Ch07 Solutions

38
CHAPTER 7 PROJECT MANAGEMENT SOLUTIONS TO DISCUSSION QUESTIONS 7-1. 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. 7-2. 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. 7-3. An activity is a task that requires a fixed amount of time and resources to complete. An immediate predecessor is an activity that must be completely finished before another activity can be started. 7-4. 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. 7-5. 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

Transcript of Ch07 Solutions

Page 1: Ch07 Solutions

CHAPTER 7

PROJECT MANAGEMENT

SOLUTIONS TO DISCUSSION QUESTIONS

7-1. 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.

7-2. 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.

7-3. An activity is a task that requires a fixed amount of time and resources to complete. An immediate predecessor is an activity that must be completely finished before another activity can be started.

7-4. 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.

7-5. 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.

7-6. The earliest activity start time is the earliest time that an activity can be started after all predecessor activities are completely finished. The earliest activity start times are determined using a forward pass through the project 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.

7-7. 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 project. 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.

7-8. 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

Page 2: Ch07 Solutions

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.

7-9. 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.

7-10. 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 that 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.

7-11. 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.

SOLUTIONS TO PROBLEMS

7-12. See file P7-12.XLS.(a)

(b)Activity Time EST EFT LST LFT Slack Critical?

A 2 0 2 13 15 13 NB 5 0 5 0 5 0 YC 1 0 1 11 12 11 ND 10 5 15 5 15 0 YE 3 15 18 15 18 0 YF 6 1 7 12 18 11 NG 8 18 26 18 26 0 Y

Critical path = B-D-E-GProject length = 26 days

B

A

C

D

E

F

GStart

Page 3: Ch07 Solutions

7-13. See file P7-13.XLS.(a)

(b)Activity Time EST EFT LST LFT Slack Critical?

A 3 0 3 0 3 0 YB 5 0 5 2 7 2 NC 7 3 10 3 10 0 YD 5 3 8 9 14 6 NE 4 10 14 10 14 0 YF 6 5 11 7 13 2 NG 2 14 16 14 16 0 YH 3 11 14 13 16 2 N

Critical path = A-C-E-GProject length = 16 weeks

A

B

C

D

EH

F

G

StartEnd

Page 4: Ch07 Solutions

7-14. See file P7-14.XLS.(a)

(b)Activity Time EST EFT LST LFT Slack Critical?

A 6 0 6 0 6 0 YB 5 0 5 0 5 0 YC 3 6 9 6 9 0 YD 2 6 8 10 12 4 NE 4 5 9 5 9 0 YF 6 5 11 6 12 1 NG 10 9 19 9 19 0 YH 7 11 18 12 19 1 N

Critical paths = A-C-G & B-E-GProject length = 19 weeks

A

B

D

C

E

H

F

G

Start

End

Page 5: Ch07 Solutions

7-15. See file P7-15.XLS.(a)

(b)Activity Time EST EFT LST LFT Slack Crit?

A 4 0 4 0 4 0 YB 6 4 10 4 10 0 YC 12 10 22 10 22 0 YD 11 10 21 13 24 3 NE 9 21 30 25 34 4 NF 8 21 29 26 34 5 NG 10 21 31 24 34 3 NH 5 22 27 24 29 2 NI 7 22 29 22 29 0 YJ 4 31 35 34 38 3 NK 9 29 38 29 38 0 Y

Critical path = A-B-C-I-KProject length = 38 days

C

B

D

H

E

J

F

K

A

End

G

I

Page 6: Ch07 Solutions

7-16. See file P7-16.XLS.(a)

(b)Activity Time EST EFT LST LST Slack Crit?

A 5 0 5 0 5 0 YB 6 5 11 6 12 1 NC 2 5 7 16 18 11 ND 9 5 14 5 14 0 YE 9 11 20 12 21 1 NF 3 14 17 18 21 4 NG 7 14 21 14 21 0 YH 4 14 18 18 22 4 NI 6 21 27 21 27 0 YJ 5 18 23 22 27 4 N

Critical path = A-D-G-IProject length = 27 days

B

C

D

E

F

JG

I

AEnd

H

Page 7: Ch07 Solutions

7-17. See file P7-17.XLS.(a)

(b & c)Act a m b Exp t Var SD EST EFT LST LFT Slack Crit?A 3 6 8 5.83 0.69 0.83 0.00 5.83 7.17 13.00 7.17 NB 2 4 4 3.67 0.11 0.33 0.00 3.67 5.33 9.00 5.33 NC 1 2 3 2.00 0.11 0.33 0.00 2.00 0.00 2.00 0.00 YD 6 7 8 7.00 0.11 0.33 2.00 9.00 2.00 9.00 0.00 YE 2 4 6 4.00 0.44 0.67 9.00 13.00 9.00 13.00 0.00 YF 6 10 14 10.00 1.78 1.33 13.00 23.00 13.00 23.00 0.00 YG 1 2 4 2.17 0.25 0.50 13.00 15.17 15.83 18.00 2.83 NH 3 6 9 6.00 1.00 1.00 23.00 29.00 23.00 29.00 0.00 YI 10 11 12 11.00 0.11 0.33 15.17 26.17 18.00 29.00 2.83 NJ 14 16 20 16.33 1.00 1.00 15.17 31.50 20.00 36.33 4.83 NK 2 8 10 7.33 1.78 1.33 29.00 36.33 29.00 36.33 0.00 Y

