Solutions Manual Contemporary Engineering Economics 7th Edition By Chan Park

Solutions Manual

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 1 Engineering Economic Decisions 1.1

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Lease o Deposit (typically one month worth of deposit) refundable when lease expires. o Monthly lease payment o Monthly maintenance fees o Monthly utility expenses Buy o Closing fees o Down payment o Monthly mortgage payments o Property taxes o Monthly utility fees o Monthly maintenance fees o Repair expenses o Homeowners’ association fee (if applicable)

1.2 •

Option 1: o Total amount at the end of two years: $1,150

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Option 2: o Loan $500 to a friend for one year and receive $600 o Deposit $500 (left over) in a back at 3% for two years: $500(1.03)(1.03) = $530.45 o Deposit $600 received from your friend at 3% per year for a year: $600(1.03) = $618 Total amount at the end of two years: $530.45 + $618 = $1,148.45 These two options are about the same. But considering the trustworthiness, you could go with Option 2.

1

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 2 Accounting Information for Engineering Economic Decisions 2.1 (2) Income statement; (1) balance sheet; (3) cash flow statement; (4) operating activities; (5) investing activities, and (6) financing activities; (7) capital account (paid-in capital) 2.2 (7), (8), (1), (11), (3), (9) 2.3 (a)

• • • •

Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 = $580,000 Current liabilities = $50,000 + $100,000 + $80,000 = $230,000 Working capital = $580,000 - $230,000 = $350,000 Shareholder’s equity = $100,000 + $150,000 + $150,000 + $70,000 = $470,000

(b) EPS = $500,000/10,000 = $50 per share (c) Par value = $15; capital surplus = $150,000/10,000 = $15

Market price = $15 + $15 = $30 per share

2.4 (a) Shareholder’s equity in 2021 = $700 - $510 = $190(M) Shareholder’s equity in 2022 = $900 - $640 = $260(M) (b) Net working capital in 2021 = $100 - $60 = $40(M) Net working capital in 2022 = $200 - $90 = $110(M) (c) The income taxes in year 2022: ($2,350 - $1,130-$420-$210) *0.35 = $206.5(M) (d) $383.50 + $420=$803.50 (M) (Cash from Operating activities = Net income + Depreciation)

Contemporary Engineering Economics, 7th ed. ©2023 2.5 (a) ROE (= Net income/Equity) ROA (= Net income + interest expense (1-tax rate)/Average total assets)

Company A 26.03%

Company B 22.29%

17.34%

12.59%

(b) Company A has performed better in terms of profitability. (c) If two companies were merged, the impact on the results of ROE could be positive under the situation where the Company A leads the acquisition using a stock swap instead of issuing new stocks for M&A cost. If Company A uses a stock swap, the stock value wouldn’t be decreased in terms of scarcity.

2.6 Inventory turnover ratio (2021) = Sales/Average inventory balance = $3,776,395 / ($202,794 + $231,313)×0.5 = 17.4 times Inventory turnover ratio (2022) = 15.6 times This ratio shows how many times the inventory of a firm is sold and replaced over a specific period. From the data, Metronix was holding more stocks of inventory than last year; having more inventories on stock is unproductive.

2.7 (b) 2.8 (b) 2.9 (d) 2.10 Given Olson’s EPS = $8 per share; Cash dividend = $4 per share; Book value per share = $80; Changes in the retained earnings = $24 million; Total debt = $240 million; Find debt ratio = total debt/total assets •

Net Income = $8 X Where X = the number of outstanding shares

•

Book value =

EPS =

Total shareholders' equity = $80 X

Contemporary Engineering Economics, 7th ed. ©2023 •

Retained earnings = Net income – Cash dividend; Net income = 8X from EPS relationship and the total cash dividend = 4X, so we rewrite 8X – 4X = $24 million, or X = 6 million shares

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From the book value per share, we know that the total shareholders’ equity = 80X, or $480 million; Total assets = Total liabilities + Total shareholders’ equity = $240 million + $480 million = $720 million

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Debt ratio = $240 million/$720 million = 0.33

