Chapter 13 - Project Management.pdf

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TEACHING SUGGESTIONS Teaching Suggestion 13.1: Importance 13.1: Importance of PERT. PERT.

PERT has rebounded and, due to PC software such as Microsoft Project, become a highly used quantitative analysis technique. It can be useful for organizations of all sizes and any individuals involved in planning and controlling projects. A good way to start this chapter is to discuss the capabilities of PERT. Students can be asked to contact a local firm (such as a builder) to ask about the use of PERT. Teaching Suggestion 13.2: Getting Students Involved with PERT.

PERT is a technique that students can apply immediately. For example, students can be asked to use PERT to plan the courses they will need to take and the timing of taking these courses until graduation. Another approach would be to have students take a typical semester and use PERT to plan the term papers, exams, and assignments that must be finished to successfully complete the semester. Teaching Suggestion 13.3: Constructing a Network.

One of the most difficult tasks of PERT or CPM is to develop an accurate network that reflects the true situation. Students should be given practice in this important aspect of network analysis as early as possible. Use the end-of-chapter problems. Students can be asked to develop their own networks. We can’t stress enough the importance of drawing networks, since many students have a conceptual problem with the task. Teaching Suggestion 13.4: Using the Beta Distribution.

PERT uses the beta distribution in estimating expected times and variances for each activity. As a matter of fact, it is questionable whether the beta distribution is appropriate. Students should be A

4

C

3

F

2

0

4

4

7

9

0

4

4

7 7

14

16 16

Start

told that other distributions such as the normal curve can be used. A discrete probability distribution can also be used to determine expected times and variances. Instead of using optimistic, most likely, and pessimistic time estimates, an entire discrete distribution can be used to determine expected times and variances. Teaching Suggestion 13.5: Finding the Critical Path.

Finding the critical path is not too difficult if the steps given in this chapter are followed. Students should be reminded that in making the forward pass all activities must be completed before any activity can be started. In the backward pass, students should be reminded that latest time is computed by making sure that the pro ject would not be delayed for any activity. This means that all activities must be completed within the original project completion time. Teaching Suggestion 13.6: Project Crashing.

In manually performing project crashing, the critical path may change. In many cases, two or more critical paths will exist after crashing. Students should be reminded of this problem. Fortunately, the linear programming approach or the use of PERT software, including QM for Windows, automatically takes care of this potential problem.

ALTERNATIVE EXAMPLES Alternative Example 13.1: Sid Orland is involved in planning a scientific research project. The activities are displayed in the following diagram. Optimistic, most likely, and pessimistic time estimates are displayed in the following table.

E

4

H

4

7

11

16

20

7

11

16

20

B

3

D

3

G

5

0

3

3

6

11

16

5

8

8

11 11

11

16

Finish

Figure for Alternative Example 13.1 189

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Activ Activit ity y

Opti Optimi mist stic ic

Most Like Likely ly

Pess Pessim imist istic ic

A B C D E F G H

3 3 2 1 4 2 4 3

4 3 3 3 4 2 5 4

5 3 4 5 4 2 6 5

Given this information, the least expensive way to reduce the pro ject using an activity on the critical path is to reduce activity G by 2 weeks, for a total cost of $1,000 ($1,000  2  $500).

SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS 13-1. PERT and CPM can answer a number of questions about a project or the activities within a project. These techniques can determine the earliest start, earliest finish, latest start, and the latest finish times for all activities within a network. Furthermore, these techniques can be used to determine the project completion data for the entire project, the slack for all activities, and those activities that are along the critical path of the network.

The activities along the critical path and the total project completion times are shown in the figure. The solution is shown below. As can be seen, the total project completion time is 20 weeks. Critical path activities are A, C, E, G, and H. Activity

A* B C* D E* F G* H*

Mean

S.D.

Variance

4 3 3 3 4 2 5 4

0.333 0.000 0.333 0.667 0.000 0.000 0.333 0.333

0.111 0.000 0.111 0.444 0.000 0.000 0.111 0.111

13-2. There are several major differences between PERT and CPM. With PERT, three estimates of activity time and completion are made. These are the optimistic, most likely, and pessimistic time estimates. From these estimates, the expected completion time and completion variance can be determined. CPM allows the use of crashing. This technique allows a manager to reduce the total project completion time by expending additional resources on activities within the network. CPM is used in determining the least-cost method of crashing a project or network. 13-3. An activity is a task that requires a fixed amount of time and resources to complete. An event is a point in time. Events mark the beginning and ending of activities. An immediate predecessor is an activity that must be completely finished before another activity can be started.

*Critical Path Activities Expected Completion Time: 20

Alternative Example 13.2: Sid Orland would like to reduce the project completion time for the problem in Alternative Example 13-1 by 2 weeks. The normal and crash times and costs are presented below.

13-4. Expected activity times and variances can be computed by making the assumption that activity times follow a beta distribution. Three time estimates are used to determine the expected activity time and variance for each activity.

TIME

COST

Activity

Immediate Predecessor

Normal

Crash

Normal

Crash

A B C D E F G H

— — A B C C D,E F,G

4 3 3 3 4 2 5 4

3 3 2 2 3 2 3 4

$2,000 3,000 5,000 5,000 8,000 2,000 3,000 4,000

$3,000 3,000 6,000 5,500 10,000 2,000 4,000 4,000

From the above table, the crash cost per week can be determined for each activity. This information is displayed in the following table. Act Activit ivity y

Crit ritical ical Pat Path?

Crash rash Cost Cost per per Week eek

A B C D E F G H

Yes

$1,000 0 or NA $1,000 $500 $2,000 0 or NA $500 0 or NA

Yes Yes Yes Yes

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13-5. The critical path consists of those activities that will cause a delay in the entire project if they themselves are delayed. These critical path activities have zero slack. If they are delayed, the entire project is delayed. Critical path analysis is a way of determining the activities along the critical path and the earliest start time, earliest finish time, latest start time, and the latest finish time for every activity. It is important to identify these activities because if they are delayed, the entire project will be delayed. 13-6. The earliest activity start time is the earliest time that an activity can be started while all previous activities are completely finished. The earliest activity start times are determined using a forward pass through the network. The latest activity start time represents the latest time that an activity can be started without delaying the entire project. Latest activity start times are determined by making a backward pass through the network. 13-7. Slack is the amount of time that an activity can be delayed without delaying the entire project. If the slack is zero, the activity cannot be delayed at all without delaying the entire pro ject. For any activity, slack can be determined by subtracting the earliest start from the latest start time, or by subtracting the earliest finish from the latest finish time. 13-8. We can determine the probability that a project will be completed by a certain date by knowing the expected project completion time and variance. The expected project completion time can be determined by adding the activity times for those activities along the critical path. The total project variance can be determined by adding the variance of those activities along the critical path. In most cases, we make the assumption that the project completion times follow a normal distribution. When this is done, we can use a standard normal table in computing the probability that a project will be completed by a certain date. 13-9. PERT/Cost is used to monitor and control project cost in addition to the time it takes to complete a particular project. This can be done by making a budget for the entire project using the activity cost estimates and by monitoring the budget as the project takes place. Using this approach we can determine the extent to which a project is incurring a cost overrun or a cost underrun. In addition, we can use the same technique to determine the extent to which a project is ahead of schedule or behind schedule. 13-10. Crashing is the process of reducing the total time it takes to complete a project by expending additional resources. In performing crashing by hand, it is necessary to identify those activities along the critical path and then to reduce those activities which cost the least to reduce or crash. This is continued until the project is crashed to the desired completion date. In doing this, however, two or more critical paths can develop in the same network. 13-11. Linear programming is very useful in CPM crashing because it is a commonly used technique and many computer programs exist that can be easily used to crash a network. In addition, there are many sensitivity and ranging techniques that are available with linear programming.

191

PROJECT MANAGEMENT

13-12. A

E

B

D

G Finish

Start

C

F

13-13.

Start

A 0 13

2 2 15

B 0 0

5 5 5

D 5 5

10 15 15

C 0 11

1 1 12

F 1 12

6 7 18

E 15 15

3 18 18 8 26 26

G 18 18

Finish

Activity

ES

EF

LS

LF

Stack

Critical Activity

A B C D E F G

0 0 0 5 15 1 18

2 5 1 15 18 7 26

13 0 11 5 15 12 18

15 5 12 15 18 18 26

13 0 11 0 0 11 0

No Yes No Yes Yes No Yes

The critical path is B–D–E–G. Project completion time is 26 days. 13-14.

