The head event slack of an activity in a network is the slack at the head.The tail event slack of an activity in a network is the slack at the tail.
As discussed earlier, the non – critical activities have some slack or float. The float of an activity is the amount of time available by which it is possible to delay its completion time without extending the overall project completion time.
For an activity i = j, let
tij = duration of activity
TE = earliest expected time
TL = latest allowable time
ESij = earliest start time of the activity
EFij = earliest finish time of the activity
LSij = latest start time of the activity
LFij = latest finish time of the activity
Total Float TFij: The total float of an activity is the difference between the latest start
time and the earliest start time of that activity.
TFij = LSij– ESij ....................(1)
TFij = (TL– TE) – tij ....................(2)
Free Float FFij: The time by which the completion of an activity can be delayed from its earliest finish time without affecting the earliest start time of the succeeding activity is called free float.
FFij = (Ej– Ei) – tij ....................(3)
FFij = Total float – Head event slack
Independent Float IFij: The amount of time by which the start of an activity can be delayed without affecting the earliest start time of any immediately following activities, assuming that the preceding activity has finished at its latest finish time.
IFij = (Ej– Li) – tij ....................(4)
IFij = Free float – Tail event slack
Where tail event slack = Li– Ei
The negative value of independent float is considered to be zero.
Critical Path: After determining the earliest and the latest scheduled times for various activities, the minimum time required to complete the project is calculated. In a network, among various paths, the longest path which determines the total time duration of the project is called the critical path. The following conditions must be satisfied in locating the critical path of a network.
An activity is said to be critical only if both the conditions are satisfied.
1. TL– TE= 0
2. TLj– tij– TEj = 0
Example : A project schedule has the following characteristics as shown in the table
i. Construct PERT network.
ii. Compute TE and TL for each activity.
iii. Find the critical path.
(i) From the data given in the problem, the activity network is constructed as shown in the following figure.
Activity Network Diagram
(ii) To determine the critical path, compute the earliest, time T Network Model E and latest time TL for each of the activity of the project. The calculations of TE and TL are as follows:
To calculate TEfor all activities,
TE1 = 0
TE2 = TE1 + t1, 2 = 0 + 4 = 4
TE3 = TE1 + t1, 3 = 0 + 1 =1
TE4 = max (TE2 + t2, 4 and TE3 + t3, 4)
= max (4 + 1 and 1 + 1) = max (5, 2)
= 5 days
TE5 = TE3 + t 3, 6 = 1 + 6 = 7
TE6 = TE5 + t 5, 6 = 7 + 4 = 11
TE7 = TE5 + t5, 7 = 7 + 8 = 15
TE8 = max (TE6 + t 6, 8 and TE7 + t7, 8)
= max (11 + 1 and 15 + 2) = max (12, 17)
= 17 days
TE9 = TE4 + t4, 9 = 5 + 5 = 10
TE10 = max (TE9 + t9, 10 and TE8 + t8, 10)
= max (10 + 7 and 17 + 5) = max (17, 22)
= 22 days
To calculate TL for all activities
TL10 = TE10 = 22
TL9 = TE10– t9,10 = 22 – 7 = 15
TL8 = TE10– t 8, 10 = 22 – 5 = 17
TL7 = TE8– t 7, 8 = 17 – 2 = 15
TL6 = TE8– t 6, 8 = 17 – 1 = 16
TL5 = min (TE6– t5, 6 and TE7– t5, 7)
= min (16 – 4 and 15 –8) = min (12, 7)
= 7 days
TL4 = TL9 – t 4, 9 = 15 – 5 =10
TL3 = min (TL4 – t3, 4 and TL55– t 3, 5)
= min (10 – 1 and 7 – 6) = min (9, 1)
= 1 day
TL2 = TL4– t2, 4 = 10 – 1 = 9
TL1 = Min (TL2– t1, 2 and TL3– t1, 3)
= Min (9 – 4 and 1 – 1) = 0
Various Activities and their Floats
(iii) From the table, we observe that the activities 1 – 3, 3 – 5, 5 – 7,7 – 8 and 8 – 10 are critical activities as their floats are zero.
Critical Path of the Project
The critical path is 1-3-5-7-8-10 (shown in double line in the above figure) with the project duration of 22 days.
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