M I M E
3 1 0
E N G I N E E R I N G
E C O N O M Y
SOLUTIONS TO PROBLEM SET #6 – PROJECT EVALUATION CRITERIA
1.
The present worth cost equivalent (PW) of each machine is determined at a discount rate
of 10 percent.
Under the repeated-projects assumption, the machines must be compared over a
period of 18 years, the smallest common multiple of their lives.
PW
A
= 9000 [1 + (P/F,10%,6) + (P/F,10%,12)] + 5000 (P/A,10%,18)
= 9000 [1 + 0.5645 + 0.3186] + 5000 (8.2014) = $57 955
PW
B
= 16 000 + (16000 - 3000) (P/F,10%,9) - 3000 (P/F,10%,18) + 4000 (P/A,10%,18)
= 16 000 + 13 000 (0.4241) - 3000 (0.1799) + 4000 (8.2014) = $53 779
Machine B is preferred
because it has the lower present worth cost equivalent.
2.
The maximum bid is the present worth equivalent (PW) of benefits over the expected
period of ownership.
The required return on investment is used as the discount rate.
PW = (15 000 - 9000) (P/A,9%,12) + 150 000 (P/F,9%,12)
= 6000 (7.1607) + 150 000 (0.3555) = $96 289
The bid should not exceed $96 289.
3.
Since the study period is 50 years, the galvanized building (g) must be replaced after the
first 30 years.
The reinforced-concrete building (c) and the first galvanized building have no
salvage value.
The second galvanized structure, which has a remaining useful life of 10 years at
the end of the 50-year study period, has a salvage value of:
310 000 (1/3) = 103 333
Present worth cost equivalent (PW) of each alternative:
PW
c
= 440 000 + [19 200 + 440 000 (0.012)] (P/A,12%,50)
= 440 000 + 24 480 (8.3045) = $643 294
PW
g
= 310 000 [1 + (P/F,12%,30)] + [30 600 + 310 000 (0.005)] (P/A,12%,50) -
103 333 (P/F,12%,50)
= 310 000 [1 + 0.0334] + 32 150 (8.3045) - 103 333 (0.00346) = $586 986
The galvanized building is preferred
because it has the lower present worth cost equivalent.
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