Operations Management Essay

EXAMPLE 4.3: Time–Cost Trade-off Procedure

The procedure for project crashing consists of the following five steps. It is explained by using the simple four-activity network shown in Exhibit 4.8. Assume that indirect costs are $10 per day for the first eight days of the project. If the project takes longer than eight days, indirect costs increase at the rate of $5 per day.

SOLUTION

1. Prepare a CPM-type network diagram. For each activity, this diagram should list:

a. Normal cost (NC): the lowest expected activity costs. (These are the lesser of the cost figures shown under each node in Exhibit 4.8.)

b. Normal time (NT): the time associated with each normal cost.

c. Crash time (CT): the shortest possible activity time.

d. Crash cost (CC): the cost associated with each crash time.

2. Determine the cost per unit of time (assume days) to expedite each activity. The relationship between activity time and cost may be shown graphically by plotting CC and CT coordinates and connecting them to the NC and NT coordinates by a concave, convex, or straight line—or some other form, depending on the actual cost structure of activity performance, as in Exhibit 4.8. For activity A, we assume a linear relationship between time and cost. This assumption is common in practice and helps us derive the cost per day to expedite because this value may be found directly by taking the slope of the line using the formula Slope = (CC − NC) ÷ (NT − CT). (When the assumption of linearity cannot be made, the cost of expediting must be determined graphically for each day the activity may be shortened.)

The calculations needed to obtain the cost of expediting the remaining activities are shown in Exhibit 4.9A.

3. Compute the critical path. For the simple network we have been using, this schedule would take 10 days. The critical path is A–B–D.

exhibit 4.8 Example of Time–Cost Trade-Off Procedure

4. page 88Shorten the critical path at the least cost. The easiest way to proceed is to start with the normal schedule, find the critical path, and reduce the path time by one day using the lowest-cost activity. Then, recompute and find the new critical path and reduce it by one day also. Repeat this procedure until the time of completion is satisfactory, or until there can be no further reduction in the project completion time. Exhibit 4.9B shows the reduction of the network one day at a time.

Working through Exhibit 4.9B might initially seem difficult. In the first line, all activities are at their normal time, and costs are at their lowest value. The critical path is A–B–D, cost for completing the project is $26, and the project completion time is 10 days.

The goal in line two is to reduce the project completion time by one day. We know it is necessary to reduce the time for one or more of the activities on the critical path. In the second column, we note that activity A can be reduced one day (from two days to on), activity B can be reduced three days (from five to two days), and activity D can be reduced two days (from three days to one). The next column tracks the cost to reduce each of the activities by a single day. For example, for activity A, it normally costs $6 to complete in two days. It could be completed in one day at a cost of $10, a $4 increase. So we indicate the cost to expedite activity A by one day is $4. For activity B, it normally costs $9 to complete in five days. It could be completed in two days at a cost of $18. Our cost to reduce B by three days is $9, or $3 per day. For C, it normally costs $5 to complete in three days. It could be completed in one day at a cost of $9; a two-day reduction would cost $4 ($2 per day). The least expensive alternative for a one-day reduction in time is to expedite activity D at a cost of $2. Total cost for the network goes up to $28 and the project completion time is reduced to nine days.

exhibit 4.9 A. Calculation of Cost per Day to Expedite Each Activity

B.Reducing the Project Completion Time One Day at a Time

*To reduce the critical path by one day, reduce either A alone or B and C together at the same time (either B or C by itself just modifies the critical path without shortening it).

†B and C must be crashed together to reduce the path by one day.

+Crashing activity B does not reduce the length of the project, so this additional cost