Solution Manual For Engineering Economy, 2025 18th Edition by William G. Sullivan Elin M Wicks Josep

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)

62 million tons per year × (0.05) = 3.1 million tons per year of reduced greenhouse gas $1.2 billion ÷ 3.1 million tons per year = $387.10 per ton

(b)

3 billion tons per year × (0.03) = 90 million tons per year of reduced greenhouse gas $1.2 billion $X billion = 3.1 million tons 90 million tons X = $34.84 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 = $0.75/1.5 Watt-hours = $0.50 per Watt-hour At a cost of $0.50 per Watt-hour, it would cost (1,000)($0.50 per Watt-hour) = $500 per kilo Watt-hour for power from a single AAA battery. This is 5,000 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)

15,000 miles per year / 25 mpg = 600 gallons per year of E20 Savings = 600 gallons per year ($4.00 − $3.37) = $378 per year

(b)

Gasoline saved = 0.20 (600 gal/yr)(1,000,000 people) = 120 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 2,000 ft2 × $120/ ft2 = $240,000. The ICF home will cost $240,000 (1.10) = $264,000. So the cost difference to construct an ICF home is $24,000.

The energy savings per month is 45% × $200 = $90 per month. Therefore, it will take $24,000/($90/month) = 267 months to pay back the extra ICF construction cost through monthly energy savings.

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1-8

Increased lifetime earnings of a college graduate = $1,200,000(0.75) = $900,000

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1-9

Strategy 1: Change oil every 3,000 miles. Cost = (15,000/3,000)($50) = $250 / year Strategy 2: Change oil every 5,000 miles. Cost = (15,000/5,000)($50) = $150 / year Savings = $100 per year

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1-10

In six months you will spend approximately (180)(2)($1.15) = $414 on bottled water. The cost of the filter is $60, so you will save $354 every six months. This amounts to $708 over a year, and you don't need to bother recycling all those plastic bottles! An up-front expenditure of $60 can indeed save a lot of money each year.

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1-11

110 gallons x $4.00 per gallon = $440 saved over 55,000 miles of driving. This comes down to $440 / 55,000 = $0.008 per mile driven. So Brand A saves 8/10 of 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 time, grade point average and energy that Pat is willing/able to exert, Problem A might be "How can Pat survive the senior year and graduate during the coming year (earn a college degree)?"

Problem B: Subject to knowledge of the job market, mobility and professional ambition, Pat's Problem B could be "How can I use my brother's entry-level job as a springboard into a higherpaying position with a career advancement opportunity (maybe no college degree)?"

(b) Problem A - Some feasible solutions for Problem A would include: (1)

Get a loan from their brother and take fewer courses per term, possibly graduating in the summer.

(2)

Quit partying and devote their extra time and limited funds to the task of graduating in the spring term (maybe Pat could get a scholarship to help with tuition, room and board).

Problem B - Some feasible solutions for Problem B would include: (1)

Work for their brother and take over the company to enable him to start another entrepreneurial venture.

(2)

Work part-time for their brother and continue to take courses over the next couple of years in order to graduate.

(3)

Work for their brother for one or two semesters to build up funds for their senior year.

While interviewing, bring up the real life working experience and request a higher starting salary.

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1-14

A Typical Discussion/Solution: (a)

One problem involves how to satisfy the hunger of three students -- assume a piping hot delicious pizza will satisfy this need. (Another problem is to learn enough about Engineering Economy to pass -- or better yet earn an “A” or a “B” -- on the final examination and ace the course. Maybe a pizza will solve this problem too?) Let’s use “hunger satisfaction with a pizza” as the problem/need definition.

(b)

Principle 1 - Develop the Alternatives i) Alternative A is to order a pizza from “Pick-Up Sticks” ii) Alternative B is to order a pizza from “Fred’s” Other options probably exist but we’ll stick to these two alternatives Principle 2 - Focus on the Differences Difference in delivery time could be an issue. A perceived difference in the quality of the ingredients used to make the pizza could be another factor to consider. 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 customers wanting to get a good deal. Does it make sense to buy a pizza having a crust that your dog enjoys, or ordering a pizza from a shop that employs only college students? Use the customer’s point of view in this situation rather than that of the owner of the pizza shop or the driver of the delivery vehicle.

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 pizza to order, 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 pizza to order. For example, variety and quality of toppings and delivery time may be extremely important to your choice. 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 The variability in quality of the pizza, its delivery time and even its price should be carefully examined in making your selection. (Advertised prices are often valid under special conditions -call first to check on this!)

Principle 7 - Revisit Your Decision After you’ve consumed your pizza and returned to studying for the final exam, were you pleased with the taste of the toppings? On the downside, was the crust like cardboard? You’ll keep these sorts of things in mind (good and bad) when you order your next pizza!

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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 cost per unit volume (Principles 1, 2, 3, 4 and 6 are integral to this comparison) Assumptions: (i) weight is directly proportional to volume (to avoid a “meringue” pizza with lots of fluff but meager substance), (ii) you and your study companions will eat the entire pizza (avoids variable amounts of discarded leftovers and hence difficult-to-predict cost of cubic inch consumed) and (iii) data provided in the Example Problem are accurate (the numbers have been confirmed by phone calls).

Analysis:

Alternative A “Pick-Up-Sticks” Volume = 20 x 20 x 1 ¼ = 500 in.3 Total Cost = $15 (1.05) + $1.50 = $17.25 Cost per in.3 = $0.035 Alternative B “Fred’s” Volume = (3.1416)(10)2 (1.75) = 550 in.3 Total Cost = $17.25 (1.05) = $18.11 Cost per in.3 = $0.033

Therefore, order the pizza from “Fred’s” to minimize total cost per cubic inch. (d)

Typical other criteria you and your friends could consider are: (i) cost per square inch of pizza (select “Pick-Up-Sticks”), (ii) minimize total cost regardless of area or volume (select “Pick-UpSticks”), and (iii) “Fred’s” can deliver in 30 minutes but “Pick-Up-Sticks” cannot deliver for one hour because one of their ovens is not working properly (select “Fred’s”).

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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. Further evaluation leads to the elimination of walking, riding a bicycle, and taking a bus as feasible alternatives.

STEP 2—Develop Your Alternatives (Principle 1 is used here.): Your problem has been reduced to either replacing or repairing your automobile. The alternatives would appear to be 1. Sell the wrecked car for $2,000 to the wholesaler and spend this money, the $1,000 insurance check, and all of your $7,000 savings account on a newer car. The total amount paid out of your savings account is $7,000, and the car will have 28,000 miles of prior use.

2. Spend the $1,000 insurance check and $1,000 of savings to fix the car. The total amount paid out of your savings is $1,000, and the car will have 58,000 miles of prior use.

3. Spend the $1,000 insurance check and $1,000 of your savings to fix the car and then sell the car for $4,500.

Spend the $4,500 plus $5,500 of additional savings to buy the newer car. The total amount paid out of savings is $6,500, and the car will have 28,000 miles.

4. Give the car to a part-time mechanic, who will repair it for $1,100 ($1,000 insurance and $100 of your savings), but will take an additional month of repair time. You will also have to rent a car for that time at $400/month (paid out of savings). The total amount paid out of savings is $500, and the car will have 58,000 miles on the odometer.

5. Same as Alternative 4, but you then sell the car for $4,500 and use this money plus $5,500 of additional savings to buy the newer car. The total amount paid out of savings is $6,000, and the newer car will have 28,000 miles of prior use.

ASSUMPTIONS: 1. The less reliable repair shop in Alternatives 4 and 5 will not take longer than one extra month to repair the car.

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

3. Interest earned on money remaining in savings is negligible.

STEP 3—Estimate the Cash Flows for Each Alternative (Principle 2 should be adhered to in this step.)

1. Alternative 1 varies from all others because the car is not to be repaired at all but merely sold. This eliminates the benefit of the $500 increase in the value of the car when it is repaired and then sold. Also this alternative leaves no money in your savings account. There is a cash flow of −$8,000 to gain a newer car valued at $10,000.

2. Alternative 2 varies from Alternative 1 because it allows the old car to be repaired. Alternative 2 differs from Alternatives 4 and 5 because it utilizes a more expensive ($500 more) and less risky repair facility. It also varies from Alternatives 3 and 5 because the car will be kept. The cash flow is −$2,000 and the repaired car can be sold for $4,500.

3. Alternative 3 gains an additional $500 by repairing the car and selling it to buy the same car as in Alternative 1. The cash flow is −$7,500 to gain the newer car valued at $10,000.

4. Alternative 4 uses the same idea as Alternative 2, but involves a less expensive repair shop. The repair shop is more risky in the quality of its end product, but will only cost $1,100 in repairs and $400 in an additional month’s rental of a car. The cash flow is −$1,500 to keep the older car valued at $4,500.

5. Alternative 5 is the same as Alternative 4, but gains an additional $500 by selling the repaired car and purchasing a newer car as in Alternatives 1 and 3. The cash flow is −$7,000 to obtain the newer car valued at $10,000.

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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 owner of the wrecked car).

The value of the car to the owner is its market value (i.e., $10,000 for the newer car and $4,500 for the repaired 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 a reliable repair shop worth?).

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

1. Alternative 1 is eliminated, because Alternative 3 gains the same end result and would also provide the car owner with $500 more cash. This is experienced with no change in the risk to the owner. (Car value = $10,000, savings = 0, total worth = $10,000.)

2. Alternative 2 is a good alternative to consider, because it spends the least amount of cash, leaving $6,000 in the bank. Alternative 2 provides the same end result as Alternative 4, but costs $500 more to repair.

Therefore, Alternative 2 is eliminated. (Car value = $4,500, savings = $6,000, total worth = $10,500.)

3. Alternative 3 is eliminated, because Alternative 5 also repairs the car but at a lower out-of-savings cost ($500 difference), and both Alternatives 3 and 5 have the same end result of buying the newer car. (Car value = $10,000, savings = $500, total worth = $10,500.)

4. Alternative 4 is a good alternative, because it saves $500 by using a cheaper repair facility, provided that the risk of a poor repair job is judged to be small. (Car value = $4,500, savings = $6,500, total worth = $11,000.)

5. Alternative 5 repairs the car at a lower cost ($500 cheaper) and eliminates the risk of breakdown by selling the car to someone else at an additional $500 gain. (Car value = $10,000, savings = $1,000, total worth = $11,000.)

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. If the original car is repaired and kept, there is a possibility that it would have a higher frequency of breakdowns (based on personal experience). If a cheaper repair facility is used, the chance of a later breakdown is even greater (based on personal experience). Buying a newer car will use up most of your savings. Also, the newer car purchased may be too expensive, based on the additional price paid (which is at least $6,000/30,000 miles = 20 cents per mile). Finally, the newer car may also have been in an accident and could have a worse repair history than the presently owned car.

Based on the information in all previous steps, Alternative 5 was actually chosen.

STEP 7—Monitor the Performance of Your Choice This step goes hand-in-hand with Principle 7 (revisit your decisions). The newer car turned out after being “test driven” for 20,000 miles to be a real beauty. 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

Imprudent use of electronic mail, for example, can involve legal issues, confidential financial data, trade secrets, regulatory issues, public relations goofs, etc. These matters are difficult to “dollarize” but add to the $30,000 annual savings cited in the problem. Surfing the web inappropriately can lead to legal prosecution for pornography violations.

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1-18

(a)

Value of metal in collection = (5,000/130 lb)(0.95)($3.50/lb) + (5,000/130 lb)(0.05)($1.00/lb) = $129.81 Each penny is worth about 2.6 cents for its metal content. The numismatic value of each coin is most likely 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

A one degree difference in thermostat setting will reduce energy use by up to five percent when the heat loss in the winter and the heat gain in the summer are roughly equal. This assumes the efficiency of the heating system is about the same as that of the cooling system.

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1-22

Assume a 50 week working year: (8 hours/day)(5 days/week)(50 weeks per year) = 2000 hours/year Savings per week = ($25/hour)(15 hours/week) = $375/week Saving per year = ($375/week)(50 weeks/year) = $18,750/year

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1-23

Annual revenue = ($0.10/kWh)(150,000 kWh/year) = $15,000/year

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1-24

$10,000/($200 per month savings) = 50 months

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1-25

Compare material cost per year.

A: $10/(5 years) = $2 per year B: $15/(8 years) = $1.875 per year On a $/year basis, Material B is more cost effective.

However, if we assume that both materials have to be purchased twice to cover the 10 year period, A: $10 X 2 = $20 for 10 years = $2/year B: $15 X 2 = $30 for 10 years = $3/year If there is no further use for the material after 10 years, Material A would be more economical.

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1-26

Annual savings = ($500,000/year)(0.03) = $15,000/year

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1-27

Cost to outsource = ($15/hour)(1000 hours) = $15,000 Cost to automate = $10,000 It is more cost effective to automate the process.

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1-28

$800/($150/year) = 5.33 years

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1-29

Compute the total cost for 1,000 non-defective units.

Cost to obtain 1 non-defective unit = ($/unit)/(% chance unit is non-defective) Supplier A = (1000 non-defective units)($20/unit)/0.98 = $20,408.16 Supplier B = (1000 non-defective units)($22/unit)/0.99 = $22,222.22 It is more cost effective to obtain the materials from Supplier A.

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1-30

Money saved each day = 10 gallons/day X $0.005/gallon = $0.05 per day Time to recover $250 investment = $250/(0.05/day) = 5000 days = 13.7 years

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1-31

$1,500 / ($5/hour) = 300 hours

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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 =

1,000,000 miles/year = 182.6 or 183 cows (365 days/year)(15 miles/day)

Annual cost = (1,000,000 miles/year)($10 / 60 miles) = $166,667 per year (b)

miles/year Annual cost of gasoline = 1,000,000 × $4/gallon = $133,333 per year 30 miles/gallon It would cost $33,334 more per year to fuel the fleet of cars with gasoline.

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2-4

Cost Rent Hauling Total

Site A = $5,000 (4)(200,000)($1.50) = $1,200,000 $1,205,000

Site B = $100,000 (3)(200,000)($1.50) = $900,000 $1,000,000

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 B.

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2-5

Present cost to company = $1,000,000 + 40 repairs × 24 hrs/repair × $2,500/hr = $3,400,000 With Ajax, the cost to your company = $2,000,000 + 40 repairs × R hrs/repair × $2,500/hr Ajax is preferred if 2,000,000 + 100,000R ≤ $3,400,000 or R ≤ 14 hours/breakdown.

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2-6

The $97 you spent on a passport is a sunk cost because you cannot get your money back. If you decide to take a trip out of the U.S. at a later date, the passport’s cost becomes part of the fixed cost of making the trip (just as the cost of new luggage would be).

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2-7

If the value of the re-machining option ($60,000) is reasonably certain, this option should be chosen.

Even if the re-machined parts can be sold for only $45,001, this option is attractive. If management is highly risk adverse (they can tolerate little or no risk), the second-hand market is the way to proceed to guarantee $15,000 on the transaction.

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2-8

The bakery makes a per unit profit of $2.50 - $1.00 = $1.50 per pastry sold. To cover the cost of the new equipment, $20,000/($1.50 per pastry sold) = 13,333.33 units.

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2-9

By automating the process, variable cost is reduced, saving $10 - $7 = $3 for every unit produced. This cost savings will breakeven with the cost to automate after $200,000/($3 per unit) = 66,667 units are produced.

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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 CT = $1,000 + $30/unit × 225 units = $7,750 and 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 ± 40.3 0.2

𝐷 = 24 (rounded up from 23.5) 𝐷 = 426 (rounded down from 426.5) 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 = 2,000 - 0.1D. Then, TR = p  D = 2,000D - 0.1D2.

The first derivative of TR with respect to D is d(TR) / dD = 2,000 - 0.2D

 This, set equal to zero, yields the D that maximizes TR. Thus,  2,000 - 0.2 D = 0

 D = 10,000 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 = 150  0.01D

CF = $50,000

cv = $40/unit

Profit = 150D  0.01D2  50,000  40D = 110D  0.01D2  50,000 d(Profit)/dD = 110  0.02D = 0

 D = 5,500 units per year, which is less than maximum anticipated demand At D = 5,500 units per year, Profit = $252,500 and p = $150  0.01(5,500) = $95/unit.

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2-14

(a)

D* = (a – Cv) / 2b = (700 – 131.50) / 0.10 = 568.5 / 0.10 = 5,685 103 board-feet Price = $700 – (0.05)(5,685) = $415.75 per 103 board feet Profit (per month) = pD – CF - CvD = $415.75(5,685) - $1,000,000 - $131.50(5,685) = 2,363,539 – 1,000,000 – 747,578 = $615,961/ month

(b)

–bD2 + (a – Cv)D – CF = 0 b= 0.05 (a – Cv) = 568.5 CF = 1,000,000 −0.05D2 + 568.5D – 1,000,000 = 0 D’1,2

= -568.5 ± [ (568.5)2 – 4(-0.05)(-1,000,000)]1/2 2 (-0.05) = -568.5 ± [ 323,192.25 – 200,000]1/2 -0.10

D1

= (-568.5 + 351) / (-0.10) = 2,175

D2

= (-568.5 - 351) / (-0.10) = 9,195

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2-15

(a)

(b)

2700 5000   Profit = 38   2  D - 1000 - 40D D D   5000 = 38D + 2700 -1000 - 40D D 5000 Profit = -2D + 1700 D 5000 d (Profit) = -2 + =0 dD D2 5000 or, D2 = = 2500 and D* = 50 units per month 2 d 2 (Profit) dD

2

=

10,000 D3

< 0 for D > 1

Therefore, D* = 50 is a point of maximum profit.

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2-16

Profit = Total revenue - Total cost = (15X - 0.2X2) - (12 + 0.3X + 0.27X2) = 14.7X - 0.47X2 - 12 dProfit = 0 = 14.7 - 0.94X dX

X = 15.64 megawatts Note:

d 2 Profit = - 0.94 thus, X = 15.64 megawatts maximizes profit dX 2

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2-17

(a)

Using Equation (2-10), 𝐷∗ =

(b)

180 − 40 = 14 units per week 2(5)

Total profit = −5(142) + (180 – 40)(14) = −980 + 140(14) = $980/wk

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2-18

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

The variable cost per pound is $20,000,000 / 40,000,000 pounds = $0.50 per pound. (a) Profit/yr

= (40,000,000 pounds / year) ($1.00 - $0.50) - $17,000,000 = $20,000,000 - $17,000,000 = $3,000,000 per year

The mine is expected to be profitable. (b) If only 17,000 tons (= 34,000,000 pounds) are produced, then Profit/yr = (34,000,000 pounds/year)($1.00 - $0.50) - $17,000,000 = 0 Because Profit =0, 17,000 tons per year is the breakeven point production level for this mine. A loss would occur for production levels < 17,000 tons/year and a profit for levels > 17,000 tons per year.

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2-19

(a)

BE = $1,500,000 / ($39.95 − $20.00) = 75,188 customers per month

(b)

New BE point = $1,500,000 / ($49.95 − $25.00) = 60,120 per month

(c)

For 63,000 subscribers per month, profit equals 63,000 ($49.95 − $25.00) − $1,500,000 = $71,850 per month This improves on the monthly loss experienced in part (a).

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2-20

(a)

D =

CF $2,000,000 = = 40,000 units per year p - cv ($90 - $40) / unit

$10,000,000 $8,000,000

$9,000,000

Profit Breakeven Point

$6,000,000 $4,000,000

$6,000,000

Loss

$2,000,000

Fixed Cost D' = 40,000 units

$0 0

20,000

40,000

60,000

80,000

100,000

Number of Units

(b)

Profit (Loss) = Total Revenue - Total Cost (90% Capacity) = 90,000 ($90) - [$2,000,000 + 90,000 ($40)] = $2,500,000 per year (100% Capacity) = [90,000($90) + 10,000($70)] - [$2,000,000 + 100,000($40)] = $2,800,000 per year

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2-21

Annual savings are at least equal to ($60/lb)(600 lb) = $36,000. So the company can spend no more than $36,000 (conservative) and still be economical. Other factors include ease of maintenance / 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

Let X represent the number of TVs sold (and for every TV sold, 2 speakers are sold). For each TV sold, a profit of $150 is realized from the TV and a profit of 2 X $50 = $100 is realized from the sale of speakers. To breakeven with the fixed costs of $120,000, X = $120,000 / ($150 + $100) = 480 televisions and 960 speakers should be sold.

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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)

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

(a)

Ownership cost = $120 + $0.60X where X = horsepower rating.

Operating cost =

$0.055

1

hp-hour X

× 9000 hp-hour

Total annual cost = 120 + 0.60X +

.

dAC dX

( X

)

= 120 + 0.6X +

X

= 0.6 - 495X -2 = 0

X = 28.72 horsepower (b)

For X to be a minimum of the function in part (a), show that d2 AC dX 2

d2 AC dX 2

= 990X-3 which is greater than 0 for all positive values of X.

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> 0.

2-26 CT = Co + Cc = knv2 +

$1,500n v

dCT 1,500 = 0 = 2kv- 2 = kv3 - 750 dv v v=3

750 k

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

Let X represent the number of new cars sold (and for every new car sold, 3 used cars are sold). For each new car sold, a profit of $2,000 is realized from the new car and a profit of 3 X $1,000 = $3,000 is realized from the sale of the used cars. To breakeven with the annual costs of $300,000, X = $300,000 / ($2,000 + $3,000) = 60 new cars and 180 used cars should be sold.

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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 B. Annual Heating Load (106 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 CR··t $

CI



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2-30

Total profit from residential = Total profit from commercial. If fixed costs are $60,000 annually, then each service type needs to cover $30,000.

