SOLUTIONS MANUAL For Engineering Economics Financial Decision Making for Engineers 7 Edition. Niall

CHAPTER 1 Solutions to Chapter-End Problems A. Key Concepts When to Use Engineering Economics: 1.1

(a) Yes - several quantifiable alternatives exist (b) Yes - if quantifiable (c) No - alternatives are not quantifiable (d) No - since it mainly involves intangible qualifications of the candidate (e) Yes - at least two alternatives are quantifiable (f) Yes - with quantifiable costs and benefits (g) No - hard to quantify (h) Yes - if the effect is quantifiable (i) Yes - quantifiable (j) No - benefits are hard to quantify (k) Yes - quantifiable (l) Yes - if quantifiable

1.2

Five examples: 1) Location of the business: renting vs. buying decision 2) Equipment: leasing vs. buying decision 3) Equipment: choosing one out of several alternatives; analysis of cost savings 4) Viability of the business: analysis of cash flow; ability to recover the capital investment 5) Effect of the corporate tax rate

1.3

Whatever the description for the items listed, it should be clear that all engineering design is founded on controlling the costs incurred.

Ethics in Decision Making: 1.4

There is no right or wrong answer to any of these questions. The student should observe, however, that there often is a difference in how we should behave compared to how we do behave. Isn’t it always wrong to cheat and plagiarize? Is so then there is a right answer to the question what should you do? You shouldn’t let anyone copy your answers

Chapter 1 - Engineering Decision Making

B. Applications 1.5

There is no right answer to which they should move into. However, they probably will move into the duplex, on the grounds that emotion has a more direct impact on our decisions than does logic.

1.6

(a) She should probably not buy the cheapest one. She cannot decide on price alone. (b) If everything else is the same, one would expect Karen to buy the cheapest one.

1.7

Ciel can (1) invest time in making sure she has the best information about all of the issues that will affect her future sales, and (2) do sensitivity analysis in her financial calculations.

1.8

It is not clear that Trevor should sell out to Venture Corp. He will get more money that way, but it may be very important to him that his brainchild continue to thrive. What Trevor has to decide is whether avoiding the displeasure he gets from Venture Corp. closing down his company is worth $1 000 000 to him. If it is, he should take the Investco offer, or if not, the Venture Corp. offer.

1.9

Possible uncertainties associated with the new technology: 1) Has the new technology proven to be viable?

2) Has the performance or quality that is expected from the technology proven to be consistent?

3) Are there any “side effects”?

4) Is this technology going to stay, or will it be replaced quickly by an alternative technology?

5) What market share might Telekom gain?

Sensitivity analysis may be able to address the second and third issues by considering a range of performance/quality/side effect issues. The first and fourth issues, however, may be difficult to address by sensitivity analysis since there is no specific quantity that can be varied.

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Chapter 1 - Engineering Decision Making

C. More Challenging Problems: 1.10

The Toyota Corolla is the right choice when the interest rate is very high, while the BMW is the right choice if the interest rate is very low.

Notes for Case in Point 1.1 1)

There is no acceptable death rate, but engineering projects proceed even when it is known that deaths will occur.

2)

No right answer. This is a matter of judgement. However, deaths by accident or otherwise will occur whether or not a particular project is approved.

3)

No right answer. It depends on the circumstances.

Notes for Mini-Case 1.1 1)

The student should observe that the offending companies usually have a good reason for staying in business, such as supplying employment, supplying a necessary good, or that efforts to clean up are taking time but should eventually be done.

2)

No right answer. Fining a company affects the shareholders, who are the real owners of the company. On the other hand, fining the management may be appropriate in large companies where the shareholders are often unaware of the effects of decisions of management.

3)

Perhaps, so there is less pollution. Perhaps not, because companies may be forced to close or move to another country, or goods may become too expensive. Some level of pollution is probably acceptable.

4)

Some companies do the socially correct thing anyhow, either for profits resulting from having a “green” image, or because of the moral values of management or owners.

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CHAPTER 2 Solutions to Chapter-End Problems A. Key Concepts Simple Interest: 2.1

P = 3000 N = 6 months i = 0.09 per year = 0.09/12 per month, or 0.09/2 per six months P + I = P + PiN = P(1 + iN) = 3000[1 + (0.09/12)(6)] = 3135 or = 3000[1 + (0.09/2)(1)] = 3135 The total amount due is $3135, which is $3000 for the principal amount and $135 in interest.

2.2

I = 150 N = 3 months i = 0.01 per month P = I/(iN) = 150/[(0.01)(3)] = 5000 A principal amount of $5000 will yield $150 in interest at the end of 3 months when the interest rate is 1% per month.

2.3

P = 2000 N = 5 years i = 0.12 per year F = P(1 + i)N = 2000(1 + 0.12)5 = 3524.68 The bank account will have a balance of $3525 at the end of 5 years.

2.4

(a) P = 21 000 i = 0.10 per year N = 2 years F = P(1 + i)N = 21000(1 + 0.10)2 = 25 410 The balance at the end of 2 years will be $25 410.

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Chapter 2 - Time Value of Money

(b) P = 2 900 i = 0.12 per year = 0.01 per month N = 2 years = 24 months F = P(1 + i)N = 2900(1 + 0.01)24 = 3682.23 The balance at the end of 24 months (2 years) will be $3682.23.

2.5

From: F = P(1 + i)N P = F/(1 + i)N = 50 000/(1 + 0.01)20 = 40977.22 Greg should invest about $40 977.

2.6

F = P(1 + i)N 50 000 = 20 000(1 + i)20 (1 + i)20 = 5/2 i = (5/2)1/20 − 1 = 0.04688 = 4.688% per quarter = 18.75% per year The investment in mutual fund would have to pay at least 18.75% nominal interest, compounded quarterly.

Cash Flow Diagrams: 2.7

Cash flow diagram: $6000

$2000

0

1

2

3

4

5

6

$900

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Chapter 2 - Time Value of Money

2.8

Showing cash flow elements separately: $10000 $10000

0

1

3

2

4

5

$15000

6

7

8

$15000

9

10

11

12

$15000

$15000

$20000

Showing net cash flow: $10000

$10000

$10000

$10000 $5000

0

1

3

2

4

$5000

5

6 $5000

7

8

9

10

11

12

$5000

$20000

2.9

The calculation of the net cash flow is summarized in the table below.

Time 0 1 2 3 4 5 6 7 8 9 10 11 12

Payment 20

20 20 20 20

Receipt 30 33 36.3 39.9 43.9 48.3 53.1 58.5 64.3 70.7 77.8 85.6

Net −20 30 33 16.3 39.9 43.9 28.3 53.1 58.5 44.3 70.7 77.8 65.6

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Chapter 2 - Time Value of Money

Cash flow diagram: $70.7 $53.1

$33

$65.5

$58.5

$39.9 $43.9 $30

$77.8

$44.3

$28.3

$16.3 0

1

2

3

4

5

6

7

8

9

10

11

12

$20

2.10

(a) functional loss (b) use-related physical loss (c) functional loss (d) time-related physical loss (e) use-related physical loss (f) use-related physical loss (g) functional loss (h) time-related physical loss

2.11

(a) market value (b) salvage value (c) scrap value (d) market value to Liam, salvage value to Jacque (e) book value

2.12

The book value of the company is $4.5 based on recent financial statements. The market value is $7 million, assuming that the bid is real and would actually be paid.

2.13

Since sewing machine technology does not change very quickly nor does the required functionality, functional loss will probably not be a major factor in the depreciation of this type of asset. Left unused, but cared for, the machine will lose some value, and hence time-related loss may be present to some extent. The greatest source of depreciation on a machine will likely be use-related and due to wear and tear on the machine as it is operated.

2.14

A switch will generally not suffer wear and tear due to use, and thus userelated physical loss is not likely to be a big factor. Nor will there likely be a physical loss due to the passage of time. The primary reason for depreciation will be functional loss—the price of a similar new unit will likely have dropped due to development of new technology and competition in the marketplace.

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Chapter 2 - Time Value of Money

2.15

The depreciation is certainly not due to use-related physical loss, or other non-physical losses in functionality. The depreciation is a time-related physical loss because it has not been used and maintained over time.

2.16

(a) BV(1) = 14 000 – (14 000 – 3000)/7 = $12 429 (b) BV(4) = 14 000 – 4  (14 000 – 3000)/7 = $7714 (c) BV(7) = 3000

2.17

(a) BV(1) = 14 000(1 – 0.2) = $11 200 (b) BV(4) = 14 000(1 – 0.2)4 = $5 734 (c) BV(7) = 14 000(1 – 0.2)7 = $2936

2.18

(a) d = 1 – (3000/14 000)1/7 = 19.75% (b) BV(4) = 14 000(1 – 0.1975)4 = $5806

2.19

Spreadsheet used for chart: Year 0 1 2 3 4 5 6 7

Straight Line 14000 12429 10857 9286 7714 6143 4571 3000

Declining Balance 14000 11200 8960 7168 5734 4588 3670 2936

14000

Book value ($)

12000 10000

Straight line

8000 6000 4000

Declining balance

2000 0 0

1

2

3

4

5

Year

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7

Chapter 2 - Time Value of Money

2.20

Spreadsheet used for chart: Year 0 1 2 3 4 5 6 7 8 9 10

d = 5% 150000 142500 135375 128606 122176 116067 110264 104751 99513 94537 89811

d = 20% 150000 120000 96000 76800 61440 49152 39322 31457 25166 20133 16106

d = 30% 150000 105000 73500 51450 36015 25211 17647 12353 8647 6053 4237

Book value ($)

150000 5%

100000

50000 20% 30% 0 0

5

10

Year

B. Applications 2.21

I = 190.67 P = 550 N = 4 1/3 = 13/3 years i = I/(PN) = 190.67/[550(13/3)] = 0.08

The simple interest rate is 8% per year.

2.22

F = P(1 + i)N 50 000 = 20 000(1 + 0.01)N (1.01)N = 5/2 N = ln(5/2)/ln(1.01) = 92.09 quarters = 23.02 years Greg would have to invest his money for about 23.02 years to reach his target.

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Chapter 2 - Time Value of Money

2.23

F = P(1 + i)N = 20 000(1 + 0.01)20 = 24 403.80 Greg would have accumulated about $24 404.

2.24

(a) P = 5000 i = 0.05 per six months F = 8000 From: F = P(1 + i)N N = ln(F/P)/ln(1 + i) = ln(8000/5000)/ln(1 + 0.05) = 9.633 The answer that we get is 9.633 (six-month) periods. But what does this mean? It means that after 9 compounding periods, the account will not yet have reached $8000. (You can verify yourself that the account will contain $7757.) Since compounding is done only every six months, we must, in fact, wait 10 compounding periods, or 5 years, for the deposit to be worth more than $8000. At that time, the account will hold $8144. (b) P = 5000 r = 0.05 (for the full year) F = 8000 i = r/m = 0.05/2 = 0.025 per six months From: F = P(1 + i)N N = ln(F/P)/ln(1 + i) = ln(8000/5000)/ln(1 + 0.025) = 19.03 We must wait 20 compounding periods, or 10 years, for the deposit to be worth more than $8000.

2.25

P = 500 F = 708.31 i = 0.01 per month From: F = P(1 + i)N N = ln(F/P)/ln(1 + i) = ln(708.31/500)/ln(1 + 0.01) = 35.001 The deposit was made 35 months ago.

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Chapter 2 - Time Value of Money

2.26

(a) P = 1000 i = 0.1 N = 20 F = P(1 + i)N = 1000(1 + 0.1)20 = 6727.50 About $6728 could be withdrawn 20 years from now. (b) F = PiN = 1000(0.1)(20) = 2000 Without compounding, the investment account would only accumulate $2000 over 20 years.

2.27

Let P = X and F = 2X. (a) By substituting F = 2X and P = X into the formula, F = P + I = P + PiN, we get 2X = X + XiN = X(1 + iN) 2 = 1 + iN iN = 1 N = 1/i = 1/0.11 = 9.0909 It will take 9.1 years. (b) From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i). By substituting F = 2X and P = X into this expression of N, N = ln(2X/X)/ln(1 + 0.11) = ln(2)/ln(1.11) = 6.642 Since compounding is done every year, the amount will not double until the seventh year. (c) Given r = 0.11 per year, the effective interest rate is i = er − 1 = 0.1163.

From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i). By substituting F = 2X and P = X into this expression of N, N = ln(2X/X)/ln(1 + 0.1163) = ln(2)/ln(1.1163) = 6.3013 Since interest is compounded continuously, the amount will double after 6.3 years.

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Chapter 2 - Time Value of Money

2.28

(a) r = 0.25 and m = 2 ie = (1 + r/m)m − 1 = (1 + 0.25/2)2 − 1 = 0.26563 The effective rate is approximately 26.6%. (b) r = 0.25 and m = 4 ie = (1 + r/m)m − 1 = (1 + 0.25/4)2 − 1 = 0.27443 The effective rate is approximately 27.4%. (c) ie = er − 1 = e0.25 − 1 = 0.28403 The effective rate is approximately 28.4%.

2.29

(a) ie = 0.15 and m = 12 From: ie = (1 + r/m)m − 1 r = m[(1 + ie)1/m − 1] = 12[(1 + 0.15)1/12 − 1] = 0.1406 The nominal rate is 14.06%. (b) ie = 0.15 and m = 365 From: ie = (1 + r/m)m − 1 r = m[(1 + ie)1/m − 1] = 365[(1 + 0.15)1/365 − 1] = 0.13979 The nominal rate is 13.98%. (c) For continuous compounding, we must solve for r in ie = er − 1: r = ln(1 + ie) = ln(1 + 0.15) = 0.13976 The nominal rate is 13.98%.

2.30

F = P(1 + i)N 14 800 = 665(1 + i)64 i = 0.04967 The rate of return on this investment was 5%.

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Chapter 2 - Time Value of Money

2.31

The present value of X is calculated as follows: F = P(1 + i)N 3500 = X(1 + 0.075)5 X = 2437.96 The value of X in 10 years is then: F = 2437.96(1 + 0.075/365)3650 = 4909.12 The present value of X is $2438. In 10 years, it will be $4909.

2.32

r = 0.02 and m = 365 ie = (1 + r/m)m − 1 = (1 + 0.02/365)365 − 1 = 0.0202 The effective interest rate is about 2.02%.

2.33

Effective interest for continuous interest account: ie = er − 1 = e0.0599 − 1 = 0.08318 = 6.173% Effective interest for daily interest account: ie = (1 + r/m)m − 1 = (1 + 0.08/365)365 − 1 = 0.08328 = 8.328% No, your money will earn less with continuous compounding.

2.34

ie(weekly) = (1 + r/m)m − 1 = (1 + 0.055/52)52 − 1 = 0.0565 = 5.65% ie(monthly) = (1 + r/m)m − 1 = (1 + 0.07/12)12 − 1 = 0.0723 = 7.23%

2.35

ie(Victory Visa) = (1 + r/m)m − 1 = (1 + 0.26/365)365 − 1 = 0.297 = 29.7% ie(Magnificent Master Card) = (1 + 0.28/52)52 − 1 = 0.322 = 32.2% ie(Amazing Express) = (1 + 0.3/12)12 − 1 = 0.345 = 34.5% Victory Visa has the lowest effective interest rate, so based on interest rate, Victory Visa seems to offer the best deal.

2.36

First, determine the effective interest rate that May used to get $2140.73 from $2000. Then, determine the nominal interest rate associated with the effective interest: F = P(1 + ie)N 2140.73 = 2000(1 + ie)1 ie = 0.070365

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Chapter 2 - Time Value of Money

ie = er − 1 0.070365 = er − 1 r = 0.068 The correct effective interest rate is then: ie = (1 + r/m)m − 1 = (1 + 0.068/12)12 − 1 = 0.07016 The correct value of $2000 a year from now is: F = P(1 + ie)N = 2000(1 + 0.07016)1 = $2140.32 2.37

The calculation of the net cash flow is summarized in the table below.

Time 0 1 2 3 4 5 6 7 8 9 10 11 12

Investment A Payment Receipt 2400 250 250 250 250 250 250 250 250 250 250 250 200 250

Net −2400 250 250 250 250 250 250 250 250 250 250 250 50

Investment B Payment Receipt 50 100 150 200 250 300 350 400 450 500 550 600

500 500 500 500 500 500

Cash flow diagram for investment A: $250 $50 0

1

2

3

4

5

6

7

8

9

10

$2400

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11

12

Net 0 50 −400 150 −300 250 −200 350 −100 450 0 550 100

Chapter 2 - Time Value of Money

Cash flow diagram for investment B: $550 $450 $350 $250 $150

$100

$50 0

1

2

3

4

5

6

7

8

9

10

11

12

$100 $200 $300 $400

Since the cash flow diagrams do not include the time factor (i.e., interest), it is difficult to say which investment may be better by just looking at the diagrams. However, one can observe that investment A offers uniform cash inflows whereas B alternates between positive and negative cash flows for the first 10 months. On the other hand, investment A requires $2400 up front, so it may not be a preferred choice for someone who does not have a lump sum of money now.

2.38

(a) The amount owed at the end of each year on a loan of $100 using 6% interest rate: Year 0 1 2 3 4 5 6 7 8 9 10

Simple Interest 100 106 112 118 124 130 136 142 148 154 160

Compound Interest 100.00 106.00 112.36 119.10 126.25 133.82 141.85 150.36 159.38 168.95 179.08

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Chapter 2 - Time Value of Money

180

Amount owed ($)

170 160

Compound interest

150 140 130

Simple interest

120 110 100 0

1

2

3

4

5

6

7

8

9

10

Year

(b) The amount owed at the end of each year on a loan of $100 using 18% interest rate: Year 0 1 2 3 4 5 6 7 8 9 10

Simple Interest Compound Interest 100 100.00 118 118.00 136 139.24 154 164.30 172 193.88 190 228.78 208 269.96 226 318.55 244 375.89 262 443.55 280 523.38

Amount owed ($)

600 500 400

Compound interest

300 200

Simple interest

100 0

1

2

3

4

5

6

7

8

Year

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10

Chapter 2 - Time Value of Money

2.39

(a) From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i).

At i = 12%: N = ln(1 000 000/0.01)/ln(1 + 0.12) = 162.54 years At i = 18%: N = ln(F/P)/ln(1 + i) = ln(1 000 000/0.01)/ln(1 + 0.18) = 111.29 years (b) The growth in values of a penny as it becomes a million dollars: Year 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180

2.40

At 12% 0.01 0.03 0.10 0.30 0.93 2.89 8.98 27.88 86.58 268.92 835.22 2 594.07 8 056.80 25 023.21 77 718.28 241 381.18 749 693.30 2 328 433.58 7 231 761.26

At 18% 0.01 0.05 0.27 1.43 7.50 39.27 205.55 1 075.82 5 630.68 29 470.04 154 241.32 807 273.70 4 225 137.79 22 113 676.39 115 739 345.70 605 760 702.48 3 170 451 901.72 16 593 623 884.84 86 848 298 654.83

From the table and the charts below, we can see that $100 will double in (a) 105 months (or 8.75 years) if interest is 8% compounded monthly (b) 13 six-month periods (6.5 years) if interest is 11% per year, compounded semi-annually (c) 5.8 years if interest is 12% per year compounded continuously Month 0 12 24 36 48 60 72 84

8% 100.00 108.30 117.29 127.02 137.57 148.98 161.35 174.74

11% 100.00 111.30 123.88 137.88 153.47 170.81 190.12 211.61

12% 100.00 112.75 127.12 143.33 161.61 182.21 205.44 231.64

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Chapter 2 - Time Value of Money

Deposit ($)

Month 96 108

8% 189.25 204.95

11% 235.53 262.15

12% 261.17 294.47

12%

300 280 260 240 220 200 180 160 140 120 100

11%

8%

0

20

40

60

80

100

120

Month

2.41

P(1 – d)n = P – n(P – S)/N 245 000(1 – d)20 = 245 000 – 20(245 000 – 10 000)/30 (1 – d)20 = 88 333.33/245 000 = 0.3605 1 – d = 0.9503 d = 4.97% The two will be equal in 20 years with a depreciation rate of 4.97%.

2.42

780 000(1 – d)20 = 60 000 (1 – d)20 = 1/13 d = 1 – (1/13)1/20 = 1 – 0.8796 = 0.1204 A depreciation rate of about 12% will produce a book value in 20 years equal to the salvage value of the press.

2.43

(a) BV(4) = 150 000 – 4[(150 000 – 25 000)/10] = 150 000 – 4(12 500) = 150 000 – 50 000 = 100 000 DC(5) = (150 000 – 25 000)/10 = 12 500 (b) BV(n) = 150 000(1 – 0.2)4 = 150 00(0.8)4 = 61 440 DC(5) = BV(4)  0.2 = 61 440(0.2) = 12 288 (c) d = 1 – (25 000/150 000)1/10 = 0.1640 = 16.4%

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Chapter 2 - Time Value of Money

C. More Challenging Problems 2.44

The present worth of each instalment: Instalment 1 2 3 4 5 6 7 8 9 10

F 100000 100000 100000 100000 100000 100000 100000 100000 100000 100000 Total

P 100000 90521 81941 74174 67143 60779 55018 49803 45082 40809 665270

Sample calculation for the third instalment, which is received at the end of the second year: P = F/(1 + r/m)N = 100 000/(1 + 0.10/12)24 = 81 941 The total present worth of the prize is $665 270, not $1 000 000.

2.45

The present worth of the lottery is $665 270. If you take $300 000 today, that leaves a present worth of $365 270. The future worth of $365 270 in 5 years (60 months) is: F = P(1 + r/m)N = 365 270(1 + 0.10/12)60 = 600 982 The payment in 5 years will be $600 982.

2.46

The first investment has an interest rate of 1% per month (compounded monthly), the second 6% per 6-month period (compounding semiannually). (a) Effective semi-annual interest rate for the first investment: ie = (1 + is)N − 1 = (1 + 0.01)6 − 1 = 0.06152 = 6.152% Effective semi-annual interest rate for the second investment is 6% as interest is already stated on that time period. (b) Effective annual interest rate for the first investment: ie = (1 + is)N − 1 = (1 + 0.01)12 − 1 = 0.1268 = 12.68% Effective annual interest rate for the second investment: ie = (1 + is)N − 1 = (1 + 0.06)2 − 1 = 0.1236 = 12.36% 19 Copyright © 2022 Pearson Canada Inc.

Chapter 2 - Time Value of Money

(c) The first investment is the preferred choice because it has the higher effective interest rate, regardless of on what period the effective rate is computed.

2.47

(a) i = 0.15/12 = 0.0125, or 1.25% per month The effective annual rate is: ie = (1 + i)m − 1 = (1 + 0.0125)12 − 1 = 0.1608 or 16.08% (b) P = 50 000 N = 12 i = 0.15/12 = 0.0125, or 1.25% per month F = P(1 + i)N = 50 000(1 + 0.0125)12 = 58 037.73 You will have $58 038 at the end of one year. (c) Adam’s Fee = 2% of F = 0.02(58037.73) = 1160.75 Realized F = 58 037.73 − 1160.75 = 56 876.97 The effective annual interest rate is: F = P(1 + i)1 56 876.97 = 50 000(1 + i) i = 56 876.97/50 000 − 1 = 0.1375 or 13.75% The effective interest rate of this investment is 13.75%.

2.48

Market equivalence does not apply as the cost of borrowing and lending is not the same. Mathematical equivalence does not hold as neither 2% nor 4% is the rate of exchange between the $100 and the $110 one year from now. Decisional equivalence holds as you are indifferent between the $100 today and the $110 one year from now.

2.49

Decisional equivalence holds since June is indifferent between the two options. Mathematical equivalence does not hold since neither 8% compounded monthly (lending) or 8% compounded daily (borrowing) is the rate of exchange representing the change in the house price ($110 000 now and $120 000 a year later is equivalent to the effective interest rate of 9.09%). Market equivalence also does not hold since the cost of borrowing and lending is not the same.

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Chapter 2 - Time Value of Money

2.50

(a) The amount of the initial deposit, P, can be found from F = P(1 + i)N with F = $3000, N = 36, and i = 0.10/12. (b) Having determined P = $2225, then we can figure out the size of the deposit at the end of years 1, 2, and 3.

If you had not invested in the fixed interest rate investment, you would have obtained interest rates of 8%, 10%, and 14% for each of the three years. The table below shows how much the initial deposit would have been worth at the end of each of the three years if you had been able to reinvest each year at the new rate. Because of the surge in interest rates in the third year, with 20/20 hindsight, you would have been better off (by about $60) not to have locked in at 10% for three years.

Deposit amount ($)

Year 0 1 2 3

Fixed Interest Rate 2225 2458 2715 3000

Varying Interest Rate 2225 2410 2662 3060

3200 3000 2800 Fixed interest rate

2600

Varying interest rate

2400 2200 0

1

2

3

Year

2.51

Interest rate i likely has its origins in commonly available interest rates present in Marlee’s financial activities such as investing or borrowing money. Interest rate j can only be determined by having Marlee choose between X and Y to determine at which interest rate Marlee is indifferent between the choice. Interest rate k probably does not exist for Marlee, since it is unlikely that she can borrow and lend money at the same interest rate. If for some reason she could, then k=j.

Also, i could be either greater or less than j.

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Chapter 2 - Time Value of Money

2.52

BV(0) = 250 000 BV(6) = 250 000  (1 − 0.3)6 = 29 412.25 The book values of the conveyor after 7, 8, 9, and 10 years are: BV(7) = 29 412.25 − 29 412.25/4  1 = 22 059.19 BV(8) = 29 412.25 − 29 412.25/4  2 = 14 706.13 BV(9) = 29 412.25 − 29 412.25/4  3 = 7353.07 BV(10) = 29 412.25 − 29 412.25/4  4 = 0

2.53

d = 1 – (S/P)1/n = 1 – (8300/12 500)1/2 = 1– 0.81486 = 0.18514 = 18.514% BVdb(5) = 12 500 (1 – 0.18514)5 = 4470.87 Enrique should expect to get about $4471 for his car three years from now.

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Chapter 2 - Time Value of Money

Notes for Case in Point 2.1 1)

Close, if the appropriate depreciation method is being used.

2)

It makes sense because it is a new technology.

3)

Because the accounting department is likely using a specific depreciation method that is not particularly accurate in this case. In particular, they may be using a depreciation method required for tax purposes.

4)

Bill Fisher is probably not doing anything wrong, but it wouldn’t hurt to check.

Notes for Mini-Case 2.1 1)

Money will always be lost over the year. If money could be gained, everybody would borrow as much money as possible to invest.

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CHAPTER 3 Solutions to Chapter-End Problems A. Key Concepts Recognizing Cash Flows: 3.1

Boarding: an annuity paid monthly, at the end of the months May to September, brought to present worth to the beginning of May. Monthly estimates obtained by historical measurements.

Breeding: annuity due, with a six-month period over 10 years Heating, water and sewage: monthly annuity, estimated from the average monthly costs over the last few years Food: annuity due over 10 years, estimated by the amount paid for food in previous years

3.2

Rent: monthly annuity for eight periods. Amount known from rental agreement.

Bus: weekly annuity. Amount estimated as five return trips, or if on a monthly bus pass, an annuity due as the cost of the pass.

Groceries: weekly annuity, amount estimated or observed weekly average Lunch: weekly annuity, amount estimated or observed weekly average Printing and copying: two single payments, one at the end of four months and one at the end of eight months, or an annuity with a four-month period. Amount estimated or observed from previous years.

Christmas presents: single payment, amount as budgeted Christmas extra cash: single payment, amount as estimated

Single Disbursements or Receipts: 3.3

P = 100 i = 8% N = 15 Using the formula: F = P(1 + i)N = 100(1 + 0.08)15 = 317.22 24 Copyright © 2022 Pearson Canada Inc.

Chapter 3 - Cash Flow Analysis

Or using the compound interest tables: F = P(F/P, 8%, 15) = 100(3.1722) = 317.22 About $317 will be in the bank account.

3.4

F = 1 000 000 i = 12% N = 30 Using the formula: P = F/(1 + i)N = 1 000 000/(1 + 0.12)30 = 33 377.92 Using the tables produces a slightly different result due to the number of significant digits in the table: P = F(P/F, 12%, 30) = 1 000 000(0.0334) = 33 400 You should invest about R 33 400.

3.5

1725(F/P, i, 5) = 3450 (F/P, i, 5) = 2 (F/P, 14%, 5) = 1.9254 (F/P, 15%, 5) = 2.0114 Solve for i using linear interpolation: i = 0.14 + (0.15 − 0.14)[(2 − 1.9254)/(2.0114 − 1.9254)] = 0.1487 = 14.87% In 10 years, you will have: F = 1725(F/P, 14.87%, 10) = 1725(1 + 0.1487)10 = $6900

Annuities: 3.6

P = 500(P/A, 0.5%, 1220) + 5000 = 500[(1 + 0.005)240 − 1]/[0.005(1 + 0.005)240] + 5000 = 69 790 + 5000 = 74 790 Morris purchased the house for £74 790.

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Chapter 3 - Cash Flow Analysis

3.7

Using the capital recovery formula: A = (P − S)(A/P, i, N) + Si = (45 000 − 25 000)(A/P, 0.15, 5) + 25 000(0.15) = 20 000(0.29832) + 3 750 = 9 716.40 The juicer would have to save a little over $18 400 per year.

3.8

F = 10 000 i = 0.06/12 = 0.005 per month N = 24 A = 10 000(A/F, 0.5%, 24) = 10 000(0.03932) = 393.20 Fred has to save a little over $393 per month.

Arithmetic Gradient Series: 3.9

(a) F = 40(F/A, 1%, 24) = 40(26.969) = 1079 (b) F = [30 + 1(A/G, 1%, 24)](F/A, 1%, 24) = [30 + 1(11.010)](26.969) = 1106

3.10

First find the annuity value of the prize by converting the gradient into an annuity: A = A’ + G(A/G, i, N) = 1000 + 1000(A/G, 15%, 20) = 1000 + 1000(5.3651) = 6365.1 Converting the annuity into a present value gives: P = A(P/A, 15%, 20) = 6365.1(6.2593) = 39 841 Therefore, the winning ticket has a present worth of about £39 800. Since 10 000 tickets are to be sold, on average each ticket is worth £39 800/10 000 = £3.98.

Geometric Gradient Series: 3.11

First find the growth adjusted interest rate: i = (1 + i)/(1 + g) − 1 = 1.2/1.1 − 1 = 0.0909 = 9.09%

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Chapter 3 - Cash Flow Analysis

We then make use of the geometric series to present worth conversion factor with A = 10 000, g = 10%, i = 0.0909, and N = 10. Since the growth adjusted interest rate is positive, we can make use of the present worth conversion factor.

P = A(P/A, g, i, N) = A(P/A, i, N)/(1 + g) = 10 000(P/A, 9.09%, 10)/(1 + 0.1) = 10 000(6.3923/1.1) = 58 110 We then use the present worth factor to calculate the worth in 10 years: F = 58 110(F/P, 20%, 10) = 58 110(6.1917) = 359 810 The savings would have accumulated to about $360 000.

3.12

First find the growth adjusted interest rate: i = (1 + i)/(1 + g) − 1 = 1.1/1.2 − 1 = −0.0833 = −8.333% We then make use of the geometric series to present worth conversion factor with A = 10 000, g = 20%, i = −0.0833, and N = 10. Since the growth adjusted interest rate is negative, we cannot make use of the present worth conversion factor.

P = A(P/A, g, i, N) = A[(1 + i)N − 1]/[i(1 + i)N][1/(1 + g)] = 10 000[(1 − 0.08333)10 − 1]/[−0.08333(1 − 0.08333)10][1/(1 + 0.2)] = 138 710 We then use the present worth factor to calculate the worth in 10 years: F = 138 710(F/P, 10%, 10) = 138 710(2.5937) = 359 790.

The savings would have accumulated to about $360 000.

Non-standard Annuities and Gradients: 3.13

Method 1: Consider the annuities as separate future payments.

Year 7 12 17 22

Present Worth 10 000(P/F, 9%, 7) = 10 000(0.54703) 10 000(P/F, 9%, 12) = 10 000(0.35553) 10 000(P/F, 9%, 17) = 10 000(0.23107) 10 000(P/F, 9%, 22) = 10 000(0.15018) Total

The investment is worth about $12 838 today.

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= 5470.3 = 3555.3 = 2310.7 = 1501.8 = 12 838.1

Chapter 3 - Cash Flow Analysis

Method 2: Convert the compounding period from yearly, to every five years. This can be done with the effective interest rate formula: ie = (1 + 0.09)5 − 1 = 53.86% The present value at the end of period 2 is then: P2 = 10 000(P/A, 53.86%, 4) and so the present worth today is: P0 = 10 000(P/A, 53.86%, 4)(P/F, 9%, 2) = 10 000(1.5253)(0.84168) = $12 838 Method 3: Convert the annuity to an equivalent yearly annuity. This can be done by considering only the first payment as a future value, and finding the equivalent annuity over the five-year period using the sinking fund factor: A = 10 000(A/F, 9%, 5) This yearly annuity is brought to present worth at the end of period 2: P2 = 10 000(A/F, 9%, 5)(P/A, 9%, 20) and so the present worth today is: P0 = 10 000(A/F, 9%, 5)(P/A, 9%, 20)(P/F, 9%, 2) = 10 000(0.16709)(9.1285)(0.84168) = $12 838 Note that each method produces the same amount, allowing for rounding.

When you have a choice in methods as in this example, your choice will depend on what you find convenient, or what is the most efficient computationally.

3.14

Amount owed in one year: P(1-year) = 2000(F/P, 0.5%, 12) = 2000(1.0616) = 2123.2 Convert this to an annuity: A = P(1-year)(A/P, 0.5%, 24) = 2123.2(0.04433) = 94.13 Payments will be $94.13 per month.

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Chapter 3 - Cash Flow Analysis

3.15

Find the effective annual interest rate first in order to determine the effective interest rate for the four-month period: ie = (1 + .12/12) 12 − 1 = 0.1268 0.1268 = (1 + i4-months)3 − 1 i4-months = (1.1268)1/3 − 1 = 0.040604 Using this interest rate, P = (P/A, 4.0806%, 20)[400 + 100(A/G, 4.0806%, 20)] = 13.5179[400 + 100(8.1904)] = 16 478.91 The present worth is $16 479.

When n Goes to Infinity: 3.16

The present worth computations for the full capacity tunnel can be found as follows: First, the $100 000 paid at the end of 10 years can be thought of as a future amount which has an equivalent annuity: A = 100 000(A/F, 8%, 10) = 100 000(0.06903) = 6903 Thus, at 8% interest, $100 000 every 10 years is equivalent to $6903 every year.

Since the tunnel will have (approximately) an infinite life, the present worth of the lining repairs can be found using the capitalized cost formula.

Added to the initial cost, the total present worth is thus: P(total) = P(initial cost) + P(lining) = 3 000 000 + 6903/0.08 = 3 086 288 Alternatively, convert to a 10-year annuity period, changing the interest rate instead of the payment amount: i = (1 + 0.08)10 − 1 = 1.1589 P = 3 000 000 + 100 000/1.1589 = $3 086 287

3.17

P = A/i = 18 000/0.07 = 257 143 GA must donate a little over $257 000.

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Chapter 3 - Cash Flow Analysis

3.18

Sum the present worth of the $350 annuity and the 10 000 future value at 5% per period over 20  2 = 40 periods.

P = 350(P/A, 5%, 40) + 10 000(P/F, 5%, 40) = 350(17.158) + 10 000(0.14205) = 7425.80 I should pay no more than (approximately) $7426 for the bond.

3.19

P = 10 000(P/F, 4%, 18) + (10 000  0.09/2)(P/A, 4%, 18) = 10 000(0.49363) + 450(12.659) = 10 632 The bond is worth about $10 632 today.

Estimating Unknowns: 3.20

Using linear interpolation: X* = 100 + (200 – 100)  [(300 000 000 – 200 000 000)/(360 000 000 – 200 000 000)] = 100 + 100(100 000 000/160 000 000) = 100 + 100(0.625) = 162.5 Trenny should be able to afford a 162.5 MW plant.

3.21

From basic principles: F = P(1 + i)N 5000 = 2300(1 + i)7 1 + i = (5000/2300)1/7 i = (5000/2300)1/7 − 1 = 0.1173 = 11.73% Or from the tables: F = P(F/P, i, N) 5000 = 2300(F/P, i, 7) (F/P, i, 7) = 5000/2300 = 2.1739 (F/P, 11%, 7) = 2.0761 (F/P, 12%, 7) = 2.2106 By linearly interpolating between the two, we get: i = 11 + (12 − 11)[(2.1739 − 2.0761)/(2.2106 − 2.0761)] = 11.73% 30 Copyright © 2022 Pearson Canada Inc.

Chapter 3 - Cash Flow Analysis

3.22

You could use either the capital recovery factor or the series present worth factor.

Method 1: Using series present worth factor 2000 = 40(P/A, 1%, N) We can solve using the formula: 2000 = 40[(1 + 0.01)N − 1]/[0.01(1 + 0.01)N] 0.5 = (1.01N − 1)/1.01N 1 = 1.01N − 0.5(1.01N) 2 = 1.01N ln(2) = N[ln(1.01)] N = ln(2)/ln(1.01) = 69.66 It will take her about 70 months or about 5.8 years.

Method 2: Using linear interpolation 2000 = 40(P/A, 1%, N) (P/A, 1%, N) = 50 (P/A, 1%, 70) = 50.169 (P/A, 1%, 65) = 47.627 Linearly interpolating: N = 65 + 5[(50 − 47.627)/(50.169 − 47.627)] = 69.97 The small difference is the error in linearly interpolating.

B. Applications 3.23

5000 = (7  20 000)(P/F, i, 18) = 140 000/(1 + i)18 (1 + i)18 = 140 000/5000 i = 281/18 − 1 = 0.20336 The annual rate of return must be 20.3%.

3.24

This problem requires the solution for N in the Sinking Fund Factor.

Starting with the definition, the explicit formula for N can be obtained through some manipulation: 31 Copyright © 2022 Pearson Canada Inc.

Chapter 3 - Cash Flow Analysis

A = F(A/F, i, N) = Fi/[(1 + i)N − 1] which leads to: (1 + i)N = (iF + A)/A Taking the logarithm of both sides and rearranging gives: N = ln[(iF + A)/A]/ln(1 + i) = ln[(0.10  50 000 + 7000)/7000]/ln(1 + 0.1) = 5.655 It will take about 5.7 years for the members to save the $50 000.

Note that this problem can also be effectively solved by trial and error calculations.

3.25

(a) 2000 + 350 + 210 = 100(P/A, 3%, N) (P/A, 3%, N) = 2560/100 = 25.6 Solve for N using linear interpolation: N = 45 + (50 − 45)(25.6 − 24.519)/(25.730 − 24.519) = 49.4632 It will take about 50 months to complete her payment. (b) A = 2560(A/P, 3%, 24) = 2560(0.05905) = 151.168 Yoko’s monthly payment will have to be $151.

3.26

(a) Deposit of $15 per week: F = (15  4)(F/A, 1.5%, 36) = 60[(1.01536 − 1)/0.015] = $2836.56 Deposit of $20 per week: F = (20  4)(F/A, 1.5%, 36) = 80  (1.01536 − 1)/0.015 = $3782.08 (b) A = 5000(A/F, 1.5%, 36) = 5000[0.015/(1.01536 − 1)] = 105.76 Rinku needs to deposit $105.76/4 = $26.44 per week.

3.27

A = (P – S)(A/P, i, N) + Si = (140 000 – 37 000)(A/P, 14%, 5) + 37 000(0.14) = 133 000(0.2918) + 37 000(0.14) = 43 989.4 The investment would have to save about $43 989 per year over its 5-year life.

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Chapter 3 - Cash Flow Analysis

3.28

Using linear interpolation: X* = 500 + (800 – 500)[(15 000 – 11 350)/(18 950 – 11 350)] = 500 + 300(3650/7600) = 500 + 300(0.48026) = 644.08 Enrique would have to invest about $644 per month.

3.29

The present worth of the mortgage is 260 000 at the end of August, last year. i = 0.12/12 = 0.01 or 1% per month Length of mortgage = 12  20 = 240 months Payments are: A = P(A/P, i, N) =260 000(A/P, 1%, 240) = 2862.82 Method 1: Find the future worth of 260 000 at the end of 5 months.

FW(mortgage, end of 5 months) = 260 000(F/P, 1%, 5) = 273 262.61 Future worth of 5 payments at end of 5 months: FW(5 payments) = 2862.82(F/A, 1%, 5) = 14603.28 Amount owing at end of January = FW(mortgage) − FW(5 payments) = 273262.61 − 14603.28 = 258659.33 They still owe $258 659.

Method 2: Find the worth of 5 payments as of end of August.

PW(5 payments) = 2862.82(P/A, 1%, 5) = 13894.52 PW(debt at end of Aug) = 260 000 PW(amount owed) = 2600 000 − 13894.52 = 246105.48 Amount owing at end of Jan = 246105.48(F/P, 1%, 5) = 2586.59 They still owe $258 659.

3.30

A' = 10 000 G = 1000 i = 0.15 N = 6 for gradient to annuity 6 for annuity to present value 2 for future value to present value 33 Copyright © 2022 Pearson Canada Inc.

Chapter 3 - Cash Flow Analysis

P = [10 000 + 1000(A/G, 15%, 6)](P/A, 15%, 6)(P/F, 15%, 2) = [10 000 + 1000(2.0971)](3.7844)(0.75614) = 34 616.29 The software is worth $34 616 today.

3.31

One way to solve this problem is to sum the future worth of each individual cash flow: Balance: F1 = 2400(F/P, 1%, 24) = 2400(1.2697) = 3047.28 Salary: F2 = 120(F/A, 1%, 24) = 120(26.973) = 3236.76 Dividends: F3 = 200(F/P, 1%, 20) = 200(1.2202) = 244.04 F4 = 200(F/P, 1%, 16) = 200(1.1726) = 234.52 F5 = 200(F/P, 1%, 12) = 200(1.1268) = 225.36 F6 = 200(F/P, 1%, 8) = 200(1.0829) = 216.58 F7 = 200(F/P, 1%, 4) = 200(1.0406) = 208.12 F8 = 200 Fee: F9 = [10 + 1(A/G, 1%, 24)](F/A, 1%, 24) = [10 + 1(11.024)](26.973) = 567.08 Total future value = F1 + F2 +  + F8 − F9 = 7045.74 Clem will have saved about $7046.

3.32

P = −30 000 − 1600(P/A, 0.5%, 24) + 40 000(P/F, 0.5%, 24) + 2000(P/A, 0.5%, 12) + 2400(P/A, 0.5%, 12)(P/F, 0.5%, 12) − [400 + 40(A/G, 0.5%, 18)](P/A, 0.5%, 18)(P/F, 0.5%, 6) = −30 000 − 1600(22.558) + 40 000(0.88721) + 2000(11.616) + 2400(11.616)(0.94192) − [400 + 40(8.3198)](17.168)(0.97052) = −30 000 − 36 092 + 35 548 + 13 232 + 26 260 − 12 210 = 6678 The present worth of this investment is $6678. Yogajothi should buy the house.

3.33

P = 15 000(P/A, 12%, 8) = 15 000(4.9676) = 74 514 Barnaby Circuit Boards could afford to spend up to about $75 000 on the wave soldering machine.

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Chapter 3 - Cash Flow Analysis

3.34

This is an arithmetic gradient with A' = 100 000 G = 10 000 per year N = 30 years i = 0.08 per year A = A' + G(A/G, 8%, 30) = 100 000 + 10 000(9.1897) = 100 000 + 91 897 = 191 897 P = A(P/A, i, N) = 191 897(P/A, 8%, 30) = 191 897(11.258) = 2 160 376 Yes, it is a good deal.

3.35

A = 100 000 g = 10% per year i = 8% per year N = 30 years i = (1 + i)/(1 + g) − 1 = 1.08/1.1 − 1 = −0.01818 For a negative interest rate, we cannot use the Series Present Worth Factor from tables, so we must use the full formula: P = 100 000[(1 − 0.01818)30 − 1]/[−0.01818(1 − 0.01818)30][1/(1 + 0.1)] = 3 670 261 No, Leon would have to pay more than $1 000 000 more.

3.36

Amount available for annuity = 20 000 − 1200 = 18 800 P = 18 800 i = 0.07/12 per month N = 55  12 = 660 months A = P(A/P, i, N) = P[i(1 + i)N]/[(1 + i)N − 1] = 18 800[(0.07/12)(1 + 0.07/12)660]/[(1 + 0.07/12)660 − 1] = 112.10 The approximate daily amount can be calculated from: Daily amount = (112.10  12)/365 = $3.68 No, Tina would not have enough money to retire. She would have about $3.68 available to spend each day.

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Chapter 3 - Cash Flow Analysis

3.37

Construction costs (in $millions): P1 = 20(P/A, 6%, 5)(P/F, 6%, 4) = 20(4.2124)(0.79209) = 66.728 Maintenance and Repair costs (in $millions): i = (1 + 0.06)/(1 + 0.01) − 1 = 0.0495  5.0% P2 = 2[(P/A, 5%, 36)/(1 + 0.01)](P/F, 6%, 9) = 2(16.547/1.01)(0.5919) = 19.3944 P = P1 + P2 = 86.1224 The present cost of the water supply project is about $86 million.

3.38

P = 10 000(P/A, 9%, 5) + 20 000(P/A, 9%, 10)(P/F, 9%, 5) = 10 000(3.8896) + 20 000(6.4176)(0.649 93) = 122 316 The present worth is €122 316.

3.39

The actual amount loaned is 500 − 45 = $455 45 = 455(A/P, i, N) (A/P, i, 12) = 45/455 = 0.0989 (A/P, 2.5%, 12) = 0.09749 (A/P, 3%, 12) = 0.10046 By linear interpolation: i = 2.5 + 0.5[(0.0989 − 0.09749)/(0.10046 − 0.09749)] = 2.73% per month The effective interest rate is: ie = (1 + 0.02737)12 − 1 = 38.3% per year

3.40

g = 0.01 i = 0.015 (monthly) ie = (1 + 0.015)2 − 1 = 0.03022 (bimonthly) i = (1 + ie)/(1 + g) − 1 = 1.0302/1.01 − 1 = 0.02 We assume that Shamsir’s monthly profit remains the same as the previous month when there is no increase. For example, his cash flows for months 1, 2, 3, and 4 would be $10 000, $10 100, $10 100, and $10 201.

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Chapter 3 - Cash Flow Analysis

Noting that there are actually two sets of geometric gradient series, identical in all aspects but different only in timing of the first cash flow, we can determine the present value as follows: P = 10 000(P/A, g, ie, 12) + 1000(P/A, g, ie, 12)(F/P, i, 1) = 10 000[(P/A, i, 12)/(1 + g)][1 + (F/P, 0.015, 1)] = 10 000[(P/A, 0.02, 12)/1.01](1 + 1.0150) = 10 000(10.575/1.01)(2.015) = 210 976 The present value of all his profit over the next 2 years is about $211 000.

3.41

The 6-month interest rate can be calculated from: 300% = 3 = (1 + i)2 − 1 (1 + i)2 = 4 1+i=2 i = 1 = 100% Then: P = 5000(P/F, 100%, 10) + 500(P/A, 100%, 10) = 5000(0.00098) + 500(0.99902) = 504.41 I should pay no more than $504 for the bond now.

3.42

A = 5000(0.015) = 75 i = 0.08/4 = 2% P = 75 + 75(P/A, 2%, 25) + 5000(P/F, 2%, 25) = 75 + 75(19.523) + 5000(0.6095) = 4586.88 You would be willing to pay up to $4587 for the bond.

C. More Challenging Problems 3.43

(a) It is not necessary to know how long the mortgages will last, as long as they are equal.

For the old plan: Using X to represent the quarterly payments, the present value of one three-month payment cycle is calculated.

P(quarterly payment) = X/(1 + i)1 = X/1.06 = 0.943X 37 Copyright © 2022 Pearson Canada Inc.

Chapter 3 - Cash Flow Analysis

For the new plan: Each payment is now 0.3X. The present value of three months of payments is P = 0.3X(P/A, 2%, 3) = 0.3X(2.8839) = 0.865X The value of three months of the old plan is greater than that of the new plan. As you prefer to minimize the value of the payments, you prefer the new plan. (b) ie = (1 + is)m − 1 = (1 + 0.06)4 − 1 = 26.24% for the old plan ie = (1 + is)m − 1 = (1 + 0.02)12 − 1 = 26.82% for the new plan The new plan has a higher effective yearly interest rate.

3.44

The cash flow series consists of G at the end of period 2, 2G at the end of period 3 and so on up to (N − 1)G at the end of the Nth period.

The first step is to convert each period’s gradient amount to its future value: F = G(1 + i)N−2 + 2G(1 + i)N−3 + ... + (N − 2)G(1 + i) + (N − 1)G Multiplying by (1 + i) and subtracting F gives: F(1 + i) = G(1 + i)N−1 + 2G(1 + i)N−2 + ... + (N − 2)G(1 + i)2 + (N − 1)G(1 + i) F(1 + i) − F = G(1 + i)N−1 + G(1 + i)N−2 + ... + G(1 + i) − (N − 1)G Fi = G[(1 + i)N−1 + (1 + i)N−2 + (1 + i)N−3 + ... + (1 + i) + 1] − NG Noting that the amount in the square brackets is the Series Compound Amount Factor: Fi = G(F/A, i, N) − NG Multiplying both sides by the Sinking Fund Factor: Fi(A/F, i , N) = G − NG(A/F, i, N) Ai = G − NG(A/F, i, N) = G[1 − N(A/F, i, N)] 1  1  Ni N A = G −  = G −  N N i i i[(1 + i) − 1] (1 + i) − 1    

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Chapter 3 - Cash Flow Analysis

3.45

The present worth of a geometric series is: A (1 + g ) A (1 + g) A P= + + ... + 2 N 1+ i (1 + i ) (1 + i )

N−1

If we divide and multiply each term in the present worth expression by (1 + g) and then simplify, we get 2 N 1 + g)  ( A  1 + g (1 + g )   P= + + ... + N  1 + g  1 + i (1 + i ) 2 1 + i ( )  

Then, substitute the growth adjusted interest rate, i: i =

1+ i 1 1+ g − 1 so that = 1+ g 1+ i 1+ i

into the present worth expression and we get: P=

A  1 1 1  + + ... +   2 1 + g  1 + i (1 + i) (1 + i)N 

The right hand side is simply the present worth of an annuity where the constant cash flow each period is A/(1 + g) and the interest rate is i. We can write this as 1 1   1  1 + i + (1 + i)2 + ... + (1 + i)N  (P/A, i, N) =A P = A 1+ g 1+ g      

so that:

(P/A, g, i, N) =

(P/A, i, N)  (1 + i)N − 1 1 = N  1+ g  i(1 + i)  1 + g

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Chapter 3 - Cash Flow Analysis

3.46

i = (1 + 0.05)/(1 + 0.5) – 1 = – 0.3 = –30% PWsales

= 1 456 988(P/A, –30%, 5)/(1.5) = 1 456 988[(0.75 – 1)/(–0.3  0.75)]/(1.5) = 16 026 550

Selling price = 16 026 550/2 = 8 013 275 Ruby should sell the company for just over 8 million dollars.

3.47

The worth of the loan at month 52 is: F(loan) = 160 000(F/P, 1%, 52) = 268 430.22 The future worth of the payments at month 52 is: F(payments) = 4000(F/A, 1%, 51)(F/P, 1%, 1) = 267 075.56 The difference is the amount of the last payment: Last Payment = F(loan) − F(payments) = 1 354.66 Clarence’s last payment will be about $1 354.66.

3.48

Only the spreadsheet for part (a) is provided. (a) Capital Amount Annual Interest Rate No. of years to repay

$50 000 8.00% 15

Payment Annual Interest Recovered Unrecovered Periods Payment Received Capital Capital 0 50000 1 5841 4000 1841 48159 2 5841 3853 1989 46170 3 5841 3694 2148 44022 4 5841 3522 2320 41702 5 5841 3336 2505 39197 6 5841 3136 2706 36491 7 5841 2919 2922 33569 8 5841 2686 3156 30413 9 5841 2433 3408 27004 10 5841 2160 3681 23323 11 5841 1866 3976 19348 12 5841 1548 4294 15054 13 5841 1204 4637 10417 14 5841 833 5008 5409 15 5841 433 5409 0 Total 50000

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Chapter 3 - Cash Flow Analysis

3.49

Once the project is started, the savings effectively reduce the costs by $50 000 to $100 000 per month.

Let X = start of project.

