Chapter 7 - Engineering Economic Analysis, 2nd Canadian Edition

Homework Solutions for Engineering Economic Analysis, 10th Edition Newnan, Lavelle, Eschenbach

1

Chapter 7: Rate of Return Analysis

7-1 $100 = $27 (P/A, i%, 10) (P/A, i%, 10) = 3.704 Performing Linear Interpolation:

(P/A, i%, 10) I 4.192 20% 3.571 25%

Rate of Return = 20% + (5%) [(4.192 − 3.704)/(4.912 − 3.571)] = 23.9%

7-2 (a) F = $5, P = $1, n = 5 F = P (1 + i)n $5 = $1 (1 + i)5 (1 + i) = 50.20 = 1.38 i* = 38% (b) For a 100% annual rate of return F = $1 (1 + 1.0)5 = $32, not $5! Note that the prices Diagonal charges do not necessarily reflect what anyone will

pay a collector for his/her stamps.


2

7-3

($175 − $35) = $12.64 (P/A, i%, 12) (P/A, i%, 12) = $140/$12.64 = 11.08 i = 1 ¼% Nominal interest rate = 12 (1 ¼%) = 15%

7-4

The rate of return exceeds 60% so the interest tables are not useful. F = P (1 + i)n $25,000 = $5,000 (1 + i)3 (1 + i) = ($25,000/$5,000)1/3 = 1.71 i* = 0.71 Rate of Return = 71%

7-5

$9,375 = $325 (P/A, i%, 36) (P/A, i%, 36) = $9,375/$325 = 28.846 From compound interest tables, i = 1.25% Nominal Interest Rate = 1.25 × 12 = 15% Effective Interest Rate = (1 + 0.0125)12 − 1 = 16.08%

$175

……….

n = 12

A = $12.64

……

A = $325

n = 36

$3,000

$12,375


3

7-6 1991 − 1626 = 365 years = n F = P (1 + i)n 12 × 109 = 24(1 + i)365 (1 + i)365 = 12 x 100/24= 5.00 × 108 This may be immediately solved on most hand calculators: i* = 5.64% Solution based on compound interest tables: (F/P, i%, 365) = 5.00 × 108 = (F/P, i%, 100) (F/P, i%, 100) (F/P, i%, 100) (F/P, i%, 65) Try i = 6% (F/P, 6%, 365) = (339.3)3 (44.14) = 17.24 × 108 (i too high) Try i = 5% (F/P, 5%, 365) = (131.5)3 (23.84) = 0.542 × 108 (i too low) Performing linear interpolation: i* = 5% + (1%) [((5 − 0.54) (108))/((17.24 − 0.54) (108))] = 5% + 4.46/16.70 = 5.27% The linear interpolation is inaccurate.

7-7

(F/A, i, 35) = 610

5800 = 172.414 and is very close to 8% from tables. (Exact = 8.003%)

7-8

(F/A, i, 20) = 7

5

10

2.5 10 = 40 and interpolating

i = 6% + (1%) 36.786 40

36.786 40.996

= 6.76% (exact value 6.774%)


4

7-9

Year Cash Flow 0 −$1,000 3 +$1,094.60 6 +$1,094.60

$1,000 = $1,094 [(P/F, i%, 6) + (P/F, i%, 9)] Try i = 20% $1,094 [(0.5787) + (0.3349)] = $1,000 Rate of Return = 20%

7-10 3,000 = 30 (P/A, i*, 120) (P/A, i*, 120) = 3,000/30 = 100 Performing Linear Interpolation:

(P/A, i%, 120) I 103.563 ¼% 100 i* 90.074 ½%

i* = 0.0025 + 0.0025 [(103.562 − 100)/(103.562 − 90.074)] = 0.00316 per month Nominal Annual Rate = 12 (0.00316) = 0.03792 = 3.79%


5

7-11 $3,000 = $119.67 (P/A, i%, 30) (P/A, i%, 30) = $3,000/$119.67 = 25.069 Performing Linear Interpolation:

(P/A, i% 30) i 25.808 1% 24.889 1.25%

i = 1% + (0.25%)((25.808−25.069)/(25.808−24.889)) = 1.201% (a) Nominal Interest Rate = 1.201 × 12 = 14.41% (b) Effective Interest Rate = (1 + 0.01201)12 − 1 = 0.154 = 15.4%

7-12

$125 = $10 (P/A, i%, 6) + $10 (P/G, i%, 6) at 12%, $10 (4.111) + $10 (8.930) = $130.4 at 15%, $10 (3.784) + $10 (7.937) = $117.2 i* = 12% + (3%) ((130.4 − 125)/(130.4−117.2)) = 13.23%

$125

$10 $20

$30 $40

$50 $60


6

7-13

$42.55 = $5 (P/A, i%, 5) + $5 (P/G, i%, 5) Try i = 15%, $5 (3.352) + $5 (5.775) = $45.64 > $42.55 Try i = 20%, $5 (2.991) + $5 (4.906) = $39.49 < $42.55 Rate of Return = 15% + (5%) [($45.64 − $42.55)/($45.64 − $39.49)] = 17.51% Exact Answer: 17.38%

7-14 The algebraic sum of the cash flows equals zero. Therefore, the rate of return is 0%.

7-15

Try i = 5% $1,000 = (?) $300 (3.546) (0.9524) = (?) $1,013.16 Try i = 6% $1,000 = (?) $300 (3.465) (0.9434) = (?) $980.66 Performing Linear Interpolation: i* = 5% + (1%) (($1,013.6 − $1,000)/($1,013.6 − $980.66)) = 5.4%

$5 $10

$15 $20 $25

$42.55

$1,000

A = $300


7

7-16 Since the rate of return exceeds 60%, the tables are useless. F = P (1 + i)n $4,500 = $500 (1 + i)4 (1 + i)4 = $4,500/$500 = 0 (1 + i) = 9¼ = 1.732 i* = 0.732 = 73.2%

7-17 (a) Using Equation (4-39): F = Pem $4,000 = $2,000er(9) 2 = er(9) 9r = In 2 = 0.693 r = 7.70% (b) Equation (4-34) ieff = er − 1 = e0.077 − 1 = 0.0800 = 8.00%

