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.
38

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*4.
The capitalized costs (CC) of the ditch-and-tunnel system (dt) and the wood-and-concrete
flume (wc) are:
CC
dt
= 500 000 + 4000 / 0.06 = $566 667
CC
wc
= 240 000 + 12 000 / 0.06 + [100 000 (A/F,6%,15)] / 0.06
= 240 000 + 200 000 + 100 000 (0.0430) / 0.06 = $511 667
The wood-and-concrete flume is preferred
because it has the lower capitalized cost.
5.
The lump-sum payment is equal to the present worth equivalent (PW) at time 3 of the six
annual payments (due at the end of years 5 to 10).
PW = 150 000 (P/A,7%,6) (P/F,7%,1)
= 150 000 (4.7665) (0.9346) = $668 216
A lump-sum payment of $668 216 is required.
6.
The discount rate used to compare the alternatives is 12%, the minimum acceptable return
on investment.
Cost of repairing machine today:
$2000
Present worth equivalent (PW) of opportunity cost (i.e. spending more on operating expenses
over the 5-year period if the machine is not repaired today):
PW = [200 + 200 (A/G,12%,5)] (P/A,12%,5)
= [200 + 200 (1.7746)] 3.6048 = $2000
The alternatives are equivalent
because they have equal present worth equivalents.
7.

This is the end of the preview.
Sign up
to
access the rest of the document.