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SOLUTION MANUAL

SOLUTIONS MANUAL for Fundamentals of Engineering Economics, 4th Edition by Chan Park Chapter 1 Engineering Economic Decisions 1.1)

Not provided For The Wall Street Journal, go to the Front page to find the section on “What’s News.” This is a section on a brief summary on major headlines of the day’s news. Quickly browse through the news summary to see if there is any news related to business investment. The best places to find the major business news on investment are sections on “BUSINESS”, “MARKETS” or “TECH.”

1.2) Not provided Some of the well-known business publications are: •

Daily Newspapers: o The Wall Street Journal o The New York Times (Business Section) o Financial Times

•

Weekly or Monthly Magazines: o o o o o

BusinessWeek Forbes Money Smart Money Fortune

Chapter 2: Time Value of Money 2.1)= I (iP = ) N (0.06)($2, 000)(5) = $600 2.2)

•

Simple interest: $20,000 = $10,000(1 + 0.075N ) (1 + 0.075N ) = 2 N=

•

1 = 13.33≈ 14 years 0.075

Compound interest: $20,000 = $10,000(1 + 0.07) N (1 + 0.07) N = 2 N = 10.24 ≈ 11years

2.3) •

Simple interest: = I iPN = (0.10)($10, 000)(5) = $5, 000

•

Compound interest: I= P[(1 + i ) N − 1]= $10, 000(1.6105 − 1)= $6,105

2.4) •

Option 1: Compound interest with 8.5%: F = $4,500(1 + 0.085)5 = $4,500(1.5037) = $6, 766.65

•

Option 2: Simple interest with 9%: $4,500(1 += 0.09 × 5) $5, 000(1.45) = $6,525

∴ Option 1 is still better.

2.5)

•

Compound interest:

= F $1, 000(1 + 0.065)5 = $1,370.09 •

Simple interest:

= F $1, 000(1 + 0.068(5)) = $1,340 The compound interest option is better.

2.6) End of Year 0 1 2 3

Principal Repayment

Interest payment

$4,620.50 $4,990.14 $5,389.35

$1,200.00 $830.36 $431.15

2.7) = P $22, 000( = P / F ,5%,5) $22, = 000(0.7835) $17, 237.58

2.8) = F $30, 000( = F / P,9%,3) $30, = 000(1.295) $38,850.87 2.9)

F = $100( F / P,10%,10) + $200( F / P,10%,8) = $688

2.10) F $250, = = 000( F / P, 6%,10) $447, 712 2.11) P = $300, 000( P / F ,8%,10) = $138,958 2.12)

i = 10.5% , two-year discount rate is (1 + 0.105) 2 = 1.221 (or 22.1%)

2.13) (a) F $5, = = 000( F / P, 7%,5) $7, 013

Remaining Balance $15,000.00 $10,379.50 $5,389.36 $0

(b) F $7, = = 250( F / P,9%,15) $26, 408 (c) F $9, = = 000( F / P, 6%,33) $61,565 (d) F $12, = = 000( F / P,5.5%,8) $18, 416

2.14) (a) P $25,500( = = P / F ,12%,8) $10, 299 (b) P $58, = = 000( P / F , 4%,12) $36, 227 (c) P $25, = = 000( P / F , 6%,9) $14, 797 (d) P $35, = = 000( P / F ,9%, 4) $24, 795 2.15) (a)

P = $12, 000( P / F ,13%, 4) = $7,360

(b)

F = $30, 000( F / P,13%,5) = $55, 273

2.16)

= F 3= P P(1 + 0.08) N log 3 = N log(1.08) = N 14.27 → 15 years 2.17)

= F 2= P P(1 + 0.06) N log 2 = N log 1.06 N = 11.896 years (or 12 years) 2.18) •

= F 2= P P(1 + 0.06) N log 2 = N log(1.06) N = 11.90 years  12 years

•

Rule of 72: 72 / 6 = 12 years

394 2.19) = F $1(1.08) = $14, 755, 694, 730, 611

2.20) = P $35, 000( P / F ,9%, 4) + $10, 000( P / F ,9%, 2) = $35, 000(0.7084) + $10, 000(0.8417) = $33, 211 2.21) = P $450, 000( = P / F ,5%,5) 450, = 000(0.7835) $352,575 2.22)

•

Simple interest (John): = I iPN = (0.1)($1, 000)(5) = $500

•

Compound interest (Susan):

I= P (1 + i ) N − 1= $1, 000 (1 + .095)5 − 1 = $574.24 •

2.23) P=

Susan’s balance will be greater by $74 (or $74.24 to be exact)

$2, 000 $800 $1, 000 + + = $3, 230.65 1.11 1.12 1.13

2.24)

