As we have understood the method of crashing in the video, let us move to the special case of Crashing. (If you have not watched the video yet, click on the "Watch Video" button above and clear your doubts about the method now.)
As title suggest, we will discuss the numerical which will have more than 1 critical path, i.e. Multiple Critical Paths.
Many may get confused in crashing of such problem which have more than 1 critical paths, as if you try to solve that you may find that:
 You are crashing a critical activity, but your project duration does not change.
 This is happening due to the other critical path (parallel critical path), as it by crashing you are trying to reduce the time, whereas parallel critical path is the longest one and it will not allow you to reduce the time.
So now, what's the solution to that?
Here is the very simple and logical solution to the same problem. We will understand now that, what we should do to tackle such problem!
Let us start by considering the following problem:
Numerical:
Table below shows the normal time & cost and crash time & cost of the various activities in a project. Find the optimum duration and minimum project cost, assuming the indirect cost of the project as Rs. 200/week.
Activity  Normal Time(weeks)  Normal Cost(Rs.)  Crash Time(weeks)  Crash Cost(Rs.) 

12  7  700  4  50 
13  5  500  3  100 
14  8  600  5  200 
25  9  800  7  225 
35  5  700  3  150 
36  6  1100  5  200 
46  7  1200  5  125 
57  2  400  1  100 
67  3  500  2  350 
Total:  6500 
Solution:
As per the look of the numerical, it looks simple to solve, and believe me, it is if you follow this simple method.
Starting by constructing the Network Diagram and finding project completion time using Critical Path Method.
If you don't know about Critical Path method (CPM), then check out  CPM Video.
Network Diagram:
Network Diagram with CPM
AS we can see from the netwrok diagram, we have more than one critical path generated, i.e. here we have two critical paths as marked in orange and blue colour, as follows:
Critical Paths:
1257 : Project completion 18 weeks
1467 : Project completion 18 weeks
As we can see above, both paths have same completion duration and the longest one as well, so both are critical paths. Such equal critical paths are called as Parallel Critical Paths.
Project completion time and cost:
$\begin{aligned} \text{Project completion time} &= 18 \ \text{days}\\ \ \\ \text{Project completion cost} &= \text{Normal cost} + \text{Indirect cost for 12 days} \\ &= 6500 + (18\ \sdot\ 200)\\ &= \text{Rs.} \ 10,100 \end{aligned}$Now,
Let us move to crashing of this numerical;
Crashing:
Evaluating cost slope for different activites:
In examination you can only copy acivites in the first column and add second column for the cost slope. For better understanding, we are providing whole table again with the addition of the last column of cost slope. Same type examination tactics are always illustrating in videos provided by Education Lessons.
Activity  Normal Time(weeks)  Normal Cost(Rs.)  Crash Time(weeks)  Crash Cost(Rs.)  Cost Slope 

12  7  700  4  50  50 
13  5  500  3  100  100 
14  8  600  5  200  200 
25  9  800  7  225  225 
35  5  700  3  150  150 
36  6  1100  5  200  200 
46  7  1200  5  125  125 
57  2  400  1  100  100 
67  3  500  2  350  350 
Total:  6500 
As you can see, cost slopes for different activities are evaluated in the table. Also, critical path activities are marked with their path colours (i.e. Orange and Blue).
In examination, for bifurcation of critical path activities use rectangular border around the activity for the first path(as mentioned in the video) and elliptical border for the other. You can use something else on your own, but use such shapes that can make different critical path activities differentiable.
We are avoiding rewriting of the table to save time in examination. Always try to complete such lengthy numerical in a single table.
Now, as per method we will be selecting the critical activity with the lowest cost slope. So, we have activity 12 with the lowest cost slope of 50 Rs.
So, crashing activity 12 directly by 4 weeks.
But wait, think once, if we try to do so, what will be result ?
We will find that, the project completion time will remain the same. This is happening because we have another parallel critical path.
We are trying to reduce total project completion time upto 3 weeks by crashing activity 12 directly by 4 weeks.
Here, I mentioned that we are going to reduce the completion time upto 3 weeks, because if we take the difference of normal & crash time of activity 12 is (74) = 3.
Instead of that, if try to crash this activity by stepbystep crashing, then also same problem persists.
So, I can conclude logically that:
 We are facing this problem while crashing only due to the other parallel critical path.
 So, if we carried out parallel crashing, then we can avoid this problem. This is because both paths will be crashed together, and both will reduce the duration.
Okay, so we have solution to problem now. And, also we are going to do the same now.
Parallel Crashing:
We have two following critical paths:
1257 : Project completion 18 weeks
1467 : Project completion 18 weeks
Already we have evaluated that, minimum cost for path 1257 is of 12 i.e, 50 Rs.
Similarly, we have to find out minimum cost slope for path 1467, as we are going to crash both paths together (parallel crashing).
We have activity 46 with minimum cost slope of Rs. 125 from 1467.
Now, let us mark this the table as follows:
Activity  Normal Time(weeks)  Normal Cost(Rs.)  Crash Time(weeks)  Crash Cost(Rs.)  Cost Slope 

