$16,000 -$16,000 -$16,000 -$16,000 $12,114

$12,000 $19,200 $11,520 $6,912 $3,456

-$28,000 -$35,200 -$27,520 -$22,912 -$19,456 $5,202

PW(12%) AW(12%)

-$72,683 -$20,163

T $11,200 $14,080 $11,008 $9,165 $7,782 -$2,081 MV (for each M1)

ATCF -$60,000 -$4,800 -$1,920 -$4,992 -$6,835 -$8,218 $10,033

ATCF -$60,000 -$4,800 -$1,920 -$4,992 -$6,835 $1,816

$3,029

M2 0 1 2 3 4 5 6 7 8

-$66,000 -$18,000 -$18,000 -$18,000 -$18,000 -$18,000 -$18,000 -$18,000 -$18,000 PW(12%) AW(12%)

$13,200 $21,120 $12,672 $7,603 $7,603 $3,802

-$31,200 -$39,120 -$30,672 -$25,603 -$25,603 -$21,802 -$18,000 -$18,000

$12,480 $15,648 $12,269 $10,241 $10,241 $8,721 $7,200 $7,200

-$66,000 -$5,520 -$2,352 -$5,731 -$7,759 -$7,759 -$9,279 -$10,800 -$10,800

-$100,165 -$20,163

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7-59 Table Entries are Before‐tax Rate of Returns on taxable bonds.

Federal Income Tax Rate After‐Tax Rate of Return on Municipal Bonds

4% 5% 6%

15% 4.71% 5.88% 7.06%

28% 5.56% 6.94% 8.33%

35% 6.15% 7.69% 9.23%

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7-60

Natural Gas‐Fired Plant Investment Capacity Factor Max Capacity Efficiency Annual o&M Cost of gas CO2 tax CO2 emitted Conversion

$1.12 billion 80% 800 MW 40% $0.01 / kWhr $8.00 per million Btu $15 / MT CO2 55 MT CO2 / billion Btu 1 kWhr = 3413 Btu

$

$ $

1,120,000,000 80% 800000 40% $0.01 0.000008 15.00 0.000000055 3413

Units $ kW $/kWhr $/Btu $ / MT CO2 MT CO2/Btu Btu/kWhr

Annual Output Hours per Year Annual Output Annual O&M Annual Cost of Gas Annual CO2 Tax

640000 8760 $ $ $

kW Hr 5606400000 kWhr 56,064,000 382,692,864 39,465,202

Total Annual Cost

$

478,222,066

Annual CO2 emitted

2631013.44 1195.9152 MT

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7-60

continued Coal‐Fired Plant Investment Capacity Factor Max Capacity Efficiency Annual o&M Cost of coal CO2 tax CO2 emitted Conversion

$1.12 billion 80% 800 MW 35% $0.02 / kWhr $3.50 / million Btu $15 / MT CO2 90 MT CO2 / billion Btu 1 kWhr = 3413 Btu

$

$ $

1,120,000,000 80% 800000 35% $0.02 0.0000035 15.00 0.00000009 3413

Annual Output Hours per Year Annual Output Annual O&M Annual Cost of Gas Annual CO2 Tax

640000 8760 5,606,400,000.00 $ 112,128,000 $ 191,346,432 $ 73,805,052

Total Annual Cost

$

Annual CO2 emitted

Units $ kW $/kWhr $/Btu $ / MT CO2 MT CO2/Btu Btu/kWhr kW hr kWhr

377,279,484 4920336.823 2236.516738 MT

After‐tax Analysis: BTCF Natural Gas: $(478,222,066) Coal: $(377,279,484)

D

TI

T

ATCF

ATCF + CR

$37,333,333

$(515,555,399)

$206,222,160

$(271,999,906)

$(390,808,664)

$37,333,333

$(414,612,818)

$165,845,127

$(211,434,357)

$(330,243,115)

AT cost of electricity Natural Gas: $ Coal: $

(0.07) (0.06)

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7-61

d2 = $150,000(0.4445) = $66,675 Select (b)

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7-62

BV2 = $150,000(1 – 0.3333 – 0.4445) = $33,330 Select (d)

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7-63

d3 = $150,000(0.1481)(0.5) = $11,108 Select (d)

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7-64

dk = ($550,000 − $25,000)/10 = $52,500 Select (c)

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7-65

BV10 = $550,000 – (10)($52,500) = $25,000 Select (b)

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7-66

dk = ($550,000 − $25,000)/10 = $52,500 BV10 = $550,000 – (10)($52,500) = $25,000 MV10 – BV10 = $35,000 − $25,000 = $10,000 Select (d)

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7-67

From Table 7-2, the GDS recovery period is 10 years (asset class 13.3).

Select (d)

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7-68

From Table 7-2, the GDS recovery period for woord products equipment (Asset Class 24.4) is 7 years.

Select (b)

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7-69

Cost Basis = $100,000 + $10,000 = $110,000 dk =

$110,000  $5,000 = $10,500 10

Select (a)

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7-70

BV10 = $110,000 – (10)($10,500) = $5,000 Select (c)

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7-71

d6 = $110,000 (0.0892) = $9,812 Select (a)

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7-72

Using the half year convention: d5* = $110,000 [0.1429 +0.2449 +0.1749 + 0.0893(0.5)] = $80,547.50 BV5 = $110,000 − $80,547.50 = $29.452.50 Select (a)

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7-73

BV4 = $16,000(1 – 2/8)4 = $5,062.50 Select (d)

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7-74

dk =

$12,000  $2,000 = $1,250 8

Select (d)

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7-75

dA*= $12,000 (0.1429 + 0.2449 + 0.1749 + 0.1249) = $8,251.20 BV4 = $12,000 − $8,251.20 = $3,748.80 Select (a)

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7-76

t = 0.05 + 0.35(0.95) = 38.25% Select (c)

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7-77

The book value at EOY 3 is $70,000 - $70,000 x [ 0.200 + 0.320 + 0.5 (0.192)] since the asset is disposed of at EOY 3. The book value is therefore $70,000 - $70,000 x ( 0.616) = $26,880. The gain on disposal will be $32,000 - $26,880 = $5,120.

Select answer (b)

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7-78

After-tax MARR = (1 − 0.4)(18%) = 10.8% Select (c)

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7-79

ATCFk = ($110,000 − $65,000) – 0.40($110,000 − $65,000 − $25,000) = $37,000 Select (e)

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7-80

BTCF5 = (R−E) + MV = [$40,000 + $30,000] + $40,000 = $110,000 Select (e)

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7-81

TI3 = $70,000 − $195,000(0.1481) = $41,120 Select (d)

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7-82

After Tax MARR = (1−0.40)(20%) = 12% PW(12%) = $70,000(1−0.40)(P/A,12%,5) = $42,000(3.6048) = $151,402 Select (c)

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7-83

0.1481 ) = $28,889 2 Deprecitaion Recapture = $50,000 − $28,889 = $21,111 BV3 = $195,000 − $195,000(0.3333 + 0.4445 +

Taxes = 0.40($21,111) = $8,444 Select (a)

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7-84

Select the alternative with the highest PW. Select (b).

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7-85

Assume repeatability. After applying the procedure of Figure 7-5, we can compare after-tax annual worth as follows. 𝐴𝑊

= −$20,000 (A/P, 10%, 4) + $5,000 = −$1,310/ yr.

𝐴𝑊

=−$36,000 (A/P, 10%, 6) + $8,000 = −$266/ yr.

Select (c) - the 5 cm. insulation.

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Solutions to Chapter 8 Problems A Note To Instructors: Because of volatile energy prices in today's world, the instructor is encouraged to vary energy prices in affected problems (e.g. the price of a gallon of gasoline) plus and minus 50 percent and ask students to determine whether this range of prices changes the recommendation in the problem. This should make for stimulating inclass discussion of the results.

8-1.

The basic question is this: “Is the extra $800 in investment cost justified by the annual energy savings of $100 per year for 10 years?” We assume that the real cost of energy does not inflate during the 10-year life (study period) of the system. Also assume that the residual value of the system is negligible after 10 years. The IRR of the difference between the two systems is: 0 = -$800 + $100 (P/A, i’%, 10) By trial and error (or Excel), we see that i’% = 4.3% > 2% so the more expensive system is cost justified.

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8-2

F15 = $1,000,000 Purchasing equivalent in today’s spending power: = $1,000,000 (P/F, 6%, 5)(P/F, 9%, 7) = $1,000,000 (0.7473)(0.5470) = $408,773

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8-3

The average rate of inflation is 10% per year, and the market place interest rate is 4% + 10% + (4%)(10%) = 14.4% per year.   P  1   F ,14.4%,23 ( F / P,10%,23)    = $13,508.41 PW2 = $1,000   0.144  0.10      

PW0 = $13,508.41 (P/F, 14.4%, 2) = $10,322.

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8-4

The future worth of their 30-year investment plan is: F = $32,000 (F/A, 7%, 30) = $32,000 (94.461) = $3,022,752 The purchasing power in today’s dollars of their savings plan is: P = $3,022,752 (P/F, 2%, 30) = $3,022,752 (0.5521) = $1,668,861 Yes, they will be able to meet their savings expectation.

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8-5

This problem can be solved with the following table (PW = Present Worth).

Year Annual Tuition (A$) PW Factor (P/F, i%, N) PW of Tuition Freshman (N=1) $12,000 (P/F, 4%, 1) $11,538 Sophomore (N=2) $12,000 (F/P,10%,1) = $13,200 (P/F, 4%,2) $12,205 Junior (N =3) $13,200 (F/P,12%,1) = $14,784 (P/F, 6%, 1)(P/F,4%,2) $12,896 Senior (N=4) $14,784 (F/P, 8%, 1) = $15,967 (P/F, 6%, 2)(P/F,4%,2) $13,139 Total PW = $49,778 Your parents would need to deposit $49,778 at the start of your senior year in high school. Notice this is for tuition only!

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8-6

f = 4% per year; ir = 8% per year; b = 0 Alternative A: Estimates are in actual dollars, so the combined (market) interest rate must be used to compute the present worth (PW). im = ir + f + (ir)(f) = 0.08 + 0.04 + (0.08)(0.04) = 0.1232 or 12.32% per year PW(12.32%) = − $120,000(P/F,12.32%,1) − $132,000(P/F,12.32%,2) − $148,000(P/F,12.32%,3) − $160,000(P/F,12.32%,4) = − $120,000(0.8903) − $132,000(0.7927) − $148,000(0.7057) − $160,000(0.6283) = − $416,444 Alternative B: Estimates are in real dollars, so the real interest rate must be used to compute the present worth (PW).

PW(9%) = − $100,000(P/A,8%,4) − $10,000(P/G,8%,4) = − $100,000(3.3121) − $10,000(4.650) = − $377,710 Alternative B has the least negative equivalent worth in the base time period (a PW value in this case since b = 0).

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8-7

(a)

$6.58 = $50(F/P, i′, 50), or i′ = −3.97% (annual average loss in purchasing power)

(b)

$1,952 = $50(F/P, i*, 50), or i* = 7.6% (which is the market interest rate). The real rate earned on the investment in stocks is (7.6% − 3.97%)/1.0397 = 3.5%.

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8-8

Use Equation (8-4) 0.12 = 0.02 + ir + 0.02 ir 0.10 = ir (1 + 0.02) ir = 0.098 or 9.8%

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8-9

The engineer's salary has increased by 6.47%, 7.18%, and 6.96% in years 2, 3, and 4, respectively.

These are annual rates of change. By using Equation 8-1, but with each year's general price inflation taken into account separately, the R$ equivalents in year 0 dollars are calculated as follows.

EOY 1 2 3 4

Salary (R$ in Year 0) $34,000 (P/F,2.4%,1) $36,200 (P/F,2.4%,1)(P/F,1.9%,1) $38,800 (P/F,2.4%,1)(P/F,1.9%,1)(P/F,3.3%,1) $41,500 (P/F,2.4%,1)(P/F,1.9%,1)(P/F,3.3%,1)(P/F,3.4%,1)

= $ 33,203 = 34,692 = 35,996 = 37,235

Note: Time reference is the beginning of the first year. For ease of computation, it is assumed that the salary is received at the end of the year.

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8-10

Hobart index-adjusted salary = (95/132) × $80,000 = $57,576 per year Hobart actual salary = (1.00 − 0.10) × $80,000 = $72,000 per year “Savings” = $72,000 − $57,576 = $14,424 per year FW (10%) = $14,424 (F/A, 8%, 6) = $105,813. Shane is not penalized at all!

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8-11

R$10 = $480M (1.75) = $840M A$10 = $1,000M = $840M (1 + f)10 (1 + f)10 = 1.19 f = 10 1.19 − 1 = 0.0176 or 1.76%

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8-12

After-tax nominal return per year = 5% (1 – 0.3) = 3.5% Approximate real return per year = 3.5% − 3% = 0.5% each year.

F (in today’s purchasing power) = $200,000 (F/P, 0.5%, 8) = $200,000 (1.0407) = $208,140

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8-13

F = $100,000(F/P, 10%, 10) = $259,370 Taxable Earnings = $159,370 After-tax F = $159,370(1 – 0.33) + $100,000 = $206,778 F (in today’s purchasing power) = $206,778 (P/F, 3%, 10) = $153,864 Your younger brother is ahead by about $43,400 with his concern over performance (total return) rather than risk avoidance (safety).

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8-14

Unit cost (8 years ago) = $89 / ft2 X = 0.92 eC = 5.4% per year eAE = 5.66% per year (a)

SB = 80,000 ft2 SA = 125,000 ft2 im = MARRc = 12% per year f = 7.69% per year

Using the power sizing technique (exponential cost estimating model) from Section 3.4.1, with an adjustment for the price increase in construction costs, we have:

S  CA = CB  A   SB 

X

(1 + eC)8

 125,000 = ($89/ft2)(80,000 ft2)    80,000  = $16,350,060

0.92

(1.054)8

Total Capital Investment = $16,350,060 (1 + 0.05 + 0.042 + 0.08 + 0.31) = $24,230,790 (b)

Note: The building is not being sold at the end of the 10 years. Therefore, working capital is not considered to be recovered at that time.

PW(12%) = −$24,230,790 – ($5)(125,000 ft2) = −$24,230,790 –

1  ( P / F ,12%,10)( F / P,5.66%,10) 0.12  0.0566

625,000(0.4416) 0.0634

= −$28,584,102 (c)

ir 

0.12 - 0.0769  0.04 or 4% per year 1.0769

Assuming the base year to be the present (b = 0), we have: AW(4%) = −$28,584,102 (A/P,4%,10) = −$3,524,420

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8-15

($100/bbl.)[(1.02)(1.055)]8 = $179.81/bbl.

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8-16

We can equate future worth's in 25 years. The FW of her savings plan will be P(1.05)25 and the FW of the $800,000 if it were to inflate at 6% per year is $800,000(1.06)25. P is what we are trying to determine.

So, we can equate these two amounts as follows: P(1.05)25 = $800,000(1.06)25 or P = $800,000(1.06 / 1.05)25.

This equals $800,000(1.009524)25 = $1,013,921. She must now put away more than $800,000, because inflation each year is greater than the interest rate on her savings account. This demonstrates the injurious effect of high inflation in an economy.

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8-17

The value of ir is desired to match the R$ of $64,000 per year. ir = (0.075 - 0.02) / (1 + 0.02) = 0.0539 (5.39%).

To determine the amount of life insurance to guarantee $64,000 (R-dollars) per year, we need to find the capitalized worth (CW) of this annuity: CW = $64,000 / 0.0539 = $1,187,384.

So, Alfred needs $1,187,384 worth of life insurance.

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8-18

(a)

Lump sum interest in 2014 = ($2.6 billion/5)(F/A, 12%, 5) – 2.6 billion = $703,456,000

(b)

 200,000  C2009 = ($2.6 billion)    160,000 

(c)

C2019 = ($3.21 billion)(F/P, 9%, 10) = $7.60 billion

0.95

= $3.21 billion

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8-19

(a)

R$28 = $690(P/F, 3.2%, 28) = $285.64

(b)

$850 = $690(1.032)N;

(c)

$285.56 = $850(1 + ir)28; ir = −3.82% This was not a good investment to have made in January of 1980.

N = 6.62 years

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8-20

(a)

Cost in year 2024 = $15,000 (F/P,8%,15) = $57,100 Cost in year 2025 = $61,668 Cost in year 2026 = $66,601 Cost in year 2027 = $71,929 Total (un−discounted dollars) = $257,298

(b)

Ieff = (1.005)12 – 1 = 0.0617 or 6.17% P2024 = $57,100 + $61,668[1 – (P/F,6.17%,3)(F/P,8%,3)]/(0.0617 – 0.08) = $234,373 A = $234,373 (A/P,0.5%,156 months) = $234,373 (0.00925) = $2,168 per month

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8-21

(a)

R$ = $7.20(1.021)20 = $10.91 per thousand cubic feet

(b)

A$ = $7.20(1.035)20 = $14.33 per thousand cubic feet

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8-22

(a)

From Equation (8-1), we see that real dollars as of year k = 0 (today) is $1,107,706(P/F, 3%, 60) = $187,978 which is still a tidy sum of purchasing power.

(b)

When f = 2% per year, we have R$0 = $1,107,706(P/F, 2%, 60) = $337,629. The impact of inflation is clear when you compare the results of Parts (a) and (b).

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8-23

(a)

 2,400 ft 2  (1.085)10 = $3,700 Cost in 10 years = ($3.75/lb)(400 lb) 2  2,200 ft  

(b)

Left to student.

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8-24

Escalation in savings is 0.5% per year (f) and the interest rate is 6% per year. By using Equation 4-30, we can calculate the present value at time zero (now): P

= ($5,000 [1- (P/F, 6%, 35)(F/P, 0.5%, 35)]) / (0.06 – 0.005) = ($5,000 [1- (0.1301)(1.1907)]) / (0.055) = $76,827

Then F (future equivalent) is $76,827 (F/P, 6%, 35) = $590,502 Students often ask about the following situation. IF the rate of inflation in the economy averages 3% per year, the spending power of F in today’s dollars (year 0) is how much?

P’ = $590,502 (P/F, 3%, 35) = $590,502 (0.3554) =$209,864

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8-25

Based on Equation (4-28), we have PW = $2.5 billion

[1  (0.0668)(3.2620)] = $48.88 billion and AW = $3.67 billion.

0.07  0.03

With inflation considered in this problem, the taxpayers can afford to increase the subsidy for the F-T technology.

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8-26

(a)

The tuition for 2017 is calculated as follows: F2017 = $5,000 (1.13)(1.046)(1.06)2(1.09)2(1.12)(1.08)(1.06)(1.056) = $10,682

(b)

The annual compounded rate of increase during this 10 year period of time is $10,682 = $5,000 (F/P, i’, 10) Or i% = 7.89% per year

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8-27

ir = 10.05% per year; f = 4.5% per year; f = ej = 5.7% per year im = 0.1005 + 0.045 + (0.1005)(0.045) = 0.15 or 15% per year ej’ = (0.15  0.057) / 1.057 = 0.0880 or 8.8% per year MV7 = (0.10)($950,000) = $95,000 PW(15%) = $950,000  $92,600(P/A, 8.8%, 7) + $95,000(P/F, 15%, 7) = $1,383,484 Annual Revenue = $1,383,484 (A/P, 15%, 7) = $332,590

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8-28

MVA = 25,000 (1.055)6 = $34,471.07 MVB = 40,000 (1.055)8 = $61,387.46 PWA

= -150,000 + 34,000 (P/A, 15, 6) – 500 (P/G,15,6) + 34,471.07 (P/F, 15, 6) = -150,000 + 34,000 (3.7845) – 500 (7.937) + 34,471.07 (0.423) = -10,393.66

AWA = PWA (A/P, 15, 6) = PWA (0.2642) = -$2,746.00 PWB

= -240,000 + 44,000 (P/A, 15, 8) + 61,387.46 (P/F, 15, 8) = -240,000 + 44,000 (4.4873) + 61,387.46 (0.3269) = -22,491.24

AWB = PWB (A/P, 15, 8) = PWB (0.2229) = -$5,013.30 Select A

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8-29

$20(F/P, 22.5%, N)  5. One approach would be to find the minimum value of N by trial and error. $10(F/P, 13.1%, N) At N = 12 years, the cost of an RA dose is $228.38 and the cost of a diabetes inhaler use is $43.81. The ratio of RA to diabetes is 5.21, so N = 12 years wil “git ‘er done.”

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8-30

Option 1: Software with 3 year upgrade agreement.

Year

(A) BTCF (A$)

0 1 2 3

−$X 0 0 0

(B) Taxable Cash Flow Depreci− Income: for ation C=A+B Income Taxes D = −t(C) −−− −−− −−− $X/3 −$X/3 $0.1133X $X/3 −$X/3 $0.1133X $X/3 −$X/3 $0.1133X

ATCF (A$) A+D −$X $0.1133X $0.1133X $0.1133X

PW1(20%) = −$X + $0.1133X(P/A,20%,3) Year

(A) BTCF (A$)

(B) Depreci− ation

1 2 3

−$20,000 − 22,000 − 24,200

−−− −−− −−−

Taxable Income: C=A+B

Cash Flow for Income Taxes D = −t(C) − $20,000 $6,800 − 22,000 7,480 − 24,200 8,228

ATCF (A$) A+D −$13,200 − 14,520 − 15,972

PW2(20%) = −$13,200(P/F,20%,1) − $14,520(P/F,20%,2) − $15,972(P/F,20%,3) = −$30,326 Set PW1 = PW2 and solve for X. −$X + $0.1133X(P/A,20%,3) = −$30,326 −$0.761X = −$30,326 X = $39,836 Therefore, $39,836 could be spent for software with a 3 year upgrade agreement (i.e., Option 1).

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8-31

The annual capital recovery cost of the autoclave is: CR (8%) = $180,000 (A/P, 8%, 8) - $18,000 (A/F, 8%, 8) = $31,320 - $1,692 = $29,628 The annual equivalent revenue, based on Equation 4-30, can be determined as follows in two steps: PW of revenue = $60,000 [1- (P/F, 8%,8)(F/P, 5%,8)] / (0.08 - 0.05) = $60,000 [1- (0.5403)(1.4775)] / 0.03 = $403,414 and the equivalent annual revenue equals $403,400 (A/P, 8%, 8) = $70,194.

The net equivalent annual worth is $70,194 - $29,628 = $40,566.

Finally, the equivalent annual worth per hour of utilization is $40,566 per year / 2,000 hours per year = $20.28 per hour. This is the expected profit per hour of utilization of the autoclave.

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8-32

In 2009 there was parity between the U.S. dollar and the Real. But in 2014 one Real is worth $0.50 U.S., so the investment is now worth $50 million and the bank has suffered a major loss. Conventional wisdom would be to cut the losses instead of chasing bad money with good money (i.e. sell out).

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8-33

(a) In two years: $1 (1.026)2 = 6.4X or, $1 = 6.4X / (1.026)2 = 6.08 units of X. (b)

In three years: $1 = (6.4X) (1.026)3 = 6.91 units of X.

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8-34

(a)

The value of 0.5 pound sterling is 90 cents, so 5 cents can be saved on each item purchased in U.K. pound sterling.

(b)

120,000 items × $0.05 = $6,000 can be saved by purchasing in the U.K.

