RYERSON UNIVERSITY
ECN 801 PRINCIPLES OF ENGINEERING ECONOMICS
INSTRUCTOR: TSOGBADRAL GALAABAATAR
TIME: 2 HOURS
Instructions:
1. Choose the option closest to your answer and record the letter you have chosen on the scanner
sheet. Use
N
(1)1+−i
PA N
ii(1)
N
(1) 1i iN
Linear Gradient Series Present Worth: PG 2 N
ii(1)
NN
1(1 )(1)gi
Geometric Gradient Series Present Worth:PA1 if gi≠
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1. Suppose 3 years ago, you paid $950 for a bond that has a par value of $1,000. The bond interest
is paid annually and annual coupon rate is 4.5%. If you want to have 5% yield on your
investment, what should be your selling price today?
a 854.34 b 878.31 c 1,135.00 d 921.45 e 957.93
The annual interest payment is $1000×0.045=$45.00.
Solving equation 950=45(P/A,5%,3)+X(P/F,5%,3) for X, we get X=$957.93.
2. Suppose 5 years ago, you purchased a corporate bond with 10-year maturity and a $1,000 par
value. The coupon rate is 8% (compounded semi-annually) and bond interest is paid semi-
annually. (You received 10 interest payments till now). Suppose the current market interest rate
is 10% (compounded semi-annually). What is the current market value of the bond?
a 1,400.00 b 1,208.19 c 922.77 d 989.14 e 1,059.67
Semi-annual effective interest rate is 4%. Semi-annual interest payment is $1000×0.04=$40.00.
Current market semi-annual (effective) interest rate is 5%. Also, there are 5 (=10-5) years left till
bond matures, which means you’ll receive 10 interest payments. The current market value is
40(P/A,5%,10)+1000(P/F,5%,10)=922.77.
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3. Net cash flows (An) and project balance of an investment with a service life of 3 years is given in
the following table. Two cash flow amounts are not given.
n Net Cash Flow (An) Project Balance
0 -4,000 -4,000
1 ( ) -1,500
2 2,495 920
3 ( ) 2,000
Using the given numbers, find the interest rate used in computing the project balance.
a 20% b 25% c 5% d 10% e 15%
Using period 1 and 2, we have the following equation -1,500(1+i)+2,495=920
Solving for I, we have i=0.05.
4. Net cash flows (An) and project balance of an investment with a service life of 4 years is given in
the following table.
n Net cash flow (An) Project Balance
0 -20,000 -20,000
1 5,000 ( )
2 6,000 -12,700
3 10,000 ( )
4 ( ) 10,000
Find the net cash flow in period 4 (A4).
a 15,127 b 13,280 c 12,986 d 14,367 e 12,700
The project balance at the end of period 1 is -20,000(1+i)+5,000. Using this value, the project
balance at the end period 2 can be expressed as (-20,000(1+i)+5,000)×(1+i). We know this
should be equal to -12,700. Denote 1+i by x. Then we have the following equation for x
(-20,000x+5,000)x=-12700.
Solving this quadratic equation gives us x1=1.1 and x2=-0.85. The corresponding interest rates
are i1=1.1-1=0.1 or i1=10% and i2=-0.85-1=-1.85 which can not hold true. Thus, the interest rate
used in this calculation is 10%.
We can calculate the project balance at the end of period 3. It is -12,700×1.1+10,000=-3,970.
Now we can solve for A4. -3,970×1.1+ A4=10,000. Thus, A4=14,367.
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5. Which of the following statement is correct for an investment with one time cash outflow (initial
investment A0<0) and cash inflows starting period 1? (Assume interest rate is positive.)
a) The discounted payback period is equal to the payback period.
b) The discounted payback period is larger or equal to the payback period.
c) The discounted payback period is smaller or equal to the payback period.
d) The discounted payback period is larger than the payback period.
e) The discounted payback period is smaller than the payback period.\
The correct answer is b).
n
For any An>0, An/(1+i) 5,000. Thus, the discounted payback period is 5.
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7. Cash flows of two independent projects are given below.
n Project A Project B
0 -5,000 -6,000
1 1,000 600
2 1,000 900
3 1,000 1,200
4 1,000 1,500
5 1,000 1,800
6 1,000 2,100
Based on Present Worth criteria, what can we say about acceptability of these projects given
i=5%.
a Accept both A and B b Accept A and reject B c Accept B and reject A d Reject both A and B
The present worth of the first project is -5,000+1,000(P/A,5%,6)=75.70>0.
The present worth of the second project is -6,000+600(P/A,5%,6)+300(P/G,5%,6)=635.82>0.
We should accept both (independent) projects.
