MTH601 Assignment No.1 (Fall 2015)Due Date: November 23, 2015

Complete Solution:

Question 1:

Necessary and sufficient conditions for critical path.

Answer:

Necessary conditions: If Earliest occurrence time equals Latest occurrence time for Tail and Head event of the activity, then it possibly can be a critical activity.

Sufficient condition:   If the necessary condition is satisfied and ES and LS of activity are same, also EF and LF of activity is same; then the activity is definitely a critical activity.

Question 2: 

a)      Determine the critical path and expected project length.

Answer:

To calculate the critical path, list all the three paths with their expected time of completion.
1-2-5-6 = 10

1-3-5-6 = 17
1-4-6 = 8

The longest path is the critical path and it consists of activities 1-3, 3-5 and 5-6 with a length Te = 4 + 6 + 7= 17.

The variance of the critical path is then VT = Variance of 1-3 + variance of 3-5 + variance of 5-6.

b) Calculate the variance and standard deviation of the project length.

Answer: