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.
B1
B2
n Cash Flow Salvage Value Cash
Flow Salvage Value
0 -$23,000 -
-$24,000 -
1 -2,300 11,500
-1,600 8,000
2 -2,300 9,500
-1,600 5,500
3 -2,300 6,500
-1,600 3,500
4 -2,300 4,500 -
-
5 -2,300 4,000 -
-
Click the icon to view the additional data about the mutually exclusive projects
Click the icon to view the interest factors for discrete compounding when
MARR=11%
per year.
(a) Assuming an infinite planning horizon, which project is a better choice at
MARR=11%?
Use 15 years as the common analysis period.The present worth for project B1 is
$________
thousand. (Round to one decimal place.)
n | B1 | B2 | ||
Cash Flow | Salvage Value | Cash Flow | Salvage Value | |
0 | -23000 | -- | -24000 | -- |
1 | -2300 | 11500 | -1600 | 8000 |
2 | -2300 | 9500 | -1600 | 5500 |
3 | -2300 | 6500 | -1600 | 3500 |
4 | -2300 | 4500 | -- | |
5 | -2300 | 4000 | -- |
Here the present worth analysis period is 15 years. The present worth of B1 can be calculated as follows
Present Worth of B1 = -23000 -2300(P/A, 11%,15) - 23,000(P/F, 11%,5) +4000(P/F, 11%,5) -23,000(P/F, 11%, 10) + 4000(P/F, 11%, 10) + 4000(P/F, 11%, 15)
= -23,000 -2300(P/A,11%,15) -19,000[(P/F,11%,5)+19,000(P/F,11%,10)]+4000(P/F,11%,15)
=-23,000 - 2300X7.19086 -19,000X(0.1606x5.8892) +4000X0.2090
=-56673.28
Present worth of B2 = -24,000-1600(P/A, 11%,15)-(24000-3500)(A/F,11%,3)(P/A,11%,12)+3500(P/F,11%,15)
= -24,000 -1600X7.19086 -20500X(0.2992x6.49235) +3500X0.209
=-74595.35
B1 has less present worth cost compared to B2, so B1 is abetter choice.
**Hope this helped. Please Upvote**
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