A) Expected time of completion for all activities = ACD = 6+3+5 = 14 B) ES for C

Question

A) Expected time of completion for all activities = ACD = 6+3+5 = 14

B) ES for C = 6

C) LS for B = 2

Computation of earliest starting and finishing times is aided by two simple rules: 1. The earliest finish time for any activity is equal to its earliest start time plus its expected duration, t: EF = ES + t 2. ES for activities at nodes with one entering arrow is equal to EF of the entering arrow. ES for activities leaving nodes with multiple entering arrows is equal to the largest EF of the entering arrow. Computation of the latest starting and finishing times is aided by the use of two rules: 1. The latest starting time for each activity is equal to its latest finishing time minus its expected duration: LS = LF - t 2. For nodes with one leaving arrow, LF for arrows entering that node equals the LS of the leaving arrow. For nodes with multiple leaving arrows, LF for arrows entering that node equals the smallest LS of leaving arrows. Finding ES and EF times involves a forward pass through the network; finding LS and LF times involves a backward pass through the network. Hence, we must begin with the EF of the last activity and use that time as the LF for the last activity. Then we obtain the LS for the last activity by subtracting its expected duration from its LF Slack = LS-ES = LF-EF Activities for which Slack = 0 are in critical path

Explanation / Answer

Answer these questions for the following simple set of project tasks. Task times are shown in hours. 4 a. What is the expected time that all five tasks will be completed? Expected time hours b. What is the earliest start for task C? Earliest start th/nd hour c. What is the latest start for task B? Latest start nd/st hour

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