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8-35

iUS = 26% per year (a)

fe = 8% per year ifm = 0.26 + 0.08 + (0.26)(0.08) = 0.3608, or 36.08% per year (IRR on project in Country A currency)

(b)

fe = −6% per year ifm = 0.26 + (− 0.06) + (0.26)(− 0.06) = 0.1844, or 18.44% per year (IRR on project in Country B currency)

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8-36

Left to student.

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8-37

(a) Exchange NCF Rate EOY (T−marks) (T−marks/$) 0 −$3,600,000 20.000 1 450,000 22.400 2 1,500,000 25.088 3 1,500,000 28.099 4 1,500,000 31.470 5 1,500,000 35.247 6 1,500,000 39.476 7 1,500,000 44.214

NCF ($) −$180,000 20,089 59,790 53,383 47,664 42,557 37,998 33,926 PW (18%) =

PW(18%) − $180,000 17,025 42,941 32,489 24,585 18,602 14,074 10,649 − $19,635

Project is not economically acceptable. (b)

IRRfm in terms of T−marks: PW(i'%) = −3,600,000 + 450,000 (P/F, i'%, 1) + 1,500,000 (P/A, i'%, 6) (P/F, i'%, 1) By linear interpolation, i'% = IRRfm = 0.2798, or 28.0% per year.

(c)

From Equation 8-7, we have: (IRR)US =

IRR fm - f e 0.28 - 0.12 = = 0.1429, or 14.29% < 18% 1.12 1+ fe

Note: This confirms our recommendation in part (a).

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8-38

100 euros × $1.24 = $124. The cost in U.S. dollars is $124 + $40 = $164. Sanjay may think he is paying too much for the jewelry, but he goes ahead with the purchase anyway. He could have converted his 100 euros into U.S. dollars, but there is a 7.5 euro commission on the transaction. Or he could have kept his 100 euros for the next trip he makes to Europe and simply charge the purchase to his credit card.

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8-39

ifm = 20% per year; fe = −2.2% per year Current exchange rate = $1 per 92 Z−Krons i US 

0.20 - (-0.022) = 22.7% 1 - 0.022

PW(22.7%) = −$168,000,000 − $32,000,000(P/F,22.7%,1) + $69,000,000(P/A,22.7%,9)(P/F,22.7%,1) = −$168,000,000 − $32,000,000(0.8150) + $69,000,000(3.7065)(0.8150) = $14,355,028 > 0 Yes, this project will meet the company's economic decision criteria.

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8-40

(a) (A)

(B)

(C)

Year

BTCF (R$)

Adjustment (1.10)Year

BTCF (A$)

1 2 3 4 5 6

− $4,000 − 4,000 − 4,000 − 4,000 − 4,000 − 4,000 0

1.100 1.210 1.331 1.464 1.611 1.772

(D)

(E)

Lease Taxable Payment Income: (A$) (C) – (D)

−$4,400 − 4,840 − 5,324 − 5,856 − 6,442 − 7,086

$80,000 60,000 50,000 50,000 50,000 50,000

$84,000 − 64,840 − 55,324 − 55,856 − 56,442 − 57,086 0

(F)

(G)

Cash Flow for Income Taxes −t(E)

ATCF (A$) C−D+F

$33,760 25,936 22,130 22,342 22,577 22,834

−$50,640 − 38,904 − 33,194 − 33,514 − 33,865 −34,252 0

(b) im = (1.05)(1.09524) – 1 = 0.15 = 15% per year PW = −$50,640(P/F,15%,1) − $38,904(P/F,15%,2) − $33,194(P/F,15%,3) − $33,514(P/F,15%,4) − $33,865(P/F,15%,5) − $34,252(P/F,15%,6) = −$146,084 EUAC = $146,084(A/P,15%,6) = $38,595

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8-41

Demand = 500 million BTU/year; Efficiency = 80%, N = 12 years f = 10% per year; b = 0 MARR = im = 18% per year 3  500 million Btu   1,000 ft of gas  Annual gas demand =   = 625,000 ft3 of gas     million Btu  0.8

A1 =

$7.50(1.1) $8.25 = 3 1000 ft 1000 ft 3

 $8.25  1  ( P / F,18%,12)(F / P,10%,12)] PW(18%) = − (625,000 ft3)  3  0.18  0.10  1000 ft 

 $8.25  = − (625,000 ft3)  (7.1176) 3   1000 ft 

= −$36,700

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8-42

(a) Cost of compressor replacement at EOY 8 (A$) = $500(1 + 0.06)8 = $797 Annual maintenance expense: A1(A$) = $100 (1.06) = $106 Annual electricity expense: A1(A$) = $680(1.10) = $748

PW(15%) = −$2,500 − $797 (P/F,15%,8) −$106 1  ( P / F ,15%,15)( F / P,6%,15) 0.15  0.06

−$748 1  ( P / F ,15%,15)( F / P,10%,15) 0.15  0.10 = $2,500 − $797 (0.3269) − $106(0.70546) − $748(0.48662) 0.09 0.05 = −$10,871 A$: AW(15%) = −$10.871(A/P,15%,15) = −$1,859

(b) ir = 0.15  0.06 = 0.085or 8.5% per year 1.06 = −$10,871 (0.1204) = −$1,309

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8-43

dk = ($150,000−0)/3 = $50,000 (A)

(B)

Year Revenues Expenses (A$) (A$)

0 1 2 3

$84,000 88,200 92,610

−$150,000 −21,800 −23,762 −25,900

(C)

(D)

(E)

(F)

(G)

BTCF (A$) A+B

Depreci− ation

Taxable Income: C−D

Cash Flow for Income Taxes −t(E)

ATCF (A$) C+F

−$150,000 62,200 64,438 66,709

−−− $50,000 50,000 50,000

−−− $12,200 14,438 16,709

−−− −$6,100 −7,219 −8,355

−$150,000 56,100 57,219 58,354

For discounting purposes, im = 26% would be used since the ATCFs are expressed in actual dollars.

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8-44

Annual revenues in year k (A$) = $360,000(1.025)k Annual expenses in year k (A$) = −$239,000(1.056)k (a)

The values in the following table are expressed in A$.

Annual EOY Revenues 0 1 $369,000 2 378,225 3 387,681 4 397,373 5 407,307 6 417,490 6

Annual Expenses − $252,384 − 266,518 − 281,442 − 297,203 − 313,847 − 331,422

BTCF −$220,000 116,616 111,707 106,239 100,170 93,460 86,068 40,000

Depr TI −−− −−− $44,000 $72,616 70,400 41,307 42,240 63,999 25,344 74,826 25,344 68,116 12,672 73,396 −−− 40,000

ATCF T(39%) (A$) −−− −$220,000 − $28,320 88,296 − 16,110 95,597 − 24,960 81,279 − 29,182 70,988 − 26,565 66,895 − 28,624 57,444 − 15,600 24,400

6

PW(10%) =

 ATCF (P/F,10%,k) = $136,557 k

k=0

Total investment that can be afforded (including new equipment) = $136,557 + $220,000 = $356,557 (b)

ATCFk (R$) = ATCFk (A$)(P/F,4.9%,k) Year, k 0 1 2 3 4 5 6 6

ATCFk (A$)

(P/F,4.9%,k)

ATCFk (R$)

− $220,000 88,296 95,597 81,279 70,988 66,895 57,444 24,400

1.0000 0.9533 0.9088 0.8663 0.8258 0.7873 0.7505 0.7505

− $220,000 84,173 86,879 70,412 58,622 52,666 43,112 18,312

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8-45

Purchase (A$ Analysis): Investment / EOY Market Value 0 − $600,000 1 2 3 4 5 6 6 101,355a EOY 0 1 2 3 4 5 6 6

Notes:

Oper., Ins. & Other Expenses (O,I,OE)

Maintenance Expense

− $27,560b − 29,214 − 30,966 − 32,824 − 34,794 − 36,881

− $34,880c − 38,019 − 41,441 − 45,171 − 49,236 − 53,667

BTCF Deprd TI − $600,000 − 62,440 $120,000 − $182,440 − 67,233 192,000 − 259,233 − 72,407 115,200 − 187,607 − 77,995 69,120 − 147,115 − 84,030 69,120 − 153,150 − 90,549 34,560 − 125,108 101,355 101,355

BTCF

− $600,000 − 62,440 − 67,233 − 72,407 − 77,995 − 84,030 − 90,549 101,355 ATCF T(34%) (A$) − $600,000 $62,030 − 410 88,139 20,906 63,786 − 8,621 50,019 − 27,976 52,071 − 31,959 42,537 − 48,012 − 34,461 66,894

a

(MV)6 = $90,000 (1.02)6 = $101,355 (O,I,OE)k = $26,000 (1.06)k c (Maint)k = $32,000 (1.09)k d Cost Basis = $600,000 b

im = 0.13208 + 0.06 + (0.13208)(0.06) = 0.20, or 20% per year 6

FW6 (A$) =

 ATCF (F/P, 20%, 6 − k) = − $1,823,920 k

k=0

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8-45

continued Lease (A$ Analysis): EOY 1 2 3 4 5 6

Leasing Costs − $300,000 − 200,000 − 200,000 − 200,000 − 200,000 − 200,000

EOY 1 2 3 4 5 6

BTCF − $362,440 − 267,233 − 272,407 − 277,995 − 284,030 − 290,549

Oper., Ins. & Other Expenses − $27,560 − 29,214 − 30,966 − 32,824 − 34,794 − 36,881

Depr 0 0 0 0 0 0

Maint.

Expense − $34,880 − 38,019 − 41,441 − 45,171 − 49,236 − 53,667

TI − $362,440 − 267,233 − 272,407 − 277,995 − 284,030 − 290,549

BTCF − $362,440 − 267,233 − 272,407 − 277,995 − 284,030 − 290,549

T(34%) $123,230 90,859 92,618 94,518 96,570 98,787

ATCF (A$) − $239,210 − 176,374 − 179,789 − 183,477 − 187,460 − 191,762

6

FW6 (A$) =

 ATCF (F/P, 20%, 6 − k) = − $1,952,551 k

k=0

Therefore choose Purchase Alternative due to smaller FW of costs.

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8-46

Assuming that EOY 1 cost for purchased components is $85,000,000 EOY 0 1 2 3 4 5

Cash Flow (w) −20,000,000 −85,000,000 −80,750,000 −76,712,500 −72,876,875 −69,233,031

Cash Flow (w/o) 0 −85,000,000 −85,000,000 −85,000,000 −85,000,000 −85,000,000

PW

−300,467,957

−306,405,977

i(r) = 0.178947368 (P/A,i(r),5) = 3.134641881 PW(w) = −20,000,000 – (85,000,000/(1−.05))[P/A,((0.12+0.05)/(1−0.05)),5] = −300,467,957.81 PW(w/o) = −85,000,000(P/A,12%,5) = −306,405,977.20 PW(Difference) = $5,938,019.39 AW(Difference) = $1,647,264.37 Assuming that EOY 1 cost for purchased components is $85,000,000 (1−0.05) EOY 0 1 2 3 4 5

Cash Flow (w) −20,000,000.00 −80,750,000.00 −76,712,500.00 −72,876,875.00 −69,233,031.25 −65,771,379.69

Cash Flow (w/o) 0.00 −85,000,000.00 −85,000,000.00 −85,000,000.00 −85,000,000.00 −85,000,000.00

PW

−286,444,559.92

−306,405,977.20

i(r) = 0.178947368 (P/A,i(r),5) = 3.134641881 PW(w) = −20,000,000 – 85,000,000[P/A,((0.12+0.05)/(1−0.05)),5] = −286,444,559.92 PW(w/o) = −85,000,000(P/A,12%,5) = −306,405,977.20 PW(Difference) = $19,961,417.28 AW(Difference) = $5,537,491.42

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8-47

Left to student.

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8-48

This is intended to be a tailor-made exercise (at the discretion of the instructor).

Assumptions: 

Salary and fringe benefits for the new analyst will be $28,000 (1.3) = $36,400 in year 1 purchasing power. This increases 6% per year thereafter.  Staff retirements occur at the end of the year. Therefore, there are no realized savings in year 1.

Savings of $16,200 in year 2, $32,400 in year 3, and $48,600 each year thereafter are expressed in real purchasing power keyed to year 0.  First−year savings on purchases are 3% of $1,000,000 (1.10) = $33,000 and this increases by 10% per year thereafter.  Contingency costs will not be considered as cash flows until they are spent (we assume they won’t be spent).

The effective income tax rate is = 38%.   There is no market value at the end of the 6−year project life.

Capital EOY Investment 0 − $80,000 1 2 3 4 5 6 EOY 0 1 2 3 4 5 6

BTCF −$80,000 −9,400 9,918 31,619 55,926 61,398 67,376

Service Contract

New Manpower Savings on Analyst Savings Purchases

− $6,000 − 6,000 − 6,000 − 6,000 − 6,000 − 6,000

−$36,400 − 38,584 − 40,899 − 43,353 − 45,954 − 48,711

Depr. −−− $16,000 25,600 15,360 9,216 9,216 4,608

$ 0 18,202 38,588 61,356 65,037 68,940

TI −−− −$25,400 −15,682 16,259 46,710 52,182 62,768

$33,000 36,300 39,930 43,923 48,315 53,147

T(38%) −−− $9,652 5,959 −6,178 −17,750 −19,829 −23,852

Total BTCF −$80,000 − 9,400 9,918 31,619 55,926 61,398 67,376

ATCF (A$) −$80,000 252 15,877 25,441 38,176 41,569 43,524

6

PW (15%) =

 ATCF (P/F, 15%, k) = $10,263 k

k=0

In view of MARR of 15%, this investment should be undertaken. The instructor may wish to ask the class to explore various “what if” questions involving changes in the assumptions listed above. For example, how much change would occur in the PW value if we assume staff retirements occur at the beginning of the year?)

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8-49

Typical solution for (*) is (P/F, 4%, 10) = 0.6756, so 32.44% of purchasing power has been lost due to inflation. This rounds to −32%.

Inflation Rate

Erosion of Money's Purchasing Power (years) 10 15 25 2% ‐18% ‐26% ‐39% 3% ‐26% ‐36% ‐52% 4% ‐32%* ‐44% ‐62%

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8-50

Starting Salary = Annual Salary Increase = Savings Interest Rate = Average Inflation Rate = Desired amount in 2027 (R$) = Desired Amount in 2027 (A$) =

$ 60,000 8.00% 7.50% 3.75% $ 500,000 $1,508,736

Year 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019 2020 2021 2022 2023 2024 2025 2026 2027

Salary (A$) $ 60,000 $ 64,800 $ 69,984 $ 75,583 $ 81,629 $ 88,160 $ 95,212 $ 102,829 $ 111,056 $ 119,940 $ 129,535 $ 139,898 $ 151,090 $ 163,177 $ 176,232 $ 190,330 $ 205,557 $ 222,001 $ 239,761 $ 258,942 $ 279,657 $ 302,030 $ 326,192 $ 352,288 $ 380,471 $ 410,909 $ 443,781 $ 479,284 $ 517,626 $ 559,036 $ 603,759

% Salary to Save Annually

Savings (A$) $ 5,184 $ 5,599 $ 6,047 $ 6,530 $ 7,053 $ 7,617 $ 8,226 $ 8,884 $ 9,595 $ 10,363 $ 11,192 $ 12,087 $ 13,054 $ 14,099 $ 15,226 $ 16,445 $ 17,760 $ 19,181 $ 20,715 $ 22,373 $ 24,162 $ 26,095 $ 28,183 $ 30,438 $ 32,873 $ 35,502 $ 38,343 $ 41,410 $ 44,723 $ 48,301 $ 52,165

8.64%

864

Bank Balance (A$) $ 5,184 $ 11,172 $ 18,056 $ 25,941 $ 34,939 $ 45,176 $ 56,791 $ 69,935 $ 84,775 $ 101,496 $ 120,300 $ 141,410 $ 165,070 $ 191,548 $ 221,141 $ 254,171 $ 290,994 $ 331,999 $ 377,615 $ 428,308 $ 484,594 $ 547,034 $ 616,244 $ 692,900 $ 777,741 $ 871,574 $ 975,284 $ 1,089,841 $ 1,216,302 $ 1,355,825 $ 1,509,677

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Difference between Desired and Actual =

8-51

$

941

f = 4.5% per year; im (after−tax) = 12% per year; t = 40%; b = 0 increase rate = 6 % per year (applies to annual expenses, replacement costs, and market value) Analysis period = 20 years; Useful life = 10 years MACRS (GDS) 5−year property class Capital investment (and cost basis, B) Market value (at end of year 10) in year 0 dollars

= −$260,000 = $50,000

Annual expenses (in year 0 dollars) = −$6,000 Annual property tax = 4% of capital investment (does not inflate) Assume like replacement at end of year 10.

Annual Property ATCF ATCF* EOY Expenses Taxes BTCF Depr.

TI T(40%) (A$) (R$) 0 -$260,000 -$260,000 -$260,000 1 -$6,360 -$10,400 -16,760 52,000 -68,760 27,504 10,744 10,281 2 -6,742 -10,400 -17,142 83,200 -100,342 40,137 22,995 21,057 3 -7,146 -10,400 -17,546 49,920 -67,466 26,986 9,440 8,272 4 -7,575 -10,400 -17,975 29,952 -47,927 19,171 1,196 1,003 5 -8,029 -10,400 -18,429 29,952 -48,381 19,352 923 741 6 -8,511 -10,400 -18,911 14,976 -33,887 13,555 -5,356 -4,113 7 -9,022 -10,400 -19,422 0 -19,422 7,769 -11,653 -8,563 8 -9,563 -10,400 -19,963 0 -19,963 7,985 -11,978 -8,423 9 -10,137 -10,400 -20,537 0 -20,537 8,215 -12,322 -8,292 10 -10,745 -10,400 -21,145 0 -21,145 8,458 -12,687 -8,170 10 89,542 89,542 -35,817 53,725 34,595 10 -465,620 -465,620 -299,826 11 -11,390 -18,625 -30,015 93,124 -123,139 49,256 19,241 11,856 12 -12,073 -18,625 -30,698 148,998 -179,696 71,878 41,180 24,282 13 -12,798 -18,625 -31,423 89,399 -120,822 48,329 16,906 9,540 14 -13,565 -18,625 -32,190 53,639 -85,829 34,332 2,142 1,157 15 -14,379 -18,625 -33,004 53,639 -86,643 34,657 1,653 854 16 -15,242 -18,625 -33,867 26,820 -60,687 24,275 -9,592 -4,743 17 -16,157 -18,625 -34,782 0 -34,782 13,913 -20,869 -9,875 18 -17,126 -18,625 -35,751 0 -35,751 14,300 -21,451 -9,713 19 -18,154 -18,625 -36,779 0 -36,779 14,712 -22,067 -9,562 20 -19,243 -18,625 -37,868 0 -37,868 15,147 -22,721 -9,421 20 160,357 160,357 -64,143 96,214 39,894

* ATCF(R$) = ATCF(A$)  1/(1.045)k ir =

012 .  0.045 = 0.0718 or 7.18% per year 1.045 20

PW =

20

 ATCF (A$) (P/F, 12%, k) =  ATCF (R$) (P/F, 7.18%, k) = −$359,665 k

k0

k

k0

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8-52

PW(maintenance costs) 

A1[1  ( P / F ,8%,5)( F / P,4%,5)] 0.08  0.04

 A1 (4.29785)

where A1 = $1,000 for the induction motor and A1 = $1,250 for the synchronous motor.

PW(electricity costs) 

A1[1  ( P / F ,8%,5)( F / P,5%,5)] 0.08  0.05

 A1 ( 4.37834 )

where A1 = $50,552 for the induction motor and A1 = $55,950 for the synchronous motor.

PWTC(induction motor) = $17,640 + $1,000(4.29785) + $50,552(4.37834) = $17,640 + $4,298 + $221,334 = $243,272 PWTC(synchronous motor) = $24,500 + $1,250(4.29785) + $55,950(4.37834) = $24,500 + $5,372 + $244,968 = $274,840 PWTC(A) = (3)($243,272) + [$17,640 + $4,298 + (350/400)($221,334)] = $945,421 PWTC(B) = (3)($274,840) + [$24,500 + $5,372 + (50/500)($244,968)] = $878,889 PWTC(C) = (3)($243,272) + [$24,500 + $5,372 + (350/500)($244,968)] = $931,166 PWTC(D) = (3)($274,840) + [$17,640 + $4,298 + (50/400)($221,334)] = $874,125 Option (D) has the lowest present worth of total costs. Thus, the recommendation is still to power the assembly line using three 500 hp synchronous motors operated at a power factor of 1.0 and one 400 hp induction motor.

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8-53

PW(electricity costs) 

A1 [1  ( P / F ,8%,8)( F / P,6%,8)] 0.08  0.06

 A1 (6.9435) where A1 = $50,552 for the induction motor and A1 = $55,950 for the synchronous motor.

PWTC(induction motor) = $17,640 + $1,000(6.5136) + $50,552(6.9435) = $17,640 + $6,514 + $351,008 = $375,162 PWTC(synchronous motor) = $24,500 + $1,250(6.5136) + $55,950(6.9435) = $24,500 + $8,142 + $388,489 = $421,131 PWTC(A) = (3)($375,162) + [$17,640 + $6,514 + (350/400)($351,008)] = $1,456,772 PWTC(B) = (3)($421,131) + [$24,500 + $8,142 + (50/500)($388,489)] = $1,334,884 PWTC(C) = (3)($375,162) + [$24,500 + $8,142 + (350/500)($388,489)] = $1,430,070 PWTC(D) = (3)($421,131) + [$17,640 + $6,514 + (50/400)($351,008)] = $1,331,423 Option (D) has the lowest present worth of total costs. Thus, the recommendation is still to power the assembly line using three 500 hp synchronous motors operated at a power factor of 1.0 and one 400 hp induction motor.

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8-54

The following four options will be considered: (A) (B) (C) (D)

Four induction motors (three at 400 hp, one at 100 hp) Three synchronous motors (two at 500 hp, one at 300 hp) Three induction motors at 400 hp plus one synchronous motor at 100 hp Two synchronous motors at 500 hp plus one induction motor at 300 hp.

PWTC(A) = (3)($364,045) + [$17,640 + $6,514 + (100/400)($339,891)] = $1,201,262 PWTC(B) = (2)($408,827) + [$24,500 + $8,142 + (300/500)($376,185)] = $1,076,007 PWTC(C) = (3)($364,045) + [$24,500 + $8,142 + (100/500)($376,185)] = $1,200,014 PWTC(D) = (2)($408,827) + [$17,640 + $6,514 + (300/400)($339,891)] = $1,096,726 Option (B) has the lowest present worth of total costs. Thus, the recommendation is to power the assembly line using three 500 hp synchronous motors: two operated at 500 hp and one at 300 hp.

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8-55

Expected cost of Machine in 2018 = $2,550(1.07)4 = $3,342.53 True percentage increzse in cost =

$3,930  $3.342.53 x 100% = 17.58% $3,343.53

Select (d)

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8-56 ir = 0.07; f = 0.09 im = 0.07 + 0.09 + (0.07)(0.09) = 0.1663 Select (a)

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8-57

By using Equation (8-1), we have R$ = $1(P/F, 2.4%, 44) = $0.3522.

Select (b)

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8-58 A$ Analysis: im – 0.098 + 0.02 + (0.098)(0.02) = 0.12 or 12% PWA(12%) = −$27,000 + $4,000(P/A,12%,5) = −$2,581 PWB(12%) = −$19,000 + $5,000(P/A,12%,5) = −$976 Neither alternative is acceptable.

