TABLE 19.13 Computing the Earliest Completion Time of a Project Earliest Start Time (EST) Expected Time (days) Earliest Completion Time (ECT) Activity A, 4 4 A2 ECT(A,) = 4 2 4 + 2 = 6 Аз ECT(A2) = 6 10 6 + 10 = 16 A4 ECT(A2) = 6 4 6 + 4 = 10 As ECT(A2) = 6 6 + 2 = 8 %3D As Max(16,10) = 16 16 + 3 = 19 A7 Max{8,19) = 19 5 19 + 5 = 24 Ag ЕСТА,) 3 24 2 24 + 2 = 26 2.

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Consider again the project with the activity network shown in Figure 19.14.
Activity A1 takes 4 days; A2, 2 days; A3, 10 days; A4, 4 days; A5, 2 days; A6, 3 days; A7,
5 days; and A8, 2 days. Compute the earliest completion time of the project.
Answer: 26 days.
We start the project with A1 because it has no predecessor activity. A1 will take 4 days, so
the ECT for A1 is 4. Once A1 is complete, we can start on A2. A2 thus has an EST of 4. A2
takes 2 days and hence has an ECT of 6.
Following the completion of A2, we can initiate A3, A4, and A5. These activities thus can
all have an EST of 6. We continue this process as shown in Table 19.13 and find that the
earliest completion time is 26 days.

TABLE 19.13 Computing the Earliest Completion Time of a Project
Earliest Start
Time (EST)
Expected
Time (days)
Earliest Completion
Time (ECT)
Activity
A,
4
4
A2
ECT(A,) = 4
2
4 + 2 = 6
Аз
ECT(A2) = 6
10
6 + 10 = 16
A4
ECT(A2) = 6
4
6 + 4 = 10
As
ECT(A2) = 6
6 + 2 = 8
%3D
As
Max(16,10) = 16
16 + 3 = 19
A7
Max{8,19) = 19
5
19 + 5 = 24
Ag
ЕСТА,) 3 24
2
24 + 2 = 26
2.
Transcribed Image Text:TABLE 19.13 Computing the Earliest Completion Time of a Project Earliest Start Time (EST) Expected Time (days) Earliest Completion Time (ECT) Activity A, 4 4 A2 ECT(A,) = 4 2 4 + 2 = 6 Аз ECT(A2) = 6 10 6 + 10 = 16 A4 ECT(A2) = 6 4 6 + 4 = 10 As ECT(A2) = 6 6 + 2 = 8 %3D As Max(16,10) = 16 16 + 3 = 19 A7 Max{8,19) = 19 5 19 + 5 = 24 Ag ЕСТА,) 3 24 2 24 + 2 = 26 2.
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