2.11 (a) Debt ratio (= Total debt/Total assets) = $19,483,000/$38,599,000 = 50.48% (b) Times-interest-earned ratio (= EBIT/Interest expense) = Not defined (c) Current ratio (= Current assets/Current liabilities) = 29,021,000/19,483,000 = 1.49 (d) Quick (acid test) ratio (= (Current assets - Inventories)/Current liabilities)) = (29,021,000-1,301,000)/19,483,000 = 1.42 (d) Inventory-turnover ratio (= Sales/Avg. inventory balance) = 61,494,000/ ((1,301,000+1,051,000)×0.5) = 52.29 (f) Days-sales-outstanding ratio (= Receivables/ (Annual sales/365)) =10,136,000/ (61,494,000/365) = 60.16 (g) Total-assets-turnover ratio (= Sales/Total assets) = 61,494,000/38,599,000 = 1.59 (h) Profit margin on sales (= Net income available to common stockholders/Sales) = 2,635,000/61,494,000 = 4.28% (i) Return on total assets (= (Net income + interest expense (1-tax rate))/Avg. total assets)

Contemporary Engineering Economics, 7th ed. ©2023 = 2,635,000/ ((38,599,000 + 33,652,000)×0.5) = 7.29% (j) Return on common equity (= (Net income available to common stockholders)/Avg. common equity) = 2,635,000/ ((7,766,000 + 5,641,000)×0.5) = 39.31% (k) Price/earnings ratio (= Price per share/Earnings per share) =13.47/ (3,350,000/1,944,000) = 7.82 (l) Book value per share (= (Total stockholders’ equity-Preferred stock)/Shares outstanding) = 7,766,000/1,944,000 = $3.99 To make an informed analysis of the firm’s financial health, we need to calculate the various financial ratios of the firm’s competitors along with the S&P 500.

2.12 Income Statement: A

B

C

D

E

F

$900,000

$585,000

$315,000

$270,000

$108,000

$162,000

Balance Sheet: ⓪

①

②

③

④

⑤

$160,000

$120,000

$320,000

$600,000

$900,000

$1,500,000

⑥

⑦

⑧

⑨

⑩

$450,000

$700,000

$100,000

$700,000

$800,000

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From Current ratio Total current assets = 2.4 × $250,000 = $600,000 ------------------------------------ ③ Plant and equipment, net = $1,500,000-$600,000=$900,000 ----------------------- ④

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From Quick ratio

Contemporary Engineering Economics, 7th ed. ©2023 Inventory = $600,000 - (1.12 × $250,000) = $320,000 ------------------------------- ② •

From Inventory Turnover Net Revenue = (($320,000 +$280,000)/2) × 6.0 =$1,800,000 Cost of goods sold = $1,800,000- $900,000= $900,000 -------- A

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From DSO Accounts receivable = 24.3333 × ($1,800,000 ÷365) = $120,000 -------------------- ① Cash = ③-(②+①) = $160,000 ⓞ

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From interest expense of income statement Bond = $450,000 ⑥ 250,000 + ⑥ = $700,000-------------------------- ⑦

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From Debt-to-Equity ratio Total Equity ⑩ = $700,000 ÷ 0.875 =$ 800,000 ------------- ⑩ Total assets or Total liabilities and equity = ⑦ + ⑩ = $1,500,000 -----⑤

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From Return on total assets Net income F = 14%× ($1,350,000) - ($45,000) (0.6) = $162,000

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From F, D =F ÷ 0.6 = $270,000, E = D× (0.4) =$108,000 C = D+45,000 = $315,000 B=$900,000-C = $585,000

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From EPS Stock Outstanding = F ÷ 4.05 = 40,000 shares Common stock = $2.50 × 40,000 = $100,000 -------------------------------- ⑧ Retained Earnings = ⑩ - ⑧ = $700,000 -------------------------------------⑨

2.13 • • •

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Accounts receivable = DSO × Sales/365 = 45 days × ($1,200)/365 days) = $147.945 Current assets = (Cash and marketable securities) + (Accounts receivable) + Inventory = $427.945 Long-term debt = (Total assets) – (Current liabilities) – (Common equities) = $427.945 + $280 – (current assets/current ratio) - $500 = ($207.945) – (427.945/3.2) = $74.212 Total assets turnover = Sales/Total assets = $1,200/ ($427.945 + $280) = 1.695 times

Contemporary Engineering Economics, 7th ed. ©2023

2.14

(a) Find Tiger’s accounts receivable.

DSO = 91.25 =

AR  AR = $50, 000 200, 000 / 365

(b) Determine the amount of current liabilities.

