Consider the two mutually exclusive projects in the table below. Salvage values represent the net proceeds (after tax) from disposal of the assets if they are sold at the end of each year. Both projects B1 and B2 will be available (or can be repeated) with the same costs and salvage values for an indefinite period. (a) Assuming an infinite planning horizon, which project is a better choice at MARR%? Use 15 years as the common analysis period. The present worth for project B1 is _____ (Round to one decimal place.) The present worth for project B2 is $_____thousand. (Round to one decimal place.)
B1 |
B2 |
|||||||
---|---|---|---|---|---|---|---|---|
n |
Cash Flow |
Salvage Value |
Cash Flow |
Salvage Value |
||||
0 |
−$15,000 |
— |
−$30,000 |
— |
||||
1 |
−2,500 |
7,500 |
−2,700 |
6,500 |
||||
2 |
−2,500 |
6,000 |
−2,700 |
3,500 |
||||
3 |
−2,500 |
3,000 |
−2,700 |
1,000 |
||||
4 |
−2,500 |
1,000 |
— |
— |
||||
5 |
−2,500 |
500 |
— |
— |
SOLUTION:
a.
Analysis period = 15 yrs
NPW of B1 = -15000 -2500*(P/A,12%,15) - (15000-500)*((P/F,12%,5) + (P/F,12%,10)) + 500*(P/F,12%,15)
= -15000 -2500*6.810864 - (14500)*(0.567427 + 0.321973) + 500*0.182696
= (-15000-17027.16) - 12896.3 + 91.348
= -32027.16 -12896.3 + 91.348
= -44832.112 ~ -44.0 thousand
NPW of B2 = -30000 -2700*(P/A,12%,15) - (30000-1000)*((P/F,12%,3) + (P/F,12%,6) +(P/F,12%,9) + (P/F,12%,12)) + 1000*(P/F,12%,15)
= -30000 -2700*6.810864 - (30000-1000)*(0.711780 + 0.506631 + 0.360610 + 0.256675) + 1000*0.182696
= (-30000-18389.3328) - (29000)*(1.835696) + 182.696
= -48389.3328 -53235.184 + 182.696
= -101448.8208 ~ -101.4 thousand
As present cost of B1 is less, it should be selected
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