A

F

C

G

Start

Finish

B

D

E

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CHAPTER 13

13-15. A 0 2

3 3 5

F 6 3 9 8 14

Start

B 0 0

7 7 7

C 3 5

4 7 9

D 7 7

2 9 9

G 14 14

3 17 17

Time (Weeks)

ES

EF

LS

LF

S

Critical Activity

A B C D E F G H

6 5 3 2 4 6 10 7

0 0 6 6 5 5 9 11

6 5 9 8 9 11 19 18

0 0 6 10 5 6 9 12

6 5 9 12 9 12 19 19

0 0 0 4 0 1 0 1

Yes Yes Yes No Yes No Yes No

Finish

E 5 9 14 9 14

There are two critical paths: A–C–G and B–E–G. Project completion time is 19 weeks.

The critical path is B–D–E–G. 13-16. C

A

Activity

G

D Start

Finish E

13-18.



40,  2  9,    3   4 0 40 P(X  4 0)  P(Z  ) 3

a. B

F

H



P(Z  0)  0.50

  4 0 40

) = P(Z  0) 3 1 0.50  0.50

b. P(X  40)  P(Z  13-17.

c. P(X  46)  P(Z  A 0 0

6 6 6

Start

B 0 0

5 5 5

C 6 6

3 9 9

D 6 10

2 8 12

E 5 5

4 9 9

F 5 6

6 11 12

G 9 9

10 19 19

46  40 3

) P(Z  2) 0.97725

d. P( X  46)  P( Z  2)  1  0.97725



0.02275

e. P( X  Due Date)  0.90 For a probability of 0.90,  z  1.28.  X   40 1.28  3

Finish

 X  40  1.28(3)  43.84. H 11 12

Start

Thus, the due date should be 43.84 weeks

7 18 19

13.19.

 A

10

D

0

10

10

20 30

0

10

10

30

F

10

K

6.7

I

11.2

62

68.7

62

68.7

 

30

40

40

51.2

30

40

50.8

62

B

7.2

0

7.2

H

15

J

7

22.8 30

40

55

55

62

40

55

55

62

C

3.2

0

3.2

19.8 23

E

7

3.2 10.2 23

30

G 30

7.3 3 7. 3

4 7. 7

55

L

2.2

55

5 7. 2

66.5 68.7

Finish

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Activity

a

m

b

t

A B C D E F G H I J K L

8 6 3 10 6 9 6 14 10 6 4 1

10 7 3 20 7 10 7 15 11 7 7 2

12 9 4 30 8 11 10 16 13 8 8 4

10.0 7.2 3.2 20.0 7.0 10.0 7.3 15.0 11.2 7.0 6.7 2.2

V

0.44 0.25 0.03 11.11 0.11 0.11 0.44 0.11 0.25 0.11 0.44 0.25

 

193

PROJECT MANAGEMENT

ES

EF

LS

LF

S

0 0 0 10.0 3.2 30.0 30.0 40.0 40.0 55.0 62.0 55.0

10.0 7.2 3.2 30.0 10.2 40.0 37.3 55.0 51.2 62.0 68.7 57.2

0 22.8 19.8 10.0 23.0 30.0 47.7 40.0 50.8 55.0 62.0 66.5

10.0 30.0 23.0 30.0 30.0 40.0 55.0 55.0 62.0 62.0 68.7 68.7

0 22.3 19.8 0 19.8 0 17.7 0 10.8 0 0 11.5

The critical path is A–D–F–H–J–K. Project completion time is 68.7 days. Project variance is 0.44  11.11  0.11  0.11  0.11  0.44  12.32. σ t   12.32  3.5 µ t   68.7

Probability of 

 P

finishing in 70 days Probability of  finishing in 80 days Probability of  finishing in 90 days

 P

 P

  70  688.7    0.644  Z   3.5       80  688.7    0.9994  Z   3.5       90  688.7    Z   3.5   0.9999

13-20. Assuming normal distribution for project completion time: a.

 P

  17 − 21    P( Z   2 )  1 − 0.9772  Z  2      0.0228

b.

 P

  20 − 21    P ( Z   0.5) 1 − 0 .6915  Z  2      0.3085

c.

 P

  23 21    P( Z   1)  0.8413  Z  2    

d.

 P

  25  21    P( Z   2 )  0.9772  Z  2    

13-21.

Activity

Total Budgeted Cost

Percentage of Completion

Value of Work Completed

Actual Cost

A B C D E F G H

$22,000 30,000 26,000 48,000 56,000 30,000 80,000 16,000

100 100 100 100 50 60 10 10

$22,000 30,000 26,000 48,000 28,000 18,000 8,000 1,600

$20,000 36,000 26,000 44,000 25,000 15,000 5,000 1,000

Activity Difference

$2,000 6,000 0 4,000 3,000 3,000 3,000 600



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After 8 weeks: Value of work completed



$181,600

Actual cost  $172,000 Cost underrun



$9,600

Using Table 13.6, $212,000 should have been spent using ES times. Using Table 13.7, with LS times, $182,000 should have been spent. Hence the project is behind schedule but there is a cost underrun on the whole. 13.22.

Activity

ES

LS

t 

A B C D E F G H I J K L M

0 1 3 4 6 14 12 14 18 18 22 22 18

0 4 3 9 6 15 18 14 21 19 22 23 24

6 2 7 3 10 11 2 11 6 4 14 8 6

Using earliest starting times.

 

Total Cost ($1,000’s)

Cost Per Month

10 14 5 6 14 13 4 6 18 12 10 16 118 146

$1,667 7,000 714 2,000 1,400 1,182 2,000 545 3,000 3,000 714 2,000 3,000

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13-22.

195

PROJECT MANAGEMENT

a. Monthly budget using earliest starting times: ACTIVITY

MONTH

A

B

C

D

E

F

G

H

I

J

K

L

M

Total

1

1667

1667

2

1667

7000

8667

3

1667

7000

8667

4

1667

714

5

1667

714

2000

4381

6

1667

714

2000

4381

7

714

2000

8

2381

1400

4114

714

1400

2114

9

714

1400

2114

10

714

1400

2114

11

1400

1400

12

1400

1400

13

1400

2000

3400

14

1400

2000

3400

15

1400

1182

545

3127

16

1400

1182

545

3127

17

1182

545

1727

18

1182

545

1727

19

1182

545

3000

3000

3000

10727

20

1182

545

3000

3000

3000

10727

21

1182

545

3000

3000

3000

10727

22

1182

545

3000

3000

3000

10727

23

1182

545

3000

714

2000

3000

10442

24

1182

545

3000

714

2000

3000

10442

25

1182

545

714

2000

4442

26

714

2000

2714

27

714

2000

2714

28

714

2000

2714

29

714

2000

2714

30

714

2000

2714

31

714

714

32

714

714

33

714

714

34

714

714

35

714

714

36

714

714

Total

10000

14000

5000

6000

14000

13000

4000

6000

18000

12000

10000

16000

18000 146000

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b. Monthly budget using latest starting times: ACTIVITY MONTH

A

B

C

D

E

F

G

H

I

J

K

L

M

Total

1

1667

1667

2

1667

1667

3

1667

1667

4

1667

5

1667

6

1667

714

2381

7000

714

9381

7000

714

9381

7

714

1400

2114

8

714

1400

2114

9

714

1400

2114

10

714

2000

1400

4114

11

2000

1400

3400

12

2000

1400

3400

13

1400

1400

14

1400

1400

15

1400

16

1400

545

1945

1182

545

3127

17

1182

545

1727

18

1182

545

1727

19

1182

2000

545

3727

20

1182

2000

545

3000

6727

21

1182

545

3000

4727

22

1182

545

3000

3000

7727

23

1182

545

3000

3000

24

1182

545

3000

714

2000

25

1182

545

3000

714

2000

3000

10442

26

1182

3000

714

2000

3000

9896

3000

714

2000

3000

8714

28

714

2000

3000

5714

29

714

2000

3000

5714

30

714

2000

3000

5714

31

714

2000

32

714

714

33

714

714

34

714

714

35

714

714

36

714

714

27

Total

10000

14000

5000

6000

14000

13000

4000

6000

18000

12000

714

10000

8442

16000

7442

2714

18000 146000

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13-23.