Select (c)

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8-59

A1 = $25,000 (1.04) = $26,000 PW =

=

$26,0001  ( P / F ,7%, )( F / P,4%,   0.07  0.04

$26,000  (1.04)   1   0.03  (1.07)  

0 = $866,667 Select (a)

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8-60

($1.60 Canadian/Euro)($0.75 US/$1.00 Canadian) = $1.20 US/Euro Select (c)

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8-61

Set up a table of the exchange rate as follows: EOY Euro / Dollar

0 (now) $1.35

1 1.31

2 1.27

3 1.24

4 1.20

5 1.16

The Euro will trade at $1.16 per Euro after 5 years.

Select (c).

Check ($1.35 / 1.035) = $1.35 / 1.1593 = $1.164 per Euro

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Solutions to Chapter 9 Problems 9-1

Defender: Old car Investment = $400 Annual maintenance = $800 Annual gas costs = (15,000 mi/yr)($3.50/gal)(gal/10mi) = $5,250 Assume SV2 = 0 AWD = -400(A/P, 15%, 2) – 800 - 5,250 = -400(0.6151) – 800 - 5,250 = - $6,296 PWD = -$10,235 FWD = -$13,536

Challenger: New car Investment = $8,000 Annual maintenance = $0 Annual gas costs = (15,000 mi/yr)($3.50/gal)(gal/30mi) = $1,750 SV2 = $5,000 AWD = -400(A/P, 15%, 2) – 800 – 1,750 + $5,000(A/F, 15%, 2) = -400(0.6151) – 800 – 1,750 + $5,000(0.4651) = - $4,345 PWD = -$7,064 FWD = -$9,342 Replace with New Car to minimize equivalent annual costs. © 2016 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9-2

AWDefender (12%) = – $43,000 (A/P, 12%, 5) – $1,800 – $2,200 (A/F, 12%, 5) = –$14,074.48 AWChallenger (12%) = – $64,200 (A/P, 12%, 10) – $2,000 – $10,000 (A/F, 12%, 10) = –$13.933.4 Select the Challenger.

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9-3

Old Crane (Defender): Using the outsider viewpoint, the investment value of the defender is its current market value plus the cost of the overhaul required to keep it in service.

Capital investment = $5,000 + $4,000 = $9,000 Annual O&M costs = $2,500 Market value (EOY 8) = $0 AW (15%) = –$9,000 (A/P, 15%,7) – $2,500 = –$4,664 New Crane (Challenger): Capital investment = $16,000 Annual O&M costs = $1,200 Market value (EOY 10) = $3,200 AW (15%) = –$16,000 (A/P, 15%, 7) – 1,200 + $3,200 (A/F, 15%, 7) = –$4,757 Retain the old crane (AWChallenger < AWDefender).

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9-4 EOY 1 2 3 4 5

Salvage Value 6,000 5,000 4,000 2,500 1,250

Depreciation Lost Interest 2,000 1,200 1,000 900 1,000 750 1,500 600 1,250 375

Op. Cost

Total Cost

PW

3,000 3,000 3,500 4,000 4,500

6,200 4,900 5,250 6,100 6,125

5,391 3,705 3,452 3,488 3,045

AW(1) = 6,200

Year 1

AW(1-2) = [5,391 + 3,705](A/P, 15%, 2) = 5,595

Year 2

AW(1-3) = [5,391 + 3,705 + 3,452] (A/P, 15%, 3) = 5,496

Year 3

AW(1-4) = [5,391 + 3,705 + 3,452 + 3,488] (A/P, 15%, 4) = 5,617

Year 4

AW(1-5) = [5,391 + 3,705 + 3,452 + 3,488 + 3,045] (A/P, 15%, 5) = 5,692

Year 5

Lowest Annual Worth is for years 1 through 3, thus replace the forklift every 3 years.

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9-5

For N = 1, EUAC = $13,200 For N = 2, EUAC = $ 6,914 For N = 3, EUAC = $ 4,825 For N = 4, EUAC = $ 4,217 For N = 5, EUAC = $ 3,854 For N = 6, EUAC = $ 3,937 For higher years, the EUAC increases, so N* = 5 years for the challenger.

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9-6

EOY EUAC 1 $62,000 2 $55,950 = $80,000(A/P, 15%, 2) − $50,000(A/F, 15%,2) + $30,000 3 $54,960 = $80,000(A/P, 15%, 2) − $50,000(A/F, 15%,2) + [$30,000(P/A, 15%, 2) + $35,000(P/F, 15%, 3)](A/P, 15%, 3) 4 $56,170 5 $56,920

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9-7

Challenger EOY 0 1 2

Market Value $42,500 $31,00 $25,000

Defender

O&M −− $10,000 $12,500

Market Value $25,000 $17,000 −−

O&M −− $14,000 −−

Challenger Year

Market Value

Loss in Value

Cost of Capital

O&M

0 1 2

$42,500.00 $31,000.00 $25,000.00

−− $11,500.00 $6,000.00

−− $6,375.00 $4,650.00

−− $10,000.00 $12,500.00

Marginal Cost −− $27,875.00 $23,150.00

EUAC −− $27,875.00 $25,677.33

Defender Year

Market Value

Loss in Value

Cost of Capital

O&M

0 1

$25,000.00 $17,000.00

−− $8,000.00

−− $3,750.00

−− $14,000.00

Marginal Cost −− $25,750.00

EUAC −− Not necessary to calculate EUAC!

Min(EUAC) of Challenger < Marginal Cost of keeping Defender for 1 additional year; therefore, Replace Immediately

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9-8

Challenger EOY

MV@EOY

0 1 2 3 4

$50,000.00 $40,000.00 $32,000.00 $24,000.00 $16,000.00

Loss in Value −− $10,000.00 $8,000.00 $8,000.00 $8,000.00

Cost of Capital −− $5,000.00 $4,000.00 $3,200.00 $2,400.00

Loss in Value −− $10,000.00 $4,000.00 $4,000.00 $4,000.00

Cost of Capital −− $3,500.00 $2,500.00 $2,100.00 $1,700.00

O&M −− $13,000.00 $15,500.00 $18,000.00 $20,500.00

Marginal Cost --$28,000.00 $27,500.00 $29,200.00 $30,900.00

EUAC −− $28,000.00 $27,761.90 $28,196.37 $28,778.93

Defender EOY

MV@EOY

0 1 2 3 4

$35,000.00 $25,000.00 $21,000.00 $17,000.00 $13,000.00

O&M −− $18,500.00 $21,000.00 $23,500.00 $26,000.00

Marginal Cost --$32,000.00 $27,500.00 $29,600.00 $31,700

EUAC −− $29,857.14

The minimum EUAC of the Challenger is less than the marginal cost of keeping the Defender for one more year. Therefore, the Defender should be replaced immediately.

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9-9

Present Machine (Defender) Expenses Year Capital Recovery Amount for Year Total EUAC 1 $4,000(0.15)+(4,000−3,000) = $1,600 $20,000 $21,600 $21,600* 2 $3,000(0.15)+(3,000−2,500) = $950 $25,000 $25,950 $23,622 3 $2,500(0.15)+(2,500−2,000) = $875 $30,000 $30,875 $25,713 *

The economic life of the "defender" is 1 year, and the EUAC during this year is $21,600.

Improved Machine (Challenger) Useful life is 12 years; Find EUAC(15%) over this period of time.

EUAC(15%) = $30,000(A/P,15%,12) − $2,000(A/F,15%,12) + $16,000 = $21,466 Because $21,466 is less than $21,600, the new improved machine should replace the present machine immediately.

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9-10

The repeatability assumption and the AW method (over one useful life cycle) are used in the comparison of the two robots. The use of repeatability as a simplified modeling approach can be supported in this case.

The estimated annual expenses for the defender are a geometric cash flow sequence (Chapter 4).

The negative estimated market value (−$1,500) indicates an expected net cost for the disposal of the asset at the end of six years.

AW D (25%)  ($38,200  $2,000)( A / P,25%,6) $1,400[1  ( P / F ,25%,6)( F / P,8%,6)] ( A / P,25%,6)  $1,500 ( A / F , 25%,6) 0.25  0.08  $15,383 

and, for the challenger, the AW over its useful life is AWC (25%)  ($51,000  $5,500)( A / P,25%,10)  [$1,000( P / A,25%,10)  $150( P / G ,25%,10)]( A / P,25%,10)  $7,000( A / F ,25%,10)  $17,035

The defender should be retained because the AW over its useful life has the least negative value ($15,382).

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9-11

The overpass can be reinforced to extend its life for 6 years and then be replaced by a new concrete overpass. An alternative is to build the new concrete overpass immediately. Coterminate the study period at 40 years.

EOY 0

Reinforce Now, Replace with New Replace Later Overpass Now (Replace Now − Reinforce) a − $40,000 − $140,000 − $100,000

1 2

0 0

3,200b 3,200

3,200 3,200

3

0

3,200

3,200

4

0

3,200

3,200

5

0

3,200

3,200

6

0

3,200

3,200

6

− $122,000c

0

122,000

7  40

0

0

0d

a

−$25,000 (reinforcement) − $15,000 (MV of scrap foregone) = −$40,000 Savings in annual expenses c $18,000(scrap value) − $140,000(investment in new overpass) = −$122,000 d The same overpass is in place during years 7 − 40, thus there is no difference in the annual expenses. b

PW(8%) = −$100,000 + $3,200(P/A,8%,6) + $122,000(P/F,8%,6) = −$8,322 < 0 It is more economical to reinforce the existing bridge and delay its replacement.

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9-12

(a)

3,200 gallons of gasoline will be required to drive a car averaging 24.5 mpg, and 2,450 gallons will be required at 32 mpg. The fuel savings will be 750 gallons over the life of the car. At $3 per gallon, this amounts to a savings of $2,250 over 100,000 miles of driving. The carbon dioxide emissions “saved” = (0.1 lb/mi)(100,000 mi) = 10,000 lb of CO2.

(b)

CO2 penalty for defender = $0.05(10,000) = $500. This places an extra burden on the defender.

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9-13

Defender: Annual expenses = $350,000 + $240,000 + $500,000 (0.05) + $12,000 = $627,000 EUACD(8%) = $180,000 (A/P, 8%, 10) − $60,000 (A/F, 8%, 10) + $627,000 = $649,680 Challenger: EUACC(8%) = $280,000 + $125,000 + $115,000 = $520,000 Decision: Sell the defender and lease a new machine to minimize EUAC.

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9-14 Year 0 1 2 3

MV 85,000 75,190 63,654 43,709

∆MV --9,810 11,536 19,945

Cost of capital --12,750 11,279 9,548

Expenses --15,675 16,380 17,117

AWk --38,235 38,681 40,965

Replace with the challenger after 1 year

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9-15

Abandonment Interval: Keep for N = 1 year: PW (9%) = −$5,700 + ($5,200 + $3,200)(P/F, 9%, 1) = $2,006 Keep for N = 2 years: PW (9%) = −$5,700 + $3,200(P/A, 9%, 2) + $4,600 (P/F, 9%, 2) = $3,801 Keep for N = 3 years: PW (9%) = −$5,700 + $3,200 (P/A, 9%, 3) + $4,000 (P/F, 9%, 3) = $5,489 Keep for N = 4 years: PW (9%) = −$5,700 + $3,200 (P/A, 9%, 4) + $3,000 (P/F, 9%, 4) = $6,792 Keep for N = 5 years: PW (9%) = −$5,700 + $3,200 (P/A, 9%, 5) = $6,747 PW (9%) is a maximum at 4 years. Therefore, the centrifuge should be retained for four years before abandonment.

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9-16 EOY, k 0 1 2 3 4 5 6 7 8

EOY, k 0 1 2 3 4 5 6 7 8

Market Value

Loss in Value

Cost of Capital

Annual Expenses

Approximate After−Tax Total (Marginal) Costa

$8,000 4,700 3,200 2,200 1,450 950 600 300 0

--$3,300 1,500 1,000 750 500 350 300 300

--$560 329 224 154 102 67 42 21

--$3,000 3,000 3,500 4,000 4,500 5,250 6,250 7,750

--$4,116 2,897 2,834 2,942 3,061 3,400 3,955 4,843

MACRS BV

Interest on Tax Creditb

Adjusted After−Tax Total (Marginal) Costc

PW(7%)

EUAC

$8,000 6,400 3,840 2,304 1,382 461 0 0 0

--$224 179 108 65 39 13 0 0

--$4,340 3,076 2,942 3,007 3,100 3,413 3,955 4,843

--$4,056 2,687 2,402 2,294 2,210 2,274 2,463 2,819

--$4,340 3,729 3,485 3,377 3,329* 3,341 3,412 3,552

a

Approx. After−Tax Total Cost = (0.6)(Loss in Value + Cost of Capital + Annual Expenses) Interest on tax credit = (0.07) (0.4) BVk−1 c Adj. After−Tax Total Cost = Approx. After−Tax Total Cost + Interest on tax credit b

*

The economic life of this equipment is 5 years.

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9-17

BV0 = $62,000(1 – 0.2 – 0.32 – 0.192 – 0.1152) = $10,714 MV0 = $12,000 Expense for year 0 repair work = $4,000 Using the format presented in Figure 9-5 with an additional row entry for the repair expense: EOY 0 0

BTCF −$12,000

Depr ---

−

---

4,000

TI ($12,000  $10,714) = $1,286 − $4,000

T (39%) $ 520

ATCF $11,498

1,560 Total

2,440 $13,938

The total after−tax investment in the defender is $13,938. This value includes the opportunity foregone by not selling the current asset for $12,000 (modified by the tax consequence of the income taxes that won’t be paid on the gain of $12,000  $10,714 = $1,286) as well as the after−tax expense of the required repair work.

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9-18

(a) Year, k 1 2 3 4 5 *

Defender EUAC through year k $15,702* 16,627 17,932 -----

Challenger EUAC through year k $20,866 20,458 20,037* 21,503 21,612

The economic life of the defender is 1 year. The economic life of the challenger is 3 years.

(b) Year 1 2 3

Defender Marginal Cost for Year $15,702 17,662 21,038

Based on this analysis, the defender should be kept for two years before being replaced by the challenger. Although the economic life of the defender is 1 year, the marginal cost of keeping the defender for the second year is less than the minimum EUAC of the challenger. (c)

Assumptions: Infinite analysis period with repetitive cycles of replacement with challenger (every three years) starting at the end of the second year.

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9-19

Keep the Defender: EOY 0 15 5 a b

BTCF −$57,000 −27,000 21,656b

Depr --6,000 ---

TI −$27,000a −33,000 21,656

T (40%) $10,800 13,200 −8,662

ATCF −$46,200 −13,800 12,994

Gain on disposal = $57,000 − $30,000 = $27,000 (if sold now).

MV5 = $18,500(1.032)5 = $21,656

Replace with Challenger: EOY 0 1−5

BTCF 0 −$36,500

EOY 0 1−5 5

Depr --0

TI --−$36,500

T (40%) --$14,600

ATCF 0 −$21,900

Incremental Cash Flows  (Def. – Chal.) − $46,200 8,100 12,994

To find the incremental IRR, solve AW(i′) = 0 = −$46,200(A/P, i′, 5) + $8,100 + $12,994(A/F, i′, 5) for i′ = 4.36%. This is less than the after-tax MARR of 9%, so the challenger should be accepted.

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9-20

(a)

Keep the Defender: EOY 0 16 6 a b

BTCF −$14,000 0 0

Depr --0 ---

TI −$14,000a 0 −10,000b

T (40%) $5,600 0 4,000

ATCF −$8,400

0 4,000

Gain on disposal = $14,000 (if sold now).

Loss on disposal for defender when sold 6 years from now.

Replace with Challenger: EOY 0 1 2 3 4 5 6 6

BTCF −$80,000 7,000 10,500 11,000 11,500 12,000 12,500 20,000

Depr --$ 16,000 25,600 15,360 9,216 9,216 4,608 ---

EOY 0 1 2 3 4 5 6 6

(b)

TI --−$ 9,000 − 15,100 − 4,360 2,284 2,784 7,892 20,000

T (40%) --$ 3,600 6,040 1,744 − 914 − 1,114 − 3,157 − 8,000

ATCF $80,000 10,600 16,540 12,744 10,586 10,886 9,343 12,000

Incremental Cash Flows  (Challenger − Defender) − $71,600 10,600 16,540 12,744 10,586 10,886 9,343 8,000

ERR Analysis ( =12%) 6

$71,600 (F/P, i'%,6) =

  ATCF (F/P,12%,6−k) k

k =1

$71,600(1+ i')6 = $105,425 i'% = ERR = 6.7% Since ERR < MARR, the defender should not be replaced with the challenger. © 2016 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9-21

Assume study period = 10 years.

Keep Pumping Station (defender) EOY 0 1−10 10

BTCF − $75,000 − 7,500 + 30,000

Depr --$3,000 ---

TI --− 10,500 − 15,000

T (50%) --5,250 7,500

ATCF −$75,000 −2,250 37,500

BV15 = $90,000 – 15($3,000) = $45,000 EUACD(5%) = $75,000(A/P, 5%, 10) − $37,500(A/F, 5%, 10) + $2,250 = $8,981 Sell Pumping Station (challenger) EOY 1−10

BTCF −$10,000

Depr ---

TI −$10,000

T (50%) +$5,000

ATCF −$5,000

EUACC(5%) = $5,000 Decision: Sell the pumping station ($5,000 < $8,981).

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9-22

Use the outsider viewpoint to determine after-tax cash flow (ATCF) of the defender and challenger.

Defender ATCF:

EOY 0 EOY 1-3 EOY 3

-$20,400 (which is -$14,000 - $6,400) -$6,800 After-Tax MV = $13,200 (which is $10,000 + $3,200)

AW(10%) = -$20,400 (A/P, 10%, 3) - $6,800 + $13,200 (A/F, 10%, 3) = - 11,015 Challenger ATCF:

EOY 0 EOY 1-20 EOY 20

-$65,000 -$4,300 MV = $10,000

AW(10%) = -$65,000 (A/P, 10%, 20) - $4,300 + $10,000 (A/F, 10%, 20) = -11,763 By a slim margin the defender should be kept at least one more year.

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9-23

Keep the Defender: EOY 0 14 5 5 a

BTCF Depr −$57,000 --− 27,000 $6,000 − 27,000 3,000 18,500 ---

TI − $30,000a − 33,000 − 30,000 18,500

T (40%) $12,000 13,200 12,000  7,400

ATCF −$45,000 − 13,800 − 15,000 11,100

T (40%) 14,600

ATCF − 21,900

Gain (if sold) = $57,000 − $27,000 = $30,000

Replace with Challenger: EOY 15

BTCF −$36,500

Depr 0

TI −$36,500

Incremental Analysis (Defender − Challenger): Capital Investment: −$45,000 Annual Expenses: Years 14 −$13,800 − (−$21,900)= $ 8,100 (Savings) Year 5 −$15,000 − (−$21,900)= $ 6,900 (Savings) Market Value (ATCF): $11,100 − $0 = $11,100 PW(i'%) = 0 = −$45,000 + $8,100(P/A, i'%, 4) + ($6,900 + $11,100)(P/F,i'%,5) By trial and error, i'% = 3.44% < MARR. Therefore, lease the challenger.

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9-24

Defender EOY BTCF Depr.

Tax Inc. Inc. Tax ATCF 0 −$57,000 −$30,000 −$27,000 $10,800 −$46,200 1 – 5 −27,000 6,000 −33,000 13,200 −13,800 (a) 5 21,656 --21,656 −8,662 12,993 (a)

MV5 = $18,500(1.032)5 = $21,656

Challenger: EOY BTCF Depr. Tax Inc. Inc. Tax ATCF 0 0 0 0 0 0 1 – 5 −$36,500 0 −$36,500 $14,600 −$21,900 5 0 0 0 0 0 Incremental cash flows: Initial Investment: Annual Expenses: Market Value:

−$46,200 – 0 = −$46,200 −$13,800 − [−$21,900] = $8,100 $12,993 – 0 = $12,993

AW = 0 = −$46,200(A/P, i’%, 5) + $8,100 + $12,993(A/F, i’%, 5).

Solving yields i’ = 4.36% < 9%. Replace the defender with the challenger.

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9-25

Use the capitalized worth method and the repeatability assumption due to the indefinitely long study period. Note that all market values are completely offset by the removal costs.

Alternative a: Do not relocate existing transformers Site A: Property tax for existing transformers (years 1 − 10) = − ($2,100 + $475)(0.02) = − $51.50 Capital investment for replacement transformers (EOY 10 and every 30 years thereafter) = − $900 − $340 = − $1,240 Property tax for replacment transformers (years 11 − ) = − ($900 + $340)(0.02) = − $24.80 Site B: Capital investment for new transformers (EOY 0 and every 30 years thereafter) = − $2,100 − $475 = − $2,575 Property tax for new transformers (years 1 − ) = −$2,575(0.02) = − $51.50 CWa(8%) = −$51.50(P/A,8%,10) −  $1,240(A / P,8%,30)  $24.80  (P/F,8%,10)   0.08 −  $2,575(A / P,8%,30)  $51.50    0.08 = −$4,629 Alternative b: Relocate existing transformers Cost to remove existing transformers from site A and reinstall at site B (EOY 0 ) = − $110 − $475 = − $585 Site A: Capital investment for replacing transformers (EOY 0 and every 30 years thereafter) = − $900 − $340 = − $1,240 Property tax for new transformers (years 1 − ) = − $1,240(0.02) = − $24.80 Site B: Property tax for relocated transformers (years 1 − 10) = − ($2,100 + $475)(0.02) = − $51.50 Capital investment for replacement transformers (EOY 10 and every 30 years thereafter) = −$2,100 − $475 = −$2,575 Property tax for replacement transformers (years 11 − ) = −$2,575(0.02) = −$51.50

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9-25

continued CWb(8%) = −$585 −  $1,240(A / P,8%,30)  $24.80  − $51.50(P/A,8%,10)   0.08 −  $2,575(A / P,8%,30)  $51.50  (P/F,8%,10)   0.08 = −$4,239 It is more economical to relocate the existing transformers (Alternative b).

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9-26

(a) EOY

MV

0 1 2 3 4

$70,000.00 $56,000.00 $44,000.00 $34,000.00 $22,000.00

Loss in Value

Cost of Capital

O&M

Marginal Cost (M.C.)

EUAC

$14,000.00 $12,000.00 $10,000.00 $12,000.00

$7,000.00 $5,600.00 $4,400.00 $3,400.00

$5,500.00 $6,800.00 $7,400.00 $9,700.00

$26,500.00 $24,400.00 $21,800.00 $25,100.00

$26,500.00 $25,500.00 $24,382,18 $24,536.85

Depr

TI

ATCF

−$11,488.00 −$7,744.00 −$4,000.00 0.00

−$28,768.00 0.00 −$19,988.00 −$18,244.00 −$18,000.00 −$16,000.00

Taxes Payable $11,507.20 0.00 $7,995.20 $7,297.60 $7,200.00 $6,400.00

Economic Life = 3 years (b) EOY

BTCF

0 0 1 2 3 4

−$40,000.00 −$12,000.00 −$8,500.00 −$10,500.00 −$14,000.00 −$16,000.00

−$28,492.80 −$12,000.00 −$504.80 −$3,202.40 −$6,800.00 −$9,600.00

BV(now) = $11,232.00; MV(now) = $40,000.00; MV(now) – BV(now) = $28,768.00

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9-27

(a) A 1 MARR = 2

3 4 5 6 7 8 9 10 11

EOY 0 1 2 3 4 5

B 7.73%

End of Year MV $ 20,000 $ 15,000 $ 11,250 $ 8,500 $ 6,500 $ 4,750

C

D

E

CR Amount

Annual Expense

PW of Annual Exp.

