Question

Given a simple project illustrated in the following table

Activity	Immediate Predecessors	Time (weeks)
1	-	2
2	-	3
3	1, 2	2
4	3	5
5	3	4

The required project completion time is 12. Assume that "Activity 1" actually finished at 3 weeks, and "Activity 2" actually finished at 2 weeks. Given the actual project progress, what is the LF (lastest time) of "Activity 4"?

Answer 1

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK	CRITICAL
1	3	0	3	0	3	0	YES
2	2	0	2	0	3	1
3	2	3	5	3	5	0	YES
4	5	5	10	5	10	0	YES
5	4	5	9	6	10	1

FORWARD PASS

We calculate the ES and EF values using a forward pass where the ES of an activity is the maximum EF of all the predecessor activities.

BACKWARD PASS

We calculate the LS and LF values using a backward pass where the LF of the activity is the minimum of all the successor activities.

SLACK

Slack is the value which is determined by subtracting EF from the LF or ES from the LS.

CRITICAL PATH

Critical path is the chain in the project network where the slack value of all the activities is 0, what this means is that any delay in these activities would result in delaying the entire project.

CRITICAL PATH = 1-3-4

DURATION OF PROJECT = 10

LF OF 4 = 10

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