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

This preview shows page 1. Sign up to view the full content.

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)
This is the end of the preview. Sign up to access the rest of the document.

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 1981–1997 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)18−1] = \$29,483 (0.03313) = \$977...
View Full Document

## This note was uploaded on 04/18/2011 for the course ENGINEER 300 taught by Professor Kashayar during the Spring '11 term at California State University Los Angeles .

Ask a homework question - tutors are online