Question

The expected time (P days) and standard deviation (Q days) of the different activities (A to F) in a project are indicated in the network diagram as (P, Q). The standard deviation of the project completion time, in days, is . (Rounded off to one decimal place)

Hint:

First find which chain of activities is the critical (longest) path, then add the squared standard deviations of only those activities.

Answer 1

Correct Answer: 3

Step 1: Find the length of every path from node 1 to node 6.
The network has two routes from the start node 1 to the end node 6.
Path 1 goes through activities A, C, D, F: \(1 \to 2 \to 4 \to 5 \to 6\).
Path 2 goes through activities B, E: \(1 \to 3 \to 6\).
The expected duration of each activity is the first number P in its (P, Q) label.

Step 2: Add up the expected durations on each path.
Path 1: \(A + C + D + F = 12 + 15 + 18 + 6 = 51\) days.
Path 2: \(B + E = 23 + 27 = 50\) days.

Step 3: Identify the critical path.
The critical path in a PERT network is the longest path, since it decides the earliest possible completion time of the whole project.
Path 1 (51 days) is longer than Path 2 (50 days), so activities A, C, D, F form the critical path.

Step 4: Add the variances of the critical activities.
For a PERT network, the variance of the project completion time is the sum of the variances of only the activities lying on the critical path. The variance of an activity is the square of its standard deviation Q.
\[ \sigma_A^2 = 0^2 = 0, \quad \sigma_C^2 = 1^2 = 1, \quad \sigma_D^2 = 2^2 = 4, \quad \sigma_F^2 = 2^2 = 4 \]
\[ \sigma_{project}^2 = 0 + 1 + 4 + 4 = 9 \]

Final Answer:
\[ \sigma_{project} = \sqrt{9} \]
\[ \boxed{\sigma_{project} = 3.0 \text{ days}} \]