Question

Consider the following project.

	Times (days)
Activity	Optimistic	Most Likely	Pessimistic	Preceding Tasks
1	8	10	13	-
2	5	6	8	-
3	13	15	21	2
4	10	12	14	1,3
5	11	20	30	4
6	4	5	8	5
7	2	3	4	5
8	4	6	10	7
9	2	3	4	8,6

1. Draw the AON Network Diagram
2. Compute the expected activity time
3. Using the expected activity times, compute the early start, early finish, late start, late finish, and total float for each activity.
4. Compute the probability of completing the critical path in 69 days.

Answer 1

solution a:

Solution b:

Solution c:

Earliest start time and earliest finish time is a forward calculation starting from activity 0.

Earliest start time for the activity is the maximum of the earliest finish time for the preceding activity. For activity 1 and 2, since there was no preceding activity, therefore their earliest start time is zero. Earliest start time for activity 3 is earliest finish time of activity 2 and for activity 4, earliest start time is maximum of the earliest finish time of the preceding activity i.e. max[earliest finish time of activity 1 or earliest finish time of activity 3] i.e. max [10 ,21]. Hence, earliest start time for activity 4 is 21 and so on.

Earliest finish time is the sum of the earliest start time and the activity duration. For activity 1, 2 and 3, earliest finish time is 0 + 10, 0+6, and 6+15 respectively.

Latest start time and latest finish time is a backward calculation starting from activity 9.

For activity 9, latest finish time is equal to earliest finish time.

Latest finish time for the activity is the minimum of the latest start time for the succeeding activity.

Latest finish time for activity 8 is latest start time for activity 9 and so on.

Latest start time is the latest finish time minus the time duration of the activity. For activity 9, latest start time is 65 – 3, for activity 8, latest start time = 62 – 6 and so on.

Latest finish time for the activity is the minimum of the latest start time for the succeeding activity.

Latest finish time for activity 8 is latest start time for activity 9 and so on.

Latest finish time for activity 5 is the minimum of latest start time for activity 6 or 7. Since, latest start time of activity 7 is minimum among activity 6 and 7, therefore, latest finish time for activity 5 is 53 and not 57.

Solution d:

Probability of completing the critical path in 69 days:

Z = [M – N] / standard deviation

M= number of days in question = 69

N = duration of critical path = 65

Standard deviation is the square root of the summation of variance of the activities along critical path

Critical path is 0-2-3-4-5-7-8-9

Therefore standard deviation = square root [0.25+1.78+0.45+10.03+0.11+1+0.11]

Therefore, standard deviation = 3.71

Therefore, Z = [69 – 65] / 3.71

Z = 1.078

Corresponding this value in normal distribution table, for Z = 1.078, probability is 14.0517 percent