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PROJECT MANAGEMENT

Project completion time is 14. This project has to be crashed to 10. This is done by the following linear programming formulation: A 0 0

C 2 2

2 2 2

2 4 4

E 4 4

Start

F 4 10

3 7 13

4 8 8

If  X i is the start time for activity i where i  C, D, E, F, G, and Finish, and Y  j is the amount of time reduced for activity  j, where j  A, B, C, D, E, F, G. H 13 13

2 15 15

Minimize Z  600Y  A  700Y  B  0Y C   75Y  D

Finish



50Y  E   1,000Y F   250Y G

subject to B

3

D

4

0 1

3 4

3 4

7 8

G 8 8

5 13 13

Y  A  1 Y  B  1 Y C   0

The critical path is A–C–E–G–H. Total time is 15 weeks.

Y  D  4

1. Activities A, C, and E all have minimum crash costs per week  of $1,000. 2. Reduce activity E by 1 week for a total cost of $1,000. There are now two critical paths. 3. The total project completion time is now 14 weeks and the new critical paths are B–D–G–H and A–C–E–G–H. 4. Activities D and E have minimum crashing costs per week for each critical path. 5. Reduce activities D and E by 1 week each for a total cost of  $3,000, including the reduction of E by 1 week. 6. The total project completion time is 13 weeks. There are two critical paths: A–C–E–G–H and B–D–G–H.

Y  E   3

13-24.

Start

A

3

D

7

0

3

3

10

0

3

3

10

B

2

E

6

0

2

2

8

G

4

2

4

4

10

10

14

10

14

C

1

F

2

0

1

1

3

11

12

12

14

Finish

Crash Cost per Week

Activity

t

m

n

C

A

3

2

1,000

1,600

$ 600

B

2

1

2,000

2,700

700

C

1

1

300

300

0

 

D

7

3

1,300

1,600

75

E

6

3

850

1,000

50

F

2

1

4,000

5,000

1,000

G

4

2

1,500

2,000

250

Y F   1 Y G  2  X Finish  10  X Finish  X G  Y G  4

 X  D  X  A  Y  A  3

 X G  X  E   Y  E   6

 X Finish  X F   Y F   2

 X G  X  D  Y  D  7

 X F   X C   Y C   1

2

All X i, Y  j  0

 X  E   X  B  Y  B



13-25. The Bender Construction Co. problem is one involving 23 separate activities. These activities, their immediate predecessors, and time estimates were given in the problem. The first results of the computer program are the expected time and variance estimates for each activity. These data are shown in the following table. Activity

Time

Variance

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

3.67 3.00 4.00 8.00 4.17 2.17 5.00 2.17 3.83 1.17 20.67 2.00 1.17 0.14 0.30 1.17 2.00 5.00 0.12 0.14 3.33 0.12 0.17

0.444 0.111 0.111 0.111 0.028 0.250 0.111 0.250 0.028 0.028 1.778 0.111 0.028 0.000 0.001 0.028 0.111 0.444 0.000 0.000 0.444 0.000 0.001

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Next, the computer determines the expected project length, variance, and data for all activities. Like the other network problems, these data include the earliest start, earliest finish, latest start, latest finish, and slack times for all activities. The data are shown in the following table. ACTIVITY TIME Activity S–F

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

ES

EF

LS

LF

0.00 0.00 0.00 0.00 3.67 4.00 8.00 13.00 7.83 3.00 0.00 15.17 20.67 21.83 21.97 21.97 23.14 25.14 30.14 30.25 30.25 30.39 33.59

3.67 3.00 4.00 8.00 7.83 6.17 13.00 15.17 11.67 4.17 20.67 17.17 21.83 21.97 22.27 23.14 25.14 30.14 30.25 30.39 33.59 30.51 33.77

9.00 16.50 14.50 3.50 12.67 18.50 11.50 16.50 16.83 19.50 0.00 18.67 20.67 21.83 24.84 21.97 23.14 25.14 30.14 33.33 30.25 33.47 33.59

12.67 19.50 18.50 11.50 16.83 20.67 16.50 18.67 20.67 20.67 20.67 20.67 21.83 21.97 25.14 23.14 25.14 30.14 30.25 33.47 33.59 33.59 33.77

Slack

9.00 16.50 14.50 3.50 9.00 14.50 3.50 3.50 9.00 16.50 0.00* 3.50 0.00* 0.00* 2.87 0.00* 0.00* 0.00* 0.00* 3.08 0.00* 3.08 0.00*

degree from their particular college or university. For every course, students should list all the immediate predecessors. Then students are asked to attempt to develop a network diagram that shows these courses and their immediate predecessors or prerequisite courses. This problem can also point out some of the limitations of the use of PERT. As students try to solve this problem using the PERT approach, they may run into several difficulties. First, it is difficult to incorporate a minimum or maximum number of courses that a student can take during a given semester. In addition, it is difficult to schedule elective courses. Some elective courses have prerequisites, while others may not. Even so, some of the overall approaches of network analysis can be helpful in terms of laying out the courses that are required and their prerequisites. Students can also be asked to think about other quantitative techniques that can be used in solving this problem. One of the most appropriate approaches would be to use linear programming to incorporate many of the constraints, such as minimum and maximum number of credit hours per semester, that are difficult or impossible to incorporate in a PERT network. 13-27. a. This project management problem can be solved using the PERT model discussed in the chapter. The results are below. As you can see, the total project completion time is about 32 weeks. The critical path consists of Tasks 3, 8, 13, and 15.

Standard Deviation Task Task Task Task Task Task Task Task

*Indicates critical path activity.

As you can see, the expected project length is about 34 weeks. The activities along the critical path are activities 11, 13, 14, 16, 17, 18, 19, 21, and 23.

1 2 3 4 5 6 7 8

0.5 0.1667 0.5 0.5 0.5 0.3333 0.5833 0.6667

Standard Deviation Task 9 Task 10 Task 11 Task 12 Task 13 Task 14 Task 15 Task 16

13-26. The overall purpose of Problem 13-26 is to have students use a network approach in attempting to solve a problem that almost all students face. The first step is for students to list all courses that they must take, including possible electives, to get a

1

9

5

20 2

10

22 15

3

6

4

7

Start

13 8

12

11

Figure for Problem 13-25: Activities for Bender Constructions

18

14

19

23 21

16

17

0.35 0.5 0.6667 0.6667 0.25 0.1667 0.5 0.6667

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Project completion time  32.05 Project standard deviation  1.003466 Early  Start

 Activity time Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task

Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

2.1667 3.5 11.8333 5.1667 3.8333 7 3.9167 7.4667 10.3167 3.8333 4 4 5.9167 1.2333 6.8333 7

0 0 0 0 0 2.1667 3.5 11.8333 11.8333 11.8333 5.1667 3.8333 19.3 15.6667 25.2167 16.9

Early Finish

Late Start

Late Finish

2.1667 3.5 11.8333 5.1667 3.8333 9.1667 7.4167 19.3 22.15 15.6667 9.1667 7.8333 25.2167 16.9 32.05 23.9

10.1333 11.8833 0 14.65 15.9833 12.3 15.3833 11.8333 14.9 19.9833 19.8167 19.8167 19.3 23.8167 25.2167 25.05

12.3 15.3833 11.8333 19.8167 19.8167 19.3 19.3 19.3 25.2167 23.8167 23.8167 23.8167 25.2167 25.05 32.05 32.05

Slack   10.1333 11.8833 0 14.65 15.9833 10.1333 11.8833 0 3.0667 8.15 14.65 15.9833 0 8.15 0 8.15

Task time computations

Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Optimistic Time

Most Likely Time

Pessimistic Time

1 3 10 4 2 6 2 5 9.9 2 2 2 5 1 5 5

2 3.5 12 5 4 7 4 7.7 10 4 4 4 6 1.1 7 7

4 4 13 7 5 8 5.5 9 12 5 6 6 6.5 2 8 9

Activity   Time 2.1667 3.5 11.8333 5.1667 3.8333 7 3.9167 7.4667 10.3167 3.8333 4 4 5.9167 1.2333 6.8333 7

PROJECT MANAGEMENT

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13-27. b. As can be seen in the following analysis, the changes do not have any impact on the critical path or the total project completion time. A summary of the analysis is below. Project completion time  32.05 Project standard deviation  1.003466 Early Start

Early Finish

Late Start

Late Finish

Slack 

0 0 0 0 0 2.1667 3.5 11.8333 11.8333 11.8333 5.1667 3.8333 19.3 11.8333 25.2167 13.0667

2.1667 3.5 11.8333 5.1667 3.8333 9.1667 7.4167 19.3 11.8333 11.8333 9.1667 7.8333 25.2167 13.0667 32.05 20.0667

