App_B Solustions to Selected Problems

B - 1

Copyright 2015 John Wiley & Sons, Inc.

APPENDIX

B Solutions to Selected Problems

CHAPTER 2: STRATEGIC MANAGEMENT AND PROJECT

SELECTION

Problem 2:

Average Rate of Return = $30,000/$200,000 = 0.15 = 15%

Problem 4:

i = 24%

Investment Yr 1 Yr 2 Yr 3 Yr 4

Cashflow (75,000)$ 20,000$ 25,000$ 30,000$ 50,000$

PVIF 1.0000 1.2400 1.5376 1.9066 2.3642

PV $ (75,000)$ 16,129$ 16,259$ 15,735$ 21,149$

NPV (5,729)$

The addition of 4% inflation makes the investment unfavorable at a hurdle rate of 20%.

Problem 6:

Year Pessimistic Most Likely Optimistic Used PVIF PV$

0 $ (65,000) $ (65,000) $ (65,000) $(65,000) 1 $(65,000)

1 $ 14,000 $ 20,000 $ 22,000 $ 20,000 1.200 $ 16,667

2 $ 19,000 $ 25,000 $ 30,000 $ 25,000 1.44 $ 17,361

3 $ 27,000 $ 30,000 $ 36,000 $ 30,000 1.728 $ 17,361

4 $ 32,000 $ 35,000 $ 39,000 $ 35,000 2.0736 $ 16,879

Rate 20% NPV $ 3,268

The column labeled Used indicates the cash flow value used to calculate the net present value.

The profitability index is the sum of the discounted cash flows divided by the initial investment. For

this problem it is the sum of the PVs for years 1-4 divided by $65,000 or:

68,268/65,000 = 1.05

Since the value it greater than one, the project should be accepted.

B - 2


Problem 8:

Year Pessimistic

Most

Likely Optimistic Inflation Used PVIF PV$

0 $ (65,000) $ (65,000) $ (65,000) - $(65,000) 1 $(65,000)

1 $ 14,000 $ 20,000 $ 22,000 2% $ 20,000 1.220 $ 16,393

2 $ 19,000 $ 25,000 $ 30,000 2% $ 25,000 1.4884 $ 16,797

3 $ 27,000 $ 30,000 $ 36,000 2% $ 30,000 1.8158 $ 16,521

4 $ 32,000 $ 35,000 $ 39,000 2% $ 35,000 2.2153 $ 15,799

Rate 20% NPV $ 510

Now the column labeled Inflation has been added. Each of these cells is individually calculated by

Crystal Ball with a normal distribution using a standard deviation of 0.33% to allow different

inflation values for each year. The result is added to the PVIF calculation used to determine the

individual PV results.

Now the forecast values for NPV look like this:

The probability that the NPV exceeds $0 is about 50%. The mean value for this distribution is -$85.

This analysis indicates that there is only a 50-50 chance that the project will qualify (meet the hurdle

rate).

B - 3


Category Weight A B C A B C

Consulting costs 20 1 2 3 20 40 60

Acquisition time 20 2 3 1 40 60 20

Disruption 10 2 1 3 20 10 30

Cultural differences 25 3 3 2 75 75 50

Skill redundencies 10 2 1 1 20 10 10

Implementation risks 10 1 2 3 10 20 30

Infrastructure 10 2 2 2 20 20 20

205 235 220Total Score

Method Options

Grade Score

Problem 10:

a) Implementation risks = 10 and cultural differences = 25

Due to the changes in weights for implementation risks and cultural differences, Method B is now the

best option.

b) Using the initial values from problem 6, make the changes so that Method A implementation

risks = 3 and Method C cultural differences = 2.




Disruption 10 2 1 3 20 10 30





225 220 235

Method Options

Grade Score

Total Score

The change for As implementation risks grade was not sufficient to replace Method C as the best

option. The grade for cultural differences at Method C had already been set to 2 in the initial

evaluation of the problem.

B - 4





Disruption 10 2 1 3 20 10 30





Tax considerations 15 3 2 1 45 30 15

220 250 250

Method Options

Grade Score

Total Score

c) Using the initial values from problem 6, insert Tax considerations = 15 and A = 3, B = 2, and

C = 1.

Due to the insertion of tax considerations, Methods B and C are now the best options.

Problem 12:

Change the grade for Location 3s rent to 3.