CA = Cash + Inventory + AR = $10, 000 + $150, 000 + $50, 000 = $210, 000 $210, 000 Current Ratio = 4.2 =  Current Liabilities = $50, 000 Current Liabilities

(c) Calculate the amount of the long-term debt.

Total Asset = Current Asset + Fixed Asset = $210, 000 + $90, 000 = $300, 000 $300, 000 = ($50, 000 + Long term debt) + $200, 000

 Long term debt = $50, 000 (d) Calculate the Return on Common Equity.

ROE =

net income $15, 000 = = 0.075  7.5% equity $200, 000

2.15 (a) Find Fisher’s accounts receivable. 𝐴𝑅 𝐷𝑆𝑂 = → 𝐴𝑅 = 𝟏𝟒𝟕. 𝟗𝟓𝑴 1,200/ 365 (b) Calculate the amount of current assets. 𝐶𝐴 = 𝑐𝑎𝑠ℎ + 𝐼𝑛𝑣. +𝐴𝑅 = 100 + 180 + 147.95 = 𝟒𝟐𝟕. 𝟗𝟓𝑴 (c) Determine the amount of current liabilities. 𝐶𝐴 427.95 𝐶𝑅 = 3.2 = = → 𝐶𝐿 = 𝟏𝟑𝟑. 𝟕𝟑𝑴 𝐶𝐿 𝐶𝐿 (d) Determine the amount of total assets. 𝑇𝐴 = 𝐶𝐴 + 𝐹𝐴 = 427.95 + 280 = 𝟕𝟎𝟕. 𝟗𝟓𝑴

Contemporary Engineering Economics, 7th ed. ©2023

(e) Calculate the amount of the long-term debt.

707.95 = (133.73 + 𝐿𝐵) + 500 → 𝐿𝐵 = 𝟕𝟒. 𝟐𝟐𝑴 (f) Calculate the profit margin. 𝑛𝑒𝑡 𝑖𝑛𝑐𝑜𝑚𝑒 358 𝑝𝑟𝑜𝑓𝑖𝑡 𝑚𝑎𝑟𝑔𝑖𝑛 = = = 𝟐𝟗. 𝟖𝟑% 𝑠𝑎𝑙𝑒𝑠 1,200 (g) Calculate the Return on Common Equity 𝑛𝑒𝑡 𝑖𝑛𝑐𝑜𝑚𝑒 358 𝑅𝑂𝐸 = = = 𝟕𝟏. 𝟔% 𝑒𝑞𝑢𝑖𝑡𝑦 500

Short Case Studies with Excel ST2.1 Not provided ST2.2 (a) Working capital = Current assets – Current liabilities Working capital requirements = Changes in current assets – Changes in current liabilities: WC req. = (+$100,000 - $20,000) – (+$30,000 - $40,000) = $90,000, indicating that additional financing is needed to fund the increase in current assets. (b) Taxable income = $1,500,000 - $650,000 - $150,000 - $20,000 = $680,000 (c) Net income = $680,000 - $272,000 = $408,000 (d) Net cash flow: • Operating activities = net income + depreciation – WC = $408,000 + $200,000 - $90,000 = $518,000 • Investing activities = equipment purchase = ($400,000) • Financing activities = borrowed fund = $200,000 • Net cash flow = $518,000 - $400,000 + $200,000 = $318,000 ST2.3

Not provided (Visit the websites and get the most recent financial statements available)

Contemporary Engineering Economics, 7th ed. ©2023

Chapter 3: Interest Rate and Economic Equivalence Types of Interest 3.1

I = (iP)N = (0.06)($2, 000)(5) = $600

3.2 •

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

3.3 •

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.

3.4 •

Simple interest (John):

I = iPN = (0.1)($1, 000)(5) = $500 •

Compound interest (Susan):

Contemporary Engineering Economics, 7th ed. ©2023 I = P (1+ i)N −1 = $1, 000 (1+ .095)5 −1 = $574.24 •

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

•

Simple interest:

3.5 I = iPN = (0.10)($10, 000)(5) = $5, 000 •

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

3.6 •

Option 1: Compound interest with 8%: F = $4, 500(1+ 0.085)5 = $5, 000(1.4693) = $6, 766.45

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Option 2: Simple interest with 9.5%: $4, 500(1+ 0.095 5) = $5, 000(1.475) = $6, 637.50

 Option 1 is still better.