10.1333 11.8833 0 14.65 15.9833 12.3 15.3833 11.8333 25.2167 23.8167 19.8167 19.8167 19.3 23.8167 25.2167 25.05

12.3 15.3833 11.8333 19.8167 19.8167 19.3 19.3 19.3 25.2167 23.8167 23.8167 23.8167 25.2167 25.05 32.05 32.05

10.1333 11.8833 0 14.65 15.9833 10.1333 11.8833 0 13.3833 11.9833 14.65 15.9833 0 11.9833 0 11.9833

 Activity Time Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

2.1667 3.5 11.8333 5.1667 3.8333 7 3.9167 7.4667 0 0 4 4 5.9167 1.2333 6.8333 7

Standard Deviation Task Task Task Task Task Task Task Task

1 2 3 4 5 6 7 8

Standard Deviation

0.5 0.1667 0.5 0.5 0.5 0.3333 0.5833 0.6667

Task Task Task Task Task Task Task Task

9 10 11 12 13 14 15 16

0 0 0.6667 0.6667 0.25 0.1667 0.5 0.6667

Task time computations Optimistic Most Pessimistic Activity   Time Likely Time Time Time Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task Task

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

1 3 10 4 2 6 2 5 0 0 2 2 5 1 5 5

2 3.5 12 5 4 7 4 7.7 0 0 4 4 6 1.1 7 7

4 4 13 7 5 8 5.5 9 0 0 6 6 6.5 2 8 9

2.1667 3.5 11.8333 5.1667 3.8333 7 3.9167 7.4667 0 0 4 4 5.9167 1.2333 6.8333 7

13-28.

a.  2

Activity

a

m

A

9

10

11

10

0.111

B

4

10

16

10

4

C

9

10

11

10

0.111

D

5

8

11

8

1

b

t

 

 

b. The critical path is AC with an expected completion time of  20. The expected completion time of BD is 18. c. The variance of AC  0.111  0.111  0.222. The variance of BD  4  1  5. d. P(Time forAC  22)  P (Z 

e. P(Time for BD  22)  P (Z 

22  20 0.222 22 18 5

)= P(Z  4.24) =1.00

) = P(Z 1.79 )  0.9 96327

f. The path BD has a very large variance. Thus, it is likely that it will take much longer than its expected time. Therefore, while it is almost certain that the critical path (AC) will be finished in 22 weeks or less, there is only a 96% chance the other path (BD) will be finished in that time.

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a.

Budget schedule based on earliest times. Costs are in $1,000s WEEK

1

2

3

4

5

6

7

8

A

1

1

1

1

1

1

1

1

B

3

3

3

3

ACTIVITY

C D

3

3

3

3

9

10

11

2

2

2

12

13

14

15

16

17

18

19

1.5 1.5

1.5

1.5

1.5

1.5

2

2

2 2

2

2

3.5

3.5

2

2

64

66

3

E F

2

2

G Total in Period

4

4

4

4

4

4

4

4

5

2

Cumulative from start

4

8

12

16

20

24

28

32

37

39

2

3.5 3.5

3.5

3.5

41 44.5

48 51.5

55 58.5

62

13

14

15

16

17

18 19

1.5

1.5

1.5

1.5

1.5 1.5

2

2

2

b. Budget schedule based on latest times. Costs are in $1,000s. WEEK ACTIVITY

A

1

2

3

4

5

6

7

8

1

1

1

1

1

1

1

1

3

3

3

3

B C D

3

3

9

10

11

2

2

2

3

3

3

12

E F

2

2

G

2

2

2

Total in Period

1

1

4

4

4

4

4

4

5

5

5

2

2

3.5

3.5

3.5

3.5

3.5 3.5

Cumulative from start

1

2

6

10

14

18

22

26

31

36

41

43

45

48.5

52

55.5

59

62.5 66

12

13

14

15

16

17

18 19

1.5 1.5

1.5

1.5

1.5

1.5

2

2

2

c.

Budget schedule based on earliest times. Costs are in $1,000s. WEEK

1

2

3

4

5

6

7

8

A

1

1

1

1

1

1

1

1

B

3

3

3

3

ACTIVITY

C D

3

3

3

3

9

10

11

2

2

2

3

E F

2

2

G Total in Period

4

4

4

4

4

4

4

4

5

2

Cumulative from start

4

8

12

16

20

24

28

32

37

39

2

3.5 3.5

41 44.5

3.5

48 5 1.5

4

1

1

3.5

5.5

1

1

55 58.5

64

65

66

3.5

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13-30. The total time to complete the project is 17 weeks. The critical path is A-E-G-H. 13-31.

there are two critical paths A-E-G-H and A-C-F-H. Each of these paths must have their times reduced by one week. The least cost way to do this is to crash H (which is no both paths) I week for an additional cost of $800.

a. Crash G 1 week at an additional cost of $700. b. The paths are A-E-G-H, A-C-F-H, and B-D-G-H. When G is crashed 1 week so the project time is 16 weeks,

SOLUTIONS TO INTERNET HOMEWORK PROBLEMS 13-32. Activity

a

m

b

A B C D E F G H I J K

3 2 1 6 2 6 1 3 10 14 2

6 4 2 7 4 10 2 6 11 16 8

8 4 3 8 6 14 4 9 12 20 10

13-33. A network for the project is shown in the figure shown at the bottom of the page.

Activity

ES

EF

LS

LF

Slack

Critical Path

A B C D E F G H I J K

0 0 0 2.00 9.00 13.00 13.00 23.00 15.17 2.00 29.00

5.83 3.67 2.00 9.00 13.00 23.00 15.17 29.00 26.17 18.33 36.33

7.17 5.33 0 2.00 9.00 13.00 15.83 23.00 18.00 20.00 29.00

13.00 9.00 2.00 9.00 13.00 23.00 18.00 29.00 29.00 36.33 36.33

7.17 5.33 0 0 0 0 2.83 0 2.83 1 8.00 0

No No Yes Yes Yes Yes No Yes No No Yes

The critical path is C–D–E–F–H–K. Project completion time is 36.33.

A 5.83 0 5.83 7.17 13

F 10 13 23 13 23

H

6

23 23

29 29

 

13-34.

Expected Time

Variance

5.83 3.67 2.00 7.00 4.00 10.00 2.17 6.00 11.00 16.33 7.33

0.69 0.11 0.11 0.11 0.44 1.78 0.25 1.00 0.11 1.00 1.78

For the project, expected time  36.33.

V t   0.11  0.11  0.44  1.78  1.00  1.78  5.22

Standard deviation



Probability of finishing project in less than 40 days:   40 − 36.33   P  Z   2.28      P

( Z     1.61)  0.9463

13-35. Before we can determine how long it will take team A to complete its programming assignment, we must develop a PERT diagram. The network showing the activities and node numbers is contained at the end of the solution for this particular problem. Once this network has been constructed, activities, and time estimates can be entered into the computer program. The first result from the computer program is a summarization of the expected time and variance for each activity. This information is shown in the table on the next page.

G 2.17 13 15.17 15.83 18 Start

B 3.67 0 3.67 5.33 9

E 9 9

4 13 13

C

2

D

7

0 0

2 2

2 2

9 9

I 11 15.17 26.17 18 29

J 2 20

Figure for Problem 13-33

16.33 18.33 36.33

2.28.