Category Weight 1 2 3 4 1 2 3 4

Class of clientele 1.000 2 3 1 3 2.000 3.000 1.000 3.000

Rent 0.900 3 2 3 3 2.700 1.800 2.700 2.700

Indoor mall 0.855 3 1 3 1 2.565 0.855 2.565 0.855

Traffic volume 0.720 3 2 3 1 2.160 1.440 2.160 0.720

9.425 7.095 8.425 7.275

Mall Options

Grade Score

Total Score

Due to the rent change, Location 3 moves up from last place to second place based upon the grades

assigned to the evaluated categories.

B - 5


CHAPTER 6: PROJECT ACTIVITY AND RISK PLANNING

Problem 2:

Threat Severity Likelihood Inability to detect RPN

#1 3 5 4 60

#2 5 6 1 30

#3 4 3 3 36

#4 7 4 6 168

The main thing that changes when using this approach is that threat #2 drops significantly from

critical to possibly ignore. This is mostly due to the lack of inability to detect. Threat #2 is

somewhat severe and the likelihood is great, but since the threat is relatively easy to detect, it can be

mitigated early and possibly even removed. Thus, this is a much more realistic evaluation of the

threats than just creating a risk matrix.

Problem 4:

B - 6


Existing tools

Cheap equipment

High-quality equipment

$1,200

$3,600

$7,200

$2,400

$14,400

p = .20

p = .10

p = .40

p = .30

p = .20

p = .10

p = .40

p = .30

p = .20

p = .10

p = .40

p = .30

$800

$2,100

$4,000

$1,300

$8,200

$900

$1,950

$3,400

$1,050

$7,300

Based on the analysis, the manufacturer should approve the purchase of the high-quality, special

equipment for $10,000. As a result, significant savings should occur.

B - 7


Problem 6:

P

r

o

b

a

b

i

l

i

t

y

5

4

3

3

1

2

2

1

1 2 3 4 5

Impact

Legend:

Critical

Monitor

Ignore

Opportunity 1:

You could accept this risk and enjoy the benefits derived from it. To increase the potential for more

impact, you could enhance the risk by providing more training.

Opportunity 2:


impact, you could further exploit the database.

Opportunity 3:


impact, you could share the data by increasing sales.

B - 8


CHAPTER 7: BUDGETING: ESTIMATING COSTS AND RISKS

Problem 2:

Tracking

Period Estimate Actual A(t)-F(t)-1 |(A(t) - F(t))-1| MAR Signal

1 179 163 -0.08939 0.089385475

2 217 240 0.105991 0.105990783 0.10 0.17

3 91 67 -0.26374 0.263736264 0.15 -1.61

4 51 78 0.529412 0.529411765 0.25 1.14

5 76 71 -0.06579 0.065789474 0.21 1.03

6 438 423 -0.03425 0.034246575 0.18 1.00

7 64 49 -0.23438 0.234375 0.19 -0.28

8 170 157 -0.07647 0.076470588 0.17 -0.74

Total -0.129

Again, the bias has reduced considerably and changed sign but the MAR is somewhat greater. Hence,

the Tracking Signal is substantially smaller and shows an acceptable level of bias on the part of this

estimator.

Problem 4:

The learning rate can only be determined in reality by trial and error.

For this problem a simple spreadsheet will be used. The learning curve factor reduces the time spent

producing the lots, from the base time of 7 hours. We also know that the total production hours for 25

units are 103.6 hours. A spreadsheet calculating the cycle time for each lot based on a learning curve

would look like this:

Unit

Base

Time Multiplier

Adjusted

Time

Trial

Rate

1 7 1.00 7.00 0.90

2 7 0.90 6.30

3 7 0.85 5.92

4 7 0.81 5.67

5 7 0.78 5.48

6 7 0.76 5.33

7 7 0.74 5.21

8 7 0.73 5.10

9 7 0.72 5.01

10 7 0.70 4.93

11 7 0.69 4.86

12 7 0.69 4.80

B - 9


13 7 0.68 4.74

14 7 0.67 4.69

15 7 0.66 4.64

16 7 0.66 4.59

17 7 0.65 4.55

18 7 0.64 4.51

19 7 0.64 4.47

20 7 0.63 4.44

21 7 0.63 4.41

22 7 0.63 4.38

23 7 0.62 4.35

24 7 0.62 4.32

25 7 0.61 4.29

Total 124.0

Note that for this example, the multiplier for year two is calculated as:

2 (log(0.90)/log(2))

or 2 (-0.045/0.301)

or 2 (-0.149)

= 0.9

Each multiplier is calculated in turn based on the unit number it represents. In each case the multiplier

is used to modify the base cycle time of 7 hours and then totaled at the bottom.