Residential = $30,000 / ($200 per job) = 150 residential jobs Commercial = $30,000 / ($500 per job) = 60 commercial jobs

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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

Let X represent the number of cakes sold (and for every cake sold, 4 dozen cookies are sold). For each cake sold, a profit of $8 is realized from the cake and a profit of 4 X $3 = $12 is realized from the sale of cookies. To breakeven with the monthly fixed costs of $2,000, X = $2,000 / ($8 + $12) = 100 cakes and 400 dozens of cookies should be sold.

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2-33

Let X represent the number of bicycles sold (and for every bicycle sold, 4 helmets are sold). For each bicycle sold, a profit of $100 is realized from the bicycle and a profit of 4 X $25 = $100 is realized from the sale of helmets. To breakeven with the fixed costs of $50,000, X = $50,000 / ($100 + $100) = 250 bicycles and 1,000 helmets should be sold.

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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)/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) = $2166.5/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

Let X represent the number of Burgers sold (and for every burger sold, 2 fries are sold). For each burger sold, a profit of $5 is realized from the burger and a profit of 2 X $2 = $4 is realized from the sale of fries. To breakeven with the fixed costs of $3,000, X = $3,000 / ($5 + $4) = 333.33 burgers and 667 fries should be sold.

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2-39

Let X represent the number of sets of five chairs sold (and for five chairs sold, 2 tables are sold). For each set of five chairs sold, a profit of 5 X $50 = $250 is realized from the chairs and a profit of 2 X $100 = $200 is realized from the sale of tables. To breakeven with the fixed costs of $40,000, X = $40,000 / ($250 + $200) = 88.89 X 5 = 444.44 chairs and 88.89 X 2 = 177.78 tables should be sold.

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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

The two alternatives are “no jig with a skilled machinist” and “use a jig with a lesser skilled machinist.”

No jig: [(1.5 min/housing)/(60 min/hr)]($25/hr)(4000 housings) = $2,500 With jig: [(2 min/1 housings)/(60 min/hr)]($15/hr)(4000 housings) + $500 = $2,500 Both methods result in the same production cost.

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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) False; (d) False; (g) False; (j) False; (m) True; (p) False; (b) False; (e) True; (h) True; (k) True; (n) True; (q) True; (c) True; (f) True; (i) True; (l) False; (o) True; (r) True;

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(s) False

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 G

H

I

J

K

L

M

N

O

P

Net Income

$40,000 $20,000

Net Income

$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 Cash Flow

Total Revenue

$300,000

Total Expense

$200,000

$100,000

0

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

$-

Volume (Demand)

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

Total Expense Net income

$100,000 $$(100,000)

0

$(200,000) 50 0 10 00 15 00 20 00 25 00 30 00 35 00 40 00 45 00 50 00 55 00

F

Cash Flow

A B C D E Fixed cost/ Demand Start point (D) = 1 mo. = $ 73,000 0 Variable Demand Increment = 2 cost/unit = $ 83 250 3 a= $ 180 4 b= $ 0.02 5 Monthly Price per Total 6 Demand Unit Revenue Total Expense Net income 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 22 3750 $ 105 $ 393,750 $ 384,250 $ 9,500 23 4000 $ 100 $ 400,000 $ 405,000 $ (5,000) 24 4250 $ 95 $ 403,750 $ 425,750 $ (22,000) 25 4500 $ 90 $ 405,000 $ 446,500 $ (41,500) 26 4750 $ 85 $ 403,750 $ 467,250 $ (63,500) 27 5000 $ 80 $ 400,000 $ 488,000 $ (88,000) 28 5250 $ 75 $ 393,750 $ 508,750 $ (115,000) 29 5500 $ 70 $ 385,000 $ 529,500 $ (144,500) 30 31 32 Summary of impact of changes in cost components on optimum 33 demand and profitable range of demand.

34 35 Percent Change CF cv D1' D2' 36 D* 37 -10% -10% 2,633 724 4541 38 0% -10% 2,633 824 4443 39 10% -10% 2,633 928 4339 40 -10% 0% 2,425 816 4036 41 0% 0% 2,425 932 3918 42 10% 0% 2,425 1060 3790 43 -10% 10% 2,218 940 3495 44 0% 10% 2,218 1092 3343 45 10% 10% 2,218 1268 3167 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

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

$ 6

Annual Heating Load (10 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

One square meter converts to 10.764 square feet. So the area of the office building in square feet is 107,640. Therefore the average annual cost of heating and cooling this area is (107,640 ft2)($3.50/ft2) = $376,740.

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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 home is: Cost or Revenue Category Captial Investment Annual Operating and Maintenance Costs Major Repair or Replacement Costs Real Estate Fees Asset Sales

Typical Cost and Revenue Elements Real estate (lot) cost; architect/engineering fees; construction costs (labor,material, other); working capital (tools, initial operating supplies, etc.); landscaping costs.

Utilities (electricity, water, gas, telephone, garbage); cable TV; painting (interior and exterior); yard upkeep (labor and materials); routine maintenance (furnace, air conditioner, hot water heater, etc.); insurance; taxes.

Roof; furnace; air conditioner; plumbing fixtures; garage door opener; driveway and sidewalks; patio; and so on.

Acquisition; selling.

Sale of home (year 10).

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3-3

The estimated cost is $12,000 + ($10/ft2)(8.5 ft)(15 ft)(500)(1.8) = $1,159,500

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3-4

(a)

(62 million tons per year) (0.05) = 3.1 million tons of greenhouse gas per year $1.2 billion = $387.10 per ton 3.1 million tons per year

(b)

(3 billion tons per year) (0.03) = 90 million tons per year $1.2 billion $X billion  3.1 million tons/year 90 million tons X  $34.84 billion

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3-5

(24,000 ft2)(60,000 Btu/ft2) = 1,440 million Btu during the heating season. This is 1,440 thousand cubic feet of natural gas, and the cost would be (1,440,000 ft3)($10.50/1000 ft3) = $15,120 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 deciding on a price to charge per square foot of leased space.

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3-6

(a)

Standard electric bill = (400 kWhr)(12 months/year)($0.10/kWhr) = $480 per year.

Green power bill = (12 months/year)($4/month) = $48 per year.

Total electric bill = $528 per year.

(b)

$528 / 4,800 kWhr = $0.11 per kWhr (a 10% increase due to green power usage)

(c)

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

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3-7

The estimated replacement cost in late 2026 can be estimated as follows: C2026 = C2021 (I2026/I2021) = $30,000 (186/140) = $39,857

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3-8

The estimated cost of the water filtration system in 2027 is: ̅ / 𝐼 2023 ̅ ) = $250,000 (298/238) = $313,025 C 2027 = C2023 (𝐼2027

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3-9 𝐼

=

.

. .

. .

.

× 100 = 153.5

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3-10

(CA / CB) = (SA / SB)x CA = $800,000 (30,000 / 20,000)0.83 = 800,000(1.4) CA = $1,120,000 for the larger warehouse

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3-11

Let

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

SA = 1.42X SB = X

 221  CB = $181,000   = $246,920  162  0.8

. X  142 CA = $246,920   = $326,879  X  Total cost with options = $326,879 + $28,000 = $354,879

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3-12

The estimated capital investment of the seven MW solar farm in four years is: $14 million (F/P, 8%, 4) = $14 million (1.3605) = $19.047 million Next, the capital investment (C) for the six MW solar farm in four years can be estimated by using Equation 3-4: C = $19.047 million (6/7)0.85 = $16.708 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 (0.8) / log 2 = -0.322 Z1 = 10 hours Z4 = 10 [(4)-0.322] = 6.4 hours

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3-18

n = log (0.85) / log2 = -0.2345 Cx = Tx / x

so

(x) Cx = Tx , or Tx = 5(15.882 hrs.) = 79.41 hours

We know that Tx = K [1-0.2345 + 2-0.2345 + 3-0.2345 + 4-0.2345 + 5-0.2345] so 79.41 = 4.031 K , or K = 19.70 hours Now with equation 3-5 we can determine Z20: Z20 = 19.70 (20-0.2345) = 9.76 hours

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3-19

(a)

∑x = 40 ∑y = 35 ∑xy = 320 ∑x2 = 360 𝑥̅ = 40/5 = 8 𝑦 = 35/5 = 7 b1 = [5(320) – 40(35)] / [5(360) – 402] = 1 b0 = [35 – (1) (40)] / 5 = -1

(b)

For R

R = (5*320) – (40*35) / [((5 *360) – (40^2)) * ((5*285) – (35^2))]^1/2 R=1

(c)

Using linear regression, for average sunlight of 9 hours, using y = x - 1, Growth = 9 - 1 = 8

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3-20

∑x = 12.5 ∑y = 2000 ∑xy = 5500 ∑x2 = 33.75

(a) B1 = (5*5500) – (12.5*200) / (5*33.75) – (1285^2) = 200 B0 = (2000 – (12.5*200))/5 = -100

So y = -100 + 200x (b)

When x = 2.25, y = -100 + 200(2.25) = 350

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3-21

X variable – Posts Y variable – Traffic Using Microsoft Excel to obtain values for parts a and b. a. b0 (intercept) = 200, b1 (slope) = 172 b. R (correlation coefficient) = 0.99919 (essentially perfect positive correlation) c. Traffic at X=22 posts = 200 + 172(22) = 3984 posts

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3-22

xi = weight of order (lbs) yi = packaging and processing costs ($) (a) y = b0 + b1x

 x = 2530 x = 253

 x = 658,900

 y = 1024

 y = 106,348

i

y = 102.4

i

2 i

2 i

 x y = 264,320 i

b1 =

i

264,320  (253)(1024) = 0.279 658,900  (253)(2530)

b0 = 102.4 – (0.279)(253) = 31.813; (b) R =

y = 31.813 + 0.279x

S xy S xx S yy

Sxy = 264,320 – (2530)(1024)/10 = 5,248 Sxx = 658,900 – (2530)2/10 = 18,810 Syy = 106,348 – (1024)2/10 = 1,490.4 R=

5248 (18,810)(1490.4)

= 0.99

(c) y = 31.813 + (0.279)(250) = $101.56

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3-23

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

0.6

 1029    = $13,915  830 

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3-24

$127(1.19)5 = $303 per square foot in five years. The total estimated cost in five years is (320,000 ft2)($303/ft2) = $96,960,000. It’s a good idea to build this facility today and then, if needed, add on the additional space five years later.

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3-25

$5million to produce 100 tons/day Cost capacity factor = 0.6 To produce 200 tons = 5 * (200/100)^(0.6) = $7.58 Million

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3-26

mW 0.8  = $451,440 Boiler Cost = $300,000 106 mW 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 2025 estimate of $320,274,240 for the plant.

WBS Element Code

Units/Factors

1.1.1-2

600

Price/Unit

Subtotal

$2,000 $

1,200,000

$

3,000,000

1.1.3 1.1

Totals

$

4,200,000

1.2-3

2.3969 a $ 110,000,000 $ 263,659,000 $ 263,659,000

1.4

4.4727 b $

5,000,000 $

80,390 c $

60 $

4,823,400

$

15,000,000

1,500 $

900,000

22,363,500 $

22,363,500

1.5.1-3 labor materials 1.5.4

600

$

1.5 1.9

3%

$ 310,944,672

TOTAL ESTIMATED COST IN 2025

$

20,723,400

$

9,328,340

$ 320,274,240

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

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

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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

3-30 A B C D E F G H I J K 1 Spacecraft Weight Cost (millions) 2 0 100 $ 600 3 1 400 $ 278 4 2 530 $ 414 5 3 750 $ 557 6 4 900 $ 689 7 5 1,130 $ 740 8 6 1,200 $ 851 9 10 SUMMARY OUTPUT 11 12 Regression Statistics 13 Multiple R 0.705850276 14 R Square 0.498224612 15 Adjusted R Square 0.397869535 16 Standard Error 151.9036833 17 Observations 7 18 19 ANOVA 20 df SS MS F Significance F 21 Regression 1 114557.2121 114557.212 4.9646179 0.076342997 22 Residual 5 115373.645 23074.729 23 Total 6 229930.8571 24 25 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% 26 Intercept 341.9170907 125.2152843 2.73063383 0.0412491 20.04148132 663.7927001 20.04148132 663.7927001 27 Weight 0.346423227 0.155476261 2.22814225 0.076343 -0.053240574 0.746087027 -0.053240574 0.746087027 28 29 30 31 RESIDUAL OUTPUT 32 33 Observation Predicted Cost Residuals 34 1 376.5594133 223.4405867 35 2 480.4863813 -202.4863813 36 3 525.5214008 -111.5214008 37 4 601.7345106 -44.73451063 38 5 653.6979946 35.30200539 39 6 733.3753367 6.624663274 40 7 757.6249626 93.37503741 41 42 43 44 45 46 47 48 49 50 0 0 0 0 0 0 0 0 0 0 0 0 0 51 10 2 0 3 0 4 0 5 0 6 0 7 0 8 0 90 , 0 0 , 1 0 , 2 0 , 3 0 1 1 1 1 52 53 54 55 56

Cost (millions)

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. Labor cost of $20/hour. The remaining items in the cost template are all based on labor cost, and an increase in labor cost (or decrease in labor cost) will be magnified throughout the solution.

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3-33

A 50% increase in training costs, will increase the factor from 10% to 15%. A 90% increase in transportation costs will increase the factor from 20% to 48%. The updated template is below:

A B C D E F G H I J

Recovery Cost Elements Factory labor Quality costs – training Total labor Factory overhead – setup costs Transportation cost Total direct charge Facility rental Total recovery cost Quantity – lot size Recovery cost per unit Outside cost per unit

Unit Elements Units Cost/Unit 100 hrs $25.00/hr

Factor Estimates Factor of Row 15%

A

150% 20%

C C

Row Total $2,500 $375 $2,875 $4,312.50 $575 $4,887.50 NA $7,762.5 400 $19.41 $22.50

Thus, the recovery cost per unit increases from $18.56 to $19.41, an increase of $0.85 per unit, or 4.6%.

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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  C2018(new design) = $900,000   150 

0.92

 450   $1,125,000   200 

0.87

 175   $750,000   100 

C2028 = $4,617,660(1.12)10 = $14,341,751 Select (c)

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0.79

= $4,617,660

3-42

1st unit = 20hrs Learning curve = 70% N = log(0.7)/log(2) = -0.5146 Now, for 10th unit T(10) = 20 * 10^(-0.5146) T(10) = 6.12 hours The reduction in time = 20 – 5.92 = 13.88 hours Select e.

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3-43

Total time = 200 * 4 = 800 hours So, Log(4) = Log(8) + n* Log(200) 0.6021 = 0.9031 + n * (2.3010) n = -0.1308 Taking antilog, Learning curve = 91.29% Select c.

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3-44

Time taken for 100th plane = 1000 hours Learning curve = 0.9 n = log(0/9)/log(2) = -0.152 Time taken for first plane = Time taken by 100th plane / (100)^(-0.152) T1 = 1000 / (100)^(-0.152) T1 = 2013 hours Option C

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3-45

1000 sqft = $120,000 Cost capacity factor = 0.8 C2 = 120,000 * (1500/1000)^0.8 C2 = $165,979 Option B

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Solutions to Chapter 4 Problems

4-1

Simple interest: I = PiN = $9,000 (0.08)(5) = $3,600 Compound interest: I = P (F/P, 8%, 5) – P = $9,000 (1.4693) − $9,000 = $4,223.70 There is a difference in the amount of interest ($4,223.70 − $3,600 = $823.70) earned because compounding allows interest from previous years to earn additional interest, whereas simple interest only considers the original principal.

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4-2

F =$10000 * (1+0.08)^30 = $100626.56

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4-3

Compound interest causes your savings to grow exponentially without additional funding.

Compound interest on your loans increases the total amount of interest repaid over time.

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4-4

P = 20000/((1+0.06)^5) = $14945.06

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4-5

N = (12 mo/yr)(10 yr) = 120 months A = $50,000 (A/F, 0.25%, 120) = $357.80

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4-6

A = $100,000 (A/P, 5%, 15) = $100,000 (0.0963) = $9,630 per year

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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 = $2,400 N = 25 I = 6%

F = A*(F/A , 6%, 25) F = 2400 * 54.8645 F = $131,674.8

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4-9

N = (12 mo/yr)(15 years) = 180 months F = $500 (F/P, 0.5%, 180) = $1,227.05

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4-10

F = A*[((1+i)^N)-1)/i] F*i = A*[((1+0.05)^20)-1] (30000*0.05) = A*1.653 A = $907.28

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4-11

Cash flow diagram (personal viewpoint)

50 51

52

53

84

85

$200,000 If person lives to age 85, his IRR can be determined as follows.

0 = −$200,000 + $9,000 (P/A, i’, 35) 22.222 = (P/A, i’, 35) From Excel, i’ = 2.77% per year If person lives to be 75 years old, we have: 0 = -$200,000 + $9,000 (P/A, i’, 25) 22.222 = (P/A, i’, 25) From Excel, i’ = 0.93% per year These are not attractive investments.

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4-12

A = $100 N = 15 I = 12%

F = 100 * ( (( 1+0.12)^15)-1) / 0.12 F = $3727

Hence, the affordable cost for purchasing new oven is <= $3727.

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4-13

Set up an equivalence relationship as follows (refer to Equation 4-3): $1,000,000 = $100,000(F/P, 7.46%, N) Or 10 = (1.0746)N Solving for N: Log 10 = N log (1.0746) N = 32 years Therefore, you can get to your goal five years sooner when i = 7.46% per year.

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4-14

F = P (F/P, i, 8) $24,760 = $10,000 (F/P, i, 8) (F/P, i, 8) = 2.4760 From Appendix C, i = 12%.

Alternatively, $24,760 = $10,000 (F/P, i, 8) = $10,000 (1 + i)8

i=

$

,

$

,

− 1 = 0.12 or 12%

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4-15

F = $15000 N= 10 I = 3%

P = F/(1+i)^N P = 15000 / (1.03)^10 P = $11161.41

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4-16

A = $30,000 I = 3% N = 10

P = 30,000 (P/A,3%,10) = $255,906

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4-17

P = $217.42 N = 4 years i = 6%

F = $217.42 (F/P, 6%, 4) = $274.49

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4-18

Monthly = $400, so A = $400 (12) = $4,800 annually N = 30 years i = 8% per year

F = $4,800 (F/A, 8%, 30) = $543,759.36

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4-19

In this problem we know P ($28,700), A ($330 per month) and N (10 years). We can calculate the value of the unknown i' % as follows:

$28,700 = $330 (P/A, i' % per month, 120 months)

Using Goal Seek in Excel, we determine i' = 0.565% per month.

Goal Seek Setup:

Cell A1 = unknown interest rate = 0.05 (starting guess) Cell B1 = 330*((1+a1)^120-1)/(a1*(1+a1)^120)

In Goal Seek, set value of cell B1 = 28700 by changing cell A1

For Instructors:

An extension to this problem takes us into Section 4.14. What is the effective annual interest rate being charged on this particular student loan?

The nominal interest rate per year is 12 (0.565%) = 6.78% per year and the effective annual interest rate (Equation 4-32) is (1.00565)^ 12 -1 = 0.07 (7%).

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4-20

F = $25,000 I = 0.5% is monthly interest So, N becomes 36

So, A = F *(A/F, 3 , 0.5) A = 25000 * (0.0254) A = $635

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4-21

(a)

N = 1997 – 2022 = 25 years $100,000 = $25,000 (1 + i)25 i = 25 4 − 1 = 0.057 or 5.7% per year.

So your friends did not really do that well at all!

(b)

$100,000 = $1,000 (F/A, i′, 25) (F/A, i′, 25) = 100 i′ = 10.1% per year In this situation they did pretty well on their mutual fund investment.

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4-22

1972 = $9000 2016 = $118500

F = (F/P , I, 44) 118500 = 9000 (F/P , I , 44) (F/P , I , 44) = 13.166 i = 6.015%

F = (F/P , I ,8) 168600 = 118500 (F/P , I ,8) 1.4227 = (F/P , I ,8) i = 4.49678%

The rate of increase over the last eight years is lower than over the previous 44 years.

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4-23

Profit from oil = 21000 * 70 = $1,470,000 Profit from natural gas = 1,500,000 * 0.4 = $600,000

Total profit per year = $2,070,000 N = 10 i = 15%

P = A* (P/A , 15%, 10) P = $2,070,000 * (5.0188) P = $10,388,916

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4-24

P0 = A (P/A, 12%, ∞) = 8.333A

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4-25

The F-equivalent amount of this series of deposits can be determined as follows:

F = $2,000 (F/A, 3%, 20) = $2,000 (26.8704) = $53,740.80

From Excel, F = −FV(0.03, 20, 2000) = $53,740.75

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4-26

N = 13, I = 4% A = $3000

P = A*(P/A, 4%, 13) P = 3000 * 9.9856 P = $29,957

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4-27

F = $20,000 (F/A, 5%,15) = $20,000 (21.5786) = $431,572

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4-28

Mortgage payment max = 1700-450 = $1250

P = 1250*[((1+0.005)^360)-1]/[0.005*((1+0.005)^360)] P = 1250 * (166.791) P = $208,489.52

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4-29

A = $848 N = 45 I = 10%

F = A* (F/A , 10%, 45) F = 848 * (718.9048) F = $609,631.27

No, the friend’s claim isn’t true.

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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 payment is doubled 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

P = $500,000 I = 5% N = 18

Case 1: When I = 5% A = P*(A/P , 5%, 18) A = 500,000 * (0.0855) A = $42,750 Total interest earned = $(42,750)(18) – $500,000 = $269,500

Case 2 : When I = 4% A = P*(A/P,4%,18) A = 500,000 * (0.0790) A = $39,500 Total interest earned = ($39,500)(18) – $500,000 = $211,000

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4-32

Since payments are made monthly, first find the monthly interest rate in effect.

