Capitulo 3 Ingenieria Economica Blank Solucionario

SOLUTIONS TO SELECTED PROBLEMS

Student: You should work the problem completely before referring to the solution.

CHAPTER 3

Solutions included for problems 1, 4, 7, 10, 13, 16, 19, 22, 25,28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, and 61

3.1 P = 100,000(260)(P/A,10%,8)(P/F,10%,2)
= 26,000,000(5.3349)(0.8264)
= $114.628 million

3.4 P = 100,000(P/A,15%,3)+ 200,000(P/A,15%,2)(P/F,15%,3)
= 100,000(2.2832) + 200,000(1.6257)(0.6575)
= $442,100

3.7 A = [0.701(5.4)(P/A,20%,2) + 0.701(6.1)(P/A,20%,2)((P/F,20%,2)](A/P,20%,4)= [3.7854(1.5278) + 4.2761(1.5278)(0.6944)](0.38629)
= $3.986 billion

3.10 A = 8000(A/P,10%,10) + 600
= 8000(0.16275) + 600
= $1902

3.13 A =15,000(F/A,8%,9)(A/F,8%,10)
= 15,000(12.4876)(0.06903)
= $12,930

3.16 A = [20,000(F/A,8%,11) + 8000(F/A,8%,7)](A/F,8%,10)
= [20,000(16.6455) +8000(8.9228)]{0.06903)
= $27,908

3.19 100,000 = A(F/A,7%,5)(F/P,7%,10)
100,000 = A(5.7507)(1.9672)
A = $8839.56

3.22 Amt year 5 = 1000(F/A,12%,4)(F/P,12%,2) +2000(P/A,12%,7)(P/F,12%,1)
= 1000(4.7793)(1.2544) + 2000(4.5638)(0.8929)
= $14,145



3.25 Move unknown deposits to year –1, amortize using A/P, andset equal to $10,000:
x(F/A,10%,2)(F/P,10%,19)(A/P,10%,15) = 10,000
x(2.1000)(6.1159)(0.13147) = 10,000
x = $5922.34

28. Find P at t = 0 and then convert to A:P = $22,994


A = 22,994(A/P,12%,8)
= 22,994(0.20130)
= $4628.69


3.31 Amt year 3 = 900(F/A,16%,4) + 3000(P/A,16%,2) – 1500(P/F,16%,3)+ 500(P/A,16%,2)(P/F,16%,3)
= 900(5.0665) + 3000(1.6052) – 1500(0.6407)
+ 500(1.6052)(0.6407)
=...