Question

Machine A and B are two mutually exclusive alternatives with different lives. Based on the data provided and the LCM approach, determine the correct equations to find PW of each machine at an interest rate of 10% per year.
Machine A Machine B
Initial cost,$ -85,000 - 100,000
Annual cost, $/year - 10,000 -8,000
Salvage value, $ 10,000 25,000
Life, years 6 3

a. PWA = -85,000 - 10,000(P/A,10%,6) +10,000(P/F,10%,6)
PWB = -100,000 -8,000(P/A,10%,6) - 75,000(P/F,10%,3)
b. PWA = -85,000 - 10,000(P/A,10%,6) +10,000(P/F,10%,6)
PWB = -100,000 -8,000(P/A, 10%,6) -75,000(P/F,10%,3)+25,000(P/F,10%,6)
c. PWA = -85,000 - 10,000(P/A,10%,6) +10,000(P/F,10%,6)
PWB = -100,000 -8,000(P/A, 10%,6) +25,000(P/F,10%,6)
d. PWA = -85,000 -10,000(P/A,10%,6) +10,000(P/F,10%,6)
PWB = -100.000 -8.0001P/A. 10% 6)-100 0001P/E 10% 31+25.000/P/E 10% 6)

Answer 1

Answer:

Correct option is B ;  PWA = -85,000 - 10,000(P/A,10%,6) +10,000(P/F,10%,6)

PWB = -100,000 -8,000(P/A, 10%,6) -75,000(P/F,10%,3)+25,000(P/F,10%,6)

Explanation:

So the period of the analysis is 6 years

PW = intial cost + annual cost x (P|A,i,n) + salvage value x (P|A,i,n)

PWA = -85000 - 10000 x (P|A,10%,6) + 10000 x (P|F,10%,6)

PWB = -100000 - 8000 x (P|A,10%,6) - 75000 x (P|F,10%,3) + 25000 x (P|F,10%,6)

Correct option is B