8. Consider the following project at interest rate 5%. Choose the correct statement about the
acceptability of the project.
n Cash flow
0 -1,300
1 700
2 50
3 100
4 200
5 400
6 800
a) The present worth of this project is a positive number; accept the project.
b) The present worth of this project is a positive number; reject the project.
c) The present worth of this project is a negative number; reject the project.
d) The present worth of this project is a negative number; accept the project.
We see a pattern of geometric gradient series starting period 2. The present worth is
5
700 1(2/1.05)−
−++× × = >1,300 50 (/,5%,1)573.330PF
1.05 0.051−
Thus, the correct answer is a)
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9. A company opened a bank savings account with i=10% for the purpose of donation. Starting
next year, the company wants to donate $10,000 to a local church each year. Suppose company
never goes bankrupt and they want to continue this donation forever. How much should they
deposit today if they want to cover their donation by yearly withdrawal from the account?
a 200,000 b 100,000 c 150,000 d 110,000 e 120,000
10,000/0.1=100,000.
10. Consider an investment in which a cash flow pattern that repeats itself every 4 years forever.
Find the capitalized equivalent amount at interest rate 5%.
250 250
150
150
100 100 100 100 100 100
…….
0 1 2 3 4 5 6 7 8 9 10
a 2,109.67 b 1,897.80 c 2,654.78 d 3,165.78 e 2,939.61
The annual equivalent worth of the first phase (first 4 periods) is
10050(/,5%,3)150(/,5%,4)(/,5%,4)146.98++ ×=( PF PF AP)
The capitalized equivalent amount is 146.98/0.05=2,939.61
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11. Find the IRR (internal rate of return) of the following investment.
n Net Cash Flows ( An)
0 -10,000
1 6,500
2 5,750
a 15% b 25% c 35% d 10% e 20%
6,5005,750
−++=10,000 2 0
1 (1)++ii
2
Let x=1/(1+i). Then above equation becomes 5,750x +6,500x-10,000. Solving for x, we get
X1=20/23 and x2=-2. The corresponding interest rates are i1=(1-x1)/x1=3/20=0.15 and
i2=(1-x2)/x2=-1.5. Since I should be greater than -1, we have interest rate of 15%.
12. For the investment project described in the previous problem (Problem 11), if MARR=20%, what
would the company do based on IRR criteria?
a) The company should accept the project.
b) The company should reject the project.
c) The company is indifferent about accepting and rejecting.
MARR=20%>15%=IRR, hence the project should be rejected. The correct answer is B.
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13. Find the IRR of the following project.
n Net Cash Flows (An)
0 -10,000
1 0
2 0
3 0
4 15,000
a 12% b 10.67% c 11.54% d 9.73% e 12.83%
4 4
PW(i)=-10,000+15,000/(1+i)=0. Thus (1+i)=1.5. Hence, i=0.1067.
14. Consider the following project. Evaluate the present worth of the project at interest rates 10%
and 20%. Then using these evaluations, find the approximate IRR using linear interpolation
method.
n Net Cash Flow (An)
0 -5,500
1 2,000
2 1,000
3 3,000
4 2,000
a 18.24% b 17.95% c 16.36% d 13.27% e 14.52%
PW(10%)=764.60>0 and PW(20%)=-438.27<0. By linear interpolation method,
x−10764.60
= = ⇒=1.7446 16.36%x
20 438.27−x
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15. Suppose the purchase price of a machine with service life of 5 years is $5,000. If the capital
recovery cost is $1,000, find salvage value given i=5%.
a 855.76 b 943.71 c 1,130.50 d 0 e 420.40
(5,000(/,5%,5)(/,5%,5)1,000−×=SPF AP)
S =− ×=(5,0001,000(/,5%,5)(/,5%,5)855.759PA FP)
16. A company has purchased a machine for $10,000. The machine has a useful life of 3 years and
has a salvage value of $2,000. By using this machine, the company will save $4,000 during the
first year, $6,000 during the second and $4,000 during the third year. If the machine is expected
to be used 3,000 hours during the first year, 2,500 hours during the second and 2,500 hours
during the third year. If the interest rate is 5%, find the equivalent net saving per machine hour.
a $0.52 b $0.61 c $0.73 d $0.97 e $0.23
The present worth of project PW=-10,000+4,000(P/F,5%,1)+ 6,000(P/F,5%,2)
+6,000(P/F,5%,3)=4,434.40
Let c is the net saving per machine hour. Then, the project present worth is
PW=3,000c(P/F,5%,1)+ 2,500c(P/F,5%,2) +2,500c(P/F,5%,3)=7,284.2c.