250 000(F/P, 1.5%, X) + 50 000(F/A, 1.5%, X) = 150 000(P/A, 1.5%, 24) 5(F/P, 1.5%, X) + (F/A, 1.5%, X) = 3(20.030) = 60.09 At X = 36: LHS = 5(1.7091) + 47.276 = 60.09 At X = 40: LHS = 5(1.814) + 54.2679 = 60.09 The project can start about 38 months from now.

3.50

(a) Cash flow diagram 0

1

2

3

4

5

6

7

8

$40000

$14000 $26000

(b) There will be a total of 9 payments. The payments can be considered in two portions: 1)

The salary costs and the constant portion of the other costs are a 9-period annuity due. The equipment and facility cost declines from $26 000 to a constant $14 000.

At the end of eight years, the payments are worth: FA = A (F/A, 7%, 9) = (40 000 + 14 000)(11.978) = 646 812 This includes the first payment, and is equivalent to treating the initial payment separate from a standard 8 period annuity: FA= 54 000(F/P, 7%, 8) + 54 000(F/A, 7%, 8)

2)

The declining costs can be modelled as an annuity due less an arithmetic gradient, after four years: FB1 = (26 000 − 14 000)(F/A, 7%, 5) = 12 000(5.7507) = 69 008 The gradient’s worth after the same four years is 41 Copyright © 2022 Pearson Canada Inc.

Chapter 3 - Cash Flow Analysis

FB2 = 3000(A/G, 7%, 5)(F/A, 7% 5) = 3000(1.8650)(5.7507) = 32 175 For both FB1 and FB2, N = 5 to conform with the standard form for an annuity and a gradient (the first cash flow for an annuity is after one year, while the first cash flow for a gradient is after two years).

The value of both cash flows after the eight years is the annuity’s value minus the gradient’s value, taken forward to the eighth year: FB = (69 008 − 32 175)(F/P, 7%, 4) = (36 833)(1.3108) = 48 281 The total worth of the project at the end of the eight years is: FA + FB = 646 812 + 48 281 = $695 093 3.51

g = 0.05 per month i = 0.01 per month i = (1 + i)/(1 + g) − 1 = 1.01/1.05 − 1 = −0.0380 P(expenses) = 15 000(P/A, g, i, 12) = 15 000[(P/A, i, 12)/(1 + g)] = 15 000 [(1.03812 − 1)/(0.038  1.03812)]/1.05 = 135 642 Let A be the amount of the monthly installment: P(grant) = A(P/A, 1%, 6)(P/F, 1%, 12) = A(5.7955)(0.88745) = 5.1432A By letting P(expenses) = P(grant), solve for A: 5.1432A = 135 642 A = $26 373 The amount of the monthly installment must be 26 373 元.

3.52

The present worth computations for the concrete pool are: P = 1 500 000 + 200 000(A/F, 5%, 10)/0.05 = 1 500 000 + 200 000(0.07951)/0.05 = $1 818 040

3.53

a) i 0  0 ; P = lim A(P / A, i 0 , N ) = A0 N →

i

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Chapter 3 - Cash Flow Analysis

b) P =

A A(1 + g ) A(1 + g )n −1 + + ... + + ...

1+ i (1 + i )2 (1 + i )n

noting that

A(1 + g )n −1 A(1 + g )n −2 (1 + g ) = (1 + i )n (1 + i )n −1 (1 + i )

and that since g  1 ,

then

(1 + g ) 1 (1 + i )

A(1 + g )n −1 A(1 + g )n −2  (1 + i )n (1 + i )n −1

Consequently P is the sum of an infinite number of monotonically increasing amounts, and takes on an infinite value.  A  =  1+ g 

c) P = lim N  N →

d) i 0  0 ; P = lim A(P / A, i 0 , N ) = A0 N →

i

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Chapter 3 - Cash Flow Analysis

Notes for Case in Point 3.1 1)

No right answer – both are to blame.

2)

On the one hand, large infrastructure projects need to be financed. On the other hand, there is a moral hazard for politicians.

3)

It is probably not reasonable for political leaders to understand the time value of money. But it is reasonable for them to have competent advisors who do.

4)

About 340 months or just over 28 years.

Notes for Mini-Case 3.1 Oil industry projects tend to be in the range of hundreds of millions to tens of billions of dollars of investment. As one might expect, very capable people think through all aspects of a project—technical, financial, environmental, political— very carefully. Oil companies take planning seriously, and their engineers are very experienced. The $90 billion of cancelled projects in 2009 does not represent incompetence, but rather something fundamental, which is that no matter how careful the planning process might be, one can never fully predict how the future will turn out. There are three key issues that have strongly affected the Canadian oil sands.

The first is that crude oil is a commodity, meaning that there are so many sources that no one can easily predict or control its price. Even if the cost of production is absolutely certain, the profitability and hence project viability depends very much on the world price, which itself is subject to many random forces. No one predicted that the price of oil would rise to a peak of US$165 per barrel in 2008 or drop in 2016 to about US$36. Both are extremes outside of the planning window imagined earlier. For conventional oil production, costs can be controlled to a degree. Exploration efforts can be reduced, for example, when prices are low. But oil sands project costs are much more difficult to cut back on once they are started, with very large capital costs to recover, and with high direct costs per barrel of oil produced. Investment decisions are thus more vulnerable to low oil prices.

The second confounding issue with the oil sands concerns social, environmental and political forces at play. Mining oil sands is very dirty work, damaging the physical sites and producing vast quantities of carbon dioxide. Early oil sands projects sometimes completely failed to recognize the costs of managing their environmental impact, and ended up with unexpected costs for administration, 44 Copyright © 2022 Pearson Canada Inc.

Chapter 3 - Cash Flow Analysis

legal bills, remediation and meeting regulations. Even in recent years where such costs are recognized and taken into account in the planning process, they are very unpredictable because of the inherent nature of social and political intervention. Otherwise benign activities can suddenly become difficult and expensive if they somehow gain media attention. This has been a critical issue in the transportation of bitumen to markets in recent years. Rail transport is costly and poses environmental risk. Pipelines traverse so much territory that they can be easily blocked by relatively small groups such as environmentalist and First Nations groups. The Northern Gateway, Keystone XL, and Energy East pipelines were all cancelled in the last few years because of activism and political controversy.

A third problem is a confluence of global trends, which has been compounded by the oil sand’s remote location. For example, in the last two decades, a key global force has been the industrialization of China. China’s industrialization has led to an increase in demand for oil and gas worldwide, causing oil companies to attempt to increase supply. This in turn has caused a global proliferation of oil and gas developments, creating a tremendous demand for manufacturing capacity, technical expertise and specialized equipment. This demand significantly increased the costs of acquiring the resources necessary to develop the oil sands. In particular, with a tight labour pool and an isolated location, costs of labour in Alberta skyrocketed – to about double previously estimated costs, far beyond even the most pessimistic expectations. Although some trends can be predicted, others defy prediction.

Even with the best planning and with the most reliable process and product, the future remains unpredictable. Economic analyses still need to be done, however, despite an uncertain future. There are sophisticated ways to deal with uncertainty about future cash flows, some of which are discussed in Chapter 12. In most cases it makes sense to carry out the economic analysis with a range of possible values for future cash flows. But, as the oil sands example shows, there are situations when even this approach does not account for what the future actually holds.

45 Copyright © 2022 Pearson Canada Inc.

CHAPTER 4 Solutions to Chapter-End Problems A. Key Concepts Relations among Projects: 4.1

(a) There are eight mutually exclusive alternatives: {T, TL, TM, TLM, L, LM, M, no new quarry} (b) There are now twelve alternatives. Let C stand for the cutter-loader. {T, TL, TLC, TM, TLM, TLMC, L, LC, LM, LMC, M, no new quarry} (c) There are now eight feasible combinations. Total first costs are in parentheses. {T ($0.9 million), TL ($2.3 million), TM ($1.9 million), L ($1.4 million), LC ($1.8 million), LM ($2.4 million), M ($1.0 million), do nothing ($0)}

4.2

If the city does not widen either King or Main, there are two projects, stoplights and no stoplights.

If the city widens King, the following five projects are available: Left Lane on King Stoplights Advanced Green on King

no no no

no yes no

yes no no

yes yes no

yes yes yes

Similarly, if the city widens Main, there are five projects: Left Lane on Main Stoplights Advanced Green on Main

no no no

no yes no

yes no no

The grand total is 12 projects.

46 Copyright © 2022 Pearson Canada Inc.

yes yes no

yes yes yes

Chapter 4 - Comparison Methods Part 1

4.3 Candidates Do nothing A B C D AB AC AD BC BD CD ABC ABD ACD BCD ABCD

Financing ok ok ok ok ok ok ok ok ok ok ok ok ok 0k

Buy Barco ok ok ok ok ok ok ok ok

Expand ok ok ok ok

ok ok ok ok ok ok

ok ok ok ok ok ok ok ok

X division ok ok ok ok ok ok ok ok ok ok ok ok

Feasible? yes yes no no no yes no no yes no no yes no no yes no

There are six feasible projects: doing nothing, A, AB, BC, ABC, BCD Present Worth Comparisons: 4.4

PW = −28 000 + 5000(P/A, 15%, 10) + 3000(P/F, 15%, 10) = −28 000 + 5000(5.0188) + 3000(0.24718) = −2164.46 The present worth of the project is about −$2164

4.5

P = −200 000 + 2000(P/A, 0.5%, 60) + 210 000(P/F, 0.5%, 60) = −200 000 + 2000(51.726) + 210 000(0.74137) = 59 140 Since the present worth of this set of cash flows is positive, Nabil should buy the house.

4.6

PW(moderate) = −6500 + [3300 − 300(A/G, 8%, 7)](P/A, 8%, 7) = −6500 + [3300 − 300(2.6937)](5.2064) = 6473.78 For extensive upgrading: g = −0.2 i = (1 + i)/(1 + g) − 1 = 1.08/0.8 − 1 = 0.35

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Chapter 4 - Comparison Methods Part 1

PW(extensive) = −10 550 + 7600(P/A, i, 7)/(1 + g) = −10 550 + 7600[(1.357 − 1)/(0.35  1.357]/0.8 = 13 271.47 Extensive upgrading is better.

4.7

At a MARR of 9%, the present worth of the contract from the first company is: P1 = 10 000(P/A, 9%, 5) + 20 000(P/A, 9%, 10)(P/F, 9%, 5) = 10 000(3.8896) + 20 000(6.4176)(0.64993) = 122 315.82 The present worth of the contract from the second company is: P2 = 10 000(P/A, 9%, 10) + 3000(A/G, 9%, 10)(P/A, 9%, 10) = (P/A, 9%, 10)[10 000 + 3000(A/G, 9%, 10)] = 6.4177[10 000 + 3000(3.7978)] = 137 296.42 The software genius should choose the contract from the second company.

4.8

Using the capitalized cost formula: PW(concrete reservoir + pipe) = 500 000 + A/i = 500 000 + 2000/0.08 = 525 000 PW(earthen dam + aqueduct) = 200 000 + 12 000/0.08 + 100 000(A/F, 8%, 15)/0.08 = 350 000 + 100 000(0.03683)/0.08 = 396 037.5 The alternative with the least present cost is the earthen dam and aqueduct alternative. It has a present cost of $396 038 and should be chosen.

Annual Worth Comparisons: 4.9

AW = −28 000(A/P, 15%, 10) + 5000 + 3000(A/F, 15%, 10) = −28 000(0.19925) + 5000 + 3000(0.04925) = −431.25 The annual worth of the project is about −$431.

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Chapter 4 - Comparison Methods Part 1

4.10

A = −200 000(A/P, 0.5%, 60) + 2000 + 210 000(A/F, 0.5%, 60) = −200 000(0.01933) + 2000 + 210 000(0.01433) = 1143 Since the annual worth of this set of cash flows is positive, Nabil should buy the house.

4.11

For the hydraulic press: g = 0.15 i = (1 + i)/(1 + g) − 1 = 1.12/1.15 − 1 = −0.026 AW(hydraulic) = −275 000(A/P, 12%, 15) + 33 000 + 19 250(A/F, 12%, 15) − 2000[(P/A, i, 15)/(1 + g)](A/P, 12%, 15) = −275 000(0.14682) + 33 000 + 19 250(0.02682) − 2000[(0.97415 − 1)/(−0.0260.97415)](0.14682)/1.15 = −275 000(0.14682) + 33 000 + 19 250(0.02682) − 2000(18.65228) = 24 716.08 AW(moulding) = −185 000(A/P, 12%, 10) + 24 500 − [1000 + 350(A/G, 12%, 10)] + 14 800(A/F,12%,10) = −185 000(0.17698) + 24 500 − [1000 + 350(3.5847)] + 14 800(0.05698) = −9652.64 The hydraulic press is the preferred alternative.

4.12

The annual cost of operating an auto is: A = 24 000(A/P, 11%, 5) + (2000 + 600 + 600 + 1000) + 400(A/G, 11%, 5) − 8000(A/F, 11%, 5) = 24 000(0.27057) + 4200 + 400(1.7922) − 8000(0.16057) = 10 126.00 If Tom was to use taxis instead, the annual cost would be A = 6600.

Tom should not buy the vehicle. He will end up saving about $3536 per year by hiring taxis.

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Chapter 4 - Comparison Methods Part 1

4.13

The annual cost of the chemical recovery system is: A = (300 000 − 75 000)(A/P, 9%, 7) + 75 000(0.09) = 225 000(0.19869) + 75 000(0.09) = 51 456 The net annual benefit is then: 52 800 − 51 456 = 1344 The net annual benefit of purchasing the chemical recovery system is about $1344 per year.

4.14

AA = 5600 − (14 000 − 2000)(A/P, 9%, 7) − 2000(0.09) = 5600 − 12 000(0.19869) − 2000(0.09) = 3035.72 AB = 5600 − (25 000 − 10 000)(A/P, 9%, 10) − 5000(0.09) = 5600 − 15 000(0.15582) − 10 000(0.09) = 2362.7 Machine A is best with an annual benefit of about $3036.

Alternatives with Unequal Lives: 4.15

(a) Twenty years is the least common multiple of the service lives. Find the present worth analysis for a twenty-year period.

Lawn Guy: P = 350 + 350(P/F, 5%, 10) + (60 + 30)(P/A, 5%, 20) = 350 + 350(0.61391) + 90(12.462) = 1 686.47 Clip Job: P = 120 + 120(P/F, 5%, 4) + 120(P/F, 5%, 8) + 120(P/F, 5%, 12) + 120(P/F, 5%, 16) + (40 + 60)(P/A, 5%, 20) = 120(1 + 0.82270 + 0.67684 + 0.55684 + 0.45811) + 100(12.462) = 1 667.96 There is very little difference between the present worth of the Lawn Guy and the present worth of the Clip Job. However, since the present worth (cost) of the “Clip Job” machine is least, it is economically preferable.

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Chapter 4 - Comparison Methods Part 1

(b) Use a four-year study period, and let X be the salvage value of the Lawn Guy after four years.

Lawn Guy: P = 350 + (30 + 60)(P/A, 5%, 4) − X(P/F, 5%, 4) = 350 + 90(3.5458) − X(0.82270) = 669.14 − 0.8227X 4.16

(a) With the least common multiple of the service lives method, the present costs are: Used: P = 475(A/P, 8%, 3)(P/A, 8%, 24) = 475(0.38803)(10.529) = 1940.65 New: P = 1250 + 1250(P/F, 8%, 8) + 1250(P/F, 8%, 16) = 1250(1 + 0.54027 + 0.29189) = 2290.2 The used refrigerator is the better buy. (b) With a service period of three years the present worth computations are: Used: P = 475 New: P = 1250 − 1000(P/F, 8%, 3) = 1250 − 1000(0.79383) = 456.17 In this case, the new refrigerator is the better buy.

Payback Period: 4.17

Payback period = 28 000/5000 = 5.6 years

4.18

Payback period = 10 000/(6567 − 2000) = 2.19 years

4.19

Payback period = 65 000 000/(12  5000  365  0.8) = 3.71 years

4.20

Payback period = 20 000/[(3000 + 1000 + 2500) − (700 + 200)] = 20 000/5600 = 3.6 years If Greene Cheese had a maximum payback period of less than 3.6 years, the project would be rejected.

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Chapter 4 - Comparison Methods Part 1

B. Applications 4.21 Possible combinations Do nothing A B C D AB AC AD BC BD CD ABC ABD ACD BCD ABCD

Budget <$200 000 <$200 000 <$200 000 <$200 000 <$200 000 <$200 000 ok <$200 000 ok <$200 000 ok >$300 000 ok >$300 000 ok >$300 000

Lead Time ok ok ok ok ok ok same ok ok same ok ok ok ok ok ok

Resources ok ok ok ok ok >100% ok ok ok ok ok >100% >100% >100% >100% >100%

Feasible? no no no no no no no no yes no yes no no no no no

There are only two mutually exclusive projects that IQ Computers should consider: (1) making products B and C, or (2) making products C and D. 4.22 Candidates 1 2 3 4 12 13 14 23 24 34 123 124 134 234 1234

Testers ok ok ok ok ok ok ok >1 ok ok >1 ok ok >1 >1

Press ok ok ok overhaul ok ok ok ok overhaul overhaul ok ok ok overhaul ok

Total cost ok ok ok ok ok ok ok ok ok >$100 000 >$100 000 ok >$100 000 >$100 000 >$100 000

Feasible? yes yes yes no yes yes yes no yes no no yes no no no

There are eight mutually exclusive projects available: {1, 2, 3, 12, 13, 14, 24, 124} 4.23

Since at least two printing lines must be available at all times, they can either upgrade one line or two lines.

If one line is to be upgraded, there are seven possible combinations of three printing stations: 52 Copyright © 2022 Pearson Canada Inc.

Chapter 4 - Comparison Methods Part 1

(1) A (2) B (3) C (4) AB (5) AC (6) BC (7) ABC

Total Cost $7000 $5000 $3000 $12000 $10000 $8000 $15000

Total Time 10 days 5 days 3 days 15 days 13 days 8 days 18 days

Feasible? yes yes yes no yes yes no

Combinations (4) and (7) are not feasible.

If two lines are to be upgraded, the following combinations must be considered: (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18) (19) (20) (21) (22)

Line 1 A A A A A B B B B C C C AC AC BC

Line 2 A B C AC BC B C AC BC C AC BC AC BC BC

Total Cost $14000 $12000 $10000 $17000 $15000 $10000 $8000 $15000 $13000 $6000 $13000 $11000 $20000 $18000 $16000

Total Time 20 days 15 days 13 days 23 days 18 days 10 days 8 days 18 days 13 days 6 days 16 days 11 days 26 days 21 days 16 days

Feasible? no no yes no no yes yes no yes yes no yes no no no

(10), (13), (14), (16), (17), and (19) are feasible.

Nottawasaga Printing can choose to upgrade one or two printing lines. If they decide to upgrade one line, there are five feasible mutually exclusive alternatives (see combinations 1, 2, 3, 5, and 6), and if they decide to upgrade two lines, there are six feasible mutually exclusive alternatives (see combinations 10, 13, 14, 16, 17, and 19).

4.24 (a) PW(Smoothie) = −15 000 + (4200 − 1200)(P/A, i, 12) + 2250(P/F, i, 12) = −15 000 + 3000(P/A, i, 12) + 2250(P/F, i, 12) PW(Creamy) = −36 000 + (10 800 − 3520)(P/A, i, 12) + 5000(P/F, i, 12) = −36 000 + 7280(P/A, i, 12) + 5000(P/F, i, 12)

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Chapter 4 - Comparison Methods Part 1

By letting PW(Smoothie) = PW(Creamy): 21 000 = 4280(P/A, i, 12) + 2750(P/F, i, 12) At i = 0.15: (P/A, i, 12) = 5.4206, (P/F, i, 12) = 0.18691, and LHS = 23 714 At i = 0.20: (P/A, i, 12) = 4.4392, (P/F, i, 12) = 0.11216, and LHS = 19 308 Using linear interpolation: i = 0.15 + (0.20 − 0.15)[(21 000 − 23 714)/(19 308 − 23 714)] = 0.180798  18.1% A MARR of 18.1% makes the two alternatives equivalent in terms of PW. (b) Since the service life differs now for the two alternatives, we cannot use a simple present worth comparison. Either repeated lives or an annual worth comparison would be appropriate. We show the annual worth method here with an assumption that the two alternatives can be repeated.

AW(Smoothie) = −15 000(A/P, 18.1%, 14) + 4200 − 1200 + 2250(A/F, 18.1%, 14) = −15 000[0.181(1.18114)/(1.18114 − 1)] + 3000 − 2250[0.181/(1.18114 − 1)] = −15 000(0.2005289) + 3000 − 2250(0.0195289) = 36.006525  $36.01 AW(Creamy) = −36 000(A/P, 18.1%, 12) + 10 800 − 3520 + 5000(A/F, 18.1%, 12) = −36 000(0.2094499) + 7280 − 5000(0.0284499) = −117.9469  −$117.95 Smoothie is better than Creamy if the service life of Smoothie is 14 years.

4.25

FW = −28 000(F/P, 15%, 10) + 5000(F/A, 15%, 10) + 3000 = −28 000(4.0456) + 5000(20.304) + 3000 = −8756.8

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Chapter 4 - Comparison Methods Part 1

The future worth of the project in 10 years is about −$8757.

Year Present worth

Cumulative

1

5000(P/F, 15%, 1) = 5000(0.86957) = 4348

4348

2

5000(P/F, 15%, 2) = 5000(0.75614) = 3781

8129

3

5000(P/F, 15%, 3) = 5000(0.65752) = 3288

11417

4

5000(P/F, 15%, 4) = 5000(0.57175) = 2859

14276

5

5000(P/F, 15%, 5) = 5000(0.49718) = 2486

16762

6

5000(P/F, 15%, 6) = 5000(0.43233) = 2162

18924

7

5000(P/F, 15%, 7) = 5000(0.37594) = 1880

20804

8

5000(P/F, 15%, 8) = 5000(0.32690) = 1635

22439

9

5000(P/F, 15%, 9) = 5000(0.28426) = 1421

23860

10

5000(P/F, 15%, 10) = 5000(0.24718) = 1236

25096

The discounted payback period is greater than 10 years, the life of the project.

4.26

(a) At a production level of 30 000 units per year, the annual worth (cost) of a wax melter from each supplier is computed as follows: Finedetail: A = 200 000(A/P, 15%, 7) + 16 500 + 2.25(30 000) − 5000(A/F, 15%, 7) = 200 000(0.24036) + 16 500 + 67 500 − 5000(0.09036) = 131 620.30 Simplicity: A = 350 000(A/P,15%,10) + 35 500 + 1.06(30 000) − 20 000(A/F,15%,10) = 350 000(0.19925) + 35 500 + 1.06(30 000) − 20 000(0.04925) = 136 053.20 Based on annual worth, the wax melter from Finedetail is preferred because it has the lowest annual cost.

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Chapter 4 - Comparison Methods Part 1

(b) At a production level of 200 000 units per year, the annual worth (cost) for the two wax melters is computed as follows: Finedetail: A = 200 000(A/P, 15%, 7) + 16 500 + 2.25(200 000) − 5000(A/F, 15%, 7) = 200 000(0.24036) + 16 500 + 67 500 − 5000(0.09036) = 514 120.27 Simplicity: A = 350 000(A/P,15%,10)+ 35 500 + 1.06(200 000) − 20 000(A/F,15%,10) = 350 000(0.19925) + 35 500 + 1.06(200 000) − 20 000(0.04925) = 316 253.18 Based on annual worth, the Simplicity wax melter is preferred because it has the lowest annual cost. (c) The annual costs associated with each wax melter for production levels between 20 000 and 200 000 units per year are: Production level 20000 30000 40000 50000 100000 150000 200000

Annual cost Finedetail Simplicity 109120 125453 131620 136053 154120 146653 176620 157253 289120 210253 401620 263253 514120 316253

The choice of supplier is quite sensitive to the number produced, with the wax melter from Finedetail being preferred for production of about 35 000 units per year or less, and the wax melter from Simplicity otherwise.

The Finedetail wax melter is roughly comparable to the Simplicity wax melter at a production level of 30 000 units per year. Moreover, if production does reach 200 000 units per year, Simplicity’s wax melter has much lower annual costs than the Finedetail wax melter. Since the 30 000 is a lower bound on production levels, and it is quite possible that production will be higher, the Simplicity wax melter is the recommended alternative.

4.27

Use the capitalized cost formula.

Concrete pool: P = 1 500 000 + 200 000(A/F, 5%, 10)/0.05 = 1 500 000 + 200 000(0.07951)/0.05 = 1 818 040 56 Copyright © 2022 Pearson Canada Inc.

Chapter 4 - Comparison Methods Part 1

Plastic pool: P = 500 000 + [100 000(A/F, 5%, 5) + 150 000(A/F, 5%, 15) + 5000]/0.05 = 500 000 + [100 000(0.18098) + 150 000(0.04634) + 5000]/0.05 = 1 100 980 The design with the lowest present cost is the pool with the plastic liner.

4.28

Using a six-year study period: PXJ3

= 4500[1 + (P/F, 10%, 3)] − 1000[(P/F, 10%, 3) + (P/F, 10%, 6)] = 4500(1 + 0.75131) − 1000(0.75131 + 0.56447) = 6565.12

PY19

= 3200[1 + (P/F, 10%, 2) + (P/F, 10%, 4)] + 300(P/A, 10%, 6) − 1000[(P/F, 10%, 2) + (P/F, 10%, 4) + (P/F, 10%, 6)] = 3200(1 + 0.82645 + 0.68301) + 300(4.3553) − 1000(0.82645 + 0.68301 + 0.56447) = 7262.93

Based on a six-year study period, Val should buy the model XJ3 display panel, which has the smallest present cost.

4.29

PXJ3

= 4500 − 1900(P/F, 10%, 2) = 4500 − 1900(0.82645) = 2929.75

PY19

= 3200 + 300(P/A, 10%, 2) − 1000[(P/F, 10%, 2) = 3200 + 300(1.7355) − 1000(0.82645) = 2894.2

Based on a two-year study period, Val should buy the model Y19 display panel, which has the smallest present cost.

4.30

Let P = first cost, A = yearly savings.

Then: Payback period = 3 = P/A  P = 3A Also: Present worth = 0 = −P + A(P/A, i%, 5) By substituting P = 3A into the present worth expression: 0 = −3A + A(P/A, i%, 5) (P/A, i%, 5) = 3 By observing the tables, Diana’s interest rate was approximately 20%.

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4.31

(a) To equate costs set: (0.08 × 100 × 500)/1000 + (2 + 1)(A/P, 1%, 1000/500) = (0.08 × 90 × 500)/1000 + (2 + 3)(A/P, 1%, N) 4 + 3(0.50757) = 3.6 + 5(A/P, 1%, N) (A/P, 1%, N) = 1.92271/5 = 0.38454 (A/P, 1%, 2) = 0.50751 (A/P, 1%, 3) = 0.34002 By linear interpolation: N = 2 + (0.50751 − 0.38454)/(0.50751 − 0.34002) = 2.734 The minimum number of required hours is approximately 2.734 × 500 = 1367 hours. (b) If costs are equated when there are no costs for changing bulbs, the results are: (0.08 × 100 × 500)/1000 + 1(A/P, 1%, 2) = (0.08 × 90 × 500)/1000 + 3(A/P, 1%, N) 4.0 + 0.50757 = 3.6 + 3(A/P, 1%, N) (A/P, 1%, N) = 0.90757/3 = 0.30252 (A/P, 1%, 3) = 0.34002 (A/P, 1%, 4) = 0.25628 N = 3 + (0.34002 − 0.30252)/(0.34006 − 0.25628) = 3.448 The minimum number of required hours is approximately 3.448 × 500 = 1640 hours. (c) Since the capital recovery factor is convex in the number of years, the linear interpolation is above the correct curve. Since the capital recovery factor is decreasing in the number of years, the linear interpolation is to the right of the correct curve. This means that the linear interpolation gives an estimate for the required number of years that is too high.

4.32

(a) PT = −100 000 + 50 000(P/A, 11%, 5) + 20 000(P/F, 11%, 5) = −100 000 + 50 000(3.6959) + 20 000(0.59345) = 96 664

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Chapter 4 - Comparison Methods Part 1

PA = −150 000 + 60 000(P/A, 11%, 5) + 30 000(P/F, 11%, 5) = −150 000 + 60 000(3.6959) + 30 000(0.59345) = 89 558 Model T has the greater PW and therefore should be taken. (b) By assuming that you can purchase each alternative as many times as necessary, we can construct new projects: T′: buy model T three times, total life 15 years A′: buy model A three times, total life 15 years X′: buy model X five times, total life 15 years We only need to compare models T and X since A is already eliminated: PT′ = −100 000[1 + (P/F, 11%, 5) + (P/F, 11%, 10)]+ 50 000(P/A, 11%, 15) + 20 000[(P/F, 11%, 5) + (P/F, 11%, 10) + (P/F, 11%, 15)] = −100 000(1 + 0.59345 + 0.35218) + 50 000(7.1909) +20 000(0.59345 + 0.35218 + 0.20900) = 188 075 PX′ = −200 000[1 + (P/F, 11%, 3) + (P/F, 11%, 6) + (P/F, 11%, 9) + (P/F, 11%, 12)] + 75 000(P/A, 11% 15) + 100 000[(P/F, 11%, 3) + (P/F, 11%, 6) + (P/F, 11%, 9) + (P/F, 11%, 12) + (P/F, 11%, 15)] = −200 000(1 + 0.73119 + 0.53464 + 0.39092 + 0.28584) + 75 000(7.1909) + 100 000(0.73119 + 0.53464 + 0.39092 + 0.28584 + 0.20900) = 165 959 Model T is still the best because it has the highest present worth.

4.33

PT = −100 000 + 50 000(P/A, 11%, 3) + 40 000(P/F, 11%, 3) = −100 000 + 50 000(2.4437) + 40 000(0.73119) = 51 433 PA = −150 000 + 60 000(P/A, 11%, 3) + 80 000(P/F, 11%, 3) = −150 000 + 60 000(2.4437) + 80 000(0.73119) = 55 117 PX = −200 000 + 75 000(P/A, 11%, 3) + 100 000(P/F, 11%, 3) = −200 000 + 75 000(2.4437) + 100 000(0.73119) = 56 397 Model X is the best, A is next, and T is worst! This is because using the study period method, the choice of salvage value is critical.

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4.34

AT = −100 000(A/P, 11%, 5) + 50 000 + 20 000(A/F, 11%, 5) = −100 000(0.27057) + 50 000 + 20 000(0.16057) = 26 154 AA = −150 000(A/P, 11%, 5) + 60 000 + 30 000(A/F, 11%, 5) = −150 000(0.27057) + 60 000 + 30 000(0.16057) = 24 232 AX = −200 000(A/P, 11%, 3) + 75 000 + 100 000(A/F, 11%, 3) = −200 000(0.40921) + 75 000 + 100 000(0.29921) = 23 079 Model T is best.

4.35

Model T: Payback period = 100 000/50 000 = 2 years Model A: Payback period = 150 000/60 000 = 2.5 years Model X: Payback period = 200 000/75 000 = 2.667 years Note that all of the following are ignored: life, salvage value, and interest rate. This will, in general, cause the ranking of alternatives to vary from the correct rankings.

4.36

(a) Machine A: A = −1 500 000(A/P, 8%, 5) + (900 000 − 600 000) + 100 000(A/F, 8%, 5) = −1 500 000(0.25046) + 300 000 + 100 000(0.17046) = −58 644 Machine B: A = −2 000 000(A/P, 8%, 10) + (1 100 000 − 800 000) + 200 000(A/F, 8%, 10) = −2 000 000(0.14903) + 300 000 + 200 000(0.06903) = 15 746 Only machine B should be purchased. (b) Least common multiple of the service lives is 10 years.

Machine A: P = −1 500 000 + (900 000 − 600 000)(P/A, 8%, 10) + (100 000 − 15 000)(P/F, 8%, 5) + 1000(P/F, 8%, 10) = −1 500 000 + 300 000(6.7101) −1 400 000(0.68058) + 100 000(0.46319) = −393 463

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Chapter 4 - Comparison Methods Part 1

Machine B: P = −2 000 000 +(1 100 000 − 800 000)(P/A, 8%, 10) + 200 000(P/F, 8%, 10) = −2 000 000 + 300 000(6.7101) + 200 000(0.46319) = 105 668 Only machine B should be purchased. (c) A: 1 500 000/300 000 = 5 years B: 2 000 000/300 000 = 6.6 years Probably neither should be purchased since the payback periods are long, more than four years.

4.37

Constant amount: F = 100 000(F/A, 10%, 10) = 100 000(15.937) = 1 594 000 Increasing amount: The growth adjusted interest rate is i = 1.1/1.05 − 1 = 4.7619%.

F = [80 000(P/A, 4.7619%, 10)/1.05](F/P, 10%, 10) = 80 000(7.8118)(2.5937)/1.05 = 1 544 000 The plan that saves the constant amount will accumulate the most money in 10 years.

4.38

Payback period for isolation: 60 000/10 000 = 6 years Payback period for curtains: 5 000/3 000 = 1.67 years The payback period for the curtains is much shorter than for the isolation solution.

No threshold for payback was given in the problem, but the payback period for the curtains is less than the standard two-year maximum for acceptable projects.

However, it is not appropriate to use the payback period to compare these alternatives. The service life of isolation is very large, so that savings will be achieved long after the payback period is passed. For the curtains, the service life is almost the same as the payback period.

4.39

Use the capitalized value formula.

For the isolation alternative: A = 10 000 − 60 000(0.11) = 3400 61 Copyright © 2022 Pearson Canada Inc.

Chapter 4 - Comparison Methods Part 1

For the curtains: A = 3000 − 5000(A/P, 11%, 2) = 3000 − 5000(0.58393) = 80.35 The isolation alternative is best with an annual worth of about $3400 compared with the curtains of about $80.

C. More Challenging Problems 4.40

Alternative 1: Using the capitalized value formula: A(first cost) = Pi = 2 000 000(0.15) = 300 000 A(maintenance) = 10 000 A(paint) = 15 000(A/F, 15%, 15) = 15 000(0.02102) = 315.3 A(total) = A(first cost) + A(maintenance) + A(paint) = 310 315.30

Alternative 2: First cost: A(first cost) = 1 250 000(0.15) + 1 000 000(P/F, 15%, 10)(0.15) = 187 500 + 150 000(0.24719) = 224 578.5 Maintenance for the first 10 years: The first 10 years has maintenance costs of $5000 per year; convert to a PW and spread over infinite life using the capitalized value formula.

A(maintenance first 10 years) = 5000(P/A, 15%, 10)(0.15) = 5000(5.0187)(0.15) = 3764.02 Maintenance after 10 years: The maintenance costs change after the renovation to be $11 000 every year. First, convert to a PW at the end of 10 years, then to PW now, then spread over infinite life using the capitalized value formula: P(maintenance after 10 years) = A/i = 11 000/(0.15) = 73 333.33 P(maintenance after 10 years, now) = 73 333.33(P/F, 15%, 10) = 73 333.33(0.2472) = 18 128 A(maintenance after 10 years) = P(maintenance after 10 years, now)(0.15) = 18 128(0.15) = 2719.2

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Chapter 4 - Comparison Methods Part 1

Painting costs: Painting costs are every 15 years, starting in 10 years.

Calculate the P(paint in 10 years), convert to P(paint now), and then to A(paint): P(paint in 10 years) = A/i = 15 000(A/F, 15%, 15)/(0.15) = 15 000(0.02102)/(0.15) = 2101 P(paint now) = 2101(P/F, 15%, 10) = 2101(0.24719) = 519.35 A(paint) = P(paint now)(0.15) = 77.90 The total annual cost for alternative 2 is: A(total) = A(first cost) + A(maintenance first 10 years) + A(maintenance after 10 years) + A(painting) = 224578.5 + 3764.02 + 2719.2 + 77.90 = 231 139.62 The annual cost of alternative 2 is less than that of alternative 1. Hence, select alternative 2.

4.41

Let

P = first cost A = annual savings PW = present worth AW = annual worth

AW = −P(A/P, 15%, 5) + 20 000 = −0.29832P + 20 000 PW = −P + 20000(P/A, 15%, 5) = −P + 20000(3.3522) = −P + 67 044 By setting PW = 3AW: −P + 67 044 = 3(−0.29832P + 20 000) = −0.89496P + 60 000 P(1 − 0.89496) = 67 044 − 60000 P = 7044/0.10504 = 67 060 The project’s first cost was about $67 060.

So that: PW = −67 060 + 67044 = −16 AW = −67 060(0.29832) + 20 000 = −5.3 Since the Present Worth and Annual Worth are each less than zero, Katie should not undertake the project.

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Chapter 4 - Comparison Methods Part 1

4.42

(a) We must use repeated lives.

PW(one CR1000) = −680 + 245(P/A, 10%, 4) − [35 + 10(A/G, 10%, 4)](P/A, 10%, 4) + 100(P/F, 10%, 4) = −680 + 245(3.1699) − [35 + 10(1.3812)](3.1699) + 100(0.68301) = 10.1973 PW(CR1000 for 12 years) = PW(one CR1000)[1 + (P/F, 10%, 4) + (P/F, 10%, 8)] = 10.1973(1 + 0.68301 + 0.46651) = 21.92 PW(one CRX) = −1100 + (440 − 60)(P/A, 10%, 6) + 250(P/F, 10%, 6) = −1100 + 380(4.3553) + 250(0.56447) = 696.1315 PW(CRX for 12 years) = PW(one CRX)[1 + (P/F, 10%, 6)] = 696.1315(1 + 0.56447) = 1089.08 CRX is the preferred choice since it has the highest PW. (b) Let S be the scrap value to be solved for. By letting PW(CR1000 for 12 years) = PW(CRX for 12 years): 1089.08 = (−58.10366 + 0.68301S)[1 + (P/F, 10%, 4) + (P/F, 10%, 8)] 1089.08 = (−58.10366 + 0.68301S)(2.14952) S = 826.88 However, the first cost of CR1000 is only $680. It is unlikely that any increase in the scrap value of CR1000 would make it the preferred choice over CRX.

4.43

The service life of A is longer than that of B.

4.44

Let Fi be the first cost, Ai be the annual savings, and Ni be the economic life, i = A, B. Two projects have the same payback period, so FA/AA = FB/AB (1).

Two projects have the same economic life, so NA = NB (2).

PWA = −FA + AA(P/A, i, NA)  PWA/AA = −FA/AA + (P/A, i, NA) PWB = −FB + AB(P/A, i, NB)  PWB/AB = −FB/AB + (P/A, i, NB)

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Chapter 4 - Comparison Methods Part 1

From (1) and (2), PWA/AA = PWB/AB. Since PWA > PWB (given information), it must be that AA > AB in order to satisfy the relationship, PWA/AA = PWB/AB. 4.45

Contracting out is already expressed as an annual cost, so it is sufficient to calculate the annual cost of the landfill alternative: A = 1 000 000(A/P, 11%, 30) + 100 000(A/F, 11%, 30) + 20 000 − 30 000(F/A, 11%, 20)(A/F, 11%, 30) = 1 000 000(0.11502) + 100 000(0.00502) + 20 000 − 30 000(64.203)(0.00502) = 125 351 Since the annual cost of the landfill site is about $4650 less per year, on economic grounds, it is the best choice. Other issues would likely affect the final decision, especially because the two alternatives are similar in price. For example, local citizens would probably not like to have a new landfill site near their properties.

4.46

The MARR at which the present worths of the two projects are the same is about 12.23%.

Year 0 1 2 3 4 5

4.47

Disbursement 1500000 50000 60000 70000 80000 90000

Automated Line Receipts 0 300000 300000 300000 300000 800000 Present worth MARR

Net Cash Flow −1500000 250000 240000 230000 220000 710000 −386544 12.2295%

Disbursement 1000000 20000 25000 30000 35000 40000

Manual Line Receipts 0 200000 200000 200000 200000 200000

Net Cash Flow −1000000 180000 175000 170000 165000 160000 −386544

As shown below, only if the salvage value for the purchased items is less than about 20% of the purchase price does Stayner Catering lose money.

This seems unlikely over only a year of usage, so Stayner Catering should proceed with the project.

Month January (beg.)

January (end) February March April May June July August September October November December

Purchase −200000

?

Labour −2000 −2000 −2000 −2000 −4000 −10000 −10000 −10000 −4000 −2000 −2000 -2000

Warehouse −3000 −3000 −3000 −3000 −3000 −6000 −6000 −6000 −3000 −3000 −3000 -3000

Revenue 2000 2000 2000 2000 10000 40000 110000 60000 30000 10000 5000 2000

PW −200000 −2970 −2941 −2912 −2883 2854 22609 87675 40633 21030 4526 0 −2662

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Salvage% 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%

Net PW −35040 −17291 458 18207 35956 53705 71454 89203 106952 124701 142450

Project present value ($)

Chapter 4 - Comparison Methods Part 1

160000 140000 120000 100000 80000 60000 40000 20000 0 -20000 0% -40000 -60000

50%

100%

Salvage value as % of first cost

4.48

The present worths of the two options are equivalent at MARR = 6.2878%. The present worth with this MARR is $38 433.

Expand: Remodel: Year Disbursement Receipt Net Indiv.PW Disbursement Receipt Net Indiv.PW 1/(1 + MARR)^N 0 850000 0 −850000 230000 0 −230000 1 25000 200000 175000 164647 9000 80000 71000 66800 0.940841752 2 30000 225000 195000 172611 11700 80000 68300 60458 0.885183203 3 35000 250000 215000 179056 15210 80000 64790 53958 0.832817316 4 40000 275000 235000 184134 19773 80000 60227 47191 0.783549303 5 45000 300000 255000 187985 25704.9 80000 54295.1 40026 0.737195899 PW 38433 PW 38433 MARR 0.062878

4.49

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

The annual worths of the two options are equivalent at MARR = 16.40925%. The annual worth with this MARR is $220 806.

Fully Automated System: Partially Automated System: Disburse. Receipt Net Indiv.PW Disburse. Receipt Net Indiv.PW 1/(1 + MARR)^N 1000000 0 −1000000 650000 0 −650000 30000 300000 270000 231940 30000 220000 190000 163217 0.859038264 30000 300000 270000 199246 30000 220000 190000 140210 0.737946739 80000 300000 220000 139463 35000 220000 185000 117276 0.633924485 30000 300000 270000 147033 35000 220000 185000 100745 0.544565389 30000 300000 270000 126307 40000 220000 180000 84204 0.467802506 80000 300000 220000 88409 40000 220000 180000 72335 0.401860253 30000 300000 270000 93208 45000 220000 175000 60412 0.345213334 30000 300000 270000 80069 45000 220000 175000 51897 0.296551463 80000 300000 220000 56045 50000 220000 170000 43307 0.254749054 30000 300000 270000 59087 50000 220000 170000 37203 0.218839185 PW 220806 PW 220806 MARR 0.1640925

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Chapter 4 - Comparison Methods Part 1

4.50 Let Q = payback period. Then P/A = Q, P = AQ PW = −P + A(P/A, i, n) = A(P/A, i, n) − AQ = A[(P/A, i, n) − Q)] Consequently, when (P/A, i, n) ≥ Q, PW ≥ 0, otherwise PW <0 Fred’s rule is sensible.

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Chapter 4 - Comparison Methods Part 1

Notes for Case in Point 4.1 1)

All of these are important.

2)

No, he should not.

3)

If he sets his MARR too low, he may not be able to distinguish truly valuable investment and overspend. If he sets his MARR too high, he may miss valuable investments and underspend.

Notes for Mini-Case 4.1 1)

It is a reasonable way of distinguishing among investments. For example, a new machine doesn't generally have associated marketing issues, and may have to be purchased even if there is a net cost. New products are very dependent on marketing issues, and clearly have to make a profit.

2)

Examples: • Marketing strategy: see if it is consistent with other products. If net effects are positive, it may be worth investing in.

• Work force: measured in number and type of people affected. If the people required can be appropriately obtained and trained in an economically feasible way, it may be worth investing in.

• Margins: measured in dollars. If the rate of return is greater than the MARR, it may be worth investing in.

• Cash flow: measured as dollar earnings per period. If it is positive within two years for a new product, or within a "reasonable" period for new equipment, it may be worth investing in.

• Quality issues: measured as defect or error rate. If reduced with the new equipment, it may be worth investing in.

• Cost avoidance: payback period. If the payback period is one year or less, it may be worth investing in.

3)

Examples: An investment is made when it should not be: • Product: A product is made that pollutes heavily and incurs unexpected legal costs.

• Equipment: Equipment is purchased without taking into account marketing information, and the task for which it was purchased disappears due to lack of demand for the product produced.

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Chapter 4 - Comparison Methods Part 1

An investment is not made when it should be: • Product: A product that requires more than two years of market development is not made in spite of high possible profits later.

• Equipment: Necessary production equipment that only has negative cash flow would never be purchased.

The sensible way to deal with such errors is to recognize that each investment has its own special characteristics, and the listed considerations should only be guidelines, not strict rules. + 490 = $55 766.

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CHAPTER 5 Solutions to Chapter-End Problems A. Key Concepts Choice of Methods: 5.1

Both the leasing costs and demand are on a monthly basis. The annual worth method would be best.

5.2

This is a relatively inexpensive purchase. If a comparison method was used at all, it would probably be payback period.

5.3

If Joan was relatively unsophisticated, a simple payback period would be adequate. If she wanted a more detailed analysis, the annual worth method would probably be best since heating has a natural yearly cycle.

5.4

Since for a large company there is probably detailed information available about the benefits and costs of new scales, IRR would be the likely choice. An alternative would be present worth.

5.5

Payback period is the likely choice since Mona probably won’t have the inclination to do a more detailed analysis. If she did want a more detailed analysis, annual worth is a likely choice.

5.6

Payback period is ideal when cash flow is tight and there is limited time to make a decision.

5.7

Lemuel would likely use IRR or present worth, as mandated by the large company he works for.

5.8

An annual worth comparing the monthly leasing costs to expected net savings would probably be the choice.

5.9

Payback period would probably be appropriate since there is a need to have capital recovered quickly.

IRR for Independent Projects: 5.10

(a) 1000 = 200(P/A, i, 7) (P/A, i, 7) = 5 (P/A, 10%, 7) = 4.8683 (P/A, 9%, 7) = 5.0329 70 Copyright © 2022 Pearson Canada Inc.

Chapter 5 - Comparison Methods Part 2

Linear interpolation: i = 9 + (5 – 5.0329)/(4.8683 – 5.0329) = 9.1998 The IRR is about 9.2%. (b) 1000 = 200(P/A, i, 6) (P/A, i, 6) = 5 (P/A, 6%, 6) = 4.9172 (P/A, 5%, 6) = 5.0756 Linear interpolation: i = 5 + (5 – 5.0756)/(4.9172 – 5.0756) = 5.48 The IRR is about 5.48%. (c) 1000 = 200(P/A, i, 100) (P/A, i, 100) = 5 This is not available in tables to infer, so trial and error calculations must be done from the basic formula: 1000 = 200[(1 + i)100 – 1]/[i(1 + i)100] Such trial and error experiments give an IRR of 20%. (d) 1000 = 200(P/A, i, 2) (P/A, i, 2) = 5 Again using trial and error in the formula: 1000 = 200[(1 + i)2 – 1]/[i(1 + i)2] The result is an IRR of –44.2%.

5.11

Setting disbursements equal to receipts (present worth) and solving for i*: 8000 = 400(P/A, i*, 30) (P/A, i*, 30) = 20 (P/A, 2%, 30) = 22.396 (P/A, 3%, 30) = 19.6 The IRR for this project is between 2% and 3%. This is less than NF’s MARR, so the windows are not a good investment.

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5.12

Setting the PW of disbursements equal to the PW of receipts: 200 000 + 40 000(P/A, i*, 6) = 40 000/i* + 20 000(P/A, i*, 5) – 5 000(A/G, i*, 5)(P/A, i*, 5) 40i* + 8i*(P/A, i*, 6) – 4i*(P/A, i*, 5) + i*(A/G, i*, 5)(P/A, i*, 5) = 8 At i* = 12%: LHS = 7.7842 At i* = 13%: LHS = 8.3319 Linearly interpolating: i* = 12 + (8 – 7.7842)/(8.3319 – 7.7842)  12.39% This is a good investment since the IRR exceeds the MARR.

5.13

The IRR for each alternative can be found by solving for i in each of the following equations: Alternative A: –100 000 + 25 000(P/F,i,1) + 25 000(P/F, i, 2) + 25 000(P/F, i, 3) + 25 000(P/F, i, 4) + 25 000(P/F, i, 5) = 0 Alternative B: –100 000 + 5 000(P/F, i, 1) + 10 000(P/F, i, 2) + 20 000(P/F, i, 3) + 40 000(P/F, i, 4) + 80 000(P/F, i, 5) = 0 Alternative C: –100 000 + 50 000(P/F, i, 1) + 50 000(P/F, i, 2) + 10 000(P/F, i, 3) = 0 Alternative D: –100 000 + 1 000 000(P/F, i, 5) = 0 Using a trial and error approach with a spreadsheet, the IRRs for the individual projects are listed below: Alternative A B C D

IRR 7.931% 11.294% 6.044% 58.489%

With a MARR of 8%, projects B and D are acceptable, project A is marginally unacceptable, and project C is unacceptable.

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Chapter 5 - Comparison Methods Part 2

5.14

Increment from do-nothing to 1: 100 000 = 160 000(P/F, i*, 1)  i* = 60% > MARR Increment is justified; we can accept contract 1.

Increment from 1 to 2: 100 000 = 140 000(P/F, i*, 1)  i* = 40% > MARR Increment is justified; we can accept contract 2.

Increment from 2 to 3: 50 000 = 55 000(P/F, i*, 1)  i* = 10% < MARR Increment from 2 to 3 cannot be justified.

Choose Contract #2.

5.15

First, the incremental IRR from do-nothing to A is found by solving for i in: (P/A, i, 5) = 100 000/50 000 = 2  IRR = 41.1% Since the IRR on A exceeds the MARR, A is the current best alternative.

B then challenges A. The IRR on the incremental investment between A and B is: –(400 000 – 100 000) + (150 000 – 50 000)(P/A, i, 5) = 0 (P/A, i, 5) = 300 000/100 000 = 3  IRR = 19.9% Because the IRR on the incremental investment is above 10%, project B becomes the current best alternative. Since there are no other projects to challenge B, it should be taken.

5.16

Since alternative A has an IRR that exceeds the MARR, it is acceptable.

The incremental IRR to the next cheapest alternative, B, is only 12%, less than the MARR, and so B is rejected. The incremental investment from A to C is 30%, so A is rejected in favour of C. The incremental investment from C to D is less than the MARR, so D is rejected. The incremental investment from C to E is equal to the MARR, and so E is accepted. The incremental investment from E to F is less than the MARR, and so E is the best choice.

5.17

(a) Setting disbursements equal to receipts at the end of year 2, taking cash on hand forward at the MARR: 8000(F/P, 6%, 1) + 8000 = 10 000(F/P, 6%, 2) + 5500(P/F, i*, 1) This could also have been written as: (10 000(P/F, 6%, 1) – 8000)*(P/F, 6%,1) – 8000 + 5500(P/F, I*, 1) Solving for I* gives I* = 4.88% = ERR 73 Copyright © 2022 Pearson Canada Inc.

Chapter 5 - Comparison Methods Part 2

(b) Setting disbursements equal to receipts at the end of the three-year period, with receipts taken forward at the MARR: 8000(F/P, I*, 2) + 8000(F/P, I*, 1) = 10 000(F/P, 6%, 3) + 5500 Solving for I* gives I* = 5.76% =approximate ERR. (Recall that the approximate ERR will always be between the accurate ERR and the MARR.) (c) This is not a good investment.

5.18

(a) 20 000 + 100 000(P/F, i, 2) = 120 000(P/F, i, 1) 1 + 5/(1 + i)2 = 6/(1 + i) (1 + i)2 + 5 = 6(1 + i) i2 + 2i + 1 + 5 = 6 + 6i i2 – 4i = 0 i(i − 4) = 0 i = 0 or 4  IRR of 0% or 400% Both IRRs are mathematically correct, as can be seen by looking at project balances at yearly intervals: End of year 0 1 2

At i = 0% 20 000 20 000 (1 + 0) – 120 000 = –100 000 –100 000(1 + 0) + 100 000 = 0

At i = 400% 20 000 20 000(1 + 4) – 120 000 = –20 000 –20 000(1 + 4) + 100 000 = 0

These results are not realistic since the money in hand for the first year cannot be invested at the same rate at the rest of the project: it will not be 0 or 400%. (b) The exact ERR can be calculated in this case by taking the $20 000 forward at the MARR for one year, and then equating the cash flows at that time.

