Question

Suppose that the following estimates of activity times (in weeks) were provided for the network shown below:

Activity	Optimistic	Most Probable	Pessimistic
A	4	6	8
B	2.5	4.5	5.5
C	5	6.5	8
D	4	5.5	8
E	6.5	7.5	9
F	2	3.5	5
G	7	9	12
H	5	6	13

What is the probability that the project will be completed. If required, round your answers to four decimal places.

Within 20 weeks?

Within 22.5 weeks?

Within 25 weeks?

Answer 1

The formula for the mean time and variance is:

Mean Time (t) = (a+4b+c)/6

Variance = [(c-a)/6]2

Activity	Optimistic (a)	Most Probable (b)	Pessimistic (c)	Mean Time (t)	Variance
A	4	6	8	6	0.444444
B	2.5	4.5	5.5	4.34	0.25
C	5	6.5	8	6.5	0.25
D	4	5.5	8	5.67	0.444444
E	6.5	7.5	9	7.58	0.17361
F	2	3.5	5	3.5	0.25
G	7	9	12	9.16	0.69444
H	5	6	13	7	1.777777

Calculation of critical path:

1. ACH=20

2. ADFH=22

3. ADG=21

4. BEG=20

5. BEFH=21

The highest value is 22, so, the critical path is ADFH and time is 22 weeks.

Variance of critical path = var. A + var. D + var. F +var. H = 2.91

Standard deviation of critical path = (Variance)2 = (2.91)2 = 8.468

Z score calculation = [Weeks (x) - mean]/sd

· Within 20 weeks

Here, x = 20, mean = 22 and sd = 8.468

So, Z = -0.2361

Therefore, the probability = 0.5793

· Within 22.5 weeks

Here, x = 22.5, mean = 22 and sd = 8.468

So, Z = 0.059

Therefore, the probability = 0.5

· Within 25 weeks

Here, x = 25, mean = 22 and sd = 8.468

So, Z = 0.354

Therefore, the probability = 0.383

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