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R. Sivakumar
P. R.Vittal

R. Sivakumar and P. R.Vittal discuss a tabular approach to project crashing

THE critical path method (CPM) was originally developed to solve scheduling problems in an industrial setting and, hence, more concerned with costs of project scheduling and how to minimise them. Therefore, an important aspect of network analysis is the cost of project.

The cost of a project is not due solely to the direct costs associated with individual activities or jobs in the project.

Normally, there are indirect expenses as well — overhead items such as managerial services, indirect supplies, equipment rentals, allocation of fixed expenses, and so forth.

Many of these are directly affected by the length of the project: The longer the time required for its completion, the higher will be the expenses.

In addition, if the project management is under contract to a government agency or some other customer to complete the programme by a given due date, often the contract will specify penalties for delay beyond this date.

The normal duration of a project is the specified completion time of the project, wherein the duration of the activities, manpower, machine power, and so on, are all fixed already under normal conditions and, therefore, the project will also incur a given cost.

However, to reduce the duration one may require, among other things, more manpower and machine power and this will increase the cost of the project. There will be a possibility of reducing the cost of the project by reducing the duration.

CPM can reduce the duration if extra resources (men, machines, money and so on) are assigned. The cost for getting the job done may increase, but if other advantages outweigh this added cost, the job should be expedited, or crashed.

On the other hand, if there is no reason to shorten a particular job — if it has a generous amount of slack — the job should be done at its normal or most efficient pace, with lesser assignment of resources.

Thus, there is no need to crash all jobs to get a project done faster; only the critical jobs need to be expedited. Which jobs to expedite and by how much are problems CPM attempts to solve.

Minimum duration and cost

Project duration is controlled by the critical path. Therefore, to reduce the project duration, the critical path duration has to be reduced keeping it as critical path. For finding this, we proceed as follows:

Step 5: Now choose the activity or combination of activities along the critical path or paths for crashing such that the cost slope is minimum.

When all the paths are critical, a combination of one activity from each path is necessary.

The following rules may be useful in finding out the combination:

Table 1— Rules for finding out combinations: i) If there are independent activities in each of the paths, a combination or combinations will emerge among these activities.

ii) If an activity is common to all the paths, that activity itself will be a combination.

iii) If an activity is not common to all the paths but only to some of the paths it can be combined with the independent activity of the other paths or combined with the common activity of the other paths.

Note: The maximum number of activities forming a combination will always be less than or equal to the number of paths.

Step 6: Repeat this procedure until no further crashing is possible. This will be possible where crashing has been exhausted along one of the critical paths. If there is an overhead cost compute the trade-off between the overhead cost vis-à-vis the crash cost for optimal crashing.

Project cost crash table

In this approach, the illustration shown in Table 1 is solved in a single table (refer Table 2).

Step 1: The network diagram shows the critical path to be 1-2, 2-3 and 3-4

Step 2: The cost slope is already given.

Step 3: Description of the project-cost crash.

The activities, the various paths along with the duration, the cost slope and the maximum crash duration days are all entered in the first six columns. The next four columns, that is, seven to 10, indicate the activity/activities selected for crashing step by step. It can be noticed that for each step that the revised crash duration available is also computed.

The activity chosen in step 1 for the number of days of reduction is calculated with reference to the difference of the duration between the critical path and the next immediate path having the highest duration (refer row six). Difference in duration between critical path 1-2-3-4 and the other two paths is one day. Hence the activity 2-3 on the critical path having the least cost slope is selected.

Step 2, for reduction, is carried out with combination of activities since all the paths are now critical. It can be noticed that activity 1-2 is common for paths 1-2-4 and 1-2-3-4 and activity 3-4 is common for 1-2-3-4 and 1-3-4. Activities 1-3, 2-3 and 2-4 are independent to each path. Therefore, applying rules 1 and 3, we get the three combinations as shown in Table 3.

When all the paths are critical, reduction of the duration is carried out by one day. However, the reduction can be carried out for more number of days provided the duration is available.

In rows as well as columns six to 10, we also capture the crash cost, the cumulative crash cost, the indirect cost, the normal cost and the total cost for each step of reduction. The optimal duration is also indicated.

Almost all the examinations problems on crashing consists of less than ten activities and less than four paths. The student may find the method of presentation discussed not only useful but also less time consuming.

Further, while the student need not draw the network at each stage, he has to exercise caution on the selection of activities when all the paths are critical.

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