K 29 29

7.33 36.33 36.33

Finish

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Activity

Time

Variance

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

4.00 5.17 7.83 3.17 7.17 4.00 3.83 4.17 2.17 2.83 4.17 5.83 8.00 4.17 4.00 4.00 3.00 4.83

0.111 0.250 0.250 0.250 0.250 0.111 0.250 0.250 0.250 0.250 0.250 0.250 0.444 0.250 0.111 0.444 0.111 0.250

We can also determine the expected project length and variance. The expected project length is 44 weeks. The variance is 2.167. In addition, we can determine the earliest start, earliest finish, latest start, latest finish, and slack times for all activities along the critical path. This information is shown in the table. As can be seen in the table, the critical path for this particular problem includes activities 1, 3, 9, 11, 12, 13, 14, 17, and 18. The solution, however, is not complete. Software Development Specialist (SDS) is not sure about the time estimates for activity 5. As indicated in the problem, these time estimates might be as high as 12, 14, and 15 weeks for the optimistic, most likely, and pessimistic times. Now, we must find out what impact this possible increase in expected times would have on the network. Fortunately, our computer program has a convenient rerun capability. We are able to go back to the original data, modify the time estimates for these activities, and resolve the problem. Doing this will result in an expected project completion time of 47.83 weeks. The variance of the project is approximately 1.92 weeks. Will this change the critical path? The answer is yes. The critical path now includes activities 1, 5, 11, 12, 13, 14, 17, and 18. Activity 5 now lies along the critical path. The earliest start, earliest finish, latest start, latest finish, and slack times for all activities with the new time estimates for activity 5 are shown below:

ACTIVITY TIME Activity

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

ES

EF

LS

LF

0.00 4.00 4.00 4.00 4.00 4.00 8.00 7.17 11.83 9.17 14.00 18.17 24.00 32.00 14.00 18.00 36.17 39.17

4.00 9.17 11.83 7.17 11.17 8.00 11.83 11.33 14.00 12.00 18.17 24.00 32.00 36.17 18.00 22.00 39.17 44.00

0.00 6.00 4.00 6.67 6.83 6.17 10.17 9.83 11.83 11.17 14.00 18.17 24.00 32.00 31.17 35.17 36.17 39.17

4.00 11.17 11.83 9.83 14.00 10.17 14.00 14.00 14.00 14.00 18.17 24.00 32.00 36.17 35.17 39.17 39.17 44.00

Slack

0.00* 2.00 0.00* 2.67 2.83 2.17 2.17 2.67 0.00* 2.00 0.00* 0.00* 0.00* 0.00* 17.17 17.17 0.00* 0.00*

ACTIVITY TIME Activity

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

*Indicates critical path activity.

ES

EF

LS

LF

0.00 4.00 4.00 4.00 4.00 4.00 8.00 7.17 11.83 9.17 17.83 22.00 27.83 35.83 17.83 21.83 40.00 43.00

4.00 9.17 11.83 7.17 17.83 8.00 11.83 11.33 14.00 12.00 22.00 27.83 35.83 40.00 21.83 25.83 43.00 47.83

0.00 9.83 7.83 10.50 4.00 10.00 14.00 13.67 15.67 15.00 17.83 22.00 27.83 35.83 35.00 39.00 40.00 43.00

4.00 15.00 15.67 13.67 17.83 14.00 17.83 17.83 17.83 17.83 22.00 27.83 35.83 40.00 39.00 43.00 43.00 47.83

*Indicates critical path activity.

Start

1

2

10

3

9

4

8

5 6

Figure for Problem 13-35

7

11

12

15

16

203

PROJECT MANAGEMENT

13

14

17

18

Finish

Slack

0.00* 5.83 3.83 6.50 0.00* 6.00 6.00 6.50 3.83 5.83 0.00* 0.00* 0.00* 0.00* 17.17 17.17 0.00* 0.00*

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13-36 a. The first step for Jim Sager is to summarize the time estimates for each of the activities, shown in the following table. Activity

Optimistic

1(A) 2(B) 3(C) 4(D) 5(E) 6(F) 7(G) 8(H) 9(I) 10(J) 11(K) 12(L) 13(M) 14(N) 15(O) 16(P) 17(Q) 18(R) 19(S) 20(T) 21(U) 22(V) 23(W)

2 5 1 8 1 3 1 5 9 1 2 3 2 8 1 4 6 1 6 3 1 9 2

Likely

Pessimistic

3 6 1 9 1 3 2 5 10 2 2 4 2 9 1 4 6 2 6 3 2 10 4

4 8 2 11 4 4 2 6 11 2 3 6 4 11 3 8 7 4 7 4 3 11 5

The next step is to compute the average or mean times and the standard deviations (S.D.) for each activity. The table below contains this information along with activity variances. Critical path activities are also shown with an asterisk (*). Activity

1(A) 2(B) 3(C) 4(D)* 5(E) 6(F) 7(G) 8(H) 9(I)* 10(J) 11(K) 12(L) 13(M) 14(N)* 15(O) 16(P) 17(Q) 18(R) 19(S)* 20(T) 21(U) 22(V) 23(W)*

Mean

3.000 6.167 1.167 9.167 1.500 3.167 1.833 5.167 10.000 1.833 2.167 4.167 2.333 9.167 1.333 4.667 6.167 2.167 6.167 3.167 2.000 10.000 3.833

*Critical path activities.

S.D.

0.333 0.500 0.167 0.500 0.500 0.167 0.167 0.167 0.333 0.167 0.167 0.500 0.333 0.500 0.333 0.667 0.167 0.500 0.167 0.167 0.333 0.333 0.500

Earliest and latest start and finish times (ES, EF, LS, and LF) can also be computed for each activity. This is shown in the table below, along with slack for each activity.

Variance

0.111 0.250 0.028 0.250 0.250 0.028 0.028 0.028 0.111 0.028 0.028 0.250 0.111 0.250 0.111 0.444 0.028 0.250 0.028 0.028 0.111 0.111 0.250

ACTIVITY TIMES Activity

1(A) 2(B) 3(C) 4(D) 5(E) 6(F) 7(G) 8(H) 9(I) 10(J) 11(K) 12(L) 13(M) 14(N) 15(O) 16(P) 17(Q) 18(R) 19(S) 20(T) 21(U) 22(V) 23(W)

ES

0.00 0.00 0.00 0.00 3.00 6.17 6.17 1.17 9.17 9.17 4.50 9.33 8.00 19.17 19.17 11.00 6.67 13.50 28.33 20.50 15.67 15.67 34.50

EF

LS

LF

Slack

3.00 6.17 1.17 9.17 4.50 9.33 8.00 6.33 19.17 11.00 6.67 13.50 10.33 28.33 20.50 15.67 12.83 15.67 34.50 23.67 17.67 25.67 38.33

15.50 12.67 17.50 0.00 18.50 18.83 22.00 18.67 9.17 26.00 20.00 22.00 23.83 19.17 30.00 27.83 22.17 26.17 28.33 31.33 32.50 28.33 34.50

18.50 18.83 18.67 9.17 20.00 22.00 23.83 23.83 19.17 27.83 22.17 26.17 26.17 28.33 31.33 32.50 28.33 28.33 34.50 34.50 34.50 38.33 38.33

15.50 12.67 17.50 0.00* 15.50 12.67 15.83 17.50 0.00* 16.83 15.50 12.67 15.83 0.00* 10.83 16.83 15.50 12.67 0.00* 10.83 16.83 12.67 0.00*

*Critical path activities.

The final network results are summarized below: Expected project length



38.3333

Variance of the critical path



0.8888

Standard deviation



0.9428

As seen above, the project will be completed in less than 40 weeks. 13-37. If activity D has already been completed, activity time for D is 0. The results are shown on the next page. As you can see, activity D (4) is still on the critical path. The project completion time is now about 29 weeks.

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Table for Problem 13-37 Activity

1(A) 2(B) 3(C) 4(D)* 5(E) 6(F) 7(G) 8(H) 9(I)* 10(J) 11(K) 12(L) 13(M) 14(N)* 15(O) 16(P) 17(Q) 18(R) 19(S)* 20(T) 21(U) 22(V) 23(W)*

Mean

3.000 6.167 1.167 0.000 1.500 3.167 1.833 5.167 10.000 1.833 2.167 4.167 2.333 9.167 1.333 4.667 6.167 2.167 6.167 3.167 2.000 10.000 3.833

S.D.

Variance

0.333 0.500 0.167 0.000 0.500 0.167 0.167 0.167 0.333 0.167 0.167 0.500 0.333 0.500 0.333 0.667 0.167 0.500 0.167 0.167 0.333 0.333 0.500

0.111 0.250 0.028 0.000 0.250 0.028 0.028 0.028 0.111 0.028 0.028 0.250 0.111 0.250 0.111 0.444 0.028 0.250 0.028 0.028 0.111 0.111 0.250

*Critical path activities.

Expected completion time is 29.167 weeks. 13-38. The results of having both activity D (4) and I (9) completed are shown below. These activities are no longer on the critical path. The project completion time is now about 26 weeks. Activity

1(A) 2(B) 3(C) 4(D) 5(E) 6(F) 7(G) 8(H) 9(I) 10(J) 11(K) 12(L) 13(M) 14(N) 15(O) 16(P) 17(Q) 18(R) 19(S) 20(T) 21(U) 22(V) 23(W)

Mean

3.000 6.167 1.167 0.000 1.500 3.167 1.833 5.167 0.000 1.833 2.167 4.167 2.333 9.167 1.333 4.667 6.167 2.167 6.167 3.167 2.000 10.000 3.833

S.D.

