Civil Engineering Reference
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b.
The FF relationship (plus lag, if any) determines the EF
2
date for activity B.
The EF date for activity A is day 10, which is the same for B (since the lag on
the FF relationship is 0). The ES
2
for activity B is then calculated as
ES
2
=
EF
2
−
Dur
= 10 − 6 = 4. The second set of early dates for activity B is (4, 10).
c.
Pick whichever date set—(
ES
1
,
EF
1
)or(
ES
2
,
EF
2
)—is later; that is (4, 10).
d.
In the backward pass, the SS relationship (minus lag if any) determines the LS
1
date for activity B, which is
LS
(for C) − 3 (lag) = 7 − 3 = 4. The LF
1
for B is
calculated as
LS
1
+
Dur
= 4 + 6 = 10. The first set of late dates for activity B
is (4, 10).
e.
The FF relationship (minus lag if any) determines the LF
2
date for activity B
as
LF
(for C) − Lag(0inthiscase)= 12 − 0 = 12. The LS
2
for B is calculated
as
LF
2
−
Dur
= 12 − 6 = 6. The second set of late dates for activity B is
(6, 12).
f.
Pick whichever date set—(
LS
1
,
LF
1
)or(
LS
2
,
LF
2
)—is earlier; that is (4, 10).
The preceding process looks complicated in theory but is actually intuitive. Total
float and free float are calculated the same way as in Chapter 4. If we re-solve example
5.1 using contiguous activities, we will get the same finish date for the project; day
20. The only difference will be that activities A and E will be completely critical. As
mentioned previously, this is not always true. Let us look at example 5.2:
Example 5.2
Perform the forward and backward passes on the precedence network shown
in Figure 5.26, on the basis of the following: interruptible activities and con-
tiguous activities.
Solution with Interruptible Activities
Activity
ES
EF
LS
LF
DRF
SRF
URF
TF
A
0
10
3
54
0
41
3
44
B
0
12
0
12
0
0
0
0
C
3
7
6
33
0
23
3
26
D
2
30
5
33
0
0
3
3
E
0
38
22
54
2
22
16
40
F
12
22
12
22
0
0
0
0
G
22
42
22
42
0
0
0
0
H
30
39
33
42
0
0
3
3
I
42
54
42
54
0
0
0
0
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