EUAC of Cumulative Ann. Exp. EUAC

$ $ $ $ $

6,546 5,758 5,098 4,554 4,159

$ 2,000 $ 3,000 $ 4,620 $ 8,000 $ 12,000

$ $ $ $ $

$ $ $ $ $

C

D

E

CR Amount

Annual Expense

PW of Annual Exp.

$ 12,630 $ 11,304 $ 10,311 $ 9,579 $ 9,073

$ 2,000 $ 3,000 $ 4,620 $ 8,000 $ 12,000

1,856 2,585 3,695 5,939 8,270

F

2,000 2,481 3,142 4,224 5,557

G

$ $ $ $ $

H

I

8,546 8,240 <- Min EUAC 8,240 8,778 9,716

(b) A 1 MARR = 2

3 4 5 6 7 8 9 10 11

EOY 0 1 2 3 4 5

B 38.15%

End of Year MV $ 20,000 $ 15,000 $ 11,250 $ 8,500 $ 6,500 $ 4,750

$ $ $ $ $

1,448 1,572 1,752 2,196 2,385

F

G

H

I

EUAC of Cumulative Ann. Exp. EUAC $ $ $ $ $

2,000 2,420 2,933 3,664 4,453

$ $ $ $ $

14,630 13,724 13,243 13,243 <- Min EUAC 13,526

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9-28 MARR

15%

Defender: Percent Change in Annual Expenses =

‐0.7%

MV 4,000 3,000 2,500 2,000

Expenses

CR Amount

Total

EUACk

19,860 24,825 29,790

1,600 950 875

21,460 25,775 30,665

$21,460.00 $23,466.98 $25,539.84

Investment Annual Expenses Economic Life MV after 12 years

$30,000 $16,000 12 $2,000

EUAC

$21,465.46

EOY k 0 1 2 3

Challenger:

If the annual expenses of the defender were to lower by 0.7%, the replacement would be delayed by at least one year.

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9-29

The defender would be the preferred alternative for annual lease rates greater than $51,818.

Effective Tax Rate = After Tax MARR =

Operating hours/yr =

40%

Study period = Annual Expense Rate of Increase / yr. =

12%

Useful Life = O&M Expenses / hr. = $ Annual Expense: other = $ Fixed Lease Rate = $

51,818

Year 0 1 2 3 4 5 6 7 8 9 10 10

Lease (10,000) (51,818) (51,818) (51,818) (51,818) (51,818) (51,818) (51,818) (51,818) (51,818) (51,818) 10,000

$ $ $ $ $ $ $ $ $ $ $ $

260

15

10

4%

85

2,400 NOTE: All estimates expressed in year 0 dollars Annual Expenses $ $ $ $ $ $ $ $ $ $

(25,480) (26,499) (27,559) (28,662) (29,808) (31,000) (32,240) (33,530) (34,871) (36,266)

Taxable Income $ $ $ $ $ $ $ $ $ $

(77,298) (78,317) (79,377) (80,479) (81,626) (82,818) (84,058) (85,348) (86,689) (88,084)

Cash Flow for Income Taxes $ $ $ $ $ $ $ $ $ $

30,919 31,327 31,751 32,192 32,650 33,127 33,623 34,139 34,676 35,234

$ $ $ $ $ $ $ $ $ $ $ $

ATCF Adjusted ATCF (10,000) $ (10,000) (46,379) $ (46,379) (46,990) $ (46,990) (47,626) $ (47,626) (48,288) $ (48,288) (48,976) $ (48,976) (49,691) $ (49,691) (50,435) $ (50,435) (51,209) $ (51,209) (52,013) $ (52,013) (52,850) $ (42,850) 10,000

PW =

$

(282,472)

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9-30

A study period of five years still favors the challenger.

E ffective T ax Ra te =

4 0%

A fter tax M ARR =

1 2%

C lass L ife =

7

D B Ra te =

20 0%

8 0 kW Co st B asi s = A nn u al E xpe nse R ate o f In crea se / yr. = M V Ra te of i ncre ase / yr.= O p er ating Hou rs / yr. =

Ye ar 0 1 2 3 4 5 5

8 0 kW Ca pita l Inve stm en t A nn ua l O & M E xpe nse s / hr .

O th er An nu al E xpe nse s De pre cia ti on yea r at stud y star t

$

21 0,0 00

MV at en d of life

4%

4 0 kW

$

90 ,0 00

$

14 0,0 00

$

80

$

35

$

3 ,2 00

$

1,0 00

5

$

30 ,0 00

0

$

3 8,0 00

2% 2 60

$ $ $ $ $ $ $

BT CF (23 0,0 00 ) (3 5,4 64 ) (3 6,8 83 ) (3 8,3 58 ) (3 9,8 92 ) (4 1,4 88 ) 7 5,0 77

NO TE : A ll estim ate s e xpr esse d in ye ar 0 do llar s

8 0 kW d( k) $ $ $

1 8,7 42 1 8,7 42 9,3 71

40 k W d (k) $ $ $ $ $

2 0,0 00 3 4,2 86 2 4,4 90 1 7,4 93 1 2,4 95

$ $ $ $ $ $ $

T axa ble In co me ( 43 ,1 45 ) ( 74 ,2 06 ) ( 89 ,9 10 ) ( 72 ,2 19 ) ( 57 ,3 85 ) ( 53 ,9 83 ) 43 ,8 41

Cash F low for In com e Ta xe s $ 1 7,2 58 $ 2 9,6 82 $ 3 5,9 64 $ 2 8,8 88 $ 2 2,9 54 $ 2 1,5 93 $ (1 7,5 36 )

$ $ $ $ $ $ $

A TCF A dj uste d AT CF ( 21 2,74 2) $ (2 12 ,74 2) ( 5,78 2) $ (5 ,78 2) (91 8) $ ( 91 8) ( 9,47 0) $ (9 ,47 0) (1 6,93 8) $ ( 16 ,93 8) (1 9,89 5) $ 37 ,64 6 5 7,54 1

PW = 80 kW Ye ar 5 BV

$

-

$

$

(2 14 ,78 0)

40 kW 3 1,2 37

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9-31

The defender would be preferred to the new challenger; however, the original challenger is still the preferred alternative.

Effective Tax Rate = After tax MARR =

200%

Capital Investment Annual O & M Expenses / hr.

Other Annual Expenses

4%

MV at end of life

12%

Class Life =

7

DB Rate = Annual Expense Rate of Increase / yr. = MV Rate of increase / yr.= Operating Hours / yr. =

Year 0 1 2 3 4 5 6 7 8 9 10 10

40 kW (per unit)

40%

Total

$

120,000

$

360,000

$

35

$

105

$

1,000

$

3,000

$

38,000

$

114,000

2% 260

$ $ $ $ $ $ $ $ $ $ $ $

BTCF (360,000) (31,512) (32,772) (34,083) (35,447) (36,865) (38,339) (39,873) (41,468) (43,126) (44,851) 138,965

NOTE: All estimates expressed in year 0 dollars Three 40 kW Cash Flow for d(k) Taxable Income Income Taxes $ $ 51,429 $ (82,941) $ 33,176 $ 88,163 $ (120,936) $ 48,374 $ 62,974 $ (97,057) $ 38,823 $ 44,981 $ (80,428) $ 32,171 $ 32,129 $ (68,994) $ 27,598 $ 32,129 $ (70,469) $ 28,187 $ 32,129 $ (72,002) $ 28,801 $ 16,065 $ (57,532) $ 23,013 $ (43,126) $ 17,251 $ (44,851) $ 17,941 $ 138,965 $ (55,586)

$ $ $ $ $ $ $ $ $ $ $ $

ATCF Adjusted ATCF (360,000) $ (360,000) 1,664 $ 1,664 15,602 $ 15,602 4,739 $ 4,739 (3,276) $ (3,276) (9,267) $ (9,267) (10,152) $ (10,152) (11,072) $ (11,072) (18,455) $ (18,455) (25,876) $ (25,876) (26,911) $ 56,468 83,379

PW =

$

(358,798)

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9-32

Machine A (Defender): PWA(12%) = −$10,400 − $2,000 (P/A,12%,5) + $2,000 (P/F,12%,5) = −$16,475 Machine B (Challenger): PWB(12%) = −$14,000 = $1,400 (P/A,12%,5) + $1,400 (P/F,12%,5) = −$18,252 Continue with Machine A. Select (a)

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9-33

Let X = repair cost for existing machine (defender) PWD(12%) = −$15,000 – X PWC(12%) = −$44,000 + $6,000 (P/A,12%,5) = −$22,371 Set PWD12%) = PWC(12%) and solve for X. −$15,000 – X = −$22,271; X = $7,371 Select (c)

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9-34

EUACA(12%) = $7,000 (A/P,12%,7) + $1,000 − $1,000 (A/F,12%,7) = $2,435 Select (e)

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9-35

EUACB(12%) = $50,000 A/P,12%,10) + $600 − $5,000 (A/F,12%,10) = $9,165 Select (d)

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9-36

If sold:

0

BTCF $40,500

d 0

TI $40,500 − $34,500 = $6,000

T (−0.29)(6,000) = −$1,740

ATCF $40,500 − $1,740 = $38,760 Select (c)

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Solutions to Chapter 10 Problems 10-1

The capital recovery amount equals $500,000 (A/P, 10%, 15) ‒ $50,000 (A/F, 10%, 15) = $500,000 (0.1315) ‒ $50,000 (0.0315) = $65,750 ‒ $1,575 = $64,175 From Equation 10-4, we can determine the modified cost-benefit ratio.

B – C = ($200,000 ‒ $80,000) / $64,175 B – C = 1.87 This is an acceptable acquisition.

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10-2

Incremental investment = $300,000(0.10) = $30,000 AW = $30,000 (A/P, 12%, 25) = $30,000 (0.1275) = $3,825 For B/C ≥ 1, the annual savings in heating and cooling expenses need to be greater than or equal to $3,825. If this savings represents 25% of the total annual expense, the total annual expenditure would have to be in the neighborhood of $3,825/0.25 = $15,300 (compare this with $3,000). This is quite high in most parts of the world, so green homes without subsidies are difficult to justify on economic grounds alone.

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10-3

(a) Sum of benefits minus disbenefits = $705,000 per year B/C ratio = $705,000 / (0.10 × $7,500,000) = 0.94 Thus, the plan is not acceptable because B/C < 1.0 (b) B/C ratio = $705,000 / (A/P, 10%, 20)($7,500,000) = $705,000 / (0.1175)($7,500,000) = 0.8 This plan is still no longer acceptable to the town.

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10-4

Conventional B/C =

240,000($0.15)(P/A, 10%, 8) 192,056  1 150,000  18,000(P/F, 10%, 8) 141,603

The project is acceptable.

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10-5

The present worth of benefits (savings) is PW(16%) = $36,000(1.04)

[1  (P/F, 16%, 8)(F/P, 4%, 8)]  $212,679 0.16  0.04

The present worth of cost is $150,000. The ratio of benefits to cost =

$212,679 = 1.42, so the project is 150,000

still acceptable.

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10-6

The benefit-cost ratio needs to be greater than 1.0 to make this project feasible. In other words, the following relationship must be true: $3,500,000/ $500,000 (P/A, 7%, N) ≥ 1.0 or, $3,500,000/ $500,000 ≥ (P/A, 7%, N) From Table C-10, we see that N ≈ 10 years. This appears to be a sound undertaking.

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10-7

PW (benefits) =

$2,500,000[1  ( P / F ,10%,30)( F / P,2.25%,30)] 0.10  0.0225 = $2,5000,000(11.4619) = $28,654,750

PW (costs) = $17,500,000* + $325,000(P/A,10%,30) + $1,250,000(A/F,10%,5)(P/A,10%,25) = $22,422,258 $28,654,750 B/C = = 1.28 > 1 $22,422,258 Therefore, the toll bridge should be constructed. * The above analysis assumes that the initial surfacing of the bridge is included in the $17,500,000.

Even if this assumption is relaxed, the toll bridge should be constructed.

B/C =

$28,654,750  1.21  1 $22,422,258 $1,250,000

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10-8

Assumption: Initial investment includes initial surfacing of the bridge. (a)

PW (Benefits) =

$3,000,000  $30,000,000 0.10

PW (Costs) = $22,500,000 +

$250,000  $1,000,000 ( A / F,10%,7 )  $1,750,000 ( A / F,10%,20) 0.1

= $26,360,250 CW (10%) = PW (benefits) − PW (costs) = $30,000,000 − $26,360,250 = $3,639,750 $30,000,000  114 . $26,360,250

(b)

B/C =

(c)

Assume repeatability for the initial design B/C =

$30,000,000 ( A / P , 10%,  )  $28,652 ,800 ( A / P , 10%,30) $26,360,250 ( A / P , 10%,  )  $22,422 ,258( A / P ,10%,30)



$40,062 $257,023

 0.16  1

The initial design (described in Problem 10-15) should be selected.

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10-9

(a)

Figures are in thousands.

Project A B C D E

Annual Benefits $1,800 5,600 8,400 2,600 6,600

Annual Costs $2,000 4,200 6,800 2,800 5,400

B/C ratio 0.90 1.33 1.24 0.93 1.22

Projects B,C and E are acceptable. Recall that these projects are independent, (b)

The best project is B, followed closely by C and E. (The instructor might suggest that students rework this problem assuming the projects are mutually exclusive. An incremental analysis then leads to the conclusion that Project C should be chosen.)

(c)

When considerable judgement regarding the value of intangible factors is present, project B should be selected because it allows for a greater degree of error in benefits estimation.

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10-10

(a) Plan A B C

PW of Costs $123,000 135,000 99,000

PW of Benefits $139,000 150,000 114000

B/C ratio 1.13 1.11 1.15

From the above analysis, all three plans (A, B, and C) should be selected. (b)

Reclassifying 10% of costs as disbenefits in the numerator of the B/C ratio produces these results: Plan A B C

PW of Costs $110,700 121,500 89,100

PW of Benefits $126,700 136,500 104,100

B/C ratio 1.14 1.12 1.17

% change +0.9 +0.9 +1.7

(c) A constant amount subtracted from the denominator and numerator of the B/C ratio does not appreciably affect the recomputed ratio.

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10-11 After ordering according to increasing costs, BEACD Alternative or Investment Considered B

E

B/C Ratio $810 = 0.9 (< 1) $900

$1,140 = 1.15 (> 1) $990

Justified?

No

Yes

(A−E)

$1,110  $1,140 = −0.5 (< 1) $1,050  $990

No

(C−E)

$1,390  $1,140 = 1.0471(> 1) $1,230  $990

Yes

(D−C)

$1,500  $1,390 = 0.917 (< 1) $1,350  $1,230

No

Therefore, Alternative C should be adopted.

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10-12

DN → A: B/C =

$80,000 ( P/A, 10%, 4)  $10,000 ( P/G, 10%, 4) $209,812 = 1.31 > 1, so select  $160,000 $160,000

system A vs. DN. B $10,000 ( P/G, 10%, 4) $43,780 = 1.0945. The incremental B-C ratio is greater than   C $40,000 $40,000 one, so keep system C.

A → C:

C → B:

B $40,000 ( P/A, 10%, 4)  $20,000 ( P/G, 10%, 4) $39,236 = 0.87 < 1.   $45,000 $45,000 C

Therefore system C is the best choice.

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10-13

The conventional B/C ratio for Alternative A is: $1,000,000  $500,000 $1,500,000   116 . $20,000,000(A / P,5%,50) + $200,000 $1,296,000 The conventional B/C ratio for Alternative B is: $800,000  $1,300,000 $2,100,000   1.20 $30,000,000(A / P,5%,50) + $100,000 $1,744 ,000 Both alternatives are acceptable. An incremental analysis is required to determine which alternative should be chosen.

Incremental Analysis: (B−A)  Capital investment  Annual O & M costs  Annual power sales  Other annual benefits B/C =

= $10,000,000 = $1,000,000 = −$200,000 = $800,000

$800,000  $200,000 $600,000   1.34  1 $10,000,000(A / P ,5%,50)  $100,000 $448,000

Select Alternative B.

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10-14

(a)

System 2 is inferior to both System 1 and System 3, so it can be dropped from consideration.

System B-C ratio

1 8.0

3 3.5

4 1.6

5 2.0

System 1 has the largest B-C ratio, but it would not be selected for this reason. We must examine the incremental B-C ratios for the four systems. Notice that the systems are rank ordered from low PW of costs to high PW of costs. System 1 is an acceptable alternative to use to start the incremental comparisons. (b)

B-C ratios for: Δ(3 – 1) = $6,000/$3,000 = 2.00. Select 3, drop 1. Δ(4 – 3) = $2,000/$6,000 = 0.33. Keep 3, drop 4. Δ(5 – 3) = $10,000/$8,000 = 1.25. Select System 5.

System 3 has the highest incremental B-C ratio, but it wouldn’t be chosen for that reason.

(c)

The last increment with a B-C ratio greater than or equal to one was System 5. This is the best alternative.

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10-15 With equal annual benefits, we can compare the capitalized worth (CW) of the two projects over an indefinitely long period of time: CW (6%) of proposal one = -$1,350,000 - $110,000 / 0.06 = -3,183,333 CW (6%) of proposal two = -$700,000 - $200,000 (A/F, 6%, 12) / 0.06 -$95,000 / 0.06 - $55,000 (P/F, 6%, 12) / 0.06 = -$700,000 - $200,000 (0.0593) / 0.06 - $95,000 / 0.06 - $55,000 (0.4970) / 0.06 = -$2,936,583.

Therefore to maximize CW, Plan 2 should be recommended because it is less costly.

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10-16 Alternative No Control Levees Small Dam

Total Equivalent Annual Cost $100,000 110,000 105,000

Annual Benefits $ 0 112,000 110,000

B/C Ratio 0.00 1.02 1.05

(a)

Maximum benefit − choose levees.

(b)

Minimum cost − choose no flood control.

(c)

Maximum (B/C) − choose the small dam.

(d)

Largest investment having incremental B/C ratio larger than 1: Rank alternatives by increasing total annual cost: No Control, Small Dam, Levees  (Small Dam − No Control):

B/C =

$110,000  22  1 $5,000

B/C =

$2,000  0.4  1 $5,000

Select the small dam over no control.  (Levees − Small Dam): Choose the small dam. (e) Largest B/C ratio − choose the small dam (which is coincidentally the correct choice). The correct choice based on incremental analysis would be to select the small dam as seen in part (d).

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10-17

B → D: B/C = $1,000/[$3,000(0.12)] = 2.78 > 1, so choose D. D → A: B/C = $1,000/[$7,000(0.12)] = 1.19 > 1, so choose A. A → C: B/C = $10,000/[$88,000(0.12)] = 0.95 < 1, so keep A.

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10-18 CR cost of Verizon = $75,000(A/P,10%,6) − $20,000(A/F,10%,6) = $14,628 CR cost of Cellgene = $175,000(A/P,10%,12) − $37,500(A/F,10%,12) = $23,935 Modified B-C ratio of Verizon = ($28,800 - $9,800)/ $14,628 = 1.3 Modified B-C ratio of Cellgene = ($38,800 − $11,300)/$23,935 = 1.15 Both are acceptable, but we must be careful to avoid the temptation of selecting the tower with the larger B-C ratio. We next examine the incremental difference of Cellgene  Verizon:  [(B - oper. expenses) / CR costs] = $8,500 / $9,307 = 0.91 < 1, So you should recommend the Verizon cell tower.

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10-19 Route* A B C

Construction Cost $185,000 220,000 290,000

Annual Maint.

Cost $2,000 3,000 4,000

Total Annual Benefits $8,500 15,000 20,800

Total PW Costs $209,467 256,701 338,934

Total PW Benefits $103,985 183,503 254,457

B/C Ratio 0.50 0.71 0.75

*Assume that a roadway must be constructed.

Sample calculation for A: Total PW Cost = $185,000 + $2,000 (P/A, 8%, 50) = $209,467 Total PW Benefits = $8,500 (P/A, 8%, 50) = $103,985 B/C Ratio = $103,985 / $209,467 = 0.50 Comparison of routes  (B − A)  (C − B)

B/C 1.68 0.86

Decision Select Alternative B Keep Alternative B

From an incremental analysis, route B is the "least objectionable" alternative. If B is chosen, it simply means that the state will be receiving less than an 8% return on its capital. If another alternative could be identified that had a B/C ratio greater than or equal to one, it should be recommended.

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10-20

(a)

DN → B: $13,000(P/A, 8%, 10)/$50,000 = 1.74; choose B. B → C: $2,300(P/A, 8%, 10)/$15,000 = 1.03; choose C. C → A: $700(P/A, 8%, 10)/$10,000 = 0.47; keep C.

(b)

DN → B: $18,000/[$50,000(A/P, 8%, 10) + $5,000] = 1.44; choose B. B → C: $2,000/[$15,000(A/P, 8%, 10) − $300] = 1.03; choose C. C → A: 0 / difference in costs = 0; keep C.

(c)

Although the incremental B-C values are not identical, the recommendation should be the same.

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Annual Benefits Flood reduction Irrigation Recreation Total Benefits

I. Improve Channel $200,000 ----$200,000

Annual Costs CR Amount O&M Total Costs

I. Improve Channel $293,750 80,000 $373,750

Dam $850,000 50,000 $900,000

Benefit − Cost Ratio

0.54

0.86

10-21

II. Dam $550,000 175,000 45,000 $770,000 II.

III. Dam and Channel $650,000 175,000 45,000 $870,000 III. Dam and Channel $1,143,750 130,000 $1,273,750 0.68

Although all three values are less than one, assume that one of the alternatives has to be chosen (i.e. doing nothing is not an acceptable alternative).

Incremental Analysis: I, II, III (ordered by increasing capital recovery amount)  (II − I) B/C =

$570,000 = 1.08 > 1, Select II. $526,250

$100,000 = 0.27 < 1, Keep II. $373,750 If one option must be selected, recommend building the dam (II).

B/C =

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10-22

(a)

The port manager's analysis is in error due to failure to consider the time value of money.

Whether or not capital would have to be borrowed, it has earning power which must be considered. He also does not consider the cost of capital relative to the recovery of invested capital.

(b)

Assuming capital to be worth 5%, and that piers will be needed indefinitely: AWold (5%) = − $40,000 (A/P, 5%, 40) − $27,000 = −$29,332 AWnew (5%) = − $600,000 (A/P, 5%, 50) − $2,000 = − $34,880 It is more economical to retain the steel pier.

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10-23

One must be chosen (DN not an option). Choose Gravity-fed unless B/C ≥ 1. B/C =

$24,500(P/F, 15%, 5) $12,181 = < 1. Stay with Gravity-fed. $13,400 $37,900  $24,500

Although annual benefits are the same, not having to re-invest in five year is an added benefit of the Vacuum-led.

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10-24

Construct Levee: AW (benefits) = $1,500,000 AW (costs) = $25,000,000 (A/P,8%,25) + $725,000 = $3,067,500 Dredge Channel: AW (benefits) = $0 AW (costs) = $15,000,000 (A/P,8%,25) + $375,000 = $1,780,500  (Levee − Channel) B/C =

$1,500,000  $0  1.17  1 $3,067,500  $1,780,500

Construct the levee.