Critical path = C-D-E-F-H-KProject length = 36.33 daysProject variance = 5.22Project std deviation = 2.29 days

(d) P(Finish <= 40 days) = 0.9457

A

C

BI

H

D

Start

E

F

G

J

K

End

Page 8: Ch07 Solutions

7-18. See file P7-18.XLS.(a)

(b and c)Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A - 3 6 8 5.83 0.69 0.83 0.00 5.83 0.00 5.83 0.00 YB A 5 8 10 7.83 0.69 0.83 5.83 13.67 5.83 13.67 0.00 YC A 5 6 8 6.17 0.25 0.50 5.83 12.00 7.50 13.67 1.67 ND B, C 1 2 4 2.17 0.25 0.50 13.67 15.83 13.67 15.83 0.00 YE D 7 11 17 11.33 2.78 1.67 15.83 27.17 18.00 29.33 2.17 NF D 7 9 12 9.17 0.69 0.83 15.83 25.00 15.83 25.00 0.00 YG D 6 8 9 7.83 0.25 0.50 15.83 23.67 17.17 25.00 1.33 NH F, G 3 4 7 4.33 0.44 0.67 25.00 29.33 25.00 29.33 0.00 YI E, F, H 3 5 7 5.00 0.44 0.67 29.33 34.33 29.33 34.33 0.00 Y

Critical path = A-B-D-F-H-IProject length = 34.33 daysProject variance = 3.22Project std deviation = 1.80 days

(d and e) P(Finish <= 34 days) = 0.4263; P(Finish >= 29 days) = 0.9985

B

C

D

I

H

F

A

G

E

Page 9: Ch07 Solutions

7-19. See file P7-19.XLS.(a)

(b and c)Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A - 8 10 12 10.00 0.44 0.67 0.00 10.00 0.00 10.00 0.00 YB - 6 7 9 7.17 0.25 0.50 0.00 7.17 22.83 30.00 22.83 NC - 3 3 4 3.17 0.03 0.17 0.00 3.17 19.83 23.00 19.83 ND A 10 20 30 20.00 11.11 3.33 10.00 30.00 10.00 30.00 0.00 YE C 6 7 8 7.00 0.11 0.33 3.17 10.17 23.00 30.00 19.83 NF B, D, E 9 10 11 10.00 0.11 0.33 30.00 40.00 30.00 40.00 0.00 YG B, D, E 6 7 10 7.33 0.44 0.67 30.00 37.33 47.67 55.00 17.67 NH F 14 15 16 15.00 0.11 0.33 40.00 55.00 40.00 55.00 0.00 YI F 10 11 13 11.17 0.25 0.50 40.00 51.17 50.83 62.00 10.83 NJ G, H 6 7 8 7.00 0.11 0.33 55.00 62.00 55.00 62.00 0.00 YK I, J 4 7 8 6.67 0.44 0.67 62.00 68.67 62.00 68.67 0.00 YL G, H 1 2 4 2.17 0.25 0.50 55.00 57.17 66.50 68.67 11.50 N

Critical path = A-D-F-H-J-KProject length = 68.67 daysProject variance = 12.33Project std deviation = 3.51 days

(d and e) P(Finish <= 70 days) = 0.6479; P(Finish >= 75 days) = 0.0357

C

F

B

H

I

G

A

E

DK

Start

J

L

End

Page 10: Ch07 Solutions

7-20. See file P7-20.XLS.(a)

(b and c)Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A - 4 8 13 8.17 2.25 1.50 0.00 8.17 0.00 8.17 0.00 YB A 4 10 15 9.83 3.36 1.83 8.17 18.00 8.17 18.00 0.00 YC B 7 14 20 13.83 4.69 2.17 18.00 31.83 23.33 37.17 5.33 ND B 9 16 19 15.33 2.78 1.67 18.00 33.33 18.00 33.33 0.00 YE B 6 9 11 8.83 0.69 0.83 18.00 26.83 24.50 33.33 6.50 NF D, E 2 4 5 3.83 0.25 0.50 33.33 37.17 33.33 37.17 0.00 YG C, F 4 7 11 7.17 1.36 1.17 37.17 44.33 37.17 44.33 0.00 YH G 3 5 9 5.33 1.00 1.00 44.33 49.67 44.33 49.67 0.00 YI G, H 2 3 4 3.00 0.11 0.33 49.67 52.67 49.67 52.67 0.00 Y