Hence, 7,284.2c=4,434.40 and c=0.6087
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17. A soft drink company has to decide whether to make bottles in house or to buy. If they choose
to buy, the can buy at market price $0.10 per bottle. They need 200,000 bottles each year. If
they choose make-in-house option, they need to buy a machine that costs $100,000 with service
life of 20 years and a salvage value of $5,000. The maintenance cost of the machine is $4,000
per year. All other additional cost of the make-in-house option is $20,000 per year. What should
the company do? Use i=5%.
a) Choose “buy” option.
b) Choose “make” option.
c) The company is indifferent between two options.
The correct answer is A (choose “buy” option). If we calculate only additional cost of the make-
in house option per bottle, it equals 20,000/200,000=$0.10 per bottle, which is the market price
of bottle. Since there are initial purchase price and maintenance cost for make-in-house option,
it is much expensive than “buy” option.
18. Consider mutually exclusive projects. Based on Present Worth criteria, which project should be
chosen if i=5%.
Net Cash Flows
n Project A Project B Project C Project D
0 -1,000 -1,000 -1,000 -1,000
1 900 -600 1,000 850
2 600 2,200 550 850
a Project A b Project B c Project C d Project D
PWA=401.36, PWB=423.96, PWC=451.25 and PWD=580.49. Project D should be selected since it
results the highest PW.
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19. Based on the Present Worth criteria, a company is indifferent between two mutually exclusive
projects. If i=5%, find the value of X.
n Project A Project B
0 -2,000 -1,500
1 1200 -1,000
2 1200 X
3 1200 2,000
a 2,078.40 b 2,314.15 c 2,196.96 d 3,067.74 e 2,403.12
The present worth of project A, PWA=-2000+1200(P/A,5%,3)=1267.84. The present worth of
project B, PWB=-724.8+X(P/F,5%,2).
PWPWAB=⇒=−+1,267.84 724.8 (/,5%,2)XPF
X FP=1,992.64(/,5%,2)2,196.88=
20. A company is considering two projects with 3 years of service life. The first project needs the
initial investment of $150,000 and an annual maintenance cost of $20,000. The second project
needs the initial investment of $200,000. If the incremental rate of return (the rate of return of
the incremental investment) is 5%, find the annual operating cost for the second project.
a 5,904 b 1,639 c 12,340 d 3,427 e 1,277
Suppose the annual operating cost of project II is X. Then the incremental investment II-I is:
n Project II-I
0 -50,000
1 20,000-X
2 20,000-X
3 20,000-X
Since the incremental rate of return is 5%, PWII-I(5%)=-50,000+(20,000-X)(P/A,5%,3)=0. Solving
for X, we get X=1,640.
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21. Cash flows of two mutually exclusive projects are given. If the IRR of the incremental
investment is 10%, find X.
n Project A Project B
0 -3,000 -3,500
1 1,500 1,500
2 1,200 X
3 1,300 1,300
a 2,000 b 1,675 c 1,920 d 1,745 e 1,805
The incremental investment, B-A, is:
n Project B-A
0 -500
1 0
2 X-1,200
3 0
2
PWB-A(10%)=-500+(X-1,200)(P/F,10%,2)=0. Thus, X=500(1.1 )+1,200=1805.
22. Let A and B be two mutually exclusive projects. Suppose the present worth of incremental
investment B-A is a positive number. Then,
a) The present worth of A is greater than the present worth of B.
b) The present worth of B is greater than the present worth of A.
c) The present worth of A is equal to the present worth of B.
Since PWB-A(i)=PWB(i)-PWA(i), PWB-A(i)>0 implies PWB(i)>PWA(i). The correct answer is B.
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23. A company is considering three projects: Project A1, A2 and A3. The internal rate return for
these projects are IRRA1=17.62%, IRRA2=16.60% and IRRA3=15.99%. The incremental rate of
returns for incremental investments A2-A1 and A3-A2 are 14.16% and 6.66% respectively.
If “do nothing” is an available choice and MARR=16.30%, which project would be chosen?
a) A1
b) A2
c) A3
d) Not enough information is given to answer the question.
If the company chooses “do nothing”, they’ll continue earning MARR rate. Thus, since
IRRA3AEWA. Project B should be selected.
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25. Suppose total annual equivalent cost (AEC) is a function of a (engineering) design variable (A) as
follows:
320
AEC A= +20
A
If you want to minimize the AEC, what is the optimal value for A?
a 1 b 2 c 3 d 4 e 5
320 320
AEC A=+≥ =20 220 160A
AA
The equality holds when 20A=320/A or A=4. Thus, AEC is minimized at A=4.
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