I am a student of CMA sir, I was on that problem, and then found you, and you clarified it even better than our book of Strategic Cost Management, Warm Appreciations, Thank you sir
Welcome! Keep watching keep learning... Visit and subscribe my channel and don't forget to recommend my channel and lectures/playlists to others also... you will find so many subjects/topics... Thanks:)
Welcome! Keep watching keep learning... Visit and subscribe my channel and don't forget to recommend my channel and lectures/playlists to others also... you will find so many subjects/topics... Thanks:)
Welcome! Keep watching keep learning... Visit and subscribe my channel and don't forget to recommend my channel and lectures/playlists to others also... you will find so many subjects/topics... Thanks... :)
Prashant Puaar ok sir..thanks..because my assigment of transportation in stepping stone method,seem i got no.occupied cell > m+n -1 ..thats why im asking..is that possible this case can occur?
Welcome... Keep watching, keep learning... Please don't forget to like and share... And also visit, subscribe and recommend my channel ruclips.net/channel/UCqedkv6AYxh3hk99EPtGlmQ to the others...
In an initial basic feasible solution the number of allocations would either be equal to (m+n-1) or less. This is also known as "rim requirement". In Linear Programming, there should be (m+n)-1 number of constraints. (m = sources and n= destinations, in case of a transportation problem) 1) The number of allocations are equal to (m+n-1). It is called non-degenerate solution and we can directly check it for optimality. 2) The number of allocations are less than (m+n-1). It is called degeneracy and we have to removing the degeneracy to reach the optimal solution because without making it non-degenerate, by using epsilon(s), we cannot check it for optimality. 3) It would be interesting to note that an optimal solution can have less than (m+n)-1 allocations (i.e. it can be a degenerate solution) because to have all opportunity costs/improvement indices (dij values in case of MODI method) non-negative (i.e. zero or greater) is the only condition to be satisfied for a solution to be optimal. So, a degenerate initial solution can be optimal, but we cannot prove it to be optimal unless we make it non-degenerate by using epsilon(s) and check it for optimality. I have obtained many such kind of solutions, particularly while dealing with unbalanced transportation problems...:)
what if the m+n-1 equals to 8 and there are 6 allocated cells. Are we gonna choose 2 independent cells ??? Also if there is 3 independent cells and all of them have same cost lets say 0, does it matter to choose one of them??
1) The answer must be yes we have to select 'one or more' deemed to be occupied cells because to check the optimality we must have (m+n-1) occupied cells, else we cannot calculate 'ui' and 'vj' values. 2) When we have two or more independent cells with same cost, it is advisable to select any of such two or more cells because it becomes very much time consuming to think further to select one better cell from such two or more. In examination we need to answer the whole paper in a stipulated time and it also creates confusion if we go for more and more logical thinking just to select a cell under such an exceptional situation... :)
It is advisable to select any of such two cells because it becomes very much time consuming to think further to select one better cell from such two. In examination we need to answer the whole paper in a stipulated time and it also creates confusion if we go for more and more logical thinking just to select a cell under such an exceptional situation... :)
hi sir , if two independent cells have a tie as both of them having the least cost out of the unoccupied cells ,which one would we choose and why ? please reply ASAP , have an exam day after.
We should select any one of such independent cell involved in such tie without wasting our time on any kind of so called logical thinking because in any examination we need to answer all the question within stipulated period of time only... You, and all other students/viewers, can find my answers to the queries which are generally common for all students, in the comment segment as my replies to the earlier queries. It'll save valuable time...
hmm sir in some questions there are more than 1 epilson are required and all the unoccupied cells are idependent cells so and as u said that least value cell are seleted but there is a situation that it is tie i m very confused ...
If the problem requires "two or more" epsilons (ε), then a cell in which an epsilon (ε) has already been placed should be treated as occupied while determining independence of cells for inserting an epsilon (ε) subsequently.
bt i want to know why u1 value will not become zero rather it changes to other numberd lyk -1,2,3 .. in optimal distribution to minimize total transportation cost using VAM method to find IBFS
In an initial basic feasible solution the number of allocations would either be equal to (m+n-1) or less. This is also known as "rim requirement". In Linear Programming, there should be (m+n)-1 number of constraints. (m = sources and n= destinations, in case of a transportation problem) 1) The number of allocations are equal to (m+n-1). It is called non-degenerate solution and we can directly check it for optimality. 2) The number of allocations are less than (m+n-1). It is called degeneracy and we have to removing the degeneracy to reach the optimal solution because without making it non-degenerate, by using epsilon(s), we cannot check it for optimality. 3) It would be interesting to note that an optimal solution can have less than (m+n)-1 allocations (i.e. it can be a degenerate solution) because to have all opportunity costs/improvement indices (dij values in case of MODI method) non-negative (i.e. zero or greater) is the only condition to be satisfied for a solution to be optimal. So, a degenerate initial solution can be optimal, but we cannot prove it to be optimal unless we make it non-degenerate by using epsilon(s) and check it for optimality. I have obtained many such kind of solutions, particularly while dealing with unbalanced transportation problems...:)
Sir u have cleared my doubts...thank u...but i wanna suggest that we can select the least independent cell from starting...i mean we don't need to find every independent cell n thn the least one...