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10-25

Design A:

$2,150,000  $21,500,000 01 . $12,000(P / A ,10%,34 )(A / P ,10%,35) CW(costs) = $17,000,000 + 0.1 $40,000(A / F,10%,7 ) $3,000,000(A / F,10%,35) + + 0.1 0.1 CW(benefits) =

= $17,272,729 Design B:

$1,900,000  $19,000,000 0.1 $17,500(P / A ,10%,24 )(A / P ,10%,25) CW(costs) = $14,000,000 + 0.1 $40,000(A / F,10%,5) $3,500,000(A / F,10%,25) + + 0.1 0.1

CW(benefits) =

= $14,595,790 Design C:

$1,750,000  $17,500,000 0.1 $20,000(P / A ,10 %,24 )( A / P ,10%,25) CW(costs) = $12,500,000 + 0.1 $40,000(A / F,10%,5) $3,750,000(A / F,10%,25) + + 0.1 0.1 CW(benefits) =

= $13,146,043 Order alternative designs by increasing investment: C, B, A.

 CW (benefits)  CW (costs) B/C Increment justified?

Current best design

C $17,500,000 13,146,043 1.33 Yes C

 (B − C) $1,500,000 1,449,747 1.03 Yes B

 (A − B) $2,500,000 2,676,939 0.93 No B

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10-26 AWB(12%) = $20,000 AWC(12%) = $50,000 (A/P,12%,6) + $6,000 = $18,160 B−C Ratio =

$20,000 = 1.1 $18,160

Select (d)

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10-27 Truck X is acceptable (results of Problem 10-22). Check B−C ratio of incremental investment required for Truck Y.  AWB(12%) = $22,000 − $20,000 = $2,000  AWC(12%) = $64,000 − $50,000) (A/P,12%,6) + ($5,000 − $6,000) = $2,405

 B−C =

$2,000 = 0.83 < 1 $2,405

Select (b) Incremental investment required by Truck Y is not justified. Purchase Truck X.

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10-28 For B−C to equal 1, AWB(12%) = AWC(12%) AWB(12%) = $600,000 (F/P,12%,2.5) (A/F,12%,5) = $126,776 PWB(12%) = $126,776 (P/A,12%,15) = $843,026 Select (c)

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10-29 PWB(12%) = $24,000 (P/A,12%,14)(P/F,12%,2) = $126,816 PWC(12%) = $60,000 + $5,000 (P/A,12%,16) = $94,870 Conventional B−C =

$126,816 = 1.337 $94,870

Select (b)

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10-30 PWB−O&M(12%) = $24,000 (P/A,12%,14) (P/F,12%,2) − $5,000 (P/A,12%,16) = $91,946 PWC(12%) = $60,000 Modified B−C =

$91,946 = 1.53 $60,000

Select (a)

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10-31 The B-C ratio for the gravity-led system is: $106,500 / [$800,000 (A/P, 8%, 30)] = 1.5, so it is acceptable.

Next consider the incremental B-C ratio for (Vacuum-led minus Gravity-fed): Δ B / Δ C = $12,500 / [$200,000 (A/P, 8%, 30)] = 0.7 This is less than 1.0, so the increment cannot be justified.

Therefore the correct choice is answer (a)

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Solutions to Chapter 11 Problems 11-1

The annual capital recovery cost is $28 million (A/P, 6%, 50) = $28 million (0.0634) = $1,775,200.

Operating and maintenance expenses add $4,200,000 to obtain the total annual cost of $5,975,200. The breakeven annual ridership is $5,975,200 per year/$5 per customer = 1,195,040 customers per year. This equates to 1,195,040 customers per year/365 days per year = 3,274 customers per day (on average).

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11-2

The unknown is now the mileage driven each year (instead of fuel cost).

EUACH = $30,000 (A/P, 3%, 5) + ($4.50/gal)(X mi/yr)/(30 mpg) EUACG = $28,000 (A/P, 3%, 5) + ($4.50/gal)(X mi/yr)/(25 mpg) Setting EUACH = EUACG, we find the breakeven mileage to be X = 14,560 miles per year.

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11-3

The annual operating expenses of long-haul tractors equipped with the various deflectors are calculated as a function of mileage driven per year, X: Windshear:

[(X mi/yr)(0.92)(0.2 gal/mi)($4/gal)] = $0.736X / yr

Blowby:

[(X mi/yr)(0.96)(0.2 gal/mi)($ 4/gal)] = $0.768X / yr

Air-vantage: [(X mi/yr)(0.90)(0.2 gal/mi)($ 4/gal)] = $0.720X / yr EUAC can now be written in terms of X. EUACW = $1,200(A/P, 8%, 10) + $14 + $0.736X = $192.80 + $0.736X EUACB = $500(A/P, 8%, 8) + $8 + $0.768X = $95 + $0.768X EUACA = $1,500(A/P, 8%, 5) + $8 + $0.720X = $383.75 + $0.720X The breakeven values can be computed between each pair of deflectors by equating their EUAC equations and solving for X. Windshear and Blowby: X = 3,056 miles per year Blowby and Air-vantage: X = 6,016 miles per year Air-vantage and Windshear: X = 11,934 miles per year The range over which each deflector is preferred is: X ≤ 3,056

Select Blowby

3,056 ≤ X ≤ 11,934

Select Windshear

11,934 ≤ X

Select Air-vantage

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11-4

(a)

Assuming repeatability, we can find the breakeven mileage as follows: ‒$32,000 (A/P, 4%, 5) + $12,000 (A/F, 4%, 5) ‒ $0.48X = ‒$27,000 (A/P, 4%, 4) + $8,000 (A/F, 4%, 4) ‒ $0.36X or, ‒4,972 – 0.48X = ‒5,554.5 – 0.36X or, X = 4,854 miles per year

(b)

EUACGasoline = $(0.48 × 20,000) + 4,972 = $14,572 EUACDiesel = $(0.36 × 20,000) + 5,554.5 = $12,754.5 Recommend the diesel truck

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11-5

(a)

AWA(12%) = −$6,000 (A/P, 12%, 10) + $1,800 − $500 + $1,200 (A/F, 12%, 10) = −$6,000 (0.1770) + $1,300 + $1,200 (0.0570) = $306 AWB(12%) = −$7,500 (A/P, 12%, 12) + $2,250 − $750 + $1,600 (A/F, 12%, 12) = −$7,500 (0.1614) + $1,500 + $1,600 (0.0414) = $356 Therefore, the initial decision is to select Alternative B.

(b)

To determine the capital investment of Alternative A (IB) so that the initial decision would be reversed, equate the AWs: AWA (12%) = AWB (12%) or, −IA (A/P, 12%, 10) + $1,300 + $1,600 (A/F, 12%, 10) = $356 or, −IA(0.1770) + $1,300 + $1,600 (0.0570) = $356 or, −IA = $5,850 The capital investment of Alternative A would have to be $5,850 or less for the initial decision to be reversed.

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11-6

(a) Let X = operating hours per year EUACA (10%) = $2,140 (A/P, 10%, 9) − $60 (A/F, 10%, 9) + (15 hp)(0.746 kW/hp)($0.06/kWh)(X)/(0.6) = $367.09 + $1.119X EUACB (10%) = $4,280 (A/P, 10%, 9) + (10 hp)(0.746 kW/hp)($0.06/kWh)(X)/(0.75) = $743.01 + $0.5968X Setting EUACA = EUACB we find X = 720 hours per year. At annual operating hours greater than 720 (including 2,000), we prefer the more efficient pump, Pump B. This is confirmed by calculating the EUAC with X = 2,000.

EUACA (10%) = $367.09 + $1.119 (2,000) = $2,605; and EUACB (10%) = $743.01 + $0.5968 (2,000) = $1,937 (b) Let A = breakeven efficiency of Pump A at 2,000 operating hours per year. We already know that EUACB (10%) = $1,937 at 2,000 hours per year.

EUACA(10%) = $1,937 = $2,410 (A/P, 10%, 9) − $60 (A/F, 10%, 9) + (15 hp)(0.746 kW/hp)($0.06/kWh)(X)/A Solving, we get A = 85.53%

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11-7

Assume repeatability. Set AWA(10%) = AWB(10%) and solve for breakeven value of X. AWA(10%) = −$5,000 (A/P,10%,5) + $1,500 + $1,900 (A/F,10%,5) = $492,22 AWB(10%) = −X((A/P,10%,7) + $1,400 + $4,00 (A/F,10%,7) = −0.2054X + $1,821.60 $492.22 = −0.2054X + $1,821.60 X = ($1,821.60 − $492.22) / (0.2054) = $6,472.15 At X = 0, AWB(10%) = $1,821.60 > $492.22, so Alt. B is preferred.

At X = $6,500, AWB(10%) = $486.50 < $492.22, so Alt. A is preferred.

Select Alt. B for 0  X  $6,472.15 MARR Polygon Chart of PW 5% vs. MARR 10% 15% 20% 20%

PWA(MARR) 5% 31k 40k 16.3k B 30 5.5k 20 10 −2.8k 0 −1

40 30 20 10 0 −10

PWB(MARR) 41.3k 21k 5.8k −6k

A −10 0 10 20 30 40

10%

−1 0 0 10 20 30 15%

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11-8 Alternative Investment Capital Annual Savings MV (after 4 years)

A 12,000 4,000 3,000

B 15,800 5,200 3,500

MARR 4% 8% 12% 16% 20%

AW(B) 1,671 1,206 730 244 ‐251

AW( C) 1,149 918 680 437 189

AW(A) 1,401 1,043 677 304 ‐77

C 8,000 3,000 1,500

AW(A) AW(B) AW( C)

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11-9

After−tax, A$ Analysis: Let X = annual before−tax revenue requirement.

EOY Investment 0 −$1,166,0001 1 2 3 4 4 441,7412

Annual Revenue X X X X

Annual Expenses

BTCF (A$) −$1,166,000 − $519,7503 −519,750+X − 545,738 −545,738+X − 573,024 −573,024+X − 601,676 −601,676+X 441,741

1

Capital Investment = (55 trucks)($21,200/truck) = $1,166,000 Market Value = MV4 = 0.35($1,166,000)(1.02)4 = $441,741 3 Annual Expenses in year k = (55 trucks)(20,000 mi/truck)($0.45/mi)(1.05)k = $495,000(1.05)k 2

EOY 0 1 2 3 4 4

BTCF (A$) Depr.

TI − $1,166,000 −−− −−− −519,750 + X $388,628 − 908,378 + X −545,738 + X 518,287 −1,064,025 + X −573,024 + X 172,685 − 745,709 + X −601,676 + X 86,401 − 688,077 + X 441,741 −−− 441,741

T(38%) −−− 345,184−0.38X 404,330−0.38X 283,369−0.38X 261,469−0.38X −167,862

ATCF (A$) − $1,166,000 −174,566+0.62X −141,408+0.62X −289,655+0.62X −340,207+0.62X 273,879

PW(15%) = 0 = − $1,166,000 − $174,566(P/F,15%,1) − $141,408 (P/F,15%,2) − $289,655(P/F,15%,3) − $340,207(P/F,15%,4) + $273,879(P/F,15%,4) + 0.62X(P/A,15%,4) 0 = − $1,653,096 + 1.77X Thus, X = $1,653,096/1.77 = $933,953 in annual revenues per year.

Breakeven Point Interpretation: The equivalent uniform annual revenue of $933,953 per year is the breakeven point between signing the contract (and purchasing the trucks, etc.), and not signing the contract (and making no change in current operations).

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11-10 (a)

(b)

 $45   365 days  Annual Revenues = (150 rooms)(0.6)    = $1,478,250  room - day   year  AW(10%) = $1,478,250 − $125,000 − $5,000,000(A/P,10%,15) + (0.2)($5,000,000)(A/F,10%,15) − $1,875,000 (A/P,10%,5) = $232,625 > 0 Yes, the project is economically feasible.

Sensitivity with respect to Decision Reversal Capital Investment: $232,625* − $5,000,000(A/P,10%,15)X = 0 X = 0.3538 or 35.38% * Market value is assumed to remain constant at $1,000,000.

Occupancy Rate:

$232,625 + 45(150)(365)(0.6)X = 0 X = −0.1574 or −15.74%

MARR: (find the IRR and calculate % change) By trial and error, IRR = 14.3%.

Therefore, the MARR must increase by 14.3/10 − 1 = 43 % The decision is most sensitive to changes in occupancy rate (requires the smallest percent change to reverse the decision). (c)

Annual worth is a linear function with respect to capital investment and occupancy rate − we can construct the plot using points from parts (a) and (b). Annual worth is non−linear with respect to the MARR, therefore additional data points are necessary. i 6% 8% 10% 12% 14%

% change −40% −20% 0 20% 40%

AW $436,125 336,363 232,625 125,925 15,863

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11-10 (c) continued

Occupancy Rate

$1,000,000 $800,000

Capital Investment

Annual Worth

$600,000 $400,000 $200,000

MARR

$0 -$200,000 -$400,000 -40

-30

-20

-10

0

10

20

30

40

-$600,000 % Change

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11-11 (a)

Let X = hours per day for new system.

EUACNew = $150,000(A/P,12%,5) − $40,000(A/F,12%,5) + ($50/h)(X)(20 days/mo)(12 mo/yr) EUACUsed = $75,000(A/P,12%,5) − $20,000(A/F,12%,5) + ($50/h)(8 h/day)(20 day/mo)(12 mo/yr) Set EUACNew = EUACUsed and solve for X = 6.76 hours per day. This corresponds to an [(8 – 6.76)/8] × 100% = 15.5% reduction in labor hours.

(b)

If the new system is expected to reduce labor hours by 17%, the new system would be recommended. This conclusion is confirmed by computing the PW of the incremental investment ($65,000) required by the new system, PWΔ = $1,437.

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11-12

Most Likely Estimates:

Capital Investment Market Value Annual Labor Cost MARR

Incremental PW

New $ 150,000 $ $ 40,000 $ $ 76,800 $ 12%

Used 85,000 20,000 96,000 12%

% Change -40% -30% -20% -10% 0% 10% 20% 30% 40%

MARR MV(New) Productivity (New) 15,686 ($3,900) ($18,353) 12,842 ($1,631) ($12,470) 10,150 $639 ($6,588) 7,598 $2,909 ($705) 5,178 $5,178 $5,178 2,882 $7,448 $11,061 700 $9,718 $16,944 (1,373) $11,988 $22,827 (3,345) $14,257 $28,710

$ $ $ $ $ $ $ $ $

$40,000 $30,000 $20,000 MARR

$10,000

MV $‐ ‐40%

‐30%

‐20%

‐10%

0%

10%

20%

30%

40%

Productivity

$(10,000) $(20,000) $(30,000)

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11-13 The entries in the following table are AW values and were generated using the following equation: AW(6%) = −$30,000,000(A/P, 6%, 40) – 300($4,000) + occupancy rate × rental fee × 300

Rental Fee $6,000 $7,000 $8,000 $9,000 $10,000 $11,000 $12,000 $13,000 $14,000 $15,000 $16,000 $17,000 $18,000

75% ($1,843,846) ($1,618,846) ($1,393,846) ($1,168,846) ($943,846) ($718,846) ($493,846) ($268,846) ($43,846) $181,154 $406,154 $631,154 $856,154

Occupancy Rate 80% 85% ($1,753,846) ($1,663,846) ($1,513,846) ($1,408,846) ($1,273,846) ($1,153,846) ($1,033,846) ($898,846) ($793,846) ($643,846) ($553,846) ($388,846) ($313,846) ($133,846) ($73,846) $121,154 $166,154 $376,154 $406,154 $631,154 $646,154 $886,154 $886,154 $1,141,154 $1,126,154 $1,396,154

90% ($1,573,846) ($1,303,846) ($1,033,846) ($763,846) ($493,846) ($223,846) $46,154 $316,154 $586,154 $856,154 $1,126,154 $1,396,154 $1,666,154

95% ($1,483,846) ($1,198,846) ($913,846) ($628,846) ($343,846) ($58,846) $226,154 $511,154 $796,154 $1,081,154 $1,366,154 $1,651,154 $1,936,154

100% ($1,393,846) ($1,093,846) ($793,846) ($493,846) ($193,846) $106,154 $406,154 $706,154 $1,006,154 $1,306,154 $1,606,154 $1,906,154 $2,206,154

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11-14 (a) Analysis at most likely estimates: PW(15%) = −$30,000 + ($20,000 − $5,000) (P/A,15%,5) + $1,000 (P/F,15%,5) = $20,780.20 Sensitivity to changes in capital investment: +5%: PW(15%) = −$30,000(1.05) + ($20,000 − $5,000)(P/A,15%,5) + $1,000(P/F,15%,5) = $19,280.20 −5%: PW*15%) = −$30,000(0.95) + ($20,000 − $5,000)(P/A,15%,5) + $1,000(P/F,15%,5) = $22,280.20 Breakeven percent change: PW(15%) = 0 = −$30,000(1 + x%) + ($20,000 − $5,000)(P/A,15%,5) + $1,000(P/F,15%,5) x = 0.693 or +69.3% increase in capital investment cost Sensitivy to changes in annual expenses: +10%: PW(15%) = −$30,000 + [$20,000 − $5,000(1.1)](P/A,15%,5) + $1,000(P/F,15%,5) = $19,104.10 −10%: PW(15%) = −$30,000 + [$20,000 − $5,000)(0.9)](P/A,15%,5) + $1,000(P/F,15%,5) = $22,456.30 Breakeven percent change: PW(15%) = 0 = −$30,000 + [$20,000 − $5,000(1 + x%)](P/A,15%,5) + $1,000(P/F,15%,5) x = 1.24 or +124% increase in annual expenses Sensitivity to changes in annual revenue: +20%: PW(15%) = −$30,000 + [$20,000(1.2) − $5,000](P/A,15%,5) + $1,000(P/F,15%,5) = $34,189 −20%: PW(15%) = −$30,000 + [$20,000(0.8) − $5,000](P/A,15%,5) + $1,000(P/F,15%,5) = $7,371.40 Breakeven percent change: PW(15%) = 0 = −$30,000 + [$20,000(1 + x%) = $5,000](P/A,15%,5) + $1,000(P/F,15%,5) x = −0.31 or – 31% (decrease in annual revenues) © 2019 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

11-14 (a) continued Sensitivity to changes in market value: +20%: PW(15%) = −$30,000 + ($20,000 − $5,000)(P/A,15%,5) + $1,000(1.2)(P/F,15%,5) = $20,879.64 −20%: PW(15%) = −$30,000 + ($20,000 − $5,000)(P/A,15%,5) + $1,000(0.8)(P/F,15%,5) = $20,680.76 Breakeven percent change: PW(15%) = 0 = −$30,000 + ($20,000 − $5,000)(P/A,15%,5) + $1,000(1 + x%)(P/F,15%,5) x = −41.79 or −417.9% (decrease in market value) Sensitivity to changes in useful life: At +20%, n = 6 PW(15%) = −$30,000 + ($20,000 − $5,000)(P/A,15%,6) + $1,000(P/F,15%,6) = $27,199.80 At −20%, n = 4 PW(15%) = −$30,000 + ($20,000 − $5,000)(P/A,15%,4) + $1,000(P/F,15%,4) = $13,396.80 Breakeven percent change: At n = 3 (−40%): PW(15%) = −$30,000 + ($20,000 − $5,000)(P/A,15%,3) + $1,000(P/F,15%,3) = $4,905.50 At n =2 (−60%): PW(15%) = −$30,000 + ($20,000 − $5,000)(P/A,15%,2) + $1,000(P/F,15%,2) = −$4,858.40 Breakeven life  2.5years, a −50% change

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11-14 (a) continued Recommendation: Proceed with the project. PW is positive for values of factors (changed individually) within stated accuracy ranges. However, if all factors are at their worst values, PW(15%) = −$1,065 [capital investment at +5%, annual expenses at +10%, annual revenues, market value, and useful life at −20%].

Capital Investment

35 PW (Thousands of Dollars)

30

Annual Revenues Useful Life Market Value

$20,780

25 20

Annual Expenses

15 10 5

−60 −50 −40 −30 −20 −10 0 10 20 30 40 50 60 70

Percent Devation Changes from Most Likely Estimate

(b) Factors can be ranked base on the breakeven percent change.

Highest need

Lowest need

Annual Revenues Useful Life Capital Investment Annual Expense Market Value

−31% −50% +69.3% +124% −418%

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11-15 Repair cost = $5,000: PW(i'%) = 0 = −$14,000 + $5,000(P/A,i'%,5) − $5,000(P/F,i'%,3) Using Excel, IRR = 13.4% Repair cost = $7,000: PW(i'%) = 0 = −$14,000 + $5,000(P/A,i'%,5) − $7,000(P/F,i'%,3) Using Excel, IRR = 9.1% Repair cost = $3,000 PW(i'%) = 0 = −$14,000 + $5,000(P/A,i'%,5) − $3,000(P/F,i'%,3) Using Excel, IRR = 17.4% The sensitivity analysis indicates that if the repairs at the end of year 3 cost $5,000 or less, it will be economical to invest in the machine. However, if the repairs cost $7,000, the IRR of the purchase is less than the MARR. A follow−up analysis would be to determine the maximum repair cost that would still result in the desired return of 10%. Let R = repair cost at the end of year 3.

PW(10%) = 0 = − $14,000 + $5,000(P/A,10%,5) − R (P/F,10%,3) 0.7513(R) = $4,954 R = $6,594 As long as the repair cost at the end of year 3 does not exceed $6,594 (which represents a 31.88% increase over the estimated cost of $5,000), the IRR of the purchase will be  10%.

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11-16 For a -30% change in the MARR, the AW(7%) = -$2,534 For a +30% change in the MARR the AW(13%) = -$11,868

-2,000 -2,534 -4,050 -5,582 -7,130 -7,000

7%

8%

9%

10%

11%

12%

13%

MARR -8,694 -10,273 -12,000

-11,868

The AW is not sensitive to changes in the MARR (i.e. the decision in problem 5-20 is not changed).

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11-17

1) Assume salvage value for each system = 0 2) The difference in user cost will be projected as a savings for system #2.