Critical path = A-B-D-F-G-H-IProject length = 52.67 daysProject variance = 11.11Project std deviation = 3.33 days

(d and e) P(Finish <= 57 days) = 0.9032; P(Finish >= 50 days) = 0.7881

C

E

B

I

H

FA D

G

Page 11: Ch07 Solutions

7-21. See file P7-21.XLS.(a)

(b and c)Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A - 3 7 13 7.33 2.78 1.67 0.00 7.33 3.00 10.33 3.00 NB - 5 10 17 10.33 4.00 2.00 0.00 10.33 0.00 10.33 0.00 YC A, B 3 5 8 5.17 0.69 0.83 10.33 15.50 10.33 15.50 0.00 YD C 5 12 14 11.17 2.25 1.50 15.50 26.67 15.50 26.67 0.00 YE C 2 5 9 5.17 1.36 1.17 15.50 20.67 21.33 26.50 5.83 NF E 2 5 15 6.17 4.69 2.17 20.67 26.83 26.50 32.67 5.83 NG F 5 8 12 8.17 1.36 1.17 26.83 35.00 32.67 40.83 5.83 NH D 6 10 12 9.67 1.00 1.00 26.67 36.33 26.67 36.33 0.00 YI F, H 3 4 8 4.50 0.69 0.83 36.33 40.83 36.33 40.83 0.00 YJ G, I 4 7 10 7.00 1.00 1.00 40.83 47.83 40.83 47.83 0.00 Y

Critical path = B-C-D-H-I-JProject length = 47.83 daysProject variance = 9.64Project std deviation = 3.10 days

(d and e) P(Finish <= 49 days) = 0.6465; P(Finish >= 54 days) = 0.0235

7-22. See file P7-22.XLS.(a) P(need >= 17 months) = 0.9772(b) P(need <= 20 months) = 0.3085(c) P(need >= 23 months) = 0.1587(d) P(need <= 25 months) = 0.9772

C

F

B

H

I

G

A

E

D

J

Start

Page 12: Ch07 Solutions

7-23. See file P7-23.XLS.(a) P(need <= 30 weeks) = 0.8944(b) P(need <= 22 weeks) = 0.2266(c) 99% completion time = 34.31(d) 85% completion time = 29.15

7-24. See file P7-24.XLS.

ActivityBudgeted cost ($)

% complete

Value of work completed

Actual cost ($)

Difference ($)

A $22,000 100% $22,000 $20,000 -$2,000B $30,000 100% $30,000 $36,000 $6,000C $26,000 100% $26,000 $26,000 $0D $48,000 100% $48,000 $44,000 -$4,000E $56,000 50% $28,000 $25,000 -$3,000F $30,000 60% $18,000 $15,000 -$3,000G $80,000 10% $8,000 $5,000 -$3,000H $16,000 10% $1,600 $1,000 -$600

Total $181,600 $172,000 -$9,600Underrun

7-25. See file P7-25.XLS for the full chart. The first 10 months are shown here.Using EST:

1 2 3 4 5 6 7 8 9 10A $1,667 $1,667 $1,667 $1,667 $1,667 $1,667B $7,000 $7,000C $714 $714 $714 $714 $714 $714 $714D $2,000 $2,000 $2,000E $1,400 $1,400 $1,400 $1,400FGHPer day = $1,667 $8,667 $8,667 $2,381 $4,381 $4,381 $4,114 $2,114 $2,114 $2,114Cumulative = $1,667 $10,333 $19,000 $21,381 $25,762 $30,143 $34,257 $36,371 $38,486 $40,600

Using LST:1 2 3 4 5 6 7 8 9 10

A $1,667 $1,667 $1,667 $1,667 $1,667 $1,667B $7,000 $7,000C $714 $714 $714 $714 $714 $714 $714D $2,000E $1,400 $1,400 $1,400 $1,400FGHPer month = $1,667 $1,667 $1,667 $2,381 $9,381 $9,381 $2,114 $2,114 $2,114 $4,114Cumulative = $1,667 $3,333 $5,000 $7,381 $16,762 $26,143 $28,257 $30,371 $32,486 $36,600

Page 13: Ch07 Solutions

7-26. See file P7-26.XLS.

ActivityBudgeted cost ($)

% complete

Value of work completed

Actual cost ($)

Difference ($)

A $10,000 100% $10,000 $13,000 $3,000B $14,000 100% $14,000 $12,000 -$2,000C $5,000 100% $5,000 $6,000 $1,000D $6,000 100% $6,000 $6,000 $0E $14,000 80% $11,200 $12,000 $800F $13,000 13% $1,690 $1,000 -$690G $4,000 100% $4,000 $4,500 $500H $6,000 20% $1,200 $500 -$700I $18,000 0% $0 $0 $0J $12,000 0% $0 $0 $0K $10,000 0% $0 $0 $0L $16,000 0% $0 $0 $0M $18,000 0% $0 $0 $0