To know and to be able to do that, first we need to learn the things right from the beginning... as a responsible teacher I ought to consider all types of students and also the fact that the proportion of beginners trying to learn from this platform is very large... So, i always try to explain the things from the very beginning and leave it to the learners to cut the process according to their understanding and ability... :)
Though your lecture is long but it covers all concept. I can find many lectures of this chapter in RUclips with short duration but each n every concept is not clear but in ur case u have cleared every concept. Now I am confident that I can solve all questions of transportation. Thank You so much sir. God bless u.
Your are one of respectable teachers for me ...
I am a student of CMA sir, I was on that problem, and then found you, and you clarified it even better than our book of Strategic Cost Management, Warm Appreciations, Thank you sir
Welcome! Keep watching keep learning... Visit and subscribe my channel and don't forget to recommend my channel and lectures/playlists to others also... you will find so many subjects/topics... Thanks:)
Woooooow, I am really amazed by your concept clarity
16:51 Sir if there is a tie situation in the least cost of unoccupied independent cells then what we will do?
this sir is xplaining so good
Dats really an amazing lecture sir
Thank you for uploadng
Teachers like u r really needed for students like us☺
A genuine lecture Thanks a lot Sir, U have cleared my doubts.
Chamatkar kr diya sir ji
I have an exam about this next week. thanks a lot for your videos about transportation forecasting methods.
How to allocate the value epsilon if there are more than one deemed occupied cells? Thank you.
great sir ...Thank you for this video
Hats off... The way u explained is literally the best 👏👏👏👏👏🙌
thank you for the help sir... it was very helpful... cleared all my doubts
Welcome! Glad to hear that... Keep watching, keep learning... Please don't forget liking, sharing, subscribing and recommending...
Thanks sir.. this was really usable for me
Welcome! Keep watching keep learning... Visit and subscribe my channel and don't forget to recommend my channel and lectures/playlists to others also... you will find so many subjects/topics... Thanks:)
Well explained sir. Thank you!
Welcome! Keep watching keep learning... Visit and subscribe my channel and don't forget to recommend my channel and lectures/playlists to others also... you will find so many subjects/topics... Thanks... :)
Really amazing
I saw many videos but didn't find anywhere
How to find independent cell
I am ca student sir
But i like ur way of reaching
Teaching*
Thank you so much.
Sir in unbalanced problem can € be assign to dummy column as in case of dummy column least value is 0??
Thank you so much sir.. well explained...
Well explained sir :) Can you plz make a tutorial on Russell's Approximation Method. Thank you :)
Instead checking independency of all cells, we can start from least value and check if that is independent to save time.
Thank you so much sir , saved my life , love from pakistan
video in Last lines isn't complete.
Wonderful explanation! Thanks bro :)
Sir can u expln the sensitivity topic of linear programming. .plzz
Good sir.. doing a good job.
Sir you are great
dear sir. what if no.occupied cell > m-n +1 ? is that we can just go to the next improvement to find optimal solution? from malaysia
Yes, but it is a rare situation that one have to face. In most of the large cases it's very difficult to satisfy (m+n-1) condition...
Prashant Puaar ok sir..thanks..because my assigment of transportation in stepping stone method,seem i got no.occupied cell > m+n -1 ..thats why im asking..is that possible this case can occur?
It's very useful
Where is stepping stone method?
Thankyou for solving my dought
Welcome...
Keep watching, keep learning...
Please don't forget to like and share...
And also visit, subscribe and recommend my channel ruclips.net/channel/UCqedkv6AYxh3hk99EPtGlmQ to the others...
Sir degeneracy hmesha initial. Bfs mein ni hoti hmlog modi method apply krte usmein bhi kayi baar degenracy aati toh inn question ko kaise krenge
If we start finding independent cell from the lowest cost then we don't need to check all the cell...can we do that sir?
Thank u sir ... From Nepal ..🤗🤗🤗
But here, we are told not to assume any cells from dummy column/row as occupied. Is it?
what if number of allocation exceeds (m+n)-1 in a transportation problem?
