## Path and lime

Consider the hypothetical project shown in Figure 8-15. Assume, for convenience, that the time units involved are days. How long will it take to complete the project? Figure 8-15: The complete network from Table 8-2.

(For the moment we will treat the expected times as if they were certain.) If we start the project on day zero, we can begin simultaneously working on activities a, b, and c, each of which have no predecessor activities. We will reach event 2 in 20 days, event 3 in 20 days, and event 4 in ten days. We have shown these times just above or below their respective nodes. They are labelled EOT (earliest occurrence time) because they represent the earliest times that the event can occur. Activity d, for example, cannot begin before event 2 has occurred, which means that all activities that precede event 2 must be completed. In this case, of course, activity a is the only predecessor of event 2.

Note that event 3 not only requires the completion of activity b, but also requires the completion of activity c, as shown by the dummy activity. (Refer to Figure 8-9 for a refresher.) The dummy requires neither time nor resources, so it does not affect the network time in any way. Event 3 does not occur until all paths leading to it have been completed. Therefore, the EOT for event 3 is equal to the time required by the longest path leading to it. The path from event 1 to event 3 requires the completion of activity b (20 days) and the completion of activities c and dummy (ten +, zero days). Because the two paths may be followed simultaneously, we can reach' event 3 in 20 days. Therefore, the earliest starting time (EST) for any activity ema-' nating from event 3 is 20 days. .

Proceeding similarly, we see that event 6 has two predecessor activities, d and e. Activity d cannot start until day 20, (EST = 20) and it requires 15 days to com-, plete. Thus, its contribution to event 6 will require a total of 35 days from the start-of the project. Activity e may also start after 20 days, the EOT for event 3, but it re-;-quires only ten days, a total of 30 days from the project start. Because event 6 rfr quires the completion of both activities d and e, the EOT for event 6 is 35 days, th% longest of the paths to it. Event 5 has an EOT of 24 days, the longest of the two paths; leading to it, and event 7, the completion event of the network, has a time of 4^, days. The EOTs are shown in Figure 8-15.

There are eight activity paths leading to event 7. They are a-d-j=20 + 15 + 8=43 days c-dummy-e-|= 10 + 0 + 10 + 8=28 days b-e-j = 20 + 10 + 8=38 days c-dummy-f =10 + 0+14 =24 days b-f=20 + 14 =34 days c-dummy-g-i= 10 + 0 + 4+18=32 days b-g-i = 20 + 4 + 18=42 days c-h-i = 10 + 11 + 18=39 days

The longest of these paths is a-d-j using 43 days, which means that 43 days is shortest time in which the entire network can be completed. This is called the cnt time of the network, and a-d-j is the critical path, usually shown as a heavy line ^ In a simple network such as our example, it is easy to find and evaluate evetf;. path between start and finish. Many real networks are considerably more compl' and finding all paths can be taxing. Using the method illustrated above, there is need to worry about the problem. Every node is characterized by the fact that one more activities lead to it. Each of these activities has an expected duration and o inates in an earlier node. As we proceed to calculate the EOT of each node, begl ning at the start, we are actually finding the critical path and time to each of the nodes in the work. Note that event 5 has an EOT (critical time) of 24 days, and its critical path is b-g rather than c-h which requires 21 days, or c-dummy-g which takes 14 days.

The number of activities directly entering an event tells us the number of paths we must evaluate to find the EOT for that event. Here, path is defined as originating at immediate predecessor events, not at the network origin. With event 5, that number is two, so we find the EOT and activity times for the two immediate predecessors. For event 7, we have three evaluations to do, the EOT of event 6 plus the duration of activity | (43 days), the EOT of event 3 plus the duration of f (34 days), and the EOT of event 5 plus the duration of 1 (42 days). There is, therefore, no need to find, list, and evaluate all possible start-to-finish paths in the network.

Although we will assume throughout this chapter that we always employ the "as-soon-as-possible" approach to scheduling tasks ("early start"), there are situations where other approaches are sometimes used. One example is the simultaneous start, where all resources are launched at the beginning. Another is the simultaneous finish, where a facility can be moved to its next location once all the tasks are finished. Of course, delay early on in a project runs the risk of delaying the overall project if some other activities inadvertently become delayed. One important reason for using an "as-late-as-possible" approach is that it delays the use of resources as much as possible, thereby optimizing the cash flow of the project, but again at some risk of delay. 