The total in this example does not equal 103.6 hours because the learning rate is incorrect. One could

insert different values and find the correct value by trial and error, or use the Solver feature of

Microsoft Excel. To use Solver, the total hours are set as the Target Cell and the Trial Rate is

designated as the cell to be changed subject to the limitation that it may not exceed 1.0 using this

technique, the spreadsheet with the correct learning rate would be:

Unit

Base

Time Multiplier

Adjusted

Time

Trial

Rate

1 7 1.00 7.00 0.85

2 7 0.85 5.95

3 7 0.77 5.41

4 7 0.72 5.06

5 7 0.69 4.80

6 7 0.66 4.60

7 7 0.63 4.44

8 7 0.61 4.30

9 7 0.60 4.18

10 7 0.58 4.08

11 7 0.57 3.99

12 7 0.56 3.91

13 7 0.55 3.84

B - 10


14 7 0.54 3.77

15 7 0.53 3.71

16 7 0.52 3.65

17 7 0.51 3.60

18 7 0.51 3.55

19 7 0.50 3.51

20 7 0.50 3.47

21 7 0.49 3.43

22 7 0.48 3.39

23 7 0.48 3.36

24 7 0.47 3.32

25 7 0.47 3.29

Total 103.6

Problem 6:

Problem 8:

Year Pessimistic Most Likely Optimistic Used PVIF PV$

0 $ (65,000) $ (65,000) $ (65,000) $(65,000) 1 $(65,000)

1 $ 14,000 $ 20,000 $ 22,000 $ 20,000 1.200 $ 16,667

2 $ 19,000 $ 25,000 $ 30,000 $ 25,000 1.44 $ 17,361

3 $ 27,000 $ 30,000 $ 36,000 $ 30,000 1.728 $ 17,361

4 $ 32,000 $ 35,000 $ 39,000 $ 35,000 2.0736 $ 16,879

Rate 20% NPV $ 3,268

Now the green (or shaded) cells in the column labeled Used (with the exception of the initial

investment of $65,000) are calculated by Crystal Ball using a triangular distribution with the end

B - 11


points selected based on the pessimistic and optimistic values given. The setup for year one looks

like this:

The net present value is then calculated for each of 1000 trials and results are displayed by

designating the NPV cell as a forecast. Typical results look like this:

By adjusting the triangular sliders it can be seen that the chance of the NPV exceeding $0 (and the

hurdle rate) is about 87%. The mean value for this distribution is $2770.

B - 12


Problem 10:

Windows

Unix

B - 13


CHAPTER 8: SCHEDULING

NOTE: Many of the AON graphics in this solutions set depict the start day of the successor activity to

be the same day as the completion of the predecessor. This is consistent with the presentation in the

text. It is not consistent with the result that would be obtained using Microsoft Project, where the

start day of the successor is always the next working day after the completion of the predecessor.

Problem 2:

Problem 4:

a) The critical path is B-E-G.

b) 23 work periods.

c and d)

Task Duration ES EF LS LF Slack

A 10 0 10 5 15 5

B 7 0 7 0 7 0

C 5 10 15 15 20 5

D 7 10 17 11 18 1

E 11 7 18 7 18 0

F 3 15 18 20 23 5

G 5 18 23 18 23 0

B - 14


Problem 6:

PDM Diagram 6a

PDM Diagram 6b

Problem 8:

B - 15


Please see note about network depiction preceding Problem 2

a) The critical path activities are A, C, E, and G.

b) The projects duration is 22 days.

c) Yes, activity B can be delayed one day without delaying the completion of the

project.

Problem 10:

a. The path with the longest expected duration is AC CB BE EF.

b. The only event with slack is D at 3 days.

c. If D were the final event in the network, then the critical path would be AC CB BD.

d. The following spreadsheet excerpt illustrates the calculation of the probability of completion

in 14 days:

Task a m b Expected Variance Std Dev.

AB 3 6 9 6.0 1.00 1.00

AC 1 4 7 4.0 1.00 1.00

CB 0 3 6 3.0 1.00 1.00

CD 3 3 3 3.0 0.00 0.00

CE 2 2 8 3.0 1.00 1.00

BD 0 0 6 1.0 1.00 1.00

BE 2 5 8 5.0 1.00 1.00

DF 4 4 10 5.0 1.00 1.00

DE 1 1 1 1.0 0.00 0.00

EF 1 4 7 4.0 1.00 1.00

B - 16


Desired

Duration

Expected

Project

Duration

Sum of

Variances

Critical

Path Z Probability

14 16.0 4.00 -1 15.9%

e. If CD slips to six days the critical path is unchanged but slack on D is reduced. If CD slips to

seven days then there are two critical paths: AC CB BE EF and AC CD DF. If CD slips to

eight days then the critical path shifts to AC CD DF and the project duration extends to 17

days.

