Please help me with this problem so it won\'t give me any issues in the future.

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Please help me with this problem so it won't give me any issues in the future. Thank you!

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. (i.e., Calculate the minimum direct cost when project is finished in 13 days, 12 days, 11 days, as well as 10 days.) 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 Crash Cost Maximum Normal Normal Act. (SlopeCrash Time Time Cost 150 100 200 200 200 150 0 100 50 40 50 0 Initial project duration 13 2x Total direct cost 1,000

Explanation / Answer

Crash cost ( slope ) is the increase in cost for crashing the activity.

Therefore in order to minimize cost, we must crash those items first which have least crash cost

It is evident from the activity diagram that :

Normal duration of path A-B-D-E-F = 2 + 3 + 4 +3 + 1 = 13

Normal duration of path A-B-C-F = 2 + 3 + 4 + 1 = 10

Special note: The diagram indicates that Normal duration of C is 6, however it is written that normal duration of C is 4. I have chosen what is written as the input for calculation i.e. Normal duration of C is 4

Therefore A-B-D-E-F is the Critical path and duration of the critical path is same as duration of the project. Therefore , normal duration of the project is 13 days

Total Normal cost of completing the project in 13 days ( when no crashing is required )

= Sum of Normal costs of all activities of the project

= 150 + 100 + 200 + 200 + 200 + 150

= 1000

DIRECT COST OF FINISHING THE PROJECT IN 13 DAYS = 1000

To finish the project in 12 days, 1 activity needs to be crashed by 1 day .

Since D has the LOWEST slope of 40 , D is chosen for crashing by 1 day to total of 12 days at a cost of 40

Therefore , direct cost of crashing the project to 12 days = 1000 ( as for 13 days ) + 40 ( crashing cost for 1 day) = 1040

DIRECT COST OF FINISHING THE PROJECT IN 12 DAYS = 1040     ( CRASH ACTIVITY D)

To finish the project in 11 days, 1 activity needs to be crashed by 1 day. It is to be noted that duration of D after crashing to 12 days is 3 days. Since D can be crashed upto 1 day , there is possibility of crashing D by maximum 2 days.

Again, since D has the minimum cost slope of 40 , we choose to crash D by another 1 day to finish the project in 11 days .

Therefore , direct cost of crashing the project to 11 days = Direct cost of crashing the project to 12 days ( as calculated ) + 40

= 1040 + 40 = 1080

DIRECT COST OF FINISHING THE PROJECT IN 11 DAYS = 1080 ( CRASH ACTIVITY D)

The crashed duration of D is now 2 and there is another possible 1 day of crashing left since maximum crash time is 1 day for D.

Here again, since D has the minimum slope, we would crash D by 1 day to crash the project to 10 days .

Therefore , direct cost of finishing the project in 10 days

= Direct cost of finishing the project in 11 days ( as calculated ) + 40

= 1080 + 40

= 1120

DIRECT OCST OF FINSIHING THE PROJECT IN 10 DAYS = 1120 ( CRASH ACTIVITY D )

DIRECT COST OF FINISHING THE PROJECT IN 13 DAYS = 1000

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