Question

The project manager of Good Public Relations gathered the data shown in the following table for a new advertising campaign. Click the icon to view the table. a. How long is the project likely to take? (Make sure to round all intermediate calculations fort, to two decimal places.) The project is expected to take 36.01 weeks. (Enter your response rounded to two decimal places.) b. What is the probability that the project will take more than 36 weeks? Refer to the standard normal table. The probability is 0.5. (Enter your response rounded to four decimal places.) c. Consider the path A-E-G-H-J. What is the probability that this path will exceed the expected project duration? Refer to the standard normal table The probability is (Enter your response rounded to four decimal places.)

Answer 1

For each activity, calculate the expected time u =(a + 4m+b)/6 For each activity, Variance oʻ =(b-a)/36. Standard deviation of each activity is square root of Variance Find the Early Start (ES), Early Finish (EF), Late Start (LS), Late Finish (LF) times and Slack for each activity. Early Start is calculated by forward pass method as latest of the predecessors Early Finish. Late Finish is calculated using reverse pass method as earliest of the successors Late Start. Early Finish = Early Start + Activity Duration Late Start=Late Finish - Activity Duration Slack = Late Finish - Early Finish, or Late Start - Early Start Activity Predecessor LS EF LF Slack Critical Activity Yes Expected Variance ES Time (u) (?) 10.17 0.25 0.00 9.17 4.69 0.00 8.00 0.44 | 0.00 2.17 0.25 9.17 9.83 0.69 10.17 6.17 0.25 20.00 3.00 0.44 20.00 4.67 1.00 26.17 3.83 0.25 23.00 5.17 | 0.69 30.83 2 .00 0.00 30.83 A,C DE DE Yes 0.00 10.17 8.67 9.17 2.17 8.00 17.83 11.33 10.17 20.00 20.00 26.17 23.17 | 23.00 26.17 30.83 32.17 26.83 30.83 36.00 34.00 32.83 0.00 8.67 2.17 8.67 0.00 0.00 10.17 17.83 10.17 20.00 20.00 26.17 26.17 30.83 36.00 36.00 36.00 Yes F G Yes 0.00 9.17 0.00 3.17 8 Yes II 2 Critical path is the path in the network having maximum expected time (-). Critical path is A-E-F-H-) Expected completion time of project is the completion time of longest path (critical path) 36.00 Variance of critical path is the sum of individual variance of each activity on critical path. Std deviation is the square root of variance. Expected duration, variance and standard deviation of Critical path Expected Completion Time (u) = 36.00 weeks Expected variance (v) = 2.89 weeks Standard Deviation (0) = v= 1.70 weeks b) Probability that the project will be completed in specified time Given Time (x) = 36 weeks z = (x - 1)/o = (36 - 36)/1.7 = 0.0000 Probability that the project will be take more than 36 weeks = NORMSDIST(0) = 0.50

c)

Expected time of path A-E-G-H-J = 10.17+9.83+3.00+4.67+5.17 = 32.83

Variance of path A-E-G-H-J = 0.25+0.69+0.44+1.00+0.69 = 3.08

Standard deviation = sqrt(3.08) = 1.76

z = (36-32.83)/1.76 = 1.80

P(z) = NORMSDIST(1.80) = 0.9643

Probability that path A-E-G-H-J will exceed the expected project duration = 1 - 0.9643 = 0.0357