$20,000 = $941.47 (P/A, i′, 24)

(P/A, i′, 24) = 21.2434

Using Appendix C we find that (P/A, 1%, 24) = 21.2434, so i′ = 1% per month.

The effective annual interest rate is (1.01)12 – 1 = 0.1268 or 12.68% per year.

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4-33

The carbon fiber automobile would average around 39 miles per gallon of gasoline. The dollar savings in fuel per year is

117,000 / 6 117,000 / 6  A =   ($4.00) = $600 30 39 

The extra sticker price that can be afforded is $600 (P/A, 20%, 6) = $1,995. This is not a whole lot of room to wiggle 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

A = $22,000

0 P=?

1

2

3

4

5

EOY

i = 15% / yr

P = $22,000 (P/A, 15%, 5) = $22,000 (3.3522) = $73,748.40 The company can justify spending up to $73,748.40 for this piece of equipment.

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4-35

$20,000

22

23

24

76 77

$250

$250

$250

EOY

$250

N = 55 A = $250 F = $20000

F = A* (F/A ,i , 55) (F/P, i, 1) I = 1.2744%

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4-36

Set up an equation for the unknown P-equivalent amount:

P - 0.3P = 0.6 ($3,500)(P/A, 9%, 15),

or 0.7P = $2,100 (P/A, 9%, 15).

0.7P = 2100 * 8.0607 P = $24182

The value of P (affordable amount) is $24,182 This type of heat pump is also a cleaner choice for the environment.

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4-37 A = 16,200,000 * (A/P , 7%, 15) A = 16,200,000 * (0.1098) A = $1,778,760

Now, the annual bond payments are less than annual savings generated. Hence, it’s a good investment.

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4-38

(A/F, 3.7%, 42) = (

. .

)

= 0.01028

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4-39

Apply the formula (Equation 4-10) to determine the (P/A, 7.75%, 10) factor: (P/A, 7.75%, 10) = [(1.0775)10-1] / [0.0775(1.0775)10] = 6.7864

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4-40

(a)

N = $1 million per year / $0.005 per cup = 200,000,000 cups per year

(b)

N=

(c)

The compounding of interest.

$5 million (A/F, 15%, 5) = 148,300,000 cups per year $0.005 per cup

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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

P = $14,500,000 N=7 I = 15%

A = P*(A/P, 15%,7) A = 14,500,000 * (0.2404) A = 3,485,800

The annual savings should be equal or more than $3,485,800.

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4-43

Fuel savings over eight years =

100,000 miles 100,000 miles  275 gallons, which is about 34.4  26 miles/gall on 28 miles/gall on

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

$800 = (34.4 gal/yr)(X/gal)(P/A, 10%, 8)

$800 = X, or X = $4.36/gal (34.4 gal)(5.3349)

If the actual cost of gasoline is less than $4.36 per gallon, the CVT is not a good deal.

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4-44

A = $10,000,000 (A/P, 7%, 6) = $10,000,000 (0.2098) = $2,098,000 per year

Total interest paid = $2,098,000 (6) − $10,000,000 = $2,588,000

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4-45

A = $500,000 (A/P, 15%, 6) = $500,000 (0.2642) = $132,100

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4-46

With traditional financing, the monthly payment will be A = $84,000 (A/P, ¾%, 60) = $1,747.20. The monthly payment with the 0% financing plan would be $84,000/60 = $1,400 per month. Total savings = ($1,747.20 − $1,400)(60) = $20,832.

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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

A = $100 / year

I = 15% N = 10 A = $5,000

F9 = A*(F/A , 15%, 10) F9 = 5000 * (20.3037) F9 = $101,518.5

F10 = $101,518.50 (F/P, 15%, 1) = $116,746.28

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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

The $30,000 is borrowed in the fall of a typical student’s education.

P = $30,000 N=4 I = 5.2%

A = 30,000 * (A/P , 5.2%, 4) A = 30,000 * (0.2833) A = $8,499

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4-51

$3,500 = $100 (P/A, 1.75%, N), so by trial and error, or exact numerical solution, we find that N = 55 months to repay the credit card value. Note: (P/A, 1.75%, 55) = 35.1345

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4-52

Sara will withdraw $40,000 for N years while her investments earn 3%. In equivalence terms we have: $40,000 = $1,000,000 (A/P, 3%, N) 0.04 = (A/P, 3%, N) N = 47 years Using the Excel function NPER (rate, pmt, pv, fv): N = NPER (0.03, -40, 000, 1000000) = 46.9

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4-53

(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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4-54

F12 = $28,000 (F/P, 0.5%, 12) = $28,000(1.0617) = $29,727.60

A = F12 (A/P, 0.5%, 48) = $29,727.60 (0.0235) = $698.60 per month

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4-55

P25 = $5,000 (P/A, 10%, 12) = $34,068.5 N = 70 – 25 = 45 F70 = $34,068.50 (F/P, 10%, 45) = $2,483,270

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4-56

Draw a cash flow diagram as follows: $5,000

0

1

2

3

4

5

6

7

8

9

10 11 12

A

Solve for A: P9 = $5,000 (P/A, 3%, 3) = $5,000 (2.8286) = $14,143 P5 = P9 (P/F, 3%, 4) = $14,143 (0.8885) = $12,566 Finally, A = P5 (A/F, 3%, 5) = $12,566 (0.1884) = $2,367.44

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4-57

Sixth year maintenance cost = (6)($250) = $1,500 EOY 1

2

3

4

5

6

7

8

9

10

11

12

1.5k 1.75k 2k 2.25k 2.5k 2.75k 3k John would need to set aside the following: P = $1,500 (P/A, 6%, 7) + $250 (P/G, 6%, 7) = $1,500 (5.5824) + $250 (15.450) = $83,736 + $38,625 = $12,236.10

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4-58

The instructor is encouraged to draw a cash flow diagram for this problem. Lisa will make 40 annual payments of $A each into her retirement account. So at age 61 she will have F = $A (F/A, 3%, 40) = $75.4012 A in her account. The amount A that will fund Lisa's withdrawals of $10,000 per 24 years will be:

75.4012 A = $10,000 (P/A, 3%, 24 ), or

75.4012 A = $10,000 (16.9355)

A = $2,246.

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4-59

The equivalence relationship is: $10,000 = $2,000 (P/F, 4%, 1) + $2,000 (P/F, 4%, 3) + $3,000 (P/F, 4%, 6) + B$ (P/F, 4%, 7) Next run the numbers to determine the value of B$ $10,000 = $2,000 (0.9615) + $2,000 (0.8890) + $3,000 (0.7903) + B$ (0.7599)

B$ = 5,169.23

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4-60

(a) EOY

EOY $2,000

0

1

2

3

4

5=N

vs.

0

1

2

3

4

5=N

X

X

X

X

X

1,000 1,500 2,000 2,500 3,000 $10,000

(b)

P0 = -$10,000 - $1,000 (P/A, 8%, 5) - $500 (P/G, 8%, 5) + $2,000 (P/F, 8%, 5) = -$16,317.5 AW = -$16,317.5 (A/P, 8%, 5) = -$4,087.53 Lease if annual fee is less than $4,086.20

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4-61. By trial-and-error you will converge on i’ = 14% per year: A = F (A/F, 14%, 40) = 0.0007F 2A = F (A/F, 14%, 35) = 0.0014F 4A = F (A/F, 14%, 30) = 0.0028F Moral: start saving early!

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4-62

(a) F15 F5 /2

1 2 3 4 5 6 7 8

9

10

11

12

13

14

15

1K 1K 1K 1K 1K 2K 2K 2K 2K 2K b) $1000 for 5 years I = 5%

F = 1000 * (F/A , 5%, 5) F = 1000 * (5.5256) F = $5525.6 So, at the end of 5 years, the maount withdrawn is 5525.6/2 = $2762.8

c) Now, N = 5-15 = 10 P = $2762.8 I = 5% A = $5000 for N = 5

F(10) = 2762*(F/P , 5% ,5) + 2000*(F/A , 5%,5) = (2762*1.2763) + (2000*5.5256) = $14,576.3406

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F(15) = 14576.3406* (F/P ,5%,5) = 14576.3406*(1.2763) = $18,603.78

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4-63

(a)

Cost in 10 years = $75,000 (F/P, 10%, 10) = $75,000 (2.5937) = $194,528

A = $194,528 (A/F, 6%, 10) = $194,528 (0.0759) = $14,765

(b)

Because $150,000 is less than $194,528, this is 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 = $15,000 (F/A, 15%, 4) = $15,000 (4.9934) = $74,901 Next the annual equivalent “return” on this investment can be determined: A = $74,901 (A/P, 15%, 10) = $74,901 (0.1993) = $14,928 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

$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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12

4-67

The cash flow diagram is shown below:

$1.25 million

2024

2025 2026 2027

2028

2029 2030 2031 2032

A

A

A

A

A

A

A

A

End of year, i = 18% per year

A

New Solution: P(2024) = $1,250,000 P2023 = 1,250,000 * (P/F, 18, 1) = 1,250,000 * (0.8475) = $1,059,375

Now, for annual beginning in 2024 and ending in 2032 (N = 9) A = 1,059,375 * (A/P , 18%, 9) = 1,059,375 * 0.2324 = $246,198.75

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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 = ($0.08/gal)(20,000 gal/mo) = $1,600 per month.

$30,000 = $1,600 (P/A, 1%, N), so (P/A, 1%, N) = 18.75

From Table C-4, N = 21 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=?

The years 7, 8, and 9 have annual payments of $10,000 at 10% compounded semiannually. This is equivalent to 6 semiannual payments of $10,000 (A/F, 5%, 2) = $4,878 for six periods beginning with period 6.5 and ending with period 9.

P6 = $4,878 (P/A, 5%, 6) = $24,759.26 P2.5 = $24,759.26 (P/F , 5%, 7) = $17,596.41 A = $17,596.41 (A/F , 5%, 6) = $2,586.67

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4-75

P0(rental income) = $13,500(P/A,9%,15) − $500(P/G,9%,15) = $13,500(8.0607) − $500(43.807) = $86,915.95

The present value of interest and rental income is $86,915.95 which is less than the original investment of the property ($95,000). Hence it is not a good investment.

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4-76

P0 (loan amount) = P0 (loan repayment) $10,000 = Z(P/G,7%,8)(P/F,7%,2) $10,000 = Z(18.789)(0.8734) $10,000 = 16.4103 Z Z = $609.37

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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)

$2,000 = $100 (P/A, 0.5%, 12) + G (P/G, 0.5%, 12) = $100 (11.6189) + G (63.214)

G = $13.26 per month beginning at the end of month 2

(b)

A = $2,000 (A/P, 0.5%, 12) = $2,000 (0.0861) = $172.20

(c)

G=

$2,000  $150(11.6189) =$4.07 per month beginning at end of month 2.

63.214

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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 (P/F,6%, 1) + 500 (P/A, 6%, 5)(P/F, 6%, 1) – G(P/G, 6%, 5)(P/F, 6%, 1) Or 2,000 = 100(0.9434) + 500 (4.2124)(0.9434) – G(7.935)(0.9434) 2,000 = 2,081.33 – 7.4859G G = $10.86

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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):

1  ( P/F, 15%, 10)(F/P, 12%, 10) = $18,000(0.2322)/0.03 = $139,320. Therefore, the 0.15  0.12 investment in the insulation is justified by a wide margin.

P = $18,000

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4-85

P = 12,000[1 – (P/F, 10%, 20)*((F/P , 2% , 20)]/(0.1 – 0.02) = $116,879 F = $116,879 * [F/P, 10% , 20] = 116,879 * (6.7275) = $786,303.472

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4-86

N = 40 years

A1 = $500

i = 4%

𝑓 ̅ = 2%

The P-equivalent of the account at 20 years old (from Equation 4-30) is:

P=

$

[

( ⁄ , %, .

)( ⁄ , %, .

)]

=

$

[ –( .

)( . .

)]

=

$

( . .

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

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

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)

= $13,502.50

4-87

P=

$

[

/ ,

%, .

)( / , %,

)]

.

= $4,684.51

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4-88

PGas =

$10,000[

(P/F, 15%, 15)(F/P, 10%, 15)] .

= $97,324

.

PMaint. = 345(15)(P/F, 15%,1) = $4,500

A = ($97,324 + $4,500)(A/P, 15%, 15) = $17,412

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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: P = $12,000 [1 – (P/F, 5% ,4)]*(F/P , 8%, 4)] / (0.05-0.08) P = $12,000 * [1-(0.8227)*(1.3605)]/(-0.03) P = $47,713.34

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4-90

F=

$3,000[

(P/F, 5%, 40)(F/P, 4%, 40)] .

.

(F/P, 5%, 40) = $672,161

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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 = $1,000 (P/A, 12%, 15) +

$300[1  ( P/F, 12%, 10)(F/P, 6%, 10)] (P/F, 12%, 5) = $8,012 0.12  0.06

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4-93

We can use Equation 4-30 to find the P-equivalent of your savings plan: P = $5,000 [(1-P/F, 10%, 20)(F/P, 3%, 20) / (0.10 -0.03)] P = $5,000 (10.456) P = $52,258

,and finally

F = $52,258 (F/P, 10%, 20) = $52,258 (6.7275) = $351,566

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4-94 5k

5k

5k

5k 4

0

1

2

3 F4

CFD (student’s viewpoint)

After two years you will owe F2 = 5k (F/P, 4%, 2) + 5k (F/P, 4%, 1) = $5k(1.0816) + 5k(1.0400) = $10,608.

At the end of year 4 you will owe F4 = 5k (F/P, 6%, 2) + 5k (F/P, 6%, 1) + $10,608 (F/P, 6%, 2) F4 = 5k (1.1236) + 5k (1.0600) + $10,608 (1.1236) F4 = $22,837.15

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4-95

Using Equation (4-30):

P1-7 = [$5000 / (0.15 – 0.08)][1 – (P/F, 15%, 7)(F/P, 8%,7) = $25,413 at time 0 P8-20 = [$5000(1.08)7 / (0.05 – 0.08)][1 – (P/F, 5%, 13)(F/P, 8%, 13)] (P/F, 15%, 7) = $47,480 at time 0 Ptotal

= $25,413 + $47,480 = $72,893

Alternative solution using iCR = (1 + i)/(1 + f) – 1: During the first seven years, i = 15%. For this time period, iCR = (1.15/1.08) – 1 = 6.48%. The initial payment for this series is $5,000.

During the next 13 years, i = 5%. For this time period, iCR = (1.05/1.08) – 1 = −2.78%. The initial payment for this series is $5,000(1.08)7 = $8,569.

P0 = ($5,000/1.08)(P/A, 6.48%, 7) + ($8,569/1.08)(P/A, −2.78%, 13)(P/F, 15%, 7) = $72,904

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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

The 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 the borrower, the company has overstated the amount of this savings.

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4-99

(a) $2500 at n=0 and i = 8% per year.

Amount after 2 years = 2500 * (F/P, 8%, 2) = 2500 * 1.1664 = $2,916 Now the interest rate becomes 12% / 4 = 3% per quarter for 3 * 4 = 12 quarters.

Amount after 5 years = 2916 * (F/P, 3%, 12) = 2916 * (1.4258) = $4,157.63 New interest rate = 8% / 12 = 0.667% monthly for (9 – 5)*12 = 48 months.

Amount after 9 years = 4157.63 * (F/P, 0.667%, 48) = 4157.63 * (1.3757) = $5,719.65 New interest rate = 6% annually Amount after 15 years = 5719.65* (F/P, 6%, 6) = 5719.65*1.4185 = $8,113.32 (b) Now, F = P*(F/P, i , 15) 8,113.32 = 2500* (F/P, i, 15) 3.2453 = (F/P, i,15) i = 8.16%

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4-100 (a)

(1 + 0.08/12)12 – 1 = 8.3%

(b)

1.1102 = (1+(APR)/12)^12 APR = 10.52%

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4-101 The total elapsed time period is 50 years or 50 × 12 = 600 months.

F = $30,000 (F/A, 0.25% per month, 600 months) = $30,000 (1,389.3231) = $41,679,693 Suppose the winner of the lottery spends $15,000 each month and saves the remaining $15,000 per month. How much will the account be worth after 50 years?

Half of the above = $20,839,846.50

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4-102 Assume monthly compounding of interest. This is a very high interest rate, bordering on usury. The interest rate per month is 28.4%/12 = 2.367%. The effective annual interest rate is (1.02367)12 – 1 = 0.3241 (32.41%). John should avoid this borrowing opportunity.

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4-103 First, draw a cash flow diagram:

A = $10,000 per year

0

7

8

9

10

11

12

13

14

15

16

17

EOY

P0

Next, solve for i/yr and P0:

i/year = (1.005)12 – 1 = 0.0617 (6.17%)

P0 = $10,000(P/A, 6.17%, 10)(P/F, 6.17%, 7) = $10,000(7.3012)(0.6576) = $48,012.69

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4-104 The future worth of monthly payments is: F

= ($75/month x 6 months)(F/A, 3.5%, 9)

= $450 (10.368) = $4,666 The amount owed after 4.5 years is $13,630 - $4,666 = $8,964

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4-105 You pay nothing on this loan for months 1-12. Your first payment in month 13 and the following 47 months will be:

A = $25,000 (A/P, 0.75%, 48) = $25,000 (0.0249) = $622.50 per month.

If you finance your loan over 60 months, your monthly payment will be:

A = $25,000 (A/P, 0.75%, 60) = $25,000 (0.0208) = $520 per month.

This is $102.50 per month lower than the 48-month loan.

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4-106 The monthly interest rate is 3% divided by 12, or ¼% per month.

In five years there are 5 x 12 = 60 months, so the future equivalent value of this account is: F = $100 (F/A, ¼%, 60) F = $100 (64.6467) F = $6,464.67

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4-107 (a)

Assume end-of-year cash flows and M = 1 P0

(b)

= $10,400 (P/A, 12%, 10) + $150,000 (P/F, 12%, 10) = $107,062

Viewpoint: the buyer $150,000 $10,400/yr.

1

2

3

4

5

6

7

8

or

12.72%

P0 (c) (d) (e) (f)

The interest rate Property taxes, upkeep expenses The security fence i = (1 + 0.12/26) 26 – 1 = 0.12718

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9

10 = N

4-108 F = $300 (F/A, 0.5%, 360) = $300 (1004.515) = $301,355.

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4-109 The quarterly nominal interest rate is 3%. This matches the quarterly deposits being made into the savings account. So we can write an equivalence relationship to determine the unknown time, N. $1,000 (F/A, 3%, N) = $42,931 (F/A, 3%, N) = 42.931

We can use Goal Seek in Excel to determine the value of N. Goal Seek Setup: Cell A1:

20 (starting guess for unknown number of deposits)

Cell B1:

=((1.03)^a1-1)/0.03

In Goal Seek, set value of cell B1 = 42.931 by changing cell A1. Doing so, we determine N = 28 months So 28 monthly deposits must be made (i.e. 7 years).

Excel Solution:

N = NPER (0.03, 1000, , -42931) = 28

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4-110 By trial and error, we need to guess what the monthly interest rate is in order to see if the stated conditions are met. Let's guess that 0.75% per month is being charged (just a lucky guess I suppose).

Then for the 10-year loan, A = $22,500 (A/P, 0.75%, 120) = $22,500 (0.0127) = $285.75 per month. For 20 years, A = $22,500 (A/P, 0.75%, 240) = $22,500 (0.0090) = $202.50 per month. This reduces the monthly payment by [($285.75 - $202.50) / $285.75] x 100 = 29%.

The total interest for the 10-year loan is 120 ($285.75) - $22,500 = $11,790. For the 20-year loan the total interest is 240 ($202.75) - $22,500 = $26,100. The increase equals ($26,100 - $11,790) / $11,790 = 1.2137, or 121%. Thus, the interest rate being charged is 9% APR or 0.75% per month with monthly compounding.

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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 Assume Pat’s first loan will be repaid at 1% per quarter for four years (no increase in the interest rate after two years). The future equivalent of the first loan is F1 = $20,000 (F/P, 1%, 16) = $20,000 (1.1726 ) = $23,452 The second loan has a future equivalent of F2 = $20,000 (F/P, 6%, 2) = $20,000 (1.1236) = $22,472 Pat will repay at the end of four years: F1 + F2 = $23,452 + $22,472 = $45,924

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4-114 A = $24,000 (A/P, 8.3%, 10) = $24,000 (0.1511) = $3,626.40

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4-115 P = $50,000 r = 10% compounded quarterly, so i = 2.5% per quarter N = 4 X 18 = 72

F = P* (F/P, 2.5%, 72) F = 50,000 * (5.9172) = $295,860

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4-116 A(monthly) = $5,500(A/F, 0.75%, 12) = $440 P = $440 (P/A, 0.75%, 12 X 16 = 192) = $44,692

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4-117 i = e0.072 – 1 = 0.0747 or 7.47% per year

F = $5,000 e0.072(2) = $5,774.42

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4-118 F18 = $15,000 (F/P,8%,18) = $15,000 * (4.2207) = $63,310.5

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4-119 r = 10% / 2 = 5% every six months, compounded continuously

A = $18,000(A/P, 5%,24) =

$18,000 $18,000 = $1,320.66  (P/A, 5%,24) 13.6296

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4-120 P = $3,500 (P/A,10%,5) = $3,500 (3.7412) = $13,094.20

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4-121 A = $400 monthly i = 6/12 = 0.5% monthly

Now at the end of 6 years,

F = A* (F/A , 0.5%, 72) = 400 * (86.4089) = $34,563.56

Now, seven years later,

F = P* (F/P , 0.5%, 84) F = 34,563 * (1.5204) = $52,550.436

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4-122 r = 10% per year, compounded continuously (a)

F = $2000 (F/A,10%,30] = $2000 (181.472) = $362,944

(b)

A = $362,944(A/P,10%,10)=

$362,944 $362,944 = $60,386  (P / A,10%,10) 6.0104

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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. 

