3 Solutions

 

CHAPTER CHAP TER 3 Solutions Manual For 

Basics of Engineering Economy, 1e Leland Blank, PhD, PE Texas A&M University and American University of Sharjah, UAE

Anthony Tarquin, PhD, PE University of Texas at El Paso

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Chapter 3 3.1 Effective 3.2 Si Simpl mple: e: F = P + Pn Pnii = 20 + 20(1)(0.18) = $23.6 million Compound: F = 20(1 + 0.18)1 = $23.6 million Difference = 23.6 – 23.6 = $0   3.3 (a) Quarter Quarter

(b) Semiannual Semiannual

3.4 (a) 2

(b) 6

(c) 1

3.5 (a) 12

(b) 3

(c) 24

3.6 (a) r = 4% 3.7 (a) Nominal

(c) Month

(b) r = 0.04*2 = 8%

(e) Continuous Continuous

(c) r = 0.04*4 = 16%

(b) Effective (c) Effective (d) Nominal

3.8 i/yr i/yr = (1 + 0.12/ 0.12/4) 4)4 –1 = 12.55% 3.9 i/yr = (1 + 0.16/4)4 –1 = 16.99% 3.10

(d) Week

0.12 = (1 + r/4)4 -1 1.12 = (1 + r/4)4  (1 + r/4) = 1.12 0.25 1 + r/4r = = 11.48% 1.0287

3. 3.11 11 i/ i/yr yr = e0.12 – 1 = 12.75% 3.12 3.1 2 r/ r/qua quart rter er = 0.06 0.06 i/quarter = e0.06 – 1 = 6.18%

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(e) Nominal

 

3.13 3.1 3 0.036/ 0.036/qua quart rter er = er  –1 er = 1.036 Take natural log of both sides: r = 0.035367 = 3.54% per quarter  r/month = 0.0354/3 = 1.18% 3.14 (a) Years

(b) Semiannual

(c) Quarter 

3.15 Time periods on i and n must be years 3.16 Time periods on i and n must be semiannual semiannual 3.17 P = 190,000 190,000(P/F (P/F,2%, ,2%,36) 36) = 190,000(0.4902) = $93,138 3.18 F = 250,000 250,000(F/P (F/P,1% ,1%,24) ,24) = 250,000(1.2697) = $317,425 3.19 F = 160,000(F/P,4%,14) = 160,000(1.7317) = $270,072 3.20 F = 192,000(F/P,1.5%,60) =192,000(2.4432) = $469,094 3.21 F = 2.3(F/P,5%,20) = 2.3(2.6533)

(millions)

= $6,102,590 3.22 P = 50,000(P/F,3%,32) = 50,000(0.3883) = $19,415 3.23 P = 80,000(P/F,3%,8) + 90,000(P/F,3%,16) = 80,000(0.7894) + 90,000(0.6232) = $119,240 3.24 i/yr = e0.14 –1 = 15.03% P = 14(P/F,15.03%,2) = 14[1/(1 + 0.1503)2]

(millions)

= $10,580,490

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3.25 i/yr = e0.10 –1 = 10.52% P = 120,000(P/F,10.52%,1) + 180,000(P/F,10.52%,2) + 250,000(P/F,10.52%,3) = 120,000[1/1.1052)1] + 180,000[1/1.1052)2] + 250,000[1/1.1052)3] = 120,000(0.90481) + 180,000(0.81869) + 250,000(0.74076) = $441,131 3.26 i/yr = e0.10 –1 = 10.52% F = 300,000(F/P,10.52%,4) + 250,000(F/P,10.52%,2) + 200,000 = 300,000(1.1052)4 + 250,000(1.1052)2 + 200,000 = $952,961 3.27 i/6 mos = (1 + 0.01)6 –1 = 6.15% P = 13,000(P/A,6.15%,5) = 13,000(4.1953) = $54,538 3.28 i/yr = (1 + 0.03)4 –1 = 12.55% P = 40,000(P/A,12.55%,5) = 40,000(3.5562) = $142,248 3.29 Monthly difference = 5296 – 3443 = $1853 F = 1853(F/A,0.5%,480) = 1853(1991.19) = $3,690,231 3.30 Cost of treatment = 10,000(F/A,1%,12) = 10,000(12.6825) = $126,825 P = 126,825(A/P,1%,60) = 126,825(0.02224) = $2820.59 3.31 Savings = 12,000 12,000 – 2000 = $10,000 per quarter  P = 10,000(P/A,3%,8) = 10,000(7.0197) = $70,197 3.32 F = 19,000(F/A,3%,13) = 19,000(15.6178) = $296,738