12  7  700  4  50  50 # 
13  5  500  3  100  100 
14  8  600  5  200  200 
25  9  800  7  225  225 
35  5  700  3  150  150 
36  6  1100  5  200  200 
46  7  1200  5  125  125 # 
57  2  400  1  100  100 
67  3  500  2  350  350 
Total:  6500 
You have to use the last table we created, and mark in that only. I am copying again so that you can understand here only, without scrolling above.
As you can see, I have use "#" mark to mark this minimum cost slopes. Let us now crash this both paths simultaneously.
We have crash time of 4 weeks for activity 12 with normal time of 7 weeks.
And for activity 46 , crash time is 5 weeks and normal time is 7 weeks
So now, question is what crash time we are gonna use, 4 or 5 weeks ?
We will use 5 weeks here. But why ???
Let me explain this:
We understand this using stepbystep crashing. If we crash both paths by 1 week, then we will find that 1 week is reduced in project duration.
As we know, the number days/weeks we are gonna crash, same will be the reduction in the total project duration.
Again, we can crash both paths by 1 week. But now, question arises  Upto how many times we can crash an activity, using stepbystep crashing ? What is the limit ?
Answer to these questions is: As I mentioned in stepbystep crashing, you can crash an activity by steps upto provided crash time.
For example, here we have:
 For activity 46 , crash time is 5 weeks .
 In the first step, we will reduce its normal time by 1 week, i.e. (71) = 6 weeks.
 Again crashing it by 1 week, i.e. (61) = 5 weeks.
 And now, we can see that the activity 46 is reduce to 5 weeks and which equal to the crash time so provided.
 So, after this we cannot crash this activity as we don't know the cost applicable to reduction in 1 or more weeks than 5 weeks. So, we stop stepbystep crashing there.
As you can see, it is very lengthy to follow stepbystep crashing in examination. So, we should go for the direct method as illustrated in the Crashing video, but with a simple trick.
We will add a additional column to our existing table as follows:
Activity  Normal Time(weeks)  Normal Cost(Rs.)  Crash Time(weeks)  Crash Cost(Rs.)  Cost Slope  Crashing limit 

12  7  700  4  50  50 #  (74)=3 
13  5  500  3  100  100  
14  8  600  5  200  200  
25  9  800  7  225  225  
35  5  700  3  150  150  
36  6  1100  5  200  200  
46  7  1200  5  125  125 #  (75)=2 
57  2  400  1  100  100  
67  3  500  2  350  350  
Total:  6500 
$Crashing \ Limit = Normal \ Time  Crash \ Time$Crashing limit: As we have understand from the concept of stepbystep crashing, we can reduce the duration of a particular acitivity upto given crash time, using crashing. So, the difference of Normal time and Crash time is the crash limit for a particular acitvity. i.e;
As you can see, crashing limit is already added for the required acitivities. We get;
Crashing limit as:
3 for 12
2 for 46
So, we will go with the minimum limit, so that both the paths will have same reduction in the duration.
Note that: You cannot use maximum limit. As you can see above, if we are gonna use maximum limit, then we are not able to define the cost of reduction in 1 extra week of activity 46.
Iteration 1:
Crashing both activities 12 & 46 directly by 2 weeks.
As you can see in the following table, I have added vertical bars "" equal to crashing limit in the last column.
This is because, we have crashed activity based on minimum crashing limit (i.e 2), but we can still take the advantage of 1 week crashing of 12 with minimum cost of 50 in the next iteration.
Activity  Normal Time(wk)  Normal Cost(Rs.)  Crash Time(wk)  Crash Cost(Rs.)  Cost Slope  Crashing limit 

