Question

It was determined on the following project that the duration for the tasks contained some uncertainty. After more research, the optimistic and pessimistic values were determined. 1. a. Calculate the expected time and variance (using the 95 percent level of confidence) for each activity. b. Determine the critical path and time for completion c. Calculate the probably that the project will finish in 30 days. Calculate the probably of completing the project in 27 days. Other than the critical path, which is the next riskiest path? What is the probability of completing that path in 27 days? Your boss wants to know what time it will take to complete if you are 99% sure (that is, 99% probable). What do you tell your boss? d. e. Activity Predecessor Opt. (days) Normal (days) Pess. (days) 13 10 4 4 1 4 5 C,D,E 8 IK H,l 4 10

Answer 1

For each activity, calculate the expected time mu = (a + 4m + b)/6 For each activity, Variance sigma^2 = (b - a)^2/36. Standard deviation of each activity is squareroot of Variance Find the Earliest start and finish and Latest start and finish times and Slack for each activity Earliest Start is calculated by forward pass method as latest of the predecessors Earliest Finish. Latest Finish is calculated using reverse pass method as earliest of the successors Latest Start. Earliest Finish = Earliest Start + Activity Duration Latest Start = Latest Finish - Activity Duration Slack = Latest Finish - Earliest Finish, or Latest Start - Earliest Start Critical path is the longest path in the network. It is also the path having 0 slack. Critical path is A-B-F-I-K Expected duration, variance and standard deviation of Critical path Expected Completion Time(mu) = 27.833 days Expected variance (v) = 5.639 days Standard Deviation (sigma) = VV = 2.375 days Probability that the project will be completed in specified time

D) Other than critical path, the next riskiest path is A-B-C-H-K

Its expected time = 27.

Variance of this path = 5.194

Std deviation = SQRT(5.194) = 2.279

For target time of 27 days, z (27-27.5)/2.279 = -0.2194

Probability = NORMSDIST(-0.2194) = 0.4132

de) For 99% probability, z = NORMSINV(0.99) = 2.3263

Expected time = 27.833 + 2.3263*2.3746 = 33.4 days