Problem 12:

Figure 8.12a shows the PDM network for the data from Table A of Problem 8-12

assuming that the data were applied as shown in Figure 8.12b.


1) The path with the longest expected duration is 2,3,4,5,7,8,9.

2) The slack for activity 1 is 11.7 days. The slack for activity 6 is 4 days.

The following table shows the calculation of the expected completion time:

Activity a m b Expected

1 8 10 13 10.2

2 5 6 8 6.2

3 13 15 21 15.7

4 10 12 14 12.0

5 11 20 30 20.2

6 4 5 8 5.3

7 2 3 4 3.0

8 4 6 10 6.3

9 2 3 4 3.0

B - 17


Expected

Project

Duration

66.4

Problem 14:

Figure 8.14a shows the original network diagram for problem 14.


B - 18


1) The critical path activities are A, D, G, and J. Activities B and E should be closely monitored as a

near critical path.

Figure 8.14b shows the impact of the projects performance to date.

2) The project will be completed in 12.5 days instead of the 13 days originally expected. The near

critical path (B, E, G, J) is now critical. Activities A, D, F, and H are now near critical activities.

Problem 16:

Using critical path analysis with the data provided gives the following table:

Activity Expected Std Dev. Variance

a 2.0 2.00 4.00

b 3.0 1.00 1.00

c 4.0 0.00 0.00

d 2.0 3.00 9.00

e 1.0 1.00 1.00

f 6.0 2.00 4.00

g 4.0 2.00 4.00

h 2.0 0.00 0.00

B - 19


Desired

Duration

Expected

Project

Duration

Sum of

Variances

Critical

Path Z Probability

12 13.0 9.00 -0.33 37.1%

13 13.0 9.00 0.00 50.0%

16 13.0 9.00 1.00 84.1%

17.3 13.0 9.00 1.44 92.5%

For this problem the variance has to be calculated from the standard deviation, and the durations

provided are assumed to be the expected durations. As can be seen there is about an 84% chance of

completing the project within the drop dead time. If a little more than a week is added to the

duration, the chance of completing the project on time rises to 92.5%.

Problem 18:

B - 20


Problem 20:

Figure 8.20 shows the network diagram for problem 20.


a) The critical path is A, D, E, G, I, J.

b) The slack on process confirmation (F) is 20 days.

c) The slack on test pension plan (C) is 61 days.

d) The slack on verify debt restriction compliance (H) is 20 days.

B - 21


Problem 22:

a)

Figure 8.22 shows the network diagram for problem 22.


b) The critical path is B, F, H.

c) Week 9.

d) If activity E requires one extra week, the time will be absorbed in free float and will not affect

any other activity. If activity E requires two extra weeks, then a second critical path will be

created for activities B, E, G. If activity E requires three weeks, negative float will be created and

the project cannot complete in nine weeks. The new completion time will rise to 10 weeks.

B - 22


Problem 24:

a)

Figure 8.24a shows the network diagram for problem 24a.

b) The critical path is B, E, G, H.

Figure 8.24c shows the network diagram solution to problem 24c.


B - 23


c & d) Given a float value of 6 weeks, activity F seems to be the best candidate to supply resources

needed to crash the project. Since the float is almost 50% of the activitys duration, using its resources

to work other activities is unlikely to convert activity F into a near-critical activity. Since activity D is

both critical and concurrent to activity F, the resources should be transferred there.

Problem 26:

Figure 8.26a shows the network, critical path and slack times.

Tabulating the calculations for expected durations and probability looks like this:


1-2 6 8 10 8 0.44 0.67

1-3 5 6 7 6 0.11 0.33

1-4 6 6 6 6 0.00 0.00

2-6 0 0 0 0 0.00 0.00

2-7 10 11 12 11 0.11 0.33

3-6 12 14 16 14 0.44 0.67

4-5 5 8 11 8 1.00 1.00

4-9 7 9 11 9 0.44 0.67

5-6 8 10 12 10 0.44 0.67

5-9 0 0 0 0 0.00 0.00

6-7 14 15 16 15 0.11 0.33

6-8 10 12 14 12 0.44 0.67

7-10 9 12 15 12 1.00 1.00

8-10 0 4 14 5 5.44 2.33

9-11 5 5 5 5 0.00 0.00

10-11 7 8 9 8 0.11 0.33

B - 24



Desired

Duration

Expected Project

Duration

Sum of Variances

Critical Path Z Probability

61.69 59.0 2.67 1.65 95.0%

The next longest path is 1-3, 3-6, 6-7, 7-10, 10-11 at 55 days. It will only be a concern if under some

circumstances; its duration exceeds the path with the longest expected duration of 59 days. Using the

same technique for calculating the probability of exceeding a particular duration gives the following

table for this path:

