I.E. 2001   OPERATIONS RESEARCH

THE CRITICAL PATH METHOD - An Example


Example:

Proceeding from Node 1 to Node 7 in the forward pass yields the ES values, and then proceeding from Node 7 to Node 1 in the backward pass yields the LC values in the table below:
 

Node
ES
LC
1
0
 Min{LC2-D12}
= Min{4-4}=0
2
 Max{ES1+D12}
= 0+4=4		 Min{LC3-D23,LC4-D24,LC5-D25} 
= Min{12-8,18-7, 21-6}=4
3
 Max{ES2+D23}
=4+8=12		 Min{LC4-D34, LC5-D35}
Min{18-0, 21-9}=12
4
 Max{ES2+D24,ES3+D34}
= Max{4+7, 12+0}=12		 Min{LC6-D46}
= Min{33-15}=18
5
 Max{ES2+D25,ES3+D35}
= Max{4+6, 12+9}=21		 Min{LC6-D56}
= Min{33-12}=21
6
 Max{ES4+D46,ES5+D56}
=Max{12+15,21+12}=33		 Min{LC7-D67}
= Min{53-20}=33
7
 Max{ES6+D67 }
=33+20 = 53		 =ES7
= 53

We now go to Steps 3 and 4 to find ES, LC and TF values for each activity...

Activity
(i-j)
Dij
ESij (=ESi)
LCij 
(=LCj)
TFij
(=LCij-ESij-Dij)
1-2
4
0
4
0
2-3
8
4
12
0
2-4
7
4
18
7
2-5
6
4
21
11
3-4
0
12
18
6
3-5
9
12
21
0
4-6
15
12
33
6
5-6
12
21
33
0
6-7
20
33
53
0

The critical activities are 1-2, 2-3, 3-5, 5-6 and 6-7 and the critical path is shown below:
 


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