3.7 End of Year

Principal Repayment

Interest payment

Remaining Balance $15,000.00

1

$4,620.50

$1,200.00

$10,379.50

2 3

$4,990.14 $5,389.35

$830.36 $431.15

$5,389.36 $0

0

Equivalence Concept 3.8

3.9

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

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

Contemporary Engineering Economics, 7th ed. ©2023 3.10 F = $100(F / P,10%,10) + $200(F / P,10%,8) = $688 3.11

$1, 000(F / P, i, 2) = $1, 200 $1, 000(1+ i)2 = $1, 200 i = 1.2 −1 i = 9.54%

Single Payments (Use of F/P or P/F Factors) 3.12

i = 10.5% , two-year discount rate is (1 + 0.105)2 = 1.221 (or 22.1%) 3.13 F = 2P = P(1+ 0.06)N log 2 = N log 1.06 N = 11.896 years (or 12 years) 3.14 F = $1(1.08)394 = $14, 755, 694, 730, 611

3.15

P = $450, 000(P / F, 5%, 5) = 450, 000(0.7835) = $352, 575

3.16 F = $250, 000(F / P, 6%,10) = $447, 712 3.17

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

Contemporary Engineering Economics, 7th ed. ©2023 3.18 P = $300, 000(P / F , 8%,10) = $138, 958 3.19 (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

3.20 (a) P = $12, 000(P / F ,13%, 4) = $7, 360 (b) F = $30, 000(F / P,13%, 5) = $55, 273

3.21 F = 3P = P(1+ 0.08)N log 3 = N log(1.08) N = 14.27 → 15 years 3.22 F = 2P = P(1+ 0.06)N •

log 2 = N log(1.06) N = 11.90 years

•

3.23

12 years

Rule of 72: 72 / 6 = 12 years

($16.50)(100)(F / P, i, 44) = $77.50(204,800) $15,872, 000 (F / P, i, 44) = = 9, 619.39 $1, 650 i = 23.18%

Contemporary Engineering Economics, 7th ed. ©2023 3.24 (a)

$18(F / P, i, 49) = $190, 500 $190, 500 (F / P, i, 49) = = 10, 583.33 $18 i = 20.82% (b)

F = $304.8 B(F / P, 20.82%,17) = $7, 592.33B

Uneven Payment Series 3.25 P= 3.26

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

P = $35, 000(P / F , 9%, 4) + $10, 000(P / F , 9%, 2) = $35, 000(0.7084) + $10, 000(0.8417) = $33, 211 3.27

$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

3.28

P=

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

3.29 $1,000 +

$1,000 $1,500 $1, 210 X + = + 1.1 1.13 1.12 1.14 X = $2,981

3.30 P=

$15, 000 $23, 000 $36, 000 $48, 000 + + + = $93, 564 1.072 1.073 1.074 1.075

Contemporary Engineering Economics, 7th ed. ©2023 3.31 F = $2, 000(F / P, 6%,10) + $2, 500(F / P, 6%,8) + $3, 000(F / P,6%,6) = $11, 822

3.32

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

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

Equal Payment Series 3.34

A = $250, 000( A / F, 5%, 5) = $250, 000(0.1810) = $45, 250 F = $5, 000(F / A,5%, 7) = $5, 000(8.1420) = $40, 710

(a)

F = $5, 000(F / A, 5%, 7)(1.05)

(b)

= $5, 000(8.1420)(1.05) = $42, 745.50

3.35 •

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

Interest payment for the second year = $1,395.80

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

Contemporary Engineering Economics, 7th ed. ©2023

3.36

F = $500(F / A, 7%,15)(1.07) = $500(25.1290)(1.07) = $13, 444.02 3.37 (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 3.38 F = $10, 000(F / A, 6%,20) = $367,856 3.39 (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 (d) F = $4, 000(F / A, 7.75%,12) = $74, 793 3.40 (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 (d) A = $12, 000( A / F ,11%,13) = $458 3.41 $50, 000( A / F , 6%,10) = $3, 793.40

