Problemas del capitulo 6 ingeniería económica dark tarkin
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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