20 000(F/P, 12%, 1) + 100 000(P/F, i, 1) = 120 000 1 + i =100 000/[1 20 000 – 20 000(1.12)] = 1.024 59 i = 2.46% = ERR

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The ERR (the correct IRR) is about 2.46%. The project should not be accepted. Calculating an accurate ERR is generally difficult because in complicated cash flows it’s difficult to tell when there is “cash in hand.” (c) For the approximate ERR, take all receipts forward at the MARR to the time of the last cash flow, and equate it to the future worth of the disbursements at an unknown interest rate. The unknown interest rate is the approximate ERR: 20 000(F/P, 12%, 2) + 100 000 = 120 000(F/P, i, 1) 1 + i = [20 000(1.2544) + 100 000]/1 20 000 = 1.0424 i = 4.24% = ERR (approximate) Note that the approximate ERR is always between the accurate ERR and the MARR. This is good in that for acceptable projects the approximate ERR always errs on the conservative side, i.e., it is always smaller than actual.

B. Applications 5.19

(a) Because CB Electonix must buy equipment, the current best is P1, the alternative with the smallest first cost.

First cost ($) Annual costs ($)

P1 0 135 000

P2 200 000 95 000

P3 850 000 0

The first challenger is P2.

P2 – P1: (200 000 – 0) + (95 000 – 135 000)(P/A, i, 10) = 0 (P/A, i, 10) = 200 000/40 000 = 5 From table lookups: (P/A, 15%,10) = 5.0187 (P/A, 20%, 10) = 4.1924 IRR is between 15 and 20%; challenge succeeds. P2 becomes the current best. Next, P3 challenges P2.

P3 – P2: (850 000 – 200 000) + (0 – 95 000)(P/A, i, 10) = 0 (P/A, i, 10) = 650 000/95 000 = 6.842

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Table lookups: (P/A, 7%, 10) = 7.0235 (P/A, 8%, 10) = 6.7100 IRR is between 7% and 8%. Challenge fails.

P2 is the best project. (b) Consider the last challenge: P3 – P2. Solve for the IRR via linear interpolation: (P/A, i, N) = 6.842 (P/A, 7%, 10) = 7.0235 (P/A, 8%, 10) = 6.7100 i = 8 + (7 – 8)[(6.842 – 6.7100)/(7.0235 – 6.7100)] = 7.5789 Below 7.58%, P3 becomes preferred.

5.20

(a) With a MARR of 16%, do projects 1, 3 and 4 as their IRRs meet or exceed 16%. (b) For MARR = 15%: The current best is alternative 1, which has the least first cost and IRR > MARR. Challenge 1 with 2: incremental IRR is 9%; challenge fails.

Challenge 1 with 3: incremental IRR is 17%; challenge succeeds.

Hence, 3 is current best.

Challenge 3 with 4: incremental IRR is 13%; challenge fails.

Pick 3. (c) For MARR = 17%: The current best is alternative 1, which has the least first cost and IRR > MARR. Challenge 1 with 2: incremental IRR is 9%; challenge fails.

Challenge 1 with 3: incremental IRR is 17%; challenge still succeeds.

Hence, 3 is current best.

Challenge 3 with 4: incremental IRR is 13%; challenge fails.

Pick 3.

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5.21

Ordered by first cost, the choices are: 1) Free machine, first cost $6000 2) Used machine, first cost $36 000 3) Owned machine, first cost $71 000 The incremental IRR from do-nothing to alternative 1: 6000 = (20 000 –15 000)(P/A, i, 6) (P/A, i, 6)= 1.2 Noting that (P/A, 40%, 6) = 2.16 and that (P/A, 50%, 6) = 1.8, the free machine is certainly acceptable.

Alternatively, accept free machine because one interprets the “requirement for a filling machine” as meaning that one of the projects must be chosen.

Look at the increment of investment between the free and used machines: an ERR method must be used since the incremental cash flows are not a simple investment.

Year 0 1 2 3 4 5 6

Free −6000 5000 5000

Used −36000 17000 14500

Increment −30000 12000 9500

5000 5000 5000 5000

12000 9500 7000 4500

7000 4500 2000 −500

FW@MARR

FW@ERR −49697

19326 13909 9317 5445 2200 Net FW 0.0003776

−500 ERR 0.0877646

The ERR is approximately 8.78%. This is less than MARR, so the incremental investment between the free and used machines is not warranted.

Look at the increment of investment between the free and new machines: since the incremental cash flows form a simple investment, the IRR approach can be taken.

Year 0 1 2 3 4 5 6

Free −6000 5000 5000 5000 5000 5000 5000

Own −71000 20000 20000 20000 20000 20000 30000

Increment −65000 15000 15000 15000 15000 15000 25000 IRR 0.12879

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PW@IRR −65000 13289 11772 10429 9239 8185 12085 Net PW 0.00433138

Chapter 5 - Comparison Methods Part 2

The IRR on the incremental investment can be determined as approximately 12.88%; hence, the investment is warranted. The new machine should be chosen.

5.22

Since this is not a simple investment, the approximate ERR should be used.

500 000(F/P, 25%, 4) + 1 200 000 = 400 000(F/P, i, 3) + 900 000(F/P, i, 2) 5(2.4414) + 12 = 4(F/P, i, 3) + 9(F/P, i, 2) 4(F/P, i, 3) + 9(F/P, i, 2) = 24.207 For i = 25: LHS = 4(1.9531) + 9(1.5625) = 21.87 For i = 30: LHS = 4(2.197) + 9(1.69) = 23.00 The approximate ERR is well above the MARR and the project should be accepted.

5.23

This problem requires the use of an explicit rate of return on the positive cash flows generated from the agreement. The cash flows associated with the project that must be invested elsewhere are the $450 000 (now) and the $650 000 at the end of the 12th year. The future value of these cash receipts (invested at MARR until the end of the 15th year) is equated to the future value of disbursements: 450 000(F/P, 20%, 15) + 650 000(F/P, 20%, 3) = 900 000(F/P, i, 10) + 900 000(F/P, i, 5) + 900 000 9(15.407) + 13(1.728) = 18(F/P, i, 10) + 18(F/P, i, 5) + 18 (F/P, i, 10) + (F/P, i, 5) = 7.9515 At i = 20%: LHS = 6.1917 + 2.4883 = 8.68 At i = 15%: LHS = 4.0455 + 2.0113 = 6.057 A linear interpolation gives: ERR = 15 + (5)(7.95 – 6.057)/(8.68 – 6.057) = 18.61% (Note that, with a spreadsheet, one can obtain an ERR = 18.77% by trial and error rather than by interpolation.)

Samiran should not accept this deal because the ERR is less than his MARR of 20%. There is only one answer using ERR; it is not the accurate IRR, but nonetheless, it will give the correct recommendation.

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5.24

Since one of the two alternatives must be chosen, the least first cost alternative (the custom automated equipment) becomes the current best.

To see if the off-the-shelf standard automated equipment should be chosen, we find the net incremental cash flows to select the off-the-shelf equipment. The least common multiple of service lives is 30 years.

Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Automated First Annual −15000 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −15000 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −15000 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400 −6400

Off-the-shelf First Annual −25000 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −25000 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625 −5625

Increment −10000 775 775 775 775 775 775 775 775 775 15775 775 775 775 775 −24225 775 775 775 775 15775 775 775 775 775 775 775 775 775 775 775 MARR 0.09

FW @MARR

FW @ERR −133785

9433 8655 7940 7284 6683 6131 5625 5160 4734 88410 3985 3656 3354 3077 −88607 2590 2376 2180 2000 37345 1683 1544 1417 1300 1192 1094 1004 921 845 775 Total FW 0.019482

ERR 0.090302

The incremental investment to the off-the-shelf system is not a simple investment. To make a decision, we will use the approximate ERR. Take the positive incremental cash flows forward to the end of 30 years at the MARR (9%) and take the negative cash flows forward to the end of 30 years at the unknown ERR. Setting FW(positive cash flows @ MARR) + FW(negative cash flows @ ERR) = 0, we find the approximate ERR to be 9.03%, which is slightly above the MARR. Since there are no other challengers, the best alternative is (marginally) the off-the-shelf material handling system.

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5.25

Since one of the two alternatives must be chosen, the least first cost alternative (the Y19) becomes the current best.

To see if the XJ3 should be chosen, we find the net incremental cash flows to select the XJ3. The length of the study period is two years.

Year 0 1 2

Y19 First/Salv. −3200

Annual

1000

−300 −300

XJ3 First/Salv −4500 1900

Increment −1300 300 1200 IRR 0.083055

PW @ IRR −1300 277 1023 Total PW 0.0050529

The incremental investment to XJ3 is a simple investment. We find the IRR to be 8.3% which is below the MARR. Since there are no other challengers, the best alternative is Y19.

5.26

The lowest first cost project is #3 (first cost of –$175 000). Check the ERR of #3 since it is a non-standard investment: 175 000(F/P, 10%, 3) – 150 000 (F/A, i*, 3) + 300 000 = 0 (F/A, i*, 3) = [175 000(1.331) + 300 000]/150 000 = 3.553  i* = ERR = 17.4% Project 3 is acceptable. Now look at the increment of investment to project #2. We still use ERR, but the disbursements and receipts change: (150 000 – 175 000)(F/P, i*, 3) – (100 000 – 150 000)(F/A, 10%, 3) + (230 000 – 300 000) = 0 –25 000(F/P, i*, 3) – (–50 000)(F/A, 10%, 3) + (– 70 000) = 0 –25 000 (F/P, i*, 3) + 50 000(F/A, 10%, 3) – 70 000 = 0 (F/P, i*, 3) = (50 000  3.31 – 70 000)/25 000 = 3.82  i* = ERR = 56.3% Project 2 is acceptable. Now look at the increment of investment to project #1. We still use ERR. (100 000 – 150 000)(F/P, i*, 3) – (75 000 – 100 000)(F/A, 10%, 3) + (200 000 – 230 000) = 0 –50 000(F/P, i*, 3) – (–25 000)(F/A, 10%, 3) + (– 30 000) = 0 –50 000(F/P, i*, 3) + 25 000(F/A, 10%, 3) – 30 000 = 0 (F/P, i*, 3) = (25 000  3.31 – 30 000)/50 000 = 1.055  i* = ERR = 1.80% The increment from #2 to #1 is not acceptable.

CCC should do project #2.

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Chapter 5 - Comparison Methods Part 2

5.27

The three proposals imply seven mutually exclusive projects: Project

Proposal

Net first cost

1 2 3 4 5 6 7

do nothing A B C AB AC BC

0 40 000 110 000 130 000 150 000 170 000 240 000

Net annual savings 0 20 000 30 000 45 000 50 000 65 000 75 000

The do-nothing alternative is the current best. First check the incremental investment from do-nothing to Project #2.

40 000 = 20 000(P/A, i*, 4) (P/A, i*, 4) = 2  i* = IRR = 34.9% Project #2 is acceptable. Now check the increment to project #3: 70 000 = 10 000(P/A, i*, 4) (P/A, i*, 4) = 7 i* = IRR < 0 Project #3 is unacceptable. Now check the increment from #2 to #4: 90 000 = 25 000(P/A, i*, 4) (P/A, i*, 4) = 3.6  i* = IRR = 4.4% Project #4 is unacceptable. Now check the increment from #2 to #5: 110 000 = 30 000(P/A, i*, 4) (P/A, i*, 4) = 3.66  i* = IRR = 3.6% Project #5 is unacceptable. Now check the increment from #2 to #6: 130 000 = 45 000(P/A, i*, 4) (P/A, i*, 4) = 2.88  i* = IRR = 14.5% Project #6 is acceptable. Now check the increment from #6 to #7: 70 000 = 10 000(P/A, i*, 4) (P/A, i*, 4) = 7  i* = IRR < 0 Project 7 is unacceptable. Project 6 is the project with the highest first cost for which all increments of investments are justified.

Kool Karavans should invest in proposals A and C.

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5.28

The total costs for the checkweigher project are: 30 000 + 5 000  5 = $55 000. The grant would amount to 55 000  0.3 = $16 500 in two payments of $8250.

The total costs for the scheduler project are: 10 000 + 12 000  5 = $70 000. The grant would amount to 70 000  0.3 = $21 000 in two payments of $10 500.

For an incremental IRR comparison, the alternative with the lowest first cost is the scheduler—its first cost is negative. Since there are no net disbursements for this project, it is accepted. The cash flows for the two projects and for the increment of investment from Scheduler to Checkweigher are shown below: Year 0 1 2 3 4 5

Checkweigher Scheduler Increment −21750 500 −22250 9000 5000 4000 9000 5000 4000 9000 5000 4000 9000 5000 4000 25250 15500 9750

The increment conforms to a simple investment. Calculating the IRR gives: 4000(P/A, i, 4) + 9750(P/F, i, 5) = 22250 16(P/A, i, 4) + 23(P/F, i, 5) = 89  i = IRR= 4.39% The increment is below the MARR, so the scheduling project should be done.

5.29

Model A has the least expensive first cost, so since one heating system must be chosen, it can be accepted. Next look at the increment from A to B. The cash flows of models A and B and the increment from A to B are shown below: Year 0 1 2 ...

9 10

Model A Model B Increment −500 −3600 −3100 −300 500 800 −300 500 800 ... ... ... −300 500 800 −300 1500 1800

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The IRR of the increment is then calculated from: 3100 = 800(P/A, i, 10) + 1000(P/F, i, 10) 8 (P/A, i, 10) + 10 (P/F, i, 10) = 31  i = IRR = 23.63% Since this is greater than the MARR, Model B is preferred over Model A. Now look at the increment to Model C. The cash flows of models B and C and the increment from B to C are shown below: Year 0 1 2 3 ...

9 10

Model B Model C Increment −3600 −4000 −400 500 1000 500 500 −3000 −3500 500 1000 500 ... ... ...

500 1000 500 1500 2000 500

The increment does not follow the pattern of a simple investment, so ERR must be used: 400(F/P, ERR, 10) + 3500(F/P, ERR, 8) = 500(F/P, 12%, 9) + 500(F/A, 12%,8) 4(F/P, ERR, 10) + 35(F/P, ERR, 8) = 5(2.7731) + 5(12.3) = 75.36  i = ERR = 8.34% The ERR is less than the MARR, so Model C is not justified. Jacob should buy model B. C. More Challenging Problems 5.30

If the projects are independent, they should both be undertaken (provided there are funds) because they each have an IRR which exceeds the MARR. If the projects are mutually exclusive, then an incremental analysis must be performed. Initially, project Y would be chosen as the current best alternative because it has the smallest first cost and it has an IRR greater than the MARR. Next, the IRR on the incremental investment between Y and X would need to be found to determine if the incremental investment meets the MARR requirements. To find the IRR on the incremental investment, the cash flows associated with each project must be available so that the difference between the two projects can be determined. Since this information is not available, a choice cannot be made.

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5.31

We know that 56 740 = –180 000 + X(P/A, 10%, 5) where X is the unknown cash flow at the end of each of the five years.

Solving for X: X = 236 740/3.7908 = 62 452 We calculate IRR from: 180 000 = X(P/A, i*, 1) Substituting in the known value for X and solving for i*: (P/A, i*, 5) = 180 000/62 452 = 2.882  i* = IRR = 21.7% The IRR is 21.7%.

5.32

We have: 20 000 = 4000(P/A, MARR, 10) (P/A, MARR, 10) = 5  MARR = 15.1% 20 000 = –P + A(P/A, 15.1%, 10) + 1000(P/F, 15.1%, 10) By substituting P = 3A: 20 000 = –3A + 5A + 1000(0.245046) 2A = 20 000 – 245.046 A = 9877.48 P = 9877.48  3 = 29 632.44 Calculate the IRR from: 29 632.44 = 9877.48(P/A, i*, 10) + 1000(P/F, i* 10)  i* = IRR = 0.312 = 31.2% The IRR for Lucy’s project is 47.8%.

5.33

From the IRR calculation: P = A(P/A, 15%, 5) + S (P/F, 15% 5) Since S = 1/2P: 2S – S(0.49718) = A(3.3522) S = A(3.3522)/1.50282 = 2.2306A From the present worth calculation: PW = –P + A(P/A, MARR, 5) + S(P/F, MARR, 5)

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Since PW = 2A, P = 2S, and S = 2.2306A: 2A = –2(2.2306A) + A(P/A, MARR, 5) + 2.2306A(P/F, MARR, 5) Dividing by A: 2 = –4.4612 + (P/A, MARR, 5) + 2.2306(P/F, MARR, 5) (P/A, MARR, 5) + 2.2306(P/F, MARR, 5) = 6.4612  MARR = 3.19% Patti’s MARR is about 3.19% 5.34

(a) A bond with a face value of $10 000 and a coupon rate of 14% pays interest of $1400 per year ($700 every six months). A bond which matures 5 years from now has 10 remaining payments. The IRR on Jerry’s investment is the interest rate that solves: 10 000(P/F, i, 10) + 700(P/A, i, 10) = 3500 14.29(P/F, i, 10) + (P/A, i, 10) = 5 By using a spreadsheet, an IRR of 25.46% (per six months) can be found.

Now, effective interest per year can be calculated: IRR = (1 + 0.2546)2 – 1 = 0.574 = 57.4% (b) Jerry’s IRR for this investment is 57.4%. If his MARR is 20%, he should buy the bond.

5.35

(a) With a present worth analysis and a least common multiple of service life of 20 years, the IRR can be found with the following analysis.

Since one of the two must be chosen, the least expensive option, the “Clip Job” becomes the current best and the “Lawn Guy” is the challenger. We need to look to see if there is potential for multiple IRRs on the incremental cash flows. Over a 20-year period (the least common multiple of service life), the cash flows for the two projects, and the incremental cash flows are as follows:

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Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Lawn Guy First Annual −350 −90 −90 −90 −90 −90 −90 −90 −90 −90 −350 −90 −90 −90 −90 −90 −90 −90 −90 −90 −90 −90

Clip Job First Annual −120 −100 −100 −100 −120 −100 −100 −100 −100 −120 −100 −100 −100 −100 −120 −100 −100 −100 −100 −120 −100 −100 −100 −100 −100

Increment −230 10 10 10 130 10 10 10 130 10 −340 10 130 10 10 10 130 10 10 10 10

(b) The incremental cash flows show that the incremental investment is not a simple investment. Hence, we need to determine if there are multiple IRRs. A plot of the PW of the incremental investment is below.

100

Present worth ($)

50 0 -50

0%

20%

40%

60%

80%

100%

-100 -150 -200 -250 Interest rate

One IRR of 3.93% is revealed, but no others. There still may be multiple IRRs, so we calculate project balances at 3.93% (rightmost column in table below). This shows that the project has positive balances and hence we still don’t know for sure if there are multiple IRRs. Therefore, we should use the approximate ERR method.

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Chapter 5 - Comparison Methods Part 2 Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Project balance −230 −229 −228 −227 −106 −100 −94 −88 39 50 −288 −289 −170 −167 −164 −160 −36 −28 −19 −10 0

(c) Applying the approximate ERR method, we take the positive cash flows on the incremental investment forward to the end of the 20-year period at the MARR of 5%. The negative cash flows are brought forward at the unknown approximate ERR. The approximate ERR is 4.7%, which is less than the MARR of 5%. Thus, the incremental investment is not justified, and the Clip Job is the preferred choice.

Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

5.36

Increment FW @ MARR FW @ ERR −230 −577 10 25 10 24 10 23 130 284 10 21 10 20 10 19 130 233 10 17 −340 −538 10 16 130 192 10 14 10 13 10 13 130 158 10 12 10 11 10 11 10 10 Total FW ERR 0.00185 0.047028

Since one of the two alternatives must be chosen, the least first cost alternative (the used refrigerator) becomes the current best.

To see if the new refrigerator should be chosen, we find the net incremental cash flows to select the new refrigerator. The least common multiple of service lives is 24 years.

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Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Automated −475

Off-the-shelf −1250

−475

−475 −1250 −475

−475

−475 −1250 −475

−475

Increment −775 0 0 475 0 0 475 0 −1250 475 0 0 475 0 0 475 −1250 0 475 0 0 475 0 0 0 MARR 0.08

FW @MARR

FW @ERR −3670

2391

1898 −3525 1507

1196

950 −2099 754

598

Total FW 0.000685

ERR 0.066939

The incremental investment to the new refrigerator is not a simple investment. To make a decision, we will use the approximate ERR. Take the positive incremental cash flows forward to the end of 24 years at the MARR (8%) and take the negative cash flows forward to the end of 24 years at the unknown ERR. Setting FW(positive cash flows @ MARR) + FW(negative cash flows @ ERR) = 0, we find the approximate ERR to be 6.69% which is slightly below the MARR. Since there are no other challengers, the best alternative is the used refrigerator.

5.37

Since the “do nothing” option is possible, we need to find the IRR on the least first cost alternative to begin. Both options have a zero first cost, so we can begin with either, say contract 1. The net cash flows of the two contracts are summarized below: Year Contract 1 Contract 2

0 15 000 16 000

1 –10 000 –9 000

2 –10 000 –9 000

3 12 000 13 000

The incremental investment from “do nothing” to contract 1 is not a simple project, so use the approximate ERR method: (15 000 – 0)(F/P, 10%, 3) + 12 000 = (10 000 – 0)[(F/P, i*, 2) + (F/P, i*, 1)] (F/P, i*, 2) + (F/P, i*, 1) = [15 000(1.3310) + 12 000]/10 000 (1 + i*)2 + (1 + i*) = 3.1965

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Chapter 5 - Comparison Methods Part 2

At i* = 30%: LHS = 2.99 At i* = 40%: LHS = 3.36 The approximate ERR is between 30% and 40%, and is clearly above the MARR. Contract 1 becomes the current best alternative.

To see if the second contract should be chosen, we find the net incremental cash flows between contract 1 to contract 2: 1000, 1000, 1000, and 1000. All cash flows are positive, so the incremental investment has an infinite IRR. Hence, contract 2 is superior.

5.38

(a) Solve for i in: –5000 + 3000(P/F, i,1) + 4000(P/F, i, 2) – 1000(P/F, i, 3) = 0 (b) The number of changes in sign of the cash flows is 2. Therefore, the maximum number of IRRs is two. (c) Year 0 1 2 3

Project balance B(i) B(0) = –5000 B(1) = B(0)(1.1458) + 3000 = –2729 B(2) = B(1)(1.1458) + 4000 = 873.1118 B(3) = B(2)(1.1458) – 1000 = 0.4115  0

(d) Due to the fact that this is not a simple investment and the fact that one project balance is positive, we cannot be sure that the IRR of 14.58% is unique.

5.39

The current policy requires P/A  5  P  5A We need –P + A(P/A, MARR, 20)  0  P  A(P/A, MARR, 20) So that: 5A = A(P/A, MARR, 20) or (P/A, MARR, 20) = 5 For MARR = 20%: (P/A, MARR, 20) = 4.6755 For MARR = 15%: (P/A, MARR, 20) = 6.2593 By linear interpolation, MARR  19% The equivalent MARR is about 19%.

5.40

First calculate the incremental IRR from do-nothing to alternative 1.

1 200 000 = 300 000(P/A, i%, 10) (P/A, i, 10) = 4  i = IRR1 = 21.4% 89 Copyright © 2022 Pearson Canada Inc.

Chapter 5 - Comparison Methods Part 2

From #1 to #2: 1 500 000 − 1 200 000 = (400 000 − 300 000)(P/A, i%, 10) 300 000 = 100 000(P/A, i%, 10) (P/A, i%, 10) = 3  i = IRR2–1 = 31.1% From #2 to #3: 2 100 000 − 1 500 000 = (500 000 − 400 000)(P/A, i%, 10) 600 000 = 100 000(P/A, i%, 10) (P/A, i%, 10) = 6  i = IRR3–2 = 10.6% Starting at the alternative with the lowest first cost, #1, the possibilities can be enumerated: If MARR > 21.4%, 1 is rejected; else it is accepted.

If 1 is accepted, 2 is also accepted.

If MARR > 10.6, 2 is the final choice; else 3 is the final choice.

Continue the process to check if there are cases that must be evaluated for MARR > 21.4%. Calculate the incremental IRR from donothing to alternative 2.

1 500 000 = 400 000(P/A, i%, 10) (P/A, i, 10) = 3.75  i = IRR2 = 23.4% Thus, design 2 is also chosen for 21.4% < MARR £ 23.4%.

Finally, calculate the incremental IRR from do-nothing to alternative 3.

2 100 000 = 500 000(P/A, i%, 10) (P/A, i, 10) = 4.2  i = IRR3 = 19.9% Note that this IRR does not affect the analysis. The previous results can be simplified and summarized as shown below.

0 £ MARR £10.6%: Choose 3 10.6% < MARR £23.4%: Choose 2 23.4% < MARR: Choose none

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Chapter 5 - Comparison Methods Part 2

5.41

We have: ERR = (12 + IRR)/2 and ERR = 18 + [(IRR – 18)/4] so that (12 + IRR)/2 = 18 + [(IRR − 18)/4] 24 + 2IRR = 72 + IRR − 18 IRR = 72 − 18 − 24 = 30 ERR = (12 + 30)/2 = 21% Let Xj be the cash flow at time j. From the IRR calculation: X0 = X1(P/F, 30%, 1) + 2000(P/F, 30%, 2) X0 = X1(0.76923) + 2000(0.59172) X0 = X1(0.76923) + 1183.44 Case 1: assume X1 is a receipt (positive): From the ERR calculation: 2000 = X0(F/P, 21%, 2) + X1(F/P, 12%, 1) 2000 = X0(1.4641) + X1(1.12) so that 2000 = [X1(0.76923) + 1183.44](1.4641) + X1(1.12) 2000 = X1(1.12622) + 1732.67 + X1(1.12) 267.3255 = X1(2.24622) X1 = $119.01 and X0 = 1274.99 Case 2: assume X1 is a disbursement (negative): From the ERR calculation: 2000 = X0(F/P, 21%, 2) + X1(F/P, 21%, 1) 2000 = X0(1.4641) + X1(1.21) so that 2000 = [X1(0.76923) + 1183.44](1.4641) + X1(1.21) 2000 = X1(1.12622) + 1732.67 + X1(1.21) 267.3255 = X1(2.33622) X1 = $114.27 This contradicts the assumption the X1 is a disbursement. There is only one possible value for X1, $119.01.

5.42

In order to calculate the exact ERR, we must calculate when there is capital available that is not bound up in the project. This capital is invested outside the project at the MARR. 91 Copyright © 2022 Pearson Canada Inc.

Chapter 5 - Comparison Methods Part 2

For year 1, $10 000 000 is invested at the MARR For year 2, the future value after 1 year of the $10 000 000 less the $8 000 000 costs at the end of year one is invested at the MARR Amount invested at MARR at end of year 1 is: 10 000 000(F/P, 10%, 1) − 8 000 000 = 10 000 000 (1.1) − 8 000 000 = 3 000 000 Clearly it is unlikely that there will be a remaining positive cash flow balance at the end of the second year.

For year 4, $15 000 000 less the future worth (at the beginning of year 4) of prior cash flows is invested at the MARR, if any The future worth of cash flows at the beginning of year four (call it α) is: α = [10 000 000 (F/P, 10%, 1) − 3 000 000] (F/P, i*, 2) + 3 000 000 (F/P, 10%, 1) (F/P, i*, 1) + 8 000 000 (F/A, i*, 3) Observing through trial and error that α < 15 000 000, there is a positive cash flow balance of 15 000 000 − α, which is then is invested at the MARR over year 4. Trial and error calculations reveals there is negative cash flow balance at the end of year 4, and no positive cash flow balances thereafter. Consequently the end of year 4 is an appropriate point in time to calculate the exact ERR: [15 000 000 − α](F/P, i*,1) − 5 000 000 − 5 000 000(P/F, i*,1) + 5 000 000/i* Calculating through spreadsheet trial and error gives i* = ERR = 21.9% 5.43

PW (Op cost) = A/(i − g) = 8*106/(0.1765 − 0.03) = $54,607,508.53 PW (Upgrades) = [A/i(effective for 5 yrs)](P/F, i, 2) = [20,000,000/((1 + 0.1765)^5 − 1)]/(1 + 0.1765)2 = $11,522,275.85 PW (first cost) = $820,000,000 PW (of all cost) = $886,129,784.38 PW (all passengers) = 10,000,000/0.1765 = $56,657,224 Ticket/visit = $15.64/ticket

5.44

No, because 15.64 > $5

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Chapter 5 - Comparison Methods Part 2

Notes for Case in Point 5.1 1)

Yes.

2)

This is a personal judgement; no right answer.

Notes for Mini-Case 5.1 1)

The payback period would appear to be more attractive.

2)

6.3 percent would be a less attractive return on investment.

3)

Such a graph shows that not discounting the revenue is highly misrepresentative of the value of the project.

4)

Long-term metal prices are more sensible to use.

5)

No.

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CHAPTER 6 Solutions to Chapter-End Problems A. Key Concepts 6.1

(b)

6.2

The unscrambled Income Statements and Balance Sheets are: Income Statement for Paradise Pond Co.: Month ending Dec. 31, 2021 (in $000s) 2020 Revenues Sales 3 000 Cost of goods sold 1 750 Gross profit 1 250

2021 3 625 2 125 1 500

Expenses Operating expenses Depreciation expense Interest expense Total expenses

75 550 125 750

100 500 150 750

Net income before taxes Income taxes Net income after taxes

500 200 300

750 300 450

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Chapter 6 - Depreciation and Financial Accounting Comparative Balance Sheets for Paradise Pond Co.: Ending 2020 and 2021 (in $000s) 2020 2021 Assets Current assets Cash 300 225 Accounts receivable 675 638 Inventories 825 938 Total current assets 1 800 1 800 Long-term assets Plant and equipment 3 300 3 900 Less accumulated depreciation 1 500 1 800 Net plant and equipment 1 800 2 100 Total long-term assets 1 800 2 100 Total assets 3 600 3 900

6.3

Liabilities Current liabilities Accounts payable Working capital loan Total current liabilities Long-term liabilities Bonds Total long-term liabilities Total liabilities

300 0 300

225 225 450

900 900 1 200

900 900 1 350

Owners’ Equity Common shares Contributed capital Retained earnings Total owners’ equity Total liabilities and owners’ equity

450 900 1 050 2 400 3 600

450 900 1 200 2 550 3 900

The answer to this question will vary depending on the imagination and creativity of the student. However, a model answer is as follows: Paradise Pond Company should use direct advertisement strategies to attract investors to invest in the company. If the company is public, introducing its products to consumers through different media channels such as magazines for particular interest groups, TV, radio and internet will enhance consumers’ recognition of the products and therefore more people will buy the company’s share and it could bring new streams of cash flow to the company. On the other hand, if the company is private, presenting the company’s products directly to investors in exhibitions could attract investors. Also, presenting a positive future outlook of the company after new developments could build better trust between the investors and the company.

The sales and marketing section of the business plan is as follows (it is assumed that the company is private and produces furniture):

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Chapter 6 - Depreciation and Financial Accounting

Sales and Marketing: The unique aspect of the Paradise Pond business is recognizing customers’ needs and fulfilling them with high quality, durable and elegant furniture products.

Our strategy is to initially focus on the market for residential furniture and eventually move towards office furniture.

Our company distinguishes itself from other competitors by manufacturing its products from original Canadian wood and customizing products based on the individual’s taste and budget.

Our main marketing strategy is building a close relationship with our clients to ensure their needs are met. We believe that listening to customers and customizing products according to their need is the key to repeat business and referrals. In order to maintain this personal relationship with clients we strongly rely on sales calls to build accounts.

To support our relationship with clients, maintain our brand in the industry and enhance our employees’ product knowledge we attend as many area conventions and trade shows as possible.

Furthermore, we have built other media channels to provide customers with our promotions and new products. Interested customers who provide their email address automatically receive our weekly news release email, featuring new products, designs and promotions. Also, we publish our ads in local newspapers in different cities in Ontario on a monthly basis. In these ads we provide fun tips about design and home decoration, in addition to profiling our products and services.

6.4 Balance Sheet for Paarl Manufacturing: As of xxx 3x, 20xx Assets Current assets Cash Accounts receivable Raw materials Work in progress Finished goods Long-term assets Property Equipment 450 000 Less accumulated depreciation 240 000 Total assets Liabilities Current liabilities Current bank account Accounts payable Accrued taxes Long-term liabilities Mortgage loan Government loan Total liabilities

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45 954 22 943 102 000 40 000 123 000 250 000 210 000 793 897 30 000 12 992 32 909 224 000 258 996 558 897

Chapter 6 - Depreciation and Financial Accounting Owners’ Equity Capital stock Retained earnings Total owners’ equity Total liabilities and owners’ equity

100 000 135 000 235 000 793 897

Income Statement for Paarl Manufacturing: Month ending xxx 3x, 20xx Income Gross income from sales 220 000 Less cost of goods sold 40 000 Net income from sales 180 000 Expenses Salaries Depreciation Maintenance Advertising Insurance Total expenses

45 000 34 000 1 500 3 400 300 84 200

Income before taxes Taxes @ 55% Net income

95 800 52 690 43 110

Equity ratio is 235 000/793 897 = 0.296. The company has too much debt, and thus is too dependent on its creditors. Consequently, it should not issue dividends but rather use earnings to reduce debt.

6.5

The answer to this question will vary depending on the imagination and creativity of the student. However, a model answer is as follows: Market analysis summary Paarl Manufacturing specializes highly in providing vinyl, aluminium window and entry door products and services to home and condo builders in Eastern Canada. Despite the economic downturns in Canada, housing has been improving significantly in recent years and therefore the need for good quality windows and door products has been a bright spot.

Target Market Paarl Manufacturing maintains a consistent high demand by satisfying the house owners’ and builders’ needs, considering the newest trends and styles in housing market. Paarl Manufacturing customers are categorized as follow: Contractors: Approximately 5000 contractors buy window and door products from Paarl Manufacturing every year to invest in house renovation projects. Considering the number of house renovations in Eastern Canada, approximately 45% of the demand is supplied by Paarl Manufacturing.

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Chapter 6 - Depreciation and Financial Accounting

In order to attract contractors, Paarl Manufacturing relies heavily on referrals and repeat business with current clients. A discount from 10% to 15% is provided for each referral.

House builders: With the rising number of new houses and condos in Toronto and surrounding cities in Eastern Canada, the need for window frames and door products has increased dramatically. Approximately 55% of the company’s gross income comes from selling products to individual house builders or condominium developers such as Tridel.

To maintain business with renowned builders, Paarl Manufacturing endeavours to customize its products according to the builders’ needs and requirements. For that, a special design department has been set up to bridge the gap between engineering and manufacturing departments.

Future outlook Considering low interest rates, a growing economy and government support for new home owners, Paarl Manufacturing predicts a steady demand trend in the new housing market. Considering steady construction and ongoing renovation projects for older houses, Paarl manufacturing predicts a 2% increase in sales.

6.6

(a) Salvador Industries: Income Statement for the Year ending June 30, 2020 Income Gross Income from sales 8 635 000 Less cost of goods sold 7 490 000 145 000 Total income 1 145 000 Expenses Depreciation Interest paid Other expenses Total expenses

70 000 240 000 100 000 410 000

Income before taxes Taxes at 40% Income before extraordinary items Extraordinary loss Net income

735 000 294 000 441 000 100 000 341 000

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Chapter 6 - Depreciation and Financial Accounting Salvador Industries: Balance Sheet as of June 30, 2020 Assets Current assets Cash Accounts receivable Inventories Prepaid services Total current assets Long-term assets Building Less accumulated depreciation Equipment less accumulated depreciation Land Total long-term assets Total assets

350 000 2 820 000 2 003 000 160 000 5 333 000 200 000 100 000 480 000 474 466

Liabilities Current liabilities Accounts payable Loan due Dec. 31, 1998 Accrued taxes Total current liabilities Long-term liabilities Mortgage Long-term loan Total long-term liabilities Total liabilities Owner’s Equity Capital stock Retained earnings Total owners’ equity Total liabilities and owners’ equity

100 000 5 534 540 000 645 534 5 978 534

921 534 50 000 29 000 1 000 534 1 200 000 318 000 1 518 000 2 518 534 1 920 000 1 540 000 3 460 000 5 978 534

(b) Current ratio = 5 333 000/1 000 534 = 5.33 Acid test ratio = (5 333 000 – 2 003 000 – 160 000)/1 000 534 = 3.17 Equity ratio = 3 460 000/5 978 534 = 0.58 Return on total assets = 341 000/5 978 534 = 5.7% The current and acid test rations show considerable security in meeting obligations. The equity ratio is slightly low, indicating reliance on debt financing, and profits are meagre. However, the extraordinary loss of $100 000 is relatively small compared to the overall activities of the company.

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Chapter 6 - Depreciation and Financial Accounting

6.7

A Financial Ratio Analysis for Fraser Phraser gives the following: Phraser’s financial ratios Current ratio Acid test Equity ratio Inventory turns ROA ROE

2020 1.84 0.78 0.50 4.31 2% 4%

2021 0.90 0.24 0.39 3.52 2% 5%

Phraser’s ability to deal with unexpected changes in cash flows has been seriously compromised over the 2020–2021 period. Phraser is increasingly relying on debt financing, has lower turnover of inventory and has a lower return on total assets. The financial position does not look promising and Phraser does not look like a solid loan applicant.

B. Applications 6.8 Balance Sheet for Movit Manufacturing: As of Dec. 31, 2021 (in $000s) Assets Current assets Cash GICs Accounts receivable Inventories Prepaid expenses Total current assets Long-term assets Land Plant and equipment Less accumulated depreciation Net plant and equipment Total long-term assets Total assets Liabilities Current liabilities Accounts payable Accrued wages Working capital loan Total current liabilities Long-term liabilities Deferred income taxes Mortgage Long-term bonds Total long-term liabilities

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2 100 450 15 000 18 000 450 36 000 3 000 18 450 10 950 7 500 10 500 46 500

7 500 2 850 4 650 15 000 2 250 9 450 4 350 16 050

Chapter 6 - Depreciation and Financial Accounting Owners’ Equity Common shares Contributed capital Retained earnings Total owners’ equity Total liabilities and owners’ equity

150 3 000 12 300 15 450 46 500

Income Statement for Movit Manufacturing: Month ending Dec. 31, 2021 (in $000s) Net sales 76 500 Cost of goods sold 57 000 Gross profit 19 500

6.9

Operating expenses Selling expenses Depreciation expense General expense Interest expense Total expenses

4 650 750 8 100 1 500 15 000

Net income before taxes Income taxes Net income after taxes

4 500 1 800 2 700

The financial ratios for Movit Manufacturing for 2021, 2022 and 2023 are: Movit’s financial ratios Current ratio Acid test Equity ratio Inventory turns ROA ROE

2021 2.40 1.17 0.33 4.25 6% 17%

2022 1.90 0.90 0.40 7.00 8% 20%

2023 1.60 0.75 0.55 12.00 10% 18%

While Movit appears to be solvent, with an increasing current ratio and acid test values, their equity ratio has been dropping, as have their inventory turns and return on total assets. This could be due to a drop in sales effectiveness over the period (drop in turns) and an increase in debt financing (drop in equity ratio and return on total assets). Despite the solvency, there is some concern about the degree to which Movit relies on debt and the apparent drop in return on total assets and inventory turns.

One reason for these changes could be increased borrowing coupled with lower sales volumes. Comparative income statements and balance sheets for the period would be useful in getting to the bottom of what is going on with Movit.

6.10

The answer to this question will vary depending on the imagination and creativity of the student.

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Chapter 6 - Depreciation and Financial Accounting

C. More Challenging Problems 6.11

(a) The direct effect of each strategy is as follows: Strategy (i) reduces the current assets.

Strategy (ii) reduces the current liabilities.

Strategy (iii) increases the total equity.

Current Ratio = (Current Assets)/(Current Liabilities) Strategy (ii) will increase, hence improve, this ratio.

Quick Ratio = (Quick Assets)/(Current Liabilities) = (Current Assets − Inventories)/(Current Liabilities) Strategy (i) or (ii) will increase, hence improve, this ratio.

Equity Ratio = (Total Equity)/(Total Assets) = (Total Equity)/(Current Assets + Long-Term Assets) = (Total Equity)/(Total Liabilities + Total Equity) Strategy (i), (ii), or (iii) will increase, hence improve, this ratio. (b) Assuming no other information is available, strategy (ii) may be seen as most effective since it can improve all three ratios.

6.12

XYZ should compute the acid-test ratio for more assurance about their financial liquidity.

Let C represent the amount of current assets. Then the amount of longterm assets is 2C and the amount of quick assets is 0.5C. Equity Ratio = (Total Equity)/(Total Assets)  0.45 = 68 000/(C + 2C) Solving for C, we get C = 50 370.37.

Current Ratio = (Current Assets)/(Current Liabilities) Solving for the current liabilities, we get: Current Liabilities = Current Assets/Current Ratio = 50 370.37/1.8 = 27 983.54 Acid-Test Ratio = (Quick Assets)/(Current Liabilities) = 0.5(50 370.37)/27 983.54 = 0.8989  0.9

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Chapter 6 - Depreciation and Financial Accounting

Since an acid-test ratio of 1 is considered adequate, the ratio of 0.9 indicates that XYZ’s quick assets are not enough to pay current liabilities.

XYZ should have some concern about liquidity.

6.13

The financial ratios for Milano Metals for the last three years: Current ratio Acid test ratio Equity ratio Inventory turnover ROA ROE

2020 1.09 0.20 0.239 2.63 1.2% 4.9%

2021 1.03 0.23 0.113 3.16 –4.7% –41.2%

2022 1.32 0.40 0.210 3.11 –2.4% –11.2%

The fortunes of Milano Metals are quite precarious. Their current ratio is terribly low⎯only if everything goes right will they have enough money to pay their creditors over the next year. Moreover, as indicated by the acid test ratio, the vast (and increasing) majority of their current assets is inventory, which may or may not be sold. (The fact that inventories are an increasing component of the current assets is suspicious in itself.)

The equity ratio reveals that survival of the company seems to be due to an extreme reliance on debt financing. This again adds to the company's precarious position since they could be put into receivership virtually at the whim of their debtors. About the only favourable aspect of the analysis is that the company made a profit last year, and one well in excess of industry average, after losses for the previous two years.

6.14

Current assets (cash) Fixed assets (bldgs & equip) Depreciation Income taxes Net income, operations Net income

increased by $100 000 to $119 000 decreased by $100 000 to $2 286 000 decreased increased decreased decreased

6.15

Current assets (cash) Current assets (accounts rec) Current assets (inventories) Total revenue Net revenue Income taxes Net income, operations Net income

increased by $50 000 to $69 000 increased by $50 000 to $829 000 decreased by $100 000 to $3 463 000 increased by $100 000 to $9 455 000 increased by $100 000 to $1 174 000 increased increased increased

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Chapter 6 - Depreciation and Financial Accounting

6.16

The financial ratios for Petit Ourson SA are as below: Petit Ourson’s financial ratios Current ratio Acid test Equity ratio Inventory turns ROA ROE

2020

2021

6.00 3.25 0.67 2.18 5% 7.5%

4.00 1.92 0.65 2.32 7% 11%

Industry norm 4.50 2.75 0.60 2.20 9% 15%

Petit Ourson appears to be financially healthy in that its financial ratios are similar to that of the industry norm for ratios except return on total assets.

The current and acid test ratio indicate that it should be able to handle some unexpected variation in working capital, although the 2020 figures for the current ratio and acid test are both down, and are below the industry norm. Petit Ourson relies a bit less heavily on debt financing than do its counterparts in industry (i.e., its equity ratio is somewhat higher than the industry average), but its return on total assets is several percentage points below industry norms. Though its financial position does not look precarious, Petit Ourson does not appear to be generating as large profits as the industry norms. Your advice would probably be to look for a company that is financially secure as well as having a higher return on assets. That way you are more likely to earn through either dividends or increased equity (value per share).

6.17

The answer to this question will vary depending on the imagination and creativity of the student.

6.18

The answer to this question will vary depending on the imagination and creativity of the student.

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Chapter 6 - Depreciation and Financial Accounting

Notes for Case in Point 6.1 1)

That they can make money from their investment

2)

It might be considered at the confluence of technology, fashion and home appliance industries.

Notes for Case in Point 6.2 1)

Strategies could include Dirk and Rahul not taking salaries, or taking low salaries, and working out of his parent’s bedroom until it became necessary to rent manufacturing and office space. Investors will be interested in anything that reduces cost without interfering with the effective operation of the business

2)

It is hard to know for sure without test marketing, but a typical customer would probably be young and affluent.

3)

There are a wide variety of choices, but paid advertising is probably not a good choice.

Notes for Case in Point 6.3 1)

A quick Google search can determine this. In 2019, according to the National Center for Education Statistics, there were 11.3 million female college students in the US, so a reasonable estimate for North America would be about 12.5 million. Since Dirk will first be offering his product on-line, he will be selling to a global market from the beginning.

2)

The company could, for example, have a person designated to follow the technology to ensure that the company is aware of any developments critical to its business. And, of course, there will be the possibility of customizing the product as different users give feedback on its use.

A variety of possibilities. One idea would be to use segmented pricing—a lower-featured version at a lower price and a full-featured version at a higher price.

3)

4)

A MARR significantly higher than his loan rate would be necessary. Also, he has very limited access to new funds. A MARR of 20-30% would not be too low.

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Chapter 6 - Depreciation and Financial Accounting

Notes for Case in Point 6.5 1)

Many creative answers are possible.

2)

Many creative answers are possible.

3)

One way to get consumer feedback is to include a warranty card that provides contact details for the consumer. Then, later, consumers could be contacted to see if they are satisfied, and if not, why not.

Notes for Mini-Case 6.1 1)

The financial ratios that can be calculated from the Balance Sheet are current ratio, acid test ratio and equity ratio

2)

Many creative answers are possible.

3)

Many creative answers are possible.

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Chapter 6 - Depreciation and Financial Accounting

Appendix 6A Solutions 6A.1 F = Fixed; V = Variable.

• Rental of restaurant space (F) • Purchase cost of tables, chairs and decorative items (F) • Food items (V) • Insurance (F) • Lights and heating (V) • Taxes (F) • Part-time staff salaries (V) • cleaning supplies (V) 6A.2 (a) Plant A:

Plant B:

Hours/unit 150 −5 −4 141 100 −15

Plant C:

85 100 15 115

Historical Baseline Adjustment for improved efficiency Part now purchased.

Projected hours per unit Historical Baseline less 5 parts at 3 hours per part each Projected hours per unit Historical Baseline Parts moved to Plant C from B Projected hours per unit

(b) Manufacturing Labour Costs: Plant A Plant B Plant C Total:

(c) Purchased Parts: Purchased Parts (including 5% allowance) New Purchased Parts (including 5% allowance) Total:

898 875 505 750 684 250 2 088 875

7 574 826.00 17 850.00 7 592 676.00

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Chapter 6 - Depreciation and Financial Accounting

(d) Tooling and Equipment: Tooling Test Equipment Total:

10 821.18 18 035.30 28 856.48

(e) Support Labour Costs: Manufacturing Support: Engineering Support: Total Support Costs:

1 120 500.00 531 200.00 1 651 700.00

(f) Total Costs: Manufacturing Labour Purchased Parts: Tooling and Equipment: Support Labour: Total Cost Estimate:

2 088 875.00 7 592 676.00 28 856.48 1 651 700.00 11 362 107.48

6A.3 (a) PREC FLEX RESL TEAM PMAT Sum B= Cost =

4 1 5 3 3 16 1.17 8 743 012

(b) The number of person-months required (effort) is 2.94 (130)1.17 = 874.3012 The cost estimate for this project at a cost of $10 000 per month is thus $8 743 012. (c) From the table above, a project that would have close to a linear relationship between effort and size will be one where the organization has considerable experience with this type of application, where the client has not prescribed the development process, when a thorough risk analysis has been undertaken, where the development team has a track record of working well together, and where there are clear-cut and well-understood processes for software development.

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Chapter 6 - Depreciation and Financial Accounting

(d) Diseconomies of effort in project size will occur when there are no precedents for this type of project in the organization, where the client has prescribed the development process, the project has not had a risk analysis carried out, the team has not worked together before (or has interpersonal friction), and where there is little in place in terms of process controls.

6A.4 The cost of the renovation will depend on the kind of roofing material he wants to use, as well as the quality of the siding. If he is changing the roof and siding, this will mean that he will also need to replace the fascia, soffit, trim, gutters and downpipes.

Typical cost estimates for roofing, siding, and fascia etc. are in either square feet or per linear foot depending on the item being estimated. In order to construct an estimate, then, you will need to estimate the surface area of the roof, the perimeter of the roof, the surface area of the house exterior and the perimeter of the windows.

Some research on the internet can provide some guidance on the costs.

Since there will be regional variations in prices, it would be good to know what the costs are locally. Below are some items to consider. Based on these rough estimates, you could estimate the cost to be between $55 000 and approximately $87 000, depending on the costs of the materials.

Basic Dimensions of house (ft)

L

42 ft

W

32 ft

Height of roof truss

H

Peak runs length of house

Hypotenuse of Roof

23.35 ft

17 ft

Area of Roof

980.50 ft 2

Height of each floor 9 ft, x 30 x 40 ft

Exterior Surface

10 800 ft 2

(two storeys)

Perimeter of Windows

Cost Item Removal of existing material, disposal & cleanup.

Garbage Bin

Min Unit Cost 0.5

120 lin ft

Max Unit Cost 2

Units

N

sq ft

980

490

1

475

475

Min Cost Max Cost 1 961

$475

$475

Asphalt Shingles (fibreglass base)

2

3

sq ft

1 961

3 922

5 883

Aluminium Counter Flashing

3

4

lin ft

1 200

3 600

4 800

Drip Edge

1

2

lin ft

1 200

1 200

2 400

Aluminium Gutters & downpipes

1.5

3

lin ft

1 200

1 800

3 600

Aluminum fascia

1.8

3

lin ft

1 200

2 160

3 600

Aluminum soffits

1.7

3

lin ft

1 200

2 040

3 600

Vents

85

150

per

2

170

300

Aluminum siding - R2000

3.6

5.5

sq ft.

10 800

38 880

59 400

Aluminum cladding of windows

1.8

3

lin ft

200

Total Estimated Cost:

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360

600

55 097

86 619

CHAPTER 7 Solutions to Chapter-End Problems A. Key Concepts Reasons for Replacement: 7.1

Probably not. If new, almost identical, printers are selling for $5000, CCC surely won’t get nearly as much as that for their year-old model. Since they use the printer for only special printing jobs, it likely has incurred little wear. The $20 000 is a sunk cost which will not have any bearing on a decision to replace it.

7.2

No. The overhaul is a periodic cost which increases the cost of keeping the defender for this year only. Evelyn should look at the EAC for the defender for a two- or three-year period to see if the challenger is still cheaper on an annual basis.

Capital Costs and Other Costs: 7.3

Only a), c) and h) are capital costs; b), e) g) and possibly i) are operating and maintenance costs; d), f) and i) are installation costs.

7.4

Only the data collection nodes, server and software would be included for calculating capital cost. The other costs are installation costs.

Capital cost = [(4500  5) + 6000 + (1190  5) + 1950]  0.3 = 10 920 The capital cost in the first year is $10 920.

Economic Life: 7.5

As an example calculation, for Item 1, year 4: Salvage value = 10 000(1 – 0.2)4 = 4096 Capital cost = (12 000 – 4096)(A/P, 8%, 4) + 4069(0.08) = 2714 Operating cost = 300 + 300(A/G, 8%, 4) = 300 + 300(1.4040) = 721.19 Total cost = 2714 + 721.2 = 3435.2 As can be seen from the spreadsheets below, Item 1 has an economic life of 8 years, Item 2 an economic life of 18 years, and Item 3 an economic life of 3 years.