Probability for path 1-3-6-7-10-11

Desired

Duration

Expected

Path

Duration

Sum of Path

Variances Z Probability

59 55.0 1.78 3.00 99.9%

Clearly the chance of exceeding 59 days is quite small. The same technique can be applied to the

next longest path 1-4, 4-5, 5-6, 6-8, 8-10, 10-11 which while relatively short has high variance:

Probability for path 1-4-5-6-8-10-11

Desired

Duration

Expected

Path

Duration

Sum of Path

Variances Z Probability

59 49.0 7.44 3.67 100.0%

Again it is clear that it is unlikely that this path will cause problems with the overall project duration.

Problem 28:

Figure 28a shows the PDM network diagram for problem 28.

B - 25


The following table tabulates the variances and probability for this project:


1 6 10 14 10 1.78 1.33

2 0 1 2 1 0.11 0.33

3 16 20 30 21 5.44 2.33

4 3 5 7 5 0.44 0.67

5 2 3 4 3 0.11 0.33

6 7 10 13 10 1.00 1.00

7 1 2 3 2 0.11 0.33

8 0 2 4 2 0.44 0.67

9 2 2 2 2 0.00 0.00

10 2 3 4 3 0.11 0.33

11 0 1 2 1 0.11 0.33

12 1 2 3 2 0.11 0.33

Desired

Duration

Expected

Project

Duration

Sum of

Variances

Critical

Path Z Probability

44 41.0 8.22 1.05 85.2%

Problem 30:

The setup for problem 30 is similar to that for problem 29. First the spreadsheet in Excel is prepared

with the calculations for the paths:

Then, similar to problem 29, triangle distributions are established to calculate the durations for all

activities except 9 (no variation in the estimate).

1 2 3 4 5 6 7 8 9 10 11 12

10 1 22 5 3 10 2 2 2 3 1 2

Paths

Activities

1-3-6

42

Project

Completion

42

1-2-4-7-9-10-12

25

1-3-9-10-12

39

1-2-4-5

19

1-2-4-7-8-11-12

23

B - 26


The resulting forecast for the duration of the project and corresponding statistics are:

Note that the probability of completing the project in 44 days has dropped to about 70%.

B - 27


To determine the probability that a path is the critical path, the simulation model can be enhanced as

follows. First, a formula can be added used Excels IF function as follows:

=IF(duration of path in question = duration of longest path,1,0)

Next the cell containing this formula can be assigned to be a Forecast cell in Crystal Ball.

Problem 32:

The Pert Entry Form in Microsoft Project is used to enter the three durations. After they are in

the Calculate Pert button is clicked to populate the Duration field with the expected durations. Note

that MSP uses the non-standard terminology Expected in lieu of Most Likely.

Using the calculated durations, the Gantt chart looks like this:

ID Task Name Duration Start Finish Predecessors

1 Start 0 days Sun 12/11/05 Sun 12/11/05

2 a 7.5 days Mon 12/12/05 Wed 12/21/05 1

3 b 8 days Mon 12/12/05 Wed 12/21/05 1

4 c 6 days Wed 12/21/05 Thu 12/29/05 2

5 d 14.5 days Thu 12/22/05 Wed 1/11/06 2,3

6 e 7 days Wed 1/11/06 Fri 1/20/06 4,5

7 f 11.5 days Fri 1/20/06 Mon 2/6/06 3,6

8 g 8 days Tue 2/7/06 Thu 2/16/06 7

9 End 0 days Thu 2/16/06 Thu 2/16/06 8

12/11

2/16

4 11 18 25 1 8 15 22 29 5 12 19 26

Dec '05 Jan '06 Feb '06

The figure shows the default Gantt chart view of the problem, with a project start day of Sunday

December 11, 2005. Note that MSP moves the beginning of the first task to the first workday of

Monday the 12th. This display shows the default calendar of 5 day 40 hr. weeks with no holidays. A

Start and End milestone have been inserted to insure that all activities have at least one

predecessor and successor.

