Question

The activities of a PERT network and their corresponding activity time estimates (in weeks), namely optimistic (\(t_o\)), most likely (\(t_m\)), and pessimistic (\(t_p\)), are given in the table below.

Activity	\(t_o\)	\(t_m\)	\(t_p\)
1-2	2	4	12
1-3	2	4	6
1-4	3	4	11
2-5	3	6	9
3-4	2	5	14
3-5	3	3	3
4-5	3	6	15
5-6	2	5	8

The expected project length is ________ weeks (in integer).

Hint:

Convert each activity's three time estimates into one expected time, then add along every path to find the longest one.

Answer 1

Correct Answer: 22

Step 1: Find the expected time for every activity.
PERT uses \(t_e = \dfrac{t_o + 4t_m + t_p}{6}\) for each activity.
Working through the table: 1-2 gives \((2+16+12)/6=5\), 1-3 gives \((2+16+6)/6=4\), 1-4 gives \((3+16+11)/6=5\), 2-5 gives \((3+24+9)/6=6\), 3-4 gives \((2+20+14)/6=6\), 3-5 gives \((3+12+3)/6=3\), 4-5 gives \((3+24+15)/6=7\), 5-6 gives \((2+20+8)/6=5\).

Step 2: List every path from the start node to the end node.
Path 1-2-5-6: \(5+6+5=16\) weeks.
Path 1-3-5-6: \(4+3+5=12\) weeks.
Path 1-4-5-6: \(5+7+5=17\) weeks.
Path 1-3-4-5-6: \(4+6+7+5=22\) weeks.

Step 3: Pick the longest path, since that fixes the project duration.
The path 1-3-4-5-6 gives the largest total at 22 weeks, so it is the critical path.

Final Answer:
The expected project length equals the critical path length, 22 weeks. \[ \boxed{22 \text{ weeks}} \]