7-18

Year Cash Flow 0 −$640 1 40 2 +$100 3 +$200 4 +$300 5 +$300

$640 = $100 (P/G, i%, 4) + $300 (P/F, i%, 5) Try i = 9% $100 (4.511) + $300 (0.6499) = $646.07 > $640 Try i = 10% $100 (4.378) + $300 (0.6209) = $624.07 < $640 Rate of Return = 9% + (1%) [(%646.07 − $640)/($646.07 − $624.07)] = 9.28%


8

7-19

Year Cash Flow 0 −$223 1 −$223 2 −$223 3 −$223 4 −$223 5 −$223 6 +$1,000 7 +$1,000 8 +$1,000 9 +$1,000 10 +$1,000

The rate of return may be computed by any conventional means. On closer

inspection one observes that each $223 increases to $1,000 in five years. $223 = $1,000 (P/F, i%, 5) (P/F, i%, 5) = $223/$1,000 = 0.2230 From interest tables, Rate of Return = 35%

7-20 Do nothing has a cash flow of zero, thus, the difference between alternatives is just

the Leaseco cash flow.

Year Leaseco – Do Nothing

0 –$1,000 1 $200 2 $200 3 $1,200 4 $1,200 5 $1,200

NPW = 0 = −1000 + 200 (P/A, ROR, 5) + 1000 (P/F, i, 2) (P/A, i, 3) and interpolating

ROR = 45% + (5%) 85.271

85.271 27.329

= 48.8%. Obviously, do nothing is rejected.


9

7-21

The easiest solution is to solve one cycle of the repeating diagram:

$120 = $80 (F/P, i%, 1) $120 = $80 (1 + i) (1 + i) = $120/$80 = 1.50 i* = 0.50 = 50% Alternative Solution: EUAB = EUAC $80 = [$200 (P/F, i%, 2) + $200 (P/F, i%, 4) + $200 (P/F, i%, 6)] (A/P, i%, 6) Try i = 50% $80 = [$200 (0.4444) + $200 (0.1975) + $200 (0.0878)] (0.5481) = $79.99 Therefore i* = 50%

$80 $80 $80 $80 $80 $80

$200 $200 $200

=

$80 $80 $80

$200 $120


10

7-22 For infinite series: A = Pi EUAC = EUAB $3,810 (i) = $250 + $250 (F/P, i%, 1) (A/F, i%, 2)* Try i = 10% $250 + $250 (1.10) (0.4762) = $381 $3,810 (0.10) = $381 i = 10%

*Alternate Equations: $3,810 (i) = $250 + $250 (P/F, i%, 1) (A/P, i%, 2) $3,810 (i) = $500 − $250 (A/G, i%, 2)

7-23

At Year 0, PW of Cost = PW of Benefits $412 + $5,000 (P/F, i%, 10) = ($1000/i) (P/F, i%, 10) Try i = 15% $412 + $5,000 (0.2472) = ($1,000/0.15) (0.2472) $1,648 = $1,648 ROR = 15%

………

n = 10

$412

Yr 0

$5,000

P’ A = $1,000

n = ∞


11

7-24 $400 = [$200 (P/A, i%, 4) − $50 (P/G, i%, 4)] (P/F, i%, 1) Try i = 7% [$200 (3.387) − $50 (4.795)] (0.9346) = 409.03 Try i = 8% [$200 (3.312) − $50 (4.650)] (0.9259) = $398.08 i* = 7% + (1%) [($409.03 − $400)/($409.03 − $398.04)] = 7.82%

7-25 The one-time $2,000 life membership fee avoids the 40-year series of beginning-of-

year membership dues that start at $200 and increase 3% annually. (a) The equation for determining the rate of return for the life membership is the

difference of the present worth of the two cash flows set to zero: 2000 – 200 – 206 (P/A, 3%, ROR, 39) = 0 (39 since beginning-of-year

payments)

(b) Use Excel where Result = 39 391800 1 (1 0.03) (1 )

206 0.03

i

i

and vary i = ROR until

zero is obtained. ROR = 14.243%

7-26

Ye

0 1 2 3 4 5 6 7 8 9 10

PW =

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14

7-28

PW of Cost = PW of Benefits $925 = $40 (P/A, i%, 10) + $1,000 (P/F, i%, 10) Try i = 5% $925 = $40 (7.722) + $1,000 (0.6139) = $922.78 (i too high) Try i = 4.5% $925 = $40 (7.913) + $1,000 (0.6439) = $960.42 (i too low) i* = 4.97%

7-29

PW of Benefits – PW of Costs = 0 $20 (P/A, i%, 40) + $1,000 (P/F, i%, 40) − $715 = 0 Try i = 3% $20 (23.115) + $1,000 (0.3066) − $715 = $53.90 i too low Try i = 3.5% $20 (21.355) + $1,000 (0.2526) − $715 = −$35.30 i too high Performing linear interpolation: i* = 3% + (0.5%) [53.90/(53.90 − (−35.30))] = 3.30% Nominal i* = 6.60%

A = $40

n = 10

$925

$1,000

$715

…

A = $40

$1,000

n = 40 semiannual periods


15

7-30

PW of Benefits – PW of Cost = $0 $30 (P/A, i%, 27) + $1,000 (P/F, i%, 27) − $875 = $0 Try i = 3 ½% $30 (17.285) + $1,000 (0.3950) − $875 = $38.55 >$0 Try i = 4% $30 (16.330) + $1,000 (0.3468) − $875 = −$38.30 < $0 i* = 3.75% Nominal rate of return = 2 (3.75%) = 7.5%

7-31

(a) For the cash flow of the bond have i = 6.8%

2 = 3.4%, so (0.034) (1000) = $34 is

paid semi-annually and $1,000 is paid at the end of the 10th year (20th pay period).

NPW = 0 = +1000 – 34 (P/A, i, 20) – 1000 (P/F, i, 20) and interpolating

i = 59.518

3% (0.5%)59.518 14.192

= 3.404% (exact value = 3.400%),

r = (2) (3.404%) = 6.808%, and ia = (1+0.03404)2 – 1 = 0.06924 or 6.924%. (b) The fee is $1,000 x 0.0075 = $7.50. So ABC Corp. receives $1,000 – $7.50 =

$992.50. NPW = 0 = 992.5 - 34 (P/A, i, 20) – 1000 (P/F, i, 20) and interpolating

i = 67.018

3% (0.5%)67.018 6.692

= 3.4546% (exact value = 3.453%),

r = (2) (3.4546%) = 6.909%, and ia = (1 + 0.034546)2 – 1 = 0.07029 or 7.029%.