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7.6

Item 1: Year 0 1 2 3 4 5 6 7 8 9 10

Salvage 10000 8000 6400 5120 4096 3277 2621 2097 1678 1342 1074

Installation 2000

Item 2: Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Salvage 20000 16000 12800 10240 8192 6554 5243 4194 3355 2684 2147 1718 1374 1100 880 704 563 450 360 288 231

Installation 2000

Item 3: Year 0 1 2 3 4 5 6

Salvage 30000 24000 19200 15360 12288 9830 7864

Installation 3000

EAC Capital

Operating 300 600 900 1200 1500 1800 2100 2400 2700 3000

4960 3652 3079 2714 2447 2238 2070 1930 1813 1714 EAC Capital

Operating 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000

2000 4000 6000 8000 10000 12000

EAC Total

300 444 585 721 854 983 1108 1230 1347 1461 EAC Operating

7760 6183 5382 4824 4393 4044 3756 3513 3307 3130 2978 2847 2732 2632 2544 2467 2399 2338 2284 2236 EAC Capital

Operating

EAC Operating

5260 4097 3664 3435 3301 3221 3178 3160 3161 3176 EAC Total

200 296 390 481 569 655 739 820 898 974 1048 1119 1188 1255 1319 1381 1441 1498 1554 1607

7960 6479 5772 5305 4962 4700 4494 4333 4205 4105 4026 3966 3920 3887 3863 3848 3839 3836 3838 3843

EAC Operating

11640 9275 8074 7236 6589 6066

EAC Total

2000 2962 3897 4808 5693 6553

13640 12236 11971 12044 12282 12619

(a) The capital and operating costs of the bottle capper are as follows: Year 0 1 2 3 4 5 6 7 8

Salvage 40000 35000 30000 25000 20000 15000 10000 5000 0

Maintenance 3000 4000 5000 6000 7000 8000 9000 10000

EAC Capital

EAC Maintenance

15400 12475 11327 10631 10122 9713 9365 9059

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3000 3472 3925 4359 4775 5172 5551 5913

EAC Total 18400 15947 15252 14990 14897 14885 14916 14972

Chapter 7 - Replacement Decisions

For example, equivalent annual costs for the capper if it is kept 2 years are: Book value of capper after 2 years = 40 000 – 2(40 000 – 0)/8 = 30 000 Notice that the installation costs are included in the original price paid (P), but not in the depreciable value of the asset.

EAC(capital costs) = (40 000 + 5000 – 30 000)(A/P, 12%, 2) + 30 000(0.12) = 12 475 EAC(maintenance costs) = [3000(P/F, 12%, 1) + 4000(P/F, 12%, 2)](A/P, 12%, 2) = 3472 EAC(total) = 12 475 + 3472 = 15 947

EAC ($/year)

(b) 20000 18000 16000 14000 12000 10000 8000 6000 4000 2000 0

EAC(total) EAC(capital)

EAC(maintenance)

1

2

3

4

5

6

7

Years

(c) The economic life of the bottle capper is 6 years.

Defender and challenger are identical: 7.7

The economic life of the pump is 8 years.

Years between replacement 1 2 3 4 5 6 7 8 9 10 11 12

EAC Capital 4100 3290 2876 2588 2365 2186 2038 1915 1811 1723 1647 1582

EAC Operating 500 643 781 914 1043 1167 1286 1401 1512 1618 1719 1817

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EAC Total 4600 3933 3657 3502 3408 3353 3325 3317 3323 3340 3366 3399

8

Chapter 7 - Replacement Decisions

Example calculation: Salvage value after 10 years can be calculated as: 10 000(1 – 0.2) 10 = 1074 EAC(capital costs) = (10 000 + 1000 – 1074)(A/P, 10%, 10) + 1074(0.1) = 9926(0.16275) + 107.4 = 1722.83 EAC(operating costs) = 500 + 300(A/G, 10%, 10) = 500 + 300(3.7255) = 1617.64 EAC(total) = 3340.47 7.8

Assumption: the new pump can be repeated. (a) Example calculation: Old pump annual costs this year are EAC = (1000 – 1000)(A/P, 10%, 1) + 1000(0.1) + 2500 = 2600 Year 1 2 3 4 5 6 7

EAC Capital 100 100 100 100 100 100 100

EAC Operating 2500 2690 2875 3052 3224 3389 3549

EAC Total 2600 2790 2975 3152 3324 3489 3649

Until the fifth year, the old pump will have lower average costs than the new pump will have at its economic life ($3317). Replace the old pump after four years. (b) For the first year: EAC = (1000 – 1000)(A/P, 10%, 1) + 1000(0.1) + 3500 = 3600 The old pump should be replaced immediately.

One Year Principle: 7.9

No – the capital costs are high compared to the operating and maintenance costs.

7.10

No – the operating and maintenance costs are decreasing.

7.11

No – the operating and maintenance costs do not have an even pattern from year to year.

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7.12

Technically, no - the operating and maintenance costs do not have an even pattern from year to year. However, on the assumption that since the asset is inaccessible, replacement would only actually be considered on a four-year basis, one could use the four-year period (with an approximately $16 000 maintenance cost) as an evaluation basis, instead of one-year, and the approach would be valid.

B. Applications 7.13

This requires calculating the service life of the car.

EAC (capital, first year) = [(15 000 – (0.7)(15 000)](A/P, 8%, 1) + (0.7)(15 000)(0.08) = 4500(1.08) + 10500(0.08) = 5700 EAC(operating, first year) = 0 EAC(total, first year) = $5700 EAC (capital, two years) = [15 000 – (0.72)(15 000)](A/P, 8%, 2) + (0.72)(15 000)(0.08) = 7650(0.56077) + 7350(0.08) = 4877.89 EAC(operating, two years) = 0 EAC(total, two years) = $4877.89 EAC (capital, three years) = [15 000 – (0.73)(15 000)](A/P, 8%, 3) + (0.73)(15 000)(0.08) = 9855(0.38803) + 5145(0.08) = 4235.64 EAC(operating, three years) = 2500(A/F, 8%, 3) = 2000(0.30803) = 770.08 EAC(total, three years) = $5005.72 Gerry should get a new car every two years.

7.14

EAC (capital, first year) = [15 000 – (0.7)(15 000)](A/P, 8%, 1) + (0.7)(15 000)(0.08) = 4500 (1.08) + 10500(0.08) = 5700 EAC(operating, first year) = 0 EAC(total, first year) = $5700 EAC (capital, two years) = [15 000 – (0.72)(15 000)](A/P, 8%, 2) + (0.72)(15 000)(0.08) = 7650(0.56077) + 7350(0.08) = 4877.89 EAC(operating, two years) = 0 EAC(total, two years) = $4877.89

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Chapter 7 - Replacement Decisions

EAC (capital, three years) = [15 000 – (0.73)(15 000)](A/P, 8%, 3) + (0.73)(15 000)(0.08) = 9855(0.38803) + 5145(0.08) = 4235.64 EAC(operating, three year) = 1500(A/F, 8%, 3) = 1500(0.30803) = 462.05 EAC(total, three years) = $4697.67 EAC (capital, four years) = [15 000 – (0.74)(15 000)](A/P, 8%, 4) + (0.74)(15 000)(0.08) = 11398.50(0.30192) + 3601.50(0.08) = 3729.56 EAC(operating, four year) = [1500(F/P, 8%, 1) + (1500)(1.5)](A/F, 8%, 4) = [1500(1.08) + 2250](0.22192) = 1098.50 EAC(total, four years) = $4828.06 Gerry should now get a new car every three years.

7.15

(a) (b) (c) (d) (e)

No - the capital costs are low but there is an uneven pattern of costs.

Yes - the capital costs are low and there is an even pattern of costs.

Yes - the capital costs are low and there is an even pattern of costs.

Yes - the capital costs are low and there is an even pattern of costs.

No - there is an even pattern of costs, but the capital costs are not low.

7.16

(a) The capital and operating costs of the roller conveyor are as follows: Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Salvage 100000 75000 56250 42188 31641 23730 17798 13348 10011 7508 5631 4224 3168 2376 1782 1336

Maintenance 6000 7200 15640 10368 12442 21930 17916 21499 32799 30959 37150 51581 53497 64196 84035

EAC Capital

EAC Maintenance

57000 42357 35508 31039 27769 25246 23242 21618 20284 19176 18248 17463 16797 16226 15735

6000 6571 9311 9539 10014 11559 12229 13039 14495 15528 16694 18326 19760 21348 23321

EAC Total 63000 48929 44819 40578 37783 36805 35470 34657 34778 34704 34942 35789 36557 37574 39056

For example, equivalent annual costs for the conveyor if it is kept after 2 years: Book value of conveyor after 2 years = 100 000(1 – 0.25)2 = 56 250 Notice that the installation costs are included in the original price paid (P), but not in the depreciable value of the asset.

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Chapter 7 - Replacement Decisions

EAC(capital costs) = (100 000 + 20 000 – 56 250)(A/P, 10%, 2) + 56 250(0.10) = 42 357 EAC(maintenance costs) = [6000(P/F, 10%, 1) + 7200(P/F, 10%, 2)](A/P, 10%, 2) = 6571 EAC(total) = 42 357 + 6571 = 48 928

EAC($/year)

(b) 70000 60000 50000 40000 30000 20000 10000 0

EAC(total) EAC(maintenance) EAC(capital)

1

3

5

7

9

11 13 15

Years

(c) The economic life of the conveyor is 8 years.

7.17

First, the EAC of keeping the robot as is for the duration of the contract is found by the EAC(total costs) for an additional life of 5 years.

Historical costs are no longer relevant for the computation of the EAC(capital costs) and, hence, they do not include the original installation costs of the robot.

3 year old Defender: Additional years Salvage 0 153600 1 122880 2 98304 3 78643 4 62915 5 50332

Maintenance 50000 55000 60500 66550 73205

EAC Capital

EAC Maintenance

53760 48759 44626 41201 38356

50000 52326 54680 57057 59452

EAC Total 103760 101085 99305 98258 97808

From the computations we can see that the EAC(total costs) of keeping the robot as is over the next 5 years is $97 808.

If the robot is moved, a $25 000 moving cost is incurred, but the operating and maintenance costs are reduced for the five-year duration of the contract. For this option, the EAC(total costs) is minimized by keeping the moved robot for the duration of the contract. The minimum EAC(total costs) of the challenger over the five years is $93 376. The robot should be moved.

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Chapter 7 - Replacement Decisions Challenger (Move Robot): Additional years Salvage 0 153600 1 122880 2 98304 3 78643 4 62915 5 50332

Maintenance 40000 44000 48400 53240 58564

EAC Capital

EAC Maintenance

82510 64137 55575 49958 45814

40000 41860 43744 45645 47561

EAC Total 122510 105997 99319 95603 93376

Since we know that the robot should be moved, the question is when? The cost of keeping the defender one more year is $103 760 (see table on defender). For two years, the cost is $101 085. We see that there is no number of additional years in the five-year horizon for which the defender is less expensive to keep. Therefore, the robot should be moved immediately.

7.18

In general, the options available to BB are to: 1) Replace the current robot at the end of its economic or useful life with a stream of new robots, or 2) Move the robot and then replace it with a stream of new robots at the end of the moved robot’s new economic life. (a) The EAC(total costs) for keeping the existing robot for the rest of its useful life is shown in the table below. The table shows that the remaining economic life is 5 years for a minimum EAC(total costs) of $97 808.

3 year old Defender: Additional years Salvage 0 153600 1 122880 2 98304 3 78643 4 62915 5 50332 6 40265 7 32212 8 25770 9 20616

Maintenance 50000 55000 60500 66550 73205 80526 88578 97436 107179

EAC Capital

EAC Maintenance

53760 48759 44626 41201 38356 35987 34009 32352 30962

50000 52326 54680 57057 59452 61859 64274 66689 69102

EAC Total 103760 101085 99305 98258 97808 97846 98282 99042 100064

(b) The EAC(total costs) of moving the robot and operating it to the end of its economic life is $91 275 (another 8 years).

Moving the existing robot (challenger): Additional years Salvage Maintenance 0 153600 1 122880 40000 2 98304 44000 3 78643 48400 4 62915 53240

EAC Capital

EAC Maintenance

82510 64137 55575 49958

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40000 41860 43744 45645

EAC Total 122510 105997 99319 95603

Chapter 7 - Replacement Decisions Moving the existing robot (challenger): Additional years Salvage Maintenance 5 50332 58564 6 40265 64420 7 32212 70862 8 25770 77949 9 20616 85744

EAC Capital 45814 42593 40018 37924 36202

EAC Maintenance 47561 49487 51419 53352 55281

EAC Total 93376 92080 91437 91275 91483

(c) The EAC(total costs) of purchasing and installing a new robot are below.

Defender, when new: Year Salvage 0 300000 1 240000 2 192000 3 153600 4 122880 5 98304 6 78643 7 62915 8 50332 9 40265 10 32212 11 25770 12 20616

Maintenance 40000 40000 40000 40000 44000 48400 53240 58564 64420 70862 77949 85744

EAC Capital

EAC Maintenance

162500 125988 109059 97984 89830 83499 78441 74331 70952 68152 65816 63857

40000 40000 40000 40000 40593 41485 42547 43714 44948 46224 47527 48845

EAC Total 202500 165988 149059 137984 130424 124984 120988 118045 115900 114376 113343 112702

(d) If a new robot is purchased, it should be kept for its full useful life of 12 years, as this is its economic life with EAC(total costs) = $112 702.

For BB, the EAC of installing a new robot is higher than the EAC for both continuing with the current robot or moving the existing robot and using them to the end of their useful lives. BB should avoid replacement for as long as possible, keeping their costs as low as possible in the meantime. This reduces to the question of whether or not BB should move the current robot, and if so, when.

Since the economic life of the current robot has an EAC(total costs) of $97 808 and the economic life of the moved robot has an EAC(total costs) of $91 275, the robot should be moved. We now need to answer the question of when it should be moved.

Looking at the cost of keeping the defender one more, two more, etc. years, we see that there is no additional life for the defender for which the EAC of that time is less than the EAC associated with the economic life of the challenger (moving the robot). Hence, the current robot should be moved immediately. Furthermore, it should be kept in operation until the end of its useful life.

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7.19

First, it is worth noting that there is a challenger available, and moreover, even if there was not, the furnace is necessary and should not be abandoned. No information is given about subsequent challengers, so it is reasonable to assume that subsequent challengers will be the same as the current challenger.

We do not know the maintenance costs for the gas furnace after the fiveyear guarantee period. However, it’s reasonable to assume that the economic life of a furnace is longer than five years, so that the annual costs over a five-year study will be no less than the annual costs over the economic life.

The salvage value of the gas furnace at the end of five years can be calculated as: BVsl(5) = 4500 – 5(4500 – 500)/10 = 4500 – 5(400) = 2500 The capital cost for the gas furnace is calculated as follows (using the capital recovery formula): Agas = (5000 – 2500)(A/P, 10%, 5) + 2500(0.1) = 2500(0.2638) + 250 = 909.50 Since there are no maintenance costs, the total cost for the new furnace will be about $910 per year. The energy savings of $500 will be included as an operating cost for the old furnace.

Capital cost for the old furnace for the next year is: Acap = (500 – 500)(A/P, 10%, 1) + 500(0.1) = 50 Note that it is not zero since $500 is being tied up in the old furnace.

Operating costs will be $300 for the maintenance contract, $200 for parts, and $500 in extra energy costs.

Po&m = 200 + 300 + 500 = 1000 Total cost for the next year will be $1050 for the defender.

By observation, the annual costs for the defender will only increase since the parts charge increases every year and the other costs remain the same per year.

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Chapter 7 - Replacement Decisions

Since the annual costs for the new furnace are more than $140 less than the old one, even if the new furnace is replaced every five years, Nico should replace his furnace immediately with a new gas furnace.

7.20

(a) We can approximate the cost of using the car by getting the annual cost and dividing by 12.

This implies that all costs are incurred at the end of each year. This will be an understatement of the monthly car cost since, in fact, some of the costs are incurred during the year. If we got the car costs on a monthly basis, we would be overstating the car costs since some of the car costs are incurred less frequently than once a month. The following table shows annual cost for the car.

End of year 4 5 6

Salvage 5900 5015 4263

EAC Capital

Maintenance 2670 2870

EAC Maintenance

1475 1370

EAC Total

2670 2765

Annual Cost/12

4145 4135

345.42 344.58

A conservative estimate of car costs is much higher than the cost of a bus pass. (b) There are many benefits of having a car that are not reflected in this calculation. These may more than offset the higher costs of having a car. Therefore, we cannot say which option is better on the basis of the information given.

7.21

(a) The required spreadsheet is as follows: Defender, when new: Year Salvage 0 100000 1 85000 2 72250 3 61413 4 52201 5 44371 6 37715 7 32058 8 27249 9 23162 10 19687 11 16734 12 14224 13 12091 14 10277 15 8735

Maintenance 15000 20000 25000 30000 45000 20000 25000 30000 35000 50000 25000 30000 35000 40000 45000

EAC Capital

EAC Maintenance

30000 27907 26112 24573 23251 22115 21139 20300 19578 18956 18420 17958 17559 17215 16918

15000 17326 19536 21631 25097 24515 24559 24955 25554 26758 26685 26800 27038 27358 27729

EAC Total 45000 45233 45648 46204 48348 46630 45698 45255 45131 45713 45105 44757 44598 44574 44647

Note that the EAC(total costs) is no longer convex over the life of the turbine. The lowest total costs occur with a replacement interval of 14 years for an EAC(total costs) of $44 574.

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Chapter 7 - Replacement Decisions

(b) A spreadsheet similar to that of part (a), except with a six-year overhaul cycle gives: Defender, when new (6-year overhaul cycle): Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Salvage 100000 85000 72250 61413 52201 44371 37715 32058 27249 23162 19687 16734 14224 12091 10277 8735

Maintenance

EAC Capital

15000 20000 25000 30000 35000 50000 20000 25000 30000 35000 40000 55000 25000 30000 35000

EAC Maintenance

30000 27907 26112 24573 23251 22115 21139 20300 19578 18956 18420 17958 17559 17215 16918

15000 17326 19536 21631 23614 26628 26029 25954 26195 26629 27178 28137 28046 28094 28240

EAC Total 45000 45233 45648 46204 46865 48744 47169 46254 45773 45585 45598 46095 45605 45310 45158

The minimum EAC is with a 15-year life, but the total EAC is $45 158, more than that for the 5-year overhaul cycle. Ener-G should not use a 6year overhaul cycle.

7.22

(a) The present worth of costs for the defender are shown in the following table.

Defender: Additional years 0 1 2 3

Salvage 1200 600 300 150

Operating 20000 20500 21013

PW Capital

PW Operating

664 961 1093

17857 34200 49156

PW Total 18521 35160 50249

(b) The challenger’s costs for three years are shown in the following table.

Challenger: Year 0 1 2 3

Salvage 20000 14000 9800 6860

Operating 13875 14361 14863

PW Capital

PW Operating

12500 17188 20117

12388 23837 34416

PW Total 24888 41024 54533

(c) We want to find the salvage value, S, such that PW(defender cost) = PW(challenger operating cost) + challenger(first cost) – S(P/F, 12%, 3) Solving this we get: S = $12 879

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Chapter 7 - Replacement Decisions

Alternate method: We find the difference between the present worths of the two projects, and project it to the end of three years: (54 533.15 – 50 249.13)(F/P, 12%, 3) = (4284.02)(1.404928) = 6018.74 Since the challenger has the higher PW of costs, adding $6018.74 to its salvage value after three years will make its PW equal to the defender’s: 6018.74 + 6860.00 = 12 878.74 The challenger’s salvage value would have to be $12 879 to equalize the two projects’ present worths.

7.23

(a) The economic life of the challenger is shown in the following table.

We see that the economic life is 11 years. The minimum equivalent annual cost is about $20 144 per year.

Challenger: Year Salvage 0 20000 1 14000 2 9800 3 6860 4 4802 5 3361 6 2353 7 1647 8 1153 9 807 10 565 11 395 12 277

Operating 13875 14361 14863 15383 15922 16479 17056 17653 18271 18910 19572 20257

PW Capital 12500 17188 20117 21948 23093 23808 24255 24534 24709 24818 24886 24929

PW Operating 12388 23837 34416 44192 53227 61576 69291 76421 83009 89098 94724 99924

PW Total 24888 41024 54533 66141 76320 85384 93546 100955 107718 113916 119611 124853

EAC Total 27875 24274 22705 21776 21172 20768 20498 20323 20216 20161 20144 20156

(b) We need to know the equivalent annual cost over its economic life for a new copy of the defender to decide whether we should replace the current defender with the challenger or with a new copy of the defender. The economic life and the equivalent annual cost for a new copy of the defender are shown below.

Defender when new: Year Salvage 0 15000 1 9846 2 6464 3 4243 4 2785 5 1828 6 1200 7 600 8 300 9 150 10 150 11 150 12 150 13 150

Operating 17250 17681 18123 18576 19041 19517 20005 20505 21017 21543 22081 22633 23199

PW Capital 8709 12347 14480 15730 16463 16892 17229 17379 17446 17452 17457 17461 17466

PW Operating 15402 29497 42397 54203 65007 74895 83944 92225 99804 106741 113089 118898 124215

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PW Total 24110 41845 56877 69933 81469 91787 101172 109604 117250 124192 130545 136359 141680

EAC Total 27004 24759 23681 23024 22600 22325 22169 22064 22005 21980 21986 22013 22056

Chapter 7 - Replacement Decisions

We see that new copies of the defender have economic lives of 10 years. The equivalent annual cost over this life is about $21 980. This is greater than the cost for a challenger. Therefore, we should replace the current defender with a challenger. (c) The current defender should be replaced immediately. The equivalent annual costs for any feasible life are greater than the minimum equivalent annual costs.

Defender at 6 Years Old: Additional years Salvage 0 1200 1 600 2 300 3 150

7.24

PW Capital

Operating 20000 20500 21013

PW Operating

664 961 1093

PW Total

17857 34200 49156

EAC Total

18521 35160 50249

20744 20804 20921

(a) There are 10 alternatives: Keep defender for X additional years 9 8 7 6 5 4 3 2 1 0

Then use challenger for Y years 0 1 2 3 4 5 6 7 8 9

EAC defender 5562 5497 5439 5391 5359 5349 5359 5407 5500 0

EAC challenger 0 5700 5495 5349 5249 5186 5150 5137 5140 5156

Weighted average 5562 5520 5451 5377 5310 5258 5220 5197 5180 5156

(b) The alternative which has the least weighted average of equivalent annual costs is to replace the defender immediately and use the challenger for nine years.

7.25

Note that it is obvious that the propane model is inherently better.

First find the EAC for the service life of the challenger.

New propane model: Year Salvage 0 58000 1 40000 2 32000 3 25600 4 20480 5 16384 6 13107 7 10486 8 8389 9 6711

Maintenance 10000 11200 12544 14049 15735 17623 19738 22107 24760

EAC Capital

EAC Maintenance

22640 17140 14620 12966 11734 10760 9965 9304 8747

10000 10577 11183 11819 12487 13187 13921 14691 15497

EAC Total 32640 27717 25803 24785 24220 23946 23886 23995 24244

The economic life for the new forklifts is seven years at an EAC of about $23 886. Compare this with the cost of the electric forklifts for the next year:

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Chapter 7 - Replacement Decisions 2 year old Defender: Additional years Salvage 0 10000 1 8000 2 6400 3 5120 4 4096 5 3277 6 2621 7 2097 8 1678 9 1342

Maintenance

EAC Capital

20000 22400 25088 28099 31470 35247 39476 44214 49519

EAC Maintenance

2800 2531 2303 2110 1946 1806 1686 1582 1493

EAC Total

20000 21154 22366 23638 24973 26374 27842 29381 30994

22800 23685 24669 25748 26919 28179 29528 30964 32487

EAC(capital) = (10 000 – 10 000  0.8)(A/P, 8%, 1) + (10 000  0.8)(0.08) = 2000(1.08) + 8000(0.08) = 2800 EAC(oper.) = 20 000 EAC(total) = 22 800 The electric forklifts’ average cost will not exceed that of the propane forklifts until the third year. Conclusion: Keep the electric forklifts, at least for another year.

7.26

The trade-in allowance of $14 000 should be treated as a $4000 reduction in the price of the new propane forklift. The modified costs are: New propane model: Year Salvage 0 54000 1 40000 2 32000 3 25600 4 20480 5 16384 6 13107 7 10486 8 8389 9 6711

Maintenance 10000 11200 12544 14049 15735 17623 19738 22107 24760

EAC Capital

EAC Maintenance

18320 14897 13068 11759 10732 9894 9197 8608 8107

10000 10577 11183 11819 12487 13187 13921 14691 15497

EAC Total 28320 25474 24251 23578 23218 23081 23118 23299 23604

Since the costs of keeping the electric forklift for an additional year is $22 800, less than the minimum EAC(total costs) for the propane model, Chatham Automotive should not replace the electric forklifts now.

7.27

(a) Sample calculations: EAC(capital, 2 additional years) = (P – S)(A/P, 10%, 2) + Si = (33 000 – 25 555.20)(0.57619) + 25 555.20(0.1)  6845.00 EAC(operating, 2 additional years) = [3400(F/P, 10%, 1) + 3900](A/F, 10%, 2) = [3400(1.1) + 3900](0.47619)  3638 The summary of the calculation results: 124 Copyright © 2022 Pearson Canada Inc.

Chapter 7 - Replacement Decisions Year 0 1 2 3

Salvage value 33 000 29 040 25 555 22 488

Op/Maint cost

Capital EAC

Operating EAC

Total EAC

3400 3900 4400

7260 6845 6476

3400 3638 3868

10 660 10 483 10 344

A new joint former has to be less than about $10 344. (b) Yes, replace the old joint former with the new one since the EAC for the economic life of the new one is lower than any of the EACs for the old one. Replace it now. (c) Since the EAC for the old asset appears to be decreasing, there may be a period for which its EAC is lower than that of the challenger. It is necessary to check if replacement should occur more than three years from now.

7.28

Reading from the first table, the challenger’s economic life is five years, with an EAC of $39 452.

Year 0 1 2 3 4 5 6 7 8 9

Salvage 90000 72000 57600 46040 36864 29491 23593 18874 15099 12080

O&M

Salvage 49000 36500 19875 15656 6742

O&M

12000 14400 17280 20736 24883 29860 35832 42998 51598

EAC Capital 31500 28570 26159 24141 22474 21086 19927 18957 18142

EAC O&M 12000 13116 14315 15601 16978 18449 20020 21694 23476

EAC Total 43500 41686 40475 39742 39452 39536 39947 40651 41618

For the defender: Year 0 1 2 3 4

17000 21320 26806 33774

EAC Capital 19850 20897 16952 15813

EAC O&M 17000 19009 21255 23762

EAC Total 36850 39906 38207 39575

Keep the defender for three more years before replacing it by the challenger. Note that the defender becomes undesirable over a two-year period, but for a three-year period is again preferred.

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Chapter 7 - Replacement Decisions

7.29

Example Calculations (row 2): EAC(Capital Cost) = (40 970 − 22 310)(A/P, 12%, 2) + (22 310)(0.12) EAC(Operating & Maintenance) = [16 500(P/F, 12%, 1) + 20 690(P/F, 12%, 2)](A/P, 12%, 2) EAC(Overhaul) = 14 000(A/P, 12%, 2) Life (years) 0 1 2 3 4

Salvage Value 55000 40970 22310 17574 7568

Operating& Maintenance 16500 20690 26013 32775

EAC Op&M 14348 15645 17104 18739

EAC Capital 16500 18449 20627 23060

EAC Overhaul 14000 16100 8612 6132 4904

EAC Total 54880 41885 31469 32604

The old cutter should be overhauled since the EAC corresponding to the economic life is less with an overhaul ($31 469) than without one ($38 207). The new cutter should replace the old after four years since the EAC for the old cutter after four years is still less ($32 604) than the EAC corresponding to the economic life for the new cutter ($39 452).

7.30

Overhaul in three years: Life (years) 0 1 2 3 4 5 6 7 8 9 10 11 12 13

Salvage Value 10000 8000 6400 5120 4096 3277 2621 2097 1678 1342 1074 859 687 550

Operating & Maintenance 500 800 1100 500 800 1100 1400 1700 2000 2300 2600 2900 3200

EAC Op&M 500 643 781 720 733 781 846 921 1000 1082 1164 1245 1325

EAC Capital 4100 3290 2876 2588 2365 2186 2038 1915 1811 1723 1647 1582 1526

EAC Overhaul

237 198 173 154 141 130 122 116 110 106

EAC Total 4600 3933 3657 3545 3297 3139 3039 2977 2942 2927 2927 2938 2957

Overhaul in 5 years: Life (years) 0 1 2 3 4 5 6 7 8 9 10 11 12

Salvage Value 10000 8000 6400 5120 4096 3277 2621 2097 1678 1342 1074 859 687

Operating & Maintenance 500 800 1100 1400 1700 500 800 1100 1400 1700 2000 2300

EAC Op&M 500 643 781 914 1043 973 954 967 999 1043 1095 1151

EAC Capital 4100 3290 2876 2588 2365 2186 2038 1915 1811 1723 1647 1582

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EAC Overhaul

143 128 116 108 101 96 91

EAC Total 4600 3933 3657 3502 3408 3301 3120 2999 2918 2867 2838 2824

Chapter 7 - Replacement Decisions Life (years) 13 14 15

Salvage Value 550 434 352

Operating & Maintenance 2600 2900 3200

EAC Op&M 1210 1271 1331

EAC Capital 1526 1477 1435

EAC Overhaul 87 84 81

EAC Total 2824 2832 2848

Example Calculations (row 5 for overhaul in 3 years): EAC(Capital Cost) = (10 000 + 1000 − 10 000  0.85)(A/P, 10%, 6) + (10 000  0.85)(0.1) EAC(Operating & Maintenance) = [500(P/F, 10%, 1) + 800(P/F, 10%, 2) + 1100(P/F, 10%, 3) + 500(P/F, 10%, 4) + 800(P/F, 10%, 5)](A/P, 10%, 5) EAC(Overhaul) = 1000(P/F, 10%, 4)(A/P, 10%, 5) The water pump should be overhauled since the EAC corresponding to the economic life is less with an overhaul ($2927 or $2824) than without one ($3317). Further, it should be overhauled in five years.

C. More Challenging Problems 7.31

The economic life is when the slopes of the two cost functions are equal.

The slopes are the derivatives of the cost functions.

500n(2n-1) = 10000n(0.8)n-1 2n-1 = 20(0.8)n-1 (n – 1)ln(2) = ln(20) + (n – 1)ln(0.8) ln2 = ln(20)/(n – 1) + ln(0.8) n = ln(20)/[ln(2) – ln(0.8)] + 1 = 4.27 The economic life of the asset is about 4.27 years.

7.32

(a) The economic life is 14 years. (b) The equivalent annual cost over 14 years is $8138 per year.

Costs and Salvage Values for Various Lives: MARR = 0.12 PW PW EAC Year Capital Operating Capital 0 5000 1 10714 2679 12000 2 14796 5250 8755 3 17711 7717 7374 4 19794 10086 6517 5 21281 12359 5904 6 22344 14541 5435 7 23103 16635 5062 8 23645 18646 4760 9 24032 20575 4510 10 24309 22427 4302 11 24506 24204 4127

EAC Operating

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3000 3106 3213 3321 3429 3537 3645 3753 3861 3969 4076

EAC Total 15000 11861 10587 9837 9332 8971 8707 8513 8372 8271 8204

Chapter 7 - Replacement Decisions Costs and Salvage Values for Various Lives: MARR = 0.12 PW PW EAC Year Capital Operating Capital 12 24647 25910 3979 13 24748 27548 3853 14 24820 29120 3745 15 24871 30628 3652 16 24908 32076 3572 17 24934 33466 3502

EAC Operating 4183 4289 4393 4497 4599 4701

EAC Total 8162 8141 8138 8149 8171 8203

(c) The economic life is now only 12 years. (d) The equivalent annual cost is now only $7152.

Costs and Salvage Values for Various Lives: MARR = 0.05 PW PW EAC EAC Year Capital Operating Capital Operating 0 5000 1 9762 2857 10250 3000 2 13390 5782 7201 3110 3 16154 8777 5932 3223 4 18260 11843 5150 3340 5 19865 14982 4588 3461 6 21088 18196 4155 3585 7 22019 21487 3805 3713 8 22729 24855 3517 3846 9 23270 28304 3274 3982 10 23682 31835 3067 4123 11 23996 35450 2889 4268 12 24235 39152 2734 4417 13 24417 42941 2599 4571 14 24556 46821 2481 4730 15 24662 50792 2376 4893 16 24742 54859 2283 5062 17 24804 59022 2200 5235 18 24850 63285 2126 5414

EAC Total 13250 10311 9155 8490 8049 7740 7519 7362 7256 7190 7157 7152 7171 7211 7269 7345 7435 7540

(e) Decreasing the MARR reduces the capital cost, thus decreasing the total costs. Also, since the capital costs decrease with increased life, and are generally smaller with smaller MARR, they will form a minimum total cost sooner; thus, the economic life will be shorter.

7.33

Note that in the table below, the true salvage value at time 0 is actually $25 000.

The spreadsheet uses the $23 000 shown as the initial cost in the capital cost formula. The $23 000 is composed of the $25 000 minus the $2000 net bonus. The salvage values for year 1 onwards are based on the true initial salvage value of $25 000, not the $23 000.

New equipment: Year Salvage 0 23000 1 20000 2 16000 3 12800 4 10240 5 8192

Maintenance 6000 6600 7260 7986 8785

EAC Capital

EAC Maintenance

4840 5205 4982 4672 4364

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6000 6288 6588 6898 7220

EAC Total 10840 11494 11570 11570 11584

Chapter 7 - Replacement Decisions New equipment: Year Salvage 6 6554 7 5243 8 4194 9 3355

Maintenance 9663 10629 11692 12862

EAC Capital 4082 3830 3608 3413

EAC Maintenance 7553 7898 8254 8623

EAC Total 11635 11728 11862 12036

Note that there is a minimum cost of $10 840 at year 1. This comes about due to the trade-in bonus of $2000 that only occurs in the first year. A more credible service life is three or four years, with an annual cost of $11 570.

The capital costs of using the defender for one more year are: EAC(capital, 1 more year) = (10 000 – 8 000)(A/P, 8%, 1) + 8000(0.08) = 2000(1.08) + 640 = 2800 The operating and maintenance costs are 9000 for a total: EAC(keep defender 1 more year) = 2800 + 9000 = 11 800 This is higher than the EAC of the challenger at its economic life (either 3 or 4 years) with an EAC of 11 570. Now we ask the question if there is a life for the defender which will have a lower EAC than the challenger at its economic life.

The EAC of capital costs for keeping the defender two more years is: EAC(capital, keep 2 more years) = (10 000 – 6400)(A/P, 8%, 2) + 6400(0.08) = 2531 The EAC of operating and maintenance costs are: EAC(op. and maint., keep 2 more years) = 9000 + 0.1  9000(P/F, 8%, 2)(A/P, 8%, 2) = 9000 + 900(0.85734)(0.56077) = 9433 The total EAC for keeping the defender for two more years is then: EAC(keep 2 more years) = 2531 + 9433 = 11 964 Given that capital costs are going down at a slower rate than the operating and maintenance costs are increasing, the costs of keeping the defender for three more years, four more years, etc., will only increase. Thus, there is no additional life for keeping the defender such that the EAC of its costs will be lower than the challenger at its economic life. Hence, we should replace immediately.

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Chapter 7 - Replacement Decisions

7.34

(a) There are 22 possible combinations. They are shown in the table.

1 2 3 4 5 6

Year defender challenger 1 challenger 2 defender challenger 1 challenger 2 defender challenger 1 challenger 2 defender challenger 1 challenger 2 defender challenger 1 challenger 2 defender challenger 1 challenger 2

1 x

2 x

3 x

4 x

x

x

x

x

x

x

x

x

x

defender challenger 1 challenger 2 defender challenger 1 challenger 2 defender challenger 1 challenger 2

x

5 x

6

7

8

9

10

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x x x

 13 14 15

 22

defender challenger 1 challenger 2

x x

x

x

x

x

x

x

x x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x x

x

x x

x

x

x

x

x

x

x

(b) The best combination is number 4. Wait two years and install the second challenger.

The defender has lower costs than the first challenger over the next year. See the first row of the table labelled “Defender Now” and the row for the minimum EAC in the seventh year for the first challenger.

Therefore, all combinations starting with number 14, that have the first challenger replacing the defender now, are eliminated.

The costs for the defender in the second year are higher than the costs for the first challenger for a life of at least three years. This means that we must consider combinations in which we replace the defender with the first challenger after one year. We would need to keep the first challenger for at least three years for this to be worthwhile. This eliminates combinations 12 and 13 that have the first challenger replace the defender in the second year and being used for one or two years only.

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Chapter 7 - Replacement Decisions

We next consider combinations that have the first challenger for at least three years. We shall also compare these with combination 4 that does not include the first challenger. The present worth of combination 11 that has the first challenger for three years starting in the second year is given by: PW(cost for defender for 1 year) + PW(costs for 1st challenger for 3 years)(P/F, 10%, 1) + PW(costs for 2nd challenger for 6 years)(P/F, 10%, 4) The present worth of combination 10 that has the first challenger for four years is obtained in a similar manner. The present worth of combination 4 that has the second challenger replace the defender after two years is obtained similarly. The present worths of costs for combinations 4, 10 and 11 are: PW(costs for combination 4) = $126 004 PW(costs for combination 10) = $138 629 PW(costs for combination 11) = $135 660 We see that keeping the first challenger for four years has higher costs than keeping it for three years. We infer from this that we need not compute the present worth of costs that have the first challenger last longer than three years. As well, going directly to the second challenger after two years has lower cost than either combination involving the first challenger. It is apparent that we need not consider combinations that include the first challenger.

We must consider combinations that have the second challenger come in after more than two years. We see that the second challenger has lower costs than any incremental years for the defender after two years. See the tables below. We conclude, therefore, that combination 4 has the lowest cost.

Defender Now: Additional Years 0 1 2 3 4 5

Salvage 4000 3000 2000 1000 500 500

Operating 20000 25000 30000 35000 40000

Defender's Cost after 1 More Year: Additional Years Salvage Operating 0 3000 1 2000 25000 2 1000 30000 3 500 35000 4 500 40000

PW Capital 1273 2347 3249 3658 3690

PW Capital 1182 2174 2624 2658

PW Operating 18182 38843 61382 85288 110125

PW Operating 22727 47521 73817 101137

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PW Total 19455 41190 64631 88946 113814

PW Total 23909 49694 76441 103796

EAC Total 21400 23733 25989 28060 30024

EAC Total 26300 28633 30738 32745

Chapter 7 - Replacement Decisions

Defender after 2 More Years: Additional Years Salvage 0 2000 1 1000 2 500 3 500 First Challenger: Year Salvage 0 25000 1 20000 2 16000 3 12800 4 10240 5 8192 6 6554 7 5243 8 4194 9 3355 10 2684 Second Challenger: Year Salvage 0 25000 1 20000 2 16000 3 12800 4 10240 5 8192 6 6554 7 5243 8 4194 9 3355 10 2684

7.35

PW Capital

Operating 30000 35000 40000

1091 1587 1624

PW Capital

Operating 16800 17640 18522 19448 20421 21442 22514 23639 24821 26062

Operating

PW Operating

PW Total

27273 56198 86251

PW Operating

EAC Total

28364 57785 87875 PW Total

31200 33295 35336 EAC Total

11818 16777 20383 23006 24913 26301 27310 28043 28577 28965

15273 29851 43767 57050 69730 81833 93386 104414 114941 124989

27091 46628 64150 80056 94643 108134 120696 132457 143518 153954

29800 26867 25796 25255 24967 24828 24792 24828 24920 25055

PW Capital

PW Operating

PW Total

EAC Total

12000 12600 13230 13892 14586 15315 16081 16885 17729 18616

11818 16777 20383 23006 24913 26301 27310 28043 28577 28965

10909 21322 31262 40750 49807 58452 66704 74581 82100 89278

22727 38099 51645 63756 74721 84753 94014 102625 110677 118243

25000 21952 20767 20113 19711 19460 19311 19236 19218 19243

(a) The costs of keeping the car for the additional years are as follows.

If you do not paint the car, you should keep it an additional five years before replacing it. The EAC(total costs) is $5880.

Car, when two years old: Additional years Salvage 0 12000 1 9000 2 6750 3 5063 4 3797 5 2848 6 2136 7 1602 8 1201

O&M 2100 2520 3024 3629 4355 5225 6271 7525

EAC Capital

EAC O&M

4440 3916 3496 3156 2881 2656 2471 2318

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2100 2298 2513 2747 3000 3274 3571 3892

EAC Total 6540 6215 6009 5903 5880 5930 6042 6210

Chapter 7 - Replacement Decisions

(b) The costs for painting the car after three years: Car with painting and salvage value increased by $2435: Additional EAC years Salvage O&M Capital 0 12000 1 9000 2100 4440 2 6750 2520 3916 3 7498 5024 2774 4 5623 3629 2774 5 4217 4355 2665 6 3163 5225 2529 7 2372 6271 2394 8 1779 7525 2271

EAC O&M

EAC Total

2100 2298 3106 3215 3395 3620 3883 4179

6540 6215 5880 5990 6060 6149 6277 6450

In the above table, the operating and maintenance costs in year 3 include the $2000 cost of painting the car, and the salvage value has been increased by $2435, then falling by 25% each year thereafter.

With the car’s value increased by $2435, the minimum EAC(total costs) is $5880 and the economic life of the car is three more years. If the painting increases the car’s value by more than $2435, the EAC(total costs) will fall below $5880 and painting the car will be the preferred choice. In addition to this, rather than keeping the car an additional five years, it will be sold at the end of three years.

7.36

Considering periods of 18 months in length, it can be seen that computers of equal power cost the following: Period –2 –1 0 1 2

Cost $320 000 $160 000 $80 000 $40 000 $20 000

Consequently, the current computer has a salvage value of 320 000(0.53) = $40 000 There are four obvious replacement plans available: 1) Do not replace the computer but keep it another three years.

2) Replace the computer now and keep the replacement for three years.

3) Replace the computer in 18 months.

4) Replace the computer now and in 18 months.

In fact, there are an infinite number of possible replacement plans since replacement could be considered at any point in time over the next three years. However, the fact that we are assuming that new models are being released every 18 months makes it natural to consider this limited set.

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Chapter 7 - Replacement Decisions

Plan 1: Do not replace the computer but keep it another three years.

Salvage value after 3 years: S = 40 000(1 – 0.5)3 = $5 000 Capital cost: P = 40 000 – 5000(P/F, 12%, 3) = 40 000 – 5000(0.71178) = $36 441 Maintenance costs: Year 1 Accumulated depreciation = 320 000 – (40 000  0.5) = 300 000 Maintenance cost = 300 000(0.1) = 30 000 Year 2 Accumulated depreciation = 320 000 – (40 000  0.52) = 310 000 Maintenance cost = 310 000(0.1) = 31 000 Year 3 Accumulated depreciation = 320 000 – (40 000  0.53) = 315 000 Maintenance cost = 315 000(0.1) = 31 500 PW(costs, plan 1) = 36 441 + 30 000(P/F, 12%, 1) + 31 000(P/F, 12%, 2) + 31 500(P/F, 12%, 3) = 36 441 + 30 000(0.89286) + 31 000(0.79719) + 31 500(0.71178) = 110 361 The present cost for retaining the defender for three more years is about $110 000.

Plan 2: Replace the computer now and keep the replacement for three years.

Cost of new computer now = $80 000 Salvage value after 3 years: S = 80 000  (1 – 0.5)3 = $10 000 Capital cost: Purchase price including installation = 80 000(1.15) = 92 000 Salvage value of old computer = 40 000 P = (92 000 – 40 000) –10 000(P/F, 12%, 3) = 52 000 – 10000(0.71178) = 44 882

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Chapter 7 - Replacement Decisions

Maintenance costs: Year 1 Accumulated depreciation = 80 000 – (80 000  0.5) = 40 000 Maintenance cost = 40 000(0.1) = 4000 Year 2 Accumulated depreciation = 80 000 – (80 000  0.52) = 60 000 Maintenance cost = 60 000(0.1) = 6000 Year 3 Accumulated depreciation = 80 000 – (80 000  0.53) = 70 000 Maintenance cost = 70 000(0.1) = 7000 PW(costs, plan 2) = 44 882 + 4000(P/F, 12%, 1) + 6000(P/F, 12%, 2) + 7000(P/F, 12%, 3) = 44 882 + 4000(0.89286) + 6000(0.79719) + 7000(0.71178) = 58 575 The present cost for replacing the defender now and keeping the challenger for three years is about $58 600.

Plan 3: Replace the current computer in 18 months.

In order to carry out calculations for a computer bought in 18 months, it is necessary to determine a MARR and a depreciation rate for an 18-month period.

MARR18 = (1 + MARR12)1.5 – 1 = 1.121.5 – 1 = 18.53% Depreciation rate: One way to calculate the depreciation rate for 18 months is to recognize that after three years or two 18-month periods, the book value must be the same, i.e., for some book value at time 0, P: P(1 – d18)2 = P(1 – d12)3 (1 – d18)2 = (1 – d12)3 d18 = 1 – (1 – d12)3/2 = 1 – (1 – 0.5)3/2 = 0.646 Salvage value of current computer in 18 months: 40 000(1 – 0.646) = $14 142 Cost of new computer in 18 months = $40 000

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Chapter 7 - Replacement Decisions

Salvage value of new computer after 18 months (three years from now): 40 000(1 – 0.646) = $14 142 Capital cost of current computer: Pcur = 40 000 – 14 142(P/F, 18.53%, 1) = 40 000 – 14 142(1/1.1853) = 28 069 Capital cost of new computer as present worth, noting installation cost of $40 000  0.15 =$6 000: Pnew

= [46 000 – 14 142(P/F, 18.53%, 1)](P/F, 18.53, 1) = [46 000 – 14 142(1/1.1853)](1/1.1853) =28 743

Total capital cost: P = 28 069 + 28 743 = $56 812 Maintenance costs: Period 1 Accumulated depreciation = 320 000 – 320 000(1 – 0.646)3 = 305 804 Maintenance cost = 305 804(0.15) = 45 871 Period 2 Accumulated depreciation = 40 000 – (40 000  0.646) = 25 840 Maintenance cost = 25 840(0.15) = 3876 PW(costs; plan 3) = 56 812 + 45 871(P/F, 18.53%, 1) + 3876(P/F, 18.53%, 2) = 56 812 + 45 871(1/1.1853) + 3876(1/1.18532) = 98 271 The present cost for replacing the defender 18 months from now is about $98 300.

Plan 4: Replace the computer now and in 18 months.

Cost of first new computer today = $80 000 Salvage value of first new computer after 18 months = 80 000(1 – 0.646) = $28 284 Cost of second new computer in 18 months = $40 000 Salvage value of second new computer after 18 months (three years from now) = 40 000(1 – 0.646) = $14 142

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Chapter 7 - Replacement Decisions

Capital cost of first new computer as present worth, noting installation cost of $80 000  0.15 = $12 000: P1 = (96 000 – 40 000) – 28 284(P/F, 18.53%, 1) = 56 000 – 28 284(1/1.1853) = 32 138 Capital cost of second new computer as present worth, noting installation cost of $40 000  0.15= $6000: P2 = [46 000 – 14 142(P/F, 18.53%, 1)](P/F, 18.53, 1) = [46 000 – 14 142(1/1.1853)](1/1.1853) =28 743 Total capital cost: P = 32 138 + 28 743 = 60 881 Maintenance costs: Period 1 Accumulated depreciation = 80 000 – 80 000(1–0.646) = 51 680 Maintenance cost = 51 680(0.15) = 7752 Period 2 Accumulated depreciation = 40 000 – (40 000  0.646) = 25 840 Maintenance cost: 25 840(0.15) = 3876 PW(costs, plan 4) = 60 881 + 7752(P/F, 18.53%, 1) + 3876(P/F, 18.53%, 2) = 100 881 + 7752(1/1.1853) + 3876(1/1.18532) = 70 180 The present cost for replacing the defender immediately and 18 months from now is about $70 000.

A summary of these alternative replacement decisions and their respective costs is shown in the following table. It can be seen that the best choice is to replace the current computer, and keep the replacement for three years (plan 2).

Plan 1 2 3 4

PW(capital cost) 36 441 44 882 56 812 60 881

PW(maintenance) 73 920 13 718 41 488 9 299

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PW(total) 110 361 58 600 98 300 70 180

Chapter 7 - Replacement Decisions

7.37

(a) There are eight feasible combinations: Truck #1 #2 #3 #4 #5 #6 EAC

Truck #1 #2 #3 #4 #5 #6 EAC

Truck kept for between time point a and point b: denoted by (a, b) Comb.1 Comb.2 Comb.3 Comb.4 (0, 5) (0, 5) (0, 4) (0, 4) (2, 5) (2, 4) (4, 5) (4, 5) (4, 5) (4, 5) (2, 5) (2, 4) (4, 5) (4, 5) (4, 5) (4, 5) 57 149

83 210

83 819

109 880

Truck kept for between time point a and point b: denoted by (a, b) Comb.5 Comb.6 Comb.7 Comb.8 (0, 2) (0, 2) (0, 2) (0, 2) (2, 5) (2, 5) (2, 4) (2, 4) (2, 5) (2, 4) (4, 5) (4, 5) (4, 5) (4, 5) (2, 5) (2, 4) (4, 5) (4, 5) (4, 5) (4, 5) 77 180 103 242 103 242 129 303

Example Calculations: EAC(0, 5) = 30 000(A/P, 12%, 5) + 7200 EAC(0, 4) = (30 000 − 30 000  0.64)(A/P, 12%, 4) + 0.12(30 000  0.64) + 7200 EAC(0, 2) = (30 000 − 30 000  0.62)(A/P, 12%, 2) + 0.12(30 000  0.62) + 7200 EAC(2, 5) = [30 000(A/P, 12%, 3) + 7200](P/F, 12%, 2) EAC(2, 4) = [(30 000 − 30 000  0.62)(A/P, 12%, 2) + 0.12(30 000  0.62) + 7200](P/F, 12%, 2) EAC(4, 5) = [30 000(A/P, 12%, 1) + 7200](P/F, 12%, 4) (b) Combination 1 has the smallest EAC, and hence, the best purchase/replacement option (buy a truck at time 0, time 2, and time 4, and once purchased, keep each truck until the end of five years).

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Chapter 7 - Replacement Decisions

(a) The economic lives are shown in the following table.

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

EAC Capital 31250 28201 25600 23372 21459 19810 18383 17144 16065 15121 14293 13564 12919 12347 11839 11385

A1 EAC Operating 30000 31829 33777 35851 38059 40412 42918 45589 48435 51469 54703 58152 61830 65753 69937 74401

EAC Total 61250 60030 59376 59223 59518 60221 61301 62733 64500 66590 68996 71716 74749 78100 81776 85786

B1 EAC Operating 30000 31667 33320 34949 36546 38102 39609 41061 42453 43780 45040 46231 47352 48403 49384 50298

EAC Capital 56250 51250 47250 44050 41490 39442 37804 36493 35444 34605 33934 33397 32968 32624 32350 32130

EAC Total 86250 82917 80570 78999 78036 77544 77412 77554 77897 78386 78975 79629 80320 81027 81734 82427

(b)

EAC ($/year)

7.38

100000 90000 80000 70000 60000 50000 40000 30000 20000 10000 0

EAC(total, B1) EAC(total, A1)

EAC(operating, A1) EAC(operating, B1) EAC(capital, B1) EAC(capital, A1)

1

3

5

7

9

11 13 15

Year

(c) We see that the economic life is longer when the MARR is higher.

Intuitively the reason for this is as follows: We are assuming that the assets under consideration are parts of sequences of repeated identical assets. Replacement of productive assets in a situation of repeated identical assets is done only to avoid rising operating costs from ageing assets. Where the MARR is high the effect of distant increases in operating costs are felt less than where the MARR is low.

Therefore, we replace assets less frequently when the MARR is high.

A more formal explanation is as follows: The economic life occurs at the point where the rate of increase in the equivalent annual operating cost equals the absolute value of the rate of decrease in the equivalent annual capital cost. We see from the diagram that the curves for the equivalent annual operating costs for the two values of 139 Copyright © 2022 Pearson Canada Inc.

Chapter 7 - Replacement Decisions

the MARR diverge. The curve for MARR = 5% becomes much steeper than the curve for MARR = 25%. The curves for the equivalent annual capital costs are essentially parallel. That is, the slopes of the curves for the equivalent annual capital costs are insensitive to the MARR. This means that equality in absolute value for the slopes of the curves for equivalent annual operating cost and equivalent annual capital cost occurs farther to the left in the diagram for MARR = 5% than for MARR = 25%.

We can see that the curves for the two equivalent annual capital costs are essentially parallel by noting that the difference between them for one-year life is the same as the difference between their asymptotic values. The formula for the equivalent annual capital cost is EAC = (P – S)(A/P, i, N) + Si For N = 1, this reduces to: EAC = (P – S)(1 + i) + Si The difference between the equivalent annual capital costs for two different values of MARR, i and i’, is simply P(i – i’).

Notice that the asymptotic value of capital recovery factor as the number of periods goes to infinity is i. We see that the asymptotic difference between the equivalent annual capital costs for two different values of the MARR is also P(i – i’).