P0 = $185(P/A,0.833%,24) = $185 

e

e 0.00833(24)  1 0.00833(24)

(e

0.00833

    1) 

= $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 F = $10,000 (F/P, 11%, 25) = $10,000 (13.5855) = $135,855

F (in today’s dollars) = $135,855 (P/F, 3%, 25) = $135,855 (0.4776) = $64,884

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4-126 (a)

Javier’s monthly mortgage payment is

$100,000 (A/P, 0.5%, 360) = $100,000 (0.0060) = $600

After adding the additional $400 for property taxes and insurance, the total condo payment is $1,000 per month.

(b)

This is an open−ended exercise for students to have fun with. Utilities could easily be in the $300−$500 per month ballpark. Condo homeowner association fees might be another $100−$200 per month. Other miscellaneous expenses could bring Javier’s total monthly expenses to $1,500−$1,700. This is a fairly expensive undertaking for Javier!

(c)

The monthly interest rate is 5.8% / 12 = 0.4833%. Javier’s monthly mortgage payment will be $100,000 (0.0083) = $830. The savings in interest paid on the mortgage between the answers in parts (a) and (c) is enormous ($66,600).

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4-127 If the rate of return on your invstment is about 14% per year, set up cash flow diagrams for starting to save at age 20 (first deposit at age 21), age 25 and age 30 to determine that

A = 0.0007F for 40 years of saving 2A = 0.0014F for 35 years of saving 4A = 0.0028F for 30 years of saving.

Here F = $1,000,000 and the coefficient of F is the appropriate (A/F, i%, N) factor. The big assumption is whether you can earn 14% per year on your investments. If not, the above statement is not true. That it is true for i = 14% is a real eye−opener for most students. Moral: start saving early!

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4-128 The equivalent value of the left-hand cash flow diagram at time 0 is: P0

= -1000 – 500(P/F, 10%, 1) + 500 (P/A, 10%, 4)(P/F, 10%, 2)

P0

= -1000 – 500(0.9048) + 500 (3.1347)(0.8187)

P0

= -169.21

The equivalent value of the right-hand cash flow diagram at time 0 is: P0’

= 500 – 500(P/F, 10%, 1) + R$ (P/F, 10%, 2) + R$ (P/F, 10%, 4) + R$ (P/F, 10%, 5)

P0’

= 500 – 452.40 + 0.8187 R$ + 0.6703 R$ + 0.6065 R$

P0’

= 47.6 + 2.0955 R$

Set P0 = P0’ and solve for R$ -169.21 = 47.6 + 2.0955 R$ -216.81 = 2.0955 R$ R$ = -103.46 The R$ amount should have been shown as downward arrows on the cash flow diagram.

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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 The initial car will be repaid after 5 years. Thus, the savings will be during the 5 years for $415.20 per month at 6% interest (compounded monthly), which is 0.5% for 60 months.

FW(0.5%) = $415.20(F/A, 0.5%, 60) = $415.20(69.7700) = $28,969 The new graduate would have saved $28,969 by the end of the 10-year period.

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4-134Currently the rent payment is $1,200 per month and the utilities cost $150 per month. The new monthly payment cannot exceed $1,200 per month, which must include property taxes and insurance.

First, calculate the amount of the $1,200 that can be dedicated to property taxes and insurance.

X*1.3 = $1,200 X = $923 Thus, the principal and interest payment cannot exceed $923 per month.

Second, calculate the present worth of the $55,000 down payment and the $923 per month payment. The interest rate will be 8/12 = 0.667% for 360 months.

PW = $55,000 + $923(P/A, 0.667%, 360) = $180,790.

Thus, the maximum purchase price for the condo is $180,790.

We note that $55,000/$180,790 is above the 20% threshold to avoid the private mortgage interest insurance. Also, we note that the $335.50 monthly savings that was going towards the down payment was not included in these calculations.

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4-135The new graduate is contributing $500 per month towards retirement. Thus, assuming monthly compounding (10/12% = 0.8333%), FW = $500(F/A, 0.833%, 120) = $102,422 The new graduate will have $102,422 in their account at the end of 10 years.

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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 P = $5000 N = 10 A = $300 i = 5%

F = P*(F/P, 5%, 10) + 300*(F/A , 5%, 10) F = 5000 * (1.6289) + 300*(12.5779) F = 8144.5 + 3773.37 F = $11917.87

Option c.

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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 7% being earned on municipal bonds and are therefore feasible investments. We need to consider rank-ordered projects (based on annual rate of profit) versus cumulative capital investment:

Project ranking

A

E

D

C

B

Profit rate per year (%)

30

25

15

12

10

Cumulative investment ($millions) 10

15

45

75

100

Because of limited funds only projects A, E and D can be undertaken. The best rejected project is C, so TTW's MARR is 12% per year. Only whole projects are possible, so C and B cannot be funded.

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5-2

Let’s calculate the present worth (PW) of the venture: PW (5%) = −80,000 + 20,000(P/A, 5%, 6) = 21,514

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5-3

The annual equivalent series of the every five year payment is: A2-6 = 15,000(A/F, 4%, 5) = 2,769.41 Note that this constitutes an indefinite series beginning in year 2. We need to find the present equivalent of that series and the initial payment of $15,000 in year 1.

P0 = 15,000 +

,

. .

(𝑃⁄𝐹 , 4%, 1) = 80,995.43

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5-4

Let X = annual savings required.

AW(20%) = 0 = X – [$3,000,000(A/P, 20%, 7) − $300,000(A/F, 20%, 7)] X = $808,980 / year Net savings per pallet = ($808,980/year) / (1,000,000 pallets/year) = $0.81 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, 15%, 5) = $500 + $500(1.7228) = $1,361.40 Next the capitalized worth can be computed: CW (15%)

= EUAC/0.15 = $1,361.40/0.15 = $9,076

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5-6

Demand charge / year =

Energy charge / year =

($92 / kW/yr)(100 hp)(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(18%) = −$80,000 + $15,000 (P/F, 18%, 1) + $20,000 (P/F, 18%, 2) + $25,000 (P/F, 18%, 3) + $30,000 (P/F, 18%, 4) + $35,000 (P/F, 18%, 5) PW(18%) = -$6,936

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5-8

The PW of the incremental investment is: PW(8%) = -$400 +

15,000

mi yr

19 mpg

-

15,000

mi yr

24 mph

$3.50 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 is roughly constant at 12,000. A four-speed transmission car will consume 400 gallons of gasoline each year, and the fuel cost is $1,600 per year. The six-speed transmission will attain (1.04)(30 mpg) = 31.2 mpg for an annual consumption of 384.6 gallons and a fuel cost of $1,538.40. The annual savings due to a six-speed transmission is $1,600  $1,538.40 = $61.60. Over a ten year life, the present worth of savings will be $61.60 (P/A, 6%, 10) = $453. This is how much extra the motorist should be willing to pay for a six-speed transmission.

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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

End of year

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8

9

10

5-11

Monthly interest payment = 0.0075($10,000) = $75 New effective monthly interest rate = 0.07/12 = 0.005833 Current value of bond = $75 (P/A, 0.5833%, 59) + $10,000 (P/F, 0.5833%, 59) = $10,829.94

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5-12

Using Equation (5-2), the value of the bond now (VN) at a higher interest rate of 5% per year can be computed as follows: VN = $10,000 (P/F, 5%, 10) + $10,000(0.04)(P/A, 5%, 10) VN = $10,000(0.6139) + $400(7.7217) VN = $9,227.68 When the desired yield is more than the interest rate being earned, the purchase price of the bond is less than its face value.

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5-13

Using now as our reference point and letting X = selling price of the bond, X = 1000 (F/P, 5%, 6) – 40(F/A, 5%, 6) = 1068.02

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5-14

$1,000,000 + $500,000 (P/F, 35%, 1) − $500,000 (P/F, 35%, 2) = ($Y / 0.35)(P/F, 35%, 2) $1,000,000 + $500,000 (0.7407) − $500,000 (0.5487) = 0.5487Y / 0.35 1,096,000 = 1.5677Y Y = $699,113

Factors: (P/F, 35%, 1) = 0.7407 (P/F, 35%, 2) = 0.5487

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5-15

From the viewpoint of WRC, they can determine the annuity equivalent of their services:

A = $360,000 + $15,000 (A/G, 12%, 100) = $484,982.

Then PW(12%) = $484,982 (P/A, 12%, 100) = $4,041,452.

WRC will receive more upfront money than the value of its services over a 100-year period, so this is a profitable agreement for WRC. They will earn more than 12% on this deal.

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5-16

(a) CW(10%) =

$1,500  $10,000 (A F ,10%,4)   ( P / F,10%,1) = $34,591 010 .

010 .  

(b) Find the value for N for which (A/P,10%,N) = 0.10 From Table C-13, N = 80 years

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5-17

AW = −$20,000,000 (A/P, 8%, 40) − $600,000 = −$2,278,000

CW = −$2,278,000 / 0.08 = −$28,475,000

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5-18 P = $150 (P/A, 2%/qtr., 60 qtr.) + $10,000 (P/F, 2%/qtr., 60 qtr.)

P = $150(34.7609) + $10,000(0.3048) P = $5,214 + $3,048 P = $8,262

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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%) = −$10,000 (F/P,15%,5) + ($8,000 − $4000)(F/A,15%,5) − $1,000 = −$10,000 (2.0114) + $4000(6.7424) + $1,000 = $7,855.60

Since FW(15%) ≥ 0, the project is acceptable.

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5-22

PW = $1,020[1 – (P/F, 10%, 20)(F/P, 2%, 20)]/(0.10 – 0.02) = $9,935

FW = $9,935(F/P, 10%, 20) = $66,838

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5-23

FW(15%) = −$150,000 (F/P, 15%, 4) + $50,000 (F/A, 15%, 4) = −$12,680

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5-24

The capital recovery amount based on Equation (5-5) is: CR(10%) = $4 million (A/P, 10%, 8) + $1 million (A/F, 10%, 8) CR(10%) = $4 million (0.1874) + $1 million (0.0874) = $0.837 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 0.48($9,137) = $4,386 per year. Fuel related operating expense is 15,000 miles/year × $0.21/mile = $3,150 per year. So the other operating expenses (insurance, tax, license and registration) are: $9,137 − $4,386 − $3,150 = $1,601 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%) = -$500 million (A/P,12%, 30) + $50 million = -$62.05 million + $50 million = -$12.05 million per year.

This is not a profitable investment.

(b)

Theta = $500 million / $50 million per year = 10 years.

A payback period of 10 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 5% interest per year is: CR(5%) = $150,000 (A/P, 5%, 8) = $23,205 per 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(15%) = −$750,000(A/P, 15%, 5) − $45,000 + X ($65 − $8.50) = 0

Solving yields X = 4,756 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, 18%, 15) + $20,000,000(P/F, 18%, 15) 0 = −$50,000,000 + 5.0916A − $76,374,000 + $1,670,000 $124,704,000 = 5.0916A A = $24,492,105 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

PW(IRR) = 0 = −$40,000 + $10,000 (P/A, IRR, 6) (P/A, IRR, 6) = $40,000 / $10,000 = 4 (P/A, 12%, 6) = 4.1114 and (P/A, 15%, 6) = 3.7845, so 12% < IRR < 15%.

Using linear interpolation we find that the IRR = 13.02%.

Using Excel, IRR = RATE(6, 10000, −40000) = 12.98%

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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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5-42

PW(IRR) = 0 = −$100,000 + $25,000 (P/A, IRR, 5) (P/A, IRR, 5) = $100,000 / $20,000 = 4 (P/A, 7%, 5) = 4.1002 and (P/A, 8%, 5) = 3.9927, so 7% < IRR < 8%.

Using linear interpolation we find that the IRR = 7.93%.

Using Excel, IRR = RATE(5, 25000, −100000) = 7.93%

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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.

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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%

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

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100%

5-46

PW(IRR) = 0 = −$50,000 + $20,000(PA, IRR, 4) + $5,000 (P/G, IRR, 4) Using Excel, IRR = 36.44%

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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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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 IRR = 28.8%

PW(i%)

5-48

$50,000 $0 -$50,000 -$100,000

5%

10%

15%

20%

25%

30%

35%

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

The simple payback period is $60,000 / $15,000 = 4 years.

The IRR can be determined from this equation:

0 = $60,000 + $15,000 (P/A, IRR, 5)

From Excel we can determine IRR = 7.93%. This is a good investment when MARR = 6%.

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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′ = 8.07% per year. Given a MARR = 6% per year, the plant would be considered profitable.

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5-51

The simple payback period is $6,500 / $1,875 = 3.47 which rounds up to 4 years.

The IRR can be determined from this equation:

0 = $6,500 + $1,875 (P/A, IRR, 7)

From Excel we can determine IRR = 21.44%.

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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 5 6 7

Cash Flow For Year $100,000 20,000 20,000 20,000 20,000 20,000 20,000 20,000

PW(8%) of Cash Flow $100,000 18,519 17,147 15,877 14,701 13,611 12,603 11,670

Cumulative PW thru Year $100,000  81,841  64,334  48,457  33,756  20,145  7,542 4,128

>0

The discounted payback period is 7 years.

PW(8%) = −$100,000 + $20,000(P/A, 8%, 10) = $34,202

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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

$60,000 (1 + ERR)5 = $20,000 (F/A, 8%, 5) ERR = 14.35%

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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 2 = 140 months or 11 years . $200/month 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

Assume the loan is being repaid over 30 years and there are 500 apartments.

Annual Rental income less expenses = ($12,000 − $4,000) × 500 = $4,000,000 AW = −$30,000,000 (A/P, 6%, 15) + $4,000,000 = $910,000

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5-62

Simple payback = $5,000 / $65 per day = 76.9 or 77 days.

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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

$ $ $ $

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

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

(% of principal)

* 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:

300

$1,993,846

Annual Profit (Loss)

$0

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

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5-69

A summary of our parameter values are as follows: Initial cost: $40,000,000 Monthly cost: $400,000 Monthly savings: $1,000,000 Interest Rate: 1% compounded monthly for 120 months

PW = -$40,000,000 - $400,000(P/A, 1%, 120) + $1,000,000(P/A, 1%, 120) = $1,820,313

The prediction is that the Present Worth will be $1,820,313.

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5-70

It is known that at the breakeven point, profit equals zero. That is, the PW of the project is equal to zero, or PW of costs = PW of revenues. The PW of costs is: PW(1%) of costs = $60,000,000 + $400,000(P/A, 1% per month, 120 months) = $87,880,209

Thus, $87,880,209 = $X (P/A, 1%, 120), where X is the breakeven financial monthly benefit needed to justify the cost of the Data Center.

X = $87,880,209 / 69.70 = $1,260,826.

Thus, provided the monthly financial benefit exceeds $1,260,826 the project will be financially successful.

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5-71

You start your economic evaluation of the proposal by summarizing the costs and benefits: Capital Improvement (Initial Cost): $60,000,000 Monthly Costs: $600,000 Monthly Benefits: $2,583,333

PW(1%) = -$60,000,000 - $600,000(P/A, 1% per month, 120 months) + $2,583,333.33(P/A, 1%, 120) = $78,239,345.

You advise the company executives that, based on PW, the project should be undertaken.

It is known that at the breakeven point, profit equals zero. That is, the PW of the project is equal to zero, or PW of costs = PW of revenues. The PW of costs is: PW(1%) of costs = $60,000,000 + $600,000(P/A, 1% per month, 120 months) = $101,820,313

Thus, $101,820,313 = $X (P/A, 1%, 120), where X is the breakeven financial monthly benefit needed to justify the cost of the Data Center.

X = $101,820,313 / 69.70 = $1,460,826.

Therefore, provided the monthly financial benefit exceeds $1,460,826 the project will be financially successful.

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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) PW (12%)  $9,000 

 $9,998.08

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

*Students may realize that the scenario described in the problem has the cost become a revenue (an unlikely situation).

A: In-house treatment

0

1

2

3

4

5 G = $5,000

A = $30,000

B: Outside treatment

0

1

2

3

4

5

$15,000

AW(A) = -$30,000 + 5,000 (A/G, 10%, 5) = −$20,949.50 AW(B) = −$15,000 (F/P, 10%, 1) = −$16,500

Outside treatment 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

The lives of the alternatives are unequal. Let us assume that repeatability does not apply, so the PW can be compared over the different time periods.

PWA = −$200,000 + $60,000 (P/A, 7%, 5) = $46,012 PWB = −$180,000 + $55,000 (P/A, 7%, 6) = $82,157.50 Choose Venture B to maximize present worth.

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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%) = -$425,000 - $2,500(P/A, 8%, 20) – $8,000 [1- (P/F, 8%, 20)(F/P, 5%, 20)] / (0.08-0.05) = -$425,000 - $2,500(9.8181) – $8,000 [1- (0.2145)(2.6533)] / (0.03) = -$564,443

Geothermal:

PW(8%)

= -$350,000 - $15,000 (P/A, 8%, 20)

= -$350,000 – $147,272 = -$497,272 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   (8,760 hr/yr) ($0.022/kWh) − 40     0.9   hp   150hp   0.746kw   (8,760 hr/yr) ($0.022/kWh) − $150,000 − 20     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   (8,760 hr/yr) ($0.022/kWh) − $100,000 − 20     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   (8,760 hr/yr) ($0.022/kWh) − 20     0.9   hp   100hp   0.746kw   (8,760 hr/yr) ($0.022/kWh) − $170,000 − 40     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.

Non−economic factors include operating considerations and licensing the plant in a given location with its unique environmental characteristics.

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6-7

Rank order MEAs:

DN → C → B → A

First check Δ(C – DN) 0 = -$300,000 (F/P, i’, 15) + $57,000 (F/A, i’, 15) + $35,000 Plug in MARR = 12% ?

0 ⇔ -$300,000 (5.4736) + $57,000(37.2797) + $35,000 0 < $517,863 so i´> 12% Select C, reject DN. Next check Δ(B – C) 0 = -$100,000 (F/P, i’, 15) + $5,000 (F/A, i’, 15) + $5,000 By inspection, ∑ neg. cash flows > ∑ pos. cash flows so reject B, keep C. Finally check Δ (A – C) 0 = -$150,000 (F/P, i’, 15) + $13,000 (F/A, i’, 15) + $15,000 ?

0 ⇔ -$150,000 (5.4736) + $13,000(37.2797) + $15,000 0 > -$321,404 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

IRRA = RATE(5, 30000, -100000) = 15.24% IRRB = RATE(5, 40000, -135000) = 14.71% Project A has a higher IRR, but this does not mean that it is the preferred alternative. Both alternatives are acceptable when the MARR = 9%. Now we must look at the IRR of the incremental investment of $35,000 required for project B and determine if the extra $10,000 each year represents a return of > 9%.

IRRB-A = RATE(5, 10000, -35000) = 13.2% > 9% Project B is the preferred alternative.

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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.

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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

IRRA = RATE(5, 18000, -75000) = 6.4% < MARR IRRB = RATE(4, 22000, -90000) < 0. Note that the sum of the inflows < the sum of the outflows.

Neither project is profitable.

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6-15

PWX = -13,000,000 + 1,745,000 (P/A, 10%, 20) + 9,000,000 (P/F, 10%, 20) = 3,193,632 PWY = -11,000,000 + 1,120,000 (P/A, 10%, 20) + 6,200,000 (P/F, 10%, 20) = -543,448 PWZ = -7,000,000 + 975,000 (P/A, 10%, 20) + 4,800,000 (P/F, 10%, 20) = 2,014,040

Building X should be built to maximize PW.

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6-16

Cost of electricity for the 90% efficient motor: [30 hp/0.90](0.746 kW/hp)($0.12/kWh)(5,000 hr/yr) = $14,920 per year

Cost of electricity of the 93% efficient motor: [30 hp/0.93](0.746 kW/hp)($0.12/kWh)(5,000 hr/yr) = $14,438.71 per year

PW90%(15%) = −$5,200 − $14,920(P/A, 15%, 8) = −$72,151

PW93%(15%) = −$7,300 − $14,438.71(P/A, 15%, 8) = −$72,091

The 93% efficient motor is the better choice.

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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: PW = (-10kW)(24 hr/day)(365 day/yr)($0.1/kWhr)(P/A, 15%, 10) = -$43,965

B:

Fluorescent lighting system: PW = -$11,000 – (4.5kW)(24 hr/day)(365 day/yr)($0.1/kWhr)(P/A, 15%, 10) = -$30,784

Assuming no salvage value for the lights, the warehouse should switch to fluorescent lighting.

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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

Assume the project is not repeatable and it is appropriate to compare PW over the different lives.

PW1 = -$100,000 + $30,000 (P/A, 5%, 5) = $29,885 PW2 = -$90,000 + $25,000 (P/A, 5%, 6) = $36,893 Select project 2 to maximize PW.

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6-21

Assume the project is not repeatable and it is appropriate to compare PW over the different lives.

PW1 = -$250,000 + $70,000 (P/A, 4%, 4) = $4,093 PW2 = -$220,000 + $65,000 (P/A, 4%, 5) = $69,367 Select project 2 to maximize PW.

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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

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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Z -14,000 1,500 2,000

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

At 75% occupancy, 150 (0.75) = 112.5 rooms will be rented each night.

Variable occupance cost = (112.5 rooms rented per night)(365 nights/year)(variable cost/rented room)

AWSK = -$450,000((A/P,15%,5) - $50,000 – (112.5)(365)($10) + $80,000(A/F,15%,5) = -$582,996 AWBK = -$320,000((A/P,15%,5) - $40,000 – (112.5)(365)($11) + $50,000(A/F,15%,5) = -$579,729

Select the Black Knight Model to maximize annual worth.

