Civil Engineering Reference
In-Depth Information
Example 5.1
Perform the CPM calculations for the network shown in Figure 5.22.
Solution
Forward Pass
1. Start the project at activity A.
ES
=
0.
EF
=
0
+
2
=
2.
2. Activity B can start as soon as activity A is finished.
ES
=
2.
EF
=
2
+
6
=
8.
3. Activity C can start 1 day after A has started.
ES
=
1,
EF
=
1
+
11
=
12.
4. Activity D can start 2 days after activity B has started.
ES
=
2
+
2
=
4. Use the later of
EF
=
4
+
7
=
11 and the
EF
(for B)
=
8. We choose
EF
=
11.
5. Activity E can start as soon as B is finished.
ES
=
8. Use the later of
EF
=
8
+
4
=
12 and
EF
(for C)
+
5-day lag
=
12
+
5
=
17. We choose
EF
=
17.
6. Activity F has no restriction for its start (dangling activity).
ES
=
0,
EF
=
0
+
8
=
8.
7. Activity G can start after both D and E are finished.
ES
(for G)
=
17 (the
later of 11 and 17). Use the later of
EF
=
17
+
3
=
20 and
EF
(for F)
+
4-day lag
=
8
+
4
=
12. We choose
EF
=
20.
8. The calculated early finish date for the project is day 20 (project dura-
tion is 20 days).
Solution
The Backward Pass
1. StartatactivityG.
LF
=
20,
LS
=
20
−
3
=
17.
2. Activity F must finish 4 days prior to the finish of G (i.e.,
LF
=
20
−
4
=
16;
LS
=
16
−
8
=
8).
3. Activity E must finish before G can start.
LF
=
17.
LS
=
17
−
4
=
13.
4. Activity D must finish before G can start.
LF
=
17.
LS
=
17
−
7
=
10.
5. Activity C must finish 5 days before the finish of E.
LF
=
17
−
5
=
12.
LS
=
12
−
11
=
1.
6. Activity B must finish before D has finished, on day 17, and before
E has started, on day 13.
LF
=
13 (the earlier). Use the earlier of
LS
=
LF
−
Dur
=
13
−
6
=
7 and 2 days before the LS for D (i.e., 8).
We choose the earlier:
LS
=
7.
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