Question

The estimated times and immediate predecessors for the activities in a project at ABC Retinal Scanning Company are given in the following table. Assume that the activity times are independent.

Activity	Immediate Predecessor	a (time weeks)	m (time weeks)	b (time weeks)
A	-	9	10	11
B	-	4	10	16
C	A	9	10	11
D	B	5	8	11

Calculate the expected time (mean) and variance for each activity, then answer the following:

a. What is the variance of the critical path? What is the variance of the other path in the network?

0.22

5.00

b. What is the project standard deviation?

0.47

c. What is the expected completion time of the critical path? What is the expected completion time of the other path in the network?

22 weeks

20 weeks

d. If the time to complete the critical path is normally distributed, what is the probability that this path will be finished in 22 weeks or less?                           

100%

e. If the time to complete the other non-critical path is normally distributed, what is the probability that this path will be finished in 22 weeks or less? 96%

96%

I believe those are the correct answers. Can you please show me the step-by-step directions for this and correct any errors?

Answer 1

a.

Activity	Optimistic time-a	Expected completion time-m	Pessimistic time-b	Expected time= (a+4*m+ b)/6	Variance, (sigma)^2= (b-a/6)^2
A	9	10	11	10.00	0.11
B	4	11	18	11.00	5.44
C	9	9	13	9.67	0.44
D	5	8	12	8.17	1.36

b.

Paths	Duration
A-C	19.67 (critical path)
B-D	19.17

c.

Paths	Duration	Variance
A-C (Critical path)	19.67	0.56
B-D	19.17	6.81

d.

step 1	variance along the critical path				0.56
step 2	mean project time (u) of critical path is=				19.67
step 3	Required completion time is				22.0
step 4	standard deviation= sqrt(variance)=				0.75
step 5	because Z= (given completion time- u)/standard deviation				3.130
step 6	P(<=z)= using NORM.S.DIST(z,true)				0.9991 or 99.91%

e.

step 1	variance along the critical path				6.81
step 2	mean project time (u) of non-critical path is=				19.17
step 3	Required completion time is				22.0
step 4	standard deviation= sqrt(variance)=				2.61
step 5	because Z= (given completion time- u)/standard deviation				1.086
step 6	P(<=z)= using NORM.S.DIST(z,true)				0.8613 or 86.13%