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6-25

Assume repeatability and compare annual worth.

AWA = -$120,000(A/P, 5%, 5) + $35,000 = $7,280 AWB = -$100,000(A/P, 5%, 6) + $30,000 = $10,300

Select project B to maximize annual worth.

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6-26

With a MARR = 15%, both the Lance and Bella locations can be eliminated from consideration (IRR < 15%).

We need to look at the ERR of the increment of capital needed to use the Lilo location instead of the Judas location.

Incremental Investment = 1,450,000- 500,000 = 950,000 Incremental Revenue = 552,500 – 225000 = 327,000 Incremental Expenses = 326,625 – 146,250 = 180,375 Incremental Market Value = 62,500 – 65,000 = - 2,500

950,000(1 + ERR)50 = (327,000 – 180,375)(F/A, 15%, 50) – 2,500 ERR = 15.07% > MARR = 15%, so the incremental investment is justified. The Lilo location should be selected.

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6-27

Assume repeatability and compare annual worth.

AWA = -$70,000(A/P, 5%, 4) + $18,000 = -$1,740 AWB = -$80,000(A/P, 5%, 3) + $20,000 = -$9,376

Neither investment makes a profit, so doing nothing is appropriate.

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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 Assume repeatability so that AWs can be directly compared (over a 12−year study period).

AWH(12%) = −$50,000(A/P,12%,2) + ($250/hr - $100/hr)(4 hr/day)(250 days/yr) + $15,000(A/F,12%,2) = $127,490.50 AWR(12%) = −$80,000(A/P,12%,4) + ($250/hr - $110/hr)(4 hr/day)(250 days/yr) + $20,000(A/F,12%,4) = $117,848

Select Handy-Cam to maximize annual worth.

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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

The question is whether or not the repeatability assumption is appropriate. Let’s assume that repeatability does not apply.

PWA = -$250,000 + $65,000(P/A, 7%, 6) = $59,822.50 PWB = -$220,000 + $60,000(P/A, 7%, 7) = $103,358 Select B to maximize PW.

FWA = -$250,000(F/P, 7%, 7) + $65,000(F/A, 7%, 6)(F/P, 7%, 1) = $96,062 FWB = -$220,000(F/P, 7%, 7) + $60,000(F/A, 7%, 7) = $165,964 Select B to maximize FW.

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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. Copyright © 2025 Pearson Education, Inc.

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

Assume repeatability so that AWs can be directly compared.

AW1 = −$120,000(A/P, 10%, 6) + $30,000 = $2,448 AW2 = −$110,000(A/P, 10%, 7) + $28,000 = $5,406

Select Franchise 2 to maximize annual worth.

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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 a 24-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%, 8) + $32,000 = $49,832

Recommend 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 CR and maintenance cost/part = = $0.6475 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

015 .  0.057 (1088 . )6 1 iCR = = 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

Assume repeatability over the 120-year study period, and compare annual worths over the useful lives of the three alternatives.

AWB = -$875,000(A/P, 10%, 8) + $150,000 + $300,000(A/F, 10%, 8) = $12,245 AWD = -$930,000(A/P, 10%, 8) + $156,000 + $400,000(A/F, 10%, 8) = $16,678 AWS = -$305,000(A/P, 10%, 4) + $85,000 + $100,000(A/F, 10%, 4) = $10,322.50

Select the Delaney model to maximize annual worth.

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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. Compare annual worths.

AWA(12%)

= -$50,000 (A/P, 12%, 20) + $10,000 (A/F, 12%, 20) - $5,000 (A/F, 12%, 5) - $5,000 - $500 (A/G, 12%, 20) = -$15,353.10

AWB(12%)

= -$20,000 (A/P, 12%, 10) - $10,000 - $1,000 (A/G, 12%, 10) + $3,000 (A/F, 12%, 10) = -$16,953.70

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 3

$300

$300

5

6

4

$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 Person A, the future worth of the account in 36 months will be

$3,000 (F/P, 1.5%, 36) = $3,000 (1.7091) = $5,127.

The future worth of Person B’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, Person A will repay $1,395 more on the credit card than Person B 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 you do not repay any interest while in school.

This is 23.5% higher than the $5,180 you repay when interest-only is paid while 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 $11,500 (if you pay $12,500 now and invest the remainder). It can grow to $11,500 (1.03) = $11,845 by January 1.

It will not cover the $12,500 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 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

Compare AW over useful life. Note that Elizabeth can be eliminated because IRRElizabeth < MARR. AWEmily = -$305,000(A/P, 12%, 4) + $85,000 + $100,000(A/F, 12%, 4) = $5,514 AWMadison = -$930,000(A/P, 12%, 8) + $156,000 + $400,000(A/F, 12%, 8) = $1,311 Select (b) – Emily.

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6-82

PWGwen = -$65,000 +($20,000 - $4,000)(P/A, 20%, 20) + $20,000(P/F, 20%,20) = $13,435.60 Select (a).

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6-83

PWSadie = 0 = -$84,000 + ($25,000 - $9,000)(P/A, IRRSadie, 20) + $20,000(P/F, IRRSadie, 20) IRRSadie = 18.64%

Select (b)

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6-84

AWMisty = -$95,000(A/P, 20%, 20) + ($32,000 - $9,500) + $30,000(A/F, 20%, 20) = $3,149 Select (b).

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6-85 Eliminate Sadie and Bonnie as IRR < MARR PWGwen = -$65,000 +($20,000 - $4,000)(P/A, 20%, 20) + $20,000(P/F, 20%,20) = $13,435.60 PWMisty = -$95,000 +($32,000 - $9,500)(P/A, 20%, 20) + $30,000(P/F, 20%,20) = $15,349 Select (d) – Misty to maximize PW.

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6-86

Assume repeatability.

AWX = −$500,000(A/P, 12%, 5) + $131,900 = -$6,800 AWY = −$250,000(A/P, 12%, 10) + $40,690 = -$3,560 AWZ = −$400,000(A/P, 12%, 20) + $44,050 = -$9,510

Select (d) – Do Nothing. None of the alternatives are profitable.

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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

dk = ($100,000 - $0)/5 = $20,000

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7-6

B = $60,000 R = 2/10 = 0.2 Using the equation dk = B (1 – R)k-1 (R) we get

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7-7

B = $90,000 Year, k 1 2 3 4 5 6

rk 0.2000 0.3200 0.1920 0.1152 0.1152 0.0576

dk = B∙rk 18,000 28,800 17,280 10,368 10,368 5,184

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7-8

Basis = $30,000 ADS is simply the straight-line method with SVN = 0 and the half-year convention.

Regular straight-line amount = $30,000 / 10 = $3,000 for years 2 thru 10.

In years 1 and 11, dk = $3,000(0.5) = $1,500.

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7-9

B = 40,000

N = 7 years

R = 2/7

The present worth of the 200% declining balance deductions is greater than that with straight-line.

200% declining balance deductions will maximize worth.

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7-10

Referring to Equation 7-9, the depreciation rate is: ($300,000 - $30,000) / (10 years x 180 days/year) = $270,000 / 1,800 rental days = $150 per day

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7-11

From Table 7-2, the GDS recovery period is 3 years. (a) Basis = $195,000 d*3 = $195,000 (0.3333 + 0.4445 + 0.1481) = $180,550.50 (b)

d4 = 0.0741 ($195,000) = $14,449.50

(c)

BV2 = $195,000 (1 − 0.3333 − 0.4445) = $43,329

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7-12

Straight-line: d4 = $50,000 / 6 = $8,333.33 200% declining balance: d4 = $50,000(1 – 2/6)4-1(2/6) = $4,938.27

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7-13

Assuming the recovery period is 5 years for both GDS and ADS.

More total depreciation is claimed using GDS for all years except N + 1.

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7-14

Straight-line: d2 = $120,000/10 = $12,000 200% declininng balance: d2 = $120,000 (1 – 2/10)2-1 (2/10) = $19,200 Using Table 7-4, an asset with a useful life of 10 years is assigned a GDS property class (recovery period) of seven years.

MACRS (7-yr recovery): d2 = $120,000 (0.2449) = $29,388

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7-15

Assume a 7 year useful life. The tax rate is 21%.

Straight line: d1 = $80,000 / 7 = 11,428.57

ATCF = + 11,428.57 (0.21) = $2,400

200% DB: d1 = $80,000 (1 – 2/7)0 (2/7) = 22,857.14 ADS: d1 = (0.5)($80,000)/7 = 5,714.29

ATCF = +22,857.14(0.21) = $4,800

ATCF = +5,714.29 (0.21) = $1,200

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7-16

$35,000  $5,000 = $0.20/unit 150,000units = (18,000 units) ($0.2/unit) = $3,600

Depreciation per unit of production = d4

BV4 = $35,000 − (60,000 units + 18,000 units)($0.20/unit) = $19,400

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7-17

Total units of production over the five year life = 100,000 cubic yards d3 = (36,000/100,000)($60,000 − $10,000) = $18,000 BV2 = $60,000 – (d1 + d2) = $60,000 – [(0.16)($50,000) + (0.24)($50,000)] = $60,000 – ($8,000 + $12,000) = $40,000

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7-18

Taxable income = gross income – all expenses (excluding capital investments) – depreciation deductions In our case we have Taxable Income = $500,000 - $400,000 - $45,000 = $55,000 The federal income tax rate is t = 21% for this tax year.

Thus, the federal income tax = 0.21 ($55,000) = $11,550

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7-19

Assume a 7 year recovery period for MACRS. The tax rate is 21%.

Straight line: d5 = ($65,000 - $5,000) / 8 = 7,500

ATCF = +7,500 (0.21) = $1,575

200% DB: d5 = $65,000 (1 – 2/8)4 (2/8) = 5,141.60

ATCF = +5,141.60 (0.21) = $1,079.74

GDS: d5 = ($65,000)(0.0893) = 5,804.50

ATCF = +5,804.50 (0.21) = $1,218.95

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7-20

The simple payback period is 3 years

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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)

(b)

PW

= -130,000 + 23,160 (P/A, 15%, 6) – 790 (P/G, 15%, 6) + 17,900 (P/F, 15%, 6) = -$40,883

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7-23

After - tax MARR 015 .   0.25 , or 25% 1- t 1  0.40

(a)

Before−tax MARR =

(b)

Year 1 2 3 4

(c)

BV8 = 0, therefore TI8 = $10,000 (Property having a 7-year recovery period is fully depreciated after N+1=8 years.)

Depreciation $12,861 22,041 15,741 11,241

Year 5 6 7 8

Depreciation $8,037 8,028 8,037 4,014

(d) EOY 0 1 2 3 4 5 6 7 8 8 (e)

(A)

(B)

(C) = (A) − (B)

(D) = −t (C)

(E) = (A) + (D)

BTCF

Depr

TI

T (40%)

ATCF

PW (15%)

−$90,000 14,144 17,816 15,296 13,496 12,215 12,211 12,215 10,606 6,000 PW(15%) =

−$90,000 12,299 13,472 10,058 7,717 6,073 5,279 4,592 3,467 1,961 −$25,082

−$90,000 --15,000 $12,861 15,000 22,041 15,000 15,741 15,000 11,241 15,000 8,037 15,000 8,028 15,000 8,037 15,000 4,014 10,000 ---

--$2,139 −7,041 −741 3,759 6,963 6,972 6,963 10,986 10,000

--−$856 2,816 296 −1,504 −2,785 −2,789 −2,785 −4,394 −4,000

No, reject the project because PW(ATCF) < 0 at MARR = 15%.

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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

AWM2(12%) = $20,163 Therefore, select M2.

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ATCF -$66,000 -$5,520 -$2,352 -$5,731 -$7,759 -$7,759 -$9,279 -$10,800 -$10,800 $0

7-25

Mortar

PW = -19000 + 79(P/F, 12%, 1) + 563(P/F, 12%, 2) + (8493 – 185)(P/F, 12%, 3) = -12,567

Xrocks PW = -9,000 (1 – 0.25)(P/A, 12%, 3) = -16,212.15 Go with purchase of Mortar, lower PW of costs.

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7-26

(a) EOY 0 1-10

BTCF −$200,000 36,000

d

TI

$20,000

$16,000

T(25%)

ATCF −$200,000 −$4,000 32,000

The IRR is found as follows: 0 = −$200,000 + $32,000(P/A, i′, 10). Solving yields i′ = 9.61%.

Because this after-tax IRR is more than 8%, the robot should be acquired. (b)

The tax advantage of higher early depreciation makes the after-tax IRR larger than in the straightline case.

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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.

PW(15%) = −$1,000,000 + $480,800 (P/A,15%,4) + $200,000 (P/F,15%,4) = $487,204 >> 0; the investment should be made.

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7-28

Assume a tax rate of 25%.

PW(20%) = -$388,845 Whether or not to make the investment will depend on the expected annual net savings from using the workstations. We can find the minimum before tax annual savings (X) to justify this investment.

(X)(1 – 0.25)(P/A, 20%, 4) = 388,845

X = 200,278

If the expected before-tax annual benefit of having these workstations is greater than $200,278 then you should 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

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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ATCF −$200,000 −23,000 −13,400 −31,320 72,720

7-30 EOY 0 1 2 3 4 5 5

(A)

(B)

(C) = (A) − (B)

(D) = −t (C)

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

(E) = (A) + (D)

ATCF −$160,000 $30,500 $38,300 $29,980 $24,988 $21,244 $37,488

*d = $130,000 · 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 BTCF d Taxable Income Income Tax ATCF 0 −$500,000 ------−$500,000 1-5 −35,000 + 41.5X $100,000 −$135,000 + $41.5X $54,000 – $16.6X 19,000 + 24.9X 6-7 −35,000 + 41.5X --−35,000 + 41.5X $14,000 – 16.6X −21,000 + 24.9X PW = 0 = −$500,000 + ($19,000 + $24.9X)(P/A,10%,5) + (−$21,000 + $24.9X)(P/A,10%,2)(P/F,10%,5) X = 3,717.24 or 3,718 units per year must be produced (and sold) for this project to be economically viable.

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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)

TI --$4,000 4,000 4,000 4,000 4,000 14,000 14,000 14,000 14,000

(D) = −t (C)

T (20%) --−$800 −800 −800 −800 −800 −2,800 −2,800 −2,800 −2,800

(E) = (A) + (D)

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 = 11.25% per year.

Choose alternative A to maximize after-tax PW. Alternative A was also chosen when a before-tax analysis was done in Problem 6-77.

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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 714,610 280,000c ----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 b 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)

EOY 0 1 2 3 3 3

(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

e MV − BV

0 −0.3664X + 284,321 −0.3664X + 315,415 −0.3664X + 273,229 −79,793 ---

3 = $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 1-5 -

BTCF $10,000

T.I. -$10,000

Income Tax $4,000 (credit)

ATCF -$6,000

EUAC of Alt. A is -$6,000 Alt. B: EOY 1-4 5

BTCF 0 -$50,000

T.I. 0 -$50,000

Income Tax 0 $20,000 (credit)

ATCF 0 -$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

Purchase Option

PW(15%) = -$36,529 AW(15%) = -$36,529 (A/P, 15%,8) = − $8,142 Leasing Option ATCF = − (1 − 0.25)(Leasing Cost) = AW(15%) For the leasing option to be more economical than the purchase option, − (0.75) (Leasing Cost) < − $8,142 Leasing Cost < $10,856 If Leasing Cost < $10,856 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

Machine A: EOY 0 1−12 12

(A)

(B)

(C) = (A) − (B)

(D) = −t (C)

BTCF −$20,000 $12,000 $4,000

Depr --$1,333.33 ---

TI --$10,666.67 0

T (40%) --−$4,266.67 0

0.1468

(E) = (A) + (D)

ATCF −$20,000 $7,733.33 $4,000

0.0468

AW = −$20,000(A/P,10%,12) + $7733.33 +$4,000(A/F,10%,12) = $4,984.53 Machine B: (A)

EOY 0 1−8

(B)

BTCF −$30,000 $18,000

Depr --$3,750

(C) = (A) − (B)

TI --$14,250

(D) = −t (C)

T (40%) --−$5,700

0.1874

AW = −$30,000(A/P,10%,8) + $12,300 = $6,678

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(E) = (A) + (D)

ATCF −$30,000 $12,300

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 $0 -120,000 8,000 84,800 84,800 142,400 200,000 1,000,000

T(20%) $0 24,000 -1,600 -16,960 -16,960 -28,480 -40,000 -200,000 PW(10%) AW(10%)

ATCF -$1,000,000 200,000 224,000 198,400 183,040 183,040 171,520 160,000 800,000 $ 344,126 $ 70,685

Design B: EOY 0 1 2 3 4 5 6 6

BTCF d -$2,000,000 400,000 $400,000 400,000 640,000 400,000 384,000 400,000 230,400 400,000 230,400 400,000 115,200 1,100,000

TI $0 -240,000 16,000 169,600 169,600 284,800 1,100,000

T(20%) $0 48,000 -3,200 -33,920 -33,920 -56,960 -220,000 PW(10%) AW(10%)

Select Design A to maximize after-tax AW.

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ATCF -$2,000,000 400,000 448,000 396,800 366,080 366,080 343,040 880,000 $ 199,725 $ 45,858

7-43

Method I – Assume repeatability

PW0(15%) = -$44,237 To have a basis for computation, assume that Method I is duplicated for years 6−10. Transform the additional PW5(15%) = -$44,237 to the present and get: PW(15%) = PW0(15%) + (P/F,15%,5)PW5(15%) = -$66,231 Method II

PW(15%) = -$61,010 Thus, Method II is the better alternative.

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7-44

Freezer 1: EOY 0 1 2 3 4 5a 5b

BTCF -$10,000 3,000 3,000 3,000 3,000 3,000 2,000

d --3,000 3,000 3,000

TI --0 0 0 3,000 3,000 0

-----

T(40%) ---

ATCF -$10,000 3,000 3,000 3,000 1,800 1,800 2,000

0 0 0 -1,200 -1,200 0

PW(12%)

Total ATCF -$10,000 3,000 3,000 3,000 1,800 3,800

$505.65

Freezer 2: Total EO 0

BTCF -$30,000

1

9,000

2 3 4 5a 5b

9,000 9,000 9,000 9,000 2,000

d --10,998.9 0 14,668.5 0 4,887.30 2,445.30 -----

TI ---

T(40%) ---

ATCF -$30,000.00

ATCF -$30,000.00

-1,998.90

799.56

9,799.56

9,799.56

-5,668.50 4,112.70 6,554.70 9,000 2,000

2,267.40 -1,645.08 -2,621.88 -3600 -800.00

11,267.40 7,354.92 6,378.12 5,400.00 1,200.00

11,267.40 7,354.92 6,378.12 6,600.00

PW(12%)

$765.42

Freezer 2 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)

Depr --$2,000

(C) = (A) − (B)

TI --−$11,500

(D) = −t (C)

T (25%) --$2,875

(E) = (A) + (D)

ATCF −$10,000 −$6,625

Straight Line Depreciation: ($10,000−0)/5 = $2,000 (B) (A)

EOY 0 1−5

BTCF −$14,000 −$8,300

(B)

Depr --$2,800

(C) = (A) − (B)

TI --−$11,100

(D) = −t (C)

T (25%) --$2,775

(E) = (A) + (D)

ATCF −$14,000 −$5,525

Depreciation: ($14,000−0)/5 = $2,800 (C) (A)

EOY 0 1−5

BTCF −$20,000 −$6,000

(B)

Depr --$4,000

(C) = (A) − (B)

TI --−$10,000

(D) = −t (C)

T (25%) --$2,500

(E) = (A) + (D)

ATCF −$20,000 −$3,500

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 (25%) --$2,625

Depreciation: ($30,000−0)/5 = $6,000 AWA = −$10,000(A/P,18%,5) − $6,625 = −$9,823 AWB = −$14,000(A/P,18%,5) − $5,525 = −$10,002 AWC = −$20,000(A/P,18%,5) − $3,500 = −$9,896 AWD= −$30,000(A/P,18%,5) − $1,875 = −$11,469 Select to automate to Degree A.

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(E) = (A) + (D)

ATCF −$30,000 −$1,875

7-46 Use a study period of 3 years Quotation I

Quotation 2I

Accept Quotation I.

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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 EOY BTCF Depr TI T(50%) NOPAT a 1 $112,000 $ 69,000 $ 43,000 − $21,500 $21,500 2 110,400 1,600 − 800 800 3 66,240 45,760 − 22,880 22,880 4 39,744 72,256 − 36,128 36,128 5 39,744 72,256 − 36,128 36,128 6 112,000 19,872 92,128 − 46,064 46,064 6 120,000b 120,000c − 60,000 60,000 a BTCF = $120,000 − $8,000 = $112,000 k b MV = $120,000 6 c Gain on disposal = MV − BV = $120,000 − 0 = $120,000 6 6 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: EVA = NOPAT − i  BV k k k−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

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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ATCF -75,000 21,500 23,025 18,040 12,887 25,548

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

Taxes Payable (40%) (C) −− $22,000 $22,000 $22,000 $22,000 $22,000

−− −$55,000 −$55,000 −$55,000 −$55,000 −$55,000

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

Payment = −$350,000 (A/P,8%,3) = −$135,800 Select the lease.

