Lowell D. Outland
IET310
Engineering Economic Analysis
IET310 Engineering Economic Analysis Fall 2011
Homework #4
1) Consider the accompanying cash flow diagram. Compare the equivalent annual worth at i=10%
(12 years). (10 points)
Using the formula 𝑃 = 𝐹(1 + 𝑖)𝑛
We Have
=100(1+.10)1+150(1+.10)2+150(1+.10)3+200(1+.10)4+200(1+.10)5+200(1+.10)6100(1+.10)7+150(1
+.10)8+150(1+.10)9+200(1+.10)10+200(1+.10)11+200(1+.10)12
=100(1.1)+150(1.21)+150(1.331)+200(1.4641)+200(1.6105)+200(1.7716)+100(1.9487)+150(2.14
36)+150(2.3579)+200(2.5937)+200(2.8531)+200(3.1384)
=110.00+181.50+199.65+292.82+322.10+354.32+194.87+321.54+353.68+518.74+570.62+627.6
8
= 4047.52(0.31863)
=1289.66
2) Calculate the equivalent annual worth of the following scheduled payments at an interest rate
of 12% (6 years). (10 points)
𝑖
Using the formulas 𝐴 = 𝐹[(1+𝑖)𝑛 −1], and 𝑃 = 𝐹(1 + 𝑖)−𝑛 I arrived at the following:
EUAB = 2000(1.0) + 3000(.4717) + 4000(.29635) + 5000(.20923) + 6000(.15741) + 7000(.12323)
=2000 + 1415.09 + 1185.39 + 1046.17 + 944.46 + 862.58
= 7453.69
EUAC = 4000(1+.12)-2 + 4000(1+.12)-4 + 1000(1+.12)-6
= 3185.78 + 2542.07 + 506.63
=6237.47
EUAW = EUAB-EUAC
=7453.69 – 6237.47
= 1216.22
Lowell D. Outland
IET310
Engineering Economic Analysis
3) Linda O’Shay deposited $30,000 in a savings account as a perpetual trust. She believes the
account will earn 7% annual interest during the first 10 years and 5% interest thereafter. The
trust is to provide a uniform end of year scholarship at the university. What uniform amount
could be used for the student scholarship each year, beginning at the end of the first year and
continuing forever? (10 points)
I used the formula F=P(1+i)n
F= 30,000(1+.07)10 = 59,014.54
+ 59014.54(1 + .05)∞ = 61965.26
61965.26 – 59014.54
=2950.72
4) A suburban taxi company is considering buying taxis with diesel engines instead of gasoline
engines. The cars average 50,000 km a year. Use an annual cash flow analysis to determine the
more economical choice if interest is 6%. (10 points)
Gasoline Vehicle
19000
2985.71
6000
Cost
yearly expenses
salvage value
Diesel Vehicle
24000
2871.43
4000
Gasoline vehicle
19000 + 2985.71(1+.06)4 – 6000(1+.06)-4
= 22769.39 – 4752.56
= 18016.82
Diesel Vehicle
24000 + 2871.43(1+.06)5 – 4000(1+.06)-5
= 27842.62 - 2989.03
= 24853.58
The most economical vehicle would be the gasoline vehicle; it would be $466.51 cheaper to
operate per year.
Lowell D. Outland
IET310
Engineering Economic Analysis
5) An industrial firm is considering purchasing several programmable controllers and automating
the company’s manufacturing operations. It is estimated that the equipment will initially cost
$120,000 and the labor to install it will cost $25,000. A service contract to maintain the
equipment will cost $5000 per year. Trained service personnel will have to be hired at an annual
salary of $50,000. Also estimated is an approximate $10,000 annual income tax savings(cash
inflow). How much will this investment in equipment and services have to increase the annual
Lowell D. Outland
IET310
Engineering Economic Analysis
revenue after taxes in order to break even? The equipment is estimated to have an operating
life of 10 years with no salvage value because of obsolescence. The firms MARR is 12%.
The initial cost of equipment and installing it will be $145,000, then there will be an additional
cost of $55,000 per year for operating expenses. There is also an annual savings of $10.000.
𝑖(1+𝑖)𝑛
Using the formulas 𝐴 = 𝑃[(1+𝑖)𝑛 −1] and 𝑃 (1 + 𝑖)𝑛
I arrived at the following:
.12(1+.12)10
145000([(1+.12)10 −1]) - 55000(1 + .12)10 + 10000(1 + .12)10
= -165425.27
The company must make $165425.27 annually to break even.