Question

A network consists of the following list. Times are given in weeks.

Activity	Preceding	Optimistic	Probable	Pessimistic
A	--	6	8	10
B	A	1	2	4
C	A	9	12	15
D	A	3	4	5
E	B	4	6	7
F	B	14	16	22
G	C, D	2	3	4
H	E, G	1	2	3
I	G	6	8	9
J	H, I	4	6	8
K	F, H, J	1	2	5

a. Draw the network diagram.

b. Calculate the expected duration and variance of each activity.

c. Calculate the expected duration and variance of the critical path

d. Calculate the probability that the project will be completed in less than 42 weeks.

Answer 1

(a) Network Diagram -

(b) Expected Duration of each activity = ( Pessimistic Duration + 4*Probable + Optimistic Duration ) / 6

Variance of Each activity = σ² = { (Pessimistic Duration - Optimistic Duration) / 6 }²

Standard Deviation of each activity = √σ² = (Pessimistic Duration - Optimistic Duration) / 6

Activity	Predecessor	Optimistic	Probable	Pessimistic	Expected	Variance	Std Dev
A	-	6	8	10	8.00	0.444	0.667
B	A	1	2	4	2.17	0.250	0.500
C	A	9	12	15	12.00	1.000	1.000
D	A	3	4	5	4.00	0.111	0.333
E	B	4	6	7	5.83	0.250	0.500
F	B	14	16	22	16.67	1.778	1.333
G	C, D	2	3	4	3.00	0.111	0.333
H	E, G	1	2	3	2.00	0.111	0.333
I	G	6	8	9	7.83	0.250	0.500
J	H, I	4	6	8	6.00	0.444	0.667
K	F, H, J	1	2	5	2.33	0.444	0.667

(c) Possible Network Path and their Duration are -

Path	Duration
ABFK	29.17
ABEHK	20.33
ABEHJK	26.33
ACGHK	27.33
ACGHJK	33.33
ACGIJK	39.17
ADGHK	19.33
ADGHJK	25.33
ADGIJK	31.17

The critical path is the path with longest duration = ACGIJK with expected duration of 39.17

Variance of critical path ACGIJK = 0.444 + 1.000 + 0.111 + 0.250 + 0.444 + 0.444 = 2.693

(d)

project duration is 39.17 time units

Standard deviation of critical path = √variance = √2.693 = 1.641

So, µ = 39.17, σ = 1.64

Probability that the project will be completed in less than 42 weeks = P(z)

z = ( x - µ )/ σ = (42 - 39.17)/1.64 = 1.73

From the z table, P(z) = 0.9582

Hence probability that the project will be completed in fewer than 30 time units is 95.82%