Variance

0.333 0.500 0.167 0.000 0.500 0.167 0.167 0.167 0.000 0.167 0.167 0.500 0.333 0.500 0.333 0.667 0.167 0.500 0.167 0.167 0.333 0.333 0.500

0.111 0.250 0.028 0.000 0.250 0.028 0.028 0.028 0.000 0.028 0.028 0.250 0.111 0.250 0.111 0.444 0.028 0.250 0.028 0.028 0.111 0.111 0.250

Critical path activities: B–F–L–R–V

Expected completion time is 25.667 weeks.

205

PROJECT MANAGEMENT

13-39. Changing the immediate predecessor activity will change the structure of the network. Fortunately, we can handle this situation. The results are shown below. Activity F (6) now goes from node 2 to node 7. Node 2 is the ending node for activity A (1). Thus activity F now has activity A as an immediate predecessor. Activity

Mean

S.D.

Variance

1(A) 2(B)* 3(C) 4(D) 5(E) 6(F)* 7(G) 8(H) 9(I) 10(J) 11(K) 12(L)* 13(M) 14(N) 15(O) 16(P) 17(Q) 18(R)* 19(S) 20(T) 21(U) 22(V)* 23(W)

3.000 6.167 1.167 0.000 1.500 3.167 1.833 5.167 0.000 1.833 2.167 4.167 2.333 9.167 1.333 4.667 6.167 2.167 6.167 3.167 2.000 10.000 3.833

0.333 0.500 0.167 0.000 0.500 0.167 0.167 0.167 0.000 0.167 0.167 0.500 0.333 0.500 0.333 0.667 0.167 0.500 0.167 0.167 0.333 0.333 0.500

0.111 0.250 0.028 0.000 0.250 0.028 0.028 0.028 0.000 0.028 0.028 0.250 0.111 0.250 0.111 0.444 0.028 0.250 0.028 0.028 0.111 0.111 0.250

*Critical path activities.

Expected completion time is 22.833 weeks.

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SOLUTIONS TO SOUTHWESTERN UNIVERSITY STADIUM CONSTRUCTION CASE 1.

B 30 60

Start

A 0 0

60 90 1 20

E 30 9 0 1 20 120 150

F 1 120 121 259 260

30 30 30

Finish

H 20 180 200 180 200

C 30 30

65 95 95

D 95 95

55 1 50 1 50

G 30 150 180 150 180

I 30 200 230 200 230

L 30 230 260 230 260

J 10 200 210 219 229

K 1 210 211 229 230

Figure 1 Network Using Activity-On-Node Notation The expected times (t ) and the variance for each activity are shown in the table.

Activity

Optimistic time

Most Likely time

Pes simistic time

Activity time (t )

Standard Deviation

Variance

A

20

30

40

30

3.333333

11.11111

B

20

65

80

60

10

100

C

50

60

100

65

8.333333

69.44444

D

30

50

100

55

11.66667

136.1111

E

25

30

35

30

1.666667

2.777778

F

1

1

1

1

0

0

G

25

30

35

30

1.666667

2.777778

H

10

20

30

20

3.333333

11.11111

I

20

25

60

30

6.666667

44.44444

J

8

10

12

10

0.6666667

0.4444445

K

1

1

1

1

0

0

L

20

25

60

30

6.666667

44.44444

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To find the critical path, the early start and finish times together with the latest times are used to find the slack as shown in the table. From this, the critical path is found.

Activity

Activity time

Early Start

A

30

0

B

60

C

65

D E

Early Finish

Late Start

Late Finish

30

0

30

0

30

90

60

120

30

10

30

95

30

95

0

8.333333

55

95

150

95

150

0

11.66667

30

90

120

120

150

30

1.666667

F

1

120

121

259

260

139

G

30

150

180

150

180

0

1.666667

H

20

180

200

180

200

0

3.333333

I

30

200

230

200

230

0

6.666667

J

10

200

210

219

229

19

0.6666667

K

1

210

211

229

230

19

0

L

20

230

260

230

260

0

6.666667

The project is expected to take 260 weeks. The critical path consists of activities A-C-D-G-H-I-L. 2. To find the probabilities, we add the variances of the critical activities and find a project variance of 319.444. The standard deviation is 17.873. Letting X  project completion time, P(X  270)  P (Z 

270  260 17.873

) = P(Z  0.56 ) =0.71 226

Thus, there is about 71% chance of finishing the project in 270 weeks. 3. To get a completion time of 250 days, we crash activity A for 10 days at a cost of $15,000. This reduces the time to 250 days. To get a completion time of 240 days, in addition to crashing A for 10 days, we crash activity D for 10 days at a cost of $19,000. The total cost of crashing is $34,000.

Slack

Standard Deviation

3.33333

0

207

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SOLUTION TO FAMILY PLANNING RESEARCH CENTER OF NIGERIA CASE This case covers three aspects of project management: 1. Critical path scheduling 2. Crashing 3. Personnel smoothing The statement by Mr. Odaga that the project will take 94 days is a red herring. That is the sum of all the task times that would be the length of the project only if all of the tasks were done serially with none in parallel. Therefore, the assignment questions would be as follows:  Network formulation. Figure 1 shows a PERT formulation of  a network based on the data on precedences and task (activity) times for each activity. The critical path is C–H–I–J–K of length 67. Table 1 shows the earliest start and finish times and the slacks for each activity, confirming this definition of the critical path. Workforce smoothing. The case asks whether the effort can be carried out with the current staff of 10. Figure 2 (on the next page) shows the network with the staffing requirements. Table 2 (on the next page) shows a blank form that can be used to insert the staffing by activity and compute the daily staffing requirements. This form is used in Table 3 and shows the staffing requirement with each activity beginning on its earliest start date. There are five days on which there are requirements for more than 10 workers. Delaying of some of the activities with slack (activities D, E, F, and G) results in the feasible schedule in Table 4 (on page 217).

Table 1 Latest and earliest starting times and slack Activity

LS

ES

Slack

A. Identify faculty B. Arrange transport C. Identify material D. Arrange accommodations E. Identify team F. Bring in team G. Transport faculty H. Print materials I. Deliver materials J. Train K. Fieldwork

8 12 0 19 13 20 19 5 15 22 37

0 0 0 5 5 12 7 5 15 22 37

8 12 0 14 8 8 12 0 0 0 0

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A 0 8

5 5 13

Start

D 5 19

3 8 22

E 5 13

7 12 20

B

7

0

7

G

3

12

19

7 19

10 22

C 0 0

5 5 5

H 5 5

10 15 15

F 12

2 14

J

15

K

30

20

22

22

37

37

67

22

37

37

67

I 15 15

PROJECT MANAGEMENT

Finish

7 22 22

Figure 1 Network for Family Planning Research

A 0 Staff

D 3 5 8 Staff 1

5 5 2

E 7 5 12 Staf f 4

Start B 0

7 7

G

3

Staff

3

7

10

C 0

5 5

Staff

2

F

2

12 Staff

14 1

J 22

15 37

Staff

0

K 37

30 67

Finish

Staff 0

Staf f 6 H 5

I 15 Staff

10 15

7 22 3

Staf f 2

Figure 2 Staffing Network for Family Planning Research

Table 2 Blank Staffing Chart DAY ACTIVITY

A B C D E F G H I Total

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20 21 22

209

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Table 3 Chart Showing Each Day’s Manpower Requirements if All Activities Are Started at ES DAY ACTIVITY

1

2

3

4

5

6

7

A

2

2

2

2

2

B

3

3

3

3

3

C

2

2

2

2

2

8

3

3

D

1

1

1

E

4

4

4

9

10

11

12

4

4

4

4

F G H

6

6

2

2

2

6

6

6

6

6

13

14

1

1

6

6

15

7

7

7

7

7

14

14

13

12

12

10

10

7

17

18

19

20 21 22

3

3

3

3

3

3

3

3

3

6

I Total

16

7

6

3

3

3

3

3

15

16

17

18

19

20 21 22

1

1

1

1

1

Table 4 Minimum Number of Personnel Needed for 22-Day Completion Time DAY ACTIVITY

1

2

3

4

5

A

2

2

2

2

2

B

3

3

3

3

3

C

2

2

2

2

2

6

7

3

3

8

9

10

11

12

13

14

4

4

4

4

4

4

4

D E F G

2

H

6

6

6

6

6

6

6

6

6

6

7

7

7

7

7

9

9

10

10

10

10

10

10

10

8

2

3

3

3

3

3

3

6

5

3

3

3

3

6

I Total

2

10

Table 5 Crashing Procedure Step

Length (Days)