Total $53,090 $55,000 $1,910Overrun

7-27. See file P7-27.XLS for the full chart. The first 10 weeks are shown here.Using EST:

1 2 3 4 5 6 7 8 9 10A $700 $700B $1,600 $1,600 $1,600C $1,000 $1,000 $1,000 $1,000D $1,400 $1,400E $800 $800 $800F $500Per day = $700 $700 $2,600 $2,600 $2,600 $1,000 $2,200 $2,200 $800 $500Cumulative = $700 $1,400 $4,000 $6,600 $9,200 $10,200 $12,400 $14,600 $15,400 $15,900

Using LST:1 2 3 4 5 6 7 8 9 10

A $700 $700B $1,600 $1,600 $1,600C $1,000 $1,000 $1,000 $1,000D $1,400 $1,400E $800 $800 $800F $500Per month = $700 $700 $1,000 $1,000 $2,600 $2,600 $2,400 $2,200 $2,200 $500Cumulative = $700 $1,400 $2,400 $3,400 $6,000 $8,600 $11,000 $13,200 $15,400 $15,900

Page 14: Ch07 Solutions

7-28. See file P7-28.XLS.

ActivityBudgeted cost ($)

% complete

Value of work completed

Actual cost ($)

Difference ($)

A $1,400 100% $1,400 $1,500 $100B $4,800 100% $4,800 $4,500 -$300C $4,000 100% $4,000 $4,000 $0D $2,800 70% $1,960 $2,800 $840E $2,400 70% $1,680 $2,000 $320F $1,500 0% $0 $0 $0

Total $13,840 $14,800 $960Overrun

7-29. Project crashing by hand. The current critical path is A–C–E–G–H. Total time is 15 weeks.1. Activity A is the cheapest critical activity to crash. Reduce activity A by 1 week at a cost of $750. There are now two critical paths: A–C–E–G–H and B–D–G–H.2. Reduce activity G by 1 week at a cost of $1,500. The total completion time is now 13 weeks.3. Since activity G can be reduced all the way down to 2 weeks, we continue to reduce it by 2 more weeks (down to a duration of 2 weeks) at a cost of $1,500 per week. The project completion time is now 11 weeks and the total crash cost is $750 + $1,500*3 = $5,250. Note: Path A–D–G–H also becomes critical when the project is crashed to 14 weeks. However, all activities on this path are also included in the two critical paths we considered: A–C–E–G–H and B–D–G–H.Note: The Excel solution is shown in file P7-29.XLS.

7-30. Let Xi be the start time of activity i and Yi be the amount of time reduced for activity i, where i = A, B, C, D, E, F, and G. Let XEnd = project completion time.

Objective: Minimize crashing cost = $600YA + $700YB + $0YC + $75YD + $50YE + $1,000YF + $250YG

Subject to:YA 1 XD ≥ XA + 3 - YA

YB 1 XE ≥ XB + 2 - YB

YC 0 XF ≥ XC + 1 - YC

YD 4 XG ≥ XD + 7 - YD

YE 3 XG ≥ XE + 6 - YE

YF 1 XEnd ≥ XF + 2 - YF

YG 2 XEnd ≥ XG + 4 - YG

XEnd 10 All Xi, Yi ≥ 0

Solution: See file P7-30.XLS for the Excel setup and solution of this LP model. The optimal solution is to crash activity D by 4 weeks and activity E by 2 weeks. The total crashing cost is $400.

7-31. See file P7-31.XLS for the Excel setup and solution of this LP model.Crash A Crash B Crash C Crash D Crash E Crash F Crash G Crash H

Solution 2.0 0.0 0.0 1.0 1.0 1.0 0.0 1.0Crash cost $300 $800 $300 $600 $500 $700 $650 $500 $2,900

Page 15: Ch07 Solutions

7-32. See file P7-32.XLS for the Excel setup and solution of this LP model.Crash A Crash B Crash C Crash D Crash E Crash F Crash G Crash H

Solution 2.0 2.0 1.0 0.0 1.0 1.0 2.0 1.0Crash cost $750 $850 $300 $500 $700 $800 $900 $1,200 $8,000

7-33. See file P7-33.XLS. The project network for Team A is as follows.

The expected time, variance, EST, EFT, LST, LFT, and slack for each activity are as follows.

Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?1 - 3 4 5 4.00 0.11 0.33 0.00 4.00 0.00 4.00 0.00 Y2 1 4 5 7 5.17 0.25 0.50 4.00 9.17 6.00 11.17 2.00 N3 1 6 8 9 7.83 0.25 0.50 4.00 11.83 4.00 11.83 0.00 Y4 1 2 3 5 3.17 0.25 0.50 4.00 7.17 6.67 9.83 2.67 N5 1 6 7 9 7.17 0.25 0.50 4.00 11.17 6.83 14.00 2.83 N6 1 3 4 5 4.00 0.11 0.33 4.00 8.00 6.17 10.17 2.17 N7 6 2 4 5 3.83 0.25 0.50 8.00 11.83 10.17 14.00 2.17 N8 4 3 4 6 4.17 0.25 0.50 7.17 11.33 9.83 14.00 2.67 N9 3 1 2 4 2.17 0.25 0.50 11.83 14.00 11.83 14.00 0.00 Y10 2 1 3 4 2.83 0.25 0.50 9.17 12.00 11.17 14.00 2.00 N11 9, 10 3 4 6 4.17 0.25 0.50 14.00 18.17 14.00 18.17 0.00 Y12 11 4 6 7 5.83 0.25 0.50 18.17 24.00 18.17 24.00 0.00 Y13 12 6 8 10 8.00 0.44 0.67 24.00 32.00 24.00 32.00 0.00 Y14 13 3 4 6 4.17 0.25 0.50 32.00 36.17 32.00 36.17 0.00 Y15 5, 7, 8, 9, 10 3 4 5 4.00 0.11 0.33 14.00 18.00 31.17 35.17 17.17 N16 15 2 4 6 4.00 0.44 0.67 18.00 22.00 35.17 39.17 17.17 N17 14 2 3 4 3.00 0.11 0.33 36.17 39.17 36.17 39.17 0.00 Y18 16, 17 3 5 6 4.83 0.25 0.50 39.17 44.00 39.17 44.00 0.00 Y

Critical path = 1-3-9-11-12-13-14-17-18Project length = 44 weeks

Page 16: Ch07 Solutions

7-34. See file P7-34.XLS. The project network for Bender Construction Co. is as follows.

The expected time, variance, EST, EFT, LST, LFT, and slack for each activity are as follows.

Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?1 — 1.0 4.00 5.00 3.67 0.444 0.67 0.00 3.67 9.00 12.67 9.00 N2 — 2.0 3.00 4.00 3.00 0.111 0.33 0.00 3.00 16.50 19.50 16.50 N3 — 3.0 4.00 5.00 4.00 0.111 0.33 0.00 4.00 14.50 18.50 14.50 N4 — 7.0 8.00 9.00 8.00 0.111 0.33 0.00 8.00 3.50 11.50 3.50 N5 1 4.0 4.00 5.00 4.17 0.028 0.17 3.67 7.83 12.67 16.83 9.00 N6 3 1.0 2.00 4.00 2.17 0.250 0.50 4.00 6.17 18.50 20.67 14.50 N7 4 4.0 5.00 6.00 5.00 0.111 0.33 8.00 13.00 11.50 16.50 3.50 N8 7 1.0 2.00 4.00 2.17 0.250 0.50 13.00 15.17 16.50 18.67 3.50 N9 5 3.0 4.00 4.00 3.83 0.028 0.17 7.83 11.67 16.83 20.67 9.00 N10 2 1.0 1.00 2.00 1.17 0.028 0.17 3.00 4.17 19.50 20.67 16.50 N11 — 18.0 20.00 26.00 20.67 1.778 1.33 0.00 20.67 0.00 20.67 0.00 Y12 8 1.0 2.00 3.00 2.00 0.111 0.33 15.17 17.17 18.67 20.67 3.50 N13 6,9,10,11,12 1.0 1.00 2.00 1.17 0.028 0.17 20.67 21.83 20.67 21.83 0.00 Y14 13 0.1 0.14 0.16 0.14 0.000 0.01 21.83 21.97 21.83 21.97 0.00 Y15 14 0.2 0.30 0.40 0.30 0.001 0.03 21.97 22.27 24.84 25.14 2.87 N16 14 1.0 1.00 2.00 1.17 0.028 0.17 21.97 23.14 21.97 23.14 0.00 Y17 16 1.0 2.00 3.00 2.00 0.111 0.33 23.14 25.14 23.14 25.14 0.00 Y18 15,17 3.0 5.00 7.00 5.00 0.444 0.67 25.14 30.14 25.14 30.14 0.00 Y19 18 0.1 0.10 0.20 0.12 0.000 0.02 30.14 30.25 30.14 30.25 0.00 Y20 19 0.1 0.14 0.16 0.14 0.000 0.01 30.25 30.39 33.33 33.47 3.08 N21 19 2.0 3.00 6.00 3.33 0.444 0.67 30.25 33.59 30.25 33.59 0.00 Y22 20 0.1 0.10 0.20 0.12 0.000 0.02 30.39 30.51 33.47 33.59 3.08 N23 21,22 0.0 0.20 0.20 0.17 0.001 0.03 33.59 33.75 33.59 33.75 0.00 Y

Critical path = 11-13-14-16-17-18-19-21-23Project length = 33.75 weeks

Page 17: Ch07 Solutions

7-35. The overall purpose of Problem 7-35 is to have students use a project 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 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 project management. As students try to solve this problem, 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 LP 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 project network.

7-36.(a) See file P7-36.XLS, sheet (a). The project network is as follows:

The expected time, variance, EST, EFT, LST, LFT, and slack for each activity are as follows.

Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A — 1.0 2.0 4.0 2.17 0.25 0.50 0.00 2.17 10.13 12.30 10.13 NB — 3.0 3.5 4.0 3.50 0.03 0.17 0.00 3.50 11.88 15.38 11.88 NC — 10.0 12.0 13.0 11.83 0.25 0.50 0.00 11.83 0.00 11.83 0.00 YD — 4.0 5.0 7.0 5.17 0.25 0.50 0.00 5.17 14.65 19.82 14.65 NE — 2.0 4.0 5.0 3.83 0.25 0.50 0.00 3.83 15.98 19.82 15.98 NF A 6.0 7.0 8.0 7.00 0.11 0.33 2.17 9.17 12.30 19.30 10.13 NG B 2.0 4.0 5.5 3.92 0.34 0.58 3.50 7.42 15.38 19.30 11.88 NH C 5.0 7.7 9.0 7.47 0.44 0.67 11.83 19.30 11.83 19.30 0.00 YI C 9.9 10.0 12.0 10.32 0.12 0.35 11.83 22.15 14.90 25.22 3.07 NJ C 2.0 4.0 5.0 3.83 0.25 0.50 11.83 15.67 19.98 23.82 8.15 NK D 2.0 4.0 6.0 4.00 0.44 0.67 5.17 9.17 19.82 23.82 14.65 NL E 2.0 4.0 6.0 4.00 0.44 0.67 3.83 7.83 19.82 23.82 15.98 N

Page 18: Ch07 Solutions

M F, G, H 5.0 6.0 6.5 5.92 0.06 0.25 19.30 25.22 19.30 25.22 0.00 YN J, K, L 1.0 1.1 2.0 1.23 0.03 0.17 15.67 16.90 23.82 25.05 8.15 NO I, M 5.0 7.0 8.0 6.83 0.25 0.50 25.22 32.05 25.22 32.05 0.00 YP N 5.0 7.0 9.0 7.00 0.44 0.67 16.90 23.90 25.05 32.05 8.15 N

Critical path = C-H-M-OProject length = 32.05 weeks

(b) If activities I and J are not required, we set their expected times and variances to zero. There is no change to the critical path or the project completion time, as shown in the following revised solution.

Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A — 1.0 2.0 4.0 2.17 0.25 0.50 0.00 2.17 10.13 12.30 10.13 NB — 3.0 3.5 4.0 3.50 0.03 0.17 0.00 3.50 11.88 15.38 11.88 NC — 10.0 12.0 13.0 11.83 0.25 0.50 0.00 11.83 0.00 11.83 0.00 YD — 4.0 5.0 7.0 5.17 0.25 0.50 0.00 5.17 14.65 19.82 14.65 NE — 2.0 4.0 5.0 3.83 0.25 0.50 0.00 3.83 15.98 19.82 15.98 NF A 6.0 7.0 8.0 7.00 0.11 0.33 2.17 9.17 12.30 19.30 10.13 NG B 2.0 4.0 5.5 3.92 0.34 0.58 3.50 7.42 15.38 19.30 11.88 NH C 5.0 7.7 9.0 7.47 0.44 0.67 11.83 19.30 11.83 19.30 0.00 YI C 9.9 10.0 12.0 0.00 0.00 0.00 11.83 11.83 25.22 25.22 13.38 NJ C 2.0 4.0 5.0 0.00 0.00 0.00 11.83 11.83 23.82 23.82 11.98 NK D 2.0 4.0 6.0 4.00 0.44 0.67 5.17 9.17 19.82 23.82 14.65 NL E 2.0 4.0 6.0 4.00 0.44 0.67 3.83 7.83 19.82 23.82 15.98 NM F, G, H 5.0 6.0 6.5 5.92 0.06 0.25 19.30 25.22 19.30 25.22 0.00 YN J, K, L 1.0 1.1 2.0 1.23 0.03 0.17 11.83 13.07 23.82 25.05 11.98 NO I, M 5.0 7.0 8.0 6.83 0.25 0.50 25.22 32.05 25.22 32.05 0.00 YP N 5.0 7.0 9.0 7.00 0.44 0.67 13.07 20.07 25.05 32.05 11.98 N

Critical path = C-H-M-OProject length = 32.05 weeks

Page 19: Ch07 Solutions

7-37.(a) See file P7-37.XLS, sheet (a). The project network is as follows:

The expected time, variance, EST, EFT, LST, LFT, and slack for each activity are as follows.

Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A - 2 3 4 3.00 0.11 0.33 0.00 3.00 15.50 18.50 15.50 NB - 5 6 8 6.17 0.25 0.50 0.00 6.17 12.67 18.83 12.67 NC - 1 1 2 1.17 0.03 0.17 0.00 1.17 32.00 33.17 32.00 ND - 8 9 11 9.17 0.25 0.50 0.00 9.17 0.00 9.17 0.00 YE A 1 1 4 1.50 0.25 0.50 3.00 4.50 18.50 20.00 15.50 NF B 3 3 4 3.17 0.03 0.17 6.17 9.33 18.83 22.00 12.67 NG B 1 2 2 1.83 0.03 0.17 6.17 8.00 22.00 23.83 15.83 NH C 5 5 6 5.17 0.03 0.17 1.17 6.33 33.17 38.33 32.00 NI D 9 10 11 10.00 0.11 0.33 9.17 19.17 9.17 19.17 0.00 YJ D 1 2 2 1.83 0.03 0.17 9.17 11.00 26.00 27.83 16.83 NK E 2 2 3 2.17 0.03 0.17 4.50 6.67 20.00 22.17 15.50 NL F 3 4 6 4.17 0.25 0.50 9.33 13.50 22.00 26.17 12.67 NM G 2 2 4 2.33 0.11 0.33 8.00 10.33 23.83 26.17 15.83 NN I 8 9 11 9.17 0.25 0.50 19.17 28.33 19.17 28.33 0.00 YO I 1 1 3 1.33 0.11 0.33 19.17 20.50 30.00 31.33 10.83 NP J 4 4 8 4.67 0.44 0.67 11.00 15.67 27.83 32.50 16.83 NQ K 6 6 7 6.17 0.03 0.17 6.67 12.83 22.17 28.33 15.50 NR L, M 1 2 4 2.17 0.25 0.50 13.50 15.67 26.17 28.33 12.67 NS N 6 6 7 6.17 0.03 0.17 28.33 34.50 28.33 34.50 0.00 YT O 3 3 4 3.17 0.03 0.17 20.50 23.67 31.33 34.50 10.83 NU P 1 2 3 2.00 0.11 0.33 15.67 17.67 32.50 34.50 16.83 NV Q, R 9 10 11 10.00 0.11 0.33 15.67 25.67 28.33 38.33 12.67 N

Critical path = D-I-N-S-WProject length = 38.33 weeksProject variance = 0.89Project std deviation = 0.94 weeks

(b) P(Finish <= 40 weeks) = 0.9617

Page 20: Ch07 Solutions

(c) If activities D and I are already completed, we set their expected times and variances to zero. The new critical path is B–F–L–R–V, and the new project time is 25.67 weeks. The revised solution is as follows (see file P7-37, sheet (b) for calculation).

Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A - 2 3 4 3.00 0.11 0.33 0.00 3.00 2.83 5.83 2.83 NB - 5 6 8 6.17 0.25 0.50 0.00 6.17 0.00 6.17 0.00 YC - 1 1 2 1.17 0.03 0.17 0.00 1.17 19.33 20.50 19.33 ND - 8 9 11 0.00 0.00 0.00 0.00 0.00 6.50 6.50 6.50 NE A 1 1 4 1.50 0.25 0.50 3.00 4.50 5.83 7.33 2.83 NF B 3 3 4 3.17 0.03 0.17 6.17 9.33 6.17 9.33 0.00 YG B 1 2 2 1.83 0.03 0.17 6.17 8.00 9.33 11.17 3.17 NH C 5 5 6 5.17 0.03 0.17 1.17 6.33 20.50 25.67 19.33 NI D 9 10 11 0.00 0.00 0.00 0.00 0.00 6.50 6.50 6.50 NJ D 1 2 2 1.83 0.03 0.17 0.00 1.83 13.33 15.17 13.33 NK E 2 2 3 2.17 0.03 0.17 4.50 6.67 7.33 9.50 2.83 NL F 3 4 6 4.17 0.25 0.50 9.33 13.50 9.33 13.50 0.00 YM G 2 2 4 2.33 0.11 0.33 8.00 10.33 11.17 13.50 3.17 NN I 8 9 11 9.17 0.25 0.50 0.00 9.17 6.50 15.67 6.50 NO I 1 1 3 1.33 0.11 0.33 0.00 1.33 17.33 18.67 17.33 NP J 4 4 8 4.67 0.44 0.67 1.83 6.50 15.17 19.83 13.33 NQ K 6 6 7 6.17 0.03 0.17 6.67 12.83 9.50 15.67 2.83 NR L, M 1 2 4 2.17 0.25 0.50 13.50 15.67 13.50 15.67 0.00 YS N 6 6 7 6.17 0.03 0.17 9.17 15.33 15.67 21.83 6.50 NT O 3 3 4 3.17 0.03 0.17 1.33 4.50 18.67 21.83 17.33 NU P 1 2 3 2.00 0.11 0.33 6.50 8.50 19.83 21.83 13.33 NV Q, R 9 10 11 10.00 0.11 0.33 15.67 25.67 15.67 25.67 0.00 Y

Critical path = B-F-L-R-VProject length = 25.67 weeksProject variance = 0.89Project std deviation = 0.94 weeks

Page 21: Ch07 Solutions

Case: Haygood Brothers Construction Co.