$875

……….

n = 2(2001 – 1998) + 1 = 27

A = $30

$1,000


16

7-32 (a) NPW = 0 = -3118 + 10000 (P/F, i, 20), so, (P/F, i, 20) = 0.3118. Next you can

solve 20(1 ) 0.3118i for i or look in the tables to find i = 0.06 or 6.0%. Next, because it is paid annually, the effective annual interest rate is 6.0%. (b) The fee is $10,000 x 0.01 = $100. So ABC Corp. receives $3,118 - $100 =

$3,018. NPW = 0 = 3018 – 10000 (P/F, i, 20), so, (P/F, i, 20) = 0.3018. Next solve

20(1 ) 0.3018i and find i = 0.06173 or 6.173%. As above ia = 6.173%.

7-33

$2,300 = $110 (P/A, i%, 24) (P/A, i%, 24) = $2,300/$110 = 20.91 From tables: 1 % < i < 1.25% On Financial Calculator: i = 1.13% per month Effective interest rate = (1 + 0.0113)12 − 1 = 0.144 = 14.4%

$3,500 - $1,200 = $2,300

……….

n = 24

A = $110


17

7-34

PW of Cost = PW of Benefits $100 (P/A, i%, 36) = $3,168 (P/A, i%, 36) = $3,168/$100 = 31.68 Performing Linear Interpolation:

(P/A, 1%, 36) i 32.871 ½% 21.447 ¾%

i* = (1/2%) + (1/4%) [(32.87 − 31.68)/(32.87 − 31.45)] = 0.71% Nominal Interest Rate = 12 (0.71%) = 8.5%

7-35

Set PW of Cost = PW of Benefits $1,845 = $50 (P/A, i%, 4) + $2,242 (P/F, i%, 4) Try i = 7% 450 (3.387) + $2,242 (0.7629) = $1,879 > $1,845 Try i = 8%

$3,168

……….

n = 36

A = $100

P = $1,845

n = 4

A = $50

F = $2,242


18

450 (3.312) + $2,242 (0.7350) = $1,813 < $1,845 Rate of Return = 7% + (1%) [($1,879 − $1,845)/($1,879 − $1,813)] = 7.52% for 6 months Nominal annual rate of return = 2 (7.52%) = 15.0% Equivalent annual rate of return = (1 + 0.0752)2 − 1 = 15.6%

7-36

(a) The monthly payments are 6000

$166.6736

(over 3 years).

NPW = 0 = 6000 -250 – 166.67 (P/A, i, 36), so, (P/A, i, 36) = 34.50 . The tables don’t go to a low enough interest rate so must solve:

36

36

(1 ) 134.50

(1 )

i

i i

by trial and error or Excel using the IRR function. Excel

yields i = 0.00232, so, ia = (1 + 0.00232)12 – 1 = 0.0282 or 2.82%. (b) The fact that the dealer would accept $5,200 cash for the car indicates its true

worth so the extra $800 is a hidden finance charge. Your payments are still based on the original $6,000 cost but you only receive a car worth only $5,200!

NPW = 0 = 5200 -250 – 166.67 (P/A, i, 36), so, (P/A, i, 36) = 29.70 and

interpolating

i = 1% + (0.25%) 30.107 29.70

30.107 28.847

= 1.081% (exact value = 1.079%), so,

ia = (1 + 0.01081)12 – 1 = 13.77% (exact value = 13.75%).


19

7-37 (a) The foregone cash rebate is like a hidden finance charge. You pay $12,000 for

the car but receive a car only worth $12,000 - $3,000 = $9,000. The monthly

payments = 12000

4 12 = $250 for 48 months. NPW = 0 = 9000 – 250 (P/A, i, 48),

so, (P/A, i, 48) = 36.0 and interpolating

i = 1% + (0.25%) 37.974 36.0

37.974 35.932

= 1.242%, so, r = (12) (1.242%) = 14.90%

and ia = (1 + 0.01242)12 -1 = 0.15965 or 15.97%. (b) Worth of car = Cost – Rebate = $18,000 - $3,000 = $15,000.

The monthly payments = 18000

4 12 = $375 for 48 months.

NPW = 0 = 15000 – 375 (P/A, i, 48), so, (P/A, i, 48) = 40.0 and interpolating

i = 0.75% + (0.25%) 40.185 40.0

40.185 37.974

= 0.771%, so, r = (12)(0.771%) = 9.65%

and ia = (1 + 0.00771)12 -1 = 0.0965 or 9.65%. (c) Worth of car = Cost – Rebate = $24,000 - $3,000 = $21,000.

The monthly payments = 24000

4 12 = $500 for 48 months.

NPW = 0 = 21000 – 500 (P/A, i, 48), so, (P/A, i, 48) = 42.0 and interpolating

i = 0.50% + (0.25%) 42.580 42.0

42.580 40.185

= 0.561%, so, r = (12)(0.561%) = 6.73%

and ia = (1 + 0.00561)12 -1 = 0.0694 or 6.94%.

7-38

First determine the monthly payments for the loan where i = 4%

12 = 0.3333%, so,

A = 6000 (A/P, 0.3333%, 36) = 36

36

(0.003333)(1 0.003333)

(1 0.003333) 1

= $177.14 .

(a) NPW = 0 = 6000 – 250 – 177.14 (P/A, i, 36), so, (P/A, i, 36) = 32.46 and interpolating

i = 0.50% + (0.25%) 32.871 32.46

32.871 31.447

= 0.572%, so,

r = (12) (0.572%) = 6.86% and ia = (1 + 0.00572)12 −1 = 0.0709 or 7.09%.


20

(b) Worth of the car = $6,000 − $800 = $5,200 but the payments are determined by the actual cost to buyer, here $6,000. Thus, the payments are the same as above.