Contemporary Engineering Economics, 7th ed. ©2023 3.42

$55, 000 = $3, 000(F / A, 6%,N ) (F / A, 6%, N ) = 18.33 N = 13years

3.43 $15, 000 = A(F / A,11%,5) A = $2, 408.57

3.44

$10, 000 = $1, 000(F / P, 7%,5) + A(F / A, 7%,5) A = $1,536.57 3.45 (a) A = $15, 000( A / P, 3.5%, 6) = $2,815.02 (b) A = $7, 500( A / P, 7.5%, 7) = $1, 416 (c) A = $2, 500( A / P, 5.25%, 5) = $581.43 (d) A = $12, 000( A / P, 6.25%,15) = $1, 255.81 3.46 (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

Contemporary Engineering Economics, 7th ed. ©2023 (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 3.47 •

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

3.48 A = $15, 000( A / P, 9%,10) = $2, 337.30

3.49 (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 (d) P = $19, 300(P / A, 7.75%,8) = $111, 970.11

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

Contemporary Engineering Economics, 7th ed. ©2023 3.50 P = $35,000( P / A,12%,10) = $197,758 Since $200,000 > $197,758, You should not purchase the equipment.

3.51 (a)

P = $9, 600, 000 + $15,100, 000(P / F , 6%,1) +$17,100, 000(P / F , 6%, 2)... + $19, 250, 000(P / F , 6%, 5) +$400, 000(P / A, 6%, 5) = $89, 970, 959 (b)

P = $7, 600, 000 + $5, 600, 000(P / A, 6%, 4) = $27, 004, 591 Since $27, 004, 591 > $25,000,000, the prorated payment option is better choice.

Linear Gradient Series 3.52

F = $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) = $165,833

3.53

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

3.54 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

Contemporary Engineering Economics, 7th ed. ©2023 3.55 A = $15, 000 − $1, 000( A / G, 8%, 12) = $10, 404.25

3.56

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

Geometric-Gradient Series 3.57

$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) 20 −20 1− (1.03) (1.06) = A1 0.06 − 0.03 = $68, 674 / year

3.58 Using the geometric gradient series present worth factor, we can establish the equivalence between the loan amount $150,000 and the balloon payment series as $150, 000 = A1 (P / A1,10%, 7%, 5) = 4.9424 A1 A1 = $30, 349.63 Payment series N 1 2 3 4 5 3.59

Payment $30,349.63 $33,384.59 $36,723.05 $40,395.35 $44,434.89

Contemporary Engineering Economics, 7th ed. ©2023 F = $8, 000(P / A1,5%, 7%, 30)(F / P, 7%, 30) = $172,895.56(7.61226) = $1, 316,126 3.60 (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 A4 = 6, 000, 000(1 − 0.055)3 = 5, 063, 451.75 P = $5, 063, 451.75(P / A4 , −5.5%,12%, 4) = $14, 269, 627.82 3.61 20

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

3.62

Contemporary Engineering Economics, 7th ed. ©2023

(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

Various Interest Factor Relationships 3.63 (a) (P / F, 8%, 67) = (P / F, 8%,50)(P / F,8%,17) = 0.0058 (P / F, 8%, 67) = (1 + 0.08)−67 = 0.0058 i 1 − (P / F , i, N ) (P / F ,8%, 42) = (P / F ,8%,40)(P / F ,8%,2) = 0.0394 0.08 ( A / P,8%, 42) = = 0.0833 1 − 0.0394

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

( A / P,8%, 42) =

0.08(1.08)42 = 0.0833 (1.08)42 −1

(c) (P / A, i, N ) = 1 − (P / F, i, N ) = 1 − (P / F ,8%,100)(P / F,8%,35) = 12.4996 i 0.08

Contemporary Engineering Economics, 7th ed. ©2023

(P / A, 8%,135) =

(1.08)135 − 1 0.08(1.08)135

= 12.4996

3.64 (a)

(F / P, i, N ) = i(F / A, i, N ) + 1 (1 + i) N −1 (1 + i) N = i +1 i = (1 + i) N − 1 + 1 = (1 + i) N

(b)

(P / F , i, N ) = 1 − (P / A, i, N )i (1 + i)−N = 1 − i =

(1 + i) N − 1 i(1 + i) N

(1 + i) N (1 + i) N − 1 − (1 + i) N (1 + i) N

= (1 + i)−N

(c)

( A / F , i, N ) = ( A / P, i, N ) − i i(1 + i) N i(1 + i) N i[(1 + i) N − 1] i = − i = − (1 + i) N − 1 (1 + i) N − 1 (1 + i) N − 1 (1 + i) N − 1 i = (1 + i) N − 1

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

i [1 − (P / F , i, N )]

i(1 + i) N i = N N (1 + i) − 1 (1 + i) 1 − N (1 + i) N (1 + i) =

i(1 + i) N (1 + i) N − 1

(e) , (f) , (g) Divide the numerator and denominator by (1 + i) N and take the limit N →.