B - 28


The tracking Gantt view can be used to display the critical path:

ID Task Name Duration

1 Start 0 days

2 a 7.5 days

3 b 8 days

4 c 6 days

5 d 14.5 days

6 e 7 days

7 f 11.5 days

8 g 8 days

9 End 0 days

12/11

0%

0%

0%

0%

0%

0%

0%

2/16

4 11 18 25 1 8 15 22 29 5 12 19 26

Dec '05 Jan '06 Feb '06

The network diagram can be displayed directly from MSP using the Network Diagram view. A

portion of it with the default format settings looks like this:

a

Start: 12/12/05 ID: 2

Finish: 12/21/05 Dur: 7.5 days

Res:

b

Start: 12/12/05 ID: 3

Finish: 12/21/05 Dur: 8 days

Res:

Start

Milestone Date: Sun 12/11/05

ID: 1

The slack values are automatically calculated by MSP. They can be revealed in a number of different

views:

ID Task Name Start Finish Late Start Late Finish Free Slack Total Slack

1 Start Sun 12/11/05 Sun 12/11/05 Mon 12/12/05 Mon 12/12/05 0 days 0 days

2 a Mon 12/12/05 Wed 12/21/05 Mon 12/12/05 Wed 12/21/05 0 days 0.5 days

3 b Mon 12/12/05 Wed 12/21/05 Mon 12/12/05 Wed 12/21/05 0 days 0 days

4 c Wed 12/21/05 Thu 12/29/05 Tue 1/3/06 Wed 1/11/06 9 days 9 days

5 d Thu 12/22/05 Wed 1/11/06 Thu 12/22/05 Wed 1/11/06 0 days 0 days

6 e Wed 1/11/06 Fri 1/20/06 Wed 1/11/06 Fri 1/20/06 0 days 0 days

7 f Fri 1/20/06 Mon 2/6/06 Fri 1/20/06 Mon 2/6/06 0 days 0 days

8 g Tue 2/7/06 Thu 2/16/06 Tue 2/7/06 Thu 2/16/06 0 days 0 days

9 End Thu 2/16/06 Thu 2/16/06 Thu 2/16/06 Thu 2/16/06 0 days 0 days

9 days

27 4 11 18 25 1 8 15 22 29 5 12 19 26 5 12 19 26 2 9 16 23

Nov '05 Dec '05 Jan '06 Feb '06 Mar '06 Apr '06

This view shows the View Detail Gantt combined with the Schedule Table. Note that the Gantt

chart also displays the slack as a green line.

B - 29


CHAPTER 9: RESOURCE ALLOCATION

Problem 2:

Activity Crash Days Crash $ Crash $/Day

A 1 $200.00 $ 200.00

B 2 $ 75.00 $ 37.50

C 3 $150.00 $ 50.00

D 2 $100.00 $ 50.00

Figure 9.2 uses the data from problems 1 and 2 to answer problem 2.

b) Task B is on the critical path and it has the lowest cost per day $37.50 to crash (additional

cost of $75.00). It enables a 13-day completion without making any other new tasks critical.

Problem 4:

Figure 9.4 shows the AON network data for problem 4.

B - 30


d) On the surface the answer is no, since the duration as calculated is one month less than the project

deadline requires. It would be prudent, however, to perform analysis as described in Chapter 8 to

determine the probability of actually achieving this schedule. Based on that analysis it may in fact be

prudent to crash some activities.

Problem 6:

Figure 9.6a shows the normal and crashed networks for problems 8.18 and 9.6.

Activity Crash Wk Crash $/Wk

C 1 $ 40.00

F 4 $ 20.00

H 3 $ 10.00

I 3 $ 30.00

Options Crash $

Option 1 H3,F4,I2 $ 170.00

Option 2 H3,F3,I1,C1 $ 160.00

Option 3 H2,F4,I1,C1 $ 170.00

Figure 9.6b shows the crashing options.

B - 31


Option 2 is the most favorable option because it satisfies the time constraint and has the lowest cost.

However, it also increases the number of critical activities (Figure 9.6a). Because two critical paths

converge at event 7, completion probabilities should consider the effects of path convergence.

Problem 8:

In all cases, crashing is applied only to critical path activities (1-2, 2-4, 4-6, and 6-7).

a) 1-2 and/or 6-7

b) 4-6, assuming that duration is equivalent to resource consumption.

c) All activities on the critical path have zero slack at present.

d) 1-2 has the most critical followers in problem 7.

e) 1-2 has the most successors in problem 7

Problem 10:

The critical path is defined by the events 1,2,3,4. The normal duration is 17 days and a normal cost of

$100 per day; a total cost of $1,700.

Activity Completion

Date

Crash

Costs Total Cost

Normal 17d $ - $ 1,700

3-4 16d $ 25 $ 1,625

1-2 15d $ 30 $ 1,555

1-2 14d $ 50 $ 1,505

3-4 13d $ 60 $ 1,465

2-3 12d $ 80 $ 1,445

Figure 9.10 shows the cost-duration history for problem 10.

