Project Management - Crashing - Example 2

the following table gives data on normal time and cost and crush time and cost for a project the indirect cost per day is rupees 10 so basically we have been given the various activities for the project and we have also been given the normal time for completion of each of the activities now the normal cost has also been given which means that for example the first activity 1 to 2 can be completed in six days at a cost of rupees 60 and we have also been given the minimum time required for completion of the activity after crashing the activity so for example the first activity 1 to 2 will take four days for completion after crashing now after crashing the cost has increased so earlier the normal cost was 60 rupees but after crushing the cost given to us for the first activity is 100 rupees so similarly we have been given the crash cost for each of the activities along with the crash time now we have to draw the network for the project find the critical path and then determine minimum total time and corresponding cost so let's first draw the network diagram so let us first draw the first node so this is 1 and then the second node let's say this is 2 so this is the activity between 1 & 2 now next is 1 2 3 so let's say this is three the next as 2 to 4 and from 2 we also have 2 to 5 so let's draw these two so let's say this is 4 and this is 5 so we have completed 1 2 2 1 2 3 2 2 4 and 2 2 5 now the next is 3 to 4 so 3 to 4 so to draw this 3 to 4 I think we should reverse the order so we should bring 5 here and 4 here so let's change that let's make this 5 and let's make this 4 and then now we can drop 3 to 4 so we have completed 3 to 4 now next is 4 to 6 and 5 to 6 so both 4 & 5 merge into 6 so we have completed four to six and five to six and X is six to seven so six and then the ending node which is seven so let's add the duration of the activities so one to two takes six days to complete one two three is 4 2 to 4 is 5 days 2 to 5 is 3 days 3 to 4 is 6 days 4 to 6 is 8 days 5 to 6 is 4 days and 6 to 7 is 3 days so we have completed the network diagram now next is to find the critical path so let's do that so let's list down all the possible paths so we have 1 2 5 6 7 then 1 2 4 6 & 7 then 1 3 4 6 7 so let's now find out the time taken for each of these paths so the first one 1 2 5 6 7 so 6 plus 3 9 plus 4 13 plus 3 16 so 16 days 1 2 4 6 7 so 1 2 4 6 7 so 6 plus 5 11 Plus 8 19 plus 3 22 and last is 1 3 4 6 7 so 1 3 4 6 7 so 4 plus 6 10 plus 8 18 plus 3 21 so this path here which is one two four six seven which is the maximum duration is the critical path so we have found the critical path and the duration of the project so the duration of the project is 22 days now next is to determine the minimum total time and corresponding cost so now we are being asked that if the normal time for completing the project is 22 days how quickly can we complete the project with all these crash times for these activities and what will be the corresponding project cost now to find the minimum total time we have to reduce the time taken by the activities on the critical path because the critical path governs the duration of the project so first we will have to identify which are the activities on the critical path and then we have to identify the activity which has the least cost of cashing and then we will crash the activity now one thing to note is that there is a very small difference between the duration of these two paths so this is turning to days and the other one is 21 days so as soon as we crash the current critical path by one day we get two paths as critical paths so let's see how we can crush this project so now one of the most important information that we need is the cost of crushing per day for each of the activities so that we can find out which activity has the least cost of crushing and then we can start crashing from that activity so let's first find out so this is cost per day so the cost after crashing the first activity is 100 rupees while the normal cost is 60 rupees so the difference is 100 minus 60 which is 40 and the difference in time is 2 days so the cost per day is 40 divided by 2 which is 20 for the second activity the cost difference is 200 minus 60 which is 140 and the time difference is 4 minus 2 which is 2 so 140 divided by 2 is 70 the third activity 150 minus 50 which is 105 minus 3 which is 2 so 100 divided by 2 is 50 for the fourth activity 65 minus 45 is 20 and 3 minus 1 is 2 so 20 divided by 2 is 10 for the activity 3 for 200 minus 90 is 110 and 6 minus 4 is 2 so 110 divided by 2 is 55 for activity 4 6 300 minus 80 which is 220 and 8 minus 4 is 4 so 2 20 divided by 4 which is again 55 so here 2 20 divided by 4 you can easily divide both the numerator and denominator by 2 so it becomes hundred and 10 divided by 2 which is the same as the previous one so 110 divided by 2 is 55 the next activity 5 6 so 100 minus 40 which is 60 divided by 4 minus 2 which is 2 so 60 divided by 2 is thirty-four next activity six seven eight t-minus 45 which is 35 divided by three minus two which is one so this again is 35 now before we start crashing the project let's first determine the total cost of the project with the normal time so with the normal time the duration of the project is twenty two days and the normal cost of performing each of the