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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

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 = −$10,000 + ($10,000)(0.10/4)(1−0.28)(P/A,i%,35) + $10,000 (P/F,i%,15) = −$10,000 + $180(P/A,i%,35) + $10,000(P/F,i%,35) i% 1% i% 2%

PW(i) $2353 0 −$500

i% = 1.80%/quarter, r = 7.2%/year, ieff. = 7.4%/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

Alt 2

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

T

-$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

$11,200 $14,080 $11,008 $9,165 $7,782 -$2,081

PW(12%)

-$72,683

AW(12%)

-$20,163

MV (for each M1)

ATCF -$60,000 -$4,800 -$1,920 -$4,992 -$6,835 -$8,218 $10,033 $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

-$100,165 -$20,163

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$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

ATCF -$60,000 -$4,800 -$1,920 -$4,992 -$6,835 $1,816

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

$

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

$ $

1,120,000,000 80% 800000 40% $0.01 0.000008 15.00 0.000000055 3413

Units $

$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

Annual CO2 emitted

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kW $/kWhr $/Btu $ / MT CO2 MT CO2/Btu Btu/kWhr

2631013.44 1195.9152 MT

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

dk = (26000 – 2000)/8 = 3000 BV5 = $26,000 – (5)($3,000) = $11,000 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 = 0.08 + 0.21(0.92) = 27.32% Select (b)

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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

d2 = $15,000 (0.4445) = $6,667.5 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

Current Price = $5000 Yearly decrease = 8% Price after 4 years = 5000 * (0.92)^4 = $3581.97

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8-3

The average rate of inflation is 10% per year, and the market place interest rate is 5% + 10% + (5%)(10%) = 15.5% per year.

PW2 = ($1,000)

1  ( P / F ,15.5%,23)( F / P,10%,23) = $12,262.36 0.155  0.10

PW0 = $12,262.36 (P/F, 15.5%, 2) = $9,192.

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8-4

Current = $100,000 Rate = 2%

The value after 5 years = 10000 * (1.02)^5 = $110,408

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8-5

Current cost = $250,000 Rate = 3% Time = 10 years Cost in 10 years = 250000 * (1.03)^10 = $335,979

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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

Current price = $20,000 Inflation = 4% N= 6 years

Price after 6 years = 20000 * (1.04)^6 = $25306.38

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8-8

Use Equation (8-4) 0.14 = 0.03 + ir + 0.03 ir 0.11 = ir (1 + 0.03) ir = 0.1068 or 10.68%

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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

Tennessee index-adjusted salary = (95/132) × $70,000 = $50,379 per year Tennessee actual salary = (1.00 − 0.11) × $70,000 = $62,300 per year “Savings” = $62,300 − $50,379 = $11,921 per year FW(10%) = $11,921(F/A, 10%, 5) = $72,779. Paul is not penalized at all!

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8-11

( 0) R$ 10 = $400M(1.65) = $660M A$10 = $920M = $660M(1+f)10

1.394 = (1+f)10 f = 10 1.394 − 1 = 0.0338 or 3.38%

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8-12

After-tax nominal return per year = 6% (1 – 0.33) = 4% Approximate real return per year = 4% − 3% = 1% each year.

F (in today’s purchasing power) = $100,000(F/P, 1%, 10) = $110,460

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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  C A = CB  A   SB 

X

(1 + eC)8

 125,000 = ($89/ft2)(80,000 ft2)    80,000 

0.92

(1.054)8

= $16,350,060 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.)[(1.03)(1.05)]7 = $173.06/bbl.

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8-16

We can equate future worth's in 20 years. The FW of her savings plan will be P(1.06)20 and the FW of the $800,000 if it were to inflate at 3% per year is $800,000(1.03)20. P is what we are trying to determine. So we can equate these two amounts as follows: P(1.06)20 = $800,000(1.03)20 or P = $800,000(1.03 / 1.06)20.

This equals $800,000(0.9717)20 = $450,541.

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8-17

The value of ir is desired to match the R$ of $100,000 per year. ir = (0.10 - 0.03) / (1 + 0.03) = 0.068 (6.8%).

To determine the amount of life insurance to guarantee $100,000 (R-dollars) per year, we need to find the capitalized worth (CW) of this annuity: CW = $100,000 / 0.068 = $1,470,588.

So Alfred needs $1,470,588 worth of life insurance.

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8-18

(a)

Lump sum interest in 2019 = ($2.4 billion/5)(F/A, 10%, 5) – 2.4 billion = $530,448,000 0.91

(b)

 200,000  C2014 = ($2.4 billion)    150,000 

(c)

C2024 = ($3.12 billion)(F/P, 9.2%, 10) = $7.52 billion

= $3.12 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$ = $7.50(1.018)22 = $11.10 per thousand cubic feet

(b)

A$ = $7.50(1.032)22 = $22.20 per thousand cubic feet

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8-22

Current cost = $200,000 Annual rate = 5% Inflation = 2% Time = 7 years Cost in 7 years = 200,000 * [(1.05)*(1.02)]^7 = $323,263.2

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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

Construction cost = $1,000,000 Rate = 4% Inflation 3%

Cost in 5 years = 1,000,000 * [(1.04)*(1.03)]^5 = $1,410,434

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8-25

Current cost = $75,000 Annual rate = 3% Inflation = 2% Cost in 4 years = 75000 * [(1.03)*(1.02)]^4 = $91,371.52

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8-26

(a)

The tuition for 2024 is calculated as follows: F2024 = $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$)

(B) Depreci− ation

0 1 2 3

−$X 0 0 0

−−− $X/3 $X/3 $X/3

Taxable Cash Flow Income: for C=A+B Income Taxes D = −t(C) −−− −−− −$X/3 $0.1133X −$X/3 $0.1133X −$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 2019 there was parity between the U.S. dollar and the Real. But in 2024 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

Current rate: $1 = 8.7 units a) So, estimated exchange rate post 2 years at 2.6% per year rate = 8.7 * (1-0.026)^2 = 8.2534 units

b) Now, if X was devauating compared to USD AT 2.6 % Dollar value = 8.7 * (1+0.026)^2 = 9.3963 units

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8-34

Canadian firm has invested CAD 200,000 Exchange rate = 075 USD/CAD So, the equivalent investment (USD) = 200,000 * 0.75 = 150,000 USD

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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 of 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 of project in Country B currency)

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8-36

US company invests in CAD 400,000 Exchange rate = 0.78 USD/CAD Inflation = 1.8% Investment in USD = 400,000 * 0.78 = $312,000 Value due to inflation = 312000 * (1.018)^6 = $347,249.2

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8-37

(a) NCF EOY (T−marks) 0 −$3,600,000 1 450,000 2 1,500,000 3 1,500,000 4 1,500,000 5 1,500,000 6 1,500,000 7 1,500,000

Exchange Rate (T−marks/$) 20.000 22.400 25.088 28.099 31.470 35.247 39.476 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 + fe 1.12

Note: This confirms our recommendation in part (a).

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8-38

Initial cost = 30,000,000 JPY Exchage rate = 0.008 USD/JPY Cost in USD = 30,000,000 * 0.008 = $240,000 Cost post 5 years = 240000 * (1.03)^5 = $278,225.8

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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

− 84,400 − 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

$21,100 16,210 13,831 13,964 14,111 14,272

−$63,300 − 48,630 − 41,493 − 41,892 − 42,331 −42,814 0

(b) im = (1.05)(1.09524) – 1 = 0.15 = 15% per year PW = −$63,300(P/F,15%,1) − $38,630(P/F,15%,2) − $41,493(P/F,15%,3) − $41,892(P/F,15%,4) − $42,331(P/F,15%,5) − $42,814(P/F,15%,6) = −$182,605 EUAC = $182,605(A/P,15%,6) = $48,244

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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):

EOY 0 1 2 3 4 5 6 6 EOY 0 1 2 3 4 5 6 6

Investment / Market Value − $600,000

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

101,355a 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

Notes: a (MV) = $90,000 (1.02)6 = $101,355 6 b (O,I,OE) = $26,000 (1.06)k k c (Maint) = $32,000 (1.09)k k d Cost Basis = $600,000

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

EOY 1 2 3 4 5 6

Leasing Costs − $300,000 − 200,000 − 200,000 − 200,000 − 200,000 − 200,000

Oper., Ins. & Other Expenses − $27,560 − 29,214 − 30,966 − 32,824 − 34,794 − 36,881

BTCF − $362,440 − 267,233 − 272,407 − 277,995 − 284,030 − 290,549

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

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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ATCF (A$) − $239,210 − 176,374 − 179,789 − 183,477 − 187,460 − 191,762

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

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

EOY 0 1 2 3 4 5 6

Depr. −−− $16,000 25,600 15,360 9,216 9,216 4,608

TI −−− −$25,400 −15,682 16,259 46,710 52,182 62,768

BTCF −$80,000 −9,400 9,918 31,619 55,926 61,398 67,376

$ 0 18,202 38,588 61,356 65,037 68,940

$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

Difference between Desired and Actual =

% 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%

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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941

864

8-51

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 =

 k0

20

ATCFk (A$) (P/F, 12%, k) =

 ATCF (R$) (P/F, 7.18%, k) = −$359,665 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

From Case Study 8.7, PW (ind) motor = $1,309,376 PW (Sync motor) = $1,171,955 Now, reduced horsepower requirement = 1850 hp a)Two sync motors + 1 induction moyor PW = $3,122,395 b)Two sync motors at original and one sync at (350HP) PW = $2,915,054

Among the combinations, the PW of combination b is much less. Hence, the ideal combination would be: 2 * Synchronous Motors (750 HP) + 1 * Synchronous Motor reduced power (350HP)

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8-55

Interest rate = 4% General inflation = 2.5% So, 1 + I = (1.04)*(1.025) I = 6.6% Option C

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8-56 Inflation rate = 4% Cost 5 years ago = $2.00 per ounce Cost now = $2.00 (1.04)^5 = $2.43 per ounce Option C

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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

Current Price = $1,500 Inflation rate = 4% Price after 2 years = 1500 * (1.04)^2 = $1,622.4 Option (d)

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Solutions to Chapter 9 Problems 9-1

The existing machine should be replaced when the EUAC of the challenger is cheaper than that of the Marginal Cost of the defender (existing machine).

The marginal cost of the defender is $2,000.

EUAC of Challenger = $25,000(A/P, 15%, t) + $500 EUAC(t = 1) = $25,000(1.15) + $500 = $29,250 EUAC(t = 5) = $25,000(0.2983) + $500 = $7,958 EUAC(t = 10) = $25,000(0.1993) + $500 = $5,483 EUAC(t = infinity) = $25,000(0.15) + $500 = $4,250 For this challenger, it will never be economical to replace the defender (until the defender’s annual maintenance costs increase significantly).

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9-2

AWDefender(20%) = $40,200(A/P, 20%, 6)  $1,400  $1,500(A/F, 20%, 6) = $13,579 AWChallenger(20%) = $56,600(A/P, 20%, 10)  $1,000 + $7,000(A/F, 20%, 10) = $14,206 Select the Defender.

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9-3

EUAC of Defender = $800 - $3,000(A/F, 12%, t) EUAC of Challenger = $15,000(A/P, 12%, t) + $300 For t = 1 EUAC of Defender = $800 - $3,000(A/F, 12%, 1) = $800 - $3,000(1) = -$2,200 EUAC of Challenger = $15,000(A/P, 12%, 1) + $300 = $15,000(1.12) + $300 = $17,100 For t = 5 EUAC of Defender = $800 - $3,000(A/F, 12%, 5) = $800 - $3,000(0.1574) = $328 EUAC of Challenger = $15,000(A/P, 12%, 5) + $300 = $15,000(0.2774) + $300 = $4,461 For t = 10 EUAC of Defender = $800 - $3,000(A/F, 12%, 5) = $800 - $3,000(0.0570) = $629 EUAC of Challenger = $15,000(A/P, 12%, 5) + $300 = $15,000(0.1770) + $300 = $2,955 Do not replace the defender. At least for the next 10 years, the defender has a better (smaller) EUAC.

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9-4

This question is asking for the economic life of the vehicle. This will occur when EUAC is at its lowest for a time period t. Note, the EUAC curve is convex; thus, when it starts to increase (as t increases), then we know we have reached the minimum EUAC. EUACt = $30,000(A/P, 8%, t) + $1,000 + $500(A/G, 8%, t) EUAC(t = 1) = $30,000(1.08) + $1,000 + $500(0) = $33,400 EUAC(t = 2) = $30,000(0.5608) + $1,000 + $500(0.4808) = $18,064 EUAC(t = 3) = $30,000(0.3880) + $1,000 + $500(0.9487) = $13,114 EUAC(t = 4) = $30,000(0.3019) + $1,000 + $500(1.4040) = $10,759 EUAC(t = 5) = $30,000(0.2505) + $1,000 + $500(1.8465) = $9,438 EUAC(t = 6) = $30,000(0.2163) + $1,000 + $500(2.2763) = $8,627 EUAC(t = 7) = $30,000(0.1921) + $1,000 + $500(2.6937) = $8,109 EUAC(t = 8) = $30,000(0.1740) + $1,000 + $500(3.0985) = $7,769 EUAC(t = 9) = $30,000(0.1601) + $1,000 + $500(3.4910) = $7,549 EUAC(t = 10) = $30,000(0.1490) + $1,000 + $500(3.8713) = $7,406 EUAC(t = 11) = $30,000(0.1401) + $1,000 + $500(4.2395) = $7,323 EUAC(t = 12) = $30,000(0.1327) + $1,000 + $500(4.5957) = $7,279 EUAC(t = 13) = $30,000(0.1265) + $1,000 + $500(4.9402) = $7,265 <<<<<< Minimum EUAC EUAC(t = 14) = $30,000(0.1213) + $1,000 + $500(5.2731) = $7,276 Thus, the economic life of this vehicle is 13 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

Let t be the year.

First cost = $50,000 + $7,000 = $57,000 Calculate the EUAC for each year.

Note, the differential in Operating Expenses is an $8,000 uniform gradient.

EUACt = $57,000(A/P, 10%, t) - St(A/F, 10%, t) + $6,000 + $8,000(A/G, 10%, t) EUAC(t = 1) = $57,000(1.1) - $35,000(1) + $6,000 + $8,000(0) = $33,700 EUAC(t = 2) = $57,000(0.5762) - $30,000(0.4762) + $6,000 + $8,000(0.4762) = $28,367 <<<<<< Minimum EUAC EUAC(t = 3) = $57,000(0.4021) - $25,000(0.3021) + $6,000 + $8,000(0.9366) = $28,860 Thus, the economic life of this machine is 2 years.

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9-7

For this question, need to compare the Marginal Cost of the Defender to the EUAC of the challenger.

Marginal Cost of Defender = $4,000 (we assume no salvage value) EUAC of Challenger, we are not given a specific t; thus, we will calculate the t for which the EUAC of the challenger is closest to the Marginal Cost of the Defender.

EUACt = $25,000(A/P, 12%, t) + $2,000 EUAC(t = 1) = $25,000(1.12) + $2,000 = $30,000 EUAC(t = 2) = $25,000(0.5917) + $2,000 = $16,793 EUAC(t = 5) = $25,000(0.2774) + $2,000 = $8,935 EUAC(t = 10) = $25,000(0.1770) + $2,000 = $6,425 EUAC(t = 15) = $25,000(0.1468) + $2,000 = $5,670 EUAC(t = 20) = $25,000(0.1339) + $2,000 = $5,348 EUAC(t = infinity) = $25,000(0.12) + $2,000 = $5,000 The current challenger will never have an EUAC smaller than the current Marginal Cost of the defender; thus, the current machinery should be kept.

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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

Let t be the year.

Calculate the EUAC for each year.

Note, the differential in maintenance costs is a $50 uniform gradient.

Assumes Salvage Value = $0 EUACt = $1,200(A/P, 6%, t) + $100 + $50(A/G, 6%, t) EUAC(t = 1) = $1,200(1.06) + $100 + $50(0) = $1,372 EUAC(t = 2) = $1,200(0.5454) + $100 + $50(0.4854) = $779 EUAC(t = 3) = $1,200(0.3741) + $100 + $50(0.9612) = $597 EUAC(t = 4) = $1,200(0.2886) + $100 + $50(1.4272) = $518 EUAC(t = 5) = $1,200(0.2374) + $100 + $50(1.8836) = $479 EUAC(t = 6) = $1,200(0.2034) + $100 + $50(2.3304) = $461 EUAC(t = 7) = $1,200(0.1791) + $100 + $50(2.7676) = $453.30 EUAC(t = 8) = $1,200(0.1610) + $100 + $50(3.1952) = $452.96 <<<<<< Minimum EUAC EUAC(t = 9) = $1,200(0.1470) + $100 + $50(3.6133) = $457 The lowest EUAC occurs in Year 8 (nearly equivalent to Year 7). replacement cycle.

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Thus, 8 years is the optimal

9-11

The existing machine should be replaced when the EUAC of the challenger is cheaper than that of the Marginal Cost of the defender (existing forklift).

The marginal cost of the defender is $1,000.

EUAC of Challenger = $15,000(A/P, 15%, t) + $500 EUAC(t = 1) = $15,000(1.15) + $500 = $17,750 EUAC(t = 5) = $15,000(0.2983) + $500 = $4,975 EUAC(t = 10) = $15,000(0.1993) + $500 = $3,490 EUAC(t = infinity) = $15,000(0.15) + $500 = $2,750 For this challenger, it will never be economical to replace the defender (until the defender’s annual maintenance costs increase significantly).

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9-12

The existing machine should be replaced when the EUAC of the challenger is cheaper than that of the Marginal Cost of the defender (existing air conditioner).

The marginal cost of the defender is $800.

EUAC of Challenger = $4,000(A/P, 15%, t) + $300 EUAC(t = 1) = $4,000(1.15) + $300 = $4,900 EUAC(t = 5) = $4,000(0.2983) + $300 = $1,493 EUAC(t = 10) = $4,000(0.1993) + $300 = $1,097 EUAC(t = infinity) = $4,000(0.15) + $300 = $900 For this challenger, it will never be economical to replace the defender (until the defender’s annual maintenance costs increase significantly).

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9-13

Defender: Annual Expenses = $300,000 + $250,000 + $500,000(0.04) + $8,000 = $578,000 EUACD(10%) = $150,000(A/P, 10%, 8) − $50,000(A/F, 10%, 8) + $578,000 = $601,740 Challenger: EUACC(10%) = $250,000 + $100,000 + $100,000 = $450,000 Decision: Sell the defender and lease a new machine to minimize EUAC.

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9-14

The existing machine should be replaced when the EUAC of the challenger is cheaper than that of the Marginal Cost of the defender (existing crane).

The marginal cost of the defender is $7,000. Note, the $30,000 market value for the existing crane could be considered the salvage value; which would make the defender worth even more (i.e., by lowering its marginal cost).

EUAC of Challenger = $100,000(A/P, 15%, t) + $2,000 EUAC(t = 1) = $100,000(1.15) + $2,000 = $117,000 EUAC(t = 5) = $100,000(0.2983) + $2,000 = $31,830 EUAC(t = 10) = $100,000(0.1993) + $2,000 = $21,930 EUAC(t = infinity) = $100,000(0.15) + $2,000 = $17,000 For this challenger, it will never be economical to replace the defender (until the defender’s annual maintenance costs increase significantly, salvage value decreases significantly, etc.).

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9-15

Abandonment Interval: Keep for N = 1 year: PW (12%) = −$9,500 + ($6,200 + $2,000)(P/F,12%,1) = −$2178 Keep for N = 2 years: PW (12%) = −$9,500 + $2,000(P/A,12%,2) + $5,200(P/F,12%,2) = −$1974 Keep for N = 3 years: PW (12%) = −$9,500 + $2,000(P/A,12%,3) + $4,000(P/F,12%,3) = −$1849 Keep for N = 4 years: PW (12%) = −$9,500 + $2,000(P/A,12%,4) + $2,200(P/F,12%,4) = −$2027 Keep for N = 5 years: PW (12%) = −$9,500 + $2,000(P/A,12%,5) = −$2290 PW (12%) is a maximum at 3 years. However, none of the PWs are positive. This tells us that the projext should be abandoned immediately.

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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) b Interest on tax credit = (0.07) (0.4) BV

k−1

c Adj. After−Tax Total Cost = Approx. After−Tax Total Cost + Interest on tax credit * 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

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

a Gain on disposal = $57,000 − $30,000 = $27,000 (if sold now). b MV = $18,500(1.032)5 = $21,656 5

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

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

a Gain on disposal = $14,000 (if sold now). b 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 ---

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

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

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.

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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

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

a 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 $6,750 −$50,250 1 – 5 −27,000 6,000 −33,000 8,250 −18,750 (a) 5 21,656 --21,656 −5,414 16,242 (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 $9,125 −$27,375 5 0 0 0 0 0 Incremental cash flows: Initial Investment: Annual Expenses: Market Value:

−$50,250 – 0 = −$50,250 −$18,750 − [−$27,375] = $8,625 $16,242– 0 = $16,242

AW = 0 = −$50,250(A/P, i’%, 5) + $8,625 + $16,242(A/F, i’%, 5).

Solving yields i’ = 4.89% < 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 Copyright © 2025 Pearson Education, Inc.

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

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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I

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 =

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$

(282,472)

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

$

-

$

40 kW 3 1,2 37

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$

(2 14 ,78 0)

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 =

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$

(358,798)

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

PW(Benefits) = $45,000(1.42)(P/A, 5%, 10) = %551,114 B – C = $551,114 / $505,000 = 1.099 > 1 The automated switchboard should be installed.

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10-2

PW(benefits) = 5000(P/A, 10%,5) + 1000(P/G, 10%, 5) + 3000(P/F, 10%, 6) = 27,509.50 B – C = 27,509.50 / 20000 = 1.375 > 1 The investment is worthwhile.

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10-3

(a) Sum of benefits minus disbenefits = $725,000 per year B/C Ratio = $725,000 / (0.10 x $6,000,000) = 1.208 Thus plan is acceptable because B/C > 1.0 (b) B/C Ratio = $725,000 / (A/P,10%,20)($6,000,000) = 1.028 This plan is still acceptable to the town.