Since activity 2-4 is not on the critical path, it is not crashed. The table indicates that the lowest-cost

crashing option was always the first crash option taken.

B - 32


Problem 12:

Figure 9.12a shows the network history of crashing the TV commercial project.

Activity Completion

Date

Crash

Costs Total Cost

Normal 19d $ - $ 1,710

1-2 18d $ 30 $ 1,650

2-3 17d $ 40 $ 1,600

1-2 16d $ 50 $ 1,560

1-2 15d $ 70 $ 1,540

1-3,2-3,2-4 14d $140 $ 1,590

Figure 9.12b shows the tabular history for Figure 9.12a.

Crashing attacks duration to compress the schedule. As slack approaches zero, the benefits of

crashing can be expected to fall below the costs required to remove time by crashing the critical path.

B - 33


Problem 14:

a. g has a slack of 1

b. e, g, and h all have 1 successor

c. e has 1 critical follower

d. g has shortest time of 4

e. f has latest start time of 29

Problem 16:

The problem with using the Normal distribution to model activity times is that with the Normal

distribution tasks are equally likely to take longer than the expected time and less than the expected

time. With many activities this is not true and it is more likely that activities will take longer than

expected time. Therefore, using distributions that can model skew in the data such as the Triangular

and Beta distributions are more appropriate for modeling project activity times.

The table below summarizes the results after modifying the model so that the activity times follow a

BetaPERT distribution with parameters 7, 10, and 15.

In comparing these results we observe that the variation (as measured by both the range and standard

deviation) went down in the BetaPERT models. Also, observe how the Min values increased in the

BetaPERT models but the Max values decreased.

B - 34


CHAPTER 10: MONITORING AND INFORMATION SYSTEMS

Problem 2:

($000)

CV = EV AC

CPI = EV/AC

SV = EV PV

SPI = EV/PV

Negative variances are unfavorable. $(000).

If an index is less than one, the variance is unfavorable.

Problem 4:

($000)

This project is seriously delayed and also over budget.

Problem 6:

($000)

AC = $550 AC = $750

Day AC PV EV Day AC PV EV

65 $ 550 $ 735 $ 678 65 $ 750 $ 735 $ 678

CV $ 128 Favorable

CV $ (72) Unfavorable

CPI 1.23 CPI 0.90

SV $ (57) Unfavorable


SPI 0.92 SPI 0.92

CSI 1.14 Favorable CSI 0.84 Unfavorable

TV (5.00) Days behind TV (5.00) Days behind

Month AC PV EV

5 $ 34 $ 42 $ 39

CV $ 5 Favorable

CPI 1.15


SPI 0.93

Month AC PV EV

17 $ 350 $ 475 $ 300

CV $ (50) Unfavorable

CPI 0.86


SPI 0.63

CSI 0.54 Unfavorable

TV (6.26) Months delayed

B - 35


The first step is to estimate EV. Starting with TV, we solve to determine SV. Once SV is known, EV

can be determined because the PV was given.

In problem 6, changing the AC value only affects cost-related measures and indices. The SV and SPI

are unaffected by a change in AC.

Problem 8:

Figure 10.8a shows the solution table for problem 8.

ETC = (BAC EV) / CPI. EAC = ETC + AC. TV = SV / (PV / Wks).

The weekly values for AC, EV, and PV are evenly prorated over the scheduled interval for each

activity.

Task Values Total $ Wk 1 Wk 2 Wk 3 Wk 4 Wk 5 Wk 6 Wk 7 Wk 8

PV 300 150 150

AC 400 200 200

EV 300 150 150

PV 200 67 67 66

AC 180 60 60 60

EV 200 67 67 66

PV 250 125 125

AC 300 150 150

EV 250 125 125

PV 600 120 120 120 120 120

AC 400 100 100 100 100

EV 120 30 30 30 30

PV 400 100 100 100 100

AC 200 100 100

EV 80 40 40

PV 1750 217 217 311 245 220 220 220 100

AC 1480 260 260 310 250 200 200 0 0

EV 950 217 217 221 155 70 70 0 0

PV 217 434 745 990 1210 1430 1650 1750

AC 260 520 830 1080 1280 1480 1480 1480

EV 217 434 655 810 880 950 950 950

CV -43 -86 -175 -270 -400 -530

SV 0 0 -90 -180 -330 -480

CPI 0.83 0.83 0.79 0.75 0.69 0.64

SPI 1.00 1.00 0.88 0.82 0.73 0.66

TV 0.00 0.00 -0.36 -0.73 -1.36 -2.01

CSI 0.83 0.83 0.69 0.61 0.50 0.43

ETC 1837 1577 1388 1253 1265 1246

EAC 2097 2097 2218 2333 2545 2726

E

Total

Cumulative

A

B

C

D

B - 36


Figure 10.8b shows the network for problem 8.

