The solutions for all assigned problems are given below.
I have also included some qualitative and quantitative questions similar to the questions in Midterm I
SampleMid1Q1. What are the main differences between NPV, IRR, and B/C?
SampleMid1Q2. What advantages are lost if the sum of the weights in a weighted scoring approach does not add to 1.0?
Why is it suggested that factors with less than 2 percent or 3 percent impact not be considered in this approach?
SampleMid1Q3. When it comes to assigning individuals to work on projects, functional managers and project managers are
often in conflict. How can the PM resolve this problem?
SampleMid1Q4. Which technique can provides almost all information provided by an interface map?
SampleMid1P1. All cash inflow/outflow in this problem occur at the end of the corresponding year. The design of a new
product is expected to take one year at a cost of 5M. There is a 0.7 probability that the project will be technically feasible. If
feasible, it can be launched with an estimated cost of $6M paid in Year 2. If launched, the useful life of the project is only
one year with no salvage value at the end of Year 3. When launched, it will be in one of the following three states: (1)
Extraordinary commercial success with probability of 0.2, earning a net after-tax cash inflow of $30M. (2) Average
commercial success with probability of 0.7, earning the net after-tax cash inflow of $15M. (3) Commercial failure earning
the net after-tax cash inflow is $5M. The discount rate is 10 percent. Compute the expected commercial value of this
project.
30M
Extraordinary
com. success
20%
15M
-6M
-5M
70%
Technically
Feasible
Invest
70%
Average com.
success
5M
10%
Commercial
failure success
Design a new
Product
Technically
Infeasible
Expected return in Year 3 if the project is technically feasible = 0.2*30+0.7*15+0.1*5
Expected return in Year 3 if the project is technically feasible = 17M
The present value of this cash flow at the end of year 2 is 17/1.1 = 15.45M
The investment at the end of Year 2 is -6M. Therefore, the net present value at the end of year 2 is 15.45 – 6 = 9.45.
But there is only 70% probability for this line. Therefore, the expected cash flow is
9.45*.7 = 6.61M. Furthermore, it is a cash flow at the end of Year 2.
The present value of this cash flow at the end of year 1 is 6.61/1.1 = 6.01M
Since the initial investment is –5, the net present value of the project at the end of year 1 is 6.01-5 = 1.01
Expected commercial value at the end of year 0 is 1.01/1.1 = 0.92
Chapter 1
Problem19.
I have solved this problem in 3 ways
1) Using statistics
2) Using Excel
3) Using a crystal ball
The analytical solution is preferred.
a) Analytical solution
The Present Value of the “Mean” of each cash flow is:
Mean I0 = -75000
Mean P1 = 20000(1.2)(-1) = 16666.7
Mean P2 = 25000(1.2)(-2) = 17361.1
Mean P3 = 30000(1.2)(-3) = 17361.1
Mean P4 = 50000(1.2)(-4) = 24112.7
The Present Value of the “Mean” of NPV is:
Mean NPV = -75000+16666.7+17361.1+17361.1+24112.7 = 501.6
The Present Value of the “Standard Deviation ” of each cash flow is:
Standard Deviation
Standard Deviation
Standard Deviation
Standard Deviation
Standard Deviation
I0 = 0
P1 = (1.2)(-1) (1000)
P2 = (1.2)(-2) (1500)
P3 = (1.2)(-3) (2000)
P4 = (1.2)(-4) (3500)
= 833.3
= 1041.7
= 1157.4
= 1687.9
The Present Value of the “Standard Deviation ” of NPV is:
Variance NPV = Variance I0 +Variance P1+ Variance P2 + Variance P3+ Variance P4
Variance NPV = 5968064
Standard Deviation of NPV = 2443
Therefore, NPV is a Normal Variable with Mean of 501.6 and Standard Deviation of 2443
Z = (0-501.6)/2443 = -0.21
If you look at the Normal table in your book, you will see that the probability associated with +0.21 is 58.3%. That
is the area to the left or +.21. Therefore the area to the right of -0.21 is the same  58.3%.