Total Cost

1. Original network 2. Crash C 5–3 3. Crash I 7–2 4. Crash H 10–9 Second critical path emerges 5. Crash A 5–2 and H 9–6 6. Crash H 6–5 and E 7–6 Third critical path emerges 7. Crash J 15–10 8. Crash K 30–20

67 65 60 59

$25,400 25,500 25,900 26,100

56 55

27,000 27,350

50 40

29,350 33,350

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Crashing the schedule. Since the objective is a 60-rather than a 67-day schedule, the team must investigate the possibilities of  crashing activities on the critical path(s) to reduce project duration using the data exhibited in the case. Table 5 shows the sequence of  crashing to get to various project lengths. Getting to 60 days is relatively easy and relatively cheap. Activity C is reduced by 2 days at a cost of $50 per day. The next cheapest alternative is activity I, which can be cut 5 days, for a total cost of $400. Therefore, Dr. Watage needs to request $500 from the Pathminder Foundation to crash the project to the 60-day duration. The instructor can also use these data to indicate to the students how further crashing would generate multiple parallel paths and necessitate use of a heuristic rule to select the activities to be cut further to shorten the network. Warning: Take up the workforce smoothing before you take up crashing. After you have smoothed out the labor and then crashed the project by 7 days, the network A through I will go from 22 to 15 days and the project will be infeasible with the 10 personnel at hand. Don’t try to redo the smoothing. Just indicate to the students that the extra money used for crashing might have been used to hire temporary help to overcome this constraint. Some students may try to do the crashing and then the smoothing and become stymied by the resulting infeasibility.

SOLUTION TO INTERNET CASE Solution to Cranston Construction Company Case Critical path scheduling is a management tool, initiated by the government and industry in 1957, which has developed into a useful method of planning, scheduling, and controlling projects, usually on a large scale. The application of the method to the Apollo project is one of the most outstanding examples of the method’s effectiveness in coordinating the activities of many different groups of people. Construction projects almost invariably have a deadline to meet with an associated penalty should the deadline not be met. It is to the benefit of the contractor to meet the deadline to avoid the penalty as well as to free his men and equipment for other projects. Unfortunately many construction managers use intuition coupled, perhaps, with simple planning techniques, and the result is less than an optimal solution to the scheduling problem. For moderate sized projects, the critical path method can be applied to an advantage using pencil and paper techniques. For larger projects, many computer programs may be used to simplify the calculations. The mathematical foundations on which the critical path method rests are quite sophisticated, but it is not necessary to master the underlying mathematics to be able to apply the principle of the method to project planning. The result is greater working efficiency and cost savings for the contractor. It is necessary to note the great importance accurate time estimates have in critical path analysis. If, at any time an activity is lengthened, the analysis should be checked to assure that the critical path has not shifted. In devising a critical path analysis for any project, it is necessity to list four things: 1. List activities necessary to complete the projects. This must be a complete list from the beginning to the end of  the project.

PROJECT MANAGEMENT

211

2. List predecessors to each activity. 3. List successors to each activity. 4. List activities concurrent with each activity. When the planner has compiled these lists, a much better grasp of  the project will enable drawing a network graph. The activities list for the Humanities Building at the University of Northern Mississippi is shown by Table 1. The events are numbered on the network graph shown by Figure 1, but in the list, each activity is given a letter designation for ease of reference. After a list of all of the necessary activities has been compiled in a project, each activity can be assigned a letter. The order of assignment is unimportant. Only the immediate predecessors and successors of each activity are listed with the understanding that if an event is a necessary prerequisite for a second event, then it is also a prerequisite for any third activity which has the second activity as a prerequisite. In the activity list the question arises as to the degree of detail necessary. It is usually profitable to list general activities at first, and construct an initial network diagram. Then it is possible to take the general activities and subnet them as necessary. Thus the overall project can be kept easily in mind, while at the same time retaining control over each activity to any degree of accuracy desired. After a list of the necessary activities to complete the project has been compiled, along with the precedence relationships for each, the network graph may be constructed. The graph shows, much more clearly, the order in which the activities must be undertaken. It also indicates the critical, or longest path in the network. It is this path that governs project completion time and thus requires the greatest management concern. On the graph are listed the expected activity times as estimated by the contractor. Using software it is also possible to make optimistic and pessimistic estimates with the expected times to get a mean value c alculated using a beta distribution. This could prove valuable, even in construction work, for activities often slip due to adverse weather, long delivery times, etc. A great deal has been written about various types of float or slack time occurring in a critical path network. The contractor is primarily interested in float as a means of indicating which pro jects can be shifted in time, to better use his resources. Those activities with no float are on the critical path and cannot be shifted. Thus all activities not on the critical path necessarily have some time which can be used prior to reaching critical events. The construction of the Humanities Building at the University of Northern Mississippi involved very high costs and was directly amenable to critical path methods. The project extended over a period of approximately one year. In a project of this length, weekly reports by the contractor would be necessary for controlling the project. In this way a troublesome delay in the critical path could be detected and circumvented. Also, the use of resources could be monitored, along with project expenditures. A useful, yet simple method of monitoring the project was introduced by Walker and Houry. This consists of  drawing a curve correlating expenditures and project duration from the expected times on the network graph, before the project begins. Then, reports from the contractor are compiled showing actual expenditures plotted against time. This provides a measure of the amount of project completion at any point in time.

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Table 1 Activities Humanities Building University of Northern Mississippi Activity

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z AA BB CC DD EE FF GG HH II

Predecessors

Excavate Tax & Ins. General Conditions Grade Beams Foundations Lower Floor Concrete Lower Floor Columns Lower Floor Frame Middle Floor Concrete Middle Floor Columns Middle Floor Frame Upper Floor Concrete Upper Floor Columns Upper Floor Frames Upper Floor Door Frames Roof Slab Elevator Lathe & Plaster Upper Floor Masonry Pent. Steel & Conc. Ceilings Paint Millwork Sitework Tile & Carpet Clean Up Tile & Marble Stairwells Hardware Lower Fl. Door Frames Lower Floor Masonry Exterior Doors Glazing & Store Front Middle Floor Dr. Frame Middle Floor Masonry

Successors

A, B, C A, B, C D, E F G H I J K L M N N N K II, O, P, Q II, O, P, Q II, O, P, Q U S, T S, T V, W GG, FF, X, Y, AA, CC R H BB H DD H H K EE, HH

Simultaneous

D, E D, E D, E F F G H GG, FF, DD, BB, I J K HH, L, R M N O, P, Q S, T, U S, T, U S, T, U AA X, W X, W V Y Y Z Z

DD, FF, GG EE, FF, DD, GG EE, FF, GG H, FF, GG HH, II, GG, FF HH, II, GG, AA P, Q Q, O O, P L, HH, BB T, U S, U S, T W, X, FF, AA, BB X, U, V, FF, GG W, U, V GG, FF, X, ZZ, CC

Z CC Z EE II Z Z II S, T, U

M, I, CC, AA, FF R, L, HH AA, M, N, II I, FF J, FF, GG DD, I I, J, K, L, M, N, P L, R, BB M, AA, CC

GG FF A

Z

D Start

B

II

EE

S

DD

O

X

HH C

E

F

G

H I

J

K

L

M

N

T

P

W Y

Q U R BB

Figure 1 Network Graph for Cranston Case

V AA CC

B, C A, C A, B E D

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Solution to Alpha Beta Gamma Record Case

PROJECT MANAGEMENT

2. The critical path has an expected length of 31.5 with variance of 0.6944. This yields a standard normal variable

1. The PERT diagram is shown on the following page. The activity times are the averages calculated from the formula

Z   ( 35  31 .5 ) /

t  (a  4m  b)/6

3. The second solution critical path has an expected length of  31.0 with variance 0.6944. This yields a standard normal value of  4.8; virtually all of the issues will be on time. 4. This question is behavioral in nature and can be answered in a multitude of ways. Factors in this analysis could include the alum’s status with the fraternity, the possibility of a reduction in printing costs from Thrift Print and the possibility of reducing the number of issues of the  Record. Depending on the factors discussed, many system-wide effects could be felt.

