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Engineering Economy Ch. 4.2

Engineering Economy Ch. 4.2 - Chapter 4-2 Example 4.5 How...

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Chapter 4-2 Example 4.5 How much is needed today to provide an annual amount of \$50,000 each year for 20 years, at 9% interest each year? P =? P = \$50,000 (P/A, 9%, 20%) P = \$50,000(9.1285) P = \$456,427 Example 4.6 IF you had \$500,000 today in an account earning 10% each year, how much could you withdraw each year for 25 years A = \$500,000 (0.1102) A= \$55,100 Example 4.7 How much will you have in 40 years if you save \$3,000 each year and your account earns 8% interest each year? F = \$3,000(F/A, 8%, 40) F = \$3,000(259.0565) F = 777,170 Example 4.8 How much would you need to set aside each year for 25 years, at 10% interest, to have accumulated \$1,000,000 at the end of the 25 years? A = 1,000,000 (A/F, 10%, 25) A = 1,000,000(0.0102)

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A = 10,200 Exercise 4.6 p/A Exercise 4.7 A = 225,000(A/P,15%,4) A = 225,000(0.35037) A = \$78,811 Exercise 4.8 A = 250,000(A/F,9%,3) A = 250,000(0.30505) A = \$76,264 Exercise 4.9 F = (100,000 + 125,000)(F/A,15%,3) F = 225,000(3.4725) F = \$781,313 Exercise 4.10 P = 9000(P/F,10%,2) + 8000(P/F,10%,3) + 5000(P/F,10%,5) P = 9000(0.8264) + 8000(0.7513) + 5000(0.6209) P = \$16,553 Exercise 4.11 P = \$22,000 (P/A,15%,5)
P = \$22,000(3.3522) P = \$73,748.40 Example 4.9,(,—Finding N \$100,000 = \$8,000(P/A,7%,N) (P/A,7%,M) = \$100,000/\$8000 = 12.5 Referred to the interest table, we’ll find that

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