MTH601 Practice Questions Solution Lecture 10

Solution of Practice Questions Lecture 10

Q1. If we have 10, 12 and 26 days as optimistic, most likely and pessimistic time for an activity, then calculate the following

1. Expected time of the activity
2. Variance and standard deviation.

Solution

a)

b)

(26 10)

2.67 7.11 p o t t t V

Standard = 7.11 =2.67 Deviation Variance

Q2. A teacher notes that when her student starts reading the optimistic (t0), most likely (tm) and pessimistic (tp) time estimates are 10, 20 and 60. Calculate the following

1. What will be the expected duration of reading?
2. What would be its variance?

Solution

(12*4) eT For Activity B:

(15*4) eT

For Activity D:

(4*5) eT

Q4. A certain project is composed of nine activities whose time estimates are given below:

Activity

Most likely

1-2 1-3 1-4 2-5 3-5 4-6 5-6 5-7 6-7

a) Draw the project network and trace all the possible paths from it. b) What is the expected project length? c) Calculate the variance and standard deviation of the project length.

Solution

a) =0.45+1+2.79+0.45=4.69

Standard = 4.69 =2.17 Deviation Variance

Q5. Activity

Most likely

1-2 1-3 1-4 2-5 3-5 4-6 5-6

If the critical path for the above project is 1-3-5-6, then find the variance and standard deviation for the critical path.

Solution

As the critical path is 1 – 3 – 5 – 6

Activity Expected Duration Variance 1-2 1-3 1-4 2-5 3-5 4-6 5-6

Project Length=4+6+7=17 Variance=sum of the variance of the critical activities =1+4+4=9 Standard Deviation=square root (variance) =3