7.39

(a) The economic lives are shown in the following table.

Notice that there is only one column for the equivalent annual capital cost. This is because the equivalent annual capital cost is the same for both assets.

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

EAC Capital 35000 31744 29053 26824 24972 23429 22141 21063 20158 19396 18754 18211 17751 17360 17028 16744

A2 EAC Operating 30000 30930 31814 32653 33446 34194 34900 35563 36184 36766 37310 37816 38288 38725 39130 39504

EAC Total 65000 62674 60868 59476 58417 57623 57041 56625 56342 56163 56064 56028 56039 56085 56158 56249

B2 EAC Operating 40000 40930 41814 42653 43446 44194 44900 45563 46184 46766 47310 47816 48288 48725 49130 49504

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EAC Total 75000 72674 70868 69476 68417 67623 67041 66625 66342 66163 66064 66028 66039 66085 66158 66249

Chapter 7 - Replacement Decisions

(b) 80000

EAC ($/year)

70000

EAC(total, B2)

60000

EAC(total, A2)

50000

EAC(operating, B2)

40000

EAC(operating, A2)

30000 20000 10000

EAC(capital, A2&B2)

0 1

4

7

10

13

16

Year

(c) We see that the economic life is unaffected by the level of operating cost. Only the rate of change in operating cost affects the economic life.

7.40

(a) The economic lives are shown in the following table.

Notice that there is only one column for the equivalent annual capital cost. This is because the equivalent annual capital cost is the same for both assets.

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

EAC Capital 35000 31744 29053 26824 24972 23429 22141 21063 20158 19396 18754 18211 17751 17360 17028 16744

A3 EAC Operating 30000 30698 31382 32052 32707 33345 33964 34565 35147 35708 36248 36767 37266 37742 38197 38631

EAC Total 65000 62442 60436 58876 57678 56774 56105 55628 55304 55104 55002 54979 55017 55102 55225 55375

B3 EAC Operating 30000 31744 33537 35375 37255 39175 41131 43119 45136 47178 49241 51322 53417 55523 57636 59752

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EAC Total 65000 63488 62590 62198 62227 62604 63272 64182 65294 66574 67995 69533 71168 72883 74664 76497

Chapter 7 - Replacement Decisions

EAC ($/year)

(b) 80000 70000 60000 50000 40000 30000 20000 10000 0

EAC(total, B3) EAC(operating, B3) EAC(total, A3) EAC(operating, A3)

EAC(capital, A3&B3)

1

4

7

10

13

16

Year

(c) We see that the economic life is shorter for Asset B3 than for Asset A3. This is because the more rapid increase in operating cost makes more frequent investment in replacing ageing assets worthwhile.

7.41

(a) The economic lives are shown in the following table.

Notice that there is only one column for the equivalent annual operating cost. This is because the equivalent annual operating cost is the same for both assets.

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

EAC Operating 30000 30698 31382 32052 32707 33345 33964 34565 35147 35708 36248 36767 37266 37742 38197 38631

A4 EAC Capital 26250 23808 21790 20118 18729 17572 16606 15797 15118 14547 14066 13658 13313 13020 12771 12558

EAC Total 56250 54506 53172 52170 51436 50916 50570 50362 50265 50255 50314 50426 50579 50762 50968 51189

B4 EAC Capital 52500 47616 43580 40235 37457 35144 33212 31594 30237 29095 28131 27317 26627 26040 25542 25116

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EAC Total 82500 78314 74962 72288 70164 68488 67176 66159 65383 64802 64379 64084 63892 63782 63739 63747

Chapter 7 - Replacement Decisions

EAC ($/year)

(b) 90000 80000 70000 60000 50000 40000 30000 20000 10000 0

EAC(total, B4) EAC(total, A4) EAC(operating, A4&B4) EAC(capital, A4) EAC(capital, B4)

1

4

7

10

13

16

Year

(c) We see that the economic life is longer where the level of capital cost is higher. This is because a greater savings in operating cost is necessary to pay for a higher investment cost.

7.42

(a) The economic life for this asset is shown in the following table.

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

Salv.

15000 12000 9600 7680 7500 6000 4800 3840 4500 3600 2880 2304 2000 1200 720 432

Oper.

2000 2200 2420 2662 2000 2200 2420 2662 2000 2800 3920 5488 4000 8000 16000

Overhaul

2500

32500

17500

PW Market cost 30000 33333 35556 36383 37589 38392 38928 38953 39302 39535 39690 39776 39888 39944 39972

PW Overhaul

1206 1206 1206 1206 8764 8764 8764 8764 10727 10727 10727 10727

PW Capital 30000 33333 35556 37589 38794 39598 40134 47718 48066 48299 48454 50503 50615 50671 50699

PW Oper.

EAC Total

1667 3194 4595 5879 6682 7419 8095 8714 9101 9553 10081 10697 11070 11694 12732

38000 23909 19060 16791 15207 14138 13380 14707 14182 13799 13528 13786 13609 13526 13567

In this table the term market cost refers to the cost due to the decline in the value of the asset. Capital cost is the sum of market cost and overhaul cost.

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Chapter 7 - Replacement Decisions

(b) We first show costs for an eight-year-old defender.

Defender that is 8 years old: Year 0 1 2 3 4 5 6 7

Salv.

4500 3600 2880 2304 2000 1200 720 432

Oper.

2000 2800 3920 5488 4000 8000 16000

Overhaul

PW Market 1500 2500 3167 3535 4018 4259 4379

17500

PW Overhaul

8439 8439 8439 8439

PW Capital

PW Oper.

PW Total

1667 3611 5880 8526 10134 12813 17278

3167 6111 9046 20501 22591 25511 30097

1500 2500 3167 11975 12457 12698 12819

EAC Total 3800 4000 4295 7919 7554 7671 8350

The present worth of cost for Plan A is given by PW(Plan A) = PW(3 additional years for defender) + PW(7 year life for first replacement)(P/F, 20%, 3) + PW(1 year life for second replacement)(P/F, 20%, 10) The present worth of cost for Plan B is obtained similarly. The numerical values for these present worths are: PW(Plan A cost) = $42 071 PW(Plan B cost) = $60 359 (c) It is necessary to take into account the age of equipment at the end of the 11 years because future costs after the 11 years are affected by the equipment’s age. The effect of the equipment’s age on future costs may not be reflected in the salvage value. The salvage value depends on the value of the equipment outside the current user’s operation. Other potential users would have to incur costs to remove the equipment from its current situation. As well, if they were to use the equipment in production elsewhere, they would incur installation costs. They might also have to modify the equipment to their own special needs. Therefore, the salvage value is not likely to reflect all of the contributions the equipment can make to the current user’s operation. Since the equipment will be only one year old under Plan A and four years old under Plan B, there is no doubt that Plan A is better than Plan B. 7.43

The present worths of costs for the new punch press and the three old ones are shown below.

Automatic Punch Press: Year Salvage 0 125 000 1 100 000 2 80 000 3 64 000 4 51 200 5 40 960

Operating 25 000 23 750 22 563 21 434 20 363

PW Capital 145 000 173 800 192 232 204 028 211 578

PW Operating 20 000 35 200 46 752 55 532 62 204

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PW Total 165 000 209 000 238 984 259 560 273 782

Chapter 7 - Replacement Decisions

Hand-Fed Presses: Additional years Salvage 0 10 000 1 9 000 2 8 000 3 7 000 4 6 000 5 5 000

Operating

PW Capital

25 000 25 000 25 000 25 000 25 000

2 800 4 880 6 416 7 542 8 362

PW Operating 20 000 36 000 48 800 59 040 67 232

PW Total

PW 3 Presses

22 800 40 880 55 216 66 582 75 594

68 400 122 640 165 648 199 747 226 781

We see that using the existing presses appears to cost about £47 000 less than using a new press for five years. This means that, if the new automatic press is to have lower costs than the three manual presses, the salvage value of the new press after five years must exceed the estimated £40 960.

The value required for equality of costs is given by S = 1.255[(new press first cost) + PW(new press operating cost) – PW(costs with the 3 existing presses)] = £184 397 Alternate method: The salvage value required for equality of costs is found by first taking the difference in project present worths forward five years, to the point in time of the final salvage value. This difference is added to the current salvage value estimate for the automatic press: (273 782 – 226 781)(F/P, 25%, 5) = 47 001(3.0518) = 143 436 S = 143 436 + 40 960 = £184 396 7.44

a)

There is no relationship to the intersection of the lines and the lowest total cost. One's eye is misled to that conclusion by the shape of the lines intersecting. The two points would only correspond by coincidence. More precisely, if the function for capital costs is C(n), where n is the number of periods, and the function for operating costs is O(n), the minimum total costs occurs when equivalently

d (C (n ) + O (n )) = 0 , or dn

d d C (n ) + O (n ) = 0 . The cost lines intersect at n* such dn dn

that C(n*) = O(n*) and the point of intersection will also be the minimum total cost when

d d C (n *) = O (n *) . Thus, only when the dn dn

slopes of the two lines are equal and opposite in sign at the same time that their values are equal will the intersection of the lines and the lowest total cost coincide.

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Chapter 7 - Replacement Decisions

b)

The situation in Table 7.2 reflects the replacement of a defender with a challenger that is not identical. If an identical challenger had been available, it would have replaced the defender at the time of lowest total cost as illustrated in Table 7.1. Also, any non-identical challengers to the defender in the past must have had a minimum EAC greater than the current costs of the defender. Consequently, this situation will persist until a successful challenger is found. This circumstance can happen in a variety of situations such as escalating machinery costs, regulatory changes, technology changes, etc.

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Chapter 7 - Replacement Decisions

Notes for Case in Point 7.1 1-4)

All of these issues are difficult to address and the answers depend on the views of the individual student.

Notes for Mini-Case 7.1 1)

There are a number of acceptable answers to this question, as long as the student recognizes that there is no simple metric for performance or environmental impact. In this case, the existing asset is being reviewed for two types of performance—storage reliability as well as environmental performance. In reality, there are many factors related to storage reliability, including ease and speed of access and verified archival storage ability.

These aspects are uncertain, and the performance metrics impact energy use and heat production, which implicate both environmental and cost performance.

2)

A sustainable business model for data storage would involve minimizing energy costs for storage. This would involve trade-offs between easy access for reading and writing and longer-term storage. Convincing clients might require achieving a certification, demonstrating archival capabilities of the storage technology, or alignment with a major organization (such as the Canadian government, IBM or Google.)

3)

What is highlighted here is the fact that legacy technological systems sometimes have better documented reliability. A firm would want both high reliability for recall and better environmental performance, if both could be obtained at minimum cost. A student could argue either (or both) sides of this and be correct.

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CHAPTER 8 Solutions to Chapter-End Problems A. Key Concepts Before- and After-Tax MARR: 8.1

Using: MARRafter-tax  MARRbefore-tax  (1 – t): (a) MARRafter-tax  0.14  (1 – 0.4) = 0.084 = 8.4% (b) MARRafter-tax  0.14  (1 – 0.5) = 0.074 = 7.0% (c) MARRafter-tax  0.14  (1 – 0.6) = 0.056 = 5.6%

8.2

Before-tax IRR: 5000(P/F, i, 1) + 10 000(P/F, i, 2) – 12 000 = 0 At i = 0.14: LHS = 80.64 At i = 0.15: LHS = –90.73 Interpolation: IRR = 14.47% Approximate after-tax IRR = 0.1447(1 – 0.4) = 0.0868 The after-tax IRR is approximately 8.68%

IRR Tax Calculations: 8.3

IRRafter-tax ≅ IRRbefore-tax × (1 – t) = 0.24(1 – 0.40) = 0.144 The after-tax IRR is approximately 14.4%. For an after-tax MARR of 18%, the project should not be approved. However, for an after-tax MARR of 14%, since the after-tax IRR is an approximation, a more detailed examination would be advisable.

8.4

The CTF is 1 – [(0.40)(0.3)(1 + i*/2)]/[(i* + 0.3)(1 + i*)].

Setting disbursements equal to receipts gives: 12 000(CTF) = [5000(P/F, i*, 1) + 10 000 (P/F, i*, 2)](1 – 0.40) Trial and error calculations give: i* = 6.226%

8.5

(a) Class 12, fully expensed in first year (b) Class 1, 3 or 6, rate 5% to 10% 148 Copyright © 2022 Pearson Canada Inc.

Chapter 8 - Taxes

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

Class 10, rate 30% Class 8, rate 20% Class 9, rate 25% Class 8, rate 20%

CTF and CSF: 8.6

(a) CSF = 1 – (0.5  0.2)/(0.09 + 0.2) = 0.6552 CTF = 1 – [(0.5)(0.2)(1 + 0.09/2)]/[(0.09 + 0.2)(1 + 0.09)] = 0.6694 (b) CSF = 1 – (0.35  0.3)/(0.12 + 0.3) = 0.75 CTF = 1 – [(0.35)(0.3)(1 + 0.12/2)]/[(0.12 + 0.3)(1 + 0.12)] = 0.7674 (c) CSF = 1 – (0.55  0.05)/(0.06 + 0.05) = 0.75 CTF = 1 – [(0.55)(0.05)(1 + 0.06/2)]/[(0.06 + 0.05)(1 + 0.06)] = 0.7571

8.7

First year CCA is on $50 000 (half of capital purchases).

Net income = 110 000 – 65 000 – 50 000(0.2) = 45 000 – 10 000 = $35 000 Taxes paid = 35 000  0.55 = $19 250

8.8

Second year CCA is on ($50 000 – $10 000) from the half that was subject to CCA during first year plus the remaining $50 000 of the original purchases: (50 000 – 10 000) + 50 000 = 90 000.

Net income = 110 000 – 65 000 – 90 000(0.2) = 45 000 – 18 000 = $27 000 Taxes paid = 27 000  0.55 = $14 850

8.9

The UCC at the end of 2016 was $26 779.

Year Adjustment Base UCC CCA@30% Remaining UCC

2012 2013 2014 2015 2016

50000 0 20000 −3000

35000 49500 44650 38255

10500 14850 13395 11477

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10000 49500 34650 41255 26779

Chapter 8 - Taxes

The following spreadsheet shows the numbers used for the chart, and the chart: Interest Rate 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

CSF 0.5000 0.6000 0.6667 0.7143 0.7500 0.7778 0.8000 0.8182 0.8333 0.8462 0.8571 0.8667 0.8750 0.8824 0.8889 0.8947 0.9000 0.9048 0.9091 0.9130 0.9167

CTF 0.5000 0.6095 0.6818 0.7329 0.7708 0.8000 0.8231 0.8418 0.8571 0.8700 0.8810 0.8903 0.8984 0.9055 0.9118 0.9173 0.9222 0.9266 0.9306 0.9342 0.9375

1.0000

CTF 0.9000

Value

8.10

CSF

0.8000 0.7000 0.6000 0.5000 0

0.2

0.4

0.6

0.8

Interest rate

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1

Chapter 8 - Taxes

8.11

Using: MARRafter-tax  MARRbefore-tax  (1 – t) 0.052  MARRbefore-tax  (1 – 0.53) MARRbefore-tax  0.052/0.47 = 11.06% Similarly, using: IRRafter-tax  IRRbefore-tax  (1 – t) 0.087  IRRbefore-tax  (1 – 0.53) IRRbefore-tax  0.087/0.47 = 18.51% Enrique will likely report a before-tax IRR of 18.5% compared to a beforetax MARR of 11%.

B. Applications 8.12

ICC’s investments are clearly in CCA class 10, with a depreciation rate (CCA rate) of 30%.

CSF = 1 – (0.45 × 0.3)/(0.10 + 0.3) = 0.6625 CTF = 1 – [(0.45) × (0.3) × (1 + 0.10/2)] / [(0.10 + 0.3) × (1 + 0.10)] = 0.645625 The present worth of the inventory system is then: PW = –2 300 000 × CTF + 880 000 × (1 – 0.45) × (P/A, 10%, 10) + 200 000 × CSF × (P/F, 10%, 10) = –2 300 000 × 0.6456 + 880 000 × 0.55 × 6.1446 + 200 000 × 0.6625 × 0.38554 = 1 540 133 The present worth of the new system is $1 540 000. ICC should make this investment.

8.13

As of March 2010: (a) Class 6, 10% (b) Class 38, 30% (c) Class 10, 30%, class 45 (45%), class 50 (55%) or class 52, 100%, depending on when it was purchased (d) Class 46, 30% (e) Cash register: class 8, rate 20%, scanner, class 12, 100%

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Chapter 8 - Taxes

8.14

P = –55 000(CTF) + 17 000(P/A, 10%, 6)(1 – 0.54) + 1000(P/F, 10%, 6)(CSF) CTF = 1 – [(0.54)(0.2)(1 + 0.1/2)]/[(0.1 + 0.2)(1 + 0.1)] = 0.6564 CSF = 1 – (0.54  0.2)/(0.1 + 0.2) = 0.64 P = –55 000(0.6564) + 17 000(4.3552)(0.46) + 1000(0.56448)(0.64) = –1 683 The after-tax present worth of the chip placer is –$1683.

8.15

Before-tax IRR: –400 000 + 85 000(P/A, i, 10) = 0 (P/A, i, 10) = 400 000/85 000 = 4.71 (P/A, 15%, 10) = 5.0187 (P/A, 20%, 10) = 4.1924 By interpolating: IRR = 16.9% After-tax IRR = 16.9(1 – 0.45) = 9.3% Quebec Widgets should not invest.

8.16

The CTF is 1 – [(0.45)(0.2)(1 + i*/2)]/[(i* + 0.2)(1 + i*)].

Setting disbursements equal to receipts gives: 400 000(CTF) = 85 000(P/A, i*, 10)(1 – 0.45) Trial and error calculations give: i* = 10.1087% The exact IRR is 10.1087%. Canadian Widgets should not invest.

0 0 0

0 5000 15 000

40 000 25 000 35 000

15 000 10 000 15 000

9

25 000 15 000 20 000

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0.3 0.3 0.2

10

7500 4500 4000

13

UCC at year end (6)(10)+(7)

7

CCA (8)*(9)

30 000 20 000 30 000

8

CCA Rate

Adjustments

10 000 10 000 20 000

6

Peduced UCC ((6)-(7)

Cost of acquisitions during the year

10 10 8

5

50% rule

4

UCC (2)+(3)(5)

3

Proceeds of disposition

2

UCC at beginning of year (Col 13 from last year

a b c

1

Class Number

8.17

32 500 20 500 31 000

Chapter 8 - Taxes

8.18

The UCC account values are shown in the following spreadsheet.

Base UCC

Year Adjustment 2001 2002 200 000 100 000 2003 0 180 000 2004 0 144 000 2005 240 000 235 200 2006 0 308 160 2007 0 246 528 2008 0 197 222 2009 60 000 187 778 2010 0 180 222 2011 0 144 178 2012 0 115 342 2013 0 92 274 2014 345 000 246 319 2015 −45 000 324 555 2016 0 259 644

CCA@20 %

Remaining UCC

20 000 36 000 28 800 47 040 61 632 49 306 39 444 37 556 36 044 28 836 23 068 18 455 49 264 64 911 51 929

0 180 000 144 000 115 200 308 160 246 528 197 222 157 778 180 222 144 178 115 342 92 274 73 819 369 555 259 644 207 715

It can be seen that the depreciation for this CCA class for 2016 is $51 929.

8.19

t = 0.40, i = 0.20, d = 0.3 CSF = 1 – (0.4  0.3)/(0.2 + 0.3) = 0.76 CTF = 1 – (0.4  0.3  1.1)/(1 + 0.2)(0.2 + 0.3) = 0.78 AW = –45 000(A/P, 20%, 5)CTF + 4500(A/F, 20%, 5)CSF – 0.22(750)(250)(1 – 0.4) = –45 000(0.33438)(0.78) + 4500(0.13438)(0.76) – 0.22(750)(250)(1 – 0.4) = –36 072 The machine has a total annual cost of about $36 072.

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Chapter 8 - Taxes

8.20

The value of the 30% UCC account at the end of 2016 was about $12 962.

Year

2008 2009 2010 2011 2012 2013 2014 2015 2016 8.21

Adjustment

25000 0 14000 0 0 28000 −5000 0

Base UCC

CCA@30%

12500 21250 21875 22313 15619 24933 26453 18517

Remaining UCC

3750 6375 6563 6694 4686 7480 7936 5555

0 21250 14875 22313 15619 10933 31453 18517 12962

CTF = 1 – [(0.35)(0.2)(1 + 0.06/2)]/[(0.06 + 0.2)(1 + 0.06)] = 0.7384 PW = 380 000(CTF) = 380 000(0.7384) = 280 588 The present worth of the crane, taking into account all future tax benefits due to CCA, is $280 588.

8.22

CTF = 1 – [(0.45)(0.2)(1 + 0.08/2)]/[(0.08 + 0.2)(1 + 0.08)] = 0.6905 CSF = 1 – (0.45  0.2)/(0.08 + 0.2) = 0.6786 First cost = –230 000(CTF) = –230 000(0.6905) = –158 810 Savings = 35 000(P/A, 8%, 12)(1 – 0.45) = 35 000(7.5361)(0.55) = 145 070 Salvage value = 30 000(P/F, 8%, 12)CSF = 30 000(0.39711)(0.6786) = 8084 PW = –158 810 + 145 070 + 8084 = –5656 The project has a present worth of –$5656; it should not be done.

8.23

CTF = 1 – [(0.35)(0.3)(1 + 0.12/2)]/[(0.12 + 0.3)(1 + 0.12)] = 0.763 CSF = 1 – (0.35  0.3)/(0.12 + 0.3) = 0.75 First cost = –65 000(A/P, 12%, 5)CTF = –65 000(0.27741)(0.763) = –13 758 Annual savings = 15 000(1 – 0.35) = 9750 154 Copyright © 2022 Pearson Canada Inc.

Chapter 8 - Taxes

Salvage value = 20 000(A/F, 12%, 5)CSF = 20 000(0.15741)(0.75) = 2361 Annual worth = –13 758 + 9750 + 2361 = –1647 The purchase will result in an annual cost of $1647 over five years.

8.24

The CTF is 1 – [(0.40)(0.2)(1 + i*/2)]/[(i* + 0.2)(1 + i*)] Setting disbursements equal to receipts gives: 17 000(CTF) = [3000(P/A, i*, 7) + 1000 (P/F, i*, 7)](1 – 0.40) Trial and error calculations give: i* = 3.8% Their exact after-tax IRR is 3.737%.

The estimated after-tax MARR is: MARRafter-tax  0.1  (1 – 0.4) = 6.0% The purchase should not be made since the after-tax IRR is less than the aftertax MARR.

8.25

CTF = 1 – [(0.45)(0.2)(1 + 0.1/2)]/[(0.1 + 0.2)(1 + 0.1)] = 0.7136 CSF = 1 – (0.45  0.2)/(0.1 + 0.2) = 0.7 AW = –200 000(A/P, 10%, 20)CTF – [7500 + 25 000(A/P, 10%, 5)](1 – 0.45) + 15 000(A/F, 10%, 20)CSF = –200 000(0.117 46)(0.7136) – [7500 + 25 000(0.26380)](0.55) + 15 000(0.01746)(0.7) = –$24 333 The after-tax annual cost of the slitter is –$24 333.

8.26

CTF = 1 – [(0.45)(0.2)(1 + 0.09/2)]/[(0.09 + 0.2)(1 + 0.09)] = 0.7025 CSF = 1 – (0.45  0.2)/(0.09 + 0.2) = 0.6897 The annual cost of the chemical recovery system is then:

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Chapter 8 - Taxes

A = –30 000(A/P, 9%, 7)CTF + 5280(1 – 0.45) + 7500(A/F, 9%, 7)CSF = –30 000(0.19869)(0.7025) + 5280(0.55) + 7500(0.10869)(0.6897) = –721 The chemical recovery system has an after-tax annual cost of about $721.

8.27

The exact after-tax IRR is calculated by trial and error to be IRRafter-tax = 8.44%. i= 7% 8% 8.44% 9% 10% CCTF(old) = 0.5926 0.6071 0.6133 0.6207 0.6333 CCTF(new ) = 0.6059 0.6217 0.6283 0.6363 0.6500 (P/A,i%,6) = 4.7665 4.6229 4.5612 4.4859 4.3553 (P/F, i%,6) = 0.6663 0.6302 0.6148 0.5963 0.5645 PW(disbursements) = −66651 −68386 −69117 −69998 −71500 PW(receipts) = 72246 70061 69117 67962 65946 PW(total) = 5595 1675 0 −2037 −5554 The before-tax IRR is calculated to be IRRbefore-tax = 18.77% i= 17% 18% 18.77% 19% 20% (P/A,i%,6) = 3.5892 3.4976 3.4292 3.4098 3.3255 (P/F, i%,6) = 0.3898 0.3704 0.3562 0.3521 0.3349 PW(disbursements) = −110000 −110000 −110000 −110000 −110000 PW(receipts) = 115472 112337 110000 109336 106463 PW(total) = 5472 2337 0 −664 −3537 IRRafter-tax  IRRbefore-tax  (1 – 0.55) = 0.1877(0.45) = 0.0845 = 8.45% Since the IRR, through either calculation, is less than the MARR, the backhoe should not be purchased.

8.28

d = 0.30, t = 0.40 Ordered by first cost, the alternatives are: 1) Free machine, first cost $6 000 2) Used machine, first cost = $36 000 3) Owned machine, first cost = $71 000

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Alternative 1: Note that the first cost is an expense of installation, and not a capital cost, so the after-tax IRR is found by solving for i in: 6000(1 – t) = (20 000 – 15 000)(P/A, i, 6)(1 – t) (P/A, i, 6) = 1.2 Noting that (P/A, 40%, 6) = 2.16 and (P/A, 50%, 6) = 1.8, the free machine is acceptable as it has an after-tax IRR of above 50%.

Next, look at the increment of investment between the free and used machines: Year 0 1 2 3 4 5 6

Increment: used – free –30000 12000 9500 7000 4500 2000 –500

This is almost a simple investment, so as an approximation, we will use an IRR approach (the negative cash flow in the sixth year is not likely to create multiple IRRs).

Taking present worths gives: –30 000(CTF) + PW(incremental annual flows)(1 – t) = 0 With a trial and error approach, we obtain an after-tax IRR of 3.18%, which is below the MARR. We now find the after-tax IRR on the incremental investment from the free machine to the new machine.

Year 0 1 2 3 4 5 6

Increment: new – free –65000 15000 15000 15000 15000 15000 25000

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This is a simple investment, so the IRR can be found as follows: –65 000CTF + PW(incremental cash flows)(1 – t) + (salvage value)CSF = 0 By trial and error we get an after-tax IRR of 9.88% that is less than the MARR. Conclusion: the free machine is best.

C. More Challenging Problems: 8.29

(a) DC(1) = (360 000 – 0)/10 = 36 000 Taxes = [1 600 000 – (1 300 000 + 36 000)]  0.5 = 132 000 They would pay about $132 000. (b) DC(1) =(360 000 – 0)/5 = 72 000 Taxes = [1 600 000 – (1 300 000 + 72 000)]  0.5 = 114 000 They would pay about $114 000. (c) DC(1) = 360 000  0.2 = 72 000 Taxes = [1 600 000 – (1 300 000 + 72 000)]  0.5 = 114 000 They would pay about $114 000. (d) DC(1) = 360 000  0.4 = 144 000 Taxes = [1 600 000 – (1 300 000 + 144 000)]  0.5 = 78 000 They would pay about $78 000. (e) DC = 360 000 Taxes = [1 600 000 – (1 300 000 + 360 000)]  0.5 = –30 000 They would pay no taxes and would have a tax credit of $30 000.

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8.30

The present worth of a year’s savings due to CCA for the assets if they were correctly recognized as Class 8 can be calculated using the CTF. CTF = 1 – [(0.5)(0.2)(1 + 0.09/2)]/[(0.09 + 0.2)(1 + 0.09)] = 0.6694 PWsavings-8 = 10 000 – 10 000(CTF) = 10 000 – 10 000(0.6694) = 3306 The present worth of a year’s savings due to CCA for the items recognized as Class 12 can be calculated directly from the tax rate: PWsavings-12 = (10 000/2)(P/F, 9%, 1) = 5000(0.91743) = 4587 The present worth of the loss per year is then PWloss = 4587 – 3306 = 1281 The present worth today of such losses over the past three years is then: PW = 1281(F/A, 9%, 3) = 1281(3.2781) = 4199 The present worth of the cost today of this mistake is about $4199.

8.31

CTF = 1 – [(0.45)(0.05)(1 + 0.15/2)]/[(0.15 + 0.05)(1 + 0.15)] = 0.8948 Alternative 1: Using the capitalized value formula: A(first cost) = Pi(CTF) = 2 000 000(0.15)(0.8948) = 268 440 A(maintenance) = 10 000(1 – 0.45) = 5500 A(paint) = 15 000(A/F, 15%, 15)(1 – 0.45)= 15 000(0.02102)(0.55) = 173.42 A(total) = 274 113 Alternative 2: A(first costs) = 1 250 000(0.15)(0.8948) + 1 000 000(P/F, 15%, 10)(0.15)(0.8948) = 167 775 + 150 000(0.24719)(0.8948) = 200 953 The first 10 years incur maintenance costs of $5000 per year; convert to a PW and spread over infinite life:

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A(maintenance first 10 years) = 5000(P/A, 15%, 10)(0.15)(1 – 0.45) = 5000(5.0187)(0.15)(0.55) = 2070.21 The maintenance costs change after the renovation to be $11 000 every year.

First convert this to a PW at the end of 10 years, then to PW now, then spread over infinite life: P(maintenance after 10 years) = A/i(1 – t) = 11 000/(0.15)(1 – 0.45) = 40 333.33 P(maintenance after 10 years, now) = 40 333.33(P/F, 15%, 10) = 40 333.33(0.2472) = 9970.40 A = P(maintenance after 10 years)(0.15) = 9970.40(0.15) = 1496 Painting costs are every 15 years, starting in 10 years. First calculate the P(paint in 10 years), convert to P(now) and then to A(now): P(paint in 10 years) = A/i(1 – t) = 15 000(A/F, 15%, 15)/(0.15)(1 – 0.45) = 15 000(0.02102)/(0.15)(0.55) = 1156 P(paint now) = 1156(P/F, 15%, 10) = 1156(0.24719) = 285.64 A = P(paint now)(0.15) = 42.85 A(total) = 200 953 + 2070.21 + 1496 + 43 = 204 562 The annual cost of alternative 2 ($204 562) is less than that of alternative 1 ($274 113). Hence, select alternative 2.

8.32

t =0.52, d = 0.2, i = 0.11 CTF = 1 – [(0.52)(0.2)(1 + 0.11/2)]/[(0.11 + 0.2)(1 + 0.11)] = 0.6811 CSF = 1 – (0.52  0.2)/(0.11 + 0.2) = 0.6645 By assuming that you can purchase each alternative as many times as necessary, we can construct new projects: T': buy model T three times, total life 15 years A': buy model A three times, total life 15 years X': buy model X five times, total life 15 years

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PT' = –100 000(0.6811)[1 + (P/F, 11%, 5) + (P/F, 11%, 10)] + 50 000(1 – 0.52)(P/A, 11%, 15) + 20 000(0.6645)[(P/F, 11%, 5) + (P/F, 11%, 10) + (P/F, 11%, 15)] = –100 000(0.6811)(1 + 0.59345 + 0.35218) + 50 000(0.48)(7.1909) + 20 000(0.6645)(0.59345 + 0.35218 + 0.20900) = 55 410 PA' = –150 000(0.6811)[1 + (P/F, 11%, 5) + (P/F, 11%, 10)] + 60 000(1 – 0.52)(P/A, 11%, 15) + 30 000(0.6645)[(P/F, 11%, 5) + (P/F, 11%, 10) + (P/F, 11%, 15)] = –150 000(0.6811)(1 + 0.59345 + 0.35218) + 60 000(0.48)(7.1909) + 30 000(0.6645)(0.59345 + 0.35218 + 0.20900) = 31 340 PX' = –200 000(0.6811)[1 + (P/F, 11%, 3) + (P/F, 11%, 6) + (P/F, 11%, 9) + (P/F, 11%, 12)] + 75 000(1 – 0.52)(P/A, 11% 15) + 100 000(0.6645)[(P/F, 11%, 3) + (P/F, 11%, 6) + (P/F, 11%, 9) + (P/F, 11%, 12) + (P/F, 11%, 15)] = –200 000(0.6811)(1 + 0.73119 + 0.53464 + 0.39092 + 0.28584) + 75 000(0.48)(7.1909) + 100 000(0.6645)(0.73119 + 0.53464 + 0.39092 + 0.28584 + 0.20900) = 1006 Model T is the best choice because it has the highest present worth.

8.33

t =0.52, d = 0.2, i = 0.11 CTF = 1 – [(0.52)(0.2)(1 + 0.11/2)]/[(0.11 + 0.2)(1 + 0.11)] = 0.6811 CSF = 1 – (0.52  0.2)/(0.11 + 0.2) = 0.6645 PT' = –100 000(0.6811) + 50 000(1 – 0.52)(P/A, 11%, 3) + 40 000(0.6645)(P/F, 11%, 3) = –100 000(0.6811) + 50 000(0.48)(2.4437) + 40 000(0.6645)( 0.73119) = 9974 PA' = –150 000(0.6811) + 60 000(1 – 0.52)(P/A, 11%, 3) + 80 000(0.6645)(P/F, 11%, 3) = –150 000(0.6811) + 60 000(0.48)(2.4437) + 80 000(0.6645) (0.73119) = 7084

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PX' = –200 000(0.6811) + 75 000(1 – 0.52)(P/A, 11% 3) + 100 000(0.6645)[(P/F, 11%, 3) = –200 000(0.6811) + 75 000(0.48)(2.4437) + 100 000(0.6645)(0.73119) = 340 Model T is the best choice because it has the highest present worth.

8.34

AWT = –100 000(0.6811)(A/P, 11%, 5) + 50 000(1 – 0.52) + 20 000(0.6645)(A/F, 11%, 5) = –100 000(0.6811)(0.27057) + 50 000(0.48) + 20 000(0.6645)(0.16057) = 7705 AWA = –150 000(0.6811)(A/P, 11%, 5) + 60 000(1 – 0.52) + 30 000(0.6645)(A/F, 11%, 5) = –150 000(0.6811)(0.27057) + 60 000(0.48) + 30 000(0.6645)(0.16057) = 4358 AWX = –200 000(0.6811)(A/P, 11%, 3) + 75 000(1 – 0.52) + 100 000(0.6645)(A/F, 11%, 3) = –200 000(0.6811)(0.40921) + 75 000(0.48) + 100 000(0.6645)(0.29921) = 140 Model T is the best choice because it has the highest annual worth.

8.35

First find the before-tax IRR for each model: –100 000 + 50 000(P/A, i%, 5) + 20 000(P/F, i%, 5) = 0  IRRT-before-tax = 43.11% i= 42% 43% 43.11% 44% 45% (P/A,i%,5) = 1.9686 1.9367 1.9334 1.9057 1.8755 (P/F, i%,5) = 0.1732 0.1672 0.1666 0.1615 0.1560 PW(disbursements) = −100000 −100000 −100000 −100000 −100000 PW(receipts) = 101892 100178 100000 98514 96897 PW(total) = 1892 178 0 −1486 −3103 –150 000 + 60 000(P/A, i%, 5) + 30 000(P/F, i%, 5) = 0  IRRA-before-tax = 31.39%

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i= 30% 31% 31.39% 32% 33% (P/A,i%,5) = 2.4356 2.3897 2.3723 2.3452 2.3021 (P/F, i%,5) = 0.2693 0.2592 0.2554 0.2495 0.2403 PW(disbursements) = −150000 −150000 −150000 −150000 −150000 PW(receipts) = 154214 151156 150000 148198 145337 PW(total) = 4214 1156 0 −1802 −4663 –200 000 + 75 000(P/A, i%, 3) + 100 000(P/F, i%, 3) = 0  IRRX-before-tax = 24.30% i= 23% 24% 24.30% 25% 26% (P/A,i%,3) = 2.0114 1.9813 1.9724 1.9520 1.9234 (P/F, i%,3) = 0.5374 0.5245 0.5207 0.5120 0.4999 PW(disbursements) = −200000 −200000 −200000 −200000 −200000 PW(receipts) = 204591 201046 200000 197600 194248 PW(total) = 4591 1046 0 −2400 −5752 Then calculate the approximate after-tax IRR: IRRT-after-tax  IRRT-before-tax  (1 – 0.52) = 0.4311(0.48) = 0.2069 = 20.69% IRRA-after-tax  IRRA-before-tax  (1 – 0.52) = 0.3139(0.48) = 0.1507 = 15.07% IRRX-after-tax  IRRX-before-tax  (1 – 0.52) = 0.2430(0.48) = 0.1167 = 11.67% 8.36

Using trial and error in a spreadsheet program results in: IRRT-after-tax = 20.64% i= CCTF(old) = CCTF(new ) = (P/A,i%,5) = (P/F, i%,5) = PW(disbursements) = PW(receipts) = PW(total) =

19% 20% 20.64% 21% 22% 0.7333 0.7400 0.7441 0.7463 0.7524 0.7546 0.7617 0.7660 0.7684 0.7747 3.0576 2.9906 2.9490 2.9260 2.8636 0.4190 0.4019 0.3913 0.3855 0.3700 −75462 −76167 −76599 −76835 −77471 79529 77722 76599 75979 74295 4067 1556 0 −857 −3176

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IRRA-after-tax = 14.67% i= 13% 14% 14.67% 15% 16% CCTF(old) = 0.6848 0.6941 0.7001 0.7029 0.7111 CCTF(new ) = 0.7030 0.7129 0.7192 0.7222 0.7310 (P/A,i%,5) = 3.5172 3.4331 3.3783 3.3522 3.2743 (P/F, i%,5) = 0.5428 0.5194 0.5043 0.4972 0.4761 PW(disbursements) = −105447 −106935 −107886 −108335 −109655 PW(receipts) = 112448 109688 107886 107025 104457 PW(total) = 7001 2753 0 −1310 −5198 IRRX-after-tax = 11.08% i= 10% 11% 11.08% 12% 13% CCTF(old) = 0.6533 0.6645 0.6653 0.6750 0.6848 CCTF(new ) = 0.6691 0.6811 0.6820 0.6924 0.7030 (P/A,i%,3) = 2.4869 2.4437 2.4405 2.4018 2.3612 (P/F, i%,3) = 0.7513 0.7312 0.7297 0.7118 0.6931 PW(disbursements) = −133818 −136228 −136406 −138482 −140595 PW(receipts) = 138613 136563 136406 134511 132465 PW(total) = 4794 335 0 −3971 −8130 8.37

First calculate the before-tax IRR of the alternative with the cheapest first cost, T. –100 000 + 50 000(P/A, i%, 5) + 20 000(P/F, i%, 5) = 0  IRRT-before-tax = 43.11% i= 42% 43% 43.11% 44% 45% (P/A,i%,5) = 1.9686 1.9367 1.9334 1.9057 1.8755 (P/F, i%,5) = 0.1732 0.1672 0.1666 0.1615 0.1560 PW(disbursements) = −100000 −100000 −100000 −100000 −100000 PW(receipts) = 101892 100178 100000 98514 96897 PW(total) = 1892 178 0 −1486 −3103 Then calculate the after-tax IRR: IRRT-after-tax  IRRT-before-tax  (1 – 0.52) = 0.4311(0.48) = 0.2069 = 20.69% This is higher than the after-tax MARR, so that alternative T is acceptable. Next check the before-tax IRR of the increment of investment from T to A: [–150 000 – (–100 000)] + (60 000 – 50 000)(P/A, i%, 5) + (30 000 – 20 000)(P/F, i%, 5) = 0 –50 000 + 10 000(P/A, i%, 5) + 10 000(P/F, i% 5) = 0  IRRA–T before-tax = 5.73% 164 Copyright © 2022 Pearson Canada Inc.

Chapter 8 - Taxes

i= (P/A,i%,5) = (P/F, i%,5) = PW(disbursements) = PW(receipts) = PW(total) =

4% 5% 5.73% 6% 6% 4.4518 4.3295 4.2432 4.2124 4.2124 0.8219 0.7835 0.7568 0.7473 0.7473 −50000 −50000 −50000 −50000 −50000 52737 51130 50000 49596 49596 2737 1130 0 −404 −404

Then calculate the after-tax IRR: IRRA–T after-tax  IRRA–T before-tax  (1 – 0.52) = 0.0573(0.48) = 0.0258 = 2.75% Alternative A is rejected because the after-tax IRR is less than the after-tax MARR. We next look at the increment of investment from T to X. We have to be careful since X has a duration of three years, while T has five years. We could use a 15-year study period, or a three-year study period using the information from Problem 8.33, or look at the increment on an annual basis. Using a threeyear study period: [–200 000 – (–100 000)] + (75 000 – 50 000)(P/A, i%, 3) + (100 000 – 40 000)(P/F, i%, 3) = 0 –100 000 + 25 000(P/A, i%, 3) + 60 000(P/F, i% 3) = 0  IRRX–T before-tax = 13.29% i= 12% 13% 13.29% 14% 15% (P/A,i%,3) = 2.4018 2.3612 2.3495 2.3216 2.2832 (P/F, i%,3) = 0.7118 0.6931 0.6877 0.6750 0.6575 PW(disbursements) = −100000 −100000 −100000 −100000 −100000 PW(receipts) = 102753 100612 100000 98539 96532 PW(total) = 2753 612 0 −1461 −3468 Then calculate the after-tax IRR: IRRX–T after-tax  IRRX–T before-tax  (1 – 0.52) = 0.1329(0.48) = 0.0598 = 6.38% Since the after-tax IRR of the incremental investment to X from T is less than the MARR, Model T should be chosen.

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8.38

d = 0.30, t = 0.40 Ordered by first cost, the alternatives are: 1) Free machine, first cost $6 000 2) Used machine, first cost = $36 000 3) Owned machine, first cost = $71 000 The incremental IRR from do-nothing to alternative 1: 6000 = (20 000 – 15 000)(P/A, i, 6) (P/A, i, 6) = 1.2 Noting that (P/A, 40%, 6) = 2.16 and (P/A, 50%, 6) = 1.8, the free machine has a before-tax IRR of greater than 50%. That is, the free machine has an approximate after-tax IRR of greater than 50%  (1 – 0.4) = 30%. The free machine is acceptable.

Next, look at the increment of investment between the free and used machines:

Year 0 1 2 3 4 5 6

Increment: used – free –30000 12000 9500 7000 4500 2000 –500

An ERR method was used since the incremental cash flows were not a simple investment. From the result of Problem 5.21, the ERR was found to be 8.78%.

Then the approximate after-tax ERR is computed as 8.78%(1 – 0.4) = 5.268% < MARR. The incremental investment between the free and used machines is not warranted.

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We now look at the increment of investment between the free and new machines: Year 0 1 2 3 4 5 6

Increment: new – free –65000 15000 15000 15000 15000 15000 25000

This is a simple investment. From the result of Problem 5.21, the IRR for this incremental investment was found to be 12.88%. The approximate after-tax IRR is then computed as 12.88%(1 – 0.4) = 7.728% < MARR. Hence, the new machine is not a good investment either.

Conclusion: the free machine is best.

8.39

a) Setting disbursements, adjusted by the tax rate, equal to receipts at the end of year 2, taking cash on hand forward at the MARR: 8000(0.5)(F/P, 6%, 1) + 8000(0.5) = 10 000(0.5)(F/P, 6%, 2) + 5500(0.5)(P/F, i*, 1) Solving for i* gives i* = 4.88% = ERR Note that the tax rate is immaterial to the calculation, as long as there are no depreciable assets involved in the cash flows. b)

Setting disbursements equal to receipts at the end of the three-year period, with receipts taken forward at the MARR: 8000(0.5)(F/P, i*, 2) + 8000(0.5)(F/P, i*, 1) = 10 000(0.50)(F/P, 6%, 3) + 5500(0.5) Solving for i* gives i* = 5.76% =approximate ERR. (Recall that the approximate ERR will always be between the accurate ERR and the MARR.)

c)

This is not a good investment.

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Notes for Mini-Case 8.1 1)

The pros of high tax include a robust social state with subsidized services.

Depending on the country, this can include long-term care, maternity leave, and daycare. Beyond this, unemployment and disability benefits are more widely available. There is also a reduction in income inequality. The cons include potentially less incentive to earn, and if capital gains are taxed, less incentive to invest. People may spend more effort to avoid taxes if taxes consume a greater percentage of earnings.

2)

The pros of low tax include putting more money in the hands of taxpayers, potentially boosting the economy. Low taxes cans also be associated with increased compliance. Cons include the potential for class stratification due to income inequality, and a limited capability for governments to fund public services.

3)

The amount of tax paid would depend on the province and assumptions about deductions. However, typical numbers for Ontario in 2020 would be about $43,000 in federal taxes and $26,500 in provincial taxes, for a total of about $69,500.

4)

The amount of tax paid would depend on the state and assumptions about deductions. However, typical numbers for New York State in 2020 would be about $41,000 in federal taxes and $12,000 in state taxes, for a total of about $53,000.

The difference in this case is about $16,500. However, it should be noted that a U.S, taxpayer may pay more for Social Security and Medicare, which are separate from the income tax scheme. Our $200,000 New York taxpayer would typically pay about $8000 for Social Security compared to about $3000 for CPP for our Ontario taxpayer.

5)

The average cost for health insurance in the U.S. is about $10,000 per person per year. The actual health costs will vary over one’s lifetime, with particular costs occurring during maternity and old age. Chronic diseases like diabetes and MS are costly, along with more critical illnesses like cancer and heart disease.

With health care considered, Canadian taxes are quite comparable to U.S. taxes.

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CHAPTER 9 Solutions to Chapter-End Problems A. Key Concepts Constant and Current Dollars: 9.1

(a) current (b) real (c) current (d) real (e) current (f) current (g) real

9.2

(a) 400(P/F, 4%, 3) = 400(0.85480) = $341.92 (b) 400(F/P, 4%, 3) = 400(1.1699) = $467.96 (c) 10(P/F, 4%, 1) = 10(0.94154) = $9.42 (d) 350 983(P/F, 4%, 10) = 350 983(0.67556) = $237 110 (e) 1(F/P, 4%, 1000) = (1.04)1000 = £1.0798  1017 (f) 1 000 000 000(P/F, 4%, 300) = 1 000 000 000/(1.04)300 = $7762.44

Time Value of Money under Inflation: 9.3

(a) 400(P/F, 4%, 3)(P/F, 4%, 3) = 400(0.85480)(0.85480) = $292.27 (b) 400(F/P, 4%, 3)(F/P, 4%, 3) = 400(1.1699)(1.1699) = €547.47 (c) 10(P/F, 4%, 1)(P/F, 4%, 1) = 10(0.94154)(0.94154) = $8.86 (d) 350 983(P/F, 4%, 10)(P/F, 4%, 10) = 350 983(0.67556)(0.67556) = $160 182 (e) 1(F/P, 4%, 1000)(F/P, 4%, 1000) = 1.042000 = £1.659  1034 (f) 1 000 000 000(P/F, 4%, 300)(P/F, 4%, 300) = 1 000 000 000/1.04600 = $0.06

9.4

(a) Using the present worth factor to convert to current (today) dollars: Real dollar amount = 10 000(P/F, 10%, 5) = 10 000(0.62092) = 6209 This is worth about $6209 in real (today) dollars.

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(b) We can first convert to real dollars using the present worth factor, and then find the present worth using the present worth factor: PW = 10 000(P/F, 10%, 5)(P/F, 10%, 5) = 10 000(0.62092)(0.62092) = $3855 (c) MARRC = (1 + MARRR)(1 + f) – 1 = (1.1)(1.1) – 1 = 0.21 The current dollar MARR is 21%. (d) PW = 10 000(P/F, 21%, 5) = 10 000(0.38554) = 3855 The present worth of the investment is $3855.

9.5

(a) The actual dollar MARR, MARRA, is given by MARRA = (1 + MARRR)(1 + f) – 1 = (1.08)(1.06) – 1 = 0.1448 = 14.48% (b) PW = 1000(P/A, 14.48%, 10) = 1000(5.1199) = 5119.9 The present worth of the annuity is about $5120.

9.6

(a) PW = –7500 + 1000(P/A, i, 12) The actual dollar internal rate of return is between 8% and 8.5% in a spread sheet approximation. Sample trial and error results are shown below: Interest rate 0.08 0.081 0.085 0.09

PW 36.08 2.81 –155.31 –339.28

(b) If we take the actual dollar internal rate of return to be 8.1%, the real internal rate of return is obtained as 1.081/1.06 – 1 = 0.0198, or about 2.0%.

9.7

(a) PW = –90 000 + 10 000(P/A, i, 10) The actual dollar internal rate of return is between 1.95% and 2.0% in a spread sheet approximation. Sample trial and error results are shown below: Interest rate 0.019 0.0195 0.02 0.0205

PW 297.65 61.29 –174.15 –408.69

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(b) If we take the actual dollar internal rate of return to be 2% the real internal rate of return is obtained as 1.02/1.05 – 1 = –0.02857, or about –2.9%. Notice that when the actual dollar internal rate of return is less than the inflation rate the real internal rate of return is negative.

9.8

(a) 1.59(F/P, 4%, 50) = 1.59(50.505) = $80.30 We would expect to pay about $80.30 for the hamburger. (b) 15 000(F/P, 4%, 50) = 15 000(50.505) = $757 575 We would expect to pay about $757 575 for the automobile. (c) 180 000(F/P, 4%, 50) = 180 000(50.505) = $9 090 900 We would expect to pay about $9 090 900 for the house.

9.9

1 000 000 = 38 000(F/P, 5%, N) 1.05N = 1 000 000/38 000 = 26.316 N = ln(26.316)/ln(1.05) = 67.025 The average person will be a millionaire in about 67 years.

B. Applications 9.10

(a) 10 000(P/F, 5%, 10)(P/F, 4%, 10) = 10 000(0.61391)(0.67556) = $4147.33 (b) [10000(P/F, 5%, 5)(P/F, 8%, 5](P/F, 5%, 5) = [10000(0.78353)(0.68058)](0.78353) = $4178.21 (c) [10000(P/F, 5%, 5)](P/F, 8%, 5)(P/F, 5%, 5) = [10000(0.78353)](0.68058)(0.78353) = $4178.21 (d) [10000(P/F, 5%, 5)(P/F, 2%, 5][(P/F, 5%, 5)(P/F, 6%, 5)] = [10000(0.78353)(0.90573)][(0.78353)(0.74726)]= $4155.10 (e) [10000(P/F, 5%, 1)(P/F, 40%, 1](P/F, 5%, 9) = [10000(0.95238)(0.71429)](0.64461) = $4385.12 Since the present worth of the $10 000 was fairly close for a variety of inflation patterns, and that the average of 4% per year was the most conservative of the calculated values, it probably is reasonable to use an average inflation rate.

9.11

Real MARR = (1 + 3)/(1 + 2.5) – 1 = 14.3%

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9.12

Years 3-5: MARRC = (1 + 0.12)(1 + 0.1) – 1 = 0.232 = 23.2% PW(end of year 2) = 25 000(P/A, 23.2%, 3) + 25 000 (in actual dollars) Year 2: MARRC = (1 + 0.12)(1 + 0.05) – 1 = 0.176 = 17.6% PW(end of year 1) = PW(end of year 2)(P/F, 17.6%, 1) + 25 000 (in actual dollars) Year 1: MARRC = (1 + 0.12)(1 + 0.03) – 1 = 0.1536 = 15.36% PW(now) = PW(end of year 1)(P/F, 15.36%, 1) = {[25 000(P/A, 23.2%, 3) + 25 000](P/F, 17.6%, 1) + 25 000}  (P/F, 15.36%, 1) = {[25 000(2.0053) + 25 000](0.85034) + 25 000}(0.86685) = 77 053 (in today’s dollars) The present worth of the contract is about $77 000.

9.13

First get the current dollar MARR: MARRC = 1.03/1.1 – 1 = 0.133 The current dollar MARR is about 13.3%.

We then use the series present worth factor: PW = 2000(P/A, 13.3%, 30) = 14 682.55 The present worth is about $14 683.

9.14

Method 1: Use present worth factor to convert to real (today) dollars, and present worth factor to convert to money today.

For example, 4th year amount = P4 = 15 000(P/F, 4%, 4)(P/F, 8%, 4) = 15 000(0.854 80)(0.73503) = 9424.55 Method 2: Combine the inflation rate and interest rate into a current dollar MARR. MARRC = (1 + MARRR)(1 + f) – 1 = 1.08/1.04 – 1 = 0.123 = 12.3% PW = 15 000(P/A, 12.3%, 10) = 15 000(5.577) = 83 655 The annuity is worth $83 655 today.