$15, 000 $23, 000 $36, 000 $48, 000 + + + = $93,564 1.07 2 1.073 1.07 4 1.075

2.25) F = $2, 000( F / P, 6%,10) + $2,500( F / P, 6%,8) + $3, 000( F / P,6%,6) = $11,822

2.26) P = $3, 000, 000 + $2, 400, 000( P / F ,8%,1) +  +$3, 000, 000( P / F ,8%,10) = $20, 734, 618

Or,

P = $3, 000, 000 + $2, 400, 000( P / A,8%,5) +$3, 000, 000( P / A,8%,5)( P / F ,8%,5) = $20, 734, 618

2.27) P = $9, 000( P / F ,8%, 2) + $6, 000( P / F ,8%,5) + $3, 000( P / F ,8%,7) = $13,550

2.28) $1,000 +

2.29)

2.30)

$1,000 $1,500 $1,210 X + = + 4 3 2 1.1 1.1 1.1 1.1 X = $2,981

$180, 000 = $20, 000( P / A,9%,5) − $10, 000( P / F ,9%,3) + X ( P / F ,9%, 6)180, 000 − 20, 000(3.8897) + 10, 000(0.7722) = X (0.5963) X = $184,350.16

$60, 000 $77, 000 $65, 000 $57, 000 45, 000 + + + + = $212,873.89 1.14 1.142 1.143 1.144 1.145

2.31) (a)

With deposits made at the end of each year

= F $3, = 000( F / A,9%,15) $88, 083

(b) With deposits made at the beginning of each year = F $3, = 000( F / A,9%,15)(1.09) $96, 010

2.32) (a) F $8, = = 000( F / A,11.75%,5) $50,571 (b) F $2, = = 000( F / A, 4.25%,12) $30, 486 (c) F $7, = = 000( F / A, 6.45%, 20) $270,309

= = 000( F / A, 7.75%,12) $74, 793 (d) F $4,

2.33) (a) A $45, = = 000( A / F ,8%,11) $2, 703 (b) A $35, = = 000( A / F , 6%,18) $1,132 (c) A $25, = = 000( A / F , 7%, 6) $3, 495 = = 000( A / F ,11%,13) $458 (d) A $12,

2.34) A $250,= = 000( A / F ,5%,5) $250, = 000(0.1810) $45, 250 (a) = F $5, 000( = F / A,5%, 7) $5, = 000(8.1420) $40, 710 F = $5, 000( F / A,5%, 7)(1.05) (b) = $5, = 000(8.1420)(1.05) $42, 745.50

2.35)

= A $250, = 000( A / F ,5%,5) $45, 244

2.36) $3, 000( F / A, 6%, N ) = $55, 000 N = 13 years

2.37) = A $15, = 000( A / F ,11%,5) $2, 408.55

2.38)

F =$500(1.04)10 + $1, 000(1.04)8 + $1, 000(1.04)6 +$1, 000(1.04) 4 + $1, 000(1.04) 2 + $1, 000 = $6, 625.47

2.39) (a) A $15, = = 000( A / P,3.5%, 6) $2,815.02 (b) A $7,500( = = A / P, 7.5%, 7) $1, 416

2.40)

•

Equal annual payment amount:

= A $20, 000( = A / P,10%,3) $20, = 000(0.4021) $8, 042

•

Loan balance calculation:

End of period 0 1 2 3

Principal Payment $0.00 $6,042.00 $6,646.20 $7,310.82

Interest Payment $0.00 $2,000.00 $1,395.80 $731.18

Remaining Balance $20,000.00 $13,958.00 $7,311.80 $0

Interest payment for the second year = $1,395.80

2.41) (a) = P $1, = 000( P / A, 7.2%,8) $5,925.29 (b) P $4,500( = = P / A,9.5%,12) $31, 427.28 (c) P $1,900( = = P / A,8.25%,13) $14,812.86 = = P / A, 7.75%,8) $111,970.11 (d) P $19,300(

2.42) (a) The capital recovery factor ( A / P, i, N ) for N 35 40

6% 0.0690 0.0665

7% 0.0772 0.0750

To find ( A / P, 6.25%, 38) , first, interpolate for N = 38 :

N 38

6% 0.0675

7% 0.0759

Then, interpolate for i = 6.25% ; (A / P, 6.25%, 38) = 0.0696 :

As compared to the value from the interest formula: (A / P, 6.25%, 38) = 0.0694

(b) The equal payment series present-worth factor ( P / A, i, 85) for i

9% 11.1038

10% 9.9970

Then, interpolate for i = 9.25% : ( P / A, 9.25%, 85) = 10.8271

As compared to the value from the interest formula: ( P / A, 9.25%, 85) = 10.8049

2.43) = F $500(= F / A, 7%,15)(1.07) $500(25.1290)(1.07) = $13, 444.02

2.44) •

Equal annual payment: A = $50, 000( A / P,12%,3) = $20,817.45

•

Interest payment for the second year: End of Year 0 1 2 3

Principal Repayment

Interest payment

$14,817.45 $16,595.54 $18,587.01

$6,000 $4,221.91 $2,230.44

Remaining Balance $50,000 $35,182.55 $18,587.01 0

2.45) P = $35,000(P / A,12%,10) = $197,758 Since $200,000 > $197,758, You should not purchase the equipment.