NPW = 0 = 5200 – 250 – 177.14 (P/A, i, 36), so, (P/A, i, 36) = 27.944 and

interpolating i = 1.25% + (0.25%) 28.847 27.944

28.847 27.661

= 1.440%, so,

r = (12) (1.440%) = 17.28% and ia = (1 + 0.01440)12 −1 = 0.1872 or 18.72%. (c) The actual value of the car seems to be the most important factor!

7-39 The amount of cash paid will be $75,000 – $50,000 = $25,000 with $50,000

financed, so, the monthly payments will be 50000 (A/P, 8%, 4) = (50000) (0.3019) = $15,095. The reduction in cost if one pays entirely in cash is $75,000 x 0.10 = $7,500, so, a 100% cash payment would be $75,000 − $7,500 = $67,500 (true value of equipment).

Year

Pay Cash

Borrow from Manufacturer

IncrementalDifference

0 -$67,500 −$25,000 −$42,500 1 −15,095 15,095 2 −15,095 15,095 3 −15,095 15,095 4 −15,095 15,095

IRR = IRR (the (1) – (2) values for the Periods 0–4) = 15.69% per year

7-40 The loan value is $120,000 −$12,000 (10% down payment) = $108,000. The loan

origination fee is $108,000 x 0.02 = $2,160, so, the loan becomes $108,000 + $2,160 = $110,160.

(a) Number of months is 30 x 12 = 360. The monthly interest rate, i = 6%

12 = 0.5%.

The monthly payment = 110160 (A/P, 0.5%, 360) but to get accuracy use

110160 110160

$660.46( / ,0.5%,360) 166.792P A

.


21

(b) The actual value received is $108,000, thus, to find the effective interest rate solve NPW = 0 = 108,000 – 660.46 (P/A, i, 360).

(P/A, i, 360) = 108000

660.46 = 163.522. Interpolating

imo = ½ % + (¼ %)[(163.522 – 166.792)/(124.282 – 166.792] = 0.51923% per month ia = (1 + 0.0051923)12 – 1 = 0.0641 or 6.41% (c) In ten years there are still 20 years left on the original loan, so, value of remaining loan at year ten = 660.46 (P/A, 0.5%, 240) = (660.46)(139.581) = $92,187.67 . To find the effective interest rate solve NPW = 108,000 – 660.46 (P/A, i, 120) – 92,187.67 (P/F, i, 120) . Interpolating imo = ½ % + (¼ %)[2156.62/(2156.62 + 18258.62] = 0.5264% (exact value 0.5236%) ia = (1 + 0.005264)12 – 1 = 0.0650 or 6.50% (exact value 6.467%)

7-41 $2,000 = $91.05 (P/A, i*, 30) (P/A, i*, 30) = $2,000/$91.05 = 21.966

(P/A, 1%, 30) I 22.396 2 20.930 2½

imo = 2% + (½%) [(22.396 − 21.966)/(22.396 − 20.930)] = 2.15% per month Nominal ROR received by finance company = 12 (2.15%) = 25.8%


22

7-42 $3,000 = $118.90 (P/A, i*, 36) (P/A, i*, 36) = $3,000/$118.90 = 26.771

(P/A, i%, 36) i 27.661 1 ½% 26.543 1 ¾%

imo = 1 ½% + ¼% [(27.661 − 26.771)/(27.661 − 26.543)] = 1.699% per month Nominal Annual ROR = 12 (1.699%) = 20.4%

7-43

PW of Benefits – PW of Cost = $0 $15,000 (P/F, i%, 4) − $9,000 − $80 (P/A, i%, 4) = $0 Try i = 12% $15,000 (0.6355) − $9,000 − $80 (3.037) = +$289.54 Try i = 15% $15,000 (0.5718) − $9,000 − $80 (2.855) = −$651.40 Performing Linear Interpolation: i* = 12% + (3%) [289.54/(289.54 + 651.40)] = 12.92%

$9,000

A = $80

$15,000


23

7-44

$240,000 = $65,000 (P/A, i%, 13) − $5,000 (P/G, i%, 13) Try i = 15% $65,000 (5.583) −$5,000 (23.135) = $247,220 > $240,000 Try i = 18% $65,000 (4.910) −$5,000 (18.877) = $224,465 < $240,000 Rate of Return = 15% + 3% [($247,220 − $240,000)/($247,220 − $224,765)] = 15.96%

7-45 (a) Total Annual Revenues = $500 (12 months) (4 apt.) = $24,000 Annual Revenues – Expenses = $24,000 − $8,000 = $16,000 To find Internal Rate of Return the Net Present Worth must be $0. NPW = $16,000 (P/A, i*, 5) + $160,000 (P/F, i*, 5) − $140,000 At i = 12%, NPW = $8,464 At i = 15%, NPW = −$6,816 IRR = 12% + (3%) [$8,464/($8,464 + $6,816)] = 13.7% (b) At 13.7% the apartment building is more attractive than the other options.

$65,000

$240,000

$5,000


24

7-46 NPW = −$300,000 + $20,000 (P/F, i*, 10) + ($67,000 − $3,000) (P/A, i*, 10) − $600 (P/G, i*, 10) Try i = 10% NPW = −$300,000 + $20,000 (0.3855) + ($64,000) (6.145) − $600 (22.891) = $87,255 > $0 The interest rate is too low. Try i = 18% NPW = −$300,000 + $20,000 (0.1911) + ($64,000) (4.494) − $600 (14.352) = −$17,173 < $0 The interest rate is too high. Try i =15% NPW = −$300,000 + $20,000 (0.2472) + ($64,000) (5.019) − $600 (16.979) = $9,130 > $0 Thus, the rate of return (IRR) is between 15% and 18%. By linear interpolation: i* = 15% + (3%) [$9,130/($9,130 − $17,173)] = 16.0%


25

7-47

The payment schedule represents a geometric gradient. There are two possibilities: i ≠ g and i = g Try the easier i = g computation first: P = A1n (1 + i)−1 where g = i = 0.10 $20,000 = $1,100 (20) (1.10)−1 = $20,000 Rate of Return i* = g = 10%

7-48 (a) When n = ∞, i = A/P = $3,180/$100,000 = 3.18% (b) (A/P, i%, 100) = $3180/$100,000 = 0.318 From interest tables, i* = 3% (c) (A/P, i%, 50) = $3, 180/$100,000 = 0.318 From interest tables, i* = 2% The saving in water truck expense is just a small part of the benefits of the pipeline.

