Software Engineering: Problems on PERT

PERT Problem 1:

Activity	O	M	P	Predecessor
A	3	4	5	-
B	2	2	2	A
C	3	5	7	A
D	1	3	5	B,C
E	3	3	3	D

Duration to the above PERT Project can be calculated using the formulae :

(O + 4M + P )/ 6

where O Represents Optimistic Time

M Represents Most Likely Time or Expected Time

P Represents Pessimistic Time

Activity	O	M	P	Predecessor	Duration
A	3	4	5	-	4
B	2	2	2	A	2
C	3	5	7	A	5
D	1	3	5	B,C	3
E	3	3	3	D	3

Network Diagram:

Now apply Node rule to all the above activities as shown in fig

Forward Pass:

ES + Duration = EF

Under Forward Pass we need to consider the MAX Value

Backward Pass:

LS = LF - Duration

Under Backward Pass we need to consider the MIN Value

Critical Path Network Diagram:

Forward Pass:

The forward pass is carried out to calculate the earliest finish dates on which each activity may be started and completed.

Activity A may start immediately. Hence, earliest date for its start is zero i.e. ES(A) = 0. It takes 4 weeks to complete its execution. Hence, earliest it can finish is week 4 i.e. EF(A) = 4.

Activity B may start immediately. Hence, earliest date for its start is 4 i.e. ES(B) = 4. It takes 2 weeks to complete its execution. Hence, earliest it can finish is week 6 i.e. EF(B) = 6.

Activity C starts as soon as activity A completes its execution. Hence, earliest week it can start its execution is week 4 i.e. ES(C) = 4. It takes 5 weeks to complete its execution. Hence, earliest it can finish is week 9 i.e. EF(C) = 9.

Activity D starts as soon as activity B and activity C completes their execution. Since, activity requires the completion of both for starting its execution, we would consider the MAX(EF(B), EF(C)). Hence, earliest week it can start its execution is week 9 i.e. ES(D) = 9. It takes 3 weeks to complete its execution. Hence, earliest it can finish is week 12 i.e. EF(D) = 12.

Activity E starts as soon as activity D completes its execution. Hence, earliest week it can start its execution is week 3 i.e. ES(E) = 12. It takes 3 weeks to complete its execution. Hence, earliest it can finish is week 15 i.e. EF(E) = 15.

Backward Pass :

The backward pass is carried out to calculate the latest dates on which each activity may be started and finished without delaying the end date of the project.

Activity E’s LS =LF - Duration latest finish date is equal to the earliest finish date of the precedent activity of finish according to the assumption i.e. LF(E) = 15. It takes 3 weeks to complete its execution. Hence, latest it can start is week 10 i.e. LS(G) = 10.

Activity H’s latest finish date is equal to the earliest finish date of the precedent activity of finish according to the assumption i.e. LF(H) = 13. It takes 2 weeks to complete its execution. Hence, latest it can start is week 11 i.e. LS(H) = 11.

The latest end date for activity C would be the latest start date of H i.e. LF(C) = 11. It takes 3 weeks to complete its execution. Hence, latest it can start is week 8 i.e. LS(C) = 8.

The latest end date for activity D would be the latest start date of H i.e. LF(D) = 11. It takes 4 weeks to complete its execution. Hence, latest it can start is week 7 i.e. LS(D) = 7.

The latest end date for activity E would be the latest start date of G i.e. LF(G) = 10. It takes 3 weeks to complete its execution. Hence, latest it can start is week 7 i.e. LS(E) = 7.

The latest end date for activity F would be the latest start date of G i.e. LF(G) = 10. It takes 10 weeks to complete its execution. Hence, latest it can start is week 0 i.e. LS(F) = 0.

The latest end date for activity A would be the latest start date of C i.e. LF(A) = 8. It takes 6 weeks to complete its execution. Hence, latest it can start is week 2 i.e. LS(A) = 2.

The latest end date for activity B would be the earliest of the latest start date of D and E i.e. LF(B) = 7. It takes 4 weeks to complete its execution. Hence, latest it can start is week 3 i.e. LS(B) = 3.