Question

An activity on a PERT network has these time estimates: optimistic = 4, most likely = 5, and pessimistic = 3. What is its expected activity time and variance?

A) 3.5, 0.028

B) 4.5,1

C) 3.5, 1

D) 4.5, 0.028

E) none of the above

Answer 1

Given, Optimistic time=4 , Most likely time=5, Pessimistic time= 3

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

Hence, E = 4 + 4*5 + 3 / 6= 4.5

Variance(V) = ((Pessimistic - Optimistic)/ 6)^2 = (3-4/6)^2 = 0.028

Hence, correct option is d).