Question

The expected times and variances for the project activities follow. Complete the table showing which activities are critical ID Predecessor te Critical? 1 None 6 Variance 3 4 2. None 22 3 None 4 2 4 Description Pilot production Select channels of distrib. Developmktg program Test market Patent Full production Ad promotion Release 4 1 1 2 5 14 5 6 7 4 10 22 13 2 3 2,5,6,7 B 2 1 ces What is the probability of completing the project in 26 periods? Hint: Use the #NORM.S DIST(2, TRUE) function in Excel to compute the probability. (Do not round Intermediate calculations. Round the final answer to 3 decimal places, ie, 0.750.) The probability of completing the project in 26 periods

Answer 1

EF = ES + Te LS = LF - Te Slack = LF - EF Activity Predecessors te Variance Std Deviation = Variance^0.5 ES LF Critical Calculation Value Calculation Value Calculation Value 1 6 3 0 6 6-6 0 6 6-6 0 NA NA NA 1.7 2.0 2 4 0 22 32 10 10 15 3 2 0 19 15 4 1 2 6 6 10 0 G[ N[ ની ન| ] ] ની નં 22 4 4 14 22 13 2 0+6 0+22 0+4 6+4 6+14 10+22 4+13 32+2 1 1.4 1.4 2.2 3.2 1.4 6 Yes No No Yes No Yes No Yes 10 20 32 5 6 32-22 19-4 10-4 32-14 32-22 32-13 34-2 18 5 10 32 32-22 19-4 10-10 32-20 32-32 32-17 34-34 12 4 10 10 32 0 7 2 4 19 32 15 3 2,5,6,7 17 34 8 1 1.0 32 32 34 0 Project completion time (m): 34 periods Critical Path = 1-4-6-8 Std deviation of Critical Path = 1.7 + 1.4 + 3.2 + 1.0 = 7.3 periods a) Let x be the expected project completion time = 26 periods s = standard deviation = 7.3 periods Actual completion time (m) = 34 periods As we know, z= (x-m) / s Putting values in the formula z = (26 - 34) / 7.3 = -8 / 7.3 = -1.0958 z score = -1.0959 Probability for z-score (-1.0958) = NORM.S.DIST(-1.0958,TRUE) = 0.137 So, probability that the project will finish in 26 periods = 0.137