Engineering Economy
Engineering Economy
8th Edition
ISBN: 9780073523439
Author: Leland T Blank Professor Emeritus, Anthony Tarquin
Publisher: McGraw-Hill Education
Question
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Chapter 3, Problem 54P
To determine

Calculate the time period.

Expert Solution & Answer
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Explanation of Solution

The present value (P) is $20,000. The first withdraw (D) is $5,000 that has taken place now. The second year’s withdrawal (D2) is $4,500 and decreases the withdrawal (d) by $500. Thereafter, interest rate (i) is 8%.

Time period (n) can be calculated as follows:

P=W+W2((1+i)n1i(1+i)n)d×1i((1+i)n1i(1+i)nn(1+i)n)20,000=5,000+4,500((1+0.08)n10.08(1+0.08)n)500×10.08((1+0.08)n10.08(1+0.08)nn(1+0.08)n)20,0005,000=4,500((1+0.08)n10.08(1+0.08)n)500×12.5((1+0.08)n10.08(1+0.08)nn(1+0.08)n)15,000=4,500((1+0.08)n10.08(1+0.08)n)500×12.5((1+0.08)n10.08(1+0.08)nn(1+0.08)n)

Substitute the time period as 5 years in the above calculation.

15,000=4,500((1+0.08)510.08(1+0.08)5)500×12.5((1+0.08)510.08(1+0.08)55(1+0.08)5)15,000=4,500(1.469310.08(1.4693))500×12.5(1.469310.08(1.4693)51.4693)15,000=4,500(0.46930.1175)500×12.5(0.46930.11753.403)15,000=4,500(3.994)500×12.5(3.9943.403)15,000=17,9733,693.7515,000>14,279.25

The calculated value is less than $15,000. Thus, it increases the time period to 5.6.

15,000=4,5000((1+0.08)5.610.08(1+0.08)5.6)500×12.5((1+0.08)5.610.08(1+0.08)5.65.6(1+0.08)5.6)15,000=4,500(1.538810.08(1.5388))500×12.5(1.538810.08(1.5388)5.61.5388)15,000=4,500(0.53880.1231)500×12.5(0.53880.12313.6392)15,000=4,500(4.3769)500×12.5(4.37693.6392)15,000=19,696.05500×12.5(0.7377)15,000=19,696.054,610.62515,00015,085.43

Since the calculated value is nearly equal to $15,000, it is confirmed that the time period is 5.6 years.

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