2.46)

P= −$3, 460 +

250 = 0  i = 7.225% i

2.47)

= P

1, 000 = $10, 000 0.1

2.48) F= F1 + F2 = $20, 000( F / A, 6%,5) + $5, 000( F / G, 6%,5) = $20, 000( F / A, 6%,5) + $5, 000( A / G, 6%,5)( F / A, 6%,5) = $20, 000(5.6371) + $5, 000(1.8836)(5.6371) = $165,833

2.49) = F $10, 000( F / A,8%,5) − $2, 000( F / G,8%,5) = $10, 000( F / A,8%,5) − $2, 000( P / G,8%,5)( F / P, 8%, 5) = $37, 001

2.50)

P= $100 + [$100( F / A,9%, 7) + $50( F / A,9%, 6) +$50( F / A,9%,4) + $50( F / A, 9%, 2)]( P / F , 9%, 7) = $991.32

2.51) = A $15, 000 − $1, 000( A / G, 8%, 12) = $10, 404.25 2.52)

= P $1, 000( P / A, 6%,5) + $250( P / G, 6%,5) = $6,196

Geometric Gradient Series 2.53)

F = $8, 000( P / A1 ,5%, 7%,30)( F / P, 7%,30) = $172,895.56(7.61226) = $1,316,126 2.54) (a) P = $6,000,000( P / A1 ,−10%,12%,7) = $21,372,076 (b) Note that the oil price increases at the annual rate of 5% while the oil production decreases at the annual rate of 10%. Therefore, the annual revenue can be expressed as follows: An = $60(1 + 0.05) n −1100, 000(1 − 0.1) n −1 = $6, 000, 000(0.945) n −1 = $6, 000, 000(1 − 0.055) n −1

This revenue series is equivalent to a decreasing geometric gradient series with g = -5.5%. So, = P $6, 000, 000( P / A1 , −5.5%,12%,7) = $23,847,897

(c) Computing the present worth of the remaining series ( A4 , A5 , A6 , A7 ) at the end of period 3 gives 3 = A4 6, 000, 000(1 − 0.055) = 5, 063, 451.75

= P $5, 063, 451.75( P / A4 , −5.5%,12%, = 4) $14, 269, 627.82

2.55)

P = ∑ An (1 + i ) − n n =1 20

= ∑ (2, 000, 000)n(1.06) n −1 (1.06) − n n =1

20 1.06 n = (2, 000, 000 /1.06)∑ n( ) 1.06 n =1 = $396, 226, 415

2.56)

(a) The withdrawal series would be Period 11 12 13 14 15

Withdrawal $15,000 $15,000(1.08) $15,000(1.08)(1.08) $15,000(1.08)(1.08)(1.08) $15,000(1.08)(1.08)(1.08)(1.08)

Amount $15,000 $16,200 $17,496 $18,896 $20,407

= P10 $15, = 000( P / A1 ,8%,9%,5) $67,556

Assuming that each deposit is made at the end of each year, then: $67,556 = A( F / A,9%,10) A = $4, 446.54

(b) P10 $15, = = 000( P / A1 ,8%, 6%,5) $73, 476 $73, 476 = A( F / A, 6%,10) A = $5,574.47

2.57) = $1, 000, 000 A= ( F / A, 6%,30) A(79.0582)  A = $12,649 should be set aside on the account a) = $1, 000, 000 A= ( P / A, 6%, 20) A(11.4699)  A = $87,185 / year b)

$1, 000, 000 = A1 ( P / A1, 3%, 6%, 20) 1 − (1.03) (1.06 ) = A1 0.06 − 0.03 = $68, 674 / year 20

−20

Equivalence Calculations 2.58) P= [$100( F / A,12%,9) + $50( F / A,12%, 7) + $50( F / A,12%,5)]( P / F ,12%,10) = $740.49

2.59) P (1.08) = + $200 $200( P / F ,8%,1) + $120( P / F ,8%, 2) + $120( P / F ,8%,3) + $300( P / F ,8%, 4) P = $373.92

2.60)

= A( P / A,15%,5) $100( P / A,15%,5) + $20( P / A,15%,3)( P / F ,15%, 2) 3.35216 A = $369.74 A = $110.30

2.61) P1 = $200 + $100( P / A,8%,5) + $50( P / F ,8%,1) +$50( P / F ,8%, 4) + $100( P / F ,8%,5) = $750.37 = P2 X= ( P / A,8%,5) $750.37 X = $187.93