Convenience, improved quality of life, increased value of the dwellings, etc., all are benefits. Thus, the pipeline appears justified.

P = $20,000

A1 = $1,100

n = 20 i = ?

g = 10%


26

7-49

Year Cash Flow 0 −$9,000 1−4 +$800 5−8 +$400 9 +$6,000

PW of Cost = PW of Benefits $9,000 = $400 (P/A, i%, 8) + $400 (P/A, i%, 4) + $6,000 (P/F, i%, 9) Try i = 3% $400 (7.020) + $400 (3.717) + $6,000 (0.7664) = $8,893 < $9,000 Try i = 2 ½% $400 (7.170) + $400 (3.762) + $6,000 (0.8007) = $9,177 > $9,000 Rate of Return = 2 ½% + (1/2%) [($9,177 − $9,000)/($9,177 − $8,893)] = 2.81%

$9,000

$800

$400 $6,000


27

7-50

PW of Cost = PW of Benefits $28,000 = $3,000 (P/A, i%, 10) + $6,000 (P/A, i%, 10) (P/F, i%, 10) + $12,000 (P/A, i%, 20) (P/F, i%, 20) Try i = 12% $3,000 (5.650) + $6,000 (5.650) (0.3220) + $12,000 (7.469) (0.1037) = $37,160 > $28,000 Try i = 15% $3,000 (5.019) + $6,000 (5.019) (0.2472) + $12,000 (6.259) (0.0611) = $27,090 < $28,000 Performing Linear Interpolation: i* = 15% − (3%) [($28,000 − $27,090)/($37,160 − $27,090)] = 15% − (3%) (910/10,070) = 14.73%

$3,000

$6,000 $12,000

n = 10 n = 10 n = 20

$28,000


28

7-51 This is a thought-provoking problem for which there is no single answer. Two

possible solutions are provided below. (a) Assuming the MS degree is obtained by attending graduate school at night while

continuing with a full-time job:

Cost: $1,500 per year for 2 years Benefit: $3,000 per year for 10 years Computation as of award of MS degree: $1,500 (F/A, i%, 2) = $3,000 (P/A, i%, 10) i* > 60 (b) Assuming the MS degree is obtained by one of year of full-time study Cost: Difference between working & going to school. Whether working or at

school there are living expenses. The cost of the degree might be $24,000 Benefit: $3,000 per year for 10 years $24,000 = $3,000 (P/A, i%, 10) i* = 4.3%

MS Degree

$1,500 $1,500

A = $3,000

n = 10


29

7-52 The problem requires an estimate for n- the expected life of the infant. Seventy or

seventy-five years might be the range of reasonable estimates. Here we will use 71 years.

The purchase of a $200 life subscription avoids the series of beginning-of-year

payments of $12.90. Based on 71 beginning-of-year payments,

$200 - $12.90 = $12.90 (P/A, i%, 70) (P/A, i%, 70) = $187.10/$ 12.90 = 14.50 6% < i* < 8%. By Calculator: i* = 6.83%

7-53

Year Case 1 (incl. Deposit) 0 −$39,264.00 1 +$599.00 2 +$599.00 3 +$599.00 4 +$599.00 5 +$599.00 6 +$599.00 7 +$599.00 8 +$599.00 9 +$599.00 10 +$599.00 11 +$599.00 12 +$599.00 … +$599.00 33 +$599.00 34 +$599.00 35 +$599.00 36 +$27,854.00 −$625.00 = +$27,229.00

IRR = 0.86% Nominal IRR = 10.32% Effective IRR =10.83%

$200

A = $12.90

n = 70

………….


30

7-54 The number of months between August 15 and January 15 is 5.

Month

Annual Permit

Semester Permit

0 –$100 –$65 1 0 0 2 0 0 3 0 0 4 0 0 5 0 –65

To solve for the monthly interest rate set the two PWs equal to each other, so,

–100 = –65 – 65 (P/F, i, 5) . Thus, 5 100 65(1 ) 0.53846

65i

.

Solving get i = 0.1318 or 13.18% and ia = (1 + 0.1318)12 – 1 = 3.418 or 342%. Unless the student is graduating in January or just doesn’t have the $100, it is clearly better to buy the permit a year at a time.

7-55 Details will vary by university, but is solved like Problem 7-54.

7-56

Quarter

Annual Payment

Quarter Payment

0 –$65,000 –$18,000 1 0 –18,000 2 0 –18,000 3 0 –18,000

To solve for the monthly interest rate set the two PWs equal to each other, so, –65000 = –18000 – 18000 (P/A, i, 3) . Thus, (P/A, i, 3) = 2.611 and interpolating

i = 7% + (1%) 2.624 2.611

2.624 2.577

= 7.28%, so, r = 4 x 0.0728 = 0.2912 or 29.1% and

ia = (1 + 0.0728)4 – 1 = 0.3246 or 32.5%. This is a high rate of return, but some firms

use an even higher hurdle rate for projects.


31

7-57 –$65,000 = –$18,000( 1 + (P/A, i , 3)) The amount that the series of future payments is worth is: –65000 + 18000 = –47000 = –18000*(P/A, i , 3) Using the end-of-period designation (default) in RATE (Excel) yields: RATE(3,18000,-47000) = 7.2766% One could also solve with quarterly payments at the beginning of the period: RATE(4,18000,-65000,0,1) = 7.2766%

7-58 Insurance payments must be paid in advance, here on the first of the month or year.

Month

Annual Basis

Monthly Basis

0 –$1,650 –$150 1 0 –150 2 0 –150 3 0 –150 4 0 –150 5 0 –150 6 0 –150 7 0 –150 8 0 –150 9 0 –150 10 0 –150 11 0 –150

To solve for the monthly interest rate set the PWs of the two cash flows equal to

each other. Thus, −1650 = −150 – 150 (P/A, i, 11), so, (P/A, i, 11) = 10.0. Interpolating

i = 1.5% + (0.25%) 10.071 10.0

10.071 9.928

= 1.624%. Next, ia = (1 + 0.01624)12 −1 = 0.2133

or 21.3%. This is a relatively high rate of return, but the student might prefer to pay monthly if there is a significant chance of wrecking the car before the year is up.