9.15

For the real rubles, the present worth is: PW(real) = 500(P/A, 1.5%, 24) = 500(20.030) = 10 015 For months 13-24: 172 Copyright © 2022 Pearson Canada Inc.

Chapter 9 - Inflation

MARRC = (1 + 0.015)(1 + 0.2) – 1 = 21.8% For months 0-12: MARRC = (1 + 0.015)(1 + 0.4) – 1 = 42.1% PW(current) = 500(P/A, 21.4%, 12)(P/F, 42.1%, 12) + 500(P/A, 42.1%, 12) = 500(4.2169)(0.01475) + 500(2.3403) = 1201.25 The total present worth of the contract is about 10 015 + 1 201 = 11 216 million rubles.

9.16

The price in 2020 relative to 2017 is 1.1(130/125) = 1.144. The percentage increase is about 14.4%.

9.17

(a) The current IRR of the project can be found by solving for i in: PW = –200 000 + (22 000 – 2 000)(P/A, i, 20) = 0 (P/A, i, 20) = 200 000/20 000 = 10 i = 0.0775 or 7.75% The current IRR of the project is 7.75%. (b) The current MARR is MARRC = MARRR + f + MARRR  f = 0.04 + 0.03 + 0.04(0.03) = 0.0712 or 7.12% (c) Yes, they should invest, as the current rate of return exceeds the current MARR.

9.18

(a) & (b) The calculation results are summarized in the following table: Year 0 1 2 19 20

Current –200 000 20 000 20 000 20 000 20 000

Real (f = 0.03) –200 000 20 600 21 218 35 070 36 122 Total:

Real (MARR = 0.04) –200 000 19 808 19 617 16 646 16 486 161 972

(c) They should invest, as the present worth of the real cash flows at the real MARR is positive ($161 972).

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9.19

(a)

Year 0 1 2 3 4 5 6 7

Real Zerts Exchange Real Dollar Cash Flow Inflation Actual Zerts Rate Actual Dollar Cash Flow Present (2018 Zerts) Factor Cash Flow Factor Cash Flow (2018 dollars) Worth ($) -1 500 000 1 -1 500 000 0.25 -375 000 -375 000 -375 000 260 870 1.15 300 000 0.22 67 500 65 534 55 328 260 870 1.32 345 000 0.2 69 863 65 852 46 938 260 870 1.52 396 750 0.18 72 308 66 172 39 820 260 870 1.75 456 263 0.16 74 838 66 493 33 782 260 870 2.01 524 702 0.15 77 458 66 816 28 659 260 870 2.31 603 407 0.13 80 169 67 140 24 313 260 870 2.66 693 918 0.12 82 975 67 466 20 627 Total: -125 532

(b) The present worth of the project is −$125 532 in dollars. It was obtained by using the real MARR and the real dollar cash flow. The real MARR is given by MARRR = (1 + MARRA)/(1 + f) – 1 = 1.22/1.03 – 1 = 0.184 We could also have found the present worth in dollars using the current dollar MARR and the actual dollar cash flow.

9.20

The present worths were computed with current dollar cash flows and the current dollar MARR. 2% Inflation Cash Flows: In-House Purchase Current Dollar Act Dollar Other PW Current Dollar PW Year Labour Cost Operat. Cost (all Op. cost) Purchase ( Purchase) 1 262 500 225 000 406 250 750 000 625 000 2 278 460 234 090 355 938 765 000 531 250 3 295 390 243 547 311 885 780 300 451 563 4 313 350 253 387 273 310 795 906 383 828 5 332 402 263 623 239 529 811 824 326 254 1 586 912 Total PW: 2 317 895 Total PW in-house, including first cost: 1 786 912

(a) $278 460 (b) $243 547 (c) Purchase PW = $2 317 895, In-house PW = $1 786 912 9.21

(a) The PW of the project is obtained by subtracting the PW of operating cost and the first cost from the PW of revenues.

Notice that the present worths are obtained using the current dollar MARR since the current dollar MARR and the real MARR are the same under the assumption of zero inflation. The present worths of operating and cost and revenues are shown below.

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Chapter 9 - Inflation Metcan No Inflation Cash Flows: Operating Year Cost Revenue 1 332 500 398 125 2 332 500 402 106 3 332 500 406 127 4 332 500 410 189 5 332 500 414 290 6 332 500 418 433 7 332 500 422 618 8 332 500 426 844 9 332 500 431 112 10 332 500 435 423 f= 0 Total: i= 0.2

PW Oper. Cost 277 083 230 903 192 419 160 349 133 624 111 354 92 795 77 329 64 441 53 701 1 393 997 PW =

PW Revenue 331 771 279 240 235 027 197 815 166 494 140 132 117 945 99 270 83 552 70 323 1 721 571 152 574

This gives a present worth for the project under the assumption of zero inflation of $152 574. (b) The present worths are shown below. They were obtained with current dollar cash flows and the current dollar MARR. Inflation rate 0 0.01 0.02 0.03 0.04

PW 152574 108775 63019 15213 –34742

Cash flows under the assumption of 1% inflation are shown below.

Metcan 1% Inflation Cash Flows: Operating Year Cost Revenue 1 332 500 398 125 2 335 825 402 106 3 339 183 406 127 4 342 575 410 189 5 346 001 414 290 6 349 461 418 433 7 352 955 422 618 8 356 485 426 844 9 360 050 431 112 10 363 650 435 423 f= 0.01 Total: i= 0.2

PW Oper. Cost 277 083 233 212 196 287 165 208 139 050 117 034 98 503 82 907 69 780 58 732 1 437 795 PW =

PW Revenue 331 771 279 240 235 027 197 815 166 494 140 132 117 945 99 270 83 552 70 323 1 721 571 108 775

(c) The project is only marginally profitable with inflation at 3%. We checked at 3.5%, and found a project present worth of –$490. That is, inflation must be less than 3.5% for this project to break even.

Inflation rates above 3.5% are well within the normal range. A conservative decision, therefore, would be to reject the project.

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9.22

See the table:

Year 0 1 2 3 4 5 6

In-house current $ 25 000 108 000 113 400 119 070 125 024 131 275 137 838

Contract current $ 55 000 55 000 55 000 63 700 63 700 63 700 Total:

PW PW in-house contract 25 000 88 525 45 082 76 189 36 952 65 573 30 289 56 435 28 754 48 572 23 569 41 803 19 319 402 097 183 965

The present worths were computed with current dollar cash flows and the current dollar MARR. The contract has the lowest costs for the company.

C. More Challenging Problems 9.23

Depends on specific circumstances

9.24

Depends on specific circumstances

9.25

(a) Comment on real yen receipts: These are shown in yen of year 1. They might have been shown in yen of year 0. Either would have been correct.

Year 0 1 2 3 4 5 6 7 8

Receipts Inflation Current Yen (year 1 yen) Factor Receipts 1 30 000 000 1 30 000 000 33 000 000 1.01 33 330 000 36 300 000 1.02 37 029 630 39 930 000 1.03 41 139 919 43 923 000 1.041 45 706 450 48 315 300 1.051 50 779 866 53 146 830 1.062 56 416 431 58 461 513 1.072 62 678 655

Rate Current Dollar Receipts Worth of Factor Receipts (Yr 0 Dollars) Receipts ($) 0.015 0.0153 459 000 445 631 376 230 0.0156 520 148 490 289 349 468 0.0159 589 442 539 423 324 610 0.0162 667 968 593 481 301 520 0.0166 756 954 652 955 280 073 0.0169 857 796 718 390 260 151 0.0172 972 071 790 383 241 646 0.0176 1 101 571 869 590 224 457 Total: 2 358 153

The present worth of receipts is $2 358 153. This was obtained using the current dollar receipts and the current dollar MARR, 22%.

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(b) The present worth of the cash outflow is $1 997 069, which was computed using current dollar cash outflow and the current dollar MARR.

Year 0 1 2 3 4 5 6 7 8

9.26

Real Dollar Cash Outflow Current Dollar (Yr 0 dollars) Cash Outflow 200 000 200 000 339 806 350 000 373 786 396 550 411 165 449 291 452 282 509 047 497 510 576 750 547 261 653 458 601 987 740 368 662 185 838 837 Total:

Present Worth of Cash Outflow ($) 200 000 286 885 266 427 247 428 229 783 213 397 198 179 184 047 170 922 1 997 069

(a) PW = –1 800 000 + 550 000(P/A, i, 5) The real Ibernian IRR, based on a spreadsheet approximation, is just over 16.0%. Sample trial and error results are shown below: Interest rate 0.1575 0.16 0.161 0.1625

PW 11414 862 –3331 –9591

(b) The current pound (CP) internal rate of return is given by IRRAP = (1 + IRRRP)(1 + fI) – 1 where IRRRP is the real IRR in pounds and fI is the inflation rate in Ibernia.

If we take the real IRR in Ibernian pounds to be 16%, we get the current pound IRR to be 0.276, or 27.6% (c) The current dollar IRR is given by IRRA = (1 + IRRRP)(1 + fI)(1 + fe) – 1 where IRRRP is the real IRR in pounds, fI is the inflation rate in Ibernia, and fe is the rate of exchange between the dollar and the Ibernian pound.

If we take the real IRR in Ibernian pounds to be 16%, we get 21.2% as the current dollar IRR.

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Alternative method: Get the current dollar IRR by converting current Ibernian cash flows into actual dollars and computing the current dollar IRR directly. Sample trial and error results are shown below: Interest rate 0.2075 0.21 0.2125 0.215

PW 24899 12176 –430 –12922

We must compare the 21.2% with Leftway’s current dollar MARR. This is obtained with the relationship, MARRC = (1 + MARRR)(1 + f) – 1 = 1.15(1.025) – 1 = 0.179 Leftway’s current dollar MARR is about 18%. This is well below the actual dollar IRR. The project is, therefore, acceptable.

9.27

(a) The real MARR is given by MARRR = (1 + MARRC)/(1 + f) – 1 = 0.113 This gives 11.3%. (b) PW = –220 000 + (50 000 + 30 000)(P/A, i, 4) + 20 000(P/F, i, 4) The real IRR is between 19% and 19.5%. This is based on a spreadsheet approximation. Sample trial and error results are shown below: Interest rate 0.185 0.19 0.1925 0.195

PW 3272.06 1060.22 –30 –1115.6

(c) The current IRR is given by IRRC = (1 + IRRR)(1 + f) – 1. If we take the real IRR turn as 19.25%, we get the current IRR as 26.4%. (d) The current IRR may be computed with a spreadsheet by first computing the current cash flows using the inflation rate. We get a current IRR to be between 26% and 26.5%. Sample trial and error results are shown below: Interest rate 0.255 0.26 0.265 0.27

PW 3735.71 1640.94 –421.57 –2452.5

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(e) The present worth may be obtained with the current cash flows and the current MARR or with real cash flows and the real MARR. PW(savings, real) = 80 000(P/A, 11.3%, 4) = 246 614.21 PW(scrap, real) = 20 000(P/F, 11.3%, 4) = 13 033.15 PW(first cost) = 220 000 PW(total) = 39 647.37 We get the present worth of about $39 650.

9.28

(a) PW = –15 500 000 + 325 000  (35 – 12 – 7.75)(P/A, i, 6) The real internal rate of return is close to 22.5%. Sample trial and error results are shown below: Interest rate 0.22 0.225 0.23 0.235

PW 195 948 9098 –173 788 –353 700

(b) We want the inflation rate that will make the real MARR equal to 22.5%.

1 + MARRR = (1 + MARRC)/(1 + f) f = (1 + MARRC)/(1 + MARRR) – 1 = 1.25/1.225 – 1 = 0.0204 We see that inflation must exceed 2.0% per year for this project to break even. (c) We can get the present worths by discounting the real cash flows with the real MARR’s that are implied by the three inflation rates, 1%, 2%, and 3%.

Inflation rate 0.01 0.02 0.03

MARRR 0.2376 0.2255 0.2136

PW –446503 –8940 441011

(d) This is a close call. We see that we need to have inflation over 2.0% per year for the project to break even. This is on the low side historically for inflation rates. But there have been countries with periods when inflation has been this low.

9.29

(a) We first get the effect of reductions in operating cost and revenue on cash flows.

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We get the present worth of these flows by discounting at the observed MARR, 20%.

No Inflation Cash Flow: Operating Year Cost Revenue 1 4 262 500 6 050 000 2 4 219 875 5 959 250 3 4 177 676 5 869 861 4 4 135 899 5 781 813 5 4 094 540 5 695 086 6 4 053 595 5 609 660 7 4 013 059 5 525 515 8 3 972 929 5 442 632 9 3 933 199 5 360 993 10 3 893 867 5 280 578 Total: i= 0.2

PW Oper. Cost 3 552 083 2 930 469 2 417 637 1 994 550 1 645 504 1 357 541 1 119 971 923 976 762 280 628 881 17 332 893 PW =

PW Revenue 5 041 667 4 138 368 3 396 910 2 788 297 2 288 727 1 878 664 1 542 070 1 265 782 1 038 996 852 843 24 232 325 -600 568

We subtract the present worth of variable costs and the first cost from the present worth of revenue. This gives a present worth of about –€601 000. (b) The internal rate of return is between 17% and 17.5%. This is both a real and a current dollar internal rate of return since we assume inflation is zero. Sample trial and error results are shown below: Interest rate 1.165 1.17 1.175 1.18

PW 262576 128455 –1836 –128438

(c) The required inflation rate is given by 1 + IRRC = (1 + IRRR)(1 + f) f = (1 + MARRC)/(1 + MARRR) – 1 = 1.2/1.175 – 1 = 0.0213 = 2.13% (d) The project probably should be rejected. The break-even requires an inflation rate of over 2.5%. While this is not high by historical standards, there have been many countries with periods with lower inflation rates.

9.30

(a) The cash flows as well as present worths of operating costs and revenues under the assumption of zero inflation are shown below.

No Inflation Cash Flows: Operating Year Cost Revenue 1 1 200 000 1 500 000 2 1 248 000 1 530 000 3 1 297 920 1 560 600 4 1 349 837 1 591 812 5 1 403 830 1 623 648

PW Oper. Cost 960 000 798 720 664 535 552 893 460 007

PW Revenue 1 200 000 979 200 799 027 652 006 532 037

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No Inflation Cash Flows: Operating Year Cost Revenue 6 1 459 983 1 656 121 7 1 518 383 1 689 244 8 1 579 118 1 723 029 9 1 642 283 1 757 489 10 1 707 974 1 792 639 f= 0 Total: i= 0.25

PW Oper. Cost 382 726 318 428 264 932 220 423 183 392 4 806 057 PW =

PW Revenue 434 142 354 260 289 076 235 886 192 483 5 668 118 -37 939

We see that the project present worth is negative under the assumption of zero inflation. (b) The present worths are shown below: Inflation rate 0 0.01 0.02

PW –37939 –18801 998

Cash flows under the assumption of 2% inflation are shown below: 2% Inflation Cash Flows: Operating Year Cost Revenue 1 1 200 000 1 500 000 2 1 272 960 1 560 600 3 1 350 356 1 623 648 4 1 432 458 1 689 244 5 1 519 551 1 757 489 6 1 611 940 1 828 492 7 1 709 946 1 902 363 8 1 813 910 1 979 218 9 1 924 196 2 059 179 10 2 041 187 2 142 369 f= 0.02 Total: i= 0.25

PW Oper. Cost 960 000 814 694 691 382 586 735 497 926 422 560 358 602 304 324 258 261 219 171 5 113 655 PW =

PW Revenue 1 200 000 998 784 831 308 691 914 575 894 479 328 398 954 332 058 276 378 230 035 6 014 654 998

(c) We see that for inflation rates lower than 2%, the project is not profitable. A conservative decision would be to reject the project. While 2% is on the low side historically, there have been countries with periods with inflation this low.

9.31

(a) Under the assumptions of fixed real prices and no inflation we have simply PW = (Revenue – Operating cost)(P/A, i, N) – (First cost) For wood veneer, we have: PW = (960 000 – 375 000)(P/A, 25%, 10) – 2 050 000 = $39 744 For plastic top, we have: PW = (1 170 000 – 405 000)(P/A, 25%, 10) – 2 700 000 = $31 435 181 Copyright © 2022 Pearson Canada Inc.

Chapter 9 - Inflation

(b) For 2% inflation we can work with real cash flows and the real MARR = 1.25/1.02 – 1 = 0.22549. For the plastic tops, the equation from part (a) is used with the new real MARR of 22.549% to find a present worth of $248 557. For wood veneer top desks, we can get the changing real values of cash flows. We can then work with these and the real MARR. These cash flows and present worths are shown below: 2% Inflation: Wood Cash Flows (real dollars) Operating PW Year Cost Revenue Oper. Cost 1 375 000 960 000 306 000 2 393 750 969 600 262 181 3 413 438 979 296 224 637 4 434 109 989 089 192 469 5 455 815 998 980 164 907 6 478 606 1 008 970 141 292 7 502 536 1 019 059 121 059 8 527 663 1 029 250 103 724 9 554 046 1 039 542 88 870 10 581 748 1 049 938 76 144 Total: 1 681 283

PW Revenue 783 360 645 614 532 089 438 527 361 416 297 865 245 488 202 322 166 745 137 425 3 810 851

We then subtract the present worth of operating costs and first cost from the present worth of revenue to get the project present worth as $79 568.

9.32

(a) We must get the annuities that are equivalent to the first costs for both systems. We then add this to the operating costs per unit.

These are shown below: Annual worth comparison for fixed prices: Inflation Real Fixed Operating Total rate (/year) MARR cost (/year) cost (/unit) cost (/unit) Manual 0% 20% 11926 10.50 11.30 Automated 0% 20% 57245 9.00 12.82

(b) The automated system has a higher first cost than the manual system. However, its operating costs are lower and stable. The effect of the higher first cost is reduced by a lower real MARR. Therefore, a higher expected inflation rate, which implies a lower real MARR, favours the automated system. As well, it is clear that a higher rate of increase in the real operating cost for the manual system favours the automated system. The manual system has a lower present value of costs for zero inflation and a 4% per year increase in manual operating costs. But the automated system is better for a 2% inflation rate and 4% rate of increase in manual operating costs and for no inflation combined with a 5% rate of increase in manual operating costs. A sample table for 2% inflation and 4% rate of increase in manual operating cost is shown below. The table contains real cash flows. The present worths are computed with the real MARR of 17.6%.

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Chapter 9 - Inflation 2% Inflation Cash Flows (real dollars): Manual Automatic PW PW Year Oper. Cost Oper. Cost Oper. Cost Oper. Cost 1 157 500 133 875 135 000 114 750 2 163 800 118 346 135 000 97 538 3 170 352 104 617 135 000 82 907 4 177 166 92 482 135 000 70 471 5 184 253 81 754 135 000 59 900 6 191 623 72 270 135 000 50 915 7 199 288 63 887 135 000 43 278 8 207 259 56 476 135 000 36 786 9 215 550 49 925 135 000 31 268 10 224 172 44 134 135 000 26 578 Total: 817 766 614 391 PW(Total)= 867 766 PW(Total)= 854 391

Since an inflation rate of 2% per year is at the low end of the typical range, the automated system appears to be the preferred alternative.

9.33

The cost of living clause clearly lowers risk for workers who are protected against inflation that is unknown at the time of the contract signing. It is not clear what the effect is for employers. There are two possibilities. (1) If revenues do not go up and down with general inflation, the contract increases their risk. This is because inflation that leads to an increase in wages without an offsetting increase in revenue would lead to lower real net revenue. (2) If revenues do go up and down with inflation, the same inflation that would lead to a rise in current dollar wage rates would also lead to an increase in revenue. Therefore, there would be no additional risk for employers in the second case.

9.34

First, note that the real MARR for Mid-Atlantic is the same, whether the money is invested in Columbo or Avalon. However, because of the different inflation rates, the actual MARR will be larger in Columbo rather than Avalon: Actual MARR in Columbo is: 0.1 + 0.04 + (0.1 × 0.04) = 0.144 Actual MARR in Avalon is: 0.1 + 0.02 + (0.1 × 0.02) = 0.122 To accumulate $200 000 = £100 000 in 10 years, the amount set aside in Columbo would be: 200 000(P/F, 0.144, 10) = 200 000 × 0.260459 = $52 092 = £26 046 The amount set aside in Avalon would be: 100 000(P/F, 0.122, 10) = 100 000 × 0.31628 = £31 628= $63 256 Consequently, the amount of $52 092 should be set aside in Columbo.

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This may seem counter-intuitive. Since pounds lose value more slowly, it would be natural to assume that Mid-Atlantic should keep its money in pounds. But since the exchange rate is fixed, and the actual MARR is larger in Columbo, Mid-Atlantic is actually better off investing its money in Columbo.

Consequently, when looking at the two investment opportunities, regardless of which one Mid-Atlantic pursues, it should exchange pounds for dollars wherever possible. Thus, although the investment and receipts in Avalon are in pounds, the equivalent amounts in dollars should then be used for calculation purposes under the assumption that pounds and dollars will be exchanged immediately as needed or received.

The present worth ten years from now of $200 000 invested in Columbo: PC = 0.3(P/A, 0.144, 5) × 200 000 = 0.3 × 3.586987 × 200 000 = $215 219 = £107 610 The present worth ten years from now of $200 000 = £100 000 invested in Avalon (using the Columbo actual MARR since the pound receipts are immediately converted to dollars): PA = 0.24(P/A, 0.144, 7) × 100 000= 0.23 × 4.534959 × 100 000 = 1.043 × 100 000 = £104 300 = $208 600 Consequently, Mid-Atlantic should make its investment in Avalon, converting all receipts from pounds to dollars as they are received.

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Notes for Mini-Case 9.1 1)

No inflation rate can change the decision of the high-pressure pipeline over the conventional pipeline, if only the capital cost is considered, since both projects would be affected in the same way. However, the difference in the savings would have been significantly smaller with a 4% inflation rate for the same actual dollar capital costs.

2)

The student should be able to demonstrate that there is a significant opportunity to control the best choice using a selected inflation rate when the cash flow structures of projects under consideration are very different.

3)

They might have done so to demonstrate that inflation was being considered. If they were well informed, they would have used actual dollars, but they may have mistakenly used real values.

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CHAPTER 10 Solutions to Chapter-End Problems A. Key Concepts Benefits and Costs: 10.1

Some examples: Cost/Benefit Noise pollution/people wake up Snowplow damage to roads/property Fewer accidents/towing Visible demonstration of taxes being used Snow in people’s driveways

10.2

How measured Public complaints—hard to quantify Observation of damage Difference in number by experiment Political benefit only Cost of removal

Some examples: Cost/Benefit: Cost of maps, etc.

Telephone costs Insurance More tourist stays in province/state Tourists more likely to return Attractions/hotels get advertising Tourists have good impression Healthy to stop and stretch/picnic

How measured: as incurred as incurred as incurred survey effectiveness survey effectiveness survey source of customers survey, but hard to quantify hard to measure

Benefit-Cost Ratios: 10.3

(a) BCR = 17 000 000/(5 000 000 + 6 000 000) = 1.546 (b) BCRM = (17 000 000 – 5 000 000)/6 000 000 = 2

10.4

(a) BCR(A) = 19 000 000/(5 000 000 + 5 000 000) = 1.9 BCR(B) = 15 000 000/(8 000 000 + 1 000 000) = 1.667 (b) BCRM(A) = (19 000 000 – 5 000 000)/5 000 000 = 1.667 BCRM(B) = (15 000 000 – 8 000 000)/1 000 000 = 7 (c) BCR(A–B) = (19 000 000 – 15 000 000)/(10 000 000 – 9 000 000) = 4 (d) PW(A) = 19 000 000 – 5 000 000 – 5 000 000 = 9 000 000 PW(B) = 15 000 000 – 8 000 000 – 1 000 000 = 6 000 000 (e) The preferred project is A, whether one uses net present worth or BCR. Both projects are found to have acceptable BCRs. Ranking 187 Copyright © 2022 Pearson Canada Inc.

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from the greatest to least PW(costs) requires calculation of the BCR for A–B. The value is greater than one, so the increment is preferred, and so project A is preferred. Note that the value of the BCR for B–A would be numerically identical, also 4.

10.5

(a) BCR(A) = 17 000 000/(5 000 000 + 6 000 000) = 1.545 BCR(B) = 17 000 000/(11 000 000 + 1 000 000) = 1.417 (b) BCRM(A) = (17 000 000 – 5 000 000)/6 000 000 = 2 BCRM(B) = (17 000 000 – 11 000 000)/1 000 000 = 6 (c) BCR(B–A) = (17 000 000 – 17 000 000)/(12 000 000 – 11 000 000) = 0 (d) PW(A) = 17 000 000 – 5 000 000 – 6 000 000 = 6 000 000 PW(B) = 17 000 000 – 11 000 000 – 1 000 000 = 5 000 000 (e) The preferred project is A, whether one uses net present worth or BCR. The incremental BCR is calculated for B–A since B has the greatest PW(costs). The BCR of zero results from equal benefits, so project A, with the lower PW(costs), is chosen.

10.6

(a) BCR(A) = 17 000 000/(5 000 000 + 6 000 000) = 1.545 BCR(B) = 15 000 000/(8 000 000 + 3 000 000) = 1.364 (b) BCRM(A) = (17 000 000 – 5 000 000)/6 000 000 = 2 BCRM(B) = (15 000 000 – 8 000 000)/3 000 000 = 2.333 (c) BCR(A–B) = (17 000 000 – 15 000 000)/(11 000 000 – 11 000 000) = not defined (d) PW(A) = 17 000 000 – 5 000 000 – 6 000 000 = 6 000 000 PW(B) = 15 000 000 – 8 000 000 – 3 000 000 = 4 000 000 (e) With an undefined incremental BCR resulting from equal costs, we chose the project with the greater PW of benefits, project A.

Market Failure: 10.7

Air pollution is an example of market failure because people who drive cars or own polluting companies benefit from the inexpensive transportation or inexpensive production of goods. However, many people are negatively affected by the air pollution. The costs to these people, and to our health system, probably exceed pollution prevention costs.

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Economic regulation by government: already being done by regulations governing emission controls for cars and factories; pedestrian-only zones in cities, alternating driving days, plant shutdowns.

Monetary incentives or deterrents: subsidized mass transit, gas taxes; pollution fines.

Seeking compensation in courts: permitting suits against air pollution.

Note, however, who would sue whom here? The reason for market failure is that there are large numbers of drivers who dump into the air, and large numbers of persons who are adversely affected. It is not worth the while for those who are adversely affected to do much about dumping since the costs of seeking compensation would exceed any compensation. This might be overcome by class action suits. The large number of persons adversely affected would have to form a coalition. This might be difficult.

The coalition would have to identify and sue all the individuals who dump.

Government provision of goods and services: government-run mass transit, e.g., GO train in Toronto.

10.8

Overfishing is an example of market failure because individual fishers benefit from taking the available fish. However, all fishers, and society at large, suffer because insufficient stock remains to restore the supply.

Economic regulation by government: already being done by regulations governing fish quotas, fishing licenses.

Monetary incentives or deterrents: payments to displaced fishers, subsidies to economic development of fishing communities.

Seeking compensation in courts: perhaps can be applied to action against native or foreign fishers.

Government provision of goods and services: government restocking programs, international treaties.

10.9

Where there is only one family affected by the Smiths’ trees, the Smiths’ refusal to top their trees does not imply that there is market failure. The Smiths’ refusal of the Browns’ compensation offer means that the trees are worth more to the Smiths than the view is worth to the Browns. There would be a social loss if the trees were cut.

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Where there are four families affected by the Smiths’ trees, the situation is not simple. The affected families would have to agree to share the cost of inducing the Smiths to top their trees. The bargaining over sharing the cost may be difficult. It is possible that the total value of the view for all four families exceeded the value of the tall trees to the Smiths. In this case, there would be market failure.

A reduction in biodiversity might cause a loss for many people, possibly for everyone on earth. All these individuals could not form a voluntary coalition to bargain with the forest companies to get them to change their cutting practices. Even if no individual sustained a large cost from a reduction in biodiversity, the total value of costs may exceed the value of clear cutting to forest companies. In this case, there would be market failure.

10.10 The technical changes reduce the need for regulation. Regulation was important when scale economies made single supplier provision of these services more efficient than multiple supplier provision. Single suppliers had monopoly power that they could have exploited. Regulation was used to prevent excessive prices, and to maintain quality standards in these industries. The existence of multiple suppliers means that the potential for monopolistic exploitation of consumers is reduced.

MARR on Government Projects: 10.11 The MARR on government projects is determined by the opportunity cost of the funds invested in the project. We usually assume that funds come from the private sector. In the private sector, the funds could come from either consumption or from investment. The tax on investment income makes the return to savers lower than the actual physical return. That is, the rate at which individuals are willing to give up current consumption to make an investment is lower than the rate the investment returns. The opportunity cost, therefore, is between the before-tax and the after-tax rates earned on investments.

10.12 The MARR on government projects is determined by the opportunity cost of the funds invested in the project. We usually assume that funds come from the private sector. In the private sector, the funds could come from either consumption or from investment. The tax on investment income makes the return to savers lower than the actual physical return. That is, the rate at which individuals are willing to give up current consumption to make an investment is lower than the rate the investment returns. The opportunity cost, therefore, is between the before-tax and the after-tax rates earned on investments. A reduction in the tax on investment income will narrow the spread of possible MARRs for government projects.

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10.13 The MARR on government projects is determined by the opportunity cost of the funds invested in the project. We usually assume that funds come from the private sector. The actual dollar return on investments in the private sector rises with an expectation of inflation. Therefore, the actual dollar MARR for government projects also rises.

B. Applications 10.14 (a) If Plant B and Plant C met Plant A’s performance of 10 cl/kilogram, Plant B would reduce dumping by 5 cl/kg and Plant C by 8 cl/kg. Total reduction in dumping would be 5(11 000) + 8(8000) = 119 000 kg per day.

The cost for Plant B would be (0.06 + 0.06 + 0.063 + 0.068 + 0.075)(11 000) = $3586 The cost for Plant C would be (0.25  7 + 0.375)(8000) = $17 000 The total cost of best practice regulation is $20 586 per day. (b) A tax of $0.20 per centilitre dumped would cause Plant A to reduce dumping by 5 cl/kg. Plant B would reduce by 6 cl/kg. Plant C would choose not to reduce dumping and to pay the tax.

Total reduction in dumping would be 5(17 000) + 6(11 000) = 151 000 cl per day. (c) The cost for Plant A would be (0.02 + 0.032 + 0.048 + 0.1 + 0.18)(17 000) = $6460 per day The cost for Plant B would be (0.06 + 0.06 + 0.063 + 0.068 + 0.075 + 0.193)(11 000) = $5709 per day.

10.15 The value of the benefit to consumers has two parts: (1) the benefit due to the lower price on power that is consumed even at the high price, and (2) the benefit due to the increased consumption. The value of the lower price on existing consumption is: 0.02(9 000 000) = $180 000.

The value of increased consumption is given by the value to users less their cost. The value per unit to users can be approximated by a point half way between the old and new price. The value must be at least $0.05/kwh, otherwise there would be no new consumption. It must be less than $0.07/kwh, otherwise the higher consumption would have existed even with the higher price.

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We get [(0.07 + 0.05)/2 – 0.05](12 250 000 – 9 000 000) = $32 500 The total benefit is then 180 000 + 32 500 = $212 500.

10.16 (a) The benefit to travellers has two parts: (1) the benefit of the zero price on crossings that would be made at $1.50 per crossing, and (2) the benefit of increased crossings. These increased crossings are valued at $0.75 per crossing—half way between the $1.50 value and zero.

This gives Benefit to users ($/year)

Total 112 500

Lower price 90 000

Increased use 22 500

(b) The owners would lose their profits. These are (Price – Cost)  (Use).

This comes to $30 000 per year. (c) The taxpayers incur the cost of building and maintaining the bridge.

This is $85 000/year. (d) (Annual benefits – Annual costs) = –$2500 (e) The situation would be made worse by charging a toll. This would transfer money from travellers to taxpayers with no net gain or loss to society as a whole. It would also reduce use. Under the assumption that use entails no cost, this reduction in use is a pure loss.

10.17 (a) Amount saved = 200(2)(75)(5.5)(52) = $8 580 000 per year (b) You could survey similar locations for an estimate. (c) You could, for example, find the average cost of the highway per kilometre per year, then divide by the average number of cars per kilometre to work out the cost per car per kilometre. In addition, to capture the benefits of reducing congestion, you could estimate the cost of a traffic jam and the amount traffic jams would be marginally reduced by the removal of one car.

10.18 Costs: There would have to be specialized equipment for collecting, sorting, and processing the various types of newspaper, cardboard, and cans. This information could be obtained by finding out what equipment would have to be purchased and what facilities built. Price estimates could then be obtained.

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There will be costs associated with recycling boxes and composters for each household. These costs could be estimated by finding the number of households and the number and type of containers used by each.

There would also be costs borne by each household for separating the waste. These costs could be estimated by determining how much time is necessary to separate waste each week, and by multiplying by an appropriate wage. If some householders feel that the cost is negligible, it may be valid to reduce the total cost by an appropriate amount.

Benefits: The newspaper, paper, and cans can be sold. These benefits could be estimated by finding out the approximate waste for each category generated by each house on a weekly basis, determining how much of it can be sold and then finding the market price for recycled materials.

The landfill site will fill more slowly, so there is a reduced cost associated with land acquisition and site management. By determining the volume of material sold and the amount of “wet” compostable material, analysts can infer the reduction in waste being brought to the landfill site on a weekly basis. This could be converted into a yearly amount and then an estimate made of the cumulative volume saved over the life of the landfill.

There may be a reduction in tipping fees as well, which can be determined by the forecasted reduction in volume of waste brought to the landfill site weekly.

10.19 (1) Reduced school cleaning and maintenance. This is partly a real benefit, and partly a transfer. It’s a benefit to the extent that the resources used for cleaning and maintenance, mostly labour, have good alternative uses on the off day. In this case, taxpayers receive the benefit. It’s a transfer from workers to taxpayers to the extent that the workers do not have good alternatives for the off days.

(2) Reduced use of school buses on the off day. This is partly a real benefit, and partly a transfer. It’s a benefit to the extent that the resources involved in school bus use have good alternative uses on the off day. Just keeping the buses in the garage yields some saving in resources. Taxpayers receive the benefit. It’s a transfer from workers, drivers, and others to taxpayers to the extent that the workers do not have good alternatives for the off days.

(3) Reduction in driving to school by parents, students, and staff on the off day. This is a real benefit received by those who avoid driving.

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reduced pollution. These are received by other drivers and by the population as a whole.

(4) Reduction in public transportation use on the off day. This is a real benefit to those who don’t have to use the transit system. If the public transit system is crowded, there will be some benefit to other users in reduced crowding. There will also be a transfer from the owners of the transit system to those who don’t have to use transit on the off days.

(5) Some high school students and school staff will seek part-time work for the off day. We have to distinguish between those who continue to have the same number of school hours, and those like bus drivers and cleaning staff whose hours are reduced. For those who continue to have the same number of total hours, there is a real benefit. This arises if having a whole day off as opposed to shorter days permits better use of the off time. The benefit will be received by whoever can make better use of the day off. This is not limited to those who seek part-time work. For those whose hours are reduced, whether or not there is a benefit, loss or a wash depends on the alternative opportunities for their time.

(6) Reduced absences by students and staff. This is mainly because some required personal activities could be scheduled for the off day.

This is a real benefit received by students if reduced absence increases effectiveness.

(7) Reduced subsidized school lunch requirements. This is a transfer from students and their families to taxpayers.

(8) Greater need for day care on the off day for working parents. About a third of elementary schools could be opened for day care. The costs would be covered by fees. This is a cost that would be borne mainly by the working parents.

(9) Learning by elementary students may be reduced because of their limited attention spans. This is a real cost borne by the students and their families.

(10) Lower school taxes. This is just a transfer to school taxpayers. The real effects have been counted already.

10.20 (a) We can compare the incremental benefit to the incremental cost.

Benefit/year = (10 km/trip)(2.5/km)(400 000 trips) = 10 000 000 pesos

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The incremental cost per year of the tunnel route over the base of the mountain route (in millions of pesos) is: Cost/year = 240 (A/P, 10%, 40) + 0.4 = 240 (0.10266) + 0.4 = 24.542240 + 0.4 = 24.9424 million pesos The incremental benefit of the tunnel route over the base of the mountain route is less than half the incremental cost. Therefore, based on the information given, we recommend the base of the mountain route. (b) The analysis does not take into account the value of time saved for travellers. (c) We have 14 942 400 pesos/year to make up with the value of time savings. At 400 000 trips per year, this is about 37.4 pesos per trip to save 6 minutes per trip. This implies a value of time of about 370 pesos/hour. We would need to know more about the travellers to say with confidence if this amount is too large to be reasonable for the opportunity cost of travellers’ time. (d) A toll would be mainly a transfer from travellers to taxpayers. It would also reduce the value of the tunnel route if waiting for the toll to be paid were required. A toll would not cause a recommendation in favour of the river route to change.

10.21 (a) Under equal emission reduction Old Gloria’s cost will be U$50 + U$45 = U$95. New Gloria’s cost will be U$40 + U$30 = U$70. The total cost will be U$165. (b) A tax on odour emission above U$25/(1000 OU/day) and below U$30/(1000 OU/day) will induce Old Gloria to reduce emission by 2000 OU/day and New Gloria to reduce by 5000 OU/day. The cost of this will be U$50/day for Old Gloria and U$105/day for New Gloria. (c) If the two plants are offered a similar amount, say U$27 (1000 OU/day) to reduce emission, Old Gloria will find it worthwhile to reduce emission by 2000 OU/day. New Gloria will find it worthwhile to reduce emission by 5000 OU/day.

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4. Reduction in accident cost.

The only incremental cost of the bridge is its first cost. (b) Some of the revenue lost by roadside business on the present road is due to the greater vehicle operating cost on the present road. The reduction in vehicle operating cost is already reflected in a separate category. Presumably the rest of the revenue will go elsewhere. This will offset the reduction of revenues to roadside businesses. There is a transfer, but not a cost or benefit. The toll is also just a transfer. This last is an approximation based on the assumption that the cost of collecting the toll is about the same as the cost of the government raising revenue in other ways. (c) The table assumes that traffic over the bridge will be the same with or without the toll. The toll is likely to reduce traffic over the bridge. (d) If the toll caused a diversion of traffic from the bridge to the present road, social benefits would be reduced. The benefits from lower vehicle costs, time costs, and accident costs would all be less.

10.23 If the only losses from floods were borne by land owners, the increase in the value of the land should reflect the reduction in flood losses. One wonders why the two numbers are not the same. Does this mean that the land owners do not capture the benefit of reduced flood losses?

10.24 (a) We must compute the present worth of costs. First costs are given as $3 750 000. Operating and maintenance costs for each year of operation are 7500 + 0.25q, where q is the number of crossings. The number of crossings at zero toll is 11 000. Thus, annual operating and maintenance costs are 7500 + 0.25(11 000) = $10 250.

We then get the present worth of this as of the end of year 1. Then we bring that figure back to the start of year 1. The present worth of operating and maintenance costs is $84 582.

We must get the present worth of the annual benefits. Annual benefits were given in the text as $498 750. We get the present worth of these benefits as of the end of year 1. We then bring this back to the start of year 1. The present worth of the benefits is $4 115 612.

The net present worth is then $281 031. (b) BCR = 4 115 612/(3 750 000 + 84 582) = 1.073 (c) BCRM = (4 115 612 – 84 582)/3 750 000 = 1.075

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10.25 (a)

First cost PW(operating & maintenance costs) Estimated increased land value Parking garage Net PW BCR (garage as cost) BCR (garage as reduction in benefit)

Road improvement ($) 15 000 000 5 000 000 26 000 000 4 000 000 2 000 000 1.083 1.1

New buses ($) 4 500 000 2 000 000 18 000 000 1 500 000 1.091

(b) Garage as cost: BCR(road–bus) = (26 – 18)/(24 – 16.5) = 1.067 Garage as benefit reduction: BCR(road–bus) = (22 – 18)/(20 – 16.5) = 1.143 (c) The benefit-cost ratio of the difference is greater than 1 under either assumption. This is consistent with the road improvement having a greater present worth than the bus project.

10.26 Tax collections are not a benefit since the gain to taxpayers at large is offset by the increased payments by the farmers who pay the higher taxes. The increased crop value and the increased land value are the same thing stated in two ways. The increased land value should approximately equal the present worth of the increased crops. One should use only one of these values to avoid double counting.

10.27 (a) BCR = AW(benefits)/AW(costs) = (150 000)/[1 750 000(A/P, 5%, 20) + 20 000] = 0.935 (b) BCR = AW(benefits)/AW(costs) = (150 000 – 20 000)/[1 750 000(A/P, 5%, 20)] = 0.926 (c) The most they would be willing to pay is $1 620 000 as seen from the summary of spreadsheet calculations.

Price ($) 1 000 000 1 200 000 1 400 000 1 620 000 1 800 000 2 000 000

BCR 1.620 1.350 1.157 1.000 0.900 0.810

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10.28 BCR = AW(benefits)/AW(costs) = (150 000)/[500 000(A/P, 9%, 8)]= 1.66 The project is worthwhile.

10.29 (a) BCR = AW(benefits)/AW(costs) = (175 000 + 50 000)/[1 500 000(A/P, 5%, 10) + 8000] = 1.11 (b) BCRM

= [AW(benefits)-AW(operating costs)]/AW(capital costs) = (175 000 + 50 000 – 8000)/[1 500 000(A/P, 5%, 10)] = 1.17

(c) Both the BCR and the BCRM have ratios exceeding 1, and thus the project appears to be a good economic choice.

10.30 (a) Current dumping of BOD5 is 20 000(1.0) + (9000)(1.5) = 33 500 kg/day.

If both plants limited their dumping to 0.81kg/steer, dumping would be (29 000)(0.81) = 23 490 kg/day.

The cost of this reduction is as follows: Plant A must reduce their dumping by 1.0 – 0.81 = 0.19kg/steer.

Plant B must reduce their dumping by 1.5 – 0.81 = 0.69kg/steer.

Plant A: [(1.5)(0.1) + (2.4)(0.09)](20 000) = RUB 7320/day Plant B: [(4.5)(0.5) + (10.5)(0.1) + (13.5)(0.09)](9000) = RUB 40 620/day The total cost is $47 940/day. (b) If there were a tax of $4.8 per kilogram dumped, Plant A would reduce dumping by 0.3 kilograms per steer since, beyond that amount, it would cost Plant A more to reduce the BOD5 than the taxes saved. Similarly, Plant B would reduce dumping by 0.5 kilograms per steer. See the answer for parts (c) and (d) for further explanation.

The total reduction in BOD5 would be: Plant A: 0.3(20 000) = 6000 kg/day.

Plant B: 0.5(9000) = 4500 kg/day.

The cost would be Plant A: (1.5 + 2.4 + 3.6)(0.1)(20 000) = $15 000/day Plant B: (4.5)(0.5)(9000) = $20 250/day The total is $35 250/day. This is a lower total cost than with regulation and the reduction in the amount dumped is greater.

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(c) & (d) The reductions in the amounts dumped would be the same under the subsidy scheme as under the tax. Therefore, the costs would be the same. The behaviour would be the same because it would cost Plant A less than $4.8 per kilogram to reduce dumping by 0.3 kilograms per steer. Greater reductions would cost more than $4.8 per kilogram. It would cost Plant B less than $ 4.8 per kilogram to reduce dumping by 0.5 kilograms per steer. Greater reductions would cost more than $4.8 kilograms per steer. Avoiding paying tax that the company would otherwise pay (at $4.8/kg) is a benefit equivalent to gaining a subsidy (at $4.8/kg). (e) The tax and subsidy schemes encourages Plant A, which has a lower cost of reducing BOD5, to reduce dumping more than Plant B, which has higher costs.

C. More Challenging Problems 10.31 (a) We first get the present worth of benefits under the toll. The benefit has two parts: (1) the value of a lower cost to visit the park for visits that would take place without the bridge, and (2) the value of increased visits.

The bridge cuts the cost of a visit from $290 to $237.50. With the toll the cost of a visit to a traveller falls from $290 to $244.50. The annual benefit to travellers from the lower cost on existing visits is 8 000  (290 – 244.5) = $364 000. We must add to this the toll that is paid on existing visits, $56 000. This is a benefit to taxpayers. Notice that the total benefit for existing visits is $420 000 = 364 000 + 56 000. This is the same result that we got in the text for the annual benefit on existing visits. The toll has taken a benefit from travellers and transferred it to taxpayers. The final social benefit from a lower cost on existing visits, which is the sum of gains by travellers and taxpayers, is unaffected by the toll.

The annual benefit from increased visits is given by the average value per increased visit times the increase in the number of visits per year.

This is given by [(290 – 244.5)/2 + 244.5]  (10 600 – 8000) = $694 850. The traveller sees a cost of $244.50 per visit. But $7 of this goes to taxpayers. The actual cost of these extra visits is $237.50 per visit.

We could either compute the cost using the $244.50 and then have a benefit to taxpayers, or we could work with the actual cost of $237.50.

We have worked with the actual cost. This is given by (10 600 – 8000)(237.50)= $617 500.

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The total annual benefit from the bridge with the toll is $77 350 for the increased visitors plus $420 000 for the existing visitors, equalling $497 350.

We must get the present worth of the annual benefits. We get the present worth of the annual benefits as of the end of year 1. We then bring this back to the start of year 1. The present worth of the benefits is $4 104 060.

We now get costs. The costs consist of first cost and operating and maintenance costs. First costs are given as $3 750 000. Operating and maintenance costs for each year of operation are 7500 + 0.25q, where q is the number of crossings. The number of crossings at the $7 toll is 10 600. Annual operating and maintenance costs are $10 150. We then get the present worth of operating and maintenance costs as of the end of year 1. Then we bring that figure back to the start of year 1. The present worth of operating and maintenance cost is $83 756.32.

The net present worth then is $4 104 060 – 3 750 000 – 83 756.32 = $270 303.50. This is about $10 000 less than without the toll. (b) The benefits of increased use have been reduced by the toll because the increase in use is less than with no toll. The toll itself is neither a social benefit nor a cost. It is a cost to users and a benefit to taxpayers. These cancel out.

10.32 (a) The tax payments are A: $66.67(25 – 16.67) = $555.55/day B: $66.67(20 – 8.33) = $777.77/day Total = $1333.33/day (b) Total cost for A is $1111.11/day. For B, $1055.56/day. (c) Under regulation the only costs were BOD5 removal costs. These were $450/day for A and $400/day for B. Thus, taxation by itself leaves both A and B worse off. (d) A total benefit of $1322.21 per day divided equally between the two plants would leave A no worse off and B better off. This is a bit less than the tax collection of $1333.33/day.

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10.33 (a) Pool: BCR = AW(benefits)/AW(costs) = 450 000/[2 500 000(A/P, 5%, 25) + 300 000] = 0.909 This is not acceptable.

Tennis courts: BCR = AW(benefits)/AW(costs) = 60 000/[200 000(A/P, 5%, 8) + 20 000] = 1.178 The tennis courts are preferable. (b) BCRM = [AW(benefits) – AW(operating costs)]/AW(capital costs) Pool: BCRM = (450 000 – 300 000)/[$2 500 000(A/P, 5%, 25)] = 0.769 This is not acceptable.

Tennis courts: BCRM = (60 000 – 20 000)/[200 000(A/P, 5%, 8)] = 1.293 The tennis courts are preferable.

10.34 (a) & (b) The benefits are reflected in the increased land values, and in the increased crop values. If the land owners capture all the benefits of the increased crop values and there are no other effects on land prices the increase in land values should equal the present worth of the increased crop values. We should not count both increased land values and increased crop values.

When you compute the present worth of increased crop values, remember that the benefits start in year 2. Tax collections are a transfer from land owners to taxpayers as a whole. They are neither a benefit nor a cost.

Road use costs have two components, the time value of drivers and the cost of operating the vehicles. Computation of the cost due to time value of drivers assumes that drivers traverse the entire road, 24 km for Route A and 16 km for Route B. We get the time spent on a trip by dividing the number of kilometres on a route by the speed in km/h. The annual cost is simply the time multiplied by the value of time and by the number of vehicles per year. For example, for Route A we have (24/100)  15  1 000 000 = 3 600 000 as the annual value of travel time. Recall that benefits start in year 2 when computing road use costs.

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Chapter 10 - Public Sector Decision Making Costs (in $000s) Construction cost PW(O&M cost) PW(resurface after 10 years) PW(road costs) PW(road use costs: vehicle time) PW(road use costs: vehicle use) PW(total cost) PW(increased crop value) Net PW PW(tax collection increase) BCR(road use costs as cost) BCR(road use costs as reduction in benefit)

Route A 53 400 464.38 1 086.53 54 951 27 862.57 55 725.15 138 539 104 484.6 –34 054.12 6278.48 0.754 0.38

Route B 75 000 348.28 823.66 76 171.94 27 862.57 44 580.12 148 615 139 312.8 –9301.83 8371.30 0.937 0.878

Computation a b c d=a+b+c e f g=d+e+f h i=h–g = h/g = (h – e – f)/d

(c) The benefit-cost ratios are consistent in that they are less than one no matter how they are computed. (d) Neither road should be built. The benefit-cost ratios are less than one.

The present worth is negative.

10.35 Dam Construction: P1 = –300 000 – 750 000(P/F, 15%, 1) – 1 500 000(P/F, 15%, 2) = –300 000 – 750 000(0.86957) – 1 500 000(0.75614) = –2 086 388 Dam Operating and Maintenance: P2 = –30 000(P/A, 15%, 50)(P/F, 15%, 2) = –30 000(6.6605)(0.75614) = –151 088 Flood Damage and Irrigation benefits: P3 =(182 510 + 200 000)(P/A, 15%, 50)(P/F, 15%, 2) = 382 510(6.6605)(0.75614) = 1 926 424 PWdam = P1 + P2 + P3 = –2 086 388 – 151 088 + 1 926 424 = –311 052 Recreation Construction: P4 = –50 000(P/F, 15%, 2) – 20 000[(P/F, 15%, 12) + (P/F, 15%, 22) + (P/F, 15%, 32) + (P/F, 15%, 42)] = –50 000(0.75614) – 20 000(0.18691 + 0.04620 + 0.01142 + 0.00298) = –42 757

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Recreation Operating and Maintenance: P5 = –15 000(P/A, 15%, 50)(P/F, 15%, 2) = –15 000(6.6605)(0.75614) = –75 544 Recreation benefits: i = (1 + i)/(1 + g) – 1 = 1.15/1.0325 – 1 = 0.1138 (P/A, 3.25%, 11.38%, 50) = [(1 + i)N – 1]/[ i(1 + i)N][1/(1 + g)] = (1.113850 – 1)/[0.1138(1.1138)50](1/1.0325) = 8.472 P6 = 27 600(P/A, 3.25%, 11.38%, 50)(P/F, 15%, 2) = 27 600(8.472)(0.75614) = 176 806 PWrec = P4 + P5 + P6 = –42 757 – 75 544 + 176 806 = 58 505 PWdam+rec = –311 052 + 58 505 = –252 547 The present worths are negative at a 15% MARR for both projects. It is not possible to state, with certainty, what the correct MARR is. The MARR is based on the assumption that all funds for this project are taken from private investment that has a pre-tax return of 15%. A next step could be to find the highest value of MARR for which the present worth is positive. If this is “close” to 15%, the project may be justified by other non-quantitative factors that have not been discussed.

10.36 The combination of projects, 2, 6, 7 and 8, maximizes the number of lifeyears saved within the constraint of $660 000 000. This is in spite of not using all of the permitted $660 000 000. This may be found by experimenting with different combinations. A useful device for guiding these experiments is to compute the number of life-years saved per dollar for each project. These are shown below.

