Operation research cost crashing

1 Network Scheduling PERT Cost Analysis: It helps us to determine. The project duration for which the cost is least. To determine the least cost for a reduction in project duration to a specified or stipulated date. The time cost risk relationship of the network technique Direct Project Cost: It is the expenditures which are directly chargeable to each and every activity of the project. Indirect Project Cost: It is the expenditures which are not to be allocated to individual activities of the project. These are associated with overhead costs such as supervision etc. Crash Cost: It is the least cost associated with a particular activity duration beyond which the activity cannot be shortened further. Crash Time: It is the minimum time to complete the job. Cost Slope:It is the least cost of shortening the activity be one unit time Cost Slope = (Crash cost -Normal cost) /(Normal time -Crash time) The project time is controlled by the activities lying on the critical path. So to minimize the cost we crash the critical activities. Particularly, critical activity with minimum cost slope must be crashed first. At each and every time check whether the critical path is unique. If not, crash the activity one by one. Total project cost = Direct Normal cost + Indirect Cost If direct cost is not given, Total project cost = Total Normal cost. Additional cost incurred = No. of days crashed X its cost slope Example Find the critical path based on normal duration. Determine the least cost 28 days schedule. Normal Crash Activity Time (Days) Cost (Rs.) Time (Days) Cost (Rs.) 1-2 6 180 4 220 2-3 14 150 9 220 3-5 12 180 7 240 1-4 9 240 6 300 4-5 12 80 11 200 5-6 8 200 6 400 Solution Cost Slope = (Crash cost -Normal cost) /(Normal time -Crash time) Critical Path is 1 -2 -3 -5 -6 Length of the Critical Path = 6 + 14 + 12 + 8 = 40 days 2 Activity Cost Slope 1-2 20 2-3 14 3-5 12 1-4 20 4-5 120 5-6 100 The critical activity with minimum cost slope is 3 -5. Therefore, we first crash the activity 3 -5 whose normal time is 12 days to crash time 7 days. The new project schedule now becomes 35 days [40 -5]. Total project cost = Total Normal cost = Rs. 1030/= Additional cost = 5 x 12 = 60 The Expenditure for 35 days schedule is Rs. 1030 + Rs. 60 = Rs. 1090/= The critical path remains the same. The next critical activity with minimum cost slope is 2 -3. By crashing so we get the new project schedule as 35 -5 = 30 days. Additional Cost incurred = 5 x 14 = 70 Total Project cost = Rs. 1090 + Rs. 70 = Rs. 1160/= The Critical path remains same. The next critical activity to be crashed is 1 -2 with cost slope 20. By crashing so, we reduce the duration of the project by two days, but the critical path is changed because 1 -2 -3 -5 -6 = 28 days and 1 -4 -5 -6 = 29 days. Thus the new critical path is 1 -4 -5 -6. To avoid this let us crash 1 -2 by 1 day. The new project schedule is 30 -1 = 29 days. Additional cost = Rs. 1 x 20 = 20/= Total project cost = Rs. 1160 + Rs. 20 = Rs. 1180/-Now we have two parallel critical paths with duration 29 days. The following are the activities yet to be crashed. They are 1 -4, 4 -5 and 5 -6. Of the 3 activities 1 -4 is with minimum cost slope. Therefore, it has to be crashed by one day to get 28 days schedule. Now, new project cost = Rs. 1180 + Rs. 20 = Rs. 1200/= 28 days project schedule costs Rs. 1200/= completed3 J E Beasley Network analysis -cost/time tradeoff This use of cost information is the CPM technique. A common assumption is to say that for each activity the completion time can lie in a range with a linear relationship holding between cost and activity completion time within this range (as illustrated below). Reducing an activity completion time is known as "crashing" the activity and to illustrate this concept consider the activity on node network we had before, for which the network diagram is reproduced below. 4 Introducing costs into this problem we have the package input shown below. For activity 1, for example, the normal cost is 100 associated with a normal time of 6 weeks, and the crash cost is 240 associated with a crash time of 4 weeks. The output from the package for this example is shown below. 5 The above calculation is based upon the normal times for each activity. If we were to adopt the crash times for each and every activity we would have: indicating that the project can take anywhere between 24 and 16 weeks depending upon the activity completion time (which can vary between the normal time and the crash time). Crashing As seen above we can have any possible project completion time between 24 and 16 weeks -but which one shall we choose? To help us answer this question let us first consider 23 weeks, i.e. we wish to move from the normal time project duration of 24 weeks to a project duration of 23 weeks. Plainly to achieve a project completion time of 23 weeks this we need to crash (reduce) the completion time of one (or more) activities -but which ones? The solution associated with 24 weeks has one critical path, namely 1-3-5-7-8-9-11. Clearly unless we crash an activity on this critical path we can never reduce the completion time of the entire project. Of these activities only 1, 5, 8 and 9 are capable of being crashed: · if we were to crash activity 1 by one week we would incur an additional cost of 70 · if we were to crash activity 5 by one week we would incur an additional cost of 40 · if we were to crash activity 8 by one week we would incur an additional cost of 20 · if we were to crash activity 9 by one week we would incur an additional cost of 40 6 Clearly the best choice is to crash the activity with the lowest incremental (additional) cost -so in this case we would choose to crash activity 8 by one week at an additional cost of 20. This will reduce the length (duration) of the critical path 1-3-5-7-8-9-11 by one week, i.e. from 24 weeks to 23 weeks. The new situation is: Suppose now we wish to reduce the project by a further week, i.e. to 22 weeks. Plainly we will adopt the same approach as before: · examine all activities in the current critical path which are capable of being crashed · choose to crash the activity with the lowest incremental cost This procedure works perfectly until we reach a situation in which there are two or more critical paths. In such cases the activity with the lowest incremental cost may not be the best choice. For example if there are two critical paths A-E-F and B-E-F in a network and activity A has the lowest incremental cost then crashing activity A will have no effect on the critical 7 path B-E-F, and hence no effect on the overall project completion time. We would still need to crash one activity on B-E-F before we could reduce the completion time for the entire project. In fact in this situation we would need to consider three options: · crash both A and B by one time unit · crash E by one time unit · crash F by one time unit to determine the best approach. Clearly once there is more than one critical path the situation becomes more complicated. Indeed we stressed the word perfectly above. More technically, choosing to crash the critical activity with the lowest incremental cost is guaranteed to be an optimal approach (i.e. we crash in the best possible way) provided we have only a single critical path. However once we encounter two or more critical paths we cannot guarantee that we can still crash the project in an optimal way. Returning for the moment to our network above the solution associated with 23 weeks has one critical path, namely 1-3-5-7-8-9-11, as before. Of these activities only 1, 5, 8 and 9 are capable of being crashed · if we were to crash activity 1 by one week we would incur an additional cost of 70 · if we were to crash activity 5 by one week we would incur an additional cost of 40 · if we were to crash activity 8 by one week we would incur an additional cost of 20 · if we were to crash activity 9 by one week we would incur an additional cost of 40 Clearly the best choice is to crash the activity with the lowest incremental (additional) cost -so in this case we would choose to crash activity 8 by one week at an additional cost of 20. This will reduce the length (duration) of the critical path 1-3-5-7-8-9-11 by one week, i.e. from 23 weeks to 22 weeks. Hence we could continue as above, and eventually we would have crashed the network down to its lowest possible completion time of 16 weeks.