Question 17.3: Computing the Latest Finishing and Starting Times Compute th...

Computing the Latest Finishing and Starting Times

Compute the latest finishing and starting times for the precedence diagram developed in Example 2.

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We must add the LS and LF times to the brackets on the diagram.

Begin by setting the LF time of the last activity equal to the EF of that activity. Thus,

LF5-6 = EF5-6 = 20 weeks

Obtain the LS time for activity 5-6 by subtracting the activity time, t, from the LF time:

LS5-6 = LF5-6t = 20 − 1 = 19

Mark these values on the diagram:

The LS time of 19 for activity 5-6 now becomes the LF time for each of the activities that precedes activity 5-6. This permits determination of the LS times for each of those activities: Subtract the activity time from the LF to obtain the LS time for the activity. The LS time for activity 3-5 is 19 − 9 = 10.

Next, the LS for activity 4-5, which is 16, becomes the LF for activity 2-4, and the LS for activity 3-5, which is 10, becomes the LF for activity 1-3. Using these values, you find the LS for each of these activities by subtracting the activity time from the LF time.

The LF for activity 1-2 is the smaller of the two LS times of the activities that 1-2 precedes. Hence, the LF time for activity 1-2 is 8. The reason you use the smaller time is that activity 1-2 must finish at a time that permits all following activities to start no later than their LS times.

Once you have determined the LF time of activity 1-2, find its LS time by subtracting the activity time of 8 from the LF time of 8. Hence, the LS time is 0.

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