Savings = $0.02/vehicle 3) There are approximately 8,760 hours/year System 1: AW method CR = $32,000(A/P,10%,10) = $5,206.40 Annual Maintenance = $75 Operation Cost =

(28 kW)(8,760 hr/yr)($0.08 / kWh) = $25,157 per year 0.78

AC#1 = $5,206.40 + $75 + $25,157 = $30,438.40 System #2: CR = $45,000(A/P,10%,15) = $5,917.50 Annual Maintenance = $100 Operation Cost =

(34 kW)(8,760 hr/yr)($0.08 / kWh) = $26,475 0.90

Let N = # vehicles using intersection Savings per vehicle = ($0.24 − $0.22) N = $0.02 N AC#2 = $5,917.50 + $100 + $26,475 – 0.2N = $32,492.50 − $0.02N For Breakeven point: $30,438.40 = $32,492.50 − $0.02N and N = 102,705 cars per year For ADT = Average Daily Traffic N=

102,705 = 282 vehicles/day (rounded to next highest integer) 365

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11-18 Let X = annual Btu requirement (in 103 Btu) AWoil (10%) = −$80,000(A/P,10%,20) + $4,000 − ($2.20/140)(X) = −$5,400 − $0.0157X AWgas (10%) = −$60,000(A/P,10%,20) + $6,000 − ($0.04/1)(X) = −$1,050 − $0.04X To find breakeven value of X, set AWoil (10%) = AWgas (10%) and solve for X. −$5,400 − $0.0157X = −$1,050 − $0.04X X = 179,012 or 179,012,000 Btu per year Now let's examine the sensitivity of the decision to changes in the annual Btu requirement. The following table and graph indicate that the conversion to natural gas is preferred if the annual Btu requirement is less than the breakeven point, else the conversion to oil is preferred. % change in X −30 −20 −10 0 10 20 30

Annual Worth Oil Gas −$7,430 −$6,125 −7,720 −6,850 −8,010 −7,575 −8,300 −8,300 −8,590 −9,025 −8,880 −9,750 −9,170 −10,475

$0 - $2,000 - $4,000

AW

Gas

- $6,000

Oil

- $8,000 - $10,000 - $12,000 -30%

-20%

-10%

0%

10%

20%

Percent Change in Annual Btu Requirement

30%

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11-19 (a)

AW(optimistic) = − $90,000(A/P,10%,12) + $35,000 + $30,000(A/F,10%,12) = $23,192 AW(most likely) = −$100,000(A/P,10%,10) + $30,000 + $20,000(A/F,10%,10) = $14,984 AW(pessimistic) = −$120,000(A/P,10%,6) + $20,000 = −$7,552

(b) Useful Life

O M P

Net Annual Cash Flow O M P $21,256 $16,256 $6,256 19,984 14,984 4,984 14,632 9,632 − 368

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11-20 Build 4−lane bridge now: PW(12%) = −$350,000 Build two−lane bridge now: Optimistic: widen bridge to four lanes in 7 years PW(12%) = −$200,000 − [$200,000 + (7)($25,000)](P/F,12%,7) = −$369,613 Most Likely: widen bridge to four lanes in 5 years PW(12%) = −$200,000 − [$200,000 + (5)($25,000)](P/F,12%,5) = −$384,405 Pessimistic: widen bridge to four lanes in 4 years PW(12%) = −$200,000 − [$200,000 + (4)($25,000)](P/F,12%,4) = −$390,650 Recommend building the 4−lane bridge now. In this problem, there is no difficulty in interpreting the results since building the 4−lane bridge now is preferred to a delay in widening the bridge for 7 years (optimistic estimate).

The advantage of pessimistic, most likely, and optimistic estimates is that the uncertainty involved is made explicit. Therefore, the information should be more useful in decision−making. However, when mixed results are obtained, significant judgement is required in reaching a decision. Mixed results would occur in this problem, for example, if the PW of a 7−year delay in widening the bridge were less than $350,000.

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11-21 Let X = miles driven per year.

EUACDart = $13,000(A/P, 10%, 5) + (X/100 mpg)($8/gal) EUACOther = $10,000(A/P, 10%, 5) + (X/50 mpg)($8/gal) Setting EUACDart = EUACOther, we can solve for X = 9,892.5 miles per year. If you are planning on driving 10,000 miles or more per year, the Dart is the most economical vehicle.

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11-22 (a) EOY 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 20

Alternative A: BTCF - $108,000,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 43,200,000

Depr --$5,400,000 10,260,000 9,234,000 8,316,000 7,484,400 6,728,400 6,372,000 6,372,000 6,382,800 6,372,000 6,382,800 6,372,000 6,382,800 6,372,000 6,382,800 3,186,000 0 0 0 0 ---

TI --- $8,860,000 - 13,720,000 - 12,694,000 - 11,776,000 - 10,944,400 - 10,188,400 - 9,832,000 - 9,832,000 - 9,842,800 - 9,832,000 - 9,842,800 - 9,832,000 - 9,842,800 - 9,832,000 - 9,842,800 - 6,646,000 - 3,460,000 - 3,460,000 - 3,460,000 - 3,460,000 43,200,000

T(40%) --$3,544,000 5,488,000 5,077,600 4,710,400 4,377,760 4,075,360 3,932,800 3,932,800 3,937,120 3,932,800 3,937,120 3,932,800 3,937,120 3,932,800 3,937,120 2,658,400 1,384,000 1,384,000 1,384,000 1,384,000 - 17,280,000

ATCF - $108,000,000 84,000 2,028,000 1,617,600 1,250,400 917,760 615,360 472,800 472,800 477,120 472,800 477,120 472,800 477,120 472,800 477,120 - 801,600 - 2,076,000 - 2,076,000 - 2,076,000 - 2,076,000 25,920,000

20

PW(10%) =  ATCFk (P/F,10%, k) = − $99,472,154 k0

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11-22 (a)

continued Alternative B: Compute ATCFs for current estimate of capital investment.

Using the ATCFs shown in the following table: 20

PW(10%) =  ATCFk (P/F,10%, k) = − $79,065,532 k0

ATCFs for Alternative B given original capital investment amount: EOY BTCF 0 - $17,000,000 1 - 12,400,000 2 - 12,400,000 3 - 12,400,000 4 - 12,400,000 5 - 15,400,000 6 - 12,400,000 7 - 12,400,000 8 - 12,400,000 9 - 12,400,000 10 - 15,400,000 11 - 12,400,000 12 - 12,400,000 13 - 12,400,000 14 - 12,400,000 15 - 15,400,000 16 - 12,400,000 17 - 12,400,000 18 - 12,400,000 19 - 12,400,000 20 - 12,400,000 20 0

Depr TI ----$ 850,000 - $13,250,000 1,615,000 - 14,015,000 1,453,500 - 13,853,500 1,309,000 - 13,709,000 1,178,100 - 16,578,100 1,059,100 - 13,459,100 1,003,000 - 13,403,000 1,003,000 - 13,403,000 1,004,700 - 13,404,700 1,003,000 - 16,403,000 1,004,700 - 13,404,700 1,003,000 - 13,403,000 1,004,700 - 13,404,700 1,003,000 - 13,403,000 1,004,700 - 16,404,700 501,500 - 12,901,500 0 - 12,400,000 0 - 12,400,000 0 - 12,400,000 0 - 12,400,000 --0

T(40%) --$5,300,000 5,606,000 5,541,400 5,483,600 6,631,240 5,383,640 5,361,200 5,361,200 5,361,880 6,561,200 5,361,880 5,361,200 5,361,880 5,361,200 6,561,880 5,160,600 4,960,000 4,960,000 4,960,000 4,960,000 0

ATCF - $17,000,000 - 7,100,000 - 6,794,000 - 6,858,600 - 6,916,400 - 8,768,760 - 7,016,360 - 7,038,800 - 7,038,800 - 7,038,120 - 8,838,800 - 7,038,120 - 7,038,800 - 7,038,120 - 7,038,800 - 8,838,120 - 7,239,400 - 7,440,000 - 7,440,000 - 7,440,000 - 7,440,000 0

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11-22 (a)

continued Alternative B (revised to include extra investment permissible to breakeven)

EOY BTCF 0 - $42,731,490 1 - 12,400,000 2 - 12,400,000 3 - 12,400,000 4 - 12,400,000 5 - 15,400,000 6 - 12,400,000 7 - 12,400,000 8 - 12,400,000 9 - 12,400,000 10 - 15,400,000 11 - 12,400,000 12 - 12,400,000 13 - 12,400,000 14 - 12,400,000 15 - 15,400,000 16 - 12,400,000 17 - 12,400,000 18 - 12,400,000 19 - 12,400,000 20 - 12,400,000 20 0

Depr TI ----$2,136,574 - $14,536,574 4,059,492 - 16,459,492 3,653,542 - 16,053,542 3,290,325 - 15,690,325 2,961,292 - 18,361,292 2,662,172 - 15,062,172 2,521,158 - 14,921,158 2,521,158 - 14,921,158 2,525,431 - 14,925,431 2,521,158 - 17,921,158 2,525,431 - 14,925,431 2,521,158 - 14,921,158 2,525,431 - 14,925,431 2,521,158 - 14,921,158 2,525,431 - 17,925,431 1,260,579 - 13,660,579 0 - 12,400,000 0 - 12,400,000 0 - 12,400,000 0 - 12,400,000 --0

T(40%) ATCF --- $42,731,490 $5,814,630 - 6,585,370 6,583,797 - 5,816,203 6,421,417 - 5,978,583 6,276,130 - 6,123,870 7,344,517 - 8,055,483 6,024,869 - 6,375,131 5,968,463 - 6,431,537 5,968,463 - 6,431,537 5,970,172 - 6,429,828 7,168,463 - 8,231,537 5,970,172 - 6,429,828 5,968,463 - 6,431,537 5,970,172 - 6,429,828 5,968,463 - 6,431,537 7,170,172 - 8,229,828 5,464,232 - 6,935,768 4,960,000 - 7,440,000 4,960,000 - 7,440,000 4,960,000 - 7,440,000 4,960,000 - 7,440,000 0 0

The above solution for Alternative B is the result of a trial and error procedure for obtaining identical present worths of ATCFs. Extra Capital = $42,731,490 − $17,000,000 = $25,731,490 This solution takes into account depreciation credits arising from the extra capital that can be invested in Alternative B to breakeven with Alternative A.

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11-22 (b)

Coterminate both alternatives at the end of year 10.

Alternative A: EOY BTCF 0 - $108,000,000 1 - 3,460,000 2 - 3,460,000 3 - 3,460,000 4 - 3,460,000 5 - 3,460,000 6 - 3,460,000 7 - 3,460,000 8 - 3,460,000 9 - 3,460,000 10 - 3,460,000 10 43,200,000

Depr --$5,400,000 10,260,000 9,234,000 8,316,000 7,484,400 6,728,400 6,372,000 6,372,000 6,382,800 3,186,000 ---

TI --- $8,860,000 - 13,720,000 - 12,694,000 - 11,776,000 - 10,944,400 - 10,188,400 - 9,832,000 - 9,832,000 - 9,842,800 - 6,646,000 4,935,600

T(40%) --$3,544,000 5,488,000 5,077,600 4,710,400 4,377,760 4,075,360 3,932,800 3,932,800 3,937,120 2,658,400 - 1,974,240

ATCF - $108,000,000 84,000 2,028,000 1,617,600 1,250,400 917,760 615,360 472,800 472,800 477,120 - 801,600 41,225,760

*MV10 = $43,200,000; BV10 = $38,264,400 10

PW(10%) =

 ATCF (P/F,10%, k) = − $87,010,230 k

k 0

Alternative B: EOY BTCF 0 - $17,000,000 1 - 12,400,000 2 - 12,400,000 3 - 12,400,000 4 - 12,400,000 5 - 15,400,000 6 - 12,400,000 7 - 12,400,000 8 - 12,400,000 9 - 12,400,000 10 - 15,400,000 10 0

Depr TI ----$850,000 - $13,250,000 1,615,000 - 14,015,000 1,453,500 - 13,853,500 1,309,000 - 13,709,000 1,178,100 - 16,578,100 1,059,100 - 13,459,100 1,003,000 - 13,403,000 1,003,000 - 13,403,000 1,004,700 - 13,404,700 501,500 - 15,901,500 --- 6,023,100

T(40%) ATCF --- $17,000,000 $5,300,000 - 7,100,000 5,606,000 - 6,794,000 5,541,400 - 6,858,600 5,483,600 - 6,916,400 6,631,240 - 8,768,760 5,383,640 - 7,016,360 5,361,200 - 7,038,800 5,361,200 - 7,038,800 5,361,880 - 7,038,120 6,360,600 - 9,039,400 2,409,240 2,409,240

*MV10 = 0; BV10 = $6,023,100 10

PW(10%) =

 ATCF (P/F,10%, k) = − $60,788,379 k

k 0

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11-22 (b)

continued If the study period is reduced to 10 years, Alternative B would still be recommended.

Sensitivity of Alternative B to cotermination at EOY 10:   $79,065,532  $60,788,379    100%  23.1% less expensive.  $79,065,532  

(c)

If our annual operating expenses of Alternative B double, the extra present worth of cost in part (a) equals: (−$2.1 million)(1 − 0.4) • (P/A, 10%, 20) = −$10,727,090.

This makes the total present worth of Alternative B equal to: −$79,065,532 − $10,727,090 = −$89,792,622.

Because −$89,792,622 > −$99,472,154, the initial decision to adopt Alternative B is not reversed.

EOY 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 20

BTCF - $17,000,000 - 14,500,000 - 14,500,000 - 14,500,000 - 14,500,000 - 17,500,000 - 14,500,000 - 14,500,000 - 14,500,000 - 14,500,000 - 17,500,000 - 14,500,000 - 14,500,000 - 14,500,000 - 14,500,000 - 17,500,000 - 14,500,000 - 14,500,000 - 14,500,000 - 14,500,000 - 14,500,000 0

Depr --$ 850,000 1,615,000 1,453,500 1,309,000 1,178,100 1,059,100 1,003,000 1,003,000 1,004,700 1,003,000 1,004,700 1,003,000 1,004,700 1,003,000 1,004,700 501,500 0 0 0 0 ---

TI --- $15,350,000 - 16,115,000 - 15,953,500 - 15,809,000 - 18,678,100 - 15,559,100 - 15,503,000 - 15,503,000 - 15,504,700 - 18,503,000 - 15,504,700 - 15,503,000 - 15,504,700 - 15,503,000 - 18,504,700 - 15,001,500 - 14,500,000 - 14,500,000 - 14,500,000 - 14,500,000 0

T (40%) --$6,140,000 6,446,000 6,381,400 6,323,600 7,471,240 6,223,640 6,201,200 6,201,200 6,201,880 7,401,200 6,201,880 6,201,200 6,201,880 6,201,200 7,401,880 6,000,600 5,800,000 5,800,000 5,800,000 5,800,000 0

ATCF - $17,000,000 - 8,360,000 - 8,054,000 - 8,118,600 - 8,176,400 - 10,028,760 - 8,276,360 - 8,298,800 - 8,298,800 - 8,298,120 - 10,098,800 - 8,298,120 - 8,298,800 - 8,298,120 - 8,298,800 - 10,098,120 - 8,499,400 - 8,700,000 - 8,700,000 - 8,700,000 - 8,700,000 0

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11-23 Left as an individual exercise.

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11-24 Single Factor Change: Extra Cost $ / gal Miles/yr

$1,200 $3.00 20,000

Fuel Economy (mpg): Gas engine % Improvement Diesel engine

25 33% 33.25

MARR

14%

AW(fuel savings) Net AW

$ $

MARR

Fuel Economy of Gas Engine (mpg) 24 25 26 12% $287.41 $262.60 $239.69 13% $279.12 $254.31 $231.41 14% $270.76 $245.95 $223.05 15% $262.32 $237.51 $214.61

27 $218.49 $210.20 $201.84 $193.40

595.49 245.95

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11-25 MARR Installation Expense ($/in.)

20%

Annual Tax and Insurance Rate

150

Useful Life Operating Hours per Year

8,760

5%

Cost of Heat Loss ($/Btu)

0.00004

Insulation Thickness (in.)

3 4 5 6 7 8

Heat Loss (Btu/hr) 4,400 3,400 2,800 2,400 2,000 1,800

Installation Expense 450 600 750 900 1050 1200

Change in Cost of Heat Loss ‐50% ‐40% ‐30% ‐20% ‐10% 0% 10% 20% 30% 40% 50%

3 $885.79 $1,039.97 $1,194.14 $1,348.32 $1,502.49 $1,656.67 $1,810.85 $1,965.02 $2,119.20 $2,273.37 $2,427.55

4 $748.89 $868.03 $987.17 $1,106.30 $1,225.44 $1,344.57 $1,463.71 $1,582.85 $1,701.98 $1,821.12 $1,940.25

20

Annual Taxes and Cost of Heat Equiv. Uniform Insurance Removal ($/yr.)

Annual Cost 22.5 1541.76 $1,656.67 30.0 1191.36 $1,344.57 37.5 981.12 $1,172.64 45.0 840.96 $1,070.78 52.5 700.80 $968.92 60.0 630.72 $937.15 Equivalent Uniform Annual Cost 5 6 $682.08 $650.30 $780.19 $734.40 $878.30 $818.49 $976.41 $902.59 $1,074.53 $986.68 $1,172.64 $1,070.78 $1,270.75 $1,154.88 $1,368.86 $1,238.97 $1,466.97 $1,323.07 $1,565.09 $1,407.16 $1,663.20 $1,491.26

7 $618.52 $688.60 $758.68 $828.76 $898.84 $968.92 $1,039.00 $1,109.08 $1,179.16 $1,249.24 $1,319.32

8 $621.79 $684.86 $747.93 $811.00 $874.08 $937.15 $1,000.22 $1,063.29 $1,126.36 $1,189.44 $1,252.51

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11-25 continued MARR Installation Expense ($/in.)

Annual Tax and Insurance Rate

20% $

150 5%

Insulation Thickness (in.)

3 4 5 6 7 8

-50% -40% -30% -20% -10% 0% 10% 20% 30% 40% 50%

Useful Life Operating Hours per Yr. Cost of Heat Loss ($/Btu)

Heat Loss (Btu/hr) 4,400 3,400 2,800 2,400 2,000 1,800

$ $ $ $ $ $ $ $ $ $ $

20 8,760 $

0.00002

Installation Annual Taxes Cost of Heat Equivalent Expense and Insurance Removal ($/yr) Annual Worth $ (450) $ (22.50) $ (770.88) $ (885.79) $ (600) $ (30.00) $ (595.68) $ (748.89) $ (750) $ (37.50) $ (490.56) $ (682.08) $ (900) $ (45.00) $ (420.48) $ (650.30) $ (1,050) $ (52.50) $ (350.40) $ (618.52) $ (1,200) $ (60.00) $ (315.36) $ (621.79)

3 (500.35) $ (577.44) $ (654.53) $ (731.61) $ (808.70) $ (885.79) $ (962.88) $ (1,039.97) $ (1,117.05) $ (1,194.14) $ (1,271.23) $

Equivalent Annual Worth 4 5 6 (451.05) $ (436.80) $ (440.06) $ (510.62) $ (485.85) $ (482.11) $ (570.19) $ (534.91) $ (524.16) $ (629.76) $ (583.97) $ (566.20) $ (689.33) $ (633.02) $ (608.25) $ (748.89) $ (682.08) $ (650.30) $ (808.46) $ (731.13) $ (692.35) $ (868.03) $ (780.19) $ (734.40) $ (927.60) $ (829.25) $ (776.44) $ (987.17) $ (878.30) $ (818.49) $ (1,046.73) $ (927.36) $ (860.54) $

7 (443.32) $ (478.36) $ (513.40) $ (548.44) $ (583.48) $ (618.52) $ (653.56) $ (688.60) $ (723.64) $ (758.68) $ (793.72) $

8 (464.11) (495.64) (527.18) (558.72) (590.25) (621.79) (653.32) (684.86) (716.40) (747.93) (779.47)

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11-26

Capital Investment Useful Life (years) Market Value Net annual cash flow MARR Annual Worth

Useful Life O ML P

O $90,000 12 $30,000 $35,000 11% $22,458

ML P $100,000 $120,000 10 6 $20,000 $0 $30,000 $20,000 11% 11% $14,216 ‐$8,365

12 10 6

Net Annual Cash Flow O ML P $35,000 $30,000 $20,000 $20,478 $15,478 $5,478 $19,216 $14,216 $4,216 $13,890 $8,890 ($1,110)

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11-27 Most Likely Estimates: Investment Market Value Annual Savings MARR Present Worth

$10,000,000 $5,000,000 $2,800,000 15% $852,706

% Change Inv. MV Ann. Sav. ‐40% $4,852,706 ($290,801) ($2,344,870) ‐30% $3,852,706 ($4,924) ($1,545,476) ‐20% $2,852,706 $280,952 ($746,082) ‐10% $1,852,706 $566,829 $53,312 0% $852,706 $852,706 $852,706 10% ($147,294) $1,138,582 $1,652,100 20% ($1,147,294) $1,424,459 $2,451,494 30% ($2,147,294) $1,710,336 $3,250,887 40% ($3,147,294) $1,996,212 $4,050,281

Breakeven Points: Investment Market Value Annual Savings MARR

$10,852,706 $5,000,000 $2,800,000 15%

Present Worth

$0.00

Investment Market Value Annual Savings

$10,000,000 $3,508,613 $2,800,000

MARR

15%

Present Worth

$0

Investment Market Value Annual Savings

$10,000,000 $5,000,000 $2,501,327

MARR

15%

Present Worth

$0

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11-28 Investment Cost (per bulb) Installation Cost (per bulb) Wattage Life (years) Disposal cost (per bulb)

$ $

$

CFL 2.00 3.00 13 3 -

$

ILB 0.50 2.00 60 1 -

Capital Recovery Electricity Disposal

$ $ $

(2,082) -

$ $ $

(2,800) -

Total Annual Cost

$

(2,082)

$

(2,800)

Present Worth

$

(10,341)

$

(13,909)

$ $

This spreadsheet was used to compile the following table of results.

Part a) Table of Results PW(disposal = $1/bulb) PW(disposal = $2/bulb) PW(disposal = $3/bulb) PW(disposal = $4/bulb) PW(disposal = $5/bulb)

$ $ $ $ $

CFL (13,153) (13,557) (13,961) (14,365) (14,769)

$ $ $ $ $

ILB Decision (49,676) CFL (49,676) CFL (49,676) CFL (49,676) CFL (49,676) CFL

Decision Reversal: Disposal Cost / CFL bulb

$

91.43

PW(Electricity = $0.04/kWh) PW(Electricity = $0.06/kWh) PW(Electricity = $0.08/kWh) PW(Electricity = $0.10/kWh) PW(Electricity = $0.12/kWh) PW(Electricity = $0.14/kWh) PW(Electricity = $0.16/kWh)

$ $ $ $ $ $ $

CFL (7,583) (8,875) (10,166) (11,458) (12,750) (14,041) (15,333)

$ $ $ $ $ $ $

ILB Decision (25,832) CFL (31,793) CFL (37,754) CFL (43,715) CFL (49,676) CFL (55,638) CFL (61,599) CFL

Decision Reversal: PW(Electricity = $0 / kWh)

$

(5,000)

$

(13,909)

Part b) Table of Results

CFL

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11-28 continued Part c) Table of Results

Assume Repeatability

AW(CFL life = 5 years) AW(CFL life = 6 years) AW(CFL life = 7 years) AW(CFL life = 8 years) AW(CFL life = 9 years) AW(CFL life = 10 years)

$ $ $ $ $ $

CFL (2,947) (2,776) (2,656) (2,567) (2,498) (2,445)

AW(CFL life = 1 year) AW(CFL life = 2 years) AW(CFL life = 3 years)

$ $ $

(5,600) (2,958) (2,082)

$ $ $ $ $ $

ILB Decision (10,000) CFL (10,000) CFL (10,000) CFL (10,000) CFL (10,000) CFL (10,000) CFL

$ $ $

(2,800) (2,800) (2,800)

ILB ILB CFL

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11-29 Difference in Purchase Difference in Resale Miles Driven / year Cost of Gas/gallon Study Period

$ (5,000) $ 2,000 15,000 4 5

Gasoline Savings

$ 1,095.65

Incremental Cash Flows

Hybrid mpg Gas mpg

Year 0 1 2 3 4 5

$ (5,000) $ 1,095.65 $ 1,095.65 $ 1,095.65 $ 1,095.65 $ 3,095.65

IRR

12.6%

46 25

This spreadsheet was used to compile the following table of results for the posed “what if” questions.