See file P7-Haygood.XLS. The project network is as follows:

(1 and 2) The expected time, variance, EST, EFT, LST, LFT, and slack for each activity are as follows.

Act Pred a m b Exp t Var SD EST EFT LST LFT Slack Crit?A - 4 5 6 5.00 0.11 0.33 0.00 5.00 0.00 5.00 0.00 YB A 2 5 8 5.00 1.00 1.00 5.00 10.00 5.00 10.00 0.00 YC B 5 7 9 7.00 0.44 0.67 10.00 17.00 10.00 17.00 0.00 YD B 4 5 6 5.00 0.11 0.33 10.00 15.00 23.00 28.00 13.00 NE C 2 4 6 4.00 0.44 0.67 17.00 21.00 17.00 21.00 0.00 YF E 3 5 9 5.33 1.00 1.00 21.00 26.33 33.67 39.00 12.67 NG E 4 5 6 5.00 0.11 0.33 21.00 26.00 34.00 39.00 13.00 NH E 3 4 7 4.33 0.44 0.67 21.00 25.33 34.67 39.00 13.67 NI E 5 7 9 7.00 0.44 0.67 21.00 28.00 21.00 28.00 0.00 YJ D,I 10 11 12 11.00 0.11 0.33 28.00 39.00 28.00 39.00 0.00 YK F,G,H,J 4 6 8 6.00 0.44 0.67 39.00 45.00 39.00 45.00 0.00 YL F,G,H,J 7 8 9 8.00 0.11 0.33 39.00 47.00 44.33 52.33 5.33 NM L 4 5 10 5.67 1.00 1.00 47.00 52.67 52.33 58.00 5.33 NN K 5 7 9 7.00 0.44 0.67 45.00 52.00 45.00 52.00 0.00 YO N 5 6 7 6.00 0.11 0.33 52.00 58.00 52.00 58.00 0.00 YP M,O 2 3 4 3.00 0.11 0.33 58.00 61.00 58.00 61.00 0.00 Y

Critical path = A-B-C-E-I-J-K-N-O-PProject length = 61.00 daysProject variance = 3.67Project std deviation = 1.91 days

(3) P(Finish <= 60 days) = 0.3008.

Page 22: Ch07 Solutions

Case: Family Planning Research Center

This case covers three aspects of project management: (1) critical path scheduling, (2) crashing, and (3) resource (personnel) leveling or smoothing.

(1) 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. The actual project network is as follows.

The EST, EFT, LST, LFT, and slack for each activity are as follows (see file P7-Family.XLS, sheet (a) for calculations).

Act Pred Time EST EFT LST LFT Slack Critical?A — 5 0 5 8 13 8 NB — 7 0 7 12 19 12 NC — 5 0 5 0 5 0 YD A 3 5 8 19 22 14 NE A 7 5 12 13 20 8 NF B, E 2 12 14 20 22 8 NG A, B 3 7 10 19 22 12 NH C 10 5 15 5 15 0 YI H 7 15 22 15 22 0 YJ D, F, G, I 15 22 37 22 37 0 YK J 30 37 67 37 67 0 Y

Critical path = C-H-I-J-KProject length = 67 days

(2) Workforce smoothing. The following table shows the staffing requirement with each activity beginning on its earliest start date [see file P7-Family.XLS, sheet (b)]. There are five days (i.e., days 6 to 10) on which the requirements are for more than 10 workers.

Start

C

B

A

D

E

G

F

H I

J K

Page 23: Ch07 Solutions

Activity 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22A 2 2 2 2 2B 3 3 3 3 3 3 3C 2 2 2 2 2D 1 1 1E 4 4 4 4 4 4 4F 1 1G 2 2 2H 6 6 6 6 6 6 6 6 6 6I 3 3 3 3 3 3 3JKTotal 7 7 7 7 7 14 14 13 12 12 10 10 7 7 6 3 3 3 3 3 3 3

If we delay some of the non-critical activities (D, E, F, and G), we get the following schedule and workforce requirements [see file P7-Family.XLS, sheet (b)]. This schedule can be accomplished with the 10 available workers.

Activity 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22A 2 2 2 2 2B 3 3 3 3 3 3 3C 2 2 2 2 2D 1 1 1E 4 4 4 4 4 4 4F 1 1G 2 2 2H 6 6 6 6 6 6 6 6 6 6I 3 3 3 3 3 3 3JKTotal 7 7 7 7 7 9 9 10 10 10 10 10 10 10 8 7 6 5 3 3 3 3

(3) See file P7-Family.XLS, sheet (c).

Crash ACrash BCrash CCrash DCrash ECrash FCrash G Crash H Crash I Crash J Crash KCrash days 0.0 0.0 2.0 0.0 0.0 0.0 0.0 0.0 5.0 0.0 0.0Crash cost $100 $150 $50 $250 $150 $1,000 $500 $200 $80 $400 $400 $500