The Islamic University of Gaza
Industrial Engineering department
Engineering Economy, EIND 4303
Instructor: Dr. Mohammad Abuhaiba, P.E.
Fall 2011
Exam date: 16/11/2011
MidTerm Exam (Open Book)
Question
Exam Duration: 1.5 hours
Grade
Maximum Grade
1
15
2
6
3
7
4
8
5
13
6
9
7
8
8
8
Total
74
1
Question #1: (15 points)
A company has determined that the price and the monthly demand of one of its products are related by
the equation
𝐷 = √500 − 𝑝
Where p is the price per unit in dollars and D is the monthly demand. The associated fixed costs are
$1100/month, and the variable costs are $100/unit.
a. What is the optimal number of units that should be produced and sold each month?
b. What are the values of D at which the company break even?
Solution:
a) D = √500 − 𝑃 (1)
𝐷2 = 500 − 𝑃 (1)  P = 500 – D2 (1)
𝑅 = 𝑃𝐷 = 𝐷 (500 − 𝐷2 ) = 500 𝐷 − 𝐷3 (2)
𝐶𝑇 = 1100 + 100 𝐷 (2)
𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑃 = 𝑅 − 𝐶𝑇 = 500 𝐷 − 𝐷3 − 1100 − 100 𝐷
= 400 D – 𝐷3 − 1100 (2)
𝑑𝑃
= 0 = 400 − 3 𝐷2 (2)
𝑑𝐷
400
𝐷2 =
 𝐷 ≅ 12 𝑢𝑛𝑖𝑡𝑠 (1)
3
b) Company breakeven when P = 0 = 400 – 400 𝐷3 − 1100
(1)
Solve for D:
D1 = 3 units
(1)
D2 = 18 units (1)
2
Question #2: (6 points)
The future cost of a machine will be $250,000 in 10 years at an APR of 7% compounded annually.
What is the present cost?
Solution:
F = $ 250,000 (1) , N = 10 years (1) , i= 7% (1)
P=?
P = F(1+i)-n = 25000 x 1.07 -10
= $12,709
Question #3: (7 points)
What is the present worth of a series of 12 monthly payments at $300 with an 9% APR compounded
monthly?
Solution:
A = $300
i=
9
12
= 0.75
(1) , N = 12 (1) , r = 9% (1),
(1)
p=?
P = 300 (P/A, 0.75, 12) (2)
= 300 x 11.4349 = $ 3430 (1)
3
Question #4: (8 points)
Calculate the future equivalent at the end of 2011, at 8% per year, of the following series of cash flows
in the figure below. Use a uniform gradient amount G in your solution.
2008
2009
2011
2010
$700
$800
$900
$1000
F=?
Solution:
(cash flows each 1 mark )
0
1
2
3
4
800
5
700
900
1000
300
300
200
100
F
0
1
2
3
4
5
0
1
2
3
4
5
1000
F = - 1000 (F/A, 8%, 4) ( F/P,8%,1) + 100(F/G,8%,4)(F/P,8%,1) (4)
= - 1000 x 4.5061 x 1.08 + 100 x 6.3264 x 1.08
= - $4183 (1)
4
Question #5: (13 points)
Determine the present equivalent value of the cash flow diagram of the figure below when the annual
interest rate, ik, varies as indicated.
P=?
$200
$200
$100
i = 6%
0
$100
i = 6%
i = 8%
i = 4%
3
2
1
i = 8%
i = 4%
5
4
6
Solution:
P = 200(P/F,6%,1) + 100(P/F,8%,1)(P/F,6%,1) +
200(P/F,4%,1)(P/F,6%,1)(P/F,8%,1)(P/F,6%,1) +
100(P/F,8%,1)(p/f,4%,2)(p/f,6%,1)(P/F,8%,1)(P/F,6%,1)
=
200
1.06
+
100
1.08 𝑋 1.06
+
200
2
1.04 𝑋 1.06 𝑋 1.08
+
100
1.082 𝑋 1.04 2 𝑋 1.062
= $505
5
Question #6: (9 points)
A small company heats its building and spends $8000 per year on natural gas for this purpose. Cost
increases of natural gas expected to be 10% per year starting one year from now (i.e. the first cash flow
is $8000 at EOY one). Their maintenance on the gas furnace is $400 per year, and this expense is
expected to increase by 10% per year starting one year from now. If the planning horizon is 10 years,
what is the total annual equivalent expense for operating and maintaining the furnace? The interest rate
is 18% per year.
Solution:
f=10% (1) ,
i=18% (1)
The cash flow (1 mark)
P = 8400 [1- (P/F,18%, 10)(F/P,10%,10)] (each term with 1 mark)
=105,000[1- 1.18 -10 x 1.1 10] = 52, 9656 (1)
A = 52965 (A/P,18%,10) (1)
= 52965 x 0.22254 = $ 11,785 (1)
6
Question #7: (8 points)
An office supply company has purchased a light duty delivery truck for $20,000. It is anticipated that the
purchase of the truck will increase the company’s revenue by $10,000 annually, while the associated
operating expenses are expected to be $2000 per year. The truck’s market value is expected to decrease
by $3000 each year it is in service. If the company plans to keep the truck for only two years, what is the
annual worth of this investment? The MARR = 15% per year.
Solution:
Cash flow (2 marks)
MARR = 15%
A = - 20(A/P,15%,2) + 8 + 14(A/F,15%,2)
= - 20 x 0.6151 + 8 + 14 x 0.4651
(3)
(2)
= $2,209 (1)
7
Question #8: (8 points)
A computer call center is going to replace all of its incandescent lamps with more energy efficient
fluorescent lighting fixtures. The total energy savings are estimated to be $2000 per year, and the cost of
purchasing and installing the fluorescent fixtures is $5000. The study period is six years, and terminal
market values for the fixtures are negligible.
a. What is the IRR of this investment?
b. What is the simple payback period of the investment?
Solution:
a) Cash flow (1 mark)
PW = -5000 + 2000(P/A,i%,6)
 (P/A,i%,6) = 2.5
(2)
(1)
(i)
30
?
35
The interpolation (2 marks)
(2.5−2.6427)
P/A
2.6427
2.5
2.3852
I= 30+5 x (2.3852−2.6427) = 32.77%
Simple payback period =
5000
2000
(1)
≅ 3 𝑦𝑒𝑎𝑟𝑠 (1)
8