Total Score
/40
1. The inflation rate over a 10-year period for an item that now costs $1000 in shown in
the following table. (20 points)
Year
1
2
3
4
5
6
7
8
9
10
Inflation rate
10%
0%
10%
0%
10%
0%
10%
0%
10%
0%
a) What will be the cost at the end of ye

Total Score
/60
1. Using AW analysis with LCM approach, determine which alternative is more
economical. Suppose that MARR is 10%/year. (20 points)
Alternative #1
Alternative #2
Initial cost
-400,000
-250,000
AOC
-4,000
-3,000
Salvage value
400,000
250,000

1. An aggie engineer has received draft cost and revenue estimates for a new exhibit
and convention center. She has asked you to perform a capitalized cost analysis at
6% per year. Here is the plan. (25 points)
Initial costs of $40 million with an expansi

Total Score
/40
1. The figure 1 shows the general cash flow diagram having geometric gradient. All
formula and table values with geometric gradient are driven from this cash flow.
Figure 1
Figure 2
Now, I want to acquire present value (in 0 year) with the

ISEN 302-502
Economic Analysis of Engineering Projects
ISEN 302-502 : Quiz #2
Name :
UIN :
E-mail:
1. The figure 1 shows the general cash flow diagram having arithmetic
gradient. All formula and table values with arithmetic gradient are driven
from this c

Total Score
/50
1. Curry borrowed $ 20,000 from his senior for his project. After 3 years, he wants to
pay it back. Suppose that interest rate is 1% per month. (30 points)
a) Draw the cash flow diagram? (10 points)
b) How much curry pay back after 3 years

NAME:_
ISEN 302
[Problem 1] - (20 points)
Tyler borrows $5000 loan. His repay it in 3 years at 10% interest per year. His payment
plan is Equal payments of compounded interest and principal made annually.
1) Fill out brackets. (10 points)
End of
year
0
1

Homework #6 Solution
1. [4.9-10 points]
Subscripts are C for contract service and B for Burling Coop installed.
PWC = -75,000(P/A,6%,3) 100,000(P/A,6%,2)(P/F,6%,3)
= -75,000(2.6730) 100,000(1.8334)(0.8396)
= $- 354,407
PWB = -150,000 60,000(P/A,6%,5)
= $-

Homework #4 Solution
1. [3.32-10 points]
Given interest rate = 12% per year compounded quarterly.
If we acquire effective interest rate per a quarter, it belongs to the case 1.
Given rate
TP
CP
M
12%
1 year
1/4year
TP/CP = 4
So, we can acquire the effecti

ISEN 302-502
Economic Analysis of Engineering Projects
Homework #3 Solution
1. [2.40 - 10 points]
Case : i g
n
P A1
1 i
First find Pg and then convert to F
Pg = 8000[10/(1 + 0.10)]
= $72,727
F = 72,727(F/P,10%,10)
= 72,727(2.5937)
= $188,632
2. [2.43 -

Homework #2 Solution
1. [Derive formulas - 20 points]
(1 i) n 1
(a) P A
n
i(1 i)
From the present worth equation in single payment case: P F
1
,
(1 i)n
We can formulate that
1
1
1
1
A
A
A
2
3
(1 i)
(1 i)
(1 i)
(1 i)n
A
1
1
1
P
1 (1 i) (1 i)2 (1 i)n1