I need assistance with the following questions in: Project Scheduling.This is a follow up to a question answered earlier. Question 1) was answered.1)  A company decides to plan a project, with activities, precedences and deterministic durations given in the following table: Activity iDuration(days)Immediate PredecessorsA32noneB21noneC30noneD45AE26A, BF28CG20E, F2) The company decided that the project in the previous problem has to be expedited in order to optimize (minimize) the project cost.  Expediting can be used to trade direct cost for indirect cost, since expediting the project reduces the indirect cost of the project by a fixed amount per day, while expediting an activity increases its total direct cost.  The Accounting Department found that indirect costs amount to $100 per day, while activity direct costs as well as their normal and expedited times are given in the table below.Activity Normal  Time(days)Expedited Time(days)NormalCost($)Expedited  Cost($)A3226200500B2120300375C3030200200D4540200800E26207001,360F28241,0001,160G2018400550 From this table you can compute the direct cost incurred by expediting an activity by one day. Questions:1) What is the optimal time to complete the project?2) What is the optimal project cost?

Question
Asked Mar 27, 2019
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I need assistance with the following questions in: Project Scheduling.

This is a follow up to a question answered earlier. Question 1) was answered.

1)  A company decides to plan a project, with activities, precedences and deterministic durations given in the following table:

 

Activity

i

Duration

(days)

Immediate

Predecessors

A

32

none

B

21

none

C

30

none

D

45

A

E

26

A, B

F

28

C

G

20

E, F

2) The company decided that the project in the previous problem has to be expedited in order to optimize (minimize) the project cost.  Expediting can be used to trade direct cost for indirect cost, since expediting the project reduces the indirect cost of the project by a fixed amount per day, while expediting an activity increases its total direct cost.  The Accounting Department found that indirect costs amount to $100 per day, while activity direct costs as well as their normal and expedited times are given in the table below.

Activity

Normal 

Time

(days)

Expedited

Time

(days)

Normal

Cost

($)

Expedited 

Cost

($)

A

32

26

200

500

B

21

20

300

375

C

30

30

200

200

D

45

40

200

800

E

26

20

700

1,360

F

28

24

1,000

1,160

G

20

18

400

550

 

From this table you can compute the direct cost incurred by expediting an activity by one day.

 

Questions:

1) What is the optimal time to complete the project?

2) What is the optimal project cost?

 

 

 

 

 

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Expert Answer

Step 1

Given the activity sequence and timings the following are the paths (in AOA notation):

A-D (length 77)

A-E-G (length 78, critical path)

B-E-G (length 67)

C-F-G (length 78, critical path)

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Step 2

Next we find the details on the crash cost for the activities in the critical path(s).

Taking critical path A-E-G we show the details in the corresponding table.

Similarly for critical path C-F-G we show in the corresponding table.

Since path G is common for both we can either crash it by 1 day costing $75. However instead of this we could choose to crash the lowest cost activities in both the critical paths (such as A and F) but the total cost would be 50 + 40 = $90 which is more than $75. So we crash G by 1 day as we save indirect cost of $100. Now we have 3 paths of the same length 77 (all become critical)

A-D (length 77 as earlier, now critical path)

A-E-G (length 77 after crashing, still critical path)

B-E-G (length 66 now)

C-F-G (length 77 after crashing, still critical path)

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Step 3

Now we construct the activity table for the newly critical path A-D we get the table as shown.

Now choosing the lowest crash cost activity A we can crash it reducing the critical paths of A-D and A-E-G by 1 day and costing $50. But in order to reduce the length of the 3rd critical path C-F-G by 1 day we choose activity F costing $40. So all the three critical paths are reduced by 1 day costing in total $(40+50) = $90 per day. Since this is less than the total indirect cost savings of ...

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