Solution Term 2 [ GE 407 ]
1. [4 Points] A manufacturer is planning to purchase a press machine. There are two makes available
in the market. The relevant values are:
Description
Purchase price, SR
Annual savings
Salvage value, SR
Useful life, years
Machine A
15,000
SR 5,000 for Year 1 to
Year 4,
SR 8,000 for Year 5 and 6
2,000
6
Machine B
11,000
SR 1,000 for first year and
increasing by SR 1,000 per year
until the end of its useful life
4,000
6
The interest rate is 15% compounded annually. Use the annual cash flow analysis to determine the
better machine.
Year
0
1
2
3
4
5
6
Machine A
-15,000
5,000
5,000
5,000
5,000
8,000
8,000+2,000
Machine B
-11,000
1,000
2,000
3,000
4,000
5,000
6,000+4,000
𝐸𝑈𝐴𝑊𝐴 = {−15,000 + 5,000(𝑃/𝐴, 15%, 4) + 8,000(𝑃/𝐴, 15%, 2)(𝑃/𝐹, 15%, 4)
+ 2,000(𝑃/𝐹, 15%, 6)}(𝐴/𝑃, 15%, 6)
= {−15,000 + 5,000(2.855) + 8,000(1.626)(0.5718) + 2,000(0.4323)}(0.2642)
= SR 2,001.995156
𝐸𝑈𝐴𝑊𝐵 = {−11,000 + 1,000(𝑃/𝐴, 15%, 6) + 1,000(𝑃/𝐺, 15%, 6)
+ 4,000(𝑃/𝐹, 15%, 6)}(𝐴/𝑃, 15%, 6)
= {−11,000 + 1,000(3.784) + 1,000(7.937) + 4,000(0.4323)}(0.2642)
= SR 647.34284
Since EUAWA > EUAWB, Machine A must be selected.
2. Two alternative mini power plants for in-site power generation are being considered. Which
project is better and why? The minimum attractive rate of return is 10%.
Alternatives
End of year
0
Project C
-150,000
Project D
-105,000
1
2
38,000
39,000
27,000
27,000
3
4
40,000
41,000
27,000
27,000
End of year
0
1
2
3
4
Difference
-45,000
11,000
12,000
13,000
14,000
𝑃Difference (𝑖) = −45,000 + 11,000(𝑃/𝐴, 𝑖, 4) + 1,000(𝑃/𝐺, 𝑖, 4)
Try 𝑖 = 5%, 𝑃Difference (5%) = −45,000 + 11,000(3.546) + 1,000(5.103) = −891 < 0
Try 𝑖 = 4%, 𝑃Difference (5%) = −45,000 + 11,000(3.630) + 1,000(5.267) = 197 > 0
Hence, 4% < ∆IRR < 5%.
Since MARR > ∆IRR, we should accept the project with the lower initial cost, which is Project D.
3. The three estimates, namely: optimistic, most likely and pessimistic values, for a project are given
below.
Optimistic value
Initial cost, SR
Benefits at the end of 1st year, SR
Benefits at the end of 2nd year, SR
Benefits at the end of 3ed year, SR
Benefits at the end of 4th year, SR
Annual maintenance cost, SR
Salvage value at the end of 4th year,
SR
16,000
6,000
14,000
15,000
16,000
2,000
20,000
Most likely
value
24,000
6,000
7,000
8,000
9,000
3,000
16,000
Pessimistic
value
30,000
1,200
6,000
7,000
8,000
4,000
12,000
If the interest rate is 10%, should the project be accepted? Use the present worth analysis.
Mean Value
Initial cost
23,666.67
Benefits at the end of 1st year
5200
Benefits at the end of 2nd year
8000
Benefits at the end of 3ed year
9000
Benefits at the end of 4th year
10000
Annual maintenance cost
3000
Salvage value at the end of 4th year
16000
𝑃 = −23,666.67 + (5200 − 3000)(𝑃/𝐹, 10%, 1) + (8000 − 3000)(𝑃/𝐹, 10%, 2) + (9000
− 3000)(𝑃/𝐹, 10%, 3) + (10000 − 3000)(𝑃/𝐹, 10%, 4) + 16000(𝑃/𝐹, 10%, 4)
= −23,666.67 + (5200 − 3000)(0.9091) + (8000 − 3000)(0.8264) + (9000 − 3000)(0.7513)
+ (10000 − 3000)(0.6830) + 16000(0.6830)
= SR 2,682.15
Since 𝑃 > 0, the project can be accepted.
4. A robot is considered for installation at a plant at a cost of SR 81,000 with no salvage value. The
company uses 𝑖 = 6%. Annual savings and useful life are random variables with the following
estimates:
Annual savings, SR
Probability
18,000
0.2
Useful life, years
Probability
20,000
0.7
5
0.8
22,000
0.1
4
0.2
Calculate the expected present worth.
Write the answer
PW= -81000+18000(P/A,6%,5) = -81000+18000* 4.2124= 5176.8
PW= -81000+18000(P/A,6%,4) = -81000+18000*3.4651= 18628.2
The Present worth of all combinations is given below:
Annual
benefits
18,000
20,000
22,000
18,000
20,000
22,000
Probability Life
Probability
PW
.2
.7
.1
.2
.7
.1
.8
.8
.8
.2
.2
.2
-5176.8
3248
11672.8
-18628.2
-11698
-4767.8
5
5
5
4
4
4
Joint
probability
.16
.56
.08
.04
.14
.02
Expected
present worth
Joint probability* PW
-828.29
1818.88
933.82
-745.13
-1637.72
-95.36
=
= -553.8
Expected present worth = -553.8 units
5. The initial cost of a project is SR 25 million. The project will run for 10 years with no salvage
value. The annual benefits are not known to the company. If the interest rate is 9%, calculate the
range of values of the annual benefits that will make the project desirable.
𝑃 = −25 + 𝑥(𝑃/𝐴, 9%, 10) = −25 + 6.418𝑥
Set 𝑃 = 0:
−25 + 6.418𝑥 = 0
𝑥=
25
= 3.895
6.418
Try annual benefits=SR 4 million, 𝑃 = −25 + 6.418(4) = 0.672 > 0.
Hence, as long as the annual benefits are greater than SR 3.895 million, the project can be
accepted.