HW-Chap-4-1 - = $869,600 A = $869,600 (A/F, 15%, 40) =...

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Engineering Economy ENGR 300 Chapter 4 4-1 (a) 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) $634 = $200 (P/A, i%, 4) ( P/A , i %, 4) = $634/$200 = 3.17 From compound interest tables, i = 10%. (c) V = $10 ( F/A , 10%, 5) – $10= $10 (6.105) - $10= $51.05 (d) $500 = x ( P/A , 10%, 4) + x ( P/G , 10%, 4) = x (3.170 + 4.378) x = $500/7.548 = $66.24 4-2 F = $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 4-13 Number of yearly investments = (59 – 20 + 1) = 40 The diagram indicates that the problem is not in the form of the uniform series compound amount factor. Thus, find F that is equivalent to $1,000,000 one year hence: F = $1,000,000 (P/F, 15%, 1) = $1,000,000 (0.8696)
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Unformatted text preview: = $869,600 A = $869,600 (A/F, 15%, 40) = $869,600 (0.00056) = $486.98 This result is very sensitive to the sinking fund factor. (A/F, 15%, 40) is actually 0.00056208, which makes A = $488.78. 4-29 Determine the required present worth of the escrow account on January 1, 1998: A = $8,000, i = 5.75%, PW = ?, n = 3 years PW = A (P/A, i %, n )= $8,000 + $8,000 (P/A, 5.75%, 3)= $8,000 + $8,000 [(1 + i )n 1]/[ i (1 + i )n] = $8,000 + $8,000 [(1.0575)3 1]/[0.0575(1.0575)3]= $29,483.00 It is necessary to have $29,483 at the end of 1997 in order to provide $8,000 at the end of 1998, 1999, 2000, and 2001. It is now necessary to determine what yearly deposits should have been over the period 19811997 to build a fund of $29,483. A = ?, i = 5.75%, F = $29,483, n = 18 years A = F (A/F, i %, n ) = $29,483 (A/F, 5.75%, 18) = $29,483 ( i ) / [(1 + i )n 1] =$29,483(0.575)/ [(1.0575)181] = $29,483 (0.03313) = $977...
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