activities has been given as four seventy so the normal cost of the project is equal to four seventy now we have also been given that the indirect cost per day is rupees ten so what that means is that depending on the critical path or the critical duration for each day there is an additional cost of ten rupees now with the normal time the critical path is twenty two days so the indirect cost will be twenty two multiplied by ten so this will add to the normal cost of the project so for seventy plus twenty two multiplied by ten so this is equal to four seventy plus two 20 which is equal to zero 7 + 2 9 + 4 + 2 6 so 6 90 rupees is the cost of the project using the normal time so here in order to crash the critical path which is 1 2 4 6 7 let's first identify the activity which has the lowest cost of crushing per day so the activities on the critical path here is one two which is this - for which is this one four six which is this one and six seven which is this one so out of these activity one two has a cost per day of twenty rupees - four has 54 six has fifty five and six seven has thirty five so the lowest is one - which is twenty so first we will crash the activity one two by one day because the difference between the critical path and the next most critical path is only one day so the first step in the crushing is crash one two by one day so basically if we crash it by one day the crash cost will be equal to twenty rupees and the project duration now will reduce by one day so the project duration is now 21 days so this is now 21 days so here if we take a consideration of all the paths and if we have crashed activity one two basically this path here 1 2 5 6 7 will also reduce by 1 day so this will become 15 days so now we have two critical paths one is 1 2 4 6 7 which is 21 days and the other one is 1 3 4 6 7 which is again 21 days so let's identify the activities which we can crash now in this situation where we have two critical paths we have to ensure that we have first the activities in such a way that the duration for both the paths have reduced for example say we crash activity 1 2 now if we crash activity 1 2 the path 1 2 4 6 7 will have a reduced duration however the path 1 3 4 6 7 we'll still have 21 days so that means that the project duration has not come down which is defecting the entire purpose of crashing so here let's identify which are the combination of activities between these two critical paths which need to be crushed so if we crash 1 2 from this path we need to crash either 1 3 or 3 4 why is this so is because 1 3 & 3 4 are not common activities between these two paths if we take a combination of 1 2 & 4 6 then 4 6 is also present in the path 1 2 4 6 7 so so that will also not be an equal crashing of both the paths so the combinations are 1 2 1 3 or 1 2 3 4 the next one is 2 4 1 3 so 2 4 1 3 or 2 4 3 4 or now the remaining activities are common so for 6 or 6 7 so for 6 and six seven so let's find out the crash cost for these combinations so 1 2 is 20 and 1 3 is 70 so 20 plus 70 is 90 so this is 91 2 is again 20 and 3/4 is 55 so 20 plus 55 is 75 2 4 is 50 and 1 3 is 70 so 50 plus 70 is 122 4 is 50 and 3/4 is 55 so 50 plus 55 is 1 0 5 4 6 is 55 and 6 7 is 35 now out of these the lowest cost of crashing is for activity 6 7 which is 35 rupees per day so let's crash the activity 6 7 now 6 7 we can crush only by one day so the normal time is 3 and the crash time is 2 so we can bring this 3 to 2 so let me quickly denote what crashing we have already done so we have brought this down to 5 and now we will bring this down to 2 so here we are crashing activity 6:7 by one day so this was first and this is second so for this the crash cost is equal to so one day so 35 rupees and the project duration is now 20 days so this has now become too now we reduced the time of activity six seven so all these three will reduce by one day so this is now 14 this is 20 and this is also 20 okay so now the third crushing so again we have the same two paths as the critical paths so again we have the same combinations now we have already exhausted the crash time for activity 6 7 so this is gone now the next lowest crash cost is for activity 4 6 so this is the activity 4 6 and the number of days that we can crash this activity for is 4 days so 8 is the normal time and crash time is 4 days so now let's confirm if we can crash this activity by 4 days and still not have the critical path changed to a different path in between so these 2 paths have duration as 20 days and the other one has 14 days so even if we reduce these 2 paths by 4 days it will go down to 16 so still this other path which has duration of 14 days will not be a critical path so we can continue with the crushing of activity for 6 by 4 days so third crushing is crash activity for 6 by 4 days so the crash cost is equal to so for x 55 so 55 x 2 will be 110 and 110 x 2 will be 220 so the cost is 220 rupees and the project duration will be equal to 20 minus 4 which is 16 days so we have crashed this activity and brought it down to 4 days and our duration of the paths is 14 because 4 6 is not in this path now the path 1 2 4 6 7 gets reduced to 16 days and so does the path 1 3 4 6 7 so this is now reduced to 16 days as well so this is no longer available now since we have the same two paths as the critical paths let's find out the lowest