7-59 Details will vary by student, but solved like Problem 7-58.


32

7-60

Year A B (B- A) 0 −$2,000 −$2,800 −$800 1− 3 +$800 +$1,100 +$300 Computed ROR 9.7% 8.7% 6.1%

The rate of return on the increment (B- A) exceeds the Minimum Attractive Rate of

Return (MARR), therefore the higher cost alternative B should be selected.

7-61

Year X Y X- Y 0 −$100 −$50 −$50 1 +$35 +$16.5 +$18.5 2 +$35 +$16.5 +$18.5 3 +$35 +$16.5 +$18.5 4 +$35 +$16.5 +$18.5 Computed ROR 15.0% 12.1% 17.8%

The ΔROR on X- Y is greater than 10%. Therefore, the increment is desirable.

Select X.

7-62

Year A B (B- A) 0 −$100.00 −$50.00 −$50.00 1- 10 +$19.93 +$11.93 +$8.00 Computed ROR 15% 20% 9.61%

ΔROR = 9.61% > MARR. Select A.


33

7-63

Year X Y X- Y 0 −$5,000 −$5,000 $0 1 −$3,000 +$2,000 −$5,000 2 +$4,000 +$2,000 +$2,000 3 +$4,000 +$2,000 +$2,000 4 +$4,000 +$2,000 +$2,000 Computed ROR 16.9% 21.9% 9.7%

Since X- Y difference between alternatives is desirable, select Alternative X.

7-64 (a) Present Worth Analysis- Maximize NPW NPWA = $746 (P/A, 8%, 5) − $2,500 = $746 (3.993) − $2,500 = +$479 NPWB = $1,664 (P/A, 8%, 5) − $6,000 = +$644 Select B. (b) Annual Cash Flow Analysis- Maximize (EUAB- EUAC) (EUAB- EAUC)A = $746 − $2,500 (A/P, 8%, 5) = $746 − $2,500 (0.2505) = +$120 (EUAB − EUAC)B = $1,664 − $6,000 (A/P, 8%, 5) = +$161 Select B. (c) Rate of Return Analysis: Compute the rate of return on the B- A increment of

investment and compare to 8% MARR.

Year A B B- A 0 −$2,500 −$6,000 −$3,5001−5 +$746 +$1,664 +$918

$3,500 =$918 (P/A, i%, 5) Try i = 8%, $918 (3.993) = $3,666 > $3,500 Try i = 10%, $918 (3.791) = $3,480 < $3,500 Δ Rate of Return = 9.8% Since ΔROR > MARR, B- A increment is desirable. Select B.


34

7-65

Year 0 1 2 3 4 5 6 7 8 Alt. A -12000 1200 Alt. B -3000 -3000 -3000 -3000 -3000 -3000 -3000 -3000 A-B -9000 3000 3000 3000 3000 3000 3000 3000 1200

IRR of A B stream = IRR (the A – B values for the Years 0–8) = 27.90% Since ΔROR > MARR (15%), choose the higher initial cost alternative, A (purchasing the equipment).

7-66

B A A- B First Cost $300,000 $615,000 $315,000 Maintenance & Operating Costs

$25,000 $10,000 −$15,000

Annual Benefit $92,000 $158,000 $66,000 Salvage Value −$5,000 $65,000 $70,000

NPW = −$315,000 + [$66,000 − (−$15,000)] (P/A, i*, 10) + $70,000 (P/F, i*, 10) = $0 Try i = 15% −$315,000 + [$66,000 − (−$15,000)] (5.019) + $70,000 (0.2472) = $108,840 ΔROR > MARR (15%) The higher cost alternative A is the more desirable alternative.


35

7-67

Year (A) Gas Station (B) Ice Cream Stand

(B- A)

0 −$80,000 −$120,000 −$40,000 1−20 +$8,000 +$11,000 +$3,000 Computed ROR 7.75% 6.63% 4.22%

The rate of return in the incremental investment (B- A) is less than the desired 6%.

In this situation the lower cost alternative (A) Gas Station should be selected.

7-68 MARR = 5% P = $30,000 n = 35 years Alternative 1: Withdraw $15,000 today and lose $15,000 Alternative 2: Wait, leave your fund in the system until retirement. Equivalency seeks to determine what future amount is equal to $15,000 now. F = P (1 + i)n = $30,000 (1.05)35 = $30,000 (5.516015) = $165,480.46 Therefore: $15,000 = $165,480.46 (1 + i)−35 $15,000 (1 + i)35 = $165,480.46 (1 + i) = [(165,480.46/$15,000)]1/35 i = 1.071 − 1 = 7.1002% > 5% Unless $15,000 can be invested with a return higher than 7.1%, it is better to wait for

35 years for the retirement fund. $15,000 now is only equivalent to $165,480.46 35 years from now if the interest rate now is 7.1% instead of the quoted 5%.


36

7-69 (a)

($2,000 − $150) = $100 (P/A, i%, 20) (P/A, i%, 20) = $1,850/$100 = 18.5 I = ¾% per month The alternatives are equivalent at a nominal 9% annual interest. (b) Take Alt 1- the $2,000- and invest the money at a higher interest rate.

7-70 (a) Salvage = 0.15 x $380,000 = $57,000 and firm’s interest rate = 12%.

Year Purchase Lease Purchase – Lease

0 –$380,000 –$60,000 –$320,000 1 0 –60,000 60,000 2 0 –60,000 60,000 3 0 –60,000 60,000 4 0 –60,000 60,000 5 0 –60,000 60,000 6 57,000 0 57,000

NPW = 0 = –320000 + 60000 (P/A, IRR , 5) + 57000 (P/A, IRR, 6) and

interpolating

IRR = 3% + (0.5%) 2538

2538 2730

= 3.24% (also 3.24% from Excel). The IRR is

well below the firm’s interest rate on the borrowed amount ($320,000) from leasing, so lease the bulldozer.

$2,000

……….

n = 20

A = $100 $150


37

(b) The firm receives $65,000 more than it spends on operating and maintenance costs.

