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PROJECT SCHEDULING WITH KNOWN ACTIVITY TIMES 375

We will write the earliest start and earliest finish times in the node to the right of
the activity letter. Using activity A as an example, we have:

Earliest start
time Earliest finish
time

A 0 5
5

Because an activity cannot be started until all immediately preceding activities have
been finished, the following rule can be used to determine the earliest start time for
each activity.
The earliest start time for an activity is equal to the largest of the earliest
finish times for all its immediate predecessors.

Let us apply the earliest start time rule to the portion of the network involving
nodes A, B, C and H, as shown in Figure 9.3. With an earliest start time of 0 and an
activity time of 6 for activity B, we show ES ¼ 0 and EF ¼ ES + t ¼ 0+ 6 ¼ 6in
the node for activity B. Looking at node C, we note that activity A is the only
immediate predecessor for activity C. The earliest finish time for activity A is 5, so
the earliest start time for activity C must be ES ¼ 5. So, with an activity time of 4,
the earliest finish time for activity C is EF ¼ ES + t ¼ 5+4 ¼ 9. Both the earliest
start time and the earliest finish time can be shown in the node for activity C (see
Figure 9.4).
Continuing with Figure 9.4, we move on to activity H and apply the earliest start
time rule for this activity. With both activities B and C as immediate predecessors,
the earliest start time for activity H must be equal to the largest of the earliest finish
times for activities B and C. So, with EF ¼ 6 for activity B and EF ¼ 9 for activity C,
we select the largest value, 9, as the earliest start time for activity H (ES ¼ 9). With

Figure 9.3 A Portion of the Souk al Bustan Shopping Centre Project Network,
Showing Activities A, B, C and H

Earliest start
time Earliest finish
time
A 0 5
5

C H
Start
4 12
B 0 6
6

Earliest start Earliest finish
time time

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