Project variance is the sum of variances of which activities?

Only the activity with maximum variance

Practice · 8 marks

PERT expected time and completion probability

Short answer

Critical path is B-D-F-G with expected project duration 22 weeks and standard deviation 2.71 weeks. Probability of completion within 24 weeks is 77% (Z=0.74). Probability within 20 weeks is 23% (Z=-0.74).

A project has the activities below with optimistic (a), most likely (m) and pessimistic (b) times in weeks:

Activity	a	m	b	Predecessors
A	2	4	6	-
B	3	5	13	-
C	4	6	8	A
D	5	8	17	A, B
E	2	3	4	C
F	3	5	7	D
G	1	2	3	E, F

(a) Calculate the expected time and variance of each activity. (b) Find the critical path and expected project duration. (c) Find the probability that the project is completed within 24 weeks and within 20 weeks.

(a) Expected time and variance

PERT uses the beta distribution:

t_e = \frac{a + 4m + b}{6}, \qquad \sigma^2 = \left(\frac{b-a}{6}\right)^2

Act.	$t_e$	$\sigma^2$	ES	EF	LS	LF	Float
A	(2+16+6)/6 = 4	0.444	0	4	2	6	2
B	(3+20+13)/6 = 6	2.778	0	6	0	6	0
C	(4+24+8)/6 = 6	0.444	4	10	11	17	7
D	(5+32+17)/6 = 9	4.000	6	15	6	15	0
E	(2+12+4)/6 = 3	0.111	10	13	17	20	7
F	(3+20+7)/6 = 5	0.444	15	20	15	20	0
G	(1+8+3)/6 = 2	0.111	20	22	20	22	0

(b) Critical path

Zero-float activities: B, D, F, G. Expected project duration $T_E = 6 + 9 + 5 + 2 = 22$ weeks.

(c) Probability

Variance of the project = sum of variances on the critical path only:

\sigma_T^2 = 2.778 + 4.000 + 0.444 + 0.111 = 7.333, \qquad \sigma_T = 2.708\ \text{weeks}

Z = \frac{T_S - T_E}{\sigma_T}

For 24 weeks: $Z = (24-22)/2.708 = 0.74$ , so $P = 0.770$ (from the standard normal table).
For 20 weeks: $Z = (20-22)/2.708 = -0.74$ , so $P = 1 - 0.770 = 0.230$ .

Answer: Critical path B-D-F-G, expected duration 22 weeks, sigma = 2.71 weeks; P(finish within 24 weeks) = 77%; P(within 20 weeks) = 23%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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PERT expected time and completion probability

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