Contemporary Engineering Economics, 7th ed. ©2023

Equivalence Calculations 3.65 P = [$100(F / A,12%, 9) + $50(F / A,12%, 7) + $50(F / A,12%, 5)](P / F ,12%,10) = $740.49

3.66 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 3.67 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 3.68 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 3.69

P = $20(P / G,10%,5) − $20(P / A,10%,12) = $0.96

$40

$60

$80

$20 0

1

2

$20

3

4

5

6

7

8

9

10

11

12

Contemporary Engineering Economics, 7th ed. ©2023 3.70 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 3.71 The original cash flow series is N

AN

N

AN

0

0

6

$900

1

$800

7

$920

2

$820

8

$300

3

$840

9

$300

4

$860 10 $300 − $500

5

$880

3.72 $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 3.73

Establishing equivalence at N = 5 $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

3.74

Computing equivalence at N = 5 X = $3, 000(F / A, 9%, 5) + $3, 000(P / A, 9%, 5) = $29, 623.08

Contemporary Engineering Economics, 7th ed. ©2023

3.75 (b), (d), and (f) 3.76 (b), (d), and (e) 3.77 A1 = ($50 + $50( A / G,10%,5) − [$50 + $50(P / F ,10%,1)]( A / P,10%,5) = $115.32 A2 = A + A( A / P,10%,5) = 1.2638 A A = $91.25 3.78 (a) 3.79 (b) 3.80 (b) $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

Solving for an Unknown Interest Rate of Unknown Interest Periods 3.81 P1 = 30, 723(P / F , i%, 5) P2 = A(P / A, i%,10)  (1+ i)10 −1  −5 $50, 000(1+ i) = $5, 000  10   i(1+ i)  i = 13.06%

Contemporary Engineering Economics, 7th ed. ©2023

3.82 2P = P(1+ i)5 21/5 = 1+ i i = 14.87% 3.83

Establishing equivalence at n = 0

$2,000(P / A,i,6) = $2,500(P / A1,−25%,i,6)

By Excel software, i = 92.36% 3.84 $40, 000 = $15, 000(F / P, i, 5) = $15, 000(1 + i)5 i = 21.67%

3.85 $1, 000, 000 = $2, 000(F / A, 6%, N ) (1+ 0.06)N −1 500 = 0.06 31 = (1+ 0.06)N log 31 = N log1.06 N = 58.93  59years 3.86 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% 3.87 Assuming that annual renewal fees are paid at the beginning of each year, (a) $15.96 + $15.96(P / A,6%,3) = $58.62 It is better to take the offer because of lower cost to renew.

Contemporary Engineering Economics, 7th ed. ©2023

(b) $57.12 = $15.96 + $15.96(P / A, i, 3) i = 7.96% 3.88 If premiums paid at the end of each year, the maximum amount to invest in the prevention program is P = $14,000(P / A,12%,5) = $50,467 .

If the premiums paid at the beginning of each year, the solution changes to P = $14, 000 + $14, 000(P / A,12%, 4) = $56, 523.

Short Case Studies ST 3.1 (a)

(b)

P = 280, 000(P / A,8%,19) = 2, 689, 007.78 280, 000(P / A, i,19) = 5, 600, 000 − 283, 770 i = 0.00709%

ST 3.2 Establish the following equivalence equation: $140,000 = $32,639 (P / A, i,9).

The interest rate makes two options equivalent is i = 18.10% by Excel.

So, if her rate of return is over 18.10%, it is a good decision.

Contemporary Engineering Economics, 7th ed. ©2023

ST 3.3 (a)

PContract = $5, 600, 000 + $7,178, 000(P / F , 6%,1) +$11, 778, 000(P / F , 6%, 2) +… +$17, 778, 000(P / F , 6%, 9) = $97,102,826.86

(b)

PBonus = $5,000,000 + $5,000,000( P / A,6%,5) + $778,000( P / A,6%,9) = $31,353,535.52  $23,000,000 It is better stay with the original plan.