Determining the Critical Path

Determining the critical Path:

Once the network diagram with single time estimates has been developed, the following computational procedure may be employed fr determining the critical path(s), event slacks, and activity floats.

1. Calculate the earliest occurance time(EOT) for each event: As seen in the network diagram above event 4 occurs when the path 1-2-4 and 1-3-4 are completed. The EOT of an event refers to the time when the event can be completed at the earliest. Looking at event ‘4’ we find that since path leading to it i.e. 1-2-4 or 1-3-4 take 15 and 20 weeks respectively, EOT of event 4 is 20 weeks.

In general, EOT of an event is duration of the longest path(fro beginning event whose EOT is set at 0) leading to that event. It may be noted in figure that an event is represented by a circle. The upper half of circle denote an event number, left quarter in lower portion of circle denote the EOT, right quarter in lower half denote the latest occurrence time.

EOT of an event obviously represent the minimum time required for completing the project. This computation is referred to as forward pass. In this computation we assume that each activity starts immediately on the occurrence of the events preceding it. Hence the starting and finishing time for various activities obtained from this computation are the earliest starting time(EST) and earliest finishing time (EFT).

General formula for EOT is:

EOT(i) = Max( EOT(k) + d(k,i)]

Where EOT(i) is the earliest occurance time of an event

EOT(k) is the earliest occurance time of event k( k preceds I, and there may be several ks.

D(k,i) is the duration of the activity (k,i)

The maximization shown is done considering all activities (k,i) leading to event node i. THE FORMULA FOR est AND eft ARE:

Est (I,J) = eot (I)

Eft(I,J) = eot (I) + D(I,J) where EST(i,j) is the earliest starting time for activity (i,j), EOT(i) is the earliest occuranec time of event (i), EFT(i,j) is the earliest finishing time for activity (I,j) and d(I,j) is the duration of the activity.

2. Calculate the latest occurance time(LOT) for each event: The LOT for an event represent the latest allowable time by which the event can occur, given the time that is allowed for the completion of the project. Normally the time for completion of project is set equal to EOT of the end event. This means that for the end event the LOT and EOT are set equal. The LOT for various events are obtained by working backwards from the end event. This procedure is called backward pass. The LOT for event ‘4’ in our illustrative project , for example, is equal to the LOT for event ‘5’, the end event, minus the duration of the activity (4-5) which connects event 4 with event 5. Since the LOT for event ‘5’ is 28weeks and the duration of the activity (4-5) is 2 weeks, the LOT for event ‘4’ is 26 weeks (28-2). This represent the latest time by which event ‘4’ should occur to enable the project to be completed in 28 weeks. Figure on right shows this.

The general formula for LOT is:

LOT(i) = Min(LOT(i) –d(I,j)]

LOT is the latest occurance time ofor I, LOT(j) is the latest occurance time for j( j follows I and there may be several js) is the duration of the activity (I,j) starting from i.

Given the LOT for various events we can calculate the latest finishing time(LFT) and latest starting time(LST) for various activities using the formuls:

LFT(I,j) = LOT (j)

LST(I,j) = LFT (I,j) – d(I,j)

Wher LFT(I,j) is the latest finishing time for activity (I,j), LOT(j) latest occurrence time for event j, LST(I,j) is the latest starting time for activity (I,j) and d(I,j) is the duration of the activity(I,j). 3. Calculate the slacks for each event: The slack for an event is defined as the difference between its LOT and EOT. The slacks for various events for our example are:

Event LOT EOT (in weeks) slack= LOT-EOT 5 28 28 0 4 26 20 6 3 18 12 6 2 13 13 0 1 0 0 0

4. Calculate the critical and slack paths

The critical path starts with the beginning event, terminates with the end event, and is marked by events which have a zero slack. This is obviously the path on which there is no slack, no cushions. Other paths are slack paths with some cushion. The critical path in the network diagram is shown by double arrow.

The critical path is the longest path from the beginning event to the end event. Since the end can be reached i.e. the project completed, only when this longest path is traversed. The minimum tiem required for completing the project is the duration on the critical path. The duration of the critical pathof our project is 28 weeks; this is the minimum time required for completing the project.

5. Compute the activity floats: Given the estimates of the activity time and event slacks, activity floats can be calculated. There are three measures of floats:

i. Total float

ii. Free Float

iii. Independent float

In figure EOT and LOT and respectively represent, earliest occurrence time, latest occurrence time, and duration. The total float of an activity is the extra time available to complete the activity if it is started as early as possible, without delaying the completion of the project. The total float for activity (2-4) is equal to:

Latest occurrence time for event ‘4’ - Earliest time occurance for event 2 - Duration of activity (2-4)

= 26 weeks - 13 weeks - 2 weeks = 11 weeks

The total float represent the float under the most favourable condition. This is so because the activity can be started at the earliest (EOT of the preceding event) and completed at the latest (the LOT of succeeding event). Obviously, the activities which do not have a float event under these conditions, the most favourable ones, are critical to project and hence lie on the critical path.

The free Float: This activity is the extra time available to complete the activity when the activity is started at the EOT of its preceding event and completed by the EOT of its succeeding event.

The free float of activity (2-4) is:

Earliest occurrence time for event ‘4’ - earliest occurrence time for event ‘2’ – duration of activity

= 20 weeks - 13 weeks - 2 weeks =5 weeks

Independent float: It is the extra time available to complete the activity when the activity is started at the LOT of its preceding event and completed by the EOT of its succeeding event. The independent float for activity (2-4) is calculated as:

Earliest occurrence time for event ‘4’ - latest occurrence time for event ‘2’ – duration of activity

= 20 weeks - 13 weeks - 2 weeks =5 weeks

Independent float represent the float under most adverse condition. Hence, when an activity has a positive independent float it means that the activity has a cushion (equal to its independent float) irrespective of what happens elsewhere. It must be noted that independent float of an activity may be negative but the total float and free float cannot be negative

Most generally, float may be represented by the following equations:

TF(I,j) = LOT (j) – EOT (i) –d(I,j)

FF(I,j) = EOT (j) – EOT(i) 0d(I,j)

IF(I,j) = EOT(j) – LOT(i) –d(I,j)

Activity(I,j) Duration Earliest Earliest Latest Latest Total Free Independent start finish start finish Float Float Float time(I,j) time(I,j) time(I,j) time(I,j) =EST(i) =EFT(j) =LST(i) =LFT(j) A(1-2) 13 0 13 0 13 0 0 0 B(1-3) 12 0 12 6 18 6 0 0 C(2-4) 2 13 15 24 26 11 5 5 D(3-4) 8 12 20 18 26 6 0 (6) E(2-5) 15 13 28 13 28 0 0 0 F(4-5) 2 20 22 26 28 6 6 0 Float for various activity