12  7  700  4  50  50 ##  3    
13  5  500  3  100  100  
14  8  600  5  200  200  
25  9  800  7  225  225  
35  5  700  3  150  150  
36  6  1100  5  200  200  
46  7  1200  5  125  125 #  2   
57  2  400  1  100  100  
67  3  500  2  350  350  
Total:  6500 
New network diagram formed as follows:
As we can see in the above diagram, both activites are crashed for 2 weeks. And, the project completion time is also reduced by 2 weeks, i.e. 16 weeks.
Critical Paths:
1257 : Project completion 16 weeks
1467 : Project completion 16 weeks
Also,
$\begin{aligned} \text{ New project completion time} =& \ 16 \ \text{weeks}\\ \ \\ \text{New project completion cost} = \ &\text{Cost obtained in last step}  \text{Indirect cost for 2 weeks}\\ &+ \text{Cost slope of acitivity 12 for 2 weeks}\ \\ &+\text{Cost slope of acitivity 46 for 2 weeks} \\ \ \\ &= 10,100  (2\ \sdot\ 200) + (2\ \sdot\ 50) + (2\ \sdot\ 125)\\ \ \\ &= \text{Rs.} \ 10,050 \end{aligned}$Here, we added cost of reduction by 2 weeks for both activities, because if we crash single activity we cannot reduce the total time of the project.
So, we have pay 100 Rs. to reduce time of acitvity 12 by 2 weeks, as well as we ahve to pay 250 Rs. to reduce time of activity 46 by 2 weeks.
Let us move to next step of crashing the same way.
Iteration 2:
Always take advantage of minimum cost slope, and crash the activity upto the limit to get minimum cost of the project.
As you can see, I have strike out the number of bars from crashing limit against 12 & 46. Strike number of weeks you have already crashed, i.e. 2 strikes for both activities.
So, I will now take advantage of 1 week of activity 12 .
Activity  Normal Time(wk)  Normal Cost(Rs.)  Crash Time(wk)  Crash Cost(Rs.)  Cost Slope  Crashing limit 

12  7  700  4  50  50 ## 
3

13  5  500  3  100  100  
14  8  600  5  200  200 ##  3    
25  9  800  7  225  225  
35  5  700  3  150  150  
36  6  1100  5  200  200  
46  7  1200  5  125  125 # 
2

57  2  400  1  100  100  
67  3  500  2  350  350  
Total:  6500 
Now, for this next iteration while we are selecting minimum cost for both paths, I will select as following:
Crashing activity 12 by 1 week for critical path 1257.
And,
Crashing activity 14 by 1 week for critical path 1467 .
Selecting crashing by 1 week, as it minimum crashing limit among 12 & 14.
Also, marking "##" for the second iteration of parallel crashing.
New network diagram formed as follows:
So, we get;
$\begin{aligned} \text{ New project completion time} =& \ 15 \ \text{days}\\ \ \\ \text{New project completion cost} = \ &\text{Cost obtained in last step}  \text{Indirect cost for 1 week}\\ &+ \text{Cost slope of acitivity 12 for 1 week}\ \\ &+\text{Cost slope of acitivity 14 for 1 week} \\ \ \\ &= 10,050  (1\ \sdot\ 200) + (1\ \sdot\ 50) + (1\ \sdot\ 200)\\ \ \\ &= \text{Rs.} \ 10,100 \end{aligned}$We can conclude that, by this is iteration2 the Project cost increases than the last iteration1.
So, we get;
$\begin{aligned} \text{Optimum project completion time} &= 16 \ weeks\\ \text{Minimum project cost} &= Rs. 10,050 \end{aligned}$Check another miscellaneous example about crashing here  Crashing Special Case  Indirect cost less than Crash Cost
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