Problem 10:

The solution to problem 10.8 as it might appear in MS-Project.

The table shows the results; however, there are several steps needed to create this data in Microsoft

Project that are not obvious to the casual observer.

1) Setup the network with predecessor/successor relationships and durations in the conventional

manner.

2) Enter the budget from the problem statement as a Fixed Cost in the Task Sheet, Cost Table View.

Note that the Baseline costs should be zero.

3) Save a Baseline by using the Save Baseline function in the Tools, Tracking menu.

a) If this has been performed correctly, the Baseline Cost in the Cost Table View should now be

equal to the budgets.

4) In the Task Information View, adjust the Percent Complete for each task using the data from the

problem statement.

5) In the Task Sheet, Cost Table View enter the actual costs for each task from the problem

statement.

6) Based on the starting date of the network, determine the date for the Friday at the end of the sixth

week.

7) In the Project, Project Information window adjust the Status Date to the Friday of the sixth week

of the project.

B - 37


8) By adjusting the status date, MSP will automatically calculate the standard earned value

information. The data can be displayed using the Task Sheet, Earned Value Table View. The

table in this manual was prepared with MSP 2000, which still used the traditional names for the

earned value data (i.e. BCWS, BCWP etc.). The titles were changed by double clicking on the

column and editing the Title field.

Problem 12:

This table shows the result of the 0-100 rule. Note that the actuals accrue in the same manner as the

previous problem. Here, the difference is that the EV accrues all at once when the task ends. The

graph below shows the sudden increases in EV that results.

B - 38


With this method, EV lags behind. For a large project well underway, the effect is small, but for this

tiny sample, the EV picture is not very useful.

B - 39


CHAPTER 11: PROJECT CONTROL

Problem 2:

Activity Actual Schedule Budget Actual CR Comment

A 4 4 60 40 1.50 Favorable, Investigate

B 3 2 50 50 1.50 Favorable, Investigate

C 2 3 30 20 1.00 OK unless time is critical

D 1 1 20 30 0.67 Unfavorable, Investigate

E 2 4 25 25 0.5 Unfavorable, Investigate

Problem 4:

The slope of the planned progress curve is 200 points divided

by 100 days, or 2 points per day. Since cost data are not

provided, the CR is simply the ratio of actual progress over

planned progress.

Based on the data in the table for problem 4, it would appear

that the project is running about 20 days behind schedule. It

is likely to require approximately 120 days (+5 days or 10

days) to complete this project.

Day Actual Planned CR

1 2 2 1.00

2 3 4 0.75

3 4 6 0.67

4 6 8 0.75

5 7 10 0.70

6 9 12 0.75

7 12 14 0.86

8 14 16 0.88

9 15 18 0.83

10 17 20 0.85

11 20 22 0.91

12 21 24 0.88

13 21 26 0.81

14 22 28 0.79

15 24 30 0.80

16 26 32 0.81

17 27 34 0.79

18 29 36 0.81

19 31 38 0.82

20 33 40 0.83

B - 40


Problem 6:

The base data for the problem looks like this:

Note that this data is incremental rather than cumulative. The cumulative data with CR calculations

looks like this:

It is interesting to note that while Team A has incurred more cost it has also achieved more progress.

Therefore, based on the Critical Ratio, the teams are identical in their achievements to date.

However, in both cases, their critical ratios are deteriorating and should be investigated for potential

corrective actions.

Problem 8:

Figure 11.8 is the network diagram for Problem 11-8.

Based on Figure 11.8 and data from the problem, activities 1-2, 2-3, and 2-4 have been completed.

These activities account for 22 days. Activity 3-5 has been 70% completed and is expected to require

B - 41


18 days. Thus 18d * 70% = 12.6 days. So, 34.6 days of progress has been accomplished. The

schedule ratio is 34.6/40 = .865.

The budget for day 40 would include all completed activities plus 100% of activity 3-5. The planned

cost is $1,455. The actual costs are $1,480. The cost ratio is 1455/1480 = .983. Thus, CR = .865 *

.983 = .850, so the project needs attention. The cost overage is 1480 1455 = 25.

App_B Solustions to Selected Problems

Documents