Ownership Period 3 4 5 6 7

IRR 2.4% 8.7% 12.6% 15.1% 16.8%

Resale Difference $ 1,000 $ 2,000 $ 3,000 $ 4,000

IRR 8.4% 12.6% 16.1% 19.2%

Cost of Gasoline $ 3.00 $ 4.00 $ 5.00

IRR 5.7% 12.6% 19.2%

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11-30 Existing Bridge: PWE(12%) = −$1,6000,000 − $20,000(P/A,12%,20) − $70,000[(P/F,12%,5) +(P/F,12%,10) + (P/F,12%,15)] = −$1,824,435 New Bridge: PWN(12%) = −X – [$24,000 + (5)($10,000)](P/A,12%,20) + ($0.25)(4000,000)(P/A,12%,20) = −X + $194,204 Set PWE(12%) = PWN(12%) and solve for X. −$1,824,435 = −X + $194,204 X = $2,018,639 Select (b)

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11-31 X = average number of vehicles per day AW(12%) = 0 = −$117,000/mile(A/P,12%,25) −

(0.03)($117,000) mile

 1,250  710accidents  + (X vehicles/day)(365 days/yr)($1,200/accident)    1,000,000vehicle  miles  X = 77.91 vehicles/day Select (a)

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11-32 AW(12%) = −$8,000(A/P,12%,7) + X($0.50 − $0.26) − $2,000 = 0 X = 15,637 Select (e)

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11-33 AW(12%) = −$16,000(A/P.12%,7) + X($0.50 − $0.16) − $4,000 = 0 X = 22,076 Select (c)

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11-34 AWA(12%) − $8,000(A/P,12%,7) + 35,000($0.50 − $0.26) − $2,000 = $4,647 AWB(12%) = −$16,000(A/F,12%,7) + 35,000($0.50 − $0.16) − $4,000 = $4,394 Select (b) – Install Machine A

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11-35 False

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11-36 False

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11-37 False

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11-38 True

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11-39 False

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11-40 Equate EUACs for both motors and solve for the unknown cost of electricity: $2,500(A/P,12%, 10)+ $400 + [90 hp(0.746kW/hp)(500 hrs./yr.)($X/kW-hr.)]/ 0.74 =$3,200(A/P,12%,10)+ $600 + [90 hp(0.746kW/hp)(500 hrs./yr.)($X/kW-hr.)]/0.89.

By solving for $X/kW-hr. we find that X = $0.0424/kW-hr. Therefore, Phillips is preferred over General Electric when X is greater than or equal $0.0424/kW-hr.and the answer is (d).

Select (d).

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11-41 The capital recovery cost of the plant expansion is: $500,000 (A/P, 18%, 5) = $159,900.

Next, the net annual cash flow from the expansion is computed on the basis of production volume (V) as follows. $2.00V -$150,000 - $159,900 = 0, or 2V = $309,900.

Thus V = 154,950 yards per year.

This breakeven production volume is less than the extra manufacturing capacity, so the company will exceed its profit expectation.

Select (c).

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11-42 Find the annual worth of A: AW (10%) = -$30,000 (A/P, 10%, 5) + $8,000 = $86 Equate the AW of A to the AW of B: $86 = -$25,000 (A/P, 10%,5) + $X $86 = -$6,595 + $X X = $6,681 So if $X > $6,681, Centrifuge B is preferred. Select answer (b).

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Solutions to Chapter 12 Problems 12-1

EUACnothing = (0.01)($1,250,000) = $12,500 $5,0001  ( P/F, 9%, 25)(F/P, 5%, 25)    EUACculvert =  $68,000   (A/P, 9%,25) = $14,649 0.09  0.05  

The EUAC of building the culvert exceeds the EUAC of a mudslide (with no culvert). Building the culvert is not economical.

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12-2

Build the 4−lane bridge now: PW = −$6,200,000 Build 2−lane bridge now, add two lanes later: Years of Expansion k 4 years 5 years 6 years 7 years

PW of Expansion Expenses − [$2,500,000 + (4)($300,000)](P/F, 10%,4) = − $2,527,100 − [$2,500,000 + (5)($300,000)](P/F, 10%,5) = − $2,483,600 − [$2,500,000 + (6)($300,000)](P/F, 10%,6) = − $2,427,350 − [$2,500,000 + (7)($300,000)](P/F, 10%,7) = − $2,360,720

E(PW) = − $4,000,000 − $2,527,100 (0.1) − $2,483,600 (0.2) − $2,427,350 (0.3) − $2,360,720 (0.4) = −$6,421,923 Decision: The four−lane bridge should be built now.

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12-3

Expected revenues = $160,000 (0.7) + $140,000 (0.3) = $154,000 Expected expenses = $75,000 (0.6) + $60,000 (0.4) = $69,000 Expected salvage value = $44,000 (0.5) + $36,000 (0.5) = $40,000 Expected PW (12%) = ‒$180,000 + ($154,000 ‒ $69,000) (P/A, 12%, 16) + $40,000 (P/F, 12%, 16) = ‒$180,000 + $85,000 (6.9740) + $40,000 (0.1631) = ‒$180,000 + $592,790 + $6,524 = $4,19,314 This is a very attractive investment.

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12-4

(a)

340,000 ft3 per day × $9 per thousand ft3 = $3,060 per day (revenue) Profit per year = (365 days/yr)($3,060/ day)(0.82) = $915,858 E(PW) = $915,858 (P/A, 12%, 15)(0.15) − $780,000 = $155,673 ≥ 0, a good investment.

(b)

E(PW) = $915,858 (P/A, 12%, 10)(0.15) − $780,000 = –$3,783 (no more a feasible investment)

(c)

Left to instructor.

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12-5

PW(7%) of the pessimistic estimate is ‒$160,000 + $45,000 (P/A, 7%, 4) = ‒$7,576 PW(7%) of the most likely estimate is ‒$160,000 + $45,000 (P/A, 7%, 6) = $54,493 PW(7%) of the optimistic estimate is ‒$160,000 + $45,000 (P/A, 7%, 9) = $133,184 The expected value of PW (7%) for this project is: E[PW(8%)] = ‒$7,576 (0.25) + $54,493 (0.6) + $133,184 (0.15) = $50,779

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12-6 Hurricane Category 5 4 3 2 1

Annual Cost of Financing the Rebuild (A) $5,082,000 $3,630,000 $2,541,000 $1,452,000 $ 726,000

Expected Annual Property Damage (B) $ 500,000 $ 1,000,000 $ 3,000,000 $ 5,000,000 $10,000,000

Expected Total Annual Cost (A) + (B) $ 5,582,000 $ 4,630,000 $ 5,541,000 $ 6,452,000 $10,726,000

(A) = Capital Investment × (A/P, 6%, 30) (B) = $100,000,000 × Probablility of storm exceeding levee height Therefore, protect the city from a category 4 hurricane.

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12-7

Expected equivalent annual costs given that one main power failure occurs and the backup generator is needed: Capital Recovery Alternative Amount* R −$30,032 S −26,092 T −32,435 *

Annual O&M Annual Cost of Backup Expenses Failure −$5,000 −($400,000)(0.04) = −$16,000 −7,000 −($400,000)(0.05) = −20,000 −4,000 −($400,000)(0.02) = −8,000

Total Annual Cost −$51,032 −53,092 −44,435

From Chapter 5, CR(10%) = −(Capital Invest.)(A/P,10%,10) + (MV)(A/F,10%,10)

To minimize total annual cost, recommend Alternative T. If two main power failures occur per year, the expected equivalent costs become: Alternative R Pr{0 failures} Pr{1 failure } Pr{2 failures}

= (0.96)(0.96) = 2(0.96)(0.04) = (0.04)(0.04)

= 0.9216 = 0.0768 = 0.0016 1.0000

Annual cost of Backup Failure = −$400,000(0.0768) + 2(−$400,000)(0.0016) = −$32,000 Alternative S Pr{0 failures} Pr{1 failure } Pr{2 failures}

= (0.95)(0.95) = 2(0.95)(0.05) = (0.05)(0.05)

= 0.9025 = 0.0950 = 0.0025 1.0000

Annual cost of Backup Failure = −$400,000(0.0950) + 2(−$400,000)(0.0025) = −$40,000 Alternative T Pr{0 failures} Pr{1 failure } Pr{2 failures}

= (0.98)(0.98) = 2(0.98)(0.02) = (0.02)(0.02)

= 0.9604 = 0.0392 = 0.0004 1.0000

Annual cost of Backup Failure = −$400,000(0.0392) + 2(−$400,000)(0.0004) = −$16,000

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12-7

continued

Alternative R S T

Capital Recovery Amount* −$30,032 −26,092 −32,435

Annual O&M Expenses −$5,000 −7,000 −4,000

Annual Cost of Backup Failure * −$32,000 − 40,000 −16,000

Total Annual Cost −$67,032 −73,092 −52,435

*

Note for each alternative, the annual cost of "failure" given 2 occurences per year is twice the cost given 1 occurrence per year. A logical result.

Again, Alternative T is selected. This answer should be obvious since Alternative T is the most reliable alternative.

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12-8

Skiing Annual Days Revenues 80 (500)(80)($10) = $400,000 100 $400,000 + (400)(20)($10) = $480,000 120 $480,000 + (300)(20)($10) = $540,000

Annual Expenses −$1,500(80) = −$120,000 −$1,500(100) = −$150,000 −$1,500(120) = −$180,000

Net Annual Receipts (NAR) $280,000 $330,000 $360,000

E(NAR) = $280,000 (0.60) + $330,000 (0.3) + $360,000 (0.10) = $303,000 i = 25% /yr.; N = 5 yr.; Project cost = $900,000 E(PW) = −$900,000 + $303,000 (P/A, 25%, 5) = −$85,142 < 0 Do not build the lift.

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12-9

Capital Investment = $900,000; Annual O&M expenses = $1,500 per day No. of People/Day Total Days Per Year 80 100 120

First 80 500 500 500

Next 20 − 400 400

Next 20 − − 300

Demand (X) Person−Days Per Year 40,000 48,000 54,000

p(X) 0.6 0.3 0.1

Annual Revenue (R) = $10 (Person−Days/Year) 80−Day Season (40,000 Person−Days) EOY BTCF Depr.

TI 0 -$900,000 ----1 280,000 $ 64,286 $215,714 2 280,000 128,571 151,429 3 280,000 128,571 151,429 4 280,000 128,571 151,429 5 280,000 128,571 151,429 6 280,000 128,571 151,429 7 280,000 128,571 151,429 8 280,000 64,286 215,714 PW80 (15%)

T(40%) ---$86,286 -60,571 -60,571 -60,571 -60,571 -60,571 -60,571 -86,286

ATCF -$900,000 193,714 219,429 219,429 219,429 219,429 219,429 219,429 193,714

= −$900,000+[$193,714 + $219,429 (P/A,15% 6)](P/F,15% 1) + $193,714 (P/F,15%, 8) = $53,920

100−Day Season (48,000 Person−Days) EOY BTCF Depr.

TI T(40%) 0 -$900,000 ------1 330,000 $ 64,286 $265,714 -$106,286 2 330,000 128,571 201,429 -80,571 3 330,000 128,571 201,429 -80,571 4 330,000 128,571 201,429 -80,571 5 330,000 128,571 201,429 -80,571 6 330,000 128,571 201,429 -80,571 7 330,000 128,571 201,429 -80,571 8 330,000 64,286 265,714 -106,286

ATCF -$900,000 223,714 249,429 249,429 249,429 249,429 249,429 249,429 223,714

PW100 (15%) = −$900,000+[$223,714 + $249,429(P/A, 15%, 6)](P/F, 15%, 1) +$223,714 (P/F, 15%, 8) = $188,545 © 2019 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12-9

continued 120−Day Season (54,000 Person−Days) EOY BTCF Depr.

TI T(40%) 0 -$900,000 ------1 360,000 $ 64,286 $295,714 -$118,286 2 360,000 128,571 231,429 -92,571 3 360,000 128,571 231,429 -92,571 4 360,000 128,571 231,429 -92,571 5 360,000 128,571 231,429 -92,571 6 360,000 128,571 231,429 -92,571 7 360,000 128,571 231,429 -92,571 8 360,000 64,286 295,714 -118,286

ATCF -$900,000 241,714 267,429 267,429 267,429 267,429 267,429 267,429 241,714

PW120 (15%) = −$900,000+[$241,714 + $267,429(P/A,15%, 6)](P/F,15,% 1) +$241,714(P/F,15%, 8) = $269,320 Days of Skiing (X) p(X) PW(ATCF) E[PW(X)] [PW(X)]2 p(X)[PW(X)]2 80 0.6 53,920 32,352 2,907,366,400 1,744,419,840 100 0.3 188,545 56,564 35,549,217,025 10,664,765,108 120 0.1 269,320 26,932 72,533,262,400 7,253,326,240 E(PW) = $115,848 E[(PW)2] = 19,662,511,188 ($)2

E(PW) =  [p(X) PW(ATCF)] = $115,848 V(PW) =  {p(X) [PW(X)]2} − [E(PW)]2 = 19,662,511,188 − (115,848)2 = 6,241,752,084 ($)2 SD(PW) = [V(PW)]1/2 = $79,005 Recommend that the lift be installed since E(PW) = $115,848 > 0 and one SD ($79,005) is only 68% of E(PW).

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12-10 Depreciation SL amount =

0.8($100,000) = $20,000 4

ADS: d1 = d5 = $20,000/2 = $10,000 d2 = d3 = d4 = $20,000 After−tax analysis Let $A1,L = cost savings in first year (before−taxes) for performance level L. PWAT(12%) = −$100,000 + (0.4)(0.2)($100,000) $ A [1  ( P / F ,12%,5)( F / P,6%,5)] + (1 − 0.4) 1, L 0.12  0.06 + 0.4($20,000)(P/A,12%,3)(P/F,12%,1) + 0.4($10,000)[(P/F,12%,1) + (P/F,12%,5)] = −$92,000 + 2.407($A1,L) + $17,157 + $5,841 = 2.407($A1,L) − $69,002, L = 1, 2, 3, 4 Performance Level (L) 1 2 3 4

p(L) 0.15 0.25 0.35 0.25

PWAT(12%) E(PWAT) − $14,854 − $2,227 15,243 3,811 37,387 13,085 74,937 18,734 Total: $33,403

E(PWAT) = $33,403; implement the project.

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12-11 (a)

PW(N) = −$418,000 + $148,000 (P/A, 15%, N) N 3 4 5 6 7 8

PW(N) −$80,086 4,540 78,126 142,106 197,739 246,120

In Millions p(N) PW(N) • p(N) [PW(N)] 2 [PW(N)]2 • p(N) 0.1 −$8,009 6,413.77 641.38 0.1 454 20.61 2.06 0.2 15,625 6,103.67 1,220.73 0.3 42,632 20,194.12 6,058.24 0.2 39,548 39,100.71 7,820.14 0.1 24,612 60,575.05 6,057.57 2 E(PW) = $114,862 E[(PW) ] = 21,800.06106($)2

V(PW) = 21,800.06106 − (114,862)2 = 8,606.78106 ($)2 SD(PW) = (8,606.78106 )1/2 = $92,773 (b)

Pr{PW < 0} = 0.1

(From work table above)

Recommend purchase of equipment: E(PW) = $114, 862 is favorable; SD(PW) = $92,773 is less than the E(PW); and Pr{PW<0} = 0.1 is low.

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12-12 (a) j 1 2 3 4 5 6 7 8 9 j 1 2 3 4 5 6 7 8 9

0 −$29,000 −29,000 −29,000 −29,000 −29,000 −29,000 −29,000 −29,000 −29,000 PW(j) −$10,551 −9,417 −6,392 −3,443 35 2,606 4,459 8,315 10,962

EOY Net Cash Flow 1 $6,000 6,000 6,000 12,000 12,000 12,000 19,000 19,000 19,000 p(j) PW(j) • p(j) 0.02 −$211 0.04 −377 0.14 −895 0.12 −413 0.30 11 0.18 469 0.06 268 0.08 665 0.06 658 E(PW) = $175

2 $17,500 19,000 23,000 20,000 24,600 28,000 22,400 27,500 31,000

PW(j) −$10,551 −9,417 −6,392 −3,443 35 2,606 4,459 8,315 10,962

In Millions [PW(j)]2 [PW(j)]2 • p(j) 111.32 2.23 88.67 3.55 40.86 5.72 11.85 1.42 − − 6.79 1.22 19.88 1.19 69.14 5.53 120.15 7.21 2 E[(PW) ] = 28.07  106 ($)2

V(PW) = 28.07 106 − (175)2 = 28.04 106 ($)2 SD(PW) = (28.04 106 )1/2 = $5,295 (b) Pr{PW > 0} = 0.30 + 0.18 +0.06 + 0.08 + 0.06 = 0.68

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12-13 (a)

im = MARR = 15%; General inflation rate (f) = 4% Increase rate for annual revenues = 6.48% $40,000[1  ( P / F ,15%, N )( F / P,6.48%, N ) 0.15  0.0648 In Millions 2 p(N) PW(N) • p(N) [PW(N)] [PW(N)]2 • p(N) 0.03 −$1,957 $4,253.39 127.60 0.10 −3,301 1,089.79 108.98 0.30 −958 10.19 3.06 0.30 7,325 596.24 178.87 0.17 8,497 2,498.30 424.71 0.10 7,365 5,424.91 542.49 2 E(PW)=$16,971 E[(PW) ] = 1,385.71  106 ($)2

PW(N) = −$100,000 + N 1 2 3 4 5 6

PW(N) −$65,218 −33,012 −3,192 24,418 49,983 73,654

V(PW) = 1,385.71  106 − (16,971)2 = 1,097.7  106 ($)2 SD(PW) = (1,097.7  106)1/2 = $33,131 (b)

Pr(PW > 0} = 0.30 + 0.17 + 0.10 = 0.57

(c)

ir =

015 .  0.04 = 0.10577 or 10.577% 1.04 6

E(AW)R$ =

 PW(N ) (A/P,10.577%,N) p(N) = $1,866 N 1

The project appears questionable. The E(PW) is positive but the SD(PW) is approximately two times the expected value. Also the Pr{PW > 0} = 0.57 is only somewhat attractive.

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12-14

Notice that the probable extra capital investment for project B is a negative consideration to the selection of this project. But let's examine the expected value and variance of cash inflows for both projects to see if they might compensate for the higher capital investment for project B. The expected cash inflow for project A is $1,840 per year, while the expected cash inflow for project B is $1,670.

The variance of project A is 86,400 $2 and the variance of project B is 1,374,100 $2, so project A has a greater expected value of cash inflows and smaller variance of cash inflows than project B. Project A appears to be the clear choice (it probably has a lower capital investment too). Note that in this problem, risk and reward do not travel in the same direction. As the risk (variance) of project B goes up, its reward (expected annual cash flow becomes lower.

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12-15

From the normal probability tables (App. E): 1.28 =

X 



=

30,000  25,000



=

5,000



so  − 3,906.25 pounds per hour and the variance =  2 = 15.26 x 106 (lbs/hr)2

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12-16 Normally distributed random variable: E(X) = $175, V(X) = 25 ($)2 Pr{X  171} = ?

Z=

X





171  175 = −0.8 25

Pr{X  171} = Pr{Z  −0.8} = 1 − Pr{Z  −0.8} = 1 − 0.2119 = 0.7881

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12-17 (a)

Ck E(Fk) (P/F,15%,k) k E(Xk) 0 −$41,167 1 −$41,167 1.0000 1 − 2,208 2 − 4,416 0.8696 2 10,600 1 10,600 0.7561 3 6,067 4 24,268 0.6575 4 4,817 5 24,085 0.5718 5 17,333 1 17,333 0.4972 2 k V(Xk) [Ck(P/F)] V[PW(Fk)] 0 1,361,111 1.0000 1,361,111 1 11,736 3.0248 35,499 2 71,111 0.5717 40,653 3 17,778 6.9169 122,969 4 6,944 18.1739 56,759 5 90,000 0.2472 22,249

PW[E(Fk)] −$41,167 − 3,840 8,015 15,956 13,772 8,618

5

E(PW) =

 PW[E(F )] = $1,354 k

k0 5

V(PW) =

 V[PW(F )] = 1,639,240 ($) k

2

k0

(b)

Assumption: The PW of the net cash flow is a normally distributed random variable with  = E(PW) and 2 = V(PW).  0  $1,354  Pr{PW  0} = 1 − Pr{PW  0} = 1 − Pr Z   1,639,240   = 1 − Pr{Z  −1.06} = 1 − 0.1446 = 0.8554

(c) Yes; if PW (at i = MARR) > 0 then the IRR > MARR. Therefore, it is correct to conclude that Pr{IRR  MARR} = Pr{PW  0}.

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12-18 Notice that these are cost alternatives.

E[PWA(15%)] = −[13,000 + 5,000 (P/A,15%,8) – 2,000 (P/F,15%,8)] = −$34,783 (500) 2 ( P / A,15%,16) + (800)2 (P/F,15%,16) = 770,746 $2 V[PWA(15%)] = 2.15 E[PWB(15%)] = −[7,500 (P/A,15%,8)] = −$33,655 (750) 2 ( P / A,15%,16) = 1,557,794 $2 2.15 Now let Y = PWA – PWB (i.e., B → A) and find Pr(Y > 0).

V[PWB(15%)] =

E(Y) = E(PWA) – E(PWB) = −1,128 V(Y) = V(PWA) + V(PWB) = 2,318,540  0  (1,128)   Pr(Y  0) = Pr  S   2 , 318 , 540  

= Pr(S  0.741) = Pr(S  −0.741)  0.23

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12-19 EOY 0 1 2 3 4

E(B – A) −$4,000 $500 −$1,500 $500 −$1,500

V (B + A) (500)2 = 250,000 (300)2 + (600)2 = 450,000 (300)2 + (600)2 = 450,000 (300)2 + (800)2 = 730,000 (300)2 + (800)2 = 730,000

At i = 15% E[PW(B – A)] = −4,000 + 500 (P/F,15%,1) – 1,500 (P/F,15%,2) + 500 (P/F,15%,3) −1,500 (P/F,15%,4) = −$5,228 V[PW(B − A)] = 250,000 + 450,00 (P/F,15%,2) + 450,000 (P/F,15%,4) + 730,000 (P/F,15%,6) + 730,000 (P/F,15%,8) = 1,401,791

 (PWA→B) = $1,183.97

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12-20 (a) Random Number

13 51 35 90

Life

RN [3 + (5−3)] 100 3.26 4.02 3.7 4.8

Nearest Whole Number

Investment −$120,000(A/P,15%.N )

Salvage SV(A/F,15%,N)

Equivalent AW

3 4 4 5

−$52,560 −$42.036 −$42,036 −$35,796

5,184 2,404 2.404 890 SUM  =

−$47,376 −$39,632 −$39,632 −$34,906 −$161,546 −$40,387

($47,376   ) 2  ($39,632   ) 2  ($39,632   ) 2  ($34,906   ) 2 3 with  = −$40,387

(b) Variance =

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12-21

Monte Carlo Simulation for the Proposed Project Random Normal Deviate

Three Random Numbers

Project Life, N = 5+ RN/999(15−5)

N to Nearest Integer

One Random Number

Annual Receipts

AW [ P(A/P, 10%, N) + A + MV(A/F, 10%, N]

RND

Investment 200,000 + RND * 10,000 P

RN

N

N

R

A

AW

1.102 0.148 2.372 −0.145 0.104 1.419 0.069 0.797 −0.393 −0.874

211,020 201,480 223,720 198,550 201,040 214,190 200,690 207,970 196,070 191,260

131 513 350 904 440 107 507 782 258 504

6.311311 10.13514 8.503504 14.04905 9.404404 6.071071 10.07508 12.82783 7.582583 10.04505

6 10 9 14 9 6 10 13 8 10

4 5 4 7 9 4 5 6 8 3

16,000 16,000 16,000 16,000 22,000 16,000 16,000 16,000 22,000 16,000 SUM

−34,450.192 −187,80.796 −24,837.792 −12,943.235 −14,900.544 −35,178.024 −18,652.263 −15,282.176 −16,743.518 −17,118.002 −208,886.54

The estimate of AW based on ten repititions of the experiment is −$208,886.54.