Year Purchase Lease Purchase –

Lease 0 –$380,000 –$60,000 –$320,000 1 65,000 –60,000 125,000 2 65,000 –60,000 125,000 3 65,000 –60,000 125,000 4 65,000 –60,000 125,000 5 65,000 –60,000 125,000 6 65,000

57,000 0 122,000

NPW = 0 = -320000 + 125000 (P/A, IRR, 5) + 122000 (P/F, IRR, 6) and

interpolating

IRR = 30% + (5%) 9778

9778 22346

= 31.5% (31.42% from Excel). Clearly, the

situation has changed. The interest rate on the borrowed amount is now well above the firm’s interest rate, so, buy the bulldozer. The rate of return for the bulldozer will clearly be largest for this cash flow and is given by

PW = 0 = −380000 + 65000 (P/A, ROR, 6) + 57000 (P/F, ROR, 6) and

interpolating

ROR = 4% + (0.5%) 5777

5777 960

= 4.43% (4.43% from Excel).

Note that the author has failed to give a practical scenario for how the $65,000 benefit can be realized if the bulldozer is purchased instead of leased!


38

7-71 (a) Salvage = $50,000 and community’s interest rate = 8%.


0 –$480,000 –$70,000 –$410,000 1 0 –70,000 70,000 2 0 –70,000 70,000 3 0 –70,000 70,000 4 0 –70,000 70,000 5 0 –70,000 70,000 6 0 –70,000 70,000 7 0 –70,000 70,000 8 0 –70,000 70,000 9 0 –70,000 70,000 10 50,000 0 50,000

NPW = 0 = −410000 +70000 (P/A, IRR , 9) + 50000 (P/A, IRR, 10) and

interpolating

IRR = 10% + (2%) 12405

12405 20940

= 10.74% (10.71% Excel). The IRR is above

the community’s interest rate on the borrowed amount ($410,000) from leasing, so buy the generator.

(b) The community spends $80,000 less on fuel and maintenance than it spends on

buying power.


0 –$480,000 –$70,000 –$410,000 1 80,000 –70,000 150,000 2 80,000 –70,000 150,000 3 80,000 –70,000 150,000 4 80,000 –70,000 150,000 5 80,000 –70,000 150,000 6 80,000 –70,000 150,000 7 80,000 –70,000 150,000 8 80,000 –70,000 150,000 9 80,000 –70,000 150,000 10 80,000

50,000 0 130,000


39

NPW = 0 = −410000 + 150000 (P/A, IRR, 9) + 130000 (P/F, IRR, 10) and interpolating

IRR = 30% + (5%) 52275

52275 3789

= 34.66% (34.63% from Excel). The interest

rate on the borrowed amount is now well above the firm’s interest rate, so buy the generator. The rate of return for the generator will clearly be largest for this cash flow and is given by

PW = 0 = −480000 + 80000 (P/A, ROR, 10) + 50000 (P/F, ROR, 10) and

interpolating

ROR = 10% + (2%) 30875

30875 11900

= 11.44% (11.42% from Excel).

Note that the author has failed to give a practical scenario for how the $80,000 benefit can be realized if the generator is purchased instead of leased!

7-72

Year A B A- B NPW at 7% NPW at 9% 0 −$9,200 −$5,000 −$4,200 −$4,200 −$4,200 1 +$1,850 +$1,750 +$100 +$93 +$92 2 +$1,850 +$1,750 +$100 +$87 +$84 3 +$1,850 +$1,750 +$100 +$82 +$77 4 +$1,850 +$1,750

−$5,000 +$5,100 +$3,891 +$3,613

5 +$1,850 +$1,750 +$100 +$71 +$65 6 +$1,850 +$1,750 +$100 +$67 +$60 7 +$1,850 +$1,750 +$100 +$62 +$55 8 +$1,850 +$1,750 +$100 +$58 +$50 Sum +$211 −$104

Δ ROR ≈ 8.3% Choose Alternative A.


40

7-73

Year Zappo Kicko Kicko − Zappo0 −$56 −$90 −$34 1 −$56 $0 +$56 2 $0 $0 $0

Compute the incremental rate of return on (Kicko − Zappo) PW of Cost = PW of Benefit $34 = $56 (P/F, i%, 1) (P/F, i%, 1) = $34/$56 = 0.6071 From interest tables, incremental rate of return > 60% (ΔROR = 64.7%), hence the

increment of investment is desirable. Buy Kicko.

7-74

Year A B A- B 0 −$9,200 −$5,000 −$4,200 1 +$1,850 +$1,750 +$100 2 +$1,850 +$1,750 +$100 3 +$1,850 +$1,750 +$100 4 +$1,850 +$1,750 −$5,000 +$100 +$5,0005 +$1,850 +$1,750 +$100 6 +$1,850 +$1,750 +$100 7 +$1,850 +$1,750 +$100 8 +$1,850 +$1,750 +$100 Sum

Rates of Return A: $9,200 = $1,850 (P/A, i%, 5) Rate of Return = 11.7% B: $5,000 = $1,750 (P/A, i%, 4) Rate of Return = 15% A−B: $4,200 = $100 (P/A, i%, 8) + $5,000 (P/F, i%, 4) ΔRORA-B = 8.3% Select A.


41

7-75

Year A B A- B 0 −$150 −$100 −$50 1- 10 +$25 +$22.25 +$2.7511- 15 +$25 $0 +$25 15 +$20 $0 +$20 Computed ROR 14.8% 18% 11.6%

Rate of Return (A- B): $50 = $2.75 (P/A, i%, 10) + $25 (P/A, i%, 5) (P/F, i%, 10) + $20 (P/F, i%, 15) Rate of Return = 11.65 Select A.

