EXAMPLE 8 Turning a Plane Around

Consider an airplane that carries both freight and passengers. The plane must have its passengers and freight unloaded and new passengers and cargo loaded before it can take off again. Also, the cabin must be cleaned before departure can occur. Thus, the job of “turning the plane around” requires completion of five tasks:

The order-requirement digraph for the problem of turning an airplane around is shown in Figure 2.17. The presence or absence of an edge in the order-requirement digraph depends on the analysis made as part of the modeling process for the problem. it seems natural that we need an arrow between task A and task C, because the passengers have to be unloaded before the cabin is cleaned. Other arrows may not seem natural—say, perhaps the arrow from task B (unload the cargo) to task E (load new passengers). This arrow may be due to government rules or union requirements.

What matters is that the mathematics of solving the problem does not depend on the reason that the order-requirement digraph looks the way that it does. The person solving the problem constructs the order-requirement digraph and then the mathematical techniques we will develop can be applied, regardless of whether or not another business faced with a similar problem might model the problem in a different way by using a different order-requirement digraph. Because we want to find the earliest completion time, it might seem that finding the shortest path in the digraph (path BD with time length ) would solve the problem. But this approach shows the danger of ignoring the relationship between the mathematical model (the digraph) and the original problem.

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The time required to complete all the tasks, A through E, must be at least as long as the time necessary to do the tasks on any particular path. Consider the path BD, which has length . Recall that here length of a path refers to the sum of the times of the tasks that lie along the path. Because task B must be done before task D can begin, the two tasks B and D cannot be completed before time 47. Hence, even if work on other tasks (such as A, C, and E) proceeds during this period, all the tasks cannot be finished before the tasks on path BD are finished. The same statement is true for every other path in the order-requirement digraph. Thus, the earliest completion time actually corresponds to the length of the longest path. in the airplane example, this earliest completion time is minutes, corresponding to the path ACE. We call ACE the critical path because the times of the tasks on this path determine the earliest completion time.

Note that if none of these tasks could go on simultaneously, the time to complete all the tasks would be minutes. However, even though some tasks may be performed simultaneously, the fact that the length of the critical path is 55 means that completion of the tasks in less than 55 minutes is not possible. Only by speeding up the times to complete the critical-path tasks themselves can a completion time less than 55 minutes be achieved.

Suppose it were desirable to speed the turnaround of the plane to less than 55 minutes. One way to do this might be to build a second jetway to help unload passengers more quickly. For example, we could unload passengers (task A) in 7 minutes instead of 13. However, reducing task A to 7 minutes does not reduce the completion time by 6 minutes, because in the new digraph (Figure 2.18) ACE is no longer the critical (longest) path. The longest path is now BE, which has a length of 52 minutes. Thus, shortening task A by 6 minutes results in only a 3-minute saving in completion time. This may mean that building a new jetway is uneconomical.

Note also that shortening the time to complete tasks that are not on the original critical path ACE will not shorten the completion time at all. Speeding up tasks on the critical path will shorten completion time of the job only up to the point where a new critical path is created. Also note that a digraph may have more than one longest path.