Crashing Example: Crash the following schedule to complete the project in 110 days. The network diagram below gives the normal durations for each activity to be completed under normal conditions without crashing.
The table given below shows all the cost-time information for the project i.e. Crash Cost, Normal Cost, Crash Time and Normal Time.
Calculating Slopes of individual activities
Slopes which show the crash cost per unit duration (days, weeks etc) for individual activities are calculated as:
Slope = (CC-NC)/(NT-CT)
Hence for individual activities this crash cost per unit duration comes out to be as below (Refer the table above for details);
SA = $100/day
SB = $200/day
SC = $600/day
SD = $60/day
SE = $120/day
SF = $300/day
Normal Cost of the Project
Normal cost of the project is the sum of normal costs of all the individual activities. In the given example the normal cost comes out to be $48,300.
Normal Duration of the Project
Normal duration of the project is the sum of the duration of all the individual activities on critical path. The normal duration of the given project under normal conditions is 140 days.
The only critical activity with the least crash cost per day is D. So we will crash it first. Before crashing make sure that:
Firstly the activity should not be crashed more than the allowed crash time limit. Secondly the activity should be crashed by duration such that it does not make the over all project duration lesser than any other path. It might create other critical paths but the activity should itself always remain on the critical path.
Crashing D by 10 days results as shown below;
Overall project duration is now reduced to 130 days and there are two critical paths now (BFE & BCDE).
Total Project Cost is now Normal Cost $48,300 plus crash cost of D for 10 days (60 * 10 = $600) thus making a total of $48,900
The next activity to be crashed would be the activity E, since it has the least crash-cost per day (slope) i.e. $120 of any of the activities on the two critical path. Activity E can be crashed by a total of 10 days. Crashing the activity E by 10 days will cost an additional (120×10) $1200.
The total cost is now $(48,900+1200) = $50,100
Total duration is 120 days
There three critical paths in total i.e. (A, BCDE, BFE)
Step 3 -This step involves crashing on multiple critical paths
This step involves a more thorough analysis of the available crashing options and selecting the most feasible one. To achieve an overall reduction in the project duration, multiple activities must be crashed. The following options are available:
Option 1: Crash A & B each by 5 days having total crash cost of $300/day
Option 2: Crash A, C & F each by 10 days having crash cost of $1000/day
The feasible one is obviously option 1 hence A&B are crashed by 5 days each costing ( 5×300) = $1500
Total project cost is now = $50,100 + $1500 = $51,600
Total project duration = 115 days
Critical paths are still the same three.
Final Step in crashing
The final step in this example is to crash the schedule by 5 more days. For this step the available options are very limited. As we go futher with crashing the crash cost per day increases. The only available crashing options are A, C and F all by 5 days because all other activities have met their maximum crashing limits and they can not be crashed any more.
The total crashing cost for 5 days of A, C and F is calculated to be 5 x 1000 = $5,000
The total cost of the project to completed in 110 days comes out to be = $56,600
And the final network diagram appears to be as follows:
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