Question

A7-3 (Algo) The expected times and variances... The expected times and variances for the project activities are given below. Complete the table showing which activities are critical ID 1 Pilot production 2 Select channels of distrib 3Develop mktg. program 4Test market 5 Patent 6 Full production 7 Ad promotion 8Release Description Predecessor None None None Variance Critical? te 6 40 18 4 15 4 16 2,5,6,7 What is the probability of completing the project in 29 periods? (Do not round intermediate calculations. Round the final answer to 3 decimal places, i.e., 0.750.) The probability of completing the project in 29 periods

Answer 1

The precedence diagram of activities as follows :

1		2	3
4	5		7
6
8

The possible paths and their corresponding expected durations as follows :

1-4-6-8 = 6 + 4 + 16 + 2 = 28

1-5-8 = 6 + 15 + 2 = 23

2-8 = 40 + 2 = 42

3-7-8 = 18 + 9 + 2 = 29

Since , 2-8 has the longest duration, it forms the critical path. The expected project duration ( which is same as expected duration of critical path ) will be 42 periods

Variance of the duration of critical path = Variance of activity 2 + Variance of activity 8 = 4 + 1 = 5

Therefore, Standard deviation of duration of activities on critical path = Square root ( Variance ) = Square root ( 5 ) = 2.236

Let z value corresponding to the probability of completing the project within 29 periods = Z1

Accordingly,

Expected project duration + Z1 x standard deviation of expected project duration = 29

Or, 40 + 2.236.Z1 = 29

Or, 2.236.Z1 = 29 – 40 = - 11

Or, Z1 = - 11/2.236 = - 4.919 ( - 4.9 rounded to 2 decimal places )

Corresponding value of probability for Z =- 4.9 as derived from standard normal distribution table is 0.0000005

PROBABILITY OF COMPLETING THE PROJECT IN 29 PERIODS = 0.0000009