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10-4

Assume an infinite life.

Plant A: B-C = $1,400,000 / [$20,000,000 (0.06)] = 1.167 > 1 Plant A is beneficial. Now we must look to the incremental investment required by Plant B. B-C = $1,200,000 / [$5,000,000 (0.06)] = 4 > 1 Plant B should be recommended.

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10-5

Assume an infinte life.

Assuming an upgrade must be made, we need to look at the B-C ratio of the incremental investment required by Upgrade B. B-C = $500,000 / [$2,000,000 (0.06)] = 4.167 > 1 Upgrade B should be chosen.

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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: $3,000,000 / $500,000(P/A, 6%, N) ≥ 1.0 Or $3,000,000 / $500,000 ≥ (P/A, 6%, N) From Table C-9, we see that N ≈ 8 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

Assume an infinite life.

Base alternative is Plan A (lowest investment cost). Look at incremental investment needed by Plan B. B-C = $1,000,000 / [$5,000,000 (0.06)] = 3.33 > 1.

Plan B is better than Plan A. Now look at incremental investment needed by Plan C. B-C = $1,000,000 / [$5,000,000 (0.06)] = 3.33 > 1.

Plan C is better than Plan B. Plan C should be implemented.

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10-9

Assume an infinte life.

Let’s first look at the lowest cost option, Project A. B-C = $1,600,000 / [$8,000,000 (0.05)] = 4 > 1.

Project A is beneficial Now look at incremental investment needed by Project B. B-C = $500,000 / [$2,000,000 (0.05)] = 5 > 1.

Project B is better than Project A. Project B should be implemented.

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10-10

Assume an infinite life.

System 1: B-C = $1,000,000 / [$5,000,000 (0.04)] = 5 > 1 System 2: B-C = $1,500,000 / [$7,000,000 (0.04)] = 5.36 > 1 Both systems are worthwhile. We must look to see if the incremental capital required by System 2 is justified by its incremental benefits.

B-C = $500,000 / [$2,000,000 (0.04)] = 6.25 > 1 System 2 should be installed.

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10-11

Assume an infinite life.

Project A: B-C = $2,000,000 / [$10,000,000 (0.05)] = 4 > 1.

Project B: B-C = $3,500,000 / [$15,000,000 (0.05)] = 4.67 > 1 Both projects are worthwhile. We must look to see if the incremental capital required by Project B is justified by its incremental benefits.

B-C = $500,000 / [$5,000,000 (0.05)] = 2 > 1 Project B should be recommended.

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10-12 Assume an infinite life.

Base alternative is Project A (lowest investment cost). Look at incremental investment needed by Project B. B-C = $500,000 / [$3,000,000 (0.05)] = 3.33 > 1.

Project B is better than Project A. Now look at incremental investment needed by Project C. B-C = $600,000 / [$3,000,000 (0.05)] = 4 > 1.

Project C is better than Project B. Project C should be implemented.

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10-13 Assume an infinite life.

Base alternative is Do Nothing. Look at incremental investment needed by Project X. B-C = $4,000,000 / [$20,000,000 (0.06)] = 3.33 > 1.

Project X is better than Do Nothing. Now look at incremental investment needed by Project Y. B-C = $400,000 / [$2,000,000 (0.06)] = 3.33 > 1.

Project Y is better than Project X. Now look at incremental investment needed by Project Z. B-C = $600,000 / [$3,000,000 (0.06)] = 3.33 > 1.

Project Z is better than Project Y. Project Z should be implemented.

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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 Assume an infinite life.

Base alternative is Do Nothing. Look at incremental investment needed by Strategy A. B-C = $3,000,000 / [$15,000,000 (0.06)] = 3.33 > 1.

Strategy A is better than Do Nothing. Now look at incremental investment needed by Strategy B. B-C = $600,000 / [$3,000,000 (0.06)] = 3.33 > 1.

Strategy B is better than Strategy A. Now look at incremental investment needed by Strategy C. B-C = $400,000 / [$2,000,000 (0.06)] = 3.33 > 1.

Strategy C is better than Strategy B. Strategy C should be implemented.

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10-21 Annual Benefits Flood reduction Irrigation Recreation Total Benefits

I. Improve Channel $200,000 ----$200,000

II.

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

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 Assume an infinite life.

Base alternative is Do Nothing. Look at incremental investment needed by Program 1.

B-C = $800,000 / [$4,000,000 (0.05)] = 4 > 1.

Program 1 is better than Do Nothing. Now look at incremental investment needed by Program 2.

B-C = $200,000 / [$1,000,000 (0.05)] = 5 > 1.

Program 2 is better than Program 1. Now look at incremental investment needed by Program 3.

B-C = $200,000 / [$1,000,000 (0.05)] = 4 > 1.

Program 3 is better than Program 2. Program 3 should be implemented.

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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

Decision: Select Design B. Copyright © 2025 Pearson Education, Inc.

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 $20 million (A/P, 5%, 40) = $20 million(0.0583) = $1,166,000.

Operating and maintenance expenses add $3,000,000 in obtaining the total annual cost of $4,166,000.

The breakeven annual ridership is $4,166,000 per year / $3 per customer = 1,388,667 customers per year.

This equates to 1,388,667 customers per year / 365 days per year = 3,805 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) + ($4.00/gal)(X mi/yr)/(30 mpg) EUACG = $28,000(A/P, 3%, 5) + ($4.00/gal)(X mi/yr)/(25 mpg) Setting EUACH = EUACG, we find the breakeven mileage to be X = 16,380 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 dirven per year, X: Windshear: [(X mi/yr)(0.92)(0.2 gal/mi)($3.00/gal)] = $0.736X / yr Blowby: [(X mi/yr)(0.96)(0.2 gal/mi)($3.00/gal)] = $0.768X / yr Air-vantage: [(X mi/yr)(0.90)(0.2 gal/mi)($3.00/gal)] = $0.720X / yr EUAC can now be written in terms of X. EUACW = $1,000(A/P, 10%, 10) + $10 + $0.736X = $172.70 + $0.736X EUACB = $400(A/P, 10%, 10) + $5 + $0.768X = $70.08 + $0.768X EUACA = $1,200(A/P, 10%, 5) + $5 + $0.720X = $321.56 + $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 = 3,207 miles per year Blowby and Air-vantage: X = 5,239 miles per year Air-vantage and Windshear: X = 9,304 miles per year The range over which each deflector is preferred is: X ≤ 3,207 Select Blowby 3,207 ≤ X ≤ 9,304 Select Windshear 9,304 ≤ X Select Air-vantage

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11-4

(a)

Assuming repeatability, we can find the breakeven mileage as follows: -$27,000 (A/P, 5%, 4) + $9,000 (A/F, 5%, 4) - $0.42X = -$21,000 (A/P, 5%, 3) + $6,000 (A/F, 5%, 3) - $0.36X X = 4,697 miles per year

(b)

EUACGasoline = $11,826 EUACDiesel = $11,208 Recommend the diesel truck

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11-5

(a)

AW1(15%) = −$4,500 (A/P, 15%, 8) + $1,600 − $400 + $800 (A/F, 15%, 8) = −$4,500(0.2229) + $1,200 + $800 (0.0729) = $255 AW2(15%) = −$6,000(A/P, 15%,10) + $1,850 − $500 + $1,200(A/F,15%,10) = −$6,000(0.1993) + $1,350 + $1,200 (0.0493) = $213 Therefore, the initial decision is to select Alternative 1. To determine the capital investment of Alternative 2 (I2) so that the initial decision would be reversed, equate the AWs: AW1(15%) = AW2(15%) $255 = −I2 (A/P, 15%, 10) + $1,350 + $1,200 (A/F, 15%, 10) $255 = −I2 (0.1993) + $1,350 + $1,200 (0.0493) I2 = $5,791 The capital investment of Alternative 2 would have to be $5,791 or less for the initial decision to be reversed.

(b)

Set AW1(15%) = AW2(15%) and solve for N assuming market values remain constant. −$4,500 (A/P,15%,N) + $1,200 + $800 (A/F,15%,N) = $213 −$4,500 (A/P, 15%, N) + $986.64 + $800 (A/F, 15%, N) = 0 By trial and error, N = 7.3 years.

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11-6

(a) Let X = operating hours per year EUACA(12%) = $2,410(A/P, 12%, 8) − $80(A/F, 12%, 8) + (15 hp)(0.746 kW/hp)($0.06/kWh)(X)/(0.6) = $478.63 + $1.119X EUACB(12%) = $4,820(A/P, 12%, 8) + (10 hp)(0.746 kW/hp)($0.06/kWh)(X)/(0.75) = $970.27 + $0.5968X Setting EUACA = EUACB we find X = 941 hours per year. At annual operating hours greater than 941 (including 2,000), we prefer the more efficient pump, Pump B. This is confirmed by calculating the EUAC with X = 2,000.

EUACA(12%) = $2,716

EUACB(12%) = $2,164

(b) Let A = breakeven efficiency of pump A at 2,000 operating hours per year. We already know that EUACB(12%) = $2,164 at 2,000 hours per year.

EUACA(12%) = $2,164 = $2,410(A/P, 12%, 8) − $80(A/F, 12%,8) + (15 hp)(0.746 kW/hp)($0.06/kWh)(X)/A Solving, we get A = 79.67%

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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 vs. MARR 5% 10% 15% 20% 20%

PWA(MARR) 5% 31k 40k 16.3k B 30 5.5k 20 10 −2.8k 0 −1 0

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 40 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.

Annual Annual BTCF EOY Investment Revenue Expenses (A$) 0 −$1,166,0001 −$1,166,000 1 X − $519,7503 −519,750+X 2 X − 545,738 −545,738+X 3 X − 573,024 −573,024+X 4 X − 601,676 −601,676+X 4 441,7412 441,741 1 Capital Investment = (55 trucks)($21,200/truck) = $1,166,000 2 Market Value = MV = 0.35($1,166,000)(1.02)4 = $441,741 4 3 Annual Expenses in year k = (55 trucks)(20,000 mi/truck)($0.45/mi)(1.05)k

= $495,000(1.05)k

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(25%) −−− 227,095−0.25X 266,006−0.25X 186,427−0.25X 172,019−0.38X −110,435

ATCF (A$) − $1,166,000 −292,655+0.75X −279,732+0.75X −386,597+0.75X −429,657+0.75X 331,306

PW(15%) = 0 = − $1,166,000 − $292,655(P/F,15%,1) − $279,732 (P/F,15%,2) − $386,597(P/F,15%,3) − $429,657(P/F,15%,4) + $331,306(P/F,15%,4) + 0.75X(P/A,15%,4) 0 = − $1,942,423 + 2.14X Thus, X = $1,942,423/2.14 = $907,674 in annual revenues per year.

Breakeven Point Interpretation: The equivalent uniform annual revenue of $907,674 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

-$600,000 % Change

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40

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%

% Change -40% -30% -20% -10% 0% 10% 20% 30% 40%

Used 85,000 20,000 96,000 12%

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%

$(10,000) $(20,000) $(30,000)

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40%

Productivity

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

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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

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) Copyright © 2025 Pearson Education, Inc.

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

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30%

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%) 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 6,561,200 - 8,838,800 5,361,880 - 7,038,120 5,361,200 - 7,038,800 5,361,880 - 7,038,120 5,361,200 - 7,038,800 6,561,880 - 8,838,120 5,160,600 - 7,239,400 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

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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

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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

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

595.49 245.95

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27 $218.49 $210.20 $201.84 $193.40

11-25 MARR Installation Expense ($/in.)

20%

Annual Tax and Insurance Rate Insulation Thickness (in.)

3 4 5 6 7 8 Change in Cost of Heat Loss -50% -40% -30% -20% -10% 0% 10% 20% 30% 40% 50%

Heat Loss (Btu/hr) 4,400 3,400 2,800 2,400 2,000 1,800

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

150

Useful Life Operating Hours per Year

8,760

5%

Cost of Heat Loss ($/Btu)

0.00004

Installation Expense 450 600 750 900 1050 1200

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

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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

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) $

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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)

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 -

Capital Recovery Electricity Disposal

$ $ $

(2,082) -

$ $ $

(2,800) -

Total Annual Cost

$

(2,082)

$

(2,800)

Present Worth

$

(10,341)

$

(13,909)

$ $

$

ILB 0.50 2.00 60 1 -

$ $

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

$ $ $ $ $ $ $

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

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) Decision Reversal: PW(Electricity = $0 / kWh)

$

(5,000)

$

(13,909)

Part b) Table of Results

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CFL

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%

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Cost of Gasoline $ 3.00 $ 4.00 $ 5.00

IRR 5.7% 12.6% 19.2%

11-30 Let X be the variable for the in-park spending per customer.

Annual Worth (Pessimistic): 0

= - $1,200,000,000(A/P, 8%, 10) - $90,000,000 + ($50 + X) * 2,000,000 = - $1,200,000,000(0.1490) - $90,000,000 + ($50 + X) * 2,000,000 = - $178,800,000 - $90,000,000 + ($50 + X) * 2,000,000 = - $268,800,000 + ($50 + X) * 2,000,000 $268,800,000 = ($50 + X) * 2,000,000 $134.40 = $50 + X $84.40 = X The breakeven for pessimistic scenario if the average ticket price is $50 (after discounts) with 2 million annual visitors for in-park spending is $84.40 per customer.

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11-31 Using the data provided, the engineer first calculates the AW of the optimistic, most likely, and pessimistic scenarios. Using a 20 year useful life.

Annual Worth (Optimistic)

= - $800,000,000(A/P, 8%, 20) - $50,000,000 + ($80 + $50)*5,000,000 = - $800,000,000(0.1019) - $50,000,000 + $650,000,000 = - $81,520,000 - $50,000,000 + $650,000,000 = $518,480,000

Annual Worth (Most Likely) = - $1,000,000,000(A/P, 8%, 20) - $70,000,000 + ($60 + $30)*3,500,000 = - $1,000,000,000(0.1019) - $70,000,000 + $315,000,000 = - $101,900,000 - $70,000,000 + $315,000,000 = $143,100,000 Annual Worth (Pessimistic) = - $1,200,000,000(A/P, 8%, 20) - $90,000,000 + ($40 + $20)*2,000,000 = - $1,200,000,000(0.1019) - $90,000,000 + $120,000,000 = - $122,280,000 - $90,000,000 + $120,000,000 = - $92,280,000 The optimistic and most likely scenarios are profitable. However, the pessimistic scenario is not profitable. The engineer now must calculate the breakeven for the pessimistic scenario with respect to the number of visitors. Letting the number of visitors be denoted by the variable X. The following formula will calculate the breakeven for the pessimistic scenario.

Annual Worth (Pessimistic): 0

= - $1,200,000,000(A/P, 8%, 20) - $90,000,000 + ($40 + $20) * X = - $1,200,000,000(0.1019) - $90,000,000 + $60 * X = - $122,280,000 - $90,000,000 + $60 * X = - $212,280,000 + $60 * X $212,280,000 = $60 * X X

= 3,538,000

The breakeven for pessimistic scenario for number of annual visitors is 3,538,000.

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11-32 Let X be the variable for the initial investment.

Annual Worth (Optimistic): 0

= - X (A/P, 8%, 10) - $50,000,000 + ($80 + $50)*5,000,000 = - X (0.1490) - $50,000,000 + $650,000,000 = - X (0.1490) + $600,000,000 X (0.1490) = $600,000,000 X = $4,026,845,638 The initial investment cost would have to be over $4 billion for the optimistic solution to breakeven.

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11-33 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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11-34 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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11-35 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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11-36 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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11-37 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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11-38 False

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11-39 False

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11-40 False

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11-41 True

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11-42 False

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11-43 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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11-44 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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11-45 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)($1,000,000) = $10,000 $2,0001  (P/F, 7%, 20)(F/P, 5%, 20)   EUACculvert =  $50,000   (A/P, 7%, 20) = $7,688 0.07  0.05  

The EUAC of building the culvert is less than the EUAC of a mudslide (with no culvert). Building the culvert is economical.

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12-2

Expected Present Worth = 120,000(0.2) + 150,000(0.3) + 100,000(0.4) + 80,000(0.1) = 117,000

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12-3

Calculate the Present Worth for each scenario separately, and then calculate the Expected Present Worth using the probability provided.

Optimistic: -200,000 + (70,000 – 30,000)(P/A, 7%, 5 years) = -35,992 Most Likely: -200,000 + (60,000 – 35,000)(P/A, 7%, 5 years) = -97,495 Pessimistic: -200,000 + (60,000 – 35,000)(P/A, 7%, 5 years) = -158,998 Expected Present Worth = -35,992(0.3) + -97,495(0.5) + -158,998(0.2) = -91,345 Since negative, the company should not pursue this new product opportunity.

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12-4

Calculate the Present Worth for each scenario separately, and then calculate the Expected Present Worth using the probability provided.

Optimistic: -100,000 + (30,000 – 10,000)(P/A, 8%, 5) = -20,146 Most Likely: -100,000 + (25,000 – 12,000)(P/A, 8%, 5) = -48,095 Pessimistic: -100,000 + (20,000 – 15,000)(P/A, 8%, 5) = -80,037 Expected Present Worth = -20,146(0.2) + -48,095(0.7) + -80,037(0.1) = -45,699 Since negative, the company should not pursue this new opportunity.

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12-5

Calculate the EUA for each scenario separately, and then calculate the Expected EUAC using the probability provided.

4 Year Life: 50,000(A/P, 12%, 4) + 2,000 = 50,000(0.3292) + 2,000 = 18,460 5 Year Life: 50,000(A/P, 12%, 5) + 2,000 = 50,000(0.2774) + 2,000 = 15,870 6 Year Life: 50,000(A/P, 12%, 6) + 2,000 = 50,000(0.2432) + 2,000 = 14,160 7 Year Life: 50,000(A/P, 12%, 7) + 2,000 = 50,000(0.2191) + 2,000 = 12,955 Expected EUAC = 18,460(0.1) + 15,870(0.2) + 14,160(0.4) + 12,955(0.3) = 14,571

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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

(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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Expected Total Annual Cost (A) + (B) $ 5,582,000 $ 4,630,000 $ 5,541,000 $ 6,452,000 $10,726,000

12-7

Calculate the Present Worth for each scenario separately, and then calculate the Expected Present Worth using the probability provided.

Optimistic: -500,000 + (120,000 – 40,000)(P/A, 5%, 10) = -500,000 + (80,000)(7.7217) = 117,736 Most Likely: -500,000 + (100,000 – 50,000)(P/A, 5%, 10) = -500,000 + (50,000)(7.7217) = -113,915 Pessimistic: -500,000 + (80,000 – 60,000)(P/A, 5%, 10) = -500,000 + (20,000)(7.7217) = -345,566 Expected Present Worth = 117,736(0.25) + -113,915(0.5) + -345,566(0.25) = -113,915 Since negative, the company should not pursue this new real estate opportunity.

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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

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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Net Annual Receipts (NAR) $280,000 $330,000 $360,000

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(25%) ---$53,929 -37,857 -37,857 -37,857 -37,857 -37,857 -37,857 -53,929

ATCF -$900,000 226,071 242,143 242,143 242,143 242,143 242,143 242,143 226,071

= −$900,000+[$226,071 + $242,143 (P/A,15% 6)](P/F,15% 1) + $226,071 (P/F,15%, 8) = $167,344

100−Day Season (48,000 Person−Days) EOY BTCF Depr.

TI 0 -$900,000 ----1 330,000 $ 64,286 $265,714 2 330,000 128,571 201,429 3 330,000 128,571 201,429 4 330,000 128,571 201,429 5 330,000 128,571 201,429 6 330,000 128,571 201,429 7 330,000 128,571 201,429 8 330,000 64,286 265,714

T(25%) ---$66,429 -50,357 -50,357 -50,357 -50,357 -50,357 -50,357 -66,429

ATCF -$900,000 263,571 279,643 279,643 279,643 279,643 279,643 279,643 263,571

PW100 (15%) = −$900,000+[$263,571 + $279,643(P/A, 15%, 6)](P/F, 15%, 1) +$263,571 (P/F, 15%, 8) = $335,618 Copyright © 2025 Pearson Education, Inc.

12-9

continued 120−Day Season (54,000 Person−Days) EOY BTCF Depr.

TI 0 -$900,000 ----1 360,000 $ 64,286 $295,714 2 360,000 128,571 231,429 3 360,000 128,571 231,429 4 360,000 128,571 231,429 5 360,000 128,571 231,429 6 360,000 128,571 231,429 7 360,000 128,571 231,429 8 360,000 64,286 295,714

T(25%) ---$73,929 -57,857 -57,857 -57,857 -57,857 -57,857 -57,857 -73,929

ATCF -$900,000 286,071 302,143 302,143 302,143 302,143 302,143 302,143 286,071

PW120 (15%) = −$900,000+[$286,071 + $302,143(P/A,15%, 6)](P/F,15,% 1) +$286,071(P/F,15%, 8) = $436,583 Days of Skiing (X) p(X) PW(ATCF) E[PW(X)] [PW(X)]2 p(X)[PW(X)]2 80 0.6 167,344 100,406 28,004,014,336 16,802,408,602 100 0.3 335,618 100,685 112,639,441,924 33,791,832,577 120 0.1 436,583 43,658 190,604,715,889 19,060,471,589 E[(PW)2] = 69,654,712,768 ($)2 E(PW) = $244,750

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 Calculate the Present Worth for each scenario separately, and then calculate the Expected Present Worth using the probability provided.

Optimistic: -150,000 + (65,000 – 20,000)(P/A, 8%, 3) = -150,000 + (45,000)(2.5771) = -34,031 Most Likely: -150,000 + (55,000 – 25,000)(P/A, 8%, 3) = -150,000 + (30,000)(2.5771) = -72,687 Pessimistic: -150,000 + (45,000 – 30,000)(P/A, 8%, 3) = -150,000 + (15,000)(2.5771) = -111,344 Expected Present Worth = -34,031(0.2) + -72,687(0.6) + -111,344(0.2) = -72,687 Since negative, the company should not pursue this new marketing opportunity.

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12-15 Calculate the Present Worth for each scenario separately, and then calculate the Expected Present Worth using the probability provided.

Optimistic: -300,000 + (90,000 – 30,000)(P/A, 6%, 7) = -300,000 + (60,000)(5.5824) = 34,944 Most Likely: -300,000 + (80,000 – 35,000)(P/A, 6%, 7) = -300,000 + (55,000)(5.5824) = 7,032 Pessimistic: -300,000 + (70,000 – 40,000)(P/A, 6%, 7) = -300,000 + (30,000)(5.5824) = -132,528 Expected Present Worth = 34,944(0.3) + 7,032(0.4) + -132,528(0.3) = -26,462 Since negative, the company should not pursue this new platform opportunity.

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12-16 Calculate the Present Worth for each scenario separately, and then calculate the Expected Present Worth using the probability provided.

Optimistic: -250,000 + (75,000 – 25,000)(P/A, 7%, 5) = -250,000 + (50,000)(4.1002) = -44,990 Most Likely: -250,000 + (65,000 – 30,000)(P/A, 7%, 5) = -250,000 + (35,000)(4.1002) = -106,493 Pessimistic: -250,000 + (55,000 – 35,000)(P/A, 7%, 5) = -250,000 + (20,000)(4.1002) = -167,996 Expected Present Worth = -44,990(0.2) + -106,493(0.5) + -167,996(0.3) = -112,643 Since negative, the company should not pursue this new manufacturing expansion opportunity.

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12-17 (a)

k E(Xk) Ck E(Fk) (P/F,15%,k) 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) V[PWA(15%)] = + (800)2 (P/F,15%,16) = 770,746 $2 2.15

E[PWB(15%)] = −[7,500 (P/A,15%,8)] = −$33,655

(750) 2 ( P / A,15%,16) V[PWB(15%)] = = 1,557,794 $2 2.15 Now let Y = PWA – PWB (i.e., B → A) and find Pr(Y > 0).

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 Calculate the Present Worth for each scenario separately, and then calculate the Expected Present Worth using the probability provided.

Optimistic: -600,000 + (130,000 – 60,000)(P/A, 6%, 10) = -600,000 + (70,000)(7.3601) = -84,793 Most Likely: -600,000 + (120,000 – 70,000)(P/A, 6%, 10) = -600,000 + (50,000)(7.3601) = -231,995 Pessimistic: -600,000 + (110,000 – 80,000)(P/A, 6%, 10) = -600,000 + (30,000)(7.3601) = -379,197 Expected Present Worth = -84,793(0.3) + -231,995(0.4) + -379,197(0.3) = -231,995 Since negative, the company should not pursue this new energy opportunity.

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12-21

Monte Carlo Simulation for the Proposed Project

The estimate of AW based on ten repititions of the experiment is −$173,773/10 = −$17,377.

An estimate of the standard deviation of AW can be found by: K

 [ AW ] 

 ( AW  E[W ]) i 1

2

i

K 1

 = 7,419.38 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 = 8% per year; Analysis Period = 8 years PWNew Product =

−$1,000,000(P/F, 8%, 2) +[0.6($200,000)+0.4($280,000)](P/A,8%,6) (P/F,8%,2) = $62,165

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 ) V[PW(15%)] = 0 + + (3k)2 (P/F,15%,2N) 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

(49 x106 )( P / A,15%,10) V[PW(15%)] = + (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.

2

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: PW(10)%) for BA = −250k (P/A,10%,5)(0.75)

12-25 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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12-28 Rebuild the Excel file, eliminating the Fixed Rate option. The average balance is approximately $1.9 million, which is more than the original case study allocation. However, the 10th percentile does decrease from the original case study percentage allocation.

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12-29 Answers will vary; however, an allocation of Mutual Fund A – Mutual Fund B – Mutual Fund C – Fixed Rate of 33.3%-33.3%-33.3%-10% or 10%-40%-40%-10% will result in a higher average retirement balance since Mutual Fund B and Mutual Fund C have the highest average returns.

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Solutions to Chapter 13 Problems 13-1

NOPAT = (1 – 0.25)($8 million − $4 million − $2 million) = $1.5 million

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13-2

ea = 0.025 + 1.28(0.084) = 0.1325 or 13.25%

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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 = 0.06 + 0.04 = 0.10 = 10%

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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 – 0.028 = 1.4(0.09 – 0.028) or,

RS = 0.028 + 1.4(0.09 – 0.028) RS = 0.1148 per year (11.48%)

The investment amount of $40 million needs to return at least 11.48% 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 Based on the use of equity money of the firm.

Buy

If (a) If (b)

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

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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ATCF −$75,000

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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13-16 Marginal Cost of Capital = investment rate of best option = 18% Weighted Average Cost of Capital = WACC = [10(0.06) +25(0.10)+40(0.08)+150(0.12)+100(0.18)]/(10+25+40+150+100) = (42.3)/(325) = 13%

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13-17 Calculate the Present Worth of each project.

Project X: -250,000 + (60,000 – 15,000)(P/A, 7%, 10) = -250,000 + (45,000)(7.0236) = 66,062 Project Y: -200,000 + (55,000 – 10,000)(P/A, 6.5%, 10) = -200,000 + (45,000)(7.1888) = 123,496 Project Z: -300,000 + (75,000 – 20,000)(P/A, 7.5%, 10) = -300,000 + (55,000)(6.8641) = 77,526 If possible, the company should pursue all projects since they are all positive. The priority order should be Project Y, Project, Z, and Project X (based on highest PW to lowest).

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13-18 Calculate the PW of each project, assume the salvage value is 0 for each alternative.

Project 1: -400,000 + (80,000 – 20,000)(P/A, 5%, 10) = -400,000 + (60,000)(7.7217) = 63,302 Project 2: -350,000 + (70,000 – 18,000)(P/A, 5%, 10) = -350,000 + (52,000)(7.7217) = 51,528 Project 3: -450,000 + (90,000 – 22,000)(P/A, 5%, 10) = -450,000 + (68,000)(7.7217) = 75,076 Project 4: -300,000 + (60,000 – 15,000)(P/A, 5%, 10) = -300,000 + (45,000)(7.7217) = 47,477 Project 5: -500,000 + (100,000 – 25,000)(P/A, 5%, 10) = -500,000 + (75,000)(7.7217) = 79,128 All projects are positive for the given interest rate. Thus, we should now consider the possible feasible combinations given the $1.2 million budget. (For brevity, only considering combinations of three projects that are feasible.)

Projects 1, 2, 3: Sum of PW $189,906*** Projects 1, 2, 4: Sum of PW $162,307 Projects 1, 3, 4: Sum of PW $185,854 Projects 1, 4, 5: Sum of PW $189,906*** Projects 2, 3, 4: Sum of PW $174,081 Projects 2, 4, 5: Sum of PW $178,132 In this situation, two combinations of projects are optimal for the Present Worth for a total present worth of $189,906; using either Projects 1, 2, and 3 or Projects 1, 4, and 5. Both combinations use the full amount of $1.2 million investment.

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13-19 Calculate the PW and IRR of each project, assume the salvage value is 0 for each alternative.

Option A: PW = $185,857, IRR = 14.116% Option B: PW = $195,833, IRR = 13.887% Option C: PW = $211,483, IRR = 14.051% Option D: PW = $211,669, IRR = 13.887% All projects are positive for the given interest rates. Thus, we should now consider the possible feasible combinations given the $900,000 budget. (For brevity, only considering combinations of three projects that are feasible.)

Projects A, B, and C: Total PW = $593,173; lowest IRR is 13.887% No combination of three projects includes Project D; and it has the lowest IRR (tied with Project B).

Thus, the best combination is doing Projects A, B, and C and fully exhausting the $900,000 budget.

The total IRR is calculated by taking the $900,000 initial investment, $180,000 annual revenue, and $44,000 annual cost; which is approximately an IRR of 14.014%.

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13-20 Calculate the PW and IRR of each project, assume the salvage value is 0 for each alternative.

Option X: PW = $31,652, IRR = 8.144% Option Y: PW = $28,487, IRR = 8.144% Option Z: PW = $34,817, IRR = 8.144% Option A: PW = $39,565, IRR = 8.144% Option B: PW = $47,478, IRR = 8.144% All projects are positive for the given interest rate. All have the same IRR. Thus, we should now consider the best possible feasible combinations given the $750,000 budget. (For brevity, only considering combinations of three projects that are feasible and use the entire $750,000 budget.)

Project Options X, A, and B use the entire budget: PW = 31,652 + 39,565 + 47,478 = 118,695 and IRR = 8.144%.

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13-21 Calculate the AW of each project, assume the salvage value is 0 for each alternative. Assume a 10 year life.

Project 1 AW: 13,354 Project 2 AW: 14,173 Project 3 AW: 14,347 Project 4 AW: 13,866 Project 5 AW: 12,719 All projects are positive for the given interest rate. Thus, we should now consider the best possible feasible combinations given the $2 million budget. (For brevity, only considering combinations of three projects that are feasible.)

Projects 1, 2, 3: Combined AW is $41,874 Projects 1, 2, 4: Combined AW is $41,393 Projects 1, 2, 5: Combined AW is $40,246 Thus, should do projects 1, 2, and 3. Will not use the full $2 million budget.

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13-22 Calculate the PW and IRR of each project, assume the salvage value is 0 for each alternative.

Project Alpha: PW = 41,605, IRR = 8.144% Project Beta: PW = 35,246, IRR = 8.144% Project Gamma: PW = 26,769, IRR = 8.144% Project Delta: PW = 16,286, IRR = 8.144% All projects are positive for the given interest rate. All have the same IRR. Thus, we should now consider the best possible feasible combinations given the $1.5 million budget. (For brevity, only considering combinations of three projects.)

Projects Alpha, Beta, Gamma: 103,620 Projects Alpha, Gamma, Delta: 84,660 Projects Alpha, Beta, Delta: 93,137 Projects Beta, Gamma, Delta: 78,301 Project Options Alpha, Beta, and Gamma are the best combination, resulting in a PW of 103,620 and IRR of 8.144%. The $1.5 million budget is not exhausted.

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13-23 Calculate the PW of each project, assume the salvage value is 0 for each alternative.

Project A PW: 93,426 Project B PW: 103,326 Project C PW: 114,005 Project D PW: 121,767 Project E PW: 126,769 All projects are positive for the given interest rate. Thus, we should now consider the best possible feasible combinations given the $1 million budget. (For brevity, only considering combinations of three projects.)

Combined PW for the following combinations: Projects A, B, C: 310,757 Projects A, B, D: 318,519 Projects A, B, E: 323,521 Projects A, C, D: 329,198 Projects A, C, E: 334,200 Projects A, D, E: 341,962 Projects B, C, D: 339,098 Projects B, C, E: 344,100 Projects B, D, E: 351,862 Note, Project combination C, D, and E is over budget.

Best combination is Projects B, D, and E with a Combined Present Worth of $351,862. It uses the entire $1 million budget.

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13-24 Calculate the IRR of each upgrade option, assume the salvage value is 0 for each alternative.

1: 8.144% 2: 8.144% 3: 8.144% 4: 8.144% 5: 8.144% All projects have the same IRR and are above their respective interest rates. Thus, we can use any combination of project upgrades that uses the full budget amount to optimize the upgrade process.

Upgrades 3 and 5 will use the full budget amount and have an IRR of 8.144%. No other combination of projects will use the full budget; thus, Upgrades 3 and 5 is the optimal solution.

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13-25 Calculate the PW and IRR of each route, assume the salvage value is 0 for each alternative and a life of 10 years.

Route 1: PW = 179,130; IRR = 13.434% Route 2: PW = 223,913; IRR = 13.434% Route 3: PW = 268,695; IRR = 13.434% Route 4: PW = 313,478; IRR = 13.434% Each IRR is the same and above the MARR. Each PW is positive. Thus, we can consider the combinations of projects that utilize the $1.3 million budget. (For brevity, only considering combinations of two projects.)

Routes 1 & 2: 403,043 Routes 1 & 3: 447,825 Routes 1 & 4: 492,608 Routes 2 & 3: 492,608 Routes 2 & 4: 537,391 Routes 3 & 4: 582,173 The best combination is Routes 3 and 4, with a PW of $582,173 and an IRR of 13.434%. The budget is fully exhausted.

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13-26 Calculate the IRR for each energy project, assume the salvage value is 0 for each alternative and a life of 10 years.

Solar: 8.144% Wind: 8.144% Hydro: 8.144% Geothermal: 9.606% All projects have a positive IRR above the MARR. Given the budget constraint, and for brevity, only feasible combinations of two projects are considered (calculated using the combination of initial costs, annual benefits and annual costs of all projects in the scenario).

Solar & Wind: Over budget, not feasible.

Solar & Hydro: 8.144% Solar & Geothermal: 8.711% Wind & Hydro: 8.144% Wind & Geothermal: 8.776% Hydro & Geothermal: 8.857% The best possible IRR is to do the Geothermal Project by itself with an IRR of 9.606%. If the company decides to pursue multiple projects, then Hydro and Geothermal should be pursued for a combined IRR of 8.857%. Either of these option combinations is below budget.

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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” Attribute A for alternatives 2 and 3 is acceptable: 70  A  100 Attribute B for all alternatives is acceptable: 6  B  10 Only alternative 2 attribute C is acceptable: Good  C 2  Excellent Attribute D for all alternatives is accceptable: 6  D  10 Only alternative 2 is acceptable because it is the only one whose attributes all lie in acceptable ranges.

(b)

No alternative dominates another.

(c)

Alternative 3 has the best value of the top ranked attribute “D”.

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14-6

(a) Dominance I vs.

Retain

II vs.

III

III vs.

Retain

Reduction in throughput time

Vendor I I vs. III vs.

Vendor II Better Worse

Better

Worse

Better

Flexibility

Worse

Worse

Better

Equal

Better

Reliability

Better

Better

Better

Equal

Better

Quality

Worse

Worse

Better

Equal

Better

Cost of System

Better

Worse

Worse

Worse

Worse

Dominance?

No

No

No

Yes

No

Attribute

Vendor II is removed from consideration. (b)

Satisficing

Attribute Reduction in throughput time Flexibility Reliability Quality Cost of System

“Worst” Acceptable Value 50% Good Good Good $350,000

Unacceptable Alternative Retain Retain Retain Retain None

Remove “Retain Existing System” from consideration (c)

Disjunctive Resolution All alternatives still available (“Retain” already eliminated) pass because all options are acceptable in at least one attribute.

(d) Lexicography Attribute Reduction in throughput time Flexibility Reliability Quality Cost of System

Number of times “greater” 0 2 1 2 4

Alternative Ranking III > I > II II = III > I I > II = III II = III > I I > III > II

Select Vendor III

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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.) $50,000

Sycon (L.A.) $45,000

Sigma (GA.) $49,500

Mc-Graw-Wesley (AZ.) $46,500

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) Wright dominates Alott – Alott is removed from further consideration. (b) Only Wright meets the minimum performance levels for all attributes. (c) All candidates would be retained under disjunctive resolution. (d) Lexicography – Based on project management skills (most important attribute), Busy is eliminated.

Looking next at general attitude, Surley is eliminated. Lastly, looking at years manufacturing experience, Wright would be 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 Assume all attributes are of equal importance.

Attribute Total years Manufacturing years Project management skills Management years General attitude Total Score

Alott 0.33 0.33 1.00 0.00 1.00 2.66

Surley 0.00 1.00 1.00 0.50 0.00 2.50

Wright would be selected.

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Busy 1.00 0.00 0.00 0.50 0.50 2.00

Wright 0.67 0.67 1.00 1.00 1.00 4.34

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 By inspection

Rankj=1 = 2.0 Rankj=2 = 1.0

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 Supplier A Score: 80(40%) + 90(30%) + 70(30%) = 80 Supplier B Score: 85(40%) + 85(30%) + 80(30%) = 83.5 Choose Supplier B

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14-20 Model A: 85(40%) + 80(30%) + 75(20%) + 90(10%) = 82 Model B: 80(40%) + 85(30%) + 80(20%) + 85(10%) = 82 Model C: 75(40%) + 90(30%) + 70(20%) + 80(10%) = 79 Model D: 90(40%) + 75(30%) + 85(20%) + 75(10%) = 83 Choose Model D

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14-21 Truck 1: 80(35%) + 90(30%) + 85(20%) + 70(15%) = 82.5 Truck 2: 85(35%) + 85(30%) + 80(20%) + 75(15%) = 82.5 Truck 3: 90(35%) + 80(30%) + 75(20%) + 80(15%) = 82.5 Truck 4: 75(35%) + 95(30%) + 90(20%) + 65(15%) = 82.5 Truck 5: 70(35%) + 80(30%) + 95(20%) + 85(15%) = 80.25 There is a tie between Truck Models 1, 2, 3, and 4. Any of these models is optimal; unless another criteria is added to distinguish between them. If you must choose between them, then Truck 3 is likely to be chosen since it scored the highest on the highest weighted criteria.

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14-22 Tool A: 85(30%) + 80(30%) + 75(25%) + 90(15%) = 81.75 Tool B: 80(30%) + 85(30%) + 85(25%) + 70(15%) = 81.25 Tool C: 75(30%) + 90(30%) + 90(25%) + 60(15%) = 81 Tool D: 90(30%) + 75(30%) + 70(25%) + 80(15%) = 79 Choose Tool A

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14-23 Project 1: 90(40%) + 70(30%) + 80(20%) + 85(10%) = 81.5 Project 2: 85(40%) + 75(30%) + 85(20%) + 80(10%) = 81.5 Project 3: 80(40%) + 80(30%) + 90(20%) + 75(10%) = 81.5 Project 4: 75(40%) + 90(30%) + 70(20%) + 90(10%) = 80 There is a tie between Projects 1, 2, and 3. Any of these models is optimal; unless another criteria is added to distinguish between them. If you must choose between them, then Project 1 is likely to be chosen since it scored the highest on the highest weighted criteria.

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14-24 Equipment 1: 85(40%) + 80(30%) + 75(20%) + 90(10%) = 82 Equipment 2: 80(40%) + 85(30%) + 70(20%) + 85(10%) = 80 Equipment 3: 75(40%) + 75(30%) + 90(20%) + 80(10%) = 78.5 Equipment 4: 90(40%) + 70(30%) + 85(20%) + 75(10%) = 81.5 Equipment 5: 70(40%) + 90(30%) + 80(20%) + 70(10%) = 78 Choose Equipment 1

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14-25 Location A: 90(35%) + 70(30%) + 85(25%) + 80(10%) = 81.75 Location B: 85(35%) + 75(30%) + 80(25%) + 85(10%) = 80.75 Location C: 80(35%) + 80(30%) + 75(25%) + 90(10%) = 79.75 Location D: 75(35%) + 85(30%) + 90(25%) + 70(10%) = 81.25 Choose Location A

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14-26 Initiative 1: 85(40%) + 80(30%) + 75(20%) + 90(10%) = 82 Initiative 2: 80(40%) + 85(30%) + 80(20%) + 85(10%) = 82 Initiative 3: 75(40%) + 90(30%) + 85(20%) + 80(10%) = 82 Initiative 4: 90(40%) + 75(30%) + 70(20%) + 75(10%) = 80 There is a tie between Initiatives 1, 2, and 3. Any of these initiatives is optimal; unless another criteria is added to distinguish between them. If you must choose between them, then Initiative 1 is likely to be chosen since it scored the highest on the highest weighted criteria.

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14-27 Provider A: 85(35%) + 80(30%) + 75(25%) + 90(10%) = 81.5 Provider B: 80(35%) + 85(30%) + 80(25%) + 85(10%) = 82 Provider C: 75(35%) + 90(30%) + 85(25%) + 80(10%) = 82.5 Provider D: 90(35%) + 75(30%) + 70(25%) + 75(10%) = 79 Choose Provider C

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14-28 Solution 1: 90(40%) + 85(30%) + 80(20%) + 75(10%) = 85 Solution 2: 85(40%) + 90(30%) + 75(20%) + 80(10%) = 84 Solution 3: 80(40%) + 75(30%) + 90(20%) + 85(10%) = 81 Solution 4: 75(40%) + 80(30%) + 85(20%) + 90(10%) = 80 Solution 5: 70(40%) + 95(30%) + 70(20%) + 95(10%) = 80 Choose Solution 1

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14-29 A B C D E F 1 Attribute Dr. Molar Dr. Feelgood Dr. Whoops Dr. Pepper 2 Cost $ 50 $ 80 $ 20 $ 40 3 Anesthesia Novocaine Acupuncture Hypnosis Laughing Gas 4 Distance 15 20 5 30 5 Office Hours 40 25 40 40 6 Quality Excellent Fair Poor Good 7 8 Quality Anesthesia 9 Excellent 4 Acupuncture 1 10 Fair 2 Hypnosis 4 11 Good 3 Laughing Gas 2 12 Poor 1 Novocaine 3 13 14 Attribute Dr. Molar Dr. Feelgood Dr. Whoops Dr. Pepper 15 Cost 0.50 0.00 1.00 0.67 16 Anesthesia 0.67 0.00 1.00 0.33 17 Distance 0.60 0.40 1.00 0.00 18 Hours 1.00 0.00 1.00 1.00 19 Quality 1.00 0.33 0.00 0.67 20 Sum = 3.77 0.73 4.00 2.67 ^ best choice 21 22 23 24 With novocaine rated as the preferred method of anesthesia, Dr. Whoops becomes 25 the dentist of choice.

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G