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PROJECT SCHEDULING WITH KNOWN ACTIVITY TIMES 377

then

LS ¼ LF t (9:2)

Beginning the backward pass with activity I, we know that the latest finish time is
LF ¼ 26 and that the activity time is t ¼ 2. Thus, the latest start time for activity I is
LS ¼ LF t ¼ 26 2 ¼ 24. We will write the LS and LF values in the node directly
below the earliest start (ES) and earliest finish (EF) times. So, for node I, we have:

I 24 26
2 24 26

Latest start Latest finish
time time

The following rule can be used to determine the latest finish time for each activity
in the network.
The latest finish time for an activity is the smallest of the latest start times
for all activities that immediately follow the activity.
Logically, this rule states that the latest time an activity can be finished equals the
earliest (smallest) value for the latest start time of following activities. Figure 9.6
shows the complete project network with the LS and LF backward pass results. We
can use the latest finish time rule to verify the LS and LF values shown for activity
H. The latest finish time for activity H must be the latest start time for activity I.
Thus, we set LF ¼ 24 for activity H. Using Equation (9.2), we find that LS ¼ LF
t ¼ 24 12 ¼ 12 as the latest start time for activity H. These values are shown in
the node for activity H in Figure 9.6.
Activity A requires a more involved application of the latest start time rule. First,
note that three activities (C, D and E) immediately follow activity A. Figure 9.6
shows that the latest start times for activities C, D and E are LS ¼ 8, LS ¼ 7 and
LS ¼ 5, respectively. The latest finish time rule for activity A states that the LF for
activity A is the smallest of the latest start times for activities C, D and E. With the
smallest value being 5 for activity E, we set the latest finish time for activity A to
LF ¼ 5. Verify this result and the other latest start times and latest finish times
shown in the nodes in Figure 9.6.
After we complete the forward and backward passes, we can determine the
amount of slack associated with each activity. Slack is the length of time an activity
can be delayed without increasing the project completion time (it is also referred to
as Float time). The amount of slack for an activity is calculated as follows:

Slack ¼ LS ES ¼ LF EF (9:3)

The slack for each For example, the slack associated with activity C is LS ES ¼ 8 5 ¼ 3 weeks.
activity indicates the So, activity C can be delayed up to three weeks, and the entire project can still be
length of time the activity
can be delayed without completed in 26 weeks. In this sense, activity C is not critical to the completion of the
increasing the overall entire project in 26 weeks. Next, we consider activity E. Using the information in
project completion time. Figure 9.6, we find that the slack is LS ES ¼ 5 5 ¼ 0. So, activity E has zero,

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