Majmaah University
GE 407 Engineering Economics
College of Engineering
Assignment 2
Due Date: April 23, 2015
Q1. A machine is engaged in production of forged parts. The purchase price of machine is
100,000 SR. The annual maintenance and operating costs are 50,000 SR. The salvage value
is 25,000 SR at the end of machine life. Life of machine is 5 years.
The rate of interest = 8%.
Calculate the cost per part due to this machine.
i)
ii)
A company will produce 10,000 parts per year with this machine.
Also what is the cost per part if machine makes 8000 parts per year.
Sol: Equivalent Uniform Annual worth (EUAW)
EUAW = -100,000(A/P, 8%, 5)-50,000+ 25000(A/F, 8%,5)
= - 100,000x 0.25046- 50000+ 25000x 0.17046
= - 70784.5 SR
Annual cost = 70784.5 SR
Ans i) cost per product = 70784.5 /10000= 7.078 SR
Ii) Cost per product = 70784.5 /8000 = 8.848 SR
Q2. A person is planning a new business. The initial investment and cash flows for the new
business are as listed below. The expected life of the business is five years. The interest rate
is 10%. Should the person go for the business? If yes why ?
and find the IRR value that the business will yield.
Period
(year)
Cash flow
(SR)
0
1
2
3
4
5
–1,00,000
30,000
30,000
30,000
30,000
30,000
Key: PW = -100,000 + 30,000(P/A,10%,5)
= -100000+ 30000*3.791
= 13,730 SR
The business makes positive present worth at discounted interest rate 10%.
Thus there exist better IRR.
When i = 15%,
PW(15%) = –1,00,000 + 30,000(P/A, 15%, 5)
= –1,00,000 + 30,000(3.3522)
= Rs. 566.
When i = 18%,
PW(18%) = –1,00,000 + 30,000(P/A, 18%, 5)
= –1,00,000 + 30,000(3.1272)
= Rs. – 6,184
Hence 𝑖 = 15 +
566∗3
566+6184
= 15.21555 %
Q3. Two mutually exclusive alternative projects with cash flow are shown below for three
years. Which project is better and why? The minimum attractive rate of return (MARR) is
10%.
End of year
0
1
2
3
3
Alternative
Project C
Project D
-380000
-415000
-38100
-27400
-39100
-27400
-40100
-27400
0
26000
Solution:
End of year
0
1
2
3
3
Alternative
Project C
Project D
-380000
-415000
-38100
-27400
-39100
-27400
-40100
-27400
0
26000
∆ (D-C)
-35000
10700
11700
12700
26000
At irr = 10%
PW(∆(D-C)= -35000+10700(P/A, irr,3)+1000(P/G,irr,3)+ 26000(P/F, irr, 3)
=-35000+10700*2.4869+1000*2.329+ 26000*.7513
= 13,472 which is positive
hence irr can be even more to bring
the ∆ PW(irr) = 0
Hence Project D is selected.
Q4. Two die making machines DMA and DMB can be used for tool room. Select the most
economical machine. Assume MARR = 21% per year
First cost ($)
Annual cost, ($
per year)
Salvage value, $
Life, years
DMA
-50,000
-100,000
DMB
-95,000
-85,000
5000
3
11000
6
Solution
DMA
-50,000
DMB
-95,000
DMB-DMA
-45000
Annual cost, ($
per year) year 1
-100,000
-85,000
15000
Year 2
-100,000
-85,000
15000
Year 3
-100,000
-85,000
15000
Year 3
(-50,000 +
5000)
0
45000
Year 4
-100,000
-85,000
15000
Year 5
-100,000
-85,000
15000
Year 6
-100,000
-85,000
15000
Year 6
5000
(Salvage
value)
11000
6000
First cost ($)
PW(i) = -45000+ 15000(P/A, i, 6) + 45000(P/F, i, 3) +6000(P/F,i,6)
At I =22%
PW(i= 22) = -45000+15000* 3.1669+ 45000*.5507 +6000*0.3033
= 29104
I =20%
PW(i) = -45000+ 15000(P/A, i, 6) + 45000(P/F, i, 3) +6000(P/F,i,6)
= -45000+15000* 3.3255+ 45000*.5787+6000*0.3349
= 32933.3
DMB is preferable. At i@20 PW(∆ i) is positive. So I is above MARR.
DMB should be selected.