2.62) Establish economic equivalent at N = 8 :

C ( F / A,8%,8) − C ( F / A,8%,2)( F / P,8%,3) = $6,000( P / A,8%,2) 10.6366C − (2.08)(1.2597)C = $6,000(1.7833) 8.0164C = $10,699.80 C = $1,334.73

2.63) The original cash flow series is

0 1 2 3 4

0 6 $900 $800 7 $920 $820 8 $300 $840 9 $300 $860 10 $300 − $500

$880

2.64) $300( F / A,10%,8) + $200( F / A,10%,3) = 2C ( F / P,10%,8) + C ( F / A,10%,7) $4,092.77 = 2C (2.1436) + C (9.4872) C = $297.13

2.65) $25, 000 + $30, 000( P / F ,10%, 6) = C ( P / A,10%,12) + $1, 000( P / A,10%, 6)( P / F ,10%, 6) $41,935 = 6.8137C + $2, 458.43 C = $5, 794

2.66) (b), (d), and (e)

2.67)

$200( F / A,8%,5) − $50( F / P,8%,1) = X ( F / A,8%,5) − ($200 + X )[( F / P,8%, 2) + ( F / P,8%,1)] $1,119.32 = X (5.8666) − ($200 + X )(2.2464) X = $433.29

2.68)

(b)

2.69)

(a)

2.70) P1 = 30, 723( P / F , i %,5) P2 = A( P / A, i %,10)  (1 + i )10 − 1  $50, 000(1 + i ) −5 = $5, 000  10   i (1 + i )  ∴ i =13.06%

2.71)

2= P P(1 + i )5 21/5 = 1 + i i = 14.87% 2.72)

Establishing equivalence at n = 0 $2,000( P / A, i,6) = $2,500( P / A1 ,−25%, i,6)

By Excel software, i = 92.36%

2.73) = $40, 000 $15, 000( F= / P, i,5) $15, 000(1 + i )5 i = 21.67%

2.74)

Option 1: $100, 000( F / A, 7%, 7)( F / P, 7%,13) = 2, 085, 484.95 Option 2: $100, 000( F / A, 7%,13) = 2, 014, 064.29

$100, 000( F / A, i, 7)( F / P, i,13) = $100, 000( F / A, i,13) i = 6.6% Short Case Studies with Excel 2.75)

The equivalent future worth of the prize payment series at the end of Year 20 (or beginning of Year 21) is

F1 = $1,952,381( F / A, 6%, 20) = $1,952,381(36.7856) = $71,819,506.51 The equivalent future worth of the lottery receipts is

= F2 ($36,100, 000 − $1,952,381)( F / P, 6%, 20) = ($36,100, 000 − $1,952,381)(3.2071) = $109,514,828.9 The resulting surplus at the end of Year 20 is = F2 − F1 $109,514,828.9 − $71,819,506.51 = $37, 695,322.4

2.76)

$1, 000( F / P,9.4%,5) + $500( F / A,9.4%,5) (1 + 0.094)5 − 1 ) 0.094 = $1, 000(1.5671) + $500(6.0326) = $4,583.4 = $1, 000((1 + 0.094)5 ) + $500(

$4,583.4( F / P,9.4%, 60) = $4,583.4((1 + 0.094)60 ) = $4,583.4(219.3) = $1, 005,141.21

The main question is whether or not the U.S. government will be able to invest the social security deposits at 9.4% interest over 60 years.

2.77) (a) It is worth $218.20M.

Season

Salary

Present Worth (6%)

2015 $ 6.50 2016 $ 9.00 2017 $ 14.50 2018 $ 25.00 2019 $ 26.00 2020 $ 26.00 2021 $ 29.00 2022 $ 29.00 2023 $ 32.00 2024 $ 32.00 2025 $ 32.00 2026 $ 29.00 2027 $ 25.00 (buy out) $ 10.00

$6.50 $8.49 $12.90 $20.99 $20.59 $19.43 $20.44 $19.29 $20.08 $18.94 $17.87 $15.28 $12.42 $4.97

Sum

$218.20

$ 325.00

(b) With the buyout option, it is better staying on the contract. Without the buyout contract, it is better being a free agent after 2020.

Season

Salary

2015 2016 2017 2018 2019 2020 2021 $ 29.00 2022 $ 29.00 2023 $ 32.00 2024 $ 32.00 2025 $ 32.00 2026 $ 29.00 2027 $ 25.00 (buy out) $ 10.00

Salary for being Free Agent

$30.00 $30.00 $30.00 $30.00 $30.00 $30.00 $30.00

Sum $ 218.00

$210.00

$ 183.00

$177.52