Chapter 9 Exercises - Answer Key PDF

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Chapter 9 Exercises – Answer key Use the following informaion to compress one-ime unReduce the schedule unil you reach the crash point of the network. For each move idenify what it per move using the least -cost method. acivity(ies) was crashed and the adjusted total cost. Note: The correct normal p...

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Chapter 9 Exercises – Answer key

1. Use the following information to compress one-time unit per move using the least -cost method. Reduce the schedule until you reach the crash point of the network. For each move identify what activity(ies) was crashed and the adjusted total cost.

Note: The correct normal project duration, critical path, and total direct cost are provided.

Duration can be reduced from 12 time periods to eight. Total direct cost goes up from $1,000 to $1,300. Steps are shown below:

3 A 12 10

B 3

F 1 X

3 D

A 1 F E 4 C 3 x The project has two paths. Path A-B-D-F takes 10 time periods and A-C-E-F takes 12 time periods. The A-C-E-F path is critical so it is the one to look at to reduce project time. (CHOOSE THE ONE WITH LESS CRASH COST (SLOPE)). Total direct cost is $1,000.

A

B

D

F

2x

3

3

1x

9

A

C

E

F

2x

4

4

1X

11

The cheapest activity to reduce is A so we reduce it by its maximum reduction of one time unit to two time periods. The A-C-E-F path remains critical at 11 time periods and direct costs go up to $1,050 since it cost $50 to crash A. ($1000 + $50)

A

B

D

F

2x

3

3

1x

A

C

E

F

2x

2x

4

1x

9

9

The next cheapest activity to reduce is C so we reduce it from four to three1 at a cost of $60. A-C-E-F remains critical, completion time goes down to ten time units and total direct cost goes up to $1,110. ($1050 + $60) C can be reduced two time periods. There is nothing wrong with reducing by two units all-at-once. It is broken into two steps here just to show the process. This also allows us to double check that the other path does not become critical with the one-unit reduction.

A

B

D

F 8

2x

3

2x

1x

A

C

E

F

2x

2x

3x

1x

8

Finally, we can reduce D by one time period off of the first path at a cost of $60 and E by one day off of the second path at a cost of $70. That reduces the completion time to eight time periods and raises the total direct cost to $1,300. We have reached the crash point for the project. No further reductions are possible. (1,110 + $60 + $60 + $70 = $1,300)

1

2. Use the information contained below to compress one-time unit per move using the least cost method. Reduce the schedule until you reach the crash point of the network. For each move identify what activity(ies) was crashed and the adjusted total cost.

Note: Choose B instead of C and E (equal costs) because it is usually smarter to crash early rather than late AND one activity instead of two activities

Project duration is reduced from 13 time periods to ten. Total direct cost goes up from $1,000 to $1,240. The steps are shown below:

2 A F 1 12 13 x

F 1 x

6 C

B 3

B 2 A E 4 D 3 x The project has two paths, A-B-C-F, which takes 12 time periods, and A-B-D-E-F, which takes 13 time periods. This gives the project a duration of 13 time periods. The total direct cost is $1,000.

A

B

C

F

2x

3

6

1x

A

B

D

E

F

2x

3

4

3

1x

12

13

The cheapest activity to reduce is D at a cost of $40 for one time period. That makes both paths critical. That means that further reductions will require either reducing the same time from an activity on both paths or finding an activity shared by both paths and reducing that activity. Total direct cost goes up from $1,000 to $1,040. 2 A 1 F 11 x

B 2 x

C 6

F 1 x

x

Activity B is on both paths and can be reduced by one time period at a cost of $100. 2 That lowers the completion time to 11 time periods and raises total direct cost to $1,140.

A

B

C

F

2x

5

1x

10 2x

A

B

D

E

F

2x

2x

3x

2x

1x

10

Finally, we can reduce C on the first path at a cost of $50 and E on the second path at a cost of $50. That reduces the completion time to 10 time units and raises total direct cost to $1,240. We have reached the crash point for the project, no further reductions are possible.

3. Use the information contained below to compress one-time unit per move using the least cost method. Reduce the schedule until you reach the crash point of the network. For each move identify what activity(ies) was crashed and the adjusted total cost.

2 Activities C on the first path and E on the second path also have a combined cost of $100. We select B first since it is usually smarter to crash earlier activities first and it is usually smarter to crash one activity rather than two.

Project duration is reduced from 16 to 12. Total direct cost increases from 1,400 to 1,640. The steps are shown below:

A

B

C

F

H

2

3

2

3

1x

A

B

D

G

F

11

A

E

F

2

5

1x

8

The project has three paths, A-B-C-F-H, A-B-D-G-H (the critical path), and A-E-H. Completion time is 16 time periods and total direct cost is $1,400.

A

B

C

F

H

2

3

2

3

1x

A

B

D

G

F

2

3

5

4x

1x 15

A

E

F

2

5

1x

11

8

We begin by reducing G by one time period at a cost of $20. That reduces the project completion time to 15 time units and raises total direct cost to $1,420.

