Engineering Economic Analysis Solutions Part 1

Chapter 4: More Interest Formulas 4-1 (a) 100 0 1 100 100 100 2 3 4 R R = $100(F/A, 10%, 4) = $100(4.641) = $464.10 (b) $50 0 1 2 $10 $15 3 4 S S = 50 (P/G, 10%, 4) = 218.90 = 50 (4.378) (c) 30 T T T 60 90 T = 30 (A/G, 10%, 5) = 54.30 T 120 T = 30 (1.810) 4-2 (a) $100 $100 $0 B = $100 (P/F, 10%, 1) + $100 (P/F, 10%, 3) + $100 (P/F, 10%, 5) = $100 (0.9091 + 0.7513 + 0.6209) = $228.13 B (b) $200 $200 $200 i=? $63 $634 = $200 (P/A, i%, 4) (P/A, i%, 4) = $634/$200 = 3.17 From compound interest tables, i = 10%. (c) $10 $10 $10 $10 V V = $10 (F/A, 10%, 5) - $10 = $10 (6.105) - $10 = $51.05 (d) 2x x 3x 4x $50 $500 $500 = x (P/A, 10%, 4) + x (P/G, 10%, 4) = x (3.170 + 4.378) x = $500/7.548 = $66.24 4-3 (a) $25 $50 $75 $50 C = $25 (P/G, 10%, 4) = $25 (4.378) = $109.45 (b) A = $140 i=? $50 $500 = $140 (P/A, i%, 6) (P/A, i%, 6) = $500/$140 = 3.571 Performing linear interpolation: (P/A, i%, 6) 3.784 3.498 i 15% 18% = 15% + (18% - 15%) ((3.487 – 3.571)/(3.784 – 3.498) = 17.24% i (c) $50 $75 $10 P F = $25 (P/G, 10%, 5) (F/P, 10%, 5) = $25 (6.862) (1.611) = $276.37 F (d) $0 P $40 A $80 A $12 A A = $40 (P/G, 10%, 4) (F/P, 10%, 1) (A/P, 10%, 4) = $40 (4.378) (1.10) (0.3155) = $60.78 A 4-4 (a) $50 W $75 $10 = $25 (P/A, 10%, 4) + $25 (P/G, 10%, 4) = $25 (3.170 + 4.378) = $188.70 W (b) $10 $15 $0 $0 x = $100 (P/G, 10%, 4) (P/F, 10%, 1) = $100 (4.378) (0.9091) = $398.00 x (c) $30 $20 $10 Y Y = $300 (P/A, 10%, 3) - $100 (P/G, 10%, 3) = $300 (2.487 – 2.329) = $513.20 (d) $10 $10 $50 Z Z = $100 (P/A, 10%, 3) - $50 (P/F, 10%, 2) = $100 (2.487) - $50 (0.8264) = $207.38 4-5 $15 $10 $25 $20 P P = $100 + $150 (P/A, 10%, 3) + $50 (P/G, 10%, 3) = $100 + $150 (2.487) + $50 (2.329) = $589.50 4-6 $40 $50 $40 $30 $30 x x = $300 (P/A, 10%, 5) + $100 (P/G, 10%, 3) + $100 (P/F, 10%, 4) = $300 (3.791) + $100 (2.329) + $100 (0.6830) = $1,438.50 4-7 $80 $50 $60 $0 P P = $10 (P/G, 15%, 5) + $40 (P/A, 15%, 4)(P/F, 15%, 1) = $10 (5.775) + $40 (2.855) (0.8696) = $157.06 4-8 B $800 $800 O B B 1.5B Receipts (upward) at time O: PW = B + $800 (P/A, 12%, 3) = B + $1,921.6 Expenditures (downward) at time O: PW = B (P/A, 12%, 2) + 1.5B (P/F, 12%, 3) Equating: B + $1,921.6 = 2.757B B = $1,921.6/2.757 = $1,093.70 4-9 F = A (F/A, 10%, n) $35.95 = 1 (F/A, 10%, n) (F/A, 10%, n) = 35.95 From the 10% interest table, n = 16. 4-10 P $1,000 = A (P/A, 3.5%, n) = $50 (P/A, 3.5%, n) (P/A, 3.5%, n) = 20 From the 3.5% interest table, n = 35. = 2.757B 4-11 $100 $100 $100 F J P’ F J = $100 (F/A, 10%, 3) = $100 (3.310)= $331 P’ = $331 (F/P, 10%, 2) = $331 (1.210)= $400.51 J = $400.51 (A/P, 10%, 3) = $400.51 (0.4021) = $161.05 Alternate Solution: One may observe that J is equivalent to the future worth of $100 after five interest periods, or: J = $100 (F/P, 10%, 5) = $100 (1.611)= $161.10 4-12 $10 $20 $30 P P’ C C C P = $100 (P/G, 10%, 4) = $100 (4.378)= $437.80 P’ = $437.80 (F/P, 10%, 5) = $437.80 (1.611) = $705.30 C = $705.30 (A/P, 10%, 3) = $705.30 (0.4021) = $283.60 4-13 G 3G 2G 4G 5G 6G 0 $500 $500 P Present Worth P of the two $500 amounts: P = $500 (P/F, 12%, 2) + $500 (P/F, 12%, 1) = $500 (0.7972) + $500 (0.7118) = $754.50 Also: P $754.50 = G (P/G, 12%, 7) = G (P/G, 12%, 7) = G (11.644) G = $754.50/11.644 = $64.80 4-14 $10 $20 $30 0 D D D P Present Worth of gradient series: P = $100 (P/G, 10%, 4) = $100 (4.378)= $437.80 D = $437.80 (A/F, 10%, 4) = $4.7.80 (0.2155) = $94.35