Question 17.P.2: Using the computing algorithm, determine the slack times for......

The task of determining ES, EF, LS, and LF times can be greatly simplified by setting up two brackets for each activity, as illustrated (Figure 1):

The bracket at the left of each activity will eventually be filled in with the earliest and latest starting times, and the bracket at the right end of each activity will be filled in with the earliest and latest finishing times (Figure 2):

This is accomplished in a two-step process. First, determine the earliest starting times and earliest finishing times, working from left to right, as shown in the following diagram (Figure 3).

Thus, activity 1-2 can start at 0. With a time of 4, it can finish at 0 + 4 = 4. This establishes the earliest start for all activities that begin at node 2. Hence, 2-5 and 2-4 can start no earlier than time 4. Activity 2-5 has an early finish of 4 + 6 = 10, and activity 2-4 has an early finish of 4 + 2 = 6. At this point, it is impossible to say what the earliest start is for 4-5; that will depend on which activity, 3-4 or 2-4, has the latest EF. Consequently, it is necessary to compute ES and EF along the lower path. Assuming an ES of 0 for activity 1-3, its EF will be 9, so activity 3-4 will have an ES of 9 and an EF of 9 + 5 = 14.

Considering that the two activities entering node 4 have EF times of 6 and 14, the earliest that activity 4-5 can start is the larger of these, which is 14. Hence, activity 4-5 has an ES of 14 and an EF of 14 + 3 = 17.

Now compare the EFs of the activities entering the final node. The larger of these, 17, is the expected project duration.

The LF and LS times for each activity can now be determined by working backward through the network (from right to left). The LF for the two activities entering node 5 is 17—the project duration. In other words, to finish the project in 17 weeks, these last two activities must both finish by that time.

In the case of activity 4-5, the LS necessary for an LF of 17 is 17 – 3 = 14. This means that both activities 2-4 and 3-4 must finish no later than 14. Hence, their LF times are 14. Activity 3-4 has an LS time of 14 – 5 = 9, making the LF of activity 1-3 equal to 9, and its LS equal to 9 – 9 = 0.

Activity 2-4, with an LF time of 14, has an LS time of 14 – 2 = 12. Activity 2-5 has an LF of 17 and therefore an LS of 17 – 6 = 11. Thus, the latest activity 2-5 can start is 11, and the latest 2-4 can start is 12 in order to finish by week 17. Since activity 1-2 precedes both of these activities, it can finish no later than the smaller of these, which is 11. Hence, activity 1-2 has an LF of 11 and an LS of 11 – 4 = 7.

The ES, EF, LF, and LS times are shown on the following network (Figure 4).

The slack time for any activity is the difference between either LF and EF or LS and ES. Thus,

The activities with zero slack times indicate the critical path. In this case the critical path is 1-3-4-5. When working problems of this nature, keep in mind the following:

• The ES time for leaving activities of nodes with multiple entering activities is the largest EF of the entering activities.
• The LF for an entering activity for nodes with multiple leaving activities is the smallest LS of the leaving activities.