2. Example
Consider the following project to
manufacture a simple mobile stone crasher.
The list of each activities, their relationship,
and the time required to complete them are
given in the table. We are interested to find
the time it will take to complete this project.
What jobs are critical to the completion of
the project in time, etc?
3. List of Activities
Activity Symbol Duration
(weeks)
Restriction
Preliminary design A 3 A < B, Cl
Engineering analysis B 1 B < Dl, F, H
Prepare layout I Cl 2 Cl < C2, Dl
prepare layout II C2 2 C2 < E
Prepare material request Dl 1 Dl < D2
Receive requested material D2 1 D2 < E
Fabricate Parts E 4 E < J
Requisition Parts F 1 F < G
Receive Parts G 2 G < J
Place subcontracts H 1 H < I
Receive subcontracted parts I 5 I < J
Assemble J 2 I < K
Inspect and test K 1
5. Then, it is necessary to find out the earliest
and latest completion time of for each
activity in the net work.
The earliest and the latest times are re-
calculated by using ‘forward pass’ and
‘backward pass’ computations, respectively.
The solution now starts by the forward pass
computation.
6. Step1. Determination of Earliest Time )
( j
E
Forward Pass Computation
The purpose of the forward pass computation is to find out
earliest start times for all the activities.
For this, it is necessary to assign some initial value to the
starting node 10.
Usually this value is taken to be zero so that the subsequent
earliest time could be interpreted as the project duration up to
that point in question.
Rules for the computation are as follows:
7. 0
10
E
j
E
]
.[ ij
i
j D
E
Max
E
Rule 1. Initial event is supposed to occur at time equal to
zero, that is,
Rule 2. Any activity can start immediately when all preceding
activities are completed.
for node j is given by
8. 0
10
E
]
.[ 20
20 i
i D
E
Max
E
10
i
Rule 3. Repeat step2 for the next eligible activity until the end
node is reached.
and
For node 20, node 10 is the only predecessor and hence
contains only one element. Therefore,
9. ]
.[ ij
i
j D
E
Max
E
3
3
0
20
,
10
10
20
D
E
E
,
,
5
2
3
21
,
20
20
21
D
E
E
4
1
3
30
,
20
20
30
D
E
E
5
1
4
50
,
30
30
50
D
E
E
5
1
4
60
,
30
30
60
D
E
E
10. 5
]
4
0
4
,
5
0
5
.[ 31
,
30
30
31
,
21
21
31
D
E
D
E
Max
E
32
E
6
1
5
32
,
31
31
32
D
E
E
The collection i consists of node 21 and 30 that are preceding node
31, Therefore,
and values of can be computed as:
Once again, for node 40 and 70;
]
.[ 31
,
31 i
i D
E
Max
E
Consider node 31, where there are two emerging
activities, i.e.
11. 7
]
7
1
6
,
7
2
5
.[ 40
,
32
32
40
,
21
21
40
D
E
D
E
Max
E
11
]
10
5
5
,
7
2
5
,
11
4
7
.[ 70
,
60
60
70
,
50
50
70
,
40
40
70
D
E
D
E
D
E
Max
E
80
E 90
E
13
2
11
80
,
70
70
80
D
E
E
14
1
13
90
,
80
80
90
D
E
E
and values of , and can be computed as:
12. From this computation, it can be inferred that this job will take 14
days to finish as this the longest path of the network.
Activities along this longest path are:– 10 – 20 – 21 – 40 70 – 80 –
90. This longest path is called the critical path. In any network, it
is not possible that there can be only one critical path.
13. Step2. Determination of Latest Time )
( i
L
Backward Pass Computation
In forward pass computation, the earliest time when a
particular activity will be completed is known.
It is also seen that some activities are not critical to
the completion of the job.
The question a manager would like to ask is: Can their
starting time be delayed so that the total completion
time is still the same?
Such a question may arise while scheduling the resources
such as manpower, equipment, finance and so on.
14. If delay is allowable, then what can be the maximum
delay? For this, the latest time for various activities
desired.
The backward pass computation procedure is used to
calculate the latest time for various activities.
15. S
i
i T
or
E
L
S
T
i
E
]
[
. ij
j
j
i D
-
L
Min
L
Rule 1. Set
Where is the scheduled date for completion and
is the earliest terminal time.
i.e. the latest time for activities is the minimum of the
latest time of all succeeding activities reducing their
activity time.
