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Unformatted text preview: HomeworkAssignment#5Solutions
PU R D U E U N I V E RSI T Y
I E 343 : E n gi n e e r i n g E con om i cs
F a l l 2010
I nst r u c to r : A . C a p pon i H om e w o r k A ssign m e n t #5
A ssign e d : 24 Se p t 2010
D u e : 01 O c t 2010 Missing name on the top of each page submitted
Not stapled together 2
Not completed on the correct template 10 2 P r ob l e m 1
(15 points)
Consider a machine that cost $15,000 and has a 5 year useful life. At the end of 5 years, it can be
sold for $3,500. The machine increases the profit for the firm by $4,000 per year.
(a) Draw cash flow diagram
3,500
$4,000
0 $4,000 $4,000 $4,000
1 2 3 4 $4,000
5 $15,000
(b) Should the machine be purchased if the MARR is 10% per year and the annual worth
method is used?
We have CR(0.1) = $15,000 (A/P, 0.1,5)  $3,500 (A/F, 0.1, 5) = $3,957$573.3 =
$3,384. Thus, we get that the annual worth is
AW(0.1) = $4,000  $3,384 = $616 > 0.
Since AW(0.1) >0, we can conclude that buying the machine is a worthwhile investment. IE343 HomeworkAssignment#5Solutions Page1 HomeworkAssignment#5Solutions
P r ob l e m 2
(15 points)
The heat loss through the exterior walls of a processing plant is estimated to cost the owner
$3,000, starting from next year. A salesman from Superfiber, Inc. claims he can reduce the heat
loss by 80% with the installation of $15,000 of Superfiber now. The cost of heat loss rises by
$200 per year after next year (gradient), and the owner plans to keep the building ten more years.
(a) Draw the cash flow diagram
The savings due to heat loss in the next year is 0.8 * 3000 + 0.8 * 200.
The savings due to heat loss in two years from now is 0.8 * 3000 + 0.8 * 400.
The savings due to heat loss in three years from now is 0.8 * 3000 + 0.8 * 600.
And so on. Hence, the cash flow diagram can be decomposed as
$ 0.8 *3,000 $0.8* 3,000 $0.8*600 $0.8 * 3,000 $0.8*400 +
0 1 2 $0.8*200 10 0 $
15,000 10
1 2 3 4 (b) What is the yearly internal rate of return?
Find the IRR via linear interpolation of 12% and 15%.
The expression for the present worth is:
PW(i) = $15,000 + $0.8 * 3,000 * (P/A, i, 10) + $0.8 * 200 * (P/G, i, 10)
We have PW(12%) = $1,800.64 and PW(15%) = $237.76 If we interpolate, we find If you complete this problem in Excel, the equation is
IRR(15,000, 24000, 2560, 2720, 2880, 3040, 3200. 3360, 3520, 3680) = 14.91%
(c) Assume the MARR = 10%. Is the installation of Superfiber a worthwhile investment,
when using the IRR an evaluation criteria?
Since the internal rate of return i is higher than MARR = 10%, then installing Superfiber
is a worthwhile investment.
IE343 HomeworkAssignment#5Solutions Page2 HomeworkAssignment#5Solutions
(10 points) P r ob l e m 3
Textbook Problem 521
This project is acceptable as it results in a positive future value. P r ob l e m 4
Textbook Problem 522 $12,000 Revenue estimated annual expenses = 1,800
EOY1 = 1,300 (10 points) 500 = 1,300 for the first year. EOYi+1= EOYi +100 FW(15%) = 10,000(FP,15%,6) + 1,300(FA,15%,6) + 100(PG,15%,6)(FP,15%,6) +
12,000
FW(15%)= 10,000(2.3131) + 1,300(8.7537) + 100(7.937)(2.3131) + 12,000=$2,085.72
Since $2,085.72 > 0, the investment result in a profit, and thus is worth pursuing it. IE343 HomeworkAssignment#5Solutions Page3 HomeworkAssignment#5Solutions
(10 points) P r ob l e m 5
Textbook Problem 525
Use the AW method to evaluate this investment. The annual worth of the investment is negative, thus this is not a good investment. (15 points) P r ob l e m 6
Textbook Problem 527
C a sh O u t f low s
Investment: C a sh I n f low s $300,000 + $600,000 + $250,000 +
$100,000 = $1,250,000 (year 0) Annual revenue:
Annual expenses: $750,000 (annually)
$475,000 (annually) Market (salvage value) +
working capital: $400,000 + $350,000 + $50,000 +
$100,000 = $900,000 (year 10) The capital recovery cost is
Since the Annual Worth at MARR = 15% is equal to $70,245 which is greater than zero, then the
company should invest a plant to manufacture the proposed new product. IE343 HomeworkAssignment#5Solutions Page4 ...
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