Question

Activity	Immediate Predecessor	Optimistic Time	Most-likely time	Pessimistic Time
A	None	3	6	9
B	A	3	5	7
C	A	4	7	12
D	B	4	8	10
E	C	5	10	16
F	D, E	3	4	5
G	D, E	3	6	8
H	F	5	6	10
I	G	5	8	11
J	H, I	3	3	3

(a). What is the expected time for each activity? (2 points)

(b.) What is the variance for each activity? (2 points)

Use both the longest path method and ES/EF; LS/LF method for the following questions

(c) What is the expected project completion time? (2 point)

(d) What are the critical activities? (1 points)

(e) What are the slack values for non-critical activities? (4 points)

(f) What is the probability that the project will be completed in 38 weeks? (2 points)

(g) What is the probability that the project will be completed in 42 weeks? (2 points)

Answer 1

Expected Duration = (Optimistic time + 4* Most likely time + Pessimistic time) / 6

Variance= ((Pessimistic time - Optimistic time)/ 6 )²

Standard Deviation = square root of Variance

For Activity A,

Expected duration = (3+4*6+9) / 6 = 6 weeks

Variance = ((9-3)/6)² = 1

Stadndard Deviation = square root (1) = 1

Remaining calculation are as below

Activity	Immediate Predecessor	Optimistic Time	Most-likely time	Pessimistic Time	Expected mean time	Variance	Std Dev	ES	EF	LS	LF	Slack	Critical
(1)	(2)	(3)	(4)	(5)	(6) = [(3)+4(4)+(5)]/6	(7)=[[(5)-(3)]/6]^2	(8)=(7)^0.5	(9)	(10)=(9)+(6)	(11)=(12)-(6)	(12)
A	None	3	6	9	6	1	1	0	6	0	6	0	Yes
B	A	3	5	7	5	0.444444444	0.666666667	6	11	10.83333333	15.83333	4.833333	No
C	A	4	7	12	7.333333333	1.777777778	1.333333333	6	13.33333333	6	13.33333	0	Yes
D	B	4	8	10	7.666666667	1	1	11	18.66666667	15.83333333	23.5	4.833333	No
E	C	5	10	16	10.16666667	3.361111111	1.833333333	13.33333	23.5	13.33333333	23.5	0	Yes
F	D, E	3	4	5	4	0.111111111	0.333333333	23.5	27.5	26.83333333	30.83333	3.333333	No
G	D, E	3	6	8	5.833333333	0.694444444	0.833333333	23.5	29.33333333	23.5	29.33333	0	Yes
H	F	5	6	10	6.5	0.694444444	0.833333333	27.5	34	30.83333333	37.33333	3.333333	No
I	G	5	8	11	8	1	1	29.33333	37.33333333	29.33333333	37.33333	0	Yes
J	H, I	3	3	3	3	0	0	37.33333	40.33333333	37.33333333	40.33333	0	Yes
Critical Path		A-C-E-G-I-J
Critical duration		40.33 weeks

f) Probability that project will be completed in 38 weeks

Variance along critical path = 1+1.7778+3.3611+0.69444+1+0 = 7.8333

Std Dev along critical path = 7.8333^0.5 = 2.79888

Z = (x- ) / = (38-40.33) / 2.7988 = -0.8325

Prob = Norm.s.dist(-0.8325,1) = 0.2056 = 20.56%

g) Probability that project will be completed in 42 weeks

Variance along critical path = 1+1.7778+3.3611+0.69444+1+0 = 7.8333

Std Dev along critical path = 7.8333^0.5 = 2.79888

Z = (x- ) / = (42-40.33) / 2.7988 = 0.59669

Prob = Norm.s.dist(0.59669,1) = 0.72464 = 72.46%