HW-Chap-4-2 - i )n1 + B (1 + i )n2 + + B Subtract equation...

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Engineering Economy ENGR 300 Chapter 4 4-35 $3,575 = $375 + $93.41 (P/A, i %, 45) ( P/A , i %, 45) = ($3,575 − $375)/$93.41 = 34.258 From compound interest tables, i = 1.25% per month. For an $800 down payment, unpaid balance is $2775. P = $2,775, n = 45 months, i = 1.25%, A = ? A = $2,775 (A/P, 1.25%, 45)* = $2,775 (0.0292) = $81.03 Effective interest rate = (1 + i)12 − 1 = (1.0125)12 – 1 = 0.161 = 16.1% per year * Note that no interpolation is required as ( A/P, 1.25%, 45) = 1/( P/A , i %, 45) = 1/34.258 = 0.0292 4-38 The solution may follow the general approach of the end-of-year derivation in the book. (1) F = B (1 + i )n +.… + B (1 + i )1 Divide equation (1) by (1 + i): (2) F (1 + i )−1 = B (1 +
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Unformatted text preview: i )n1 + B (1 + i )n2 + + B Subtract equation (2) from equation (1): (1) (2) F F (1 + i )1 = B [(1 + i )n 1] Multiply both sides by (1 + i): F (1 + i ) F = B [(1 + i )n+1 (1 + i)] So the equation is: F = B[(1 + i )n+1 (1 + i)]/i Applied to the numerical values: F = 100/0.08 [(1 + 0.08)7 (1.08)] = $792.28 4-40 (a) (b) S = $50 ( P/G , 10%, 4) = $50 (4.378) = $218.90 (c) T = $30 ( A/G , 10%, 5) = $30 (1.810) = $54.30 4-48 P = $100 (P/A, 10%, 4) + $100 (P/G, 10%, 4) = $100 (3.170 + 4.378) = $754.80 Also: P = 4B (P/A, 10%, 4) B (P/G, 10%, 4) Thus, 4B (3.170) B (4.378) = $754.80 B = $754.80/8.30 = $90.94...
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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 .

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