374 Chapter 8 Scheduling 10. A project is shown below, and the project manager i

Question

374 Chapter 8 Scheduling 10. A project is shown below, and the project manager is 11. Use Ms Project for the follow asked to develop an optimal time-cost solution: project: activities Estimated Crash Costweek Activity ID Activity Predecessors Duration Time for Activity ID Predecessor (Days) (Days) Crash Study feasibility Gather information 8 $500 Consider 4 $300 alternatives Define problem $600 B, D Find a solution 1 $800 Prototype A, D, E Initial test a. Determine the critical path, total time, and slack. Final test b. Rank the critical path activities in order of lowest i Measure crashing cost, and determine the number of days performance each can be crashed. Production c. Shorten the project one day at a time, and check after each reduction to see what path is critical Show the crashing sequence. a. Find the critical path. b. How long will the project take to complete Note: Another path may equal the length of the shortened c. Use fast tracking to complete the project one wed critical path and no additional improvement is feasible. early. This is the optimum time-cost solution,

Explanation / Answer

Q 10 .

a)

critical path is the longest path in the table. so just connect the activities and see which path is highest. what to connect? if B's predeccessor is A , then there will be a link bwtween A -> B . since A does not have a predeccessor , it is the start time

critical path - A -> B -> E -> F will be the highest cost = 10+9+2 = 21 . you can draw other paths and check.

slack -

The slack is the difference between the duration of the critical path and the duration of each alternative path. When an activity appears on more than one path, the slack is the smallest difference.

list all possible paths

A->c->D->E->F

A -> B -> E -> F (critical path)

activity slack

A 0 (in CP)

B 0 (in CP)

C 2

D 2

E 0

F 0

b) crashing out

path activity crash cost

A->B A 0

B 500

B->E B 500

E 600

E->F E 600

F 800

crashing cost = one activity of each path ( here A , B , E since they;re min)

crashing cost = 0 + 500 + 600 = 1100

number of days each can be crashed -

A == 6 - 6 = 0

B = 10 - 8 = 2

similarly for others.

C ) if we shorten activities A, B... each by one day at one time, then -

when A is shortened no change . old critical path remains.

but if we short B by one day then, critical path becomes A - > C ->D-> E-> F.

for C by one day,, old CP remains.

for D, old CP

for E old CP

for F old CP

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Q11 similar as abovee .

for fast tracking , aply similar method as Q 10 c)

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thank you

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