Question

The estimated times and immediate predecessors for the activities in a project at Caesar Douglas’s screening company are given in the following table. Assume that the activity times are independent. All times are in weeks.

Activity	Predecessors	Optimistic Time (a)	Most Likely Time (m)	Pessimistic Time (b)
A	--	9	10	11
B	--	4	10	16
C	A	9	10	11
D	B	5	8	11

1. What is the critical path based on expected times? You may use any method to determine this?

Critical Path is A-C

2. What is the probability of completing the project in 22 weeks or less?

Answer 1

Answer

answer: activity optimistic most likely pessimistic A 9 4 9 5 / 10 10 10 8 time variance estimates 10 0.11111 10 4 10 0.11111 8 1 11 16 11 11 C (0 + 4.m + P)/6 formula: * time estimate= O --> optimistic time P --> pessimistic time m --> most likely time ((b-a)/6))2 * variance = b --> pessimistic time values a --> optimistic time values

activity of node diagram; 0 A 10 10 C 19 0 10 10 10 10 19 19 finish OB 10 10 D 18 11 10 11 11 8 19 a) critical path: A-C completion time: 19 weeks Variance along the critical path: b) probability of completing the project in 22 weeks. 0.45 standard deviation = No.2 mean (m) х z =(x - m)/SD z= (22-19)/0.45 6.667 z= value of z from standard normal distribution table is: thus, the probability is 100%

value of L = 1. hence; probability = 100% Z=6.7

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