# Problemas del capitulo 6 ingeniería económica dark tarkin

Páginas: 2 (415 palabras) Publicado: 1 de noviembre de 2011
SOLUTIONS TO SELECTED PROBLEMS

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

CHAPTER 6

Solutions included for problems 1, 4, 7, 10, 13, 16, 19, 22, 25,28, and 31

6.1 The estimate obtained from the three-year AW would not be valid, because the AW
calculated over one life cycle is valid only for the entire cycle, not part of the cycle.Here the asset would be used for only a part of its three-year life cycle.

6.4 AWcentrifuge = -250,000(A/P,10%,6) – 31,000 + 40,000(A/F,10,6)
= \$-83,218AWbelt = -170,000(A/P,10%,4) – 35,000 – 26,000(P/F,10%,2)(A/P,10%,4)
+ 10,000(A/F,10%,4)
= \$-93,549
Selectcentrifuge.

6.7 AWX = -85,000(A/P,12%,3) – 30,000 + 40,000(A/F,12%,3)
= \$-53,536

AWY = -97,000(A/P,12%,3) – 27,000 + 48,000(A/F,12%,3)
= \$-53,161Select robot Y by a small margin.

6.10 AWC = -40,000(A/P,15%,3) – 10,000 + 12,000(A/F,15%,3)
= \$-24,063

AWD = -65,000(A/P,15%,6) – 12,000 + 25,000(A/F,15%,6)= \$-26,320
Select machine C.

6.13 AWland = -110,000(A/P,12%,3) – 95,000 + 15,000(A/F,12%,3)
= \$-136,353

AWincin =-800,000(A/P,12%,6) – 60,000 + 250,000(A/F,12%,6)
= \$-223,777

AWcontract = \$-190,000

Use land application.
6.16 AW100 = 100,000(A/P,10%,100)= \$10,001

AW∞ = 100,000(0.10)
= \$10,000

Difference is \$1.

6.19 AW = -100,000(0.08) –50,000(A/F,8%,5)
= -100,000(0.08)–50,000(0.17046)
= \$-16,523

6.22 Find P in year –1, move to year 9, and then multiply by i. Amounts are in \$1000.

P-1 = [100(P/A,12%,7) – 10(P/G,12%,7)](F/P,12%,10)...

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