An estimate of the standard deviation of AW can be found by: K

 [ AW ] 

 ( AW  E[W ]) i 1

2

i

K 1

 = 7,992.784 Due to reasons of space only 10 of the simulation trials are presented here. It can be noticed that these estimates are within the range of the real value and will get even closer with more trials.

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12-22 Although the present worth of Alternative 2 is higher than the present worth of Alternative 1, it may be wise to select Alternative 1 since the probability of present worth being greater than zero for Alternative 1 is greater and there is less variability in present worth for Alternative 1.

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12-23 Given: MARR = 10% per year; analysis period = 8 years PWNew Product = −$1,000,000(P/F, 10%, 2) + [0.6($200,000) + 0.4($280,000)](P/A, 10%, 6)(P/F, 10%, 2) = $8,619 PWDo Nothing = $0; Select new product.

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12-24 E[PW(15%)] = −300k + 100k (P/A,15%,N) + 20k (P/F,15%,N) (7 k ) 2 ( P / A,15%,2 N ) + (3k)2 (P/F,15%,2N) V[PW(15%)] = 0 + 2.15 Pr (IRR  15%)  0.90 or Pr(PW  0 | i = 15%)  0.90 Step 1: Try N = 4 years, E[PW(15%)] = −300k + 285.5k + 11.4k = −3.1k V[PW(15%)] =

(49 x106 )( P / A,15%,8) + (9x106) (P/F,15%,8) 2.15

= (102.27 x 106) + (2.94 x 106) = 105.21 x 106 ; (  = $10,257)

 0  (3.1k )  Pr (PW  0 | i = 15%) = Pr  S   10.257k   = Pr (S ≥ 0.30)  0.38 Step 2: Try N = 5 years, E[PW(15%)] = −300 + 335,2 + 9.9 = 45.1k V[PW(15%)] =

(49 x106 )( P / A,15%,10) + (9x106) (P/F,15%,10) 2.15

= (114.38 x 106) + (2.22 x 106) = 116.6 x 106 ; (  = $10,798)

 0  (45.1k )  Pr (PW  0 | i = 15%) = Pr  S   10.798k   = Pr (S ≥ −4.18)  0.99 Hence, N = 5 years is the smallest value permissible.

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12-25

: PW(10)%) for BA = −250k (P/A,10%,5)(0.75)

Start at

−150k (P/A,10%,5)(0.25) – 5,500k + 2,000k (P/F,10%,5) = −5,111,14k Therefore, select RA If at

2

PW(10%) for RA = −1,000k (P/A,10%,5)(0.60) −1,200k (P/A,10%,5)(0.50) – 25k = −4,119.06k

Bottom branch at

1

: PWBUILD(10%) = [−200k(P/A,10%,5) −4,119.06k (P/F,10%,50}(0.2) – 150k (P/A,10%,10)(0.6) −300k (P/A,10%,10)(0.2) – 10,500k + 17,00k (P/F,10%,10) = −$5,531.33k

Top branch at

1

: PWRENT(10%) = −900k [1−(P/F,10%,10)(F/P,7%,10)](0.55) −700k [10(P/F,10%,1)](0.45) – 25k = −$6,875,73k

Therefore, choose to build.

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12-26 A B C D E F 1 MARR = 12% Useful Life Probability Capital 2 Investment = $ 521,000 14 0.3 Annual 3 Savings = $ 48,600 15 0.4 Increased 4 Revenue = $ 31,000 16 0.3 5 6 7 Useful Life prob (N) PW E(PW) PW^2 p(N)[PW]^2 8 14 0.3 $ 6,602 $ 1,981 4.359E+07 1.308E+07 9 15 0.4 $ 21,145 $ 8,458 4.471E+08 1.788E+08 10 16 0.3 $ 34,129 $ 10,239 1.165E+09 3.494E+08 11 Totals = $ 20,677 $ 541,360,618 12 13 E(PW) $ 20,677 14 V(PW) = $ 113,806,908 15 SD(PW) = $ 10,668 16 17 18 The revised estimates for useful life have resulted in an increased E (PW) and a reduced SD(PW).

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12-27

Left as an individual exercise.

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Solutions to Chapter 13 Problems 13-1

NOPAT = (1 – 0.3)($14 million − $8 million − $2.6 million) = $2.38 million

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13-2

ea = 0.0225 + 1.82(0.084) = 0.1754 or 17.54% Note: Considering the long-term average market premium of 8.4%. Refer to Paragraph 13.3.1.

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13-3

Recommendations differ for Projects A and C. These differences occur because the WACC is indifferent to market volatility, whereas the CAPM strives to take market risk into account. This is the basic difference shown by the slope of the CAPM line in the diagram on page 548.

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13-4

Rs = 0.065 + 0.048 = 0.113 = 11.3%

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13-5

In this case, (P/A, i′, ) = 12, so 1/i′ = 12 and i′ = 0.0833. The after-tax return on this stock is 8.33%.

With a higher P/E ratio, the implied IRR will drop because investors are more concerned about growth in the price of the common stock and are willing to trade earnings for growth.

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13-6

Refer to Equation (13-2) to determine the cost of this equity.

RS – 0.02 = 1.6 (0.093 – 0.02) or,

RS = 0.02 + 1.6 (0.073)

or,

RS = 0.1368 per year (13.68%)

The investment amount of $44 million needs to return at least 13.68% per year to be financially viable.

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13-7 (a) Keep Old: Capital recovery = ($6,000−$1,000)(NP,15%,3) + $1,800(0.15) = $2,110 Operating Disbursements Total Annual Cost =

720 −−−−−−−−−−−−− $2,830

Lease: 30 days($30) + 3,000 miles ($0.40) = $2,100 Thus, leasing a truck is better. (b) The annual cost of having to operate without a truck, $2,000, is less than the minimum cost alternative in (a). Hence, it is better to operate without a truck.

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13-8 (a) Keep Old Yr. 0 1−3 3

BTCF −$6,000 −720 +1,800

Depr. − −$1,000 −

TI (−)+$1,000(1) −1,720 −200(2)

T (t=0.40) (+)−$400 +688 +80

ATCF −$5,600 −32 +1,880

Lease Yr. 1−3

BTCF −$2,100

Depr. −

TI (−)$2,100

T (t=0.40) (+)$840

ATCF −$1,260

(1) Cap. gain if we sell = $6,000 − $5,000 = $1,000 (2) Cap. loss if we sell = [$5,000 − 3($1,000) − $1,800 = −$200 PW(Keep Old): −$5,600 − $32(P/A,5%,3) + $1 ,880(P/F,5%,3) = −$4,063 PW(Lease): −$1,260(P/A,5%,3) = −$3,431 Thus leasing is better. (b) Presumably the extra cost of having to operate without a truck would be tax deductible.

Thus, the extra cost after income taxes would be $2,000 − 40%($2,000) = $1 ,200/yr.

PW(without truck) = −$1 ,200(P/A,5%,3) = −$3,268 Thus, operating without a truck is still more economical.

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13-9 (a)

Purchase: Cash Inflows = $20,000 CR Cost = ($56,000−$10,000)(AIP,10%,3) + $10,000(0.10) = $19,497 −−−−−−−−−−−−−− Net AW $ 503 Lease: Cash Inflows = $20,000 Lease Payments (converted to end−of−year) = $24,200 −−−−−−−−−−−−− Net AW −$4,200 It is better to purchase the lathe.

(b)

If the value of the lathe (as measured by cash inflows) is only $18,000, the lathe should not be acquired by purchase or lease.

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13-10 Buy

If (a) If (b)

Based on the use of equity money of the firm.

Yr.

BTCF

Depr.

0 0 1−20 −$ 100,000 −$100,000 20 − 2,000,000 20 − 1,500,000

Taxable Inc.

t=0.40 Income Taxes

−$200,000

+$ 80,000

+ 500,000*

− 200,000

Depr.

Taxable Inc.

−−

−$125.000

t=0.40 Income Taxes +$50,000

ATCF 0 −$ 20,000 − 2,000,000 − 1,700,000

*Capital gain = $2,000,000 − $1,500,000 = $500,000 Lease

Yr.

BTCF

1−20 −$125,000

ATCF −$75,000

Annual Worths Buy and (a) $0 salvage: −$20,000−$2,000,000(A/F,5%,20) = −$80.400 < −$75.000 Therefore, Lease.

Buy and (b) $500,000 salvage: −$20,000 − $1,700,000(A/F,5%,20) = −$71.340> −$75,000 Thus leasing is better for case a, but buying is better for case b.

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13-11 (1)

Yr.

BTCF

Depr.

Taxable Income

t = 0.25 Taxes

ATCF

Beginning Lease

Purchase

Of each Year

−$ 35,000

−$ 35,000 +$ 8,750 −$ 26,250

0

−$100,000

−$100,000

1−5

−$1,000 −$20,000(1) −$21,000

5

+$5,250

+ 4,250

0

+0

Depr. = ($100,000−$0)/5 = $20,000 Annual Worths Lease: (Adjusted to end−of−year) =−$26,250(F/P,1 0%,1) = −$28,875 Purchase: − $100,000(A/P, 1 0%,5) +$4,250 = −$22,130 Thus, purchasing is better if the life is 5 years.

To illustrate the iterative calculations to find the breakeven life, here are the figures for a 3−year life Purchase Last 3 Years

0 1−3 3

−$100,000 −$1,000 0

−− −$20,000 −−

−− −$21,000 −$40,000(2)

−− +$5,250 +$10,000

−− +$4,250 +$10,000

(2) Capital Loss = Book Value − Selling Price = [$100,000−3($20,000)]−$0 = $40,000 Annual Worths Lease: (see above) = −$28,875 Purchase, last 3 yrs .= − $100,000(A/P,10%,3) + $4,250 + $10,000(A/F, 1 0%,3) = −$32,939 Thus, leasing is better if the life is 3 years. It appears that a breakeven life is 4 years (to the nearest whole year).

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13-12

A B C

EOY

Capital Annual Investment Net Cash Income

PW(10%)

A1 A2 A3 B1 B2 C1 C2

−500000 −650,000 −700,000 −600,000 −675,000 −800,000 −1,000,000

53014 25,906 6,629 45,183 13,195 121,690 75,305

90000 110,000 115,000 105,000 112,000 150,000 175,000

Maximize: PW = 53,014XA1 + 25,906XA2+ 6629XA3 +45,183XB1 + 13,195XB2 + 121,690XC1 + 75,305XC2 Subject to: 500,000 XA1 + 650,000 XA2 + 700000 XA3 + 600,000 XB1 + 675,000 XB2 +800,000XC1 + 1,000,000XC2 ≤ 2,100,000 XA1 + XA2 + XA3 ≤ 1 XB1 + XB2 ≤ 1 XC1 + XC2 ≤ 1 XA1 + XA2 XA3 + XB1 XB2 + XC1 XC2 = 0 or 1 Solution: XA1 = 1; XB1 = 1; XC1 = 1 Objective function value = $219.89

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13-13 MEC 1 2 3 4 5 6 7

A 0 1 0 1 1 0 0

Proposal B 0 0 0 1 0 1 0

C 0 0 1 0 0 1 1

D 0 0 0 0 1 0 1

Investment 0 $100,000 $120,000 $120,000 $130,000 $140,000 $150,000*

*MEC #7 is not feasible due to budget limitations For the objective function coefficients PWA(15%)= −$100,000 + $40,000(P/A, 15%,3) + $20,000(P/F, 15%,3) = $4,478 PWB(15%) = −20,000 + $6,000(P/F, 15%, 1) + $10,000(P/A, 15%,2)(P/F, 15%, 1) = −$645 PWC(15%) = −120,000 + $25,000(P/F, 15%, 1) + $50,000(P/F, 15%,2) + 85,000(P/F,15%,3) = −$4,729 Integer L.P. Setup: Maximize PW = 4,478XA − 645XB − 4,568XC − 4,729XD Subject to:

XA + XC ≤ 1 XB + XD ≤ 1 100,000XA+ 20,000XB + 120,000XC + 30,000XD ≤ 140,000 XA , XB ,Xc, XD = 0 or 1

Solving yields: XA = 1, XB = 0, Xc =0 and XD = 0 Objective function value = $4,478

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13-14

Maximize PW = 0.12 XA + 2.47 XB+ 1.85 Xc Subject to 4 XA + 4.5 XB + Xc ≤ 5 XA+ XB ≤ 1 Xc ≤ XA 7,000 XA + 9,000 XB + 3,000 Xc ≤ 10,000 XA, XB, Xc = 0 or 1 Solving yields: XA = 0, XB =1, XC = 0 Objective function = $2.47

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Solutions to Chapter 14 Problems 14-1

Left to student.

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14-2

This is left to the individual student. The problem can be expanded to include more attributes, such as “early career earnings” and/or “reputation of the university.”

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14-3

Left to student.

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14-4

Left to individual student.

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14-5

(a)

Assume “ideal” means “maximum” Satisficing For Attribute I: Only alternatives A and B are acceptable: 60 ≤ I ≤ 100 For Attribute II: All three alternatives are acceptable: 5 ≤ II ≤ 10 For Attribute III: All three alternatives are acceptable: 6 ≤ II ≤ 10 For Attribute IV: Only alternatives A and B are acceptable: Fair ≤ IV ≤ Excellent

(b)

Dominance Alternative A dominates both Alternative B and Alternative C. Thus, Alternative A gets selected.

(c)

Lexicography Using the top ranked Attribute II: indifferent between Alternative A and Alternative C. Then, using the next preferred Attribute IV: Alternative A gets selected.

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14-6

(a)

Dominance

Vendor I vs Vendor II vs Attribute Vendor II Vendor III Vendor II vs Retain Vendor III vs Retain A Worse Better Better Better B Equal Better Better Better C Worse Worse Better Better D Worse Worse Better Better E Worse Worse Worse Worse Dominance?

Yes No No No Vendor I is removed from consideration. (b)

Satisficing Attribute A B C D E

Minimum Acceptable Value 60% Good Fair Fair $400,000

Unacceptable Alternative None Retain Retain Retain None

Remove ‘Retain Existing System’ from consideration. (c)

Disjunctive Resolution Two alternatives, namely Vendor II and Vendor III are available (Vendor I and ‘Retain’ have been eliminated) as these alternatives are acceptable in atleast one attribute.

(d)

Lexicography

Attribute A B C D E

No. of times ‘greater’ 3 2 1 1 0

Rank Vendor II > Vendor I > Vendor III Vendor I = Vendor II > Vendor III Vendor III > Vendor II > Vendor I Vendor III > Vendor II > Vendor I Vendor III > Vendor II > Vendor I

Select Vendor II

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14-7

Left to student.

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14-8

(a) Dominance: Paired Comparison Attribute A vs. B B vs. C A vs. C 1 better Worse worse 2 worse better better 3 Worse worse better * 4 worse better better 5 worse better better Dominance? no no no No alternatives can be eliminated based on the dominance method. *Assume that knowing the safety value is better than any unknown value.

Satisficing: Attribute

Feasible Range

1 2 3 4 5

$80,000 - $100,000 Fair - Excellent Good - Excellent 94 - 99% Fair - Excellent

Unacceptable Alternatives none none Alternative A has unknown value * none none

*if the same assumption is used (as in the dominance model) alternative A would be eliminated using the satisficing model Lexicography: Paired comparisons - using given weighting: 5 > 1 > 4 > 3 > 2 Attribute 1 2 3 4 5 **

Ordinal Ranking ** 3 0 1 2 4

Ranking C>A>B B>A>C C>B>A B>A>C B>A=C

4 is most important rank.

The selection, based on highest ranked attribute (#5), would be Alternative B. Because all alternatives meet at least one acceptability range, no alternatives are rejected.

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14-8

(a)

continued

Non-dimensional scaling Attribute

Value

Rating Procedure

1

$100,000 $140,000 $180,000 Excellent Good Fair Excellent Good Not known 99% 98% 94% Excellent Good

($180,000-Cost)/$80,000

2 3 4 5

Attribute 1 2 3 4 5

Relative rank - 1 2 Relative rank - 1 2 Reliability % - 94 99-94 Relative rank - 1 2

Dimensionless Value 1.0 0.5 0.0 1.0 0.5 0.0 1.0 0.5 0.0 1.0 0.8 0.0 1.0 0.0

Non-Dimensional Value A B C 0.5 0.0 1.0 0.5 1.0 0.0 0.0 0.5 1.0 0.8 1.0 0.0 0.0 1.0 0.0

Additive Weighting (using given weights) Attribute 1. Initial Cost 2. Maintenance 3. Safety 4. Reliability 5. Prod. Quality

Weight 0.25 0.10 0.15 0.20 0.30

A B 0.5(0.25)=0.125 0.0(0.25)=0 0.5(0.10)=0.05 1.0(0.10)=0.1 0.0(0.15)=0.00 0.5(0.15)=0.075 0.8(0.20)=0.16 1.0(0.20)=0.2 0.0(0.30)=0.00 1.0(0.30)=0.3

C 1.0(0.25)=0.25 0.0(0.10)=0.0 1.0(0.15)=0.15 0.0(0.20)=0.0 0.0(0.30)=0

Using Additive Weighting Alternative B would be selected. (b)

If two (or more) attributes are dependent, including them in the analysis will result in the same ranking of alternatives for these attributes, essentially double (or triple, or more) counting those decision elements. This can be a false representation of the decision maker's true attitudes. When attributes are entirely dependent, they are more appropriately modeled as a single attribute and weighted appropriately (note that in these cases, the weight of these combined attributes may not necessarily be the sum of the individual weights of the original attributes). Once this is accomplished, the decision analysis proceeds as in part (a).

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14-9

(a) Attribute Social Climate Starting Salary Career Adv. Weather/Sports

Relative Rank 1.00 0.50 0.33 0.25 2.08

Normalized Rank 1/2.08 = 0.481 0.5/2.08 = 0.240 0.33/2.08 = 0.159 0.25/2.08 = 0.120 1.00

(b) Alternatives Attribute Starting Salary

Apex (N.Y.) $35,000

Sycon (L.A.) $30,000

Sigma (GA.) $34,500

Mc-Graw-Wesley (AZ.) $31,500

Dimensionless Equivalent (DE)

1.0

0.0

0.9

0.3

DE = Worst Outcome - Outcome Being Made Dimensionless Worst Outcome - Best Outcome (c) Attribute

Normalized Apex Weight

Sycon

Sigma

Mc-Graw Wesley

Social Climate

0.48

1x0.48

1x0.48

0.5x0.48

0x0.48

Starting Salary

0.24

1x0.24

1x0.24

0.9x0.24

0.3x0.25

Career Adv.

0.16

0x0.16

0x0.16

0.6x0.16

1x0.16

0.12

0x0.12

0x0.12

0.33x0.12

0.67x0.12

Sum

0.72

0.63

0.59

0.31

Weather/ Sports

Using lexicography we conclude that social climate is the most important attribute and Apex is selected.

Additive weighting also selects Apex.

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14-10 (a) Dominance: Candidate C is removed from further consideration. (b) Satisficing Attribute 1 2 3 4 5

Minimum Acceptable Value 7 years Yes Yes 1 year Good

Unacceptable Candidate None (Candidate C already removed) None (Candidate C already removed) Candidate A None (Candidate C already removed) Candidate A

Candidate A stands removed. (c) Disjunctive Resolution Three candidates, namely candidates B, D, and E are available (Candidates A and C have been removed) as these alternatives are acceptable in at least one attribute. (d) Lexicography Based on leadership skills (most important attribute): all three candidates B, D, and E still stand out; Looking next at project management skills: all three candidates B, D, and E still stand out; Looking next at general attitude: both candidates B and D still stand out, but candidate E is removed; Thereafter, considering previous management experience: Candidate D gets selected.

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14-11 Left to student.

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14-12 (a) (b)

Left to student – no unique answer.

Select a mathematical model similar to additive weighting. Let each judge set his/her own weightings and develop a score for each contestant. Then, sum the three scores for each contestant. The contestant with the highest total score is the winner.

This method will allow each judge to be as subjective about each attribute as he/she desires while making the final selection objective.

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14-13 Job Attribute 1.

2.

3.

4.

5.

Total years experience Project management skills Leadership skills Previous management experience (years) General attitude Total Score

A 0.67 1 0 1 0 2.67

Candidate B C 0.33 0 1 0 1 1 0.5 0 2 1 4.83 2

D 1 1 1 1 2 6

Candidate D should be selected.

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14-14

i 1 2 3 4

Wi 1.0 ?

0.8 0.7

Rank 1 4 2 ? By inspection, Ranki=4=3 10

n

Vj   WiX ij i1

V2 =

2.3 = (1.0) (0.7) + W2 (1.0) + (0.8) (0.5) + (0.7) (1.0) 2.3 = W2 + 0.7 + 0.4 + 0.7

W2 = 0.50 V1 =

2.69 = (1.0) (1.0) + (0.5) (0.8) + (0.8) X3,1 + (0.7) (0.7) 2.69 = 1.0 + 0.4 + 0.8 X3,1 + 0.49

X3,1 = 2.69 - 1.0 - 0.4 - 0.49 = 1.0 0.8 Vj normalized = V1 = 2.69/2.69 = 1.00 V2 = 2.30/2.69 = 0.86 Rankj=1 = 2.0 Rankj=2 = 1.0

By inspection

Filling in blanks, i 1 2 3 4

Wi 1.0 0.5 0.8 0.7

Rank 1 4 2 3

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14-4

continued

Keep Existing Tool

Purchase new machine Tool

Rank X1j

2.0 1.0

1.0 0.7

Rank X2j

2.0 0.8

1.0 1.0

Rank X3j

1.0 1.0

2.0 0.5

Rank X4j

2.0 0.7

1.0 1.0

Vj Vj norm

2.69 1.0

2.30 0.86

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14-15 Left to student.

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14-16 Left to student.

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14-17 Left to student.

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14-18 Left to student.

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14-19 A B C D E F 1 Attribute Dr. Molar Dr. Feelgood Dr. Whoops Dr. Pepper $ 50 $ 80 $ 20 $ 40 2 Cost Novocaine Acupuncture Hypnosis Laughing Gas 3 Anesthesia 15 20 5 30 4 Distance 40 25 40 40 5 Office Hours Excellent Fair Poor Good 6 Quality 7 8 Quality Anesthesia 4 Acupuncture 1 9 Excellent 2 Hypnosis 4 10 Fair 3 Laughing Gas 2 11 Good 1 Novocaine 3 12 Poor 13 14 Attribute Dr. Molar Dr. Feelgood Dr. Whoops Dr. Pepper 0.50 0.00 1.00 0.67 15 Cost 0.67 0.00 1.00 0.33 16 Anesthesia 0.60 0.40 1.00 0.00 17 Distance 1.00 0.00 1.00 1.00 18 Hours 1.00 0.33 0.00 0.67 19 Quality 3.77 0.73 4.00 2.67 20 Sum = ^ best choice 21 22 23 24 With novocaine rated as the preferred method of anesthesia, Dr. Whoops becomes 25 the dentist of choice.

G

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