7-76 This is an unusual problem with an extremely high rate of return. Available interest

tables obviously are useless. One may write: PW of Cost = PW of Benefits $0.5 = $3.5 (1 + i)−1 + $0.9 (1 + i)−2 + $3.9 (1 + i)−3 + $8.6 (1 + i)−4 + … For high interest rates only the first few terms of the series are significant: Try i = 650% PW of Benefits = $3.5/(1 + 6.5) + $0.9/(1 + 6.5)2 + $3.9/(1 + 6.5)3 + $8.6/(1 + 6.5)4 +

… = 0.467 + 0.016 + 0.009 + 0.003 = 0.495 Try i = 640% PW of Benefits = $3.5/(1 + 6.4) + $0.9/(1 + 6.4)2 + $3.9/(1 + 6.4)3 + $8.6/(1 + 6.4)4 +

… = 0.473 + 0.016 + 0.010 + 0.003 = 0.502 i* = 642% (Calculator Solution: i = 642.9%)


42

7-77

$52,000.00 income 3.00% income gradient 10.00% % deposit 15 horizon (years) 4.00% savings rate

Year Salary Deposit Cumulative Savings

1 $52,000.00 $5,200.00 $5,200.00 2 53,560.00 5,356.00 10,764.00 3 55,166.80 5,516.68 16,711.24 4 56,821.80 5,682.18 23,061.87 5 58,526.46 5,852.65 29,836.99 6 60,282.25 6,028.23 37,058.70 7 62,090.72 6,209.07 44,750.12 8 63,953.44 6,395.34 52,935.46 9 65,872.04 6,587.20 61,640.09 10 67,848.21 6,784.82 70,890.51 11 69,883.65 6,988.37 80,714.50 12 71,980.16 7,198.02 91,141.09 13 74,139.57 7,413.96 102,200.6914 76,363.75 7,636.38 113,925.1015 78,654.67 7,865.47 126,347.57

For any row: Salary = (1 + 0.03)*(Previous year’s Salary) Deposit = (Percent Deposit)*(Current year’s Salary) Savings = (1 + 0.04)*(Previous year’s Savings) + Current year’s Deposit Amount saved is $126,347.57 in 15 years.

7-78



43


1 $55,000.00 $5,500.00 $5,500.00 2 56,100.00 5,610.00 11,385.00 3 57,222.00 5,722.20 17,676.45 4 58,366.44 5,836.64 24,396.92 5 59,533.77 5,953.38 31,570.14 6 60,724.44 6,072.44 39,221.09 7 61,938.93 6,193.89 47,376.04 8 63,177.71 6,317.77 56,062.61 9 64,441.27 6,444.13 65,309.87 10 65,730.09 6,573.01 75,148.37 11 67,044.69 6,704.47 85,610.26 12 68,385.59 6,838.56 96,729.33 13 69,753.30 6,975.33 108,541.1314 71,148.36 7,114.84 121,083.0215 72,571.33 7,257.13 134,394.3016 74,022.76 7,402.28 148,516.3017 75,503.21 7,550.32 163,492.4318 77,013.28 7,701.33 179,368.3819 78,553.54 7,855.35 196,192.1520 80,124.61 8,012.46 214,014.2221 81,727.11 8,172.71 232,887.6522 83,361.65 8,336.16 252,868.1923 85,028.88 8,502.89 274,014.4924 86,729.46 8,672.95 296,388.1625 88,464.05 8,846.40 320,053.9726 90,233.33 9,023.33 345,080.0127 92,038.00 9,203.80 371,537.8128 93,878.76 9,387.88 399,502.5729 95,756.33 9,575.63 429,053.3330 97,671.46 9,767.15 460,273.1531 99,624.89 9,962.49 493,249.2932 101,617.38 10,161.74 528,073.4933 103,649.73 10,364.97 564,842.1434 105,722.73 10,572.27 603,656.5235 107,837.18 10,783.72 644,623.0736 109,993.93 10,999.39 687,853.6137 112,193.80 11,219.38 733,465.6738 114,437.68 11,443.77 781,582.7239 116,726.43 11,672.64 832,334.5040 119,060.96 11,906.10 885,857.33

For any row: Salary = (1 + 0.02)*(Previous year’s Salary) Deposit = (Percent Deposit)*(Current year’s Salary)


44

Savings = (1 + 0.05)*(Previous year’s Savings) + Current year’s Deposit Amount saved is $885,857.33 in 40 years.

7-79



1 $55,000.00 $6,209.50 $6,209.50 2 56,100.00 6,333.69 12,853.67 3 57,222.00 6,460.36 19,956.71 4 58,366.44 6,589.57 27,544.12 5 59,533.77 6,721.36 35,642.69 6 60,724.44 6,855.79 44,280.61 7 61,938.93 6,992.91 53,487.55 8 63,177.71 7,132.76 63,294.69 9 64,441.27 7,275.42 73,734.84 10 65,730.09 7,420.93 84,842.51 11 67,044.69 7,569.35 96,653.98 12 68,385.59 7,720.73 109,207.41 13 69,753.30 7,875.15 122,542.93 14 71,148.36 8,032.65 136,702.73 15 72,571.33 8,193.30 151,731.17 16 74,022.76 8,357.17 167,674.90 17 75,503.21 8,524.31 184,582.96 18 77,013.28 8,694.80 202,506.90 19 78,553.54 8,868.70 221,500.94 20 80,124.61 9,046.07 241,622.06 21 81,727.11 9,226.99 262,930.15 22 83,361.65 9,411.53 285,488.19 23 85,028.88 9,599.76 309,362.36 24 86,729.46 9,791.76 334,622.23 25 88,464.05 9,987.59 361,340.94 26 90,233.33 10,187.34 389,595.33 27 92,038.00 10,391.09 419,466.18 28 93,878.76 10,598.91 451,038.40 29 95,756.33 10,810.89 484,401.21 30 97,671.46 11,027.11 519,648.38 31 99,624.89 11,247.65 556,878.45


45

32 101,617.38 11,472.60 596,194.97 33 103,649.73 11,702.05 637,706.78 34 105,722.73 11,936.10 681,528.21 35 107,837.18 12,174.82 727,779.44 36 109,993.93 12,418.31 776,586.73 37 112,193.80 12,666.68 828,082.74 38 114,437.68 12,920.01 882,406.90 39 116,726.43 13,178.41 939,705.66 40 119,060.96 13,441.98 1,000,132.92

For any row: Salary = (1 + 0.02)*(Previous year’s salary) Deposit = (Percent deposit)*(Current year’s salary) Savings = (1 + 0.05)*(Previous year’s savings) + Current year’s deposit To solve, just vary the percent deposit to get $1M in savings for year 40. Amount saved is $1,000,132.92 in 40 years at 11.29%.

7-80 Details will vary by student, but solved like Problem 7-79.


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Chapter 7 - Engineering Economic Analysis, 2nd Canadian Edition

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Transcript of Chapter 7 - Engineering Economic Analysis, 2nd Canadian Edition