cost of crushing so this 75 rupees CV should be the lowest cost now this means that we have to crush both 1 2 & 3 4 now 1 2 has already been crushed by one day and now the crush time available is 1 day and 4 3 4 we have 2 days available now we have to crush the combination of the two so the minimum common time between the two activities is one which can be used for crushing so let's crush activity 1 2 and three four by one day so forth is crash one two and three four by one day so this will become 4 and this will become 5 and the cost is 20 plus 55 so 75 so crash cost is equal to 75 rupees and the new project duration is equal to 15 days so now 1 2 is available in the first path so this goes down by 1 day 2 13 days 3 4 is not available on the first path so that remains unaffected for the second path 1 2 4 6 7 1 2 is available so this becomes 15 and for a third path 1 3 4 6 7 3 4 is available in this path so this becomes again 15 days so now again let's find the critical path so again both these paths are critical so now let's take the same combinations so this one is now not available now let's find the activity which has the next lowest cost of crashing so that is 1 2 and 1/3 which is 90 rupees but now 1 2 has already been completely crashed so we can't use this so this is not available for crushing now the next lowest is one zero five rupees which is 2 4 3 4 now to 4 can be crashed by 2 days & 3 4 can be crashed by 1 day so let's crash 2 4 & 3 4 by 1 day so let me put it here so this is the 5th crashing so crash 2 4 & 3 4 by 1 day so the crash cost is equal to 105 rupees and the project duration now becomes 14 days so 2 4 so this has been crashed by 1 day and 3/4 this has also been crashed by another one day now this path 1 2 5 6 7 does not have either 2 4 or 3 4 so the duration remains the same 1 2 4 6 7 has 2 4 there which has been crashed by 1 day so the duration becomes 14 days the next 1 1 3 4 6 7 has 3 4 & 3 4 has been crashed by 1 day so the duration again becomes 14 days so now again for crushing this project let's find out the critical path so again the same paths 1 2 4 6 7 & 1 3 4 6 7 are critical so now let's find out which other combination of activities can be crashed so this one is now no longer available so we have already used activity 3 4 2 its entirety so we can't further crash activity 3 4 so this one is also out of question now now the next activity which is remaining is 2 4 1 3 so 2 4 can be crashed by 1 day and 1 3 and 1/3 can be crashed by 2 days so now let's crush activities 2 4 & 1 3 by 1 day because to 4 can only be crashed by 1 day so 6th is crash 2 4 & 1 3 by 1 day so crash cost is equal to 120 rupees and project duration will be equal to 13 days so here the second path has the activity 2 4 so this goes down to 13 days and the third path this 1 3 4 6 7 has the activity 1 3 so this goes down by 1 day as well so now this has been reduced to 3 and 2 4 has also been reduced to 3 days so now for the next crushing basically we now have 3 parts or all the parts which are on the critical path so if we have to crash we have to crash all the 3 parts to reduce the type now let's evaluate the first path so first path is 1 2 5 6 7 so activity 1 2 has already been completely crashed so we cannot crash this activity actually 2 5 still has 2 days that can be crashed so there is scope there 5 6 still has 2 days and 6 7 has already been crashed so there is definitely some scope for crushing the path 1 2 5 6 7 now let's look at the next path 1 2 4 6 7 so 1 2 has already been completely crashed - 4 has been completely crashed as well 4 6 has also been completely crashed and 6 7 has also been completely crashed so basically this path we cannot crash any further so the duration of this path will always remain 13 days so what that means that even though we reduce the duration of the other paths we will not be able to reduce the duration of this path and so this is the maximum we can crush the overall project to benefit in terms of the reduced timelines of the project so now there are two questions which we were asked I'll just make some area here so the two questions were determine the minimum total time and the corresponding cost so the minimum total time is 30 days now the cost of the project will be equal to the normal cost which is the addition of this cost plus the crash crushing cost which is the cost that we found here so 20 + 35 + 2 xx + 75 plus 105 + 120 + the indirect cost so we were given that the indirect cost is 10 rupees per day for the project so let's find this out so normal cost is 470 which was given to us earlier plus the crushing cost so 20 plus 35 is 55 plus 220 is 275 + 75 is 350 plus 105 is 455 plus 120 is 575 so 470 plus 575 plus the indirect cost is 10 per day and the reduced duration of the project is 13 days so 10 multiplied by 13 so this is 470 + 575 + 10 x 13 so let me pull my calculator here so 10 x 13 is 130 plus 5 75 plus 470 so this is equal to 1 1 7 5 so the total cost of completion is 1 1 7 5 rupees and the minimum total time is 13 days

Channel: maxus knowledge

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Keywords: Schedule compression technique, compressing a project, time schedule trade off, crashing of a project, crashing, Project crashing example

Id: 4Ea30vkEBFk

Length: 34min 15sec (2055 seconds)

Published: Fri Jul 05 2013