P(NPV≥0) = 58.3
If you solve the problem using Crystal Ball – see below – you will get P(NPV≥0) = 59.6.
If you solve the problem using Crystal Ball – see below – you will get P(NPV≥0) = 56.9.
The analytical solution is a rigorous reliable solution.
b) Solution to Problem 19 using random
Year
0
1
2
3
4
Distribution
Constant
N(20000,1000)
N(25000,1500)
N(30000,2000)
N(50000,3500)
Cash Flow
-75000
19277.38608
23617.89846
27145.2099
45474.30949
PV index
1
0.8333333
0.6944444
0.5787037
0.4822531
NPV
PV
-75000
16064.49
16401.32
15709.03
21930.13
-4895.03
This is the result for only one replication. Ti see the functions please double click on the Table.
The histogram of NPV for 1000 replications is shown below
Histogram
100
90
80
70
Frequency
Bin
Frequency Cumulative Cum. Freq.
-6941.4
1
-6436.4
0
1
0.001
-5931.39
3
4
0.004
-5426.39
4
8
0.008
-4921.38
5
13
0.013
-4416.38
6
19
0.019
-3911.38
11
30
0.03
-3406.37
23
53
0.053
-2901.37
28
81
0.081
-2396.36
37
118
0.118
-1891.36
47
165
0.165
-1386.36
58
223
0.223
-881.351
72
295
0.295
-376.347
62
357
0.357
0
74
431
0.431
633.6613
77
508
0.508
1138.665
87
595
0.595
1643.67
85
680
0.68
2148.674
73
753
0.753
2653.678
57
810
0.81
3158.682
49
859
0.859
3663.686
42
901
0.901
4168.69
25
926
0.926
4673.694
27
953
0.953
5178.698
19
972
0.972
5683.702
12
984
0.984
6188.706
7
991
0.991
6693.71
5
996
0.996
7198.714
1
997
0.997
7703.719
0
997
0.997
8208.723
1
998
0.998
More
2
1000
1
60
50
40
30
Frequency
20
10
0
Bin
As we can see. P(NPV≤0) = 43.1%  P(NPV≥0) = 56.9%
c) Solution to Problem 19 using Crystal Ball
o
o
Cells B5 through B8 were defined as assumption cells. Cell B5 was defined as an Assumption Cell with a
normal distribution with a mean of 20,000 and standard deviation of 1,000, and
Cell B9 was defined as a Forecast Cell.
The results of simulating this situation 1,000 times in Crystal Ball are shown below. According to the results, the
expected NPV is $589 while there is a 59.6% probability that the NPV will be positive.
The distribution of potential NPVs is generated with the simulation approach providing the decision maker with
insight into the uncertainty and risk associated with the situation. In this case, we see the NPV ranges from a low
of -$6,742 to a high of $8,935.
A
B
1 Req Rate of Return
20.0%
2
Year
Cash Flow
3
4
0
-$75,000
5
1
$20,000
6
2
$25,000
7
3
$30,000
8
4
$50,000
NPV
$502
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Problem 23.
C
D
E
Assumption Cells
Forecast Cell
F
G
Factor
Class of Clientele
Rent
Indoor Mall
Volume
Weight Normal Weight
100
28.78
90
25.90
85.5
24.60
72
20.72
347.5
100
1
fair
good
good
good
Factor
Class of Clientele
Rent
Indoor Mall
Volume
Normal Weight
28.777
25.899
24.604
20.719
1
2
3
3
3
271
Loca tion
2
3
good
poor
fair
poor
poor
good
fair
good
1
Loca tion
2
3
3
1
2
1
1
3
2
3
204
191
3
4
4
good
good
poor
poor
4
3
3
1
1
209
271
2
Chapter 3
Problem 24. Prepare an action plan using MSP with the steps that must be completed before Vern Toomey
can contact outsourcing vendors. If Vern starts on August 1, 2005, how long will it take to get ready to
contact outsourcing vendors?
Based on the information in the Gantt chart it would be Tuesday, August 23, '05 before Vern would be able to
contact vendors for the outsourcing proposals,
Chapter 4
Problem 13. As can be seen from the following Table, the number of hours needed to prepare the slides (given
an 85% learning rate) would be 50 hours in session 1, 42.5 hours in session 2, 36.1 hours in session 4, and
30.7 in session 8. This enables the total cost per session to be calculated … and thus the total for the eight
sessions.
Session
1
2
3
4
5
6
7
8
Total
Hours
50
42.5
38.6
36.1
34.3
32.8
31.7
30.7
V.C.
$5,000
$4,250
$3,865
$3,613
$3,428
$3,285
$3,168
$3,071
Fixed Cost Total Cost Cum Cost Cum Rev
$600
$5,600
$5,600
$5,172
$600
$4,850
$10,450
$10,344
$600
$4,465
$14,915
$15,516
$600
$4,213
$19,127
$20,688
$600
$4,028
$23,155
$25,860
$600
$3,885
$27,040
$31,032
$600
$3,768
$30,809
$36,204
$600
$3,671
$34,479
$41,376
$34,479
The total cost for the eight sessions is $34,479 to which your firm adds a 20 percent profit margin to give a bid
price of $41,375. The per session price, therefore, is $5,172. Comparing the cumulative revenue with the
cumulative cost, your firm will break even in the third session.
Problem 14.
If you compute MAD and TS
t
1
2
3
4
5
6
7
8
Ft
179
217
91
51
76
438
64
170
At
163
240
67
78
71
423
49
157
At-Ft
-16
23
-24
27
-5
-15
-15
-13
|At-Ft|
16
23
24
27
5
15
15
13
MAD
16
19.5
21.0
22.5
19.0
18.3
17.9
17.3
TS
-1
0.36
-0.81
0.44
0.26
-0.55
-1.40
-2.20
Column1
TS
1
0.5
0
-0.5
1
2
3
4
5
6
7
8
TS
-1
-1.5
-2
-2.5
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
If you have computed MAR and TS
Period
1
2
3
4
5
6
7
8
Total
Estimate
179
217
91
51
76
438
64
170
Actual
163
240
67
78
71
423
49
157
(At/Et) - 1 |(At/Et) - 1|
-8.9%
8.9%
10.6%
10.6%
-26.4%
26.4%
52.9%
52.9%
-6.6%
6.6%
-3.4%
3.4%
-23.4%
23.4%
-7.6%
7.6%
-12.9%
139.9%
MAR
Tracking
Signal
0.098
0.153
0.247
0.211
0.181
0.189
0.175
0.170
-1.615
1.142
1.027
1.005
-0.276
-0.735
-4.75
6.75
-14
-15
19.10
364.79
-0.29
1.15
51
-24
27
-38
8
The mean bias is -1.61% (-0.129/8) and the MAR is 0.175. Thus, the new model is slightly less accurate, but
has significantly less bias.
Problem 15.
alternative.
Alternative
a
b
c
d
As shown in the table below, alternative d has the lowest expected cost and is thus the best
Rainy
0.3
6
2
1
5
Cloudy
0.2
3
4
2
4
Sunny
0.5
4
5
7
3
Expected
Cost
4.4
3.9
4.2
3.8
16. The worst possible outcomes for Alternatives a, b, c and d are 6, 5, 7 and 5 respectively.
Alternative
a
b
c
d
Rainy
0.3
6
2
1
5
Cloudy
0.2
3
4
2
4
Sunny
0.5
4
5
7
3
Worst
Case
6
5
7
5
Alternatives b and d have the best (lowest) worst outcome.
Problem 17. In the spreadsheet shown below the project cost, nonengineering labor hours, material cost, and
downtime hours were all modeled using a triangular distribution. The engineering labor hours were modeled
using a uniform distribution over the range of 510 to 690 hours. As shown in the spreadsheet below, the
project has a 92.37% chance of meeting the firm’s NPV hurdle.
Hours
Proj Cost
Eng Labor
Non Eng Labor
Matl
Downtime
Total Cost
600
1633
112
Cost
$1,033,333
48,000
57,167
130,000
55,833
1,324,333