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);

S_{A} = $100/day

S_{B }= $200/day

S_{C} = $600/day

S_{D }= $60/day

S_{E} = $120/day

S_{F }= $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.

## Solution

### Step 1

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

### Step 2

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.

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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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Hi!

Its urgent. How did you calculated the Normal duration of the project under normal conditions is 140 days? Can you show me how?

What do you mean by “sum of the durations of all the individual activities on critical path”. I have calculated in all way. Unable to figure out how this 140 comes.

Thank you

Dear Shenny,

Do you know how to calculate the critical path?

By definition critical path is the longest path along the duration of the project. For example in the given project above, activity C can not be started until activity B is completed in first 20 days, activity D starts only when activity C is completed in 40 days (i.e. total 60 days), activity E can be only started when activity D is completed in 30 days (total 90 days of the project) and the last activity E takes 50 days to complete. Thus the sequence of interdependent activities B,C,D and E can not be completed in less than 20+40+30+50=140 days. now if we complete activity A in 120 days, or activities B-F-E in 130 days, the project still requires 10 more days for the B-C-D-E sequence to complete. Thus B-C-D-E is the longest path in the project and hence it is critical path. And the normal duration of the project can in no case be less than 140 days. so normal duration is 140. I hope it will be clear now. In case you still have any confusions, you can contact back. Regards by Assad Iqbal, Department of Computer Science, Bahria University Islamabad