Total cost ($s) 18 000 000

Life years saved (per year) 14

Total cost per life year saved ($s) 1 285 714

2. Flashing lights and gates

30 000 000

20

1 500 000

3. Widen bridge #1

48 000 000

14

3 428 571

4. Widen bridge #2

28 000 000

10

2 800 000

5. Widen bridge #3

44 000 000

18

2 444 444

6. Pole density reduction

120 000 000

96

1 250 000

7. Runaway lane #1

240 000 000

206

1 165 049

8. Runaway lane #2

240 000 000

156

1 538 462

Project 1. Flashing lights

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10.37 All computations are for one kilometre. (a) The present worth of road widening costs per kilometre has two parts, the first cost and the present worth of the annuity equivalent to operating and maintenance costs. These sum to $243 078. (b) We get the reduction in the number of accidents when 6-metre lanes are replaced with 7.5-metre lanes. This reduces the number of accidents to 4.5 accidents per km per year. We multiply this by the cost per accident. This gives an annual value. We then get the present worth of this annuity. The present worth is $102 117. (c) Note that if we considered only material costs, the net present worth would be negative since costs from part (a) exceed the benefit of the reduction in material damages from part (b). To get the minimum value for avoiding a serious personal injury, we find the reduction in the number of serious injuries. This is 0.45 accidents per kilometre per year. We then get the annual worth of the difference between $243 078 and $102 117. This is $15 529. We divide 0.45 into $15 529. The answer is $34 510.

10.38 (a) The present worths (in millions of yuan) are: PW(A) = –200 + (26.4 – 2.4)(P/A,10%,25) = –200 + (26.4 – 2.4)(9.077) = 17.84 PW(B) = –256 + (33.6 – 2.4)(9.077) = 27.2 PW(C) = –416 + (56.8 – 4.0)(9.077) = 63.28 Project C is the best choice. (b) BCR(A) = (26.4  9.077)/(200 + 2.4  9.077) = 239.6/221.76 = 1.080 BCR(B) = (4.2  9.077)/(32 + 0.3  9.077) = 304.96/277.76 = 1.098 BCR(C) = (7.1  9.077)/(52 + 0.5  9.077) = 515.6/452.32 = 1.140 (c) First consider the increment from Project A to Project B: BCR(B–A) = [(33.6 – 26.4)  9.077]/[(256 – 200) + (2.4 – 2.4)  9.077] = 9.6 Since the ratio is greater than 1, B is better than A. Now consider the increment from Project B to Project C: BCR(C–B) = [(56.8 – 33.6)  9.077]/[416 – 256 + (4 – 2.4)  9.077] = 9.68 This confirms that Project C is the best choice.

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(d) We can see that the additional first cost does not change the best project choice because the present worth of Project C, now 39 280 000 yuan, remains the highest. But if we look at the new benefit-cost ratio of Project C, we find that it is less than that of Project B. BCR(revised) = 515.6/476.32 = 1.082 However, the incremental BCR from B to C remains greater than one.

BCR(C–B) = [(56.8 – 33.6)  9.077]/[416 + 24 – 256 + (4 – 2.4)  9.077] = 1.06 10.39 The incremental cost of starting with the four-lane road = 12 000 000 – 18 000 000(P/F, i, 10) + 10 000(P/A, i, 10) For years 1-5, there is time savings of 0.25 – 0.2 = 0.05 hours per car with the four-lane road alternative, which translates to $2 savings per car.

Similarly, for years 6-10, there is $3.43 savings per car with the four-lane road alternative.

The incremental benefit of starting with the four-lane road is: {[5000 + 200(A/G, i, 5)](2) + [6000 + 200(A/G, i, 5)](3.43)(P/ F, i, 5)}  (P/A, i, 5) (a) For MARR = 5%: P(cost) = 12 000 000 – 18 000 000(P/F, 5%, 10) + 10 000(P/A, 5%, 10) = 12 000 000 – 18 000 000(0.61391) + 10 000(7.7217) = 12 000 000 – 11 050 380 + 77217 = 1 026 837 P(benefit) = {[5000 + 200(A/G, 5%, 5)](2) + [6000 + 200(A/G, 5%, 5)](3.43)(P/ F, 5%, 5)}  (P/A, 5%, 5) = {[5000 + 200(1.90011)](2) + [6000 + 200(1.90011)](3.43)(0.78353)}(4.31431) = 10 760 + 17 146(4.31431) = 84 733 The incremental benefit of the four-lane road does not exceed the incremental cost with the MARR = 5%.

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(b) For MARR = 20%: P(cost) = 12 000 000 – 18 000 000(P/F, 20%, 10) + 10 000(P/A, 20%, 10) = 12 000 000 – 18 000 000(0.16151) + 10 000(4.1925) = 12 000 000 – 2 907 180 + 41 925 = 9 134 745 P(benefit) = {[5000 + 200(A/G, 20%, 5)](2) + [6000 + 200(A/G, 20%, 5)](3.43)(P/ F, 20%, 5)}  (P/A, 20%, 5) = {[5000 + 200(1.6405)](2) + [6000 + 200(1.6405)](3.43)(0.40188)}(2.9906) = 10 656 + 8723(2.9906) = 36 743 The incremental benefit of the four-lane road does not exceed the incremental cost with the MARR = 20%. (c) It is reasonable to assume that the opportunity cost of funds for the used in the road upgrading is between 5% and 20%. If the alternative to upgrading the road is private consumption, the opportunity cost is 5%. Private individuals deciding between consumption and saving face a choice between the return on a safe investment like a government savings bond, at 5%, and the benefit of consumption now. If the alternative to upgrading the road is private investment, the opportunity cost is what could be earned before tax on that investment, 20%. At either 5% or at 20%, the immediate building of the four-lane road is not preferred.

10.40 (a) Equivalent annual cost of Alternative A (in $000s) at 5%: = 420(A/P, 0.05, 15) + 52.5 = 420(0.09634) + 52.5 = 92.96 Equivalent annual cost of Alternative B (in $000s) at 5%: = 315(A/P, 0.05, 15) + 74 = 315(0.09634) + 74 = 104.35 (b) Equivalent annual cost of Alternative A (in $000s) at 20%: = 420(A/P, 0.20, 15) + 52.5 = 420(0.21388) + 52.5 = 142.33 Equivalent annual cost of Alternative B (in $000s) at 20%: = 315(A/P, 0.20, 15) + 74 = 315(0.21388) + 74 = 141.37 (c) Alternative A has lower equivalent annual cost at 5%. Project B has a marginally lower equivalent annual cost at 20%. For B to be preferred, we would have to assume that virtually all of the resources used in the project came from private investment. That over a third of the cost 206 Copyright © 2022 Pearson Canada Inc.

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for the project will come from increased dumping fees suggests that at least some of the resources will be from reduced private consumption. Alternative A is, therefore, preferred.

10.41 (a) The social benefits are a reduction in the health costs associated with exposure and the reduced environmental costs that accrue when the ban takes place. The costs are calculated using the base case of having the controls in place. The social costs are the lost value of the pesticide use to its users and to the manufacturer. (b) No, although it might seem so initially. The social costs and benefits of the ban both need to be computed starting from the same basecase. Since the benefits were computed comparing the ban to the pre-existing regulations, the costs must be computed on this basis as well. The social benefits to users and to the producers lost due to the ban already account for the costs of production and compliance, so reduction in production and compliance costs should not be treated as a savings.

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Notes for Case in Point 10.1 1-3) All of these questions depend on the judgement of the student.

Notes for Mini-Case 10.1 1)

Students could construct a number of cogent arguments here. The essential thing to notice is that access to food happens in the south for a number of reasons, including government support of infrastructure and shipping. It’s also fair to note that historical structural inequality has led to lower capital available to Indigenous communities to invest in local farming.

2)

This is a tricky conversation to manage, not least because the entire question is one of an extreme power difference between the two groups, measured by population, wealth, or access to the RCMP. While there are many interesting answers to this question, this is a false trade-off discussion, as the framing suggests that the only reasonable choice for mitigating climate change comes from overruling Indigenous Peoples’ rights to land. In fact, there are many excellent options for reducing carbon emissions.

3)

A number of proposals are possible here, including video conferencing, exchange visits between students to northern communities, and expanded environmental education that includes social implications as part of sustainability.

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CHAPTER 11 Solutions to Chapter-End Problems A. Key Concepts Project Life Cycle: 11.1

The purpose of a charter is to clearly and concisely document the agreed upon scope, objectives, and deliverables of a project. It is important because it serves as a communications device to stakeholders with different backgrounds and different interests. It helps ensure that all the project stakeholders “are on the same page.” It reduces the chance of miscommunication, confusion, and the possibility of conflicting stakeholder interests.

11.2

The purpose of project planning is to clarify who is accountable for every aspect of the project, the approach being taken, the major deliverables, and the timing of the key decisions and review points. It provides a road map for the project.

It is important because it is used to understand what needs to be done, when it needs to be done, and what resources will be necessary. It also provides a mechanism to measure project status and provides an early warning of problem areas, allowing risk management and the running of "what if" scenarios to mitigate any risks that may have arisen. Without adequate planning, projects may overrun on resources and put the project benefits in jeopardy.

11.3

One possible sequence is as follows (other sequences are possible because there are several steps in each stage, and they may occur concurrently): • • • • • • • • • • •

Identify stakeholders Define deliverables Form project team Identify project activities Develop schedule Estimate costs Approval of project plan Track project expenses Manage schedule changes Make progress reports to stakeholders Deliverables accepted by client

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11.4

The activities have been divided into stages as follows: Initiation: • Identifying who the project steering committee should be and how often they should meet Planning: • Establishing the project team members • Looking up industry standard costs for materials • Project funding is approved • A discussion of whether a programming team has worked together before on a project • Looking into whether your company has done a project like this before Execution: • Selecting a contractor to lay optical cabling • Pouring concrete footings Monitoring and Controlling • A meeting to discuss the implications of late delivery for a key piece of equipment needed to complete a project activity • Paying a bill and allocating its cost to the project Closure: • Documenting an unexpected complication in the project for future reference.

Work Breakdown Structure: 11.5

The purpose of a WBS is to define and organize the work to be done as part of a project. The WBS breaks the project deliverables into specific and measurable pieces of work that can be used to assign responsibility, to allocate resources, and to monitor the project. The WBS is useful because it makes the work to be done concrete so that the project team knows exactly what is to be accomplished within each deliverable. It allows for better estimating of cost, risk, and time because you can start from the smaller tasks and roll them up to the level of the entire project. It is also useful to double check all the deliverables with the stakeholders and make sure there is nothing missing or double-counted.

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11.6

The work packages have been divided as follows:

Food Guests Facilities Menu Planning Guest List Utensils Caterer Invitations/RSVP Seating/Tables Selection Decorations

Reunion Event Door Prizes Business Card Draw Name Tags Coat Check

Supervision Co-ordination Bill Payment Purchase Orders Budget

11.7

It is important that they not overlap so that work is identified only once in the WBS. If work packages overlap, then the amount of work, and thus resources will be overestimated. Also, overlapping work packages will make assigning responsibilities more difficult.

11.8

One possible answer is as follows. The project is to plan and hold a research workshop for approximately 30 people. Some of the attendees will make presentations on their research. There will be a keynote speaker.

The initial WBS is below: 0.0 Research Workshop 1.0 Planning 1.1 Select time and location 1.2 Select venue 1.3 Budget 1.4 Schedule talks 1.5 Plan meals 2.0 Room and Equipment 2.1 Tables and chairs 2.2 Data projector 2.3 Computer 3.0 Speakers 3.1 Call for presentations 3.2 Collect responses 3.3 Post tentative schedule 3.4 Post final schedule 4.0 Keynote Speaker 4.1 Select speaker 4.2 Make travel arrangements 4.3 Purchase gift

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5.0 Food 5.1 Budget 5.2 Plan meals 5.3 Initial booking 5.4 Food selection 5.5 Final booking 5.6 Pay bill After reviewing the initial WBS for whether a) it consists of distinct deliverables, b) it has mutually exclusive and collectively exhaustive items and c) work items are not larger than 80 hours of work, several changes needed to be made.

First, the “budget” item appears in several places. It should be broken down into food budget, travel budget (for the keynote speaker), and facilities budget.

Booking the venue is missing, and scheduling the talks and planning for food appears in two places. The speakers will bring their own computers for the presentation. The work packages appear to be under 80 hours each, so no changes are needed for this best practice. Below is a revised WBS (many other revisions are possible).

0.0 Research Workshop 1.0 Planning 1.1 Select time and location 1.2 Select venue 1.3 Overall budget 1.4 Select organizing team and responsibilities 2.0 Room and Equipment 2.2 Book room 2.2 Arrange for sufficient tables and chairs 2.3 Book data projector 2.5 Arrange for notepaper, pens 3.0 Speakers 3.1 Call for presentations 3.2 Collect responses 3.3 Post tentative schedule 3.4 Post final schedule 3.5 Registration (day of workshop) 4.0 Keynote Speaker 4.1 Select speaker 4.2 Make travel arrangements 4.3 Purchase gift 4.4 Give gift

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5.0 Food 5.1 Food budget 5.2 Plan meals 5.3 Initial booking 5.5 Final booking (1 week prior) 5.6 Pay bill Gantt Charts: 11.9

The Gantt chart is below. The project is expected to take 18 days to complete.

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11.10 The project is expected to take 22 weeks to complete.

11.11 a)

The Gantt chart is below:

Weeks 0

5

10

15

Activity 1 Activity 2 Activity 3 Activity 4 Activity 5 Activity 6 Activity 7 Activity 8 Activity 9 Activity 10

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b)

The project is expected to take 18 weeks to complete.

11.12 If activity 2 will take 7 weeks rather than 6 weeks, activities 4 and 5 will be delayed by one week. Since activity 4 is delayed one week, activity 6 will also be affected by one week, and hence activity 10 will be delayed, as well. Since activity 5 is delayed by one week, activity 7 will be delayed by one week (but will not delay the start of activity 10). This means that the project will be delayed one week to 19 weeks.

Weeks 0

5

10

15

Activity 1 Activity 2 Activity 3 Activity 4 Activity 5 Activity 6 Activity 7 Activity 8 Activity 9 Activity 10 PERT/CPM Networks: 11.13

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11.14

11.15 a)

b)

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11.16 a)

(Note that there are several ways this network can be correctly written.)

Here is a second solution:

b)

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Critical Path: 11.17 a)

This is the latest of the completion times for activities 4 and 5. Both must be complete for the project to be considered complete. b)

The earliest start time for activity 3 depends on when activities 1 and 2 are complete. Activity 1 takes the longest, completing by day 7. Even through activity 2 is done by day 4, activity 3 cannot start until both are done, which is day 7.

c)

The latest finish time for activity 1 depends on the latest start times for its successor activities (3 and 4). The latest start time for activity 4 is 10 days, and for activity 3 is 7 days. Therefore, activity 1 can finish no later than day 7 in order not to delay the project.

d)

The critical path is activities 1-3-5.

e)

The project should take 18 days to complete.

11.18 a)

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b)

The earliest activity 4 can begin is the earliest time at which both activities 2 and 3 are completed. Activity 2 can be completed at the earliest by week 15, and activity 3 by 12 weeks. Therefore activity 4 must wait until week 15 to start.

c)

The latest finish time for activity 2 depends on how late activities 4 and 5 can start. Activity 5 can start as late as week 17 and still have the project complete on time. Activity 4 must begin no later than week 15. Therefore, since both follow activity 3, activity 3 must finish no later than week 15 in order not to delay the project.

d)

The critical path is activities 1-2-4-6.

e)

The expected project completion time is 22 weeks.

11.19 a)

See diagram: Activity 1 T=3 days

Activity 6

Activity 3

ES=0

EF=3

LS=0

LF=3

T=4 days

ES=3

EF=7

LS=10 LF=14

T=4 ES=7 EF=11 days LS=14 LF=18

Start

Activity 2 T =2 days

Activity 4

ES=0

EF=2

LS=1

LF=3

T=4 days

ES=3

EF=7

LS=3

LF=7

T= 8 days

Activity 7

Activity 9

ES=7

EF=15

LS=7

LF=15

T= 1 ES=18 EF=19 day LS=18 LF=19

Activity 5 T =2 days

b) c) d)

Activity 8

ES=3

EF=5

LS=5

LF=7

T =3 ES=15 EF=18 days LS=15 LF=18

The project is expected to take 19 days.

The critical path activities are 1-4-7-8-9.

Activities 2, 3, 5, and 6 have slack of 1, 6, 2, and 6 days, respectively.

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11.20 a)

b)

The project is expected to take 18 weeks.

c)

The critical path is activities 2-4-6-10.

d)

Activities 1, 3, 5, 7, 8, and 9 have slack of 2, 12, 1, 1, 12, and 13 weeks respectively.

Project Crashing: 11.21 a)

See below. The critical path is activities 1-2-5-6 (29 days). Crashing activity 1 by three days will not create a new critical path, and will shorten the completion time of the project by 3 days.

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b)

See below. The critical path is activities 3-4-5-6 (28) days. Activity 1 has 10 days of slack. Therefore, crashing it by 3 days will not affect the completion time of the project. (The LS of activity 3 should be zero not 5).

11.22 a)

See below. The critical path is activities 1-3-5-6 (35 weeks). Activities 2 and 4 have 11 weeks of slack each. Crashing activity 1 by 7 weeks will not introduce new critical paths, and will reduce the slack in activity 2 by 7 weeks.

b)

The critical path is activities 1-3-5-6 (35 weeks). The slack in activity is 2 weeks and 1 week for activity 4. If activity 1 is crashed by 7 weeks, a new critical path emerges: activities 1-2-4-6. The slack in activity 2 is now 0 weeks, down by 3 weeks.

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11.23 a)

b)

Activities 2 and 4 are on the critical path since they have zero slack. Activity 2 costs (70 000 − 25 000)/((7-4) = $16 667 per week to crash, and can be crashed by up to 3 weeks. Activity 4 costs (40 000 − 20 000)/(5 − 4) = $20 000 per week to crash, and can be crashed by up to 1 week. The least cost way to crash the project by 1 week is to crash activity 2 by this amount. the additional cost is $16 667.

11.24 a), b)

a) The critical path is 3-5-6-7. b) The project will take 32 days to complete. c) The normal cost of the project is $70 000. d) The critical path activities are 3,5,6 and 7. The cost per day to crash activity 3 is (16 000)/4 = $4000 per day. The cost per day for activity 5 is 8000/4 = $2000 per day. The cost per day for activity 6 is ($16 000)/6 = $2667 per day, and for activity 7 is 8000/1 = $8000 per day. The least cost per day is $2000, which corresponds to activity 5.

B. Applications 11.25 What is the purpose of project monitoring and controlling? Why are these activities important?

The purpose of project monitoring and controlling is to track and communicate the project’s progress. It will identify when a project’s performance deviates significantly from the plan so that appropriate corrective actions can be taken.

Executing a project without monitoring and controlling is like driving a car with your eyes closed. You could find yourself heading over a cliff before you realize that slowing down and turning left could have kept you comfortably on the road.

Monitoring and controlling permit a manager to assess how well the project is proceeding compared to the plan. Controlling permits a manager to make changes to the execution of the project to put it back on track or to modify the plan if that is the appropriate course of action.

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What is the purpose of project closure? Why is it important?

The purpose of project closure is to formally conclude the project. Closure is important because it permits an opportunity to leverage all of the lessons learned from the project; loose ends are “wrapped up”, staff evaluations completed, and deliverables formally transitioned to support or operations. Resources allocated to the project will be terminated, and the closure document acts as a communications vehicle so that all stakeholders are aware of the results.

11.26 a)

The three examples are construction, software development, and process redesign. Many others can be found on the web.

Civil Engineering – Construction Example Market Demands or Perceived Needs Conceptual Planning and feasibility Study Design and Engineering Procurement and Construction Startup for Occupancy Software Engineering – Software Development Requirements Analysis and Planning System Design and Specification Coding and Verification Testing and Integration Implementation Maintenance Process Improvement Project Define Goals of Project Develop performance metrics for process Analyze current process Design improvements Pilot testing of new process and feedback Education/Training Rollout Post Implementation Assessment

Life Cycle Stage Initiation Planning Planning Execution/Monitoring and Controlling Closure Life Cycle Stage Initiation Planning Execution Execution/Monitoring and Controlling Closure (maintenance is post-closure) Life Cycle Stage Initiation Planning Planning Planning Execution/Monitoring and Controlling Execution Execution Closure

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b)

Construction example resources: People: architects, engineers, construction labourers, project manager, lawyers. Equipment: computer hardware and software for planning, construction machinery such as cranes, trucks, backhauls. Materials: concrete, steel, wood, furniture, heating and cooling equipment. Money: salaries, capital funding for construction costs.

Software engineering example resources: People: software engineers, programmers, project manager. Equipment: computers, printers, peripherals, test system equipment. Materials: paper, computer supplies. Money: staff salaries, materials.

Process improvement example: People: industrial or management engineers, project manager, domain expert staff. Equipment: computer systems and software, materials, process equipment as needed. Materials: paper, computer and other supplies. Money: salaries, funding for materials.

11.27 Initiation Planning

Execution

Monitoring and Controlling Closure

Decision to attend university made Decision to stay in residence made Agreement regarding any financial support from family struck Evaluate alternative residences and select preferred Create budget for year Plan what items you will need Evaluate transportation alternatives, select preference Send paperwork to university residence to confirm selection Purchase items needed for residence Pay residence fees Purchase transportation ticket(s) (if applicable) Pack Travel to residence Fill out necessary paperwork; obtain keys Move luggage in; unpack Meet some neighbours Evaluate facilities; identify gaps Confirm residence space Check list of needed items as items are being purchased Review checklist as packing Ensure that all luggage items have been loaded, unloaded, and moved into residence Call home to indicate move has been completed Arrange for any forgotten items to be sent 224 Copyright © 2022 Pearson Canada Inc.

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11.28 Initiation

Planning

Execution

Monitoring and Controlling

Closure

Idea to take a trip Create rough plan of locations to visit, including costs, and timing Decide whether you have enough funding to go (or not) Create list of all locations you wish to visit and benefits of each Evaluate transportation and accommodation alternatives and costs Carry out cost/benefit analysis to determine subset of locations to see and modes of transportation to take Determine what visa requirements are and whether you need special travel inoculations Create list of clothing/equipment you will need Determine gap.

Construct itinerary evaluating feasibility and costs Purchase transportation tickets Make accommodation reservations Get visas, inoculations (if necessary) Purchase/borrow any necessary equipment Arrange housing Pack Review itinerary Confirm reservations Check list of required equipment as it is being purchased/borrowed Monitor costs as tickets, reservations are made Check list of equipment/clothing while packing Confirm itinerary is complete Leave on trip

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11.29

Level 0 Trip

Level 1

Level 2

Time estimate (mins)

1.0 1.1Identify alternatives 1.2 Read information and compare 1.3 Select preferred residence 1.4 Send in confirmation paperwork 1.5 Create budget 1.6 Pay fees 1.7 Arrival paperwork

60 30 10 15

2.1 Identify alternatives 2.2 Compare and select alternative 2.3 Purchase tickets 2.4 Travel

30 120 60 720-840

3.1 List of essential items 3.2 List of other items 3.3 Gap analysis 3.4 Acquisition of items 3.5 Pack 3.6 Check list Unpack

45 60 15 180-600 180 10 120

Meet neighbours

60-300

45 15 30

2.0 Travel

3.0 Materials

4.0 Other

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11.30 a)

Day 0

5

10

15

20

Activity 1 Activity 2 Activity 3 Activity 4 Activity 5 Activity 6 Activity 7 Activity 8 Activity 9 b)

The project is expected to take 19 days to complete.

c)

If activity 6 is delayed by 3 days, it will not delay the project completion because its successor (activity 9) must wait for 7 to complete. Currently, activity 7 completes on day 15. Activity 6 starts on day 10 and ends on day 14.

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11.31 a) and b)

c)

The critical path is activities 1-3-5-7-8-9-11. The project should take 24 weeks to complete.

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11.32 Refer to the solution of problem 11.31 for the CPM representation.

The cost of crashing each of the critical path activities is as follows: Activity 1 3 5 7 8 9 11

Crashing Cost/Week ($000) 30/(6 − 4) = 15 8/1 = 8 13/2.5 = 5.2 6/0.5 = 12 22/2 = 11 15/1 = 15 N/A

To crash the project by 2 weeks, we start with the least expensive activity to crash. That is activity 5. It can be crashed up to 2.5 weeks at a cost of $5200 per week. Crashing activity 5 by 2 weeks does not cause any new critical paths to arise, so activity 5 should be shortened to 11 weeks. The cost is $10 400.

11.33 a)

b)

Using the data from problem 11.24, the least cost activity to crash is activity 5.

It can be crashed by 2 days, at a cost of $2000 per day, or a total of $4000.

The total project cost will then be $74 000. The revised CPM diagram is below:

Activity 5 can be crashed one more week (for a total of 3 weeks) at an additional cost of $2000. The project will then cost $76 000.

C. More Challenging Problems 11.34 Risks are either controllable or not controllable. For risks we cannot control, contingency plans help deal with the event to help mitigate consequences. For controllable risks, planning ahead can reduce the likelihood of the events.

Contingency plans help mitigate the impact.

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Construction project risks: cold temperatures delay construction activities, contamination of construction site stops work, water table is too high for water containment to be effective.

To mitigate the likelihood or impact of these risks: Cold temperatures are not controllable. If the project is being done in a region that experiences cold temperatures regularly, some additional time might be allowed in the schedule to allow for this. A thorough assessment of the site prior to construction will reveal whether contamination or water table issues are present. Effective planning can reduce the likelihood of these events.

Software development project: project team has dysfunctional relationship which delays deliverables, scope of project has enlarged threatening the completion date, one of the key project team members leaves her job.

In constructing the software development schedule, the background of the programmers is something that should be examined. If the team has had little experience working together, this should be built into the effort estimation computations. The risk of scope creep is minimized by having a very clear scope statement. Having good documentation and, for example, frequent design reviews can help mitigate the impact of the departure of a team member. Crosstrained staff can also help.

Process improvement project: staff are resisting and undermining the change process, a major step (and stakeholder) in the process has been omitted, the project is dragging on.

Frequent consultation and communication with affected staff are important to reduce the likelihood of resistance. Training is also important. Documentation and process evaluation should catch missing steps in the process; if an event does occur, then communication and rectification are needed. A project that is dragging on could be because its scope is unclear, or milestones have not been set. The project manager should rectify either if they occur.

A lessons-learned document at the close of each project will help with planning for future projects. This can help reduce the likelihood of controllable risks.

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11.35

House Construction Gantt Chart days

0

10

20

30

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

The project is expected to take 69 days

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40

50

60

70

Chapter 11 - Project Management

11.36 a)

b)

The house is expected to take 69 days to complete.

c)

The critical path is activities 1-2-3-5-11-13-15 -16.

d)

The windows were supposed to be ready by day 27 of the project. A 21-day delay will extend the project by 20 days: there is 1 day of slack in this activity; the remaining days will delay the entire project.

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11.37

Days 0

10

20

30

40

50

60

70

80

Activity 1 Activity 2 Activity 3 Activity 4 Activity 5 Activity 6 Activity 7 Activity 8 Activity 9 Activity 10 Activity 11 Activity 12 Activity 13 Activity 14 The expected completion time is 70 days.

11.38 a)

b)

The critical path is activities 1-4-5-7-9-14. The project is expected to take 70 days to complete.

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11.39 If the QA specialist can only work on one major activity at a time, then QA on the core modules (activity 8) and the QA on the auxiliary modules (activity 9) cannot overlap in time. In other words, they must occur in sequence. At the moment, the earliest that activity 8 can start is day 36, and the latest is day 56. Activity 9 can start at the earliest day 56 and the latest day 56 (it is on the critical path).

Adding a precedence relationship between activities 8 and 9 will ensure the sequential completion of the QA activities.

Making this change alters the LS and LF times for activities 6 and 8. The revised times are shown in the PERT/CPM network below:

11.40 Activity 1. Finalize new product plans 2. Complete drawings 3. Write Specifications 4. Market Assessment 5. Develop Prototype 6. Develop Marketing Strategy 7. Purchase Raw Materials 8. Setup equipment 9. Produce Test Batch 10. Launch Marketing Campaign 11. Evaluate test batch 12. Test Market Feedback 13. Engineering Changes 14. Full scale production

Earliest Start 0 4 4 4 16 13 33 33 47 25 54 54 75 82

Duration (days) 4 12 8 9 17 12 14 10 7 1 7 21 7 3

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a)

Gantt Chart for Problem 11 - 40 (days)

0

20

40

60

80

100

1.

Finalize new product plans 2.

Complete drawings 3.

Write Specifications 4.

Market Assessment 5.

Develop Prototype 6.

Develop Marketing Strategy 7.

Purchase Raw Materials 8.

Setup equipment 9.

Produce Test Batch 10. Launch Marketing Campaign 11. Evaluate test batch 12. Test Market Feedback 13. Engineering Changes 14. Full scale production

b)

Full scale production is expected to begin on day 82.

11.41 The shortest time in which the project can be completed depends on the activities on the critical path. As a starting point, look at crashing all the activities on the critical path as much as possible. The resulting CPM diagram is below.

Crashing all the activities on the critical path as much as possible does not make any other sequence of activities critical. Therefore, the shortest completion time is 15 weeks. The crashing cost is then $94 000 (30 + 8 + 13 + 6 + 22 + 15 = 94, the cost to crash each critical path activity as much as possible).

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Notes for Case in Point 11.1 1)

Here are a few of the activities that could be involved with this project: - Design of the transit system: routes, signalling - Issuing permits - Legal work - Hiring staff, engineers - Purchasing equipment - Installation - Testing

2)

The delivery of rail cars is part of the CPM. It is required to be done before testing of the system starts. If purchase is delayed, then the other activities dependent on it will be delayed, and therefore, the total project time will be delayed.

3)

They could have ordered the rail cars when the project was in its design stage, or they could have required stronger commitments from the vendor.

4) The project managers could have considered the activities that could run in parallel with each other.

Notes for Mini-Case 11.1 1)

Politicians are generally driven by a variety of influences, including moral principles, their party platform, the desire to be respected and re-elected, and financial support from local businesses and others. The overall cost of LEED certification on projects is probably the most important factor in this case, as stated, because of the immediate political impacts. For example, local contractors might be inhibited from bidding on public buildings at all because of the cost and paperwork burden of meeting LEED standards. This could alienate political supporters and potentially limit prospects for business growth in Nanaimo. Also, some voters may be more focused on how much money is spent today, rather than abstract future benefits tomorrow. Spending lots of money on “gold-standard” buildings could create political risk.

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2)

It is not necessarily true that anything went wrong. Society is still learning how to balance environmental responsibilities with economic interests. Nanaimo took a bold leap and learned in the process what can work and what cannot work.

3)

There is no right answer to this question. A range of 5 to 15% is sometimes suggested. Generally speaking, a larger project will require a smaller percentage.

4)

It all depends on what costs are recognised, and who gains the benefits. LEED certification certainly imposes higher initial costs, which can be of predominant importance to developers. A green building does usually reduce energy costs over time, but the benefits of this may not fully offset the initial costs, depending on the discounting assumptions used. However, many other benefits are incurred, such as the health and quality of life for the people that work in a LEED certified building, which are usually not calculated into the economic value. Moreover, green buildings benefit society as a whole, which is why governments, like Nanaimo, require LEED certification by policy or regulation.

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CHAPTER 12 Solutions to Chapter-End Problems A. Key Concepts Sensitivity Graphs: 12.1

(a) - First cost - Annual operating and maintenance costs - Annual savings or revenue - Salvage value - Service life (b) - First cost - Annual operating and maintenance costs - Annual revenue - Shipping cost - Inflation rate - Exchange rate - Tariff

12.2

(a) For example, for Canada, the base case can be April 2016, in which the average inflation rate was 1.7%.

Range of variation observed over last 10 years: 0.5% to 3%; this should be reasonable for a relatively short time period (b) For example, for Canada, as of April 2016: Base figure: $1.30US per $1CAN. Range of variation observed in last two years: $1.22 to $1.45; this should be reasonable for a relatively short time period. (c) Base figure: use the average annual savings of the equipment you already have.

Range of variation: a variation range of 5 to 10% should be reasonable since the new equipment is similar to the old equipment; you can probably get good estimate for the range of variation based on the information on the old equipment. (d) Base figure: use the average annual revenue of a similar internetbased business (if it exists).

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Chapter 12 - Dealing With Risk

Range of variation: since the nature of internet-based business is highly unpredictable at this stage, one should employ a large range of variation such as 5 to 50%. (e) Base figure: use the book value computed by declining-balance method with depreciation rate of 30%.

Range of variation: this depends on the age of the computer, but this could be anywhere between 0 and 50% of the purchase price.

12.3

(a) Break-even analysis for multiple projects: to see the effect of the future demand on the annual worth of leasing the trucks of different sizes (b) Break-even analysis for a single project: to see the effect of the uncertain heating expenses on the annual worth of the business (c) Break-even analysis for a single project: to see the effect of the future demand on the annual worth of the business (d) Sensitivity graphs: to see the effect of the uncertain construction cost on the total cost (e) Break-even analysis for a single project or scenario analysis: breakeven analysis for examining the effect of the growth rate and the length of growth separately; scenario analysis is appropriate to examine the growth rate and the length of growth together

12.4

The number of years, N, to save F $ by putting aside A $ per year at an annual interest rate i is N = ln[(iF + A)/A]/ln(1 + i).

With 5% and 10% decreases and increases in the savings per year and the interest rate, Kelowna Go-Karts will take the following number of years to save $50 000: Parameter Savings per year Interest rate

−10% 6300 9.00%

−5% Base Case 6650 7000 9.50% 10.00%

5% 7350 10.50%

10% 7700 11.00%

Number of years to save $50 000: With changes to savings per year With changes to interest rate

−10% 6.13 5.76

−5% 5.88 5.71

5% 5.44 5.60

10% 5.25 5.56

0%

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5.66 5.66

Years

Chapter 12 - Dealing With Risk

6.20 6.10 6.00 5.90 5.80 5.70 5.60 5.50 5.40 5.30 5.20 -10%

Interest rate Savings

-5%

0%

5%

10%

% change from base case

Break-Even Analysis: 12.5

(a) AW = −21 500(A/P, 10%, 10) + 10(1500) − 5(1500) + 21 500(1 − 0.2)10(A/F, 10%, 10) = −21 500(0.16275) + 15 000 − 7500 + 21 500(0.8)10  (0.06275) = $4145.74 (b) Break-even graph:

Annual worth ($)

8000 6000 4000 2000 0 0

500

1000

1500

2000

-2000 -4000 Operating hours

The break-even level of operating hours is 670.85 hours.

12.6

(a) First, the IRR for the incremental investment from “do nothing” to A is found by solving for i in: 100 000(P/A, i, 5) = 50 000  (P/A, i, 5) = 2  IRR = 41.1%

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A is the current best alternative. The IRR on the incremental investment between A and B is found by solving for i in: (400 000 – 100 000) + (150 000 – 50 000)(P/A, i, N) = 0 (P/A, i, N) = 300 000/100 000 = 3 This gives an IRR of 19.9%, thus the incremental investment in B is justified. (b) Preference for A over “do nothing” remains unchanged. The incremental investment between A and B is of concern. If the savings per year due to B are X, then the IRR in the incremental investment is found by solving for i in: (P/A, i%, N) = 300 000/(X – 50 000) The IRR for various values of X is: X 120 000 125 000 130 000 135 000 140 000 145 000 150 000

(300 000)/(X−50 000) 4.286 4.000 3.750 3.529 3.333 3.158 3.000

IRR 5.38% 8.00% 10.42% 12.86% 15.23% 17.57% 19.90%

IRR on increment

With a MARR of 10%, A is preferred for X (annual savings of B) below about $129 000 and B is preferred for X above $129 000. The amount $129 000 is the break-even annual savings for B. A diagram of savings due to B versus IRR on the incremental investment is:

19% 17% 15% 13% 11% 9% 120000 7% 5%

130000

140000

150000

Annual savings of B

12.7

From trial and error with a spreadsheet, the break-even interest rate is 11.2%. Model T is preferred for a MARR of 16%.

PW(T) = –100 000 + 50 000(P/A, i, 5) + 20 000(P/F, i, 5) PW(A) = –150 000 + 62 000(P/A, i, 5) + 30 000(P/F, i, 5) 242 Copyright © 2022 Pearson Canada Inc.

Present worth ($)

Chapter 12 - Dealing With Risk

140000 120000 100000 80000 60000

Model T

40000

Model A 6%

8% 10% 12% 14% 16% 18% 20%

Interest rate

12.8

Since the two models have unequal lives, it is easiest to compare them based on an annual worth computation.

The annual worth of each for a variety of interest rates is: Depreciation AW Rate (Model A) Model A 0.3 3365 0.4 3669 0.5 3887 0.6 4033

AW Model B 3809 3809 3809 3809

Sample computation for d = 40% depreciation rate for Model A: AW(model A) = 8000(A/P, 11%, 3) + 1000 – 8000(1 – d)3 = 8000(0.3982) + 1000 – 8000(0.6)3 = 3669 And for Model B with straight line depreciation at $2500 per year: AW(model B) = 10 000(A/P, 11%, 4) + 700 – (10 000 – 2500  4)(A/F, 11%, 4) = 10 000 (0.3108) + 700 = 3809 Since the annual worth of Model A is lower than that of Model B over the range of depreciation rates Julia has estimated for Model A, she should pick Model A. From the above table, we can interpolate the break-even depreciation rate to be 46%. Below this rate, Model A is preferred; above this rate, Model B is preferred. She is indifferent with a depreciation rate of 46%. A break-even chart is as follows:

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Annual worth ($)

Chapter 12 - Dealing With Risk

4100 4000 3900 3800 3700 3600 3500 3400 3300

Model A Model B

0.3

0.4

0.5

0.6

Depreciation rate for A

Probability and Expected Value: 12.9

(a) E(return on investment) = Pr{7%}(10 000  0.07) + Pr{10%}(10 000  0.1) + Pr{15%}(10 000  0.15) = 0.65(700) + 0.25(1000) + 0.1(1500) = 855 The expected return from this investment is $855.

12.10 (a) E(loss) = Pr{capacity 30}(loss of 20) + Pr{capacity 40}(loss of 10) + Pr{capacity 50}(loss of 0) + Pr{capacity 60}(loss of 0) = 0.2(20) + 0.4(10) + 0.3(0) + 0.1(0) = 8 calls per hour The expected loss of customers due to the lack of processing capacity is 8 per hour.

12.11 (a) E(number of defects, A1) = 0.3(0) + 0.28(1) + 0.15(2) + 0.15(3) + 0.1(4) + 0.02(5) = 1.53/100 units E(number of defects, X1000) = 0.25(0) + 0.33(1) + 0.26(2) + 0.1(3) + 0.05(4) + 0.01(5) = 1.4/100 units According to the expected number of defects, X1000 seems to be slightly better than A1.

12.12 (a) E(cost in a summer month) = 0.4(800) + 0.25(2  800) + 0.2(3  800) + 0.1(3  800 + 1500) + 0.05(3  800 + 2  1500) = $1860 per month 244 Copyright © 2022 Pearson Canada Inc.

Chapter 12 - Dealing With Risk

E(cost in a non-summer month) = 0.45(0) + 0.4(800) + 0.15(2  800) = $560 per month CB Electronix should consider getting the complete coverage policy because the expected cost is over $500 even in the non-summer months.

Decision Trees: 12.13 (a) p(customer getting a bad board) = 2/5 = 0.4 (b) E(send stock) = p(good board)(0) + p(bad board)(–10 000) = 0.6(0) + 0.4(–10 000) = –4000 The expected cost of sending the customer one of the five boards is $4000. (c) Tree diagram: Incremental Cost ($) E(2) = 4000 Send Stock E(1) = 4000

Good Board 0.6

0

2 Bad Board

1

0.4 Send from Other Plant

10 000

5000

E(1) = E(2) = 4000 because E(2) < 5000 Randall should send the customer one from stock.

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12.14 Decision Tree diagram: Payoff ($)

Increase Production

E(2) = 447 500

Demand High 0.4

750 000

2

Demand Same 0.35

350 000

Demand Low 0.25

100 000

Demand High 0.4

500 000

3

Demand Same 0.35

400 000

E(3) = 390 000

Demand Low 0.25

200 000

E(1) = 447 500 1 Do Not Increase Production

E(2)

E(3) E(1)

= Pr{high demand}(750 000) + Pr{medium demand}(350 000) + Pr{low demand}(100 000) = 0.4(750 000) + 0.35(350 000) + 0.25(100 000) = 447 500 = 0.4(500 000) + 0.35(400 000) + 0.25(200 000) = 390 000 = E(2) = 447 500 because E(2) > E(3)

SJCF should increase their production. (b) The cumulative risk profile for the two decision alternatives are in the figure below.

Outcome dominance does not exist because one decision is not better than the other for all outcomes (e.g., for the outcome demand low, “do not increase production” is better, but for the outcome demand high, “increase production” is better.)

First degree stochastic dominance does not exist. This can be seen from the cumulative risk profiles below, where Pr(demand ≤ X) is greater for the “increase production” decision for all values of X up to $500, but not for all values of X.

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1.0 0.9 0.8 0.7

P(x)

0.6 0.5

Increase Do Not Increase

0.4 0.3 0.2 0.1 0.0 0

50

100

150

200

250

300

350

400

450

500

550

600

650

700

750

800

X , Payoff ($000's)

12.15 Tree diagram: Expected Gain in Market Share (%)

Combined Rate

E(2) = 15.9

Rapid Growth 0.3

30

2

Steady Growth 0.4

15

Slow Growth 0.3

3

Rapid Growth 0.3

15

3

Steady Growth 0.4

10

E(3) = 10

Slow Growth 0.3

5

E(1) = 15.9 1 Add-On Rates

E(2) = Pr{rapid growth}(30) + Pr{steady growth}(15) + Pr{slow growth}(3) = 0.3(30) + 0.4(15) + 0.3(3) = 15.9 E(3) = 0.3(15) + 0.4(10) + 0.3(5) = 10 E(1) = E(2) = 15.9 because E(2) > E(3) LOTell should introduce the combined rate.

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(b) Several types of dominance reasoning can be applied: meanvariance, outcome or stochastic dominance. Each is covered below: Mean-variance dominance: Var(combined rate) = Pr{rapid growth}(30 − 6.9)2 + Pr{steady growth}(15 − 6.9)2 + Pr{slow growth}(3 − 6.9)2 = 0.3(533.61) + 0.4(65.61) + 0.3(15.21) = 31 $2 Var(add-on rates) = Pr{rapid growth}(15 – 5.5)2 + Pr{steady growth}(10 − 5.5)2 + Pr{slow growth}(5 – 5.5)2 = 8 $2 One of the two decisions cannot be eliminated with mean-variance reasoning because the combined rate decision has the higher expected gain in market share, but the add-on rates decision has the lower (better) variance.

Outcome dominance: Observe that the “add-on rate” decision has a better outcome for slow growth, but the “combined rate” decision has a better outcome for both steady and rapid growth. Since one of the decisions is not better than the other for all possible outcomes, outcome dominance cannot be used to eliminate either decision.

Stochastic Dominance: The Cumulative risk profiles (CDFs) for the two decisions are shown in the figure below. The “combined rate” decision is dominated for market share outcomes up to 5% (i.e., has a higher probability that demand is less than or equal to X for X ≤ 5%), but the “add-on rate” decision is dominated for outcomes over 5% market share. Thus, neither decision dominates the other.

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1.0 0.9 0.8 0.7

P(x)

0.6 0.5

Add-on rates

Combined rates

0.4 0.3 0.2 0.1 0.0 0

5

10

15

20

25

30

35

X , Gain in market share (%)

B. Applications 12.16 The present worth of the software when the base annual savings are = 10 000, the gradient is G = 1000, and interest rate is i = 15%: PW = [10 000 + 1000(A/G, 15%, 6)](P/A, 15%, 6)(P/F, 15%, 2) = 34 617 The other computations are as follows: −15%

−7.5%

Base Case

7.5%

15%

Base savings Savings gradient Interest rate

8500 850 12.75%

9250 925 13.88%

10000 1000 15.00%

10750 1075 16.13%

11500 1150 17.25%

Present Worth: Changes to base savings per year Changes to savings gradient Changes to interest rate

−15% 30325 33717 38484

−7.5% 32471 34167 36485

0% 34617 34617 34617

7.5% 36764 35068 32872

15% 38910 35518 31238

Parameter

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Chapter 12 - Dealing With Risk

Present worth ($)

The sensitivity graph is: 40000

Base savings

37500 Savings gradient

35000 32500

Interest rate

30000 -15.0% -7.5%

0.0%

7.5% 15.0%

% change from base case

12.17 First, note that all cost figures are given in real dollars, as they do not take into account the effect of inflation.

The municipality uses an actual interest rate of 7% when inflation is expected to be 3%, and hence, their real MARR is: MARRR = (1 + MARRA)/(1 + f) – 1 = 1.07/1.03 – 1 = 0.0388 or 3.88% Since all costs are based on current estimates, it is simplest to work with real dollars and the real MARR for the analysis. Note that even if the City’s estimates of inflation change by 5% or 10%, the real MARR they use will not change and hence the present worth of the project (in real dollars) will be unaffected by the inflation rate. (a) The present worth of the project is $360 204.

Parameter Annual Construction Costs (in $000s) Annual Maint. and Rep. costs (in $000s) Inflation Rate

−10% 18000 1800 3.30%

−5% Base Case 19000 20000 1900 2000 3.15% 3.00%

5% 21000 2100 3.15%

10% 22000 2200 3.30%

Present Worth (in $000s) Changes to annual construction costs Changes to annual maint. costs Changes to inflation rate

−10% 95949 100927 103619

−5% 99784 102273 103619

5% 107454 104965 103619

10% 111289 106310 103619

0% 103619 103619 103619

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(b) The sensitivity graph is as follows:

Present worth (in $000s)

115000 110000 105000 100000 95000 90000 -10%

-5%

0%

5%

10%

% change to base case

The present worth of the project is most sensitive to changes in the annual maintenance and repair costs.

12.18 The present worth of the pool (in $000s) with the “base case” costs and MARR is: PW = 6000 + [400(A/F, 5%, 10)]/0.05 With one at a time 5% and 10% variations in the first cost, maintenance costs and the MARR, the present worth of costs are: Parameter Construction costs (in $000s) Refinishing costs (in $000s) MARR

−10% 5400 360 4.50%

−5% Base Case 5700 6000 380 400 4.75% 5.00%

5% 6300 420 5.25%

10% 6600 440 5.50%

Present worth of costs (in $000s) Changes to construction costs Changes to refinishing costs Changes to MARR

−10% 6036 6572 6723

−5% 6336 6604 6677

5% 6936 6668 6599

10% 7236 6700 6565

0% 6636 6636 6636

Present worth (in €000s)

The sensitivity graph shows that the present worth is most sensitive to the first cost of the pool: 7400 7200 7000 6800 6600 6400 6200 6000 -10%

Construction costs Refinishing costs MARR

-5%

0%

5%

10%

% change from base case

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12.19 Sample Equivalent Monthly Worth (EMW) Computation (for base case): EMW = (Monthly Revenues) – (Monthly Costs) = 300  2 – 100 – [(6000 – 3000)(A/P, 8%/12, 24) + 3000(0.08/12)] – 600(A/P, 8%/12, 6) = 242 −20% 80 240 0.533%

−10% Base Case 90 100 270 300 0.600% 0.667%

10% 110 330 0.733%

20% 120 360 0.800%

Sensitivity Graph Information Changes to Advertising Costs Changes to Number of Deliveries Changes in Interest Rate

−10% 262 122 245

−5% 252 182 243

5% 232 302 241

10% 222 362 239

Equivalent monthly worth ($)

Parameter Monthly Advertising Costs Number of Customers per month Interest rate (% per month)

0.0% 242 242 242

400 Number of deliveries

300 Interest rate

200

Advertising costs

100 -10%

-5%

0%

5%

10%

% change from base case

The monthly worth is most sensitive to changes in the number of deliveries per month.

12.20 (a) A summary of the costs for Vendor A’s device is below at the base value and for 5% and 10% increases and decreases.

Also given is the annual worth of costs for Vendor A’s device. A sample computation for the base case is: AW = 200 000(A/P, 15%, 7) + 10 000 + 6500 + (0.05 + 0.95 + 1.25)(50 000) – 5000(A/F, 15%, 7) = 176 620 Parameter First Cost Annual Maintenance Cost Maintenance cost/unit Labour cost/unit Annual other costs Other Costs/unit Salvage Value

−10% −5% Base Case 5% 10% 180 000 190 000 200 000 210 000 220 000 9 000 9 500 10 000 10 500 11 000 0.045 0.048 0.050 0.053 0.055 1.13 1.19 1.25 1.31 1.38 5 850 6 175 6 500 6 825 7 150 0.86 0.90 0.95 1.00 1.05 4 500 4 750 5 000 5 250 5 500

Annual costs Changes in First Cost Changes in Fixed Costs Changes in Variable costs Changes in Salvage Value

−10% 171 813 175 620 176 370 176 665

−5% 174 217 176 120 176 495 176 643

0% 176 620 176 620 176 620 176 620

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5% 179 024 177 120 176 745 176 598

10% 181 427 177 620 176 870 176 575

Chapter 12 - Dealing With Risk

A sensitivity graph indicating the sensitivity of the annual worth of Vendor A’s device to changes in the costs shows that the annual worth is most sensitive to changes in the first cost and then to changes to the fixed costs (maintenance and other).

182 000

First cost

Annual cost ($)

180 000 Fixed costs

178 000

Variable costs

176 000

Salvage value

174 000 172 000 170 000 -10%

-5%

0%

5%

10%

% change from base case

(b) Vendor B’s computations and sensitivity graph are below. As with Vendor A, the annual worth is most sensitive to the first cost and to the fixed annual costs. Also observe that at expected annual production levels, Vendor B’s device has a lower annual worth.

Parameter

−10%

−5%

Base Case

5%

10%

First Cost Annual Maintenance Cost Maintenance cost/unit Labour cost/unit Annual other costs Other Costs/unit Salvage Value

315 000 332 500 18 000 19 000 0.009 0.010 0.45 0.48 13 950 14 725 0.50 0.52 18 000 19 000

350 000 367 500 385 000 20 000 21 000 22 000 0.010 0.011 0.011 0.50 0.53 0.55 15 500 16 275 17 050 0.55 0.58 0.61 20 000 21 000 22 000

Annual cost Changes in First Cost Changes in Fixed Costs Changes in Variable costs Changes in Salvage Value

−10% 150 279 155 253 157 203 157 352

0% 157 253 157 253 157 253 157 253

−5% 153 766 156 253 157 228 157 302

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5% 160 740 158 253 157 278 157 204

10% 164 227 159 253 157 303 157 155

Annual cost ($)

Chapter 12 - Dealing With Risk

165 000

First cost

160 000

Fixed costs Variable costs Salvage value

155 000

150 000 -10%

-5%

0%

5%

10%

% change from base case

12.21 (a) Let D = the number of deliveries per month. Solve for D in: 0 = EMW = (Monthly Revenues) – (Monthly Costs) = 2D – 100 – [(6000 – 3000)(A/P, 8%/12, 24) + 3000(0.08/12)] – 600(A/P, 8%/12, 6) D = 179 The break-even number of deliveries is 179.

Monthly worth ($)

300 200 100 0 - 100

50

100

150

200

250

300

- 200 - 300 Number of deliveries per month

(b) Solving for the interest rate is most easily done with a trial and error approach on a spreadsheet. The break-even interest rate is 10.29% per month, or 223.5% per year.

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Monthly worth ($)

Chapter 12 - Dealing With Risk

300 200 100 0 - 100 0% - 200 - 300 - 400 - 500 - 600

5%

10%

15%

20%

25%

Interest rate per month

12.22 (a) A summary of the costs and benefits and the annual worth computations for Alternative A is as follows: Costs and Benefits: Initial hardware and installation costs Annual Benefits

−10% 83 250 58 500

−5% Base Case 87 875 92 500 61 750 65 000

5% 10% 97 125 101 750 68 250 71 500

Annual Worths: Initial hardware and installation costs Annual Benefits

−10% 11 516 3 173

−5% 10 595 6 423

5% 8 751 12 923

0% 9 673 9 673

10% 7 830 16 173

A sample computation for the annual worth of the base case is: AW(base case) = 65 000 – [(138 750 + 92 500)(A/P, 15%, 10) + 9250] = 9673 and for the annual worth if the initial hardware and installation costs are 10% higher than the base case: AW(base case) = 65 000 – [(138 750 + 101 750)(A/P, 15%, 10) + 9250] = 7830

Annual worth ($)

A sensitivity graph of these results is: 18 000 16 000 14 000 12 000 10 000 8 000 6 000 4 000 2 000 0 -10%

Annual benefits

Initial hardware & installation costs

-5%

0%

5%

10%

% change from base case

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The annual worth is most sensitive to changes in the annual benefits of the network. (b) A summary of the costs and benefits and the annual worth computations for Alternative B is as follows: Costs and Benefits ($): Initial hardware and installation costs Annual Benefits

−10% −5% Base Case 5% 10% 94 950 100 225 105 500 110 775 116 050 66 600 70 300 74 000 77 700 81 400

Annual Worths ($) Initial hardware and installation costs Annual Benefits

−10% 12 999 3 497

−5% 11 948 7 197

0% 10 897 10 897

5% 9 846 14 597

10% 8 795 18 297

Annual worth ($)

And the corresponding sensitivity graph is: 20 000 18 000 16 000 14 000 12 000 10 000 8 000 6 000 4 000 2 000 0 -10%

Annual benefits

Hardware & installation costs

-5%

0%

5%

10%

% change from base case

12.23 A graph showing the number of years required to save $50 000 as a function of the amount saved per year is below.

Number of years to save €50 000

With annual savings of A and an interest rate i = 0.15, the number of years, N, to save F = $50 000 is N = ln[(iF + A)/A]/ln(1 + i).