Rule 3.Repeat rule 2 until starting activity reached.
Rule 2.
18. 7
4
11
D
-
L
}
D
-
{L
Min.
L 40,70
40
j
40,
j
j
40
6
1
7
D
-
L
}
D
-
{L
Min.
L 32,40
32
j
32,
j
j
32
5
1
6
D
-
L
}
D
-
{L
Min.
L 32,32
31
j
31,
j
j
31
(j contains node 70)
(j contains node 40)
(j contains node 32)
Now consider node21, for this node, there are two succeeding
activities, namely 21 – 40, and 21 – 31. Hence,
19. 5
5
0
5
5
2
7
.
.
31
,
21
31
40
,
21
40
Min
D
L
D
L
Min
}
D
-
{L
Min.
L j
21,
j
40)
(31,
j
21
3
3
1
4
3
2
5
.
.
30
,
20
30
21
,
20
21
Min
D
L
D
L
Min
}
D
-
{L
Min.
L j
21,
j
30)
(21,
j
20
4
4
1
5
4
1
5
5
0
5
.
.
31
,
60
31
31
,
50
31
31
,
30
31
Min
D
L
D
L
D
L
Min
}
D
-
{L
Min.
L j
30,
j
60)
50,
(31,
j
30
Similarly, for node 20, and 30,
20. 0
3
D
-
L
}
D
-
{L
Min.
L 10,20
j
10,
j
j
10
3
20
0
L10
E
L i
i
0
L10
and like the other one, for node 10,
The minimum value of
is no surprising result. Since, started with
, it is always possible to have
If this is not so, it means that some error is made in calculations of
forward pass or backward pass values.
22. Recall that path 10 – 20 – 21 – 40 – 70 – 80 – 90
was defined as the critical path of this network.
Along this path, it is observed that the latest
and earliest times are the same implying that any
activity along this path cannot be delayed
without affecting the duration of the project.
23. Step 3. Computation of Float )
( f
L
By definition, for activity 60 – 70, the float is one day
1
5
6
60
60
E
L
This float represents the amount by which this particular activity
can be delayed without affecting the total time of the project.
24. Also, by definition, free float, if any will exist only on
the activities merge points.
To illustrate the concept of free float, consider path
10 – 20 – 30 – 50 – 70, total float on activity 50 - 70 is
four days and since this is the last activity prior to
merging two activities, this float is free float also.
Similarly, consider the activity 30-50 which has a total
float of 4 days but has zero free float because 4 day
of free float is due to the activity 50-70.
25. If activity 30-50 is delayed up to four days,
the early start time of no activity in the
network will be affected.
Therefore, the concept of free float clearly
states that the use of free float time will
not influence any succeeding activity float
time.
26. Step 4. To Identify Critical Path
Identifying the critical path is a byproduct
of boundary time calculations. A critical
activity has no leeway in scheduling and
consequently zero total float. It is important
to note that the value of slack, associated
with an event, determines how critical that
event is. The less the slack, the more critical
an event is.
27. The earlier calculation shows that the path or paths
which have zero float are called the critical ones or in
other words, a critical path is the one which connects
the events having zero total float or a minimum slack
time.
If this logic is extended further more, it would
provide a guide rule to determine the next most
critical path, and so on.
Such information will be useful for managers in the
control of project. In this example, path 10 – 20 – 30 –
60 – 70 – 80 - 90 happens to be next to critical path
because it has float of one day on many of its
activities.
28. Activity Duration
Start Finish
Total Float
Earliest Latest Earliest Latest
(1) (2) (3) (6)
A 3 0 0 3 3 0
B 1 3 4 4 5 1
C1 2 3 3 5 5 0
C2 2 5 5 7 7 0
D1 1 5 5 6 6 0
D2 1 6 6 7 7 0
E 4 7 7 11 11 0
F 1 4 8 5 9 4
G 2 5 9 7 11 4
H 1 4 5 5 6 1
I 5 5 6 10 11 1
J 2 11 11 13 13 0
K 1 13 13 14 14 0
i
E (2)
-
(6)
(4) (2)
(3)
(5)
(3)
-
(4)
(7)
ij
D
j)
-
(i
Boundary Times Duration for the Start and Finish of Activities