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3.33 i/yr = (1 + 0.05)2 –1 = 10.25% A = 9,000,000(A/P,10.25%,10) = 9,000,000(0.16450) = $1,480,476 3.34 i/quarter = (1 + 0.01)3 –1 = 3.03% P = [140,000 + 140,000(0.20)](P/A,3.03%,12) = 168,000(9.9362) = $1,669,282 3.35 Income/month = 10(75)(300) = $225,000 P = 225,000(P/A,0.25%, 24) = 225,000(23.2660) = $5,234,850 3.36 25,000 = 40,000(P/F,1%,n) (P/F,1%,n) = 0.6250 n is between 47 and 48 (spreadsheet) Therefore, n = 48 months 3.37 Solve for F at the end of year 4 and then convert to A: F = 835,000(F/P,1%,24) + 1,100,000 = 835,000(1.2697) + 1,100,000 = $2,160,200 A = 2,160,200(A/F,1%,48) = 2,160,200(0.01633) = $35,276 3.38 F = 7500(F/A,1.5%,24) = 7500(28.6335) = $214,751 3.39 840,000 = (30 + 5)(500)(P/A,0.5%,n) 840,000 = 17,500(P/A,0.5%,n) (P/A,0.5%,n) = 48.0000 From 0.5% interest tables, n is slightly greater than 55 Therefore, n = 56 months 3.40 i/mo = e0.01 –1 = 1.005% A = 9900 – 100(A/G,1.005%,36) = 9900 – 100(16.42318) = $8258

(formula)

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3.41 Monthly power savings = 4*24*30*0.08 = $230.40 10,000 = 230.40(P/A,1%,n) (P/A,1%,n) = 43.40278 From 1% table, n is between 55 and 60 months n = 58 (by trial and error)

3.42 i/year = (1 + 0.00833)12 –1 = 10.47% A = 1.8(A/P,10.47%,200) = $0.19 trillion (formula) 3.43 i/year = (1 + 0.01)12 –1 = 12.68% P = 300,000(P/A,12.68%,5) – 25,000(P/G,12.68%,5) = 300,000(3.5449) – 25,000(6.24853) = $907,247 3.44 A = 40,000 + 1000(A/G,3%,16) 1000(A/G,3%,16) = 40,000 + 1000(6.8742) = $46,874 3.45 290,000(P/F,0.5%,48) = 4000(P/A,0.5%,48) + G(P/G,0.5%,48) 290,000(0.7871) = 4000(42.5803) + G(959.9188) 959.9188G = 57,938 G = $60.36 3.46 Pg = 100,000[1 – (1.01/1.015)24]/(0.015 – 0.01) = 100,000(22.35297) = $2,235,297 3.47 i = e0.10 –1 = 10.52% Pg = 8000[1 – (1.10/1.1052)7]/(0.1052 – 0.10) = 8000(6.2450) = $49,960 3.48 Convert all cash flows into present worth and then amortize: i/yr = (1 + 0.0112 –1 = 12.68% Pg = 90,000[1 – (1.03/1.1268)10]/(0.1268 – 0.03) = $551,073 P = 800,000 + Pg = 800,000 + 551,073 = $1,351,073 A = 1,351,073(A/P,12.68%,10) = 1,351,073(0.18194) = $245,813

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3.49 Move all cash flows to year 10 and check balances: Option 1: F = 6697.44(F/P,0.01083%,120) – 100(F/A,0.01083%,120) = 6697.44(3.6423) – 100(243.9789) = 24,394.08 – 24,397.89 = $-3.86 (zero balance) Option 2: Credit bal: F = 6697.44(F/P,0.01083%,120) – 50(F/A,0.01083%,120) = 6697.44(3.6423) – 50(243.9789) = 24,394.08 – 12,198.95 = $12,195.14 Savings bal: F = 50(F/A,0.5%,120) = 50(163.8793) = $8193.97 3.50 Move chemical cost to end of interest period and amortize pump: Chemical cost/mo = 10(30) = $300 A = 950(A/P,1%,36) + 300 = 950(0.03321) + 300 = $331.55 3.51 Find A per six months and then divide by 6: A/semi = 201,500(A/P,6%,6) = 201,500(0.20336) = $40,977 A/mo = 40,977/6 = $6829.51 3.52 Move deposits deposits to to end of compoundin compounding g period and then find find F: F = 6000(F/A,5%,20) = = 6000(33.0660) $198,396 3.53 Move cash flow to end of quarter and find F: F = 9000(F/A,2%,12) = 9000(13.4121) = $120,709 3.54 Move cash flow to end of interest period and then find P: Cost/quarter = 100,000(0.019)(3) = $5700 P = 5700(P/A,3%,12) = 5700(9.9540) = $56,737

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3.55 Move cash flow to end of interest period and then find P: Fuel savings = 800(0.50) = $400 per month = 400(12) = $4800 per year  P = 4800(P/A,12%,3) = 4800(2.4018) = $11,529 Problems for Test Review and FE Exam Practice

3. 3.56 56 Answ Answer er iiss (d) (d) 3. 3.57 57 Answ Answer er iiss (c) (c) 3. 3.58 58 Answ Answer er iiss (d) (d) 3. 3.59 59 Answ Answer er iiss (b) (b) 3.60 3.6 0 0.1268 0.1268 = (1 + r/1 r/12) 2)12 –1 r = 0.12 = 12% per year, compounded monthly i = 1% per month Answer is (c) 3. 3.61 61 Answ Answer er iiss (d) (d) 3.62 3.6 2 P = 40,000 40,000(P (P/F /F,5% ,5%,8) ,8) = 27,074 Answer is (a) 3.63 A = 500,000 500,000(A/F (A/F,7%, ,7%,12) 12) = $27,950 Answer is (c)  

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