A

B

C

F

H

2

3

2

3

1x

A

B

D

G

F

2

3

5

4x

1x 15

A

E

F

2

5

1x

11

8

Next, we reduce D by one time period at a cost of $40. That reduces the project completion time to 14 and raises total direct cost to $1,460.

A

B

C

F

H

2

2x

2

3

1x 10

A

B

D

G

F

2

2x

4x

4x

1x

A

E

13

F 1x 8

Next, we reduce B by one time period at a cost of $80. That reduces the project completion time to 13 and raises total direct cost to $1,540.

2

5

A

B

C

F

H

1x

2x

2

3

1x 9

A

B

D

G

F

1x

2x

4x

4x

1x 12

A

E

F

1x

5

1x

7

Finally, we reduce A by one time period at a cost of $100. That reduces the project completion time to 12 and raises total direct cost to $1,640. No further reductions are possible. The crash point has been reached.

4. Given the data and information that follow, compute the total direct cost for each project duration. If the indirect costs for each project duration are $90 (15 time units), $70 (14), $50 (13), $40 (12), and $30 (11), compute the total project cost for each duration. What is the optimum cost-time schedule for the project? What is this cost?

Total cost at the original duration of 15 time periods is $730 total direct cost plus $90 total indirect cost equals $820. That total cost. goes to $810 for 14 time periods, $820 for 13 time periods, $1,000

for 12 time periods, and $1,180 for 11 time periods. Total costs are minimized at 14 time periods and a total cost of $810. The steps are shown below:

A

C

F

I

5

4x

2

3

A

D

G

I

5

2

5

3

B

E

H

I

3

5

2x

3

14

15

13

There are three paths, A-C-F-I, A-D-G-I, and B-E-H-I. The second, A-D-G-I, is critical at fifteen time periods. However, the first and third paths are 14 and 13 respectively, so other paths will rapidly become critical and will have to be monitored closely. Total direct cost is $730, indirect cost is $90, for a total cost of $820.

A

C

F

I

5

4x

2

3

A

D

G

I

5

1x

5

3

B

E

H

I

3

5

2x

3

14

14

13

Reducing D by one time period at a cost of $10 reduces the completion time to 14. The first path, A-C-F-I becomes critical. Total direct cost rises to $740, indirect cost falls to $70, and total cost falls to $810.

A

C

F

I

4x

4x

2

3

A

D

G

I

4x

1x

5

3

B

E

H

I

3

5

2x

3

13

13

13

Reducing A by one reduces the completion time to 13 and all three paths are now critical. Direct cost goes up to $770, indirect cost falls to $50, and total cost rises $820. Since total cost has started to rise, we know that 14 time periods is optimum. Several more crashes are shown below just to illustrate that total direct cost continues to rise faster than indirect costs fall so total cost continues to rise. However, it is reasonable for a student to stop at this point.

A

C

F

I

4x

4x

1x

3

A

D

G

I

4x

1x

4x

3

B

E

H

I

2

5

2x

3

12

12

12

With all three paths critical, each path must be reduced. It is cheaper to reduce B for $60, F for $100, and G for $30. Completion time goes down to 12 and total cost goes up to $1,000.

A

C

F

I

4x

4x

1x

2x

A

D

G

I

11

11 4x

1x

4x

2x

B

E

H

I

2

5

11 2x

2x

Reducing I by one time period, at a cost of $200, reduces all three paths. This reduces the completion time to 11 and raises total cost to $1,190. We have reached the crash point for the network. No further reductions are possible.

5. Use the following information to compress one time unit per move using the least-cost method. Assume the total indirect cost for the project is $700 and there is a savings of $50 per time unit reduced. Record the total direct, indirect, and project costs for each duration. What is the optimum cost-time schedule for the project? What is the cost?

Note: The correct normal project duration and total direct cost are provided.

Total cost at 14 time units is $1,200 total direct cost plus $700 total indirect cost equals $1,900. This drops to $1,870 at 13 time units, $1,860 at 12 time units, and $1,910 at 11 time units. Total cost is minimized at 12 time units and total cost of $1,860. The steps are shown below:

A

B

D

G

2x

3

3

2x 10

A

C

E

G

2x

5

5

2x 14

A

C

F

G

2x

5

4

2x

13

The project has three paths, A-B-D-G, A-C-E-G, and A-C-F-G. The second one, A-C-E-G is critical at 14 time periods but the third, A-C-F-G, at 13 is almost critical. Direct cost is $1,200, indirect cost is $700, and total cost is $1,900.

A

B

D

G

2x

3

3

2x 10

A

C

E

G

2x

5

4x

2x

A

C

F

G

2x

5

4

2x 13

13

Reducing E by one at a cost of $20 reduces the time to 13 and makes the third path critical as well. Direct cost goes up to $1,220, indirect cost goes down to $650, and total cost goes down to $1,870.

A

B

D

G

2x

3

3

2x

A

C

E

G

2x

4x

4x

2x 12

A

C

F

G

2x

4x

4

2x 12

10

Reducing C by one at a cost of $40 lowers the completion time to 12. Direct cost goes up to $1,260, indirect cost goes down to $600, and total cost goes down to $1,860. No more activities on the critical A-C-E-G path can be reduced and 12 is the crash point for the schedule....