The network below represents a project being analyzed by the critical path metho

Question

The network below represents a project being analyzed by the critical path method. Activity durations are A=5, B=2, C=12, D=3, E=5, F=1, G=7, H=2, I=10, and J=6.

a. What task must be on the critical path, regardless of activity durations? b. What is the duration of path A-B-E-H-J? c. What is the critical path of this network? d. What is the length of the critical path? e. What is slack time at activity H? f. What is the Late Finish time of activity H? g. If activity C were delayed by two time units, what would happen to the project duration?

Explanation / Answer

a. The last task of the network will have no slack regardless of activity duration and hence will always be on critical path b. 5 + 2 + 5 + 2 + 6 = 20 c. Critical Path: A-B-G-I-J d. The length of the critical path is 5 + 2 + 7 + 10 + 6 = 30 e. Slack Time of Activity H = 5 f. Latest Finish Time of Activity H = 24 g. The project duration would not change because the slack time of C is 5 so it can be delayed up to 5 days without affecting the project Activity Predecessor Activity Time ES EF LS LF Slack Critical Activity A None 5 0 5 0 5 0 Yes B A 2 5 7 5 7 0 Yes C A 12 5 17 10 22 5 No D None 3 0 3 11 14 11 No E B 5 7 12 17 22 10 No F D 1 3 4 13 14 10 No G B 7 7 14 7 14 0 Yes H C,E 2 17 19 22 24 5 No I G,F 10 14 24 14 24 0 Yes J H,I 6 24 30 24 30 0 Yes ES = E of Tail Event EF = ES + Activity Duration LF = L of Head Event LS = LF - Activity Duration Slack - LS - ES E = Earliest occurrence time for the event L = Latest allowable time for the occurrence of event

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