ISEN302 Solutions - INEN 302-501 SPR 06 HW#2 SOLUTION 1.12...

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Unformatted text preview: INEN 302-501 SPR. 06 HW#2 SOLUTION 1.12 Rate of return = (2.3/6)(100) = 38.3% 1.13 Profit = 8(0.28) = $2,240,000 1.17 Equivalent cost now: P + 0.1P = 16,000 1.1P = 16,000 P = $14,545.45 1.26 (a) Simple interest total amount = 1,750,000(0.075)(5) = $656,250 Compound interest total = total amount due after 4 years – amount borrowed = 1,750,000(1 + 0.08) 4 – 1,750,000 = 2,380856 – 1,750,000 = $630,856 (b) The company should borrow 1 year from now for a savings of $656,250 – $630,856 = $25,394 1.36 WACC = (0.25)(0.18) + (0.75)(0.10) = 12% Therefore, MARR = 12% Select the last three projects: 12.4%, 14%, and 19% 1.39 The cash flow diagram is: 1 2 3 4 5 6 7 8 F= ? $9000 $3000 $10,000 i = 10% 1.42 Time to double = 72/9 = 8 years Time to quadruple = (8)(2) = 16 years Chapter 2 2.6) F = 150,000(F/P,18%,7) = 150,000(3.1855) = $477,825 2.12 P = 9000(P/F,10%,2) + 8000(P/F,10%,3) + 5000(P/F,10%,5) = 9000(0.8264) + 8000(0.7513) + 5000(0.6209) = $16,553 2.17 A = 3.4(A/P,20%,8) = 3.4(0.26061) = $886,074 2.18) P = (280,000-90,000)(P/A,10%,5) = 190,000(3.7908) = $720,252 2.20 F = (458-360)(20,000)(0.90)(F/A,8%,5) = 1,764,000(5.8666) = $10,348,682 2.25 (a) 1. Interpolate between n = 32 and n = 34: 1/2 = x/0.0014 x = 0.0007 (P/F,18%,33) = 0.0050 – 0.0007 = 0.0043 2. Interpolate between n = 50 and n = 55: 4/5 = x/0.0654 x = 0.05232 (A/G,12%,54) = 8.1597 + 0.05232 = 8.2120 (b) 1. (P/F,18%,33) = 1/(1+0.18) 33 = 0.0042 2. (A/G,12%,54) = {(1/0.12) – 54/[(1+0.12) 54 –1} = 8.2143 2.31 (a) CF 3 = 280,000 – 2(50,000) = $180,000 (b) A = 280,000 – 50,000(A/G,12%,5) = 280,000 – 50,000(1.7746) = $191,270 2.36 Convert future to present and then solve for G using P/G factor: 6000(P/F,15%,4) = 2000(P/A,15%,4) – G(P/G,15%,4) 6000(0.5718) = 2000(2.8550) – G(3.7864) G = $601.94 2.38 A = [4 + 0.5(A/G,16%,5)] – [1 –0.1(A/G,16%,5) = [4 + 0.5(1.7060)] – [1 –0.1(1.7060)] = $4,023,600 2.39 For n = 1: {1 – [(1+0.04) 1 /(1+0.10) 1 }]}/(0.10 –0.04) = 0.9091 For n = 2: {1 – [(1+0.04) 2 /(1+0.10) 2 }]}/(0.10 –0.04) = 1.7686 For n = 3: {1 – [(1+0.04) 3 /(1+0.10) 3 }]}/(0.10 –0.04) = 2.5812 2.40 For g = i, P = 60,000(0.1)[15/(1 + 0.04)] = $86,538 2.42 Find P and then convert to A. P = 5,000,000(0.01){1 – [(1+0.20) 5 /(1+0.10) 5 }]}/(0.10 – 0.20) = 50,000{5.4505} = $272,525 A = 272,525(A/P,10%,5) = 272,525(0.26380) = $71,892 2.45 Find A in year 1 and then find next value. 900,000 = A{1 – [(1+0.05) 5 /(1+0.15) 5 }]}/(0.15 – 0.05) 900,000 = A{3.6546) A = $246,263 in year 1 Cost in year 2 = 246,263(1.05) = $258,576 2.51 2,400,000 = 760,000(P/A,i,5) (P/A,i,5) = 3.15789 i = 17.6% (Excel) 2.54 400,000 = 320,000 + 50,000(A/G,i,5) (A/G,i,5) = 1.6000 Interpolate between i = 22% and i = 24% i = 22.6% 2.57 160,000 = 30,000(P/A,12%,n) (P/A,12%,n) = 5.3333 From 12% table, n is between 9 and 10 years; therefore, n = 10 years 2.58 2,000,000 = 100,000(P/A,4%,n) (P/A,4%,n) = 20.000 From 4% table, n is between 40 and 45 years; by spreadsheet, 42 > n > 41 Therefore, n = 41 years H/W 3 Solutions...
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This note was uploaded on 03/22/2010 for the course MEEN ISEN 302 taught by Professor Kim during the Spring '10 term at Texas A&M University–Commerce.

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ISEN302 Solutions - INEN 302-501 SPR 06 HW#2 SOLUTION 1.12...

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