8 7 6 5 5000 4

6000

7000

8000

9000

Amount saved per year (€)

The break-even annual savings amount is $8190 (this was obtained with trial and error with the spreadsheet table using the amount saved per year as the input variable). The Go-Kart Klub will need to raise $1190 each year (in addition to the $7000) if they wish to save $50 000 in 5 years.

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12.24 First the PW of acquiring the moulder and keeping it over its 6-year life is the sum of its capital costs, and operating and maintenance costs: PW(capital) = 27 000 – BVdb(6)(P/F, 15%, 6) = 27 000 – 20 000(1 – 0.4)6(0.4323) = 26 596.61 The present worth of the operating and maintenance costs over the 6-year life can be obtained with the use of a geometric gradient series to present worth conversion factor: PW(op. and maint.) = 30 000(P/A, i, 6)/(1 + g) where i = (1 + i)/(1 + g) – 1 = 1.15/1.05 – 1 = 0.09524 PW(op. and maint.) = 30 000(4.4167)/1.05 = 126 191.43 This gives a present worth of cost for the 6-year planned life of the moulding equipment of $150 788 = 126 191.43 + 24 596.61.

The equivalent annual cost can be found from: EAC(moulder) = 152 788(A/P, 15%, 6) = 152 788(0.2642) = 40 366.59

Cost per unit (£)

The cost per piece is then EAC(moulder)/production volume. Using this, we can construct the break-even analysis. The break-even quantity is 212 500. Below this annual production quantity, Bountiful should continue to purchase from outside. Above this quantity, they should buy the moulding equipment.

0.23 0.22 0.21 0.20 0.19 0.18 0.17 0.16 0.15

Purchase cost In-house cost

180

200

220

240

Production volume (1000s of pieces/yr)

12.25 Under Canadian tax rules, assuming a CCA rate of 30%, first, we need to calculate the Tax Benefit Factor: TBF = 1 – td/(i + d) = 1 – (0.35)(0.3)/(0.12 + 0.30) = 0.75 AW(first cost) = [–65 000 + (65 000/2xTBF + 65 000/2(P/F, 12%, 1))(A/P, 12%, 5) = –13 758 257 Copyright © 2022 Pearson Canada Inc.

Chapter 12 - Dealing With Risk

AW(savings) = (Annual Savings)(1 – 0.35) AW(salvage) = 20 000(1 − TBF)(A/F, 12%, 5) = 20 000(0.75)(0.15741) = 2361 AW(truck) = –13 758 + 2361 + (Annual Savings)(0.65)

Annual worth ($)

With a spreadsheet, the annual worth of the truck for annual revenues from $12 000 to $22 000 can be calculated. The break-even revenue per year is $17 535.

4000 3000 2000 1000 0 -1000 12 -2000 -3000 -4000

14

16

18

20

22

Revenues (in $000s/yr)

12.26 (a) EAC calculation summary: Service Life

6

7

8

9

10

Year 1 2 3 4 5 6 7 8 9 10

EAC(total) 20 067 17 519 16 733 16 385 16 209 16 117

EAC(total) 19 114 16 621 15 887 15 588 15 459 15 413 15 412

EAC(total) 18 400 15 947 15 252 14 990 14 897 14 885 14 916 14 972

EAC(total) 17 844 15 423 14 758 14 525 14 460 14 474 14 531 14 610 14 703

EAC(total) 17 400 15 004 14 363 14 153 14 110 14 146 14 222 14 321 14 432 14 549

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(b) Sensitivity graph:

Economic life (years)

8

6

4

2

0 6

7

8

9

10

Service life (years)

The sensitivity graph indicates that the economic life is somewhat sensitive to the length of service life. For a service life of 6 or 7 years, it is economical to keep the bottle-capping machine until the end of its service life. For a service life longer than 7 years, it is more economical to keep the machine for a shorter period than the service life.

12.27 The annual worth of the Clip Job for annual maintenance costs varying from $30 to $80 is as follows: Annual Maintenance 30 40 50 60 70 80

Annual Worth 103.84 113.84 123.84 133.84 143.84 153.84

For example, with annual maintenance costs of $50, the annual worth of the Clip Job is: AW(clip job) = 120(A/P, 5%, 4) + 40 + 50 = 120(0.282) + 90 = 123.84 The annual worth of the Lawn Guy is: AW(lawn guy) = 350(A/P, 5%, 10) + 60 + 30 = 350(0.1295) + 90 = 135.33

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Annual worth ($)

And the maintenance cost for the Clip Job which makes Sam indifferent between the two lawnmowers is $61.50. A break-even graph illustrates the break-even maintenance costs: 160

Clip Job

140

Lawn Guy

120 100 30

40

50

60

70

80

Clip Job annual maintenance costs

Since Sam estimates the maintenance costs to be about $60 per year, and the break-even maintenance cost is close to this amount, we would recommend that Sam purchase the Lawn Guy if he is risk averse and wants to avoid the possibility of $80 maintenance costs for the Clip Job. 12.28 The annual worth of the car for salvage values varying from $6000 to $24 000 are: Salvage Value 6000 9000 12000 15000 18000 21000 24000

Annual Worth 10647 10165 9684 9202 8720 8239 7757

For example, the annual worth with a salvage value for $9000 is: AW(car) = 24 000(A/P, 11%, 5) + (2000 + 600 + 800 + 1000) + 400(A/G, 11%, 5) – 9000(A/F, 11%, 5) = 24 000(0.27057) + 4400 + 400(1.7923) – 9000(0.16057) = 10 165 The annual costs of taking taxis are $8000 (=$7000 + $1000). Based on an annual worth comparison, Ganesh should not buy the car. The salvage value of the car would have to exceed the break-even value of $22 486 before the car will have equal or lower costs than that of taking taxis.

Taking into account only the financial aspects of the decision, Ganesh should not buy the car unless he feels there are other benefits (e.g., convenience of having a car) that have not been taken into account.

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11000

Annual worth ($)

Car 10000 9000

Taxi

8000 7000 6000 6000

9000

12000

15000

18000

21000

24000

Salvage value of car

12.29 E(monthly savings) = 0.25(800 000) + 0.25(1 000 000) + 0.25(1 200 000) + 0.25(1 400 000) = 1 100 000 PW = 1 100 000(P/A, 1%, 24) = 1 100 000(21.243) = 23 367 000 The present worth of the expected monthly savings is about $23 367 000.

12.30 E(revenue) = 0.1(2.95) + 0.35(3.25) + 0.3(3.50) + 0.15(4.00) + 0.1(5.00) = $3.58 per parcel Regional’s monthly capacity is 60 000 parcels. So the present worth of the expected revenue over 12 months is: PW = (3.58  60 000 − 8000)(P/A, 1%, 12) = (206 800)(11.255) = $2 327 534 12.31 E(life) = p(4 years)(48 months) + p(5 years)(60 months) + p(6 years)(72 months) = 0.4(48) + 0.4(60) + 0.2(72) = 57.6 months Using the expected life of 57.6 months, the present worth of the monthly expenses is computed as follows: PW = (90 + 30 + 20)(P/A, 10/12%, 57.6) = 140[(1 + 0.008333)57.6 – 1]/[0.008333(1 + 0.008333)57.6] = (140)(45.599) = 6383.86 The present worth of the monthly expenses is about €6384 before a major repair.

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12.32 First, find the annual cost for each scenario: AC(optimistic) = 75 000 AC(expected) = 240 000(A/F, 15%, 2) = 240 000(0.46512) = 111 628.80 AC(pessimistic) = 500 000(A/F, 15%, 3) = 500 000(0.28798) = 143 990 E(annual cost) = p(optimistic)(75 000) + p(expected)(111 628.80) + p(pessimistic)(143 990) = 0.15(75 000) + 0.5(111 628.80) + 0.35(143990) = 117 460.90 The expected annual cost of the vitamin C project is $117 461.

12.33 Tree diagram: Payoff ($) E(2) = 150 Buy Ticket E(1) = 150

Win House 0.001

249 900

2 Win Nothing

1

0.999 Do Not Buy Ticket

−100

0

E(2) = p(not win)(−100) + p(win)(250 000 − 100) = 0.999(−100) + 0.001(249 900) = 150 E(1) = E(2) = 150 because E(2) > 0 Buying a ticket has a greater expected value than not buying a ticket.

12.34 (a) If E(2) = 0.999X + 0.001(250 000 − X) = 0.998X + 250 < 0, then not buying is preferred.

0.998X + 250 < 0 0.998X < −250 X < −250.50 Hence, if the ticket costs more than $250.50, not buying is the preferred option.

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(b) If E(2) = (−100)(1 − X) + 249 900X < 0, then not buying is preferred. (−100)(1 − X) + 249 900X < 0 −100 + 100X + 249 900X = −100 + 250 000X < 0 250 000X < 100 X < 0.00039984 Hence, if the probability of winning is less than 0.04%, not buying is the preferred option.

12.35 (a) PW(high performance) = −550 000 + 500 000(P/A, 12%, 5) = −550 000 + 500 000(3.6048) = 1 252 400 PW(medium performance) = −550 000 + 250 000(P/A, 12%, 5) = 351 200 PW(low performance) = −550 000 + 150 000(P/A, 12%, 5) = −9280 (b) Tree diagram: Probabilities do not add up to 100%.

New Design

Annual Savings ($)

Present Worth ($)

E(2) = 252 532

Performance Low 0.35

150 000

−9280

2

Performance Med. 0.55

250 000

351 200

Performance High 0.05

500 000

1 252 400

0

0

E(1) = 252 532 1

No New Design

E(2) = p(high performance)(1 252 400) + p(medium performance)(351 200) + p(low performance)(−9280) = 0.05(1 252 400) + 0.55(351 200) + 0.35(−9280) = 252 532 E(1) = E(2) = 252 532 because E(2) > 0 BB should approve the development of the new robot.

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12.36 Tree diagram:

Partnership with China

E(2) = 113.75

E(3) = 156.25 Acceptable Quality Shipping 0.75 No Delay 3 0.6 Poor Quality 0.25

2

Acceptable Quality Delay E(1) = 113.75

0.4

0.75

4

Poor Quality E(4) = 50

1

Gain in Annual Profit (in $000s) 200

25

100 −100

0.25

No Partnership

0

E(3) = p(acceptable quality)(200 000) + p(poor quality)(25 000) = 0.75(200 000) + 0.25(25 000) = 156 250 E(4) = p(acceptable quality)(100 000) + p(poor quality)(−100 000) = 0.75(100 000) + 0.25(−100 000) = 50 000 E(2) = p(no shipping delay)E(3) + p(shipping delay)E(4) = 0.6(156 250) + 0.4(50 000) = 113 750 E(1) = E(2) = 113 750 because E(2) > 0 The partnership with China is still recommended as a result of analyzing shipping delay and quality control possibility.

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12.37 Tree diagram: Cost ($)

Minor Modification

Not Favourable 0.66

Severe Failure 0.15 Minor Failure

5

No Failure

4

0.4

2

90 000 55 000

E(6) = 98 000

Severe Failure 0.05

230 000

6

Minor Failure 0.3

115 000

No Failure 0.65

80 000

Major Modification

E(1) = 78 545 Favourable Result

1

205 000

0.45

E(4) = 93 250

E(2) = 78 545 Test

E(5) = 93 250

50 000

0.34

E(3) = 94 750 No Test

3

Severe Failure

150 000

0.55 Minor Failure

35 000

0.35 No Failure

0.1 E(5) = p(severe failure)(test, modification, and failure costs) + p(minor failure)(test, modification, and failure costs) + p(no failure)(test and modification costs) = 0.15(50 000 + 5000 + 150 000) + 0.45(50 000 + 5000 + 35 000) +0.4(50 000 + 5000) = 93 250

0

E(6) = 0.05(50 000 + 30 000 + 150 000) + 0.3(50 000 + 30 000 + 35 000) +0.65(50 000 + 30 000) = 98 000 E(4) = E(5) = 93 250 because E(5) < E(6) E(2) = p(not favourable)E(4) + p(favourable)(test costs) = 0.66(93 250) + 0.34(50 000) = 78 545 E(3) = p(severe failure)(failure costs) + p(minor failure)(failure costs) + p(no failure)(no costs) = 0.55(150 000) + 0.35(35 000) + 0.1(0) = 94 750 E(1) = E(2) = 78 545 because E(2) < E(3) 265 Copyright © 2022 Pearson Canada Inc.

Chapter 12 - Dealing With Risk

According to decision tree analysis, Rockies should test the upgraded system, and if the result is not favourable, Rockies should apply the minor modification to the system.

C. More Challenging Problems 12.38 (a) The break-even installation cost for Alternative A is $141 045. A break-even chart is shown below:

Annual worth ($)

15000 10000 5000 0 70000 -5000

90000

110000 130000 150000 170000

-10000 Installation cost ($)

The break-even installation cost is outside the range Merry Metalworks has estimated.

Annual worth ($)

(b) The break-even annual cost for Alternative A is $55 327. A breakeven chart is shown below: 30000 25000 20000 15000 10000 5000 0 50000 -5000 -10000

60000

70000

Annual benefits ($)

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80000

Chapter 12 - Dealing With Risk

12.39 (a) The break-even installation cost for Alternative B is $160 190, which is well above the range of estimated costs.

A break-even chart showing various annual worths is below:

Annual worth ($)

20000 15000 10000 5000

0 80000 -5000

100000 120000 140000 160000 180000 Installation cost ($)

(b) The break-even annual benefits for Alternative B are $63 103. This is within the range of benefits Merry Metalworks has specified, so there is some risk that Alternative B will yield a negative present worth. The break-even chart showing various annual benefits is below:

Annual worth ($)

30000 15000 0 40000 -15000

50000

60000

70000

80000

90000

-30000 Annual benefits ($)

12.40 (a) The break-even production volume with a per-unit revenue of $3.25 for Vendor A is 64 120 (obtained by interpolation).

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Chapter 12 - Dealing With Risk

Annual cost or revenue ($)

This can be seen in the following break-even graph: Annual sales revenue

240000

Annual cost

200000 160000 120000

80000 30000 40000 50000 60000 70000 80000

Production volume (boards/yr)

Annual cost or revenue ($)

(b) For Vendor B, the break-even volume is 47 375 boards/year.

280000

Annual sales revenue

240000 200000

Annual cost

160000 120000 80000 30000 40000 50000 60000 70000 80000 Production volume (boards/yr)

12.41 Since the lease is an annual amount, it is easiest to compare the two alternatives based on annual costs. This solution assumes Canadian tax rules and a 20% CCA rate.

First, the annual worth of the lease decision is $5500 per year, after taxes this is $5500(1 – t) where t is the tax rate. The annual worth of costs for the buy decision depends on Kelly’s tax rate.

Tax rate 0.20 0.25 0.30 0.35 0.40

AW(buy) 4180 4060 3949 3843 3742

TBF 0.143 0.152 0.158 0.163 0.167

AW(lease) 4400 4125 3850 3575 3300

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Chapter 12 - Dealing With Risk

For example, at a tax rate of 30% the annual worth of costs for the two alternatives, taking taxes into account is: AW(buy) = [15 000 − (15 000/2xTBF + 15 000/2xTBF(P/F, 8%, 1))](A/P, 8%, 5) + [1000 + 400(A/G, 8%, 5)](1 – t) – 2500(1 − TBF) = [15000 − (750 × 0.158 + 750 × 0.158 × 0.54027](0.25046) ) + [1000 + 400(1.846 47)](1 – 0.3) – 2500(0.842) = 3949

Annual worth ($)

AW(lease) = 5500(1 – t) = 3850 We can see from the table that Kelly should lease if his tax rate is below 27% and buy if he expects his tax rate to be above 27%.

4600 4400 4200 4000 3800 3600 3400 3200 3000 0.20

Buy Lease

0.25

0.30

0.35

0.40

Tax rate

12.42 Break-even graph: AW(Small) = –(6000 + 1500)(A/P, 12%, 5) + (annual savings per employee)  (#employees) AW(Large) = –(10 000 + 3500)(A/P, 12%, 5) + (annual savings per employee)  (#employees)

Annual worth ($)

30000 25000

Large-scale

20000 15000 10000

Small-scale

5000 0 40

60

80

100

Number of Employees

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Chapter 12 - Dealing With Risk

The small-scale version would be a better choice if Western Insurance wasn’t expecting much growth in the next 5 years. However, they are growing. If the growth continues steadily for the next 5 years, the largescale version seems to be a better choice since it has a greater annual worth than the small-scale version does as long as the average number of employees is greater than 60.

12.43 (a) From Problem 12.32: AC(optimistic) = 75 000 AC(expected) = 111 628.80 AC(pessimistic) = 143 990 AW(optimistic, public accept) = 1 000 000(A/F, 15%, 1) − AC(optimistic) = 1 000 000(1) − 75 000 = 925 000 AW(optimistic, not accept) = 200 000(A/F, 15%, 1) − AC(optimistic) = 125 000 AW(expected, public accept) = 1 000 000(A/F, 15%, 2) − AC(expected) = 1 000 000(0.46512) − 111 628.80 = 353 491.20 AW(expected, not accept) = 200 000(A/F, 15%, 2) − AC(expected) = −18 604.80 AW(pessimistic, public accept) = 1 000 000(A/F, 15%, 3) − AC(pessimistic) = 1 000 000(0.28798) − 143 990 = 143 990 AW(pessimistic, not accept) = 200 000(A/F, 15%, 3) − AC(pessimistic) = −86 394

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(b) Tree diagram: E(3) = 391 400 Research Optimistic 3 0.15

Annual Worth ($) Public Accept 0.333 Public Not Accept 0.667

E(2) = 107 975

Develop New Product

2

Public Accept Expected 0.5

0.333

4 E(4) = 105 303

Public Not Accept

E(1) = 107 975 1

0.667 Public Accept

Pessimistic

5 E(5) = −9676

0.333 Public Not Accept 0.667

925 000

125 000

353 491

−18 605

143 990

−86 394

Do Not Develop New Product

0

E(3) = p(public accept)(925 000) + p(not accept)(125 000) = 0.333(925 000) + 0.667(125 000) = 391 400 E(4) = 0.333(353 491.20) + 0.667(−18 604.80) = 105 303.17 E(5) = 0.333(143 990) + 0.667(−86 394) = −9676.13 E(2) = p(optimistic)E(3) + p(expected)E(4) + p(pessimistic)E(5) = 0.15(391 400) + 0.5(105 303.17 + 0.35(−9676.13) = 107 974.94 E(1) = E(2) = 107 974.94 because E(2) > 0 Pharma-Excel should proceed with the development of the new product.

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Chapter 12 - Dealing With Risk

12.44 Tree diagram:

E(2) = 40 000 Repair

No Extra Expenses 0.75

2

E(4) = 65 000

0.25 1

5

E(1) = 40 000 E(6) = 36 600

E(3) = 30 520 Replace

3

No Modification 0.6

No Future Need 0.3

−40 000

Future Need 0.7

110 000

No Future Need 0.3

−140 000

Future Need 0.7

10 000

4

E(5) = −35 000

Extra Expenses

Net Value ($)

6

E(7) = 21 400 Modification 7 0.4

E(8)

No Future = −18 000 Need 8 0.3 Future Need 0.7 E(9) No Future = −22 000 Need 9 0.3 Future Need 0.7

Do Nothing

Adopt 0.8

0

Not Adopt 0.2

−90 000

Adopt 0.8

0

Not Adopt 0.2

−110 000

60 000

40 000

0

Based on the tree diagram, the net value for each terminal position (from top to bottom of the tree) is calculated as follows: Terminal position 1 = cost of repair = −40 000 Terminal position 2 = PW(benefit) − (cost of repair) = 110 000 Terminal position 3 = (cost of repair) + (extra expenses) = −140 000 Terminal position 4 = PW(benefit) − (costs of repair & extra expenses) = 10 000 Terminal position 5 = complete recovery of investment = 0 Terminal position 6 = cost of replacement = −90 000 Terminal position 7 = PW(benefit) − (cost of replacement) = 60 000 Terminal position 8 = complete recovery of investment = 0 Terminal position 9 = (cost of replacement) + (cost of modification) = −110 000 Terminal position 10 = PW(benefit) − (replacement & modification costs) = 40 000 Terminal position 11 = do nothing = 0 E(8) = p(adopt)(0) + p(not adopt)(−90 000) = −18 000 272 Copyright © 2022 Pearson Canada Inc.

Chapter 12 - Dealing With Risk

E(9) = p(adopt}(0) + p(not adopt)(−110 000) = −22 000 E(4) = p(not needed)(−40 000) + p(needed)(110 000) = 65 000 E(5) = p(not needed)(−140 000) + p(needed)(10 000) = −35 000 E(6) = p(not needed)E(8) + p(needed)(60 000) = 36 600 E(7) = p(not needed)E(9) + p(needed)(40 000) = 21 400 E(2) = p(no extra expenses)E(4) + p(extra expenses)E(5) = 40 000 E(3) = p(no modification)E(6) + p(modification)E(7) = 30 520 E(1) = E(2) = 40 000 because E(2) > E(3) and E(2) > 0 BBB should repair the production line.

12.45 The Result of Sensitivity Analysis Probability Asian demand  Probability Asian demand  E(4) E(5) E(6) E(7) E(2) = 0.6E(4) + 0.4E(5) E(3) = 0.6E(6) + 0.4E(7) E(1) = max{E(2), E(3)} Partnership or no partnership?

0.1 0.9 0.8755 −0.85 0.525 0.28 0.185 0.427 0.427 no

0.2 0.8 1 −0.7 0.55 0.23 0.32 0.422 0.422 no

0.3 0.7 1.125 −0.55 0.575 0.24 0.455 0.441 0.455 partner

0.4 0.6 1.25 −0.4 0.6 0.22 0.59 0.448 0.59 partner

0.5 0.5 1.375 −0.25 0.725 0.2 0.725 0.515 0.725 partner

As long as the probability that Asian demand increases is greater than or equal to 0.3, then the partnership with China seems to be the preferred option. Also, the higher the probability, the more confidence in the partnership option.

12.46 (a) E(2) = p(rapid growth)E(3) + p(steady growth)E(4) + p(slow growth)E(5) = 16p1 + 8p2 + 0p3 = 16p1 + 8p2 (b) Graphically, all possible values of p1 and p2 are represented by the area that satisfy p1 + p2  1, p1  0, and p2  0. Furthermore, the values of p1 and p2 that also satisfy 16p1 + 8p2 < 10 are represented by the shaded region in the graph.

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Chapter 12 - Dealing With Risk

p2 1.25 16p1 + 8p2 = 10 1

(0.25, 0.75)

p1 + p2 = 1 Area = 1/2(0.375)(0.75) = 0.14

Area = 0.5-0.14 = 0.36 0

0.625

1

p1

From the graph, we can identify the estimate of probabilities that imply introducing the package now is the preferred option (see the shaded region). For example, (p1, p2) = (0.25, 0.5) would lead to the decision to introduce now. Note that whenever the values of p1 and p2 do not add up to 1, then it implies that p3 = 1 − p1 − p2. Also from the graph, we have the information that roughly 0.36/0.5 = 72% of all possible (p1, p2) values indicate that introducing the package now is the preferred option. Another observation is that, for (p1, p2) values with high p1 value (e.g., p1 > 0.625), the option to wait for the survey result is preferred. The possible interpretation for this decision is, if the market growth is likely to be rapid, then waiting for 3 months would not jeopardize LOTell’s opportunity to gain more market share even in the presence of the competitors.

12.47 US Salary US Salary Inflation Number of staff: Turnover Rate: Recruiting and Training Costs Salary Salary Inflation Post-training productivity Productivity while training Project Management Costs Conversion to $ MARR Outsourcing setup: PW(Benefits of Outsourcing)

75000 5.00% 20 25.00%per year 0.7 per person 600000 per year 18.21% 70.00% 50.00% 15.00% 40 0.25% 400,000 2,935

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Chapter 12 - Dealing With Risk

Parameter Values Base −10% Case 10% 4.50% 5.00% 5.50% 11.25% 12.50% 13.75% 22.50% 25.00% 27.50% 63.00% 70.00% 77.00% 45.00% 50.00% 55.00% 63.00% 70.00% 77.00% 13.50% 15.00% 16.50% 22.50% 25.00% 27.50% 36.00 40.00 44.00 360,000 400,000 440,000

PW(Benefits of Outsourcing) Base −10% Case 10% 937,992 1,064,322 1,193,606 1,260,899 1,064,322 855,758 1,203,150 1,064,322 918,041 1,116,097 1,064,322 1,012,547 969,083 1,064,322 1,142,244 596,900 1,064,322 1,446,758 1,095,200 1,064,322 1,033,443 1,247,302 1,064,322 901,575 505,063 1,064,322 1,521,897 1,104,322 1,064,322 1,024,322

It appears that the productivity rate and the exchange rate have the greatest impact on the PW of outsourcing. Next is the salary growth rate in India.

The exchange rate needs to go to about 33.01INR per USD for the PW(benefits) to drop to zero.

The productivity needs to drop to about 55.86% for the PW(benefits) to drop to zero.

The growth rate needs to be about 18.2% for the PW(benefits) to drop to zero.

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Chapter 12 - Dealing With Risk

Notes for Case in Point 12.1 1-4)

These are all difficult questions and the answers depend on the views of the individual student.

Notes for Mini-Case 12.1 1)

10% does make sense as an abstract deviation for sensitivity analysis purposes because it provides a way to compare the effect of changes to different parameters. It can also be easily scaled to approximate other particular variation values. Although a change of 10% of the assumed value is a sensible amount for many parameters, for cost of borrowing, which is already a percentage, a 1% change is simply more meaningful.

Again, it can be easily scaled to approximate other particular variations.

2)

No. Although a 10% change in revenue had the largest effect on profit, it doesn't take into account how likely a 10% change in revenue is compared to the chance of other parameters varying by 10%. For example, even though the effect of a 10% change on cost of sales is less, it may be much more likely to occur, making it the larger risk.

3)

You would have to research the likelihood of each of the relevant parameters varying by a significant amount. You might focus particularly on the risks of changes to revenue, cost of sales, and cost of borrowing, but all of the parameters would merit some deeper evaluation.

4)

Examples include: researching land values, establishing building costs early, determining expected usage through surveys, etc., researching similar facility fee methods and revenues, getting an early commitment on subsidies, etc.

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CHAPTER 13 Solutions to Chapter-End Problems 13.1

#1 is best for C since it cannot be dominated. #2 is best for D since it cannot be dominated. #3 is not dominated. #4, 5, 6, 7, 8, and 10 are all dominated by #2. #9 is not dominated.

Thus, #4, 5, 6, 7, 8, and 10 can be eliminated from further consideration.

13.2

Completed pairwise comparison matrix:  1 12 2 1  4 2  9 4

1 1

4 2

1 2

 1  4 1  2  1 1

9

The priority weights are: [0.064 0.131 0.263 0.542]T. Normalized PCM 0.063 0.067 0.125 0.133 0.250 0.267 0.563 0.533

0.067 0.133 0.267 0.533

0.060 0.134 0.269 0.537

13.3

(a) First, sort by one criterion, e.g., C1:

Crit C1 C2 C3 C4 C5

L1 81900 25500 65 8.6 3

L9 36500 10300 12.2 3 6

L2 31800 11600 45 6.3 7

L4 31100 10500 35 14.1 6

L8 28200 7400 30 6 10

L10 24400 7100 40 3.7 9

L5 23000 10200 40 12.1 5

Average 0.064 0.131 0.263 0.542

L11 18400 4700 43 3.7 5

L6 16100 3500 10 11.8 4

L12 13900 3100 25 5.8 4

L7 13200 3700 32 3.9 9

L3 11500 7100 28 4.5 4

Then proceed to check for dominance. It can be seen that L10 is dominated by L8 (recall that low capital cost, C3, is desirable).

Similarly, L5 is dominated by L4, L7 by L8, L11 by L8, and L12 by L6.

The rest, namely L1, L2, L3, L4, L6, L8, and L9, are efficient.

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Chapter 13 Solutions to Chapter-End Problems

(b) The decision matrix: Criteria Weight L1 L9 L2 L4 L8 L6 L3 C1 81900 36500 31800 31100 28200 16100 11500 Normalized 1.5 10.0 3.6 2.9 2.8 2.4 0.7 0.0 C2 25500 10300 11600 10500 7400 3500 7100 Normalized 2.0 10.0 3.1 3.7 3.2 1.8 0.0 1.6 C3 65 12.2 45 35 30 10 28 Normalized 2.5 0.0 9.6 3.6 5.5 6.4 10.0 6.7 C4 8.6 3 6.3 14.1 6 11.8 4.5 Normalized 3.0 5.0 0.0 3.0 10.0 2.7 7.9 1.4 C5 1.0 3 6 7 6 10 4 4 Score 53.1 41.5 36.7 60.2 41.1 53.8 28.1

The best subway route from this data is L4. (c) The revised decision matrix: Criteria Weight L1 L9 L2 L4 L8 L6 L3 C1 81900 36500 31800 31100 28200 16100 11500 Normalized 2.0 10.0 3.6 2.9 2.8 2.4 0.7 0.0 C2 25500 10300 11600 10500 7400 3500 7100 Normalized 2.0 10.0 3.1 3.7 3.2 1.8 0.0 1.6 C3 65 12.2 45 35 30 10 28 Normalized 2.0 0.0 9.6 3.6 5.5 6.4 10.0 6.7 C4 8.6 3 6.3 14.1 6 11.8 4.5 Normalized 2.0 5.0 0.0 3.0 10.0 2.7 7.9 1.4 C5 2.0 3 6 7 6 10 4 4 Score 56.1 44.5 40.3 54.8 46.4 45.2 27.4

Yes. In this case, L1 is slightly better.

13.4

Many solutions for this problem are possible. In the decision matrix below, four different weighting schemes were applied.

Weightings (a) (b) (c) (d) Cost 0.5 0.5 1.0 3.0 Service 0.5 0.5 1.0 0.5 Reliability 3.0 5.0 2.0 2.0 Speed 2.0 1.25 1.5 1.25 Quality 2.0 1.25 1.5 1.25 Flexibility 1.0 1.0 1.0 1.0 Size 0.5 0.25 1.0 0.5 Other 0.5 0.25 1.0 0.5 Scores: (a) Balanced weighting (b) Emphasizing reliability (c) More evenly distributed (d) Emphasizing cost

1

2

2 5 5 8 7 10 5 5 63.5 59.75 59.5 52.25

7 3 5 8 4 2 5 5 51.0 49.5 50.0 54.5

Alternatives 3 4 0 6 3 5 5 3 3 5 8 5 8 5 8 5 2 6 51.5 45.0 50.75 40.75 47.5 48.0 38.25 49.5

5

6

10 5 5 10 10 5 0 0 67.5 62.5 60.0 72.5

3 3 10 5 8 5 5 5 69.0 76.75 60.5 56.75

Alternative 6 was always best, except when cost was heavily weighted, in which case, alternative 5 was slightly better.

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Chapter 13 Solutions to Chapter-End Problems

13.5

(a) Completed pairwise comparison matrix: 1 1   15  3

1 1 3

1

1

1

 2  2  1

5

1

3

1

2

2

3

(b) Completed pairwise comparison matrix: 1 1  9  17  1  13

13.6

9 1 1

3

6 1

7 3 1 4 1

2

3 1  2  1 2 1 

1 1 1

6 4

1 2

(a) The priority weights are: [0.292 0.199 0.276 0.232]T. PCM 1 1 1 1 0.2 3 3 0.5 Normalized PCM 0.192 0.182 0.192 0.182 0.038 0.545 0.577 0.091

5 0.333 1 0.5

0.333 2 2 1

0.732 0.049 0.146 0.073

0.063 0.375 0.375 0.188

Average 0.292 0.199 0.276 0.232

(b) The priority weights are: [0.397 0.093 0.092 0.257 0.161]T. PCM 1 9 0.111 1 0.142 0.333 1 6 0.333 1 Normalized PCM 0.387 0.519 0.043 0.058 0.055 0.019 0.387 0.346 0.129 0.058

7 3 1 4 0.5

1 0.167 0.25 1 2

3 1 2 0.5 1

0.452 0.194 0.065 0.258 0.032

0.226 0.038 0.057 0.226 0.453

0.400 0.133 0.267 0.067 0.133

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Average 0.397 0.093 0.092 0.257 0.161

Chapter 13 Solutions to Chapter-End Problems

13.7

(a) Job opportunities 2, 4, 5, 6 and 9 should no longer be considered. (b) Create a decision matrix, only with the efficient jobs: Criterion Pay Home Studies Size

Weight 4 2.5 2 1.5

#1 Spinoff 1700 2 3 5

#3 Soutel 2200 80 4 150

#7 Ring 2200 250 5 300

#8 Jones 2700 500 3 20

As Francis did not provide subjective ratings, one may rely on normalized ratings. The least and most preferred alternatives rate 0 and 10, respectively; other alternatives are ranked linearly between these extremes.

The normalized decision matrix: Criterion Pay Home Studies Size

Weight 4 2.5 2 1.5 Scores:

#1 Spinoff 0.0 10.0 6.0 10.0 52.0

#3 Soutel 5.0 8.4 8.0 5.1 64.7

#7 Ring 5.0 5.0 10.0 0.0 52.6

#8 Jones 10.0 0.0 6.0 9.5 66.2

With the given information, job #8 scores the highest and is therefore the best.

13.8

No.

13.9

According to the decision matrix, the best consultant is D, with an overall score of 57.5.

Weight Criterion Cost Reliability Familiarity Location Quality

A 2 3 1 1 3 Score:

5 3.3 0 5 10 55

Alternative Consultants B C D 5 7.5 0 3.3 10 3.3 5 5 10 10 0 7.5 3.3 0 10 45 50 57.5

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E 10 0 5 10 3.3 45

Chapter 13 Solutions to Chapter-End Problems

13.10 Select Consultant

Cost

Reliability

Familiarity

Location

Quality

ABCDE

ABCDE

ABCDE

ABCDE

ABCDE

13.11 Pairwise comparison matrix for cost:

1 1  3 1  5  5

1 1 3 1 5

5

5 1 5 3 1 5 1 1 5 3 5

1 3

   1  3  1 5 1 

1 5

1 5

13.12 Pairwise comparison matrix for reliability:

1 1  5  1  13

1 1 5 1

1 5

1 5

1 1 5 1

1 3

1 7

1 3

1 5

1

3 3 7  3 1

13.13 Pairwise comparison matrix for familiarity:

1 15 5 1  5 1  9 5 5 1

1 5

1 9

1 1 5 1

1 5 1 5

1 1 5

 1  1  5 1 

1 5

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Chapter 13 Solutions to Chapter-End Problems

13.14 Pairwise comparison matrix for location: 1 4   14  2  4

4 12 1 8 2 1 1 16 8 1 6 1 2

1

4

1 8

2

 1  1  8  1 2 1  1

4

13.15 Pairwise comparison matrix for quality: 1 1  7  19  1  17

7 9 1 5 1 1 5 7 9 1 5

1 1 1

7 9

1 1

7

7 1  1  5  7 1 

13.16 The priority weights for cost are: [0.121 0.121 0.241 0.046 0.470]T. PCM 1 1 1 1 3 3 0.2 0.2 5 5 Normalized PCM 0.098 0.098 0.098 0.098 0.294 0.294 0.020 0.020 0.490 0.490

0.333 0.333 1 0.2 3

5 5 5 1 5

0.2 0.2 0.333 0.2 1

0.068 0.068 0.205 0.041 0.616

0.238 0.238 0.238 0.048 0.238

0.103 0.103 0.172 0.103 0.517

Average 0.121 0.121 0.241 0.046 0.470

13.17 The priority weights for reliability are: [0.130 0.130 0.557 0.130 0.052]T. PCM 1 1 1 1 5 5 1 1 0.333 0.333 Normalized PCM 0.120 0.120 0.120 0.120 0.600 0.600 0.120 0.120 0.040 0.040

0.2 0.2 1 0.2 0.143

1 1 5 1 0.333

3 3 7 3 1

0.115 0.115 0.574 0.115 0.082

0.120 0.120 0.600 0.120 0.040

0.176 0.176 0.412 0.176 0.059

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Average 0.130 0.130 0.557 0.130 0.052

Chapter 13 Solutions to Chapter-End Problems

13.18 The priority weights for familiarity are: [0.036 0.137 0.137 0.555 0.137]T. PCM 1 0.2 5 1 5 1 9 5 5 1 Normalized PCM 0.040 0.240 0.200 0.122 0.200 0.122 0.360 0.610 0.200 0.122

0.2 1 1 5 1

0.111 0.2 0.2 1 0.2

0.2 1 1 5 1

0.240 0.122 0.122 0.610 0.122

0.065 0.117 0.117 0.584 0.117

0.240 0.122 0.122 0.610 0.122

Average 0.036 0.137 0.137 0.555 0.137

13.19 The priority weights for location are: [0.100 0.340 0.035 0.185 0.340]T. PCM 1 0.25 4 1 0.25 0.125 2 0.5 4 1 Normalized PCM 0.089 0.087 0.356 0.348 0.022 0.043 0.178 0.174 0.356 0.348

4 8 1 6 8

0.5 2 0.167 1 2

0.25 1 0.125 0.5 1

0.148 0.296 0.037 0.222 0.296

0.088 0.353 0.029 0.176 0.353

0.087 0.348 0.043 0.174 0.348

Average 0.100 0.340 0.035 0.185 0.340

13.20 The priority weights for quality are: [0.402 0.083 0.030 0.402 0.083]T. PCM 1 7 0.143 1 0.111 0.2 1 7 0.143 1 Normalized PCM 0.417 0.432 0.060 0.062 0.046 0.012 0.417 0.432 0.060 0.062

9 5 1 9 5

1 0.143 0.111 1 0.143

7 1 0.2 7 1

0.310 0.172 0.034 0.310 0.172

0.417 0.060 0.046 0.417 0.060

0.432 0.062 0.012 0.432 0.062

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Average 0.402 0.083 0.030 0.402 0.083

Chapter 13 Solutions to Chapter-End Problems

13.21 Priority matrix:  0.121  0.121   0.241  0.046 0.470

0.402  0.083  0.030   0.402  0.052 0.137 0.340 0.083 

0.130 0.130 0.557 0.130

0.036 0.137 0.137 0.555

0.100 0.340 0.035 0.185

13.22 Completed pairwise comparison matrix for criteria:

1 3   13 1  3  1

3 1 7 1 1 7 1 1 6 1 7 1

3

3 6 1 1 7

1 1  1  7  1 7 1 

The priority weights for criteria are: [0.186 0.383 0.053 0.055 0.323]T. Normalized PCM 0.176 0.126 0.529 0.378 0.059 0.054 0.059 0.063 0.176 0.378

0.158 0.368 0.053 0.053 0.368

0.167 0.333 0.056 0.056 0.389

0.304 0.304 0.043 0.043 0.304

Average 0.186 0.383 0.053 0.055 0.323

13.23 The consultant’s scores are [0.210 0.125 0.277 0.228 0.160]T. Consultant C is best.  0.121  0.121   0.241  0.046 0.470

0.402  0.186  0.210  0.083  0.383  0.125  0.030   0.053  = 0.277       0.402  0.055  0.228  0.052 0.137 0.340 0.083  0.323  0.160 

0.130 0.130 0.557 0.130

0.036 0.137 0.137 0.555

0.100 0.340 0.035 0.185

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Chapter 13 Solutions to Chapter-End Problems

13.24 Choose Job

Pay

Distance

Studies

Size

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

13.25 (a) Pairwise comparison matrix for pay:  1 15 5 1  5 1  7 5

1

5

1 1 5

  5  1  5  1

1

7

1

(b) The priority weights for pay are: [0.051 0.171 0.171 0.606]T. Normalized PCM 0.056 0.028 0.278 0.139 0.278 0.139 0.389 0.694

0.028 0.139 0.139 0.694

0.093 0.130 0.130 0.648

Average 0.051 0.171 0.171 0.606

13.26 (a) Pairwise comparison matrix for distance from home: 1 1  5  17 1  9

5 1 1 1

5

7

7 9 5 7 1 2  1 1 2

(b) The priority weights for distance from home are: [0.617 0.259 0.077 0.047]T. Normalized PCM 0.688 0.788 0.138 0.158 0.098 0.032 0.076 0.023

0.519 0.370 0.074 0.037

0.474 0.368 0.105 0.053

Average 0.617 0.259 0.077 0.047

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Chapter 13 Solutions to Chapter-End Problems

13.27 (a) Pairwise comparison matrix for studies: 1 2  3  1

1

2

1 2 1

2

1 1 2 2 1 3  1 1 3 1

3

(b) The priority weights for studies are: [0.141 0.263 0.455 0.141]T. Normalized PCM 0.143 0.125 0.286 0.250 0.429 0.500 0.143 0.125

0.154 0.231 0.462 0.154

0.143 0.286 0.429 0.143

Average 0.141 0.263 0.455 0.141

13.28 (a) Pairwise comparison matrix for size: 1 1  7  19  1

7 9 1 2 1 1 2 7 9

1 1  7  1  9  1

(b) The priority weights for size are: [0.442 0.072 0.045 0.442]T. Normalized PCM 0.444 0.452 0.063 0.065 0.049 0.032 0.444 0.452

0.429 0.095 0.048 0.429

0.444 0.063 0.049 0.444

Average 0.442 0.072 0.045 0.442

13.29 The priority weights put together make up the priority matrix:  0.051  0.171   0.171  0.606

0.617 0.141 0.259 0.263 0.077 0.455 0.047 0.141

0.442  0.072  0.045   0.442 

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Chapter 13 Solutions to Chapter-End Problems

13.30 The priority weights for criteria are: [0.550 0.249 0.118 0.083]T. Normalized PCM 0.577 0.643 0.192 0.214 0.115 0.071 0.115 0.071

0.526 0.316 0.105 0.053

0.455 0.273 0.182 0.091

Average 0.550 0.249 0.118 0.083

13.31 To calculate the priority weights for the goal, multiply the priority matrix by the criteria weights:  0.051  0.171   0.171  0.606

0.617 0.141 0.442  0.550  0.235  0.259 0.263 0.072  0.249  0.196   = 0.077 0.455 0.045  0.118   0.171      0.047 0.141 0.442  0.083  0.398 

From this, it can be seen that the best job is Jones Mines, followed by Spinoff Consulting.

13.32 Yes, Alternative 1 could become the preferred choice when an identical or near identical copy of another alterative is added. This phenomenon, called rank reversal, has generated strong criticism of the AHP process.

The proof is by example.

The calculations for the example PCM's given are made as follow: 0.25 0.25   0.75 0.75 

Normalized PCM for Criterion 1: 

0.75 0.75   0.25 0.25 

Normalized PCM for Criterion 2: 

0.75 0.75   0.25 0.25 

Normalized PCM for Goal: 

The score calculation is then as follows: 0.75 0.25  0.75  0.375  0.25 0.75   0.25  = 0.625       

Since 0.625 > 0.375, Alternative 2 is preferred.

If an Alternative 3, identical to Alternative 2, is added to these calculations:

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Chapter 13 Solutions to Chapter-End Problems

1 PCM for Criterion 1: 3  3

1

3

1 1

3 1  1 

1

 1 3 3 PCM for Criterion 2:  13 1 1  13 1 1

Normalizing gives:  0.142857 0.142857 0.142857  Normalized PCM for Criterion 1:  0.428571 0.428571 0.428571  0.428571 0.428571 0.428571  0.6 0.6 0.6  Normalized PCM for Criterion 1: 0.2 0.2 0.2   0.2 0.2 0.2 

The normalized PCM for the Goal remains the same.

The score calculation is then as follows: 0.142857 0.6  0.707143   0.428571 0.6    0.75  = 0.371429     0.25     0.371429   0.428571 0.6    

Here, 0.707143 > 0,371429, and Alternative 1 is preferred.

13A.1

PCM

PCM2

PCM3

PCM4

0.25 0.1111 0.5 0.25 1 1 1 1

0.063 0.125 0.250 0.563

Normalized 0.043 0.091 0.087 0.182 0.261 0.364 0.609 0.364

0.047 0.106 0.424 0.424

Average 0.061 0.125 0.325 0.490

4 2.5278 0.8611 0.5972 8.25 5.25 1.75 1.2222 23 15 4.5 3.1944 36 21.5 7.75 4.75

0.056 0.116 0.323 0.505

0.057 0.119 0.339 0.486

0.058 0.118 0.303 0.521

0.061 0.125 0.327 0.486

0.058 0.119 0.323 0.500

17.875 11.292 3.7222 2.5347 36.75 23.181 7.6597 5.2014 99.75 62.361 20.944 14 152.75 96 32.25 21.875

0.058 0.120 0.325 0.497

0.059 0.120 0.323 0.498

0.058 0.119 0.324 0.499

0.058 0.119 0.321 0.502

0.058 0.119 0.323 0.499

78.16 49.139 16.372 11.066 160.56 100.94 33.639 22.74 434.25 273.07 91.063 61.618 670.63 422.25 140.31 95.097

0.058 0.120 0.323 0.499

0.058 0.119 0.323 0.499

0.058 0.120 0.324 0.499

0.058 0.119 0.323 0.499

0.058 0.119 0.323 0.499

1 2 4 9

0.5 1 3 7

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Chapter 13 Solutions to Chapter-End Problems

This gives the principal eigenvector w = [0.058 0.119 0.323 0.499] T. To determine max, we must now solve Aw = maxw for one row of A. Selecting, for example, the last row of A results in the following expression: [9 7 1 1]  [0.058 0.119 0.323 0.499] T = max  0.499 Or equivalently, max = (9  0.058 + 7  0.119 + 1  0.323 + 1  0.499)/0.499 = 4.366 The consistency index can then be calculated from CI = (max – n)/(n – 1) = (4.366 – 4)/3 = 0.1209 As seen in Table 12A.1, the RI for a matrix of 4 rows and columns is 0.90.

We can then calculated the consistency ratio as: CR = CI/RI = 0.1209/0.90 = 0.1343 Clearly, the CR is significantly greater than 0.1. It can thus be concluded that the original PCM is not acceptably consistent.

13A.2

PCM

1 3 0.333 0.333 1

0.333 1 0.143 0.167 1

3 7 1 1 7

3 6 1 1 7

1 1 0.143 0.143 1

0.176 0.529 0.059 0.059 0.176

Normalized 0.126 0.158 0.167 0.378 0.368 0.333 0.054 0.053 0.056 0.063 0.053 0.056 0.378 0.368 0.389

0.304 0.304 0.043 0.043 0.304

Average 0.186 0.383 0.053 0.055 0.323

PCM2

5 11.33 1.571 1.643 9.667

2.595 5 0.706 0.73 4.5

18.33 36 5 5.167 31

18 35 4.857 5 30

3.19 6.857 0.905 0.929 5

0.171 0.388 0.054 0.056 0.331

0.192 0.370 0.052 0.054 0.333

0.192 0.377 0.052 0.054 0.325

0.194 0.377 0.052 0.054 0.323

0.189 0.406 0.054 0.055 0.296

0.188 0.384 0.053 0.055 0.321

PCM3

28.09 56.86 7.881 8.151 48.5

13.07 26.61 3.659 3.778 22.15

91.83 188 25.85 26.71 156.5

89.24 183 25.14 25.98 152

15.98 33.33 4.591 4.754 27.88

0.188 0.380 0.053 0.055 0.324

0.189 0.384 0.053 0.055 0.320

0.188 0.385 0.053 0.055 0.320

0.188 0.385 0.053 0.055 0.320

0.185 0.385 0.053 0.055 0.322

0.187 0.384 0.053 0.055 0.321

PCM4

143.6 293.7 40.45 41.8 245.7

66.4 136.3 18.76 19.39 113.9

468.7 961.2 132.4 136.9 804.2

455.6 934.6 128.7 133.1 782.1

83 169.8 23.41 24.21 142.6

0.188 0.384 0.053 0.055 0.321

0.187 0.384 0.053 0.055 0.321

0.187 0.384 0.053 0.055 0.321

0.187 0.384 0.053 0.055 0.321

0.187 0.383 0.053 0.055 0.322

0.187 0.384 0.053 0.055 0.321

This gives the principal eigenvector w = [0.187 0.384 0.053 0.055 0.321] T. 289 Copyright © 2022 Pearson Canada Inc.

Chapter 13 Solutions to Chapter-End Problems

To determine max, we must now solve Aw = maxw for one row of A. Selecting, for example, the first row of A results in the following expression: [1 0.333 3 3 1]  [0.187 0.384 0.053 0.055 0.321] T = max  0.187 Or equivalently, max = (1  0.187 + 0.333  0.384 + 3  0.053 + 3  0.055 + 1  0.321)/0.1870 = 5.113 The consistency index can then be calculated from CI = (max – n)/(n – 1) = (5.113 – 5)/4 = 0.0333 As seen in Table 12A.1, the RI for a matrix of 5 rows and columns is 1.12.

We can then calculate the consistency ratio as: CR = CI/RI = 0.0333/1.12 = 0.0297 Clearly, the CR is significantly less than 0.1. It can thus be concluded that the original PCM is acceptably consistent.

13A.3

PCM

PCM2

PCM3

PCM4

5 3 2 1

0.577 0.192 0.115 0.115

Normalized 0.643 0.526 0.214 0.316 0.071 0.105 0.071 0.053

0.455 0.273 0.182 0.091

Average 0.550 0.249 0.118 0.083

4 9.3333 21.5 29 1.8667 4 9.1667 13.667 0.9111 1.9333 4 6 0.6111 1.4333 3 4

0.541 0.253 0.123 0.083

0.559 0.240 0.116 0.086

0.571 0.243 0.106 0.080

0.551 0.259 0.114 0.076

0.555 0.249 0.115 0.081

17.211 38.167 84 120 7.7667 17.211 37.333 53.333 3.5556 8 17.356 24.356 2.4889 5.6 12.356 17.356

0.555 0.250 0.115 0.080

0.553 0.250 0.116 0.081

0.556 0.247 0.115 0.082

0.558 0.248 0.113 0.081

0.556 0.249 0.115 0.081

70.733 157.8 344.56 488.56 31.637 70.733 154.47 218.47 14.564 32.57 71.311 100.84 10.298 22.97 50.278 71.311

0.556 0.249 0.114 0.081

0.555 0.249 0.115 0.081

0.555 0.249 0.115 0.081

0.556 0.248 0.115 0.081

0.556 0.249 0.115 0.081

1 0.333 0.2 0.2

3 1 0.333 0.333

5 3 1 0.5

This gives the principal eigenvector w = [0.556 0.249 0.115 0.081] T. These values in this eigenvector are quite close to the priority weights calculated in Problem 13.30.

290 Copyright © 2022 Pearson Canada Inc.

Chapter 13 Solutions to Chapter-End Problems

To determine max, we must now solve Aw = maxw for one row of A. Selecting, for example, the first row of A results in the following expression: [1 3 5 5]  [0.556 0.249 0.115 0.081] T = max  0.556 Or equivalently, max = (1  0.556 + 3  0.249 + 5  0.115 + 5  0.081)/0.556 = 4.106 The consistency index can then be calculated from CI = (max – n)/(n – 1) = (4.106 – 4)/3 = 0.0354 As seen in Table 12A.1, the RI for a matrix of 4 rows and columns is 0.90.

We can then calculate the consistency ratio as: CR = CI/RI = 0.0354/0.90 = 0.0393 Clearly, the CR is significantly less than 0.1. It can thus be concluded that the original PCM is acceptably consistent.

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Chapter 13 Solutions to Chapter-End Problems

Notes for Mini-Case 13.1 1)

This will vary from student to student.

2)

It will be easy for most students.

3)

Most students should readily identify alternatives, and many will identify criteria.

4)

There may be some difficulty coming up with both ratings and weightings.

5)

Students will generally come up with the same decision they actually made.

If not, it is generally a result of difficulties in establishing the weights and ratings.

292 Copyright © 2022 Pearson Canada Inc.