Table 1 Mean and Variance for Variable Length Activities Activity

Mean

Variance

A B C D H I J L Q

2 2 1 1 1 3 3 2 1

0.1111 0.4444 0.0278 0.1111 0.0069 0.4444 0.4444 0.1111 0.0278

PERT Networks: Thrift Print and Kwik Print showing expected values

Start

D 1

K 2

C 1

E 2

J 3

L 2

M 2

R 0

N 1

O 0.5

Q 1

V 1

W 1

S 1

T 3

U 1

X 1

AA 1

BB 4

G 2 B 2

F 1 H 1

I 3

P 0.5

Y 1

Z 4

A. Thrift Print Total Completion time  31.5 days

Start

A 2

D 1

K 2

C 1

E 2

J 3

L 2

M 2

R 0

N 1

O 0.5

Q 1

V 1

W3

S 1

T 5

U 1

X 0.5

AA 1

G 2 B 2

F 1 H 1

0 .6944  4 .2

corresponding to the 99.99 percentile of the normal distribution.

where a is the minimum, m is the most likely, and b is the maximum activity time. These are shown in Table 1 for those activities whose times might vary. Also shown are the variances of these activity times calculated from v  [(b  a)/6]2. The activities not shown in Table 1 are deterministic with variance zero.

A 2

213

I 3

B. Kwik Prin t Total Completion time  31 days

P 0.5

Y 1

Z 1

BB 3

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SOLUTION TO HAYGOOD BROTHERS CONSTRUCTION COMPANY CASE Activity

AB BC CD CE DF FG FH FI FJ JK KL KM MN LO OP PQ

 

a

m

b

te

4 2 5 4 2 3 4 3 5 10 4 7 4 5 5 2

5 5 7 5 4 5 5 4 7 11 6 8 5 7 6 3

6 8 9 6 6 9 6 7 9 12 8 9 10 9 7 4

5 5 7 5 4 513 5 413 7 11 6 8 523 7 6 3

2

 

 

1 9

1 4 9 1 9 4 9

1 1 9 4 9 4 9 1 9 4 9 1 9

1

Event

T E 

T L

AB BC CD CE DF FG FH FI FJ JK KL KM MN LO OP PQ

0 5 10 10 17 21 21 21 21 28 39 39 47 45 52 58

0 5 10 23 17 33–13 34 34 –23 21 28 39 44–13 52 –13 45 52 58

4 9 1 9 1 9

 z 

The critical path is A–B–C–D–F–J–K–L–O–P–Q (61 days). A delay in the completion of an event on the critical path will delay the entire project by an equal amount of time.

T E  T S 

σ

T   E 



61 60 1.92

 

Slack

0 0 0 13 0 12–23 13 13–23 0 0 0 5–13 5–13 0 0 0

 0.52

 P (T  )  30.15 %  s

Solution to Shale Oil Company Internet Case Study 1. Determine the expected shutdown time, and the probability the shutdown will be completed one week earlier. 2. What are the probabilities that Shale finishes the maintenance project one day, two days, three days, four days, five days, or six days earlier? From the precedence data supplied in the problem, we can develop the following network:

3

Start

1

2

8

16

21

9

17

23

4

10

18

22

5

12

24

11

19

27

25 6

14 20 13

7

26 15

28

29

Finish

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The following table indicates the expected times, variances, and slacks needed to complete the rest of the problem. Task

Optimistic

Most likely

Pessimistic

E(t )

1

1

2

2.5

1.92

2

1.5

2

2.5

2

3

2

3

4

3

4

1

2

3

2

5

1

2

4

2.17

6

2

2.5

3

2.5

7

2

4

5

3.83

8

1

2

3

2

9

1

1.5

2

10

1

1.5

2

11

2

12

15

13

1

14

2.5

ES

 

LS

LF

Slack

0

1.92

0

1.92

0

.17

1.92

3.92

1.92

3.92

0

.33

3.92

6.92

3.92

6.92

0

.33

3.92

5.92

22.5

24.5

3.92

6.08

10.25

12.42

3.92

6.42

13.42

15.92

10

3.92

7.75

29.58

33.42

25.67

.33

6.92

8.92

6.92

8.92

1.5

.17

5.92

7.42

26.67

28.17

20.75

1.5

.17

5.92

7.42

24.5

26

18.58

3

2.5

30

20.83

1.5

2

1.5

3

5

8

5.17

15

3

8

15

8.33

16

14

21

28

17

1

5

10

5.17

18

2

5

10

19

5

10

20

10

21

4

22

0.25

EF

0.5 .17 0.5

6.333

0

6.08

8.58

19.92

22.42

13.83

6.08

26.92

12.42

33.25

6.33

.17

6.42

7.92

15.92

17.42

10

.83

6.42

11.58

28.08

33.25

21.67

2

7.75

16.08

33.42

41.75

25.67

2.33

8.92

29.92

8.92

29.92

1.5

7.42

12.58

28.17

33.33

20.75

5.33

1.33

7.42

12.75

26

31.33

18.58

20

10.83

2.5

8.58

19.42

22.42

33.25

13.83

15

25

15.83

2.5

7.92

23.75

17.42

33.25

10

5

8

5.33

.67

29.92

35.25

29.92

35.25

0

1

2

3

2

.33

12.75

14.75

31.33

33.33

18.58

23

1

2

2.5

1.92

0.25

14.75

16.67

33.33

35.25

18.58

24

1

2

3

2

.33

26.92

28.92

33.25

35.25

6.33

25

1

2

3

2

.33

23.75

25.75

33.25

35.25

9.5

26

2

4

6

4

.67

16.08

20.08

41.75

45.75

25.67

27

1.5

2

2.5

2

.17

35.25

37.25

35.25

37.25

0

28

1

3

5

3

.67

37.25

40.25

37.25

40.25

0

29

3

5

10

1.17

40.25

45.75

40.25

45.75

0

20

21

5.5

From the table, we can see that the expected shutdown time is 45.75 or 46 days. There are 9 activities on the critical path.

.17

18.58

2.5

0

Activities on the critical path Task



 

2

1

0.25

0.0625

2

0.17

0.0289

3

0.33

0.1089

8

0.33

0.1089

16

2.33

5.4289

21

0.67

0.4489

27

0.17

0.0289

28

0.67

0.4489

29

1.17

1.3689

Variance for critical path:

8.0337

REVISED M13_REND6289_10_IM_C13.QXD

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11:32 AM

CHAPTER 13

Page 216

P R O J E C T  M A N A G E M E N T

Therefore,    2.834. As an approximation, we can use the customary equation for the Normal Distribution: z

Solution to Bay Community Hospital Internet Case Study 1.

The CPM network is as follows:

Due date  E ( t ) 

(Note: This might be a good time to discuss the difference between a continuous and a discrete probability distribution, and the appropriate procedure for using a continuous distribution as an approximation to a discrete, if you have not already done so.) Finish Time

Z 

One day early



Two days early



Three days early



Four days early



Five days early



Six days early



Seven days early



 

Probability

0.353

36.3%

0.706

24.0

1.058

14.5

1.411

7.9

1.764

3.9*

2.117

1.7

2.470

0.7

*The appropriate procedure for using the Normal distribution gives 3.0%—roughly a 30% difference.

There is, by the approximate procedure used, a 3.9% probability of finishing five days or one week early. 3. Shale Oil is considering increasing the budget to shorten the shutdown. How do you suggest the company proceed? In order to shorten the shutdown, Shale Oil would have to determine the costs of decreasing the activities on the critical path. This is the vessel and column branch of the network which is typically the longest section in a shutdown. The cost of reducing activity time by one time unit for each activity in this branch would have to be calculated. The activity with the lowest of these costs could then be acted upon. Perhaps the repairs to the vessels and columns could be expedited with workers from some of the other branches with high slack time. However, delivery on materials could be an overriding factor.

Start

A 0 3

2 2 5

C 2 5

3 5 8

B 0 1

4 4 5

D 4 6

4 8 10

E 0 0

8 8 8

Finish

F 8 8

2 10 10

The times in the network are the expected times shown in Exhibit 1 of the Case. The completion time is 10 weeks with critical path e, f. 2. If activity e on the critical path is reduced by one week using express truck, the completion time becomes 9 weeks with two critical paths: e, f and b, c, f. The completion time can be reduced to 8 weeks by resorting to air shipment in activity e and using overtime in activity c. 3. The cost of air shipment ($750) and overtime ($600) would increase the cost by $1,350. However, $300 could be saved by allowing activity a (not on any of the critical paths) to take 3 weeks yielding a net cost increase of $1,050.