In an initial basic feasible solution the number of allocations would either be equal to (m+n-1) or less. This is also known as "rim requirement". In Linear Programming, there should be (m+n)-1 number of constraints. (m = sources and n= destinations, in case of a transportation problem)
1) The number of allocations are equal to (m+n-1). It is called non-degenerate solution and we can directly check it for optimality.
2) The number of allocations are less than (m+n-1). It is called degeneracy and we have to removing the degeneracy to reach the optimal solution because without making it non-degenerate, by using epsilon(s), we cannot check it for optimality.
3) It would be interesting to note that an optimal solution can have less than (m+n)-1 allocations (i.e. it can be a degenerate solution) because to have all opportunity costs/improvement indices (dij values in case of MODI method) non-negative (i.e. zero or greater) is the only condition to be satisfied for a solution to be optimal. So, a degenerate initial solution can be optimal, but we cannot prove it to be optimal unless we make it non-degenerate by using epsilon(s) and check it for optimality. I have obtained many such kind of solutions, particularly while dealing with unbalanced transportation problems...:)
thank u sir. it was really helpful .
what if the m+n-1 equals to 8 and there are 6 allocated cells. Are we gonna choose 2 independent cells ??? Also if there is 3 independent cells and all of them have same cost lets say 0, does it matter to choose one of them??
1) The answer must be yes we have to select 'one or more' deemed to be occupied cells because to check the optimality we must have (m+n-1) occupied cells, else we cannot calculate 'ui' and 'vj' values.
2) When we have two or more independent cells with same cost, it is advisable to select any of such two or more cells because it becomes very much time consuming to think further to select one better cell from such two or more. In examination we need to answer the whole paper in a stipulated time and it also creates confusion if we go for more and more logical thinking just to select a cell under such an exceptional situation... :)
Sir what if in the question there are two independent cells of the same value of least cost? Which shall we choose?
It is advisable to select any of such two cells because it becomes very much time consuming to think further to select one better cell from such two. In examination we need to answer the whole paper in a stipulated time and it also creates confusion if we go for more and more logical thinking just to select a cell under such an exceptional situation... :)
hi sir , if two independent cells have a tie as both of them having the least cost out of the unoccupied cells ,which one would we choose and why ?
please reply ASAP , have an exam day after.
We should select any one of such independent cell involved in such tie without wasting our time on any kind of so called logical thinking because in any examination we need to answer all the question within stipulated period of time only... You, and all other students/viewers, can find my answers to the queries which are generally common for all students, in the comment segment as my replies to the earlier queries. It'll save valuable time...
U r great sir.... Love you
👍👍👍👍
hmm
sir in some questions there are more than 1 epilson are required
and all the unoccupied cells are idependent cells so
and as u said that least value cell are seleted but there is a situation that it is tie
i m very confused ...
plz reply fast
my exam ahead ...
If the problem requires "two or more" epsilons (ε), then a cell in which an epsilon (ε) has already been placed should be treated as occupied while determining independence of cells for inserting an epsilon (ε) subsequently.
thank you sir ..
God bless you .
Thnx Sir
Thank you very much sir
Sound does not clear
Thanks...
Thank you sir
bt i want to know why u1 value will not become zero rather it changes to other numberd lyk -1,2,3 ..
in optimal distribution to minimize total transportation cost using VAM method to find IBFS
Plss give the solution for when x>(m+n)-1
In an initial basic feasible solution the number of allocations would either be equal to (m+n-1) or less. This is also known as "rim requirement". In Linear Programming, there should be (m+n)-1 number of constraints. (m = sources and n= destinations, in case of a transportation problem)
1) The number of allocations are equal to (m+n-1). It is called non-degenerate solution and we can directly check it for optimality.
2) The number of allocations are less than (m+n-1). It is called degeneracy and we have to removing the degeneracy to reach the optimal solution because without making it non-degenerate, by using epsilon(s), we cannot check it for optimality.
3) It would be interesting to note that an optimal solution can have less than (m+n)-1 allocations (i.e. it can be a degenerate solution) because to have all opportunity costs/improvement indices (dij values in case of MODI method) non-negative (i.e. zero or greater) is the only condition to be satisfied for a solution to be optimal. So, a degenerate initial solution can be optimal, but we cannot prove it to be optimal unless we make it non-degenerate by using epsilon(s) and check it for optimality. I have obtained many such kind of solutions, particularly while dealing with unbalanced transportation problems...:)
Sir u have cleared my doubts...thank u...but i wanna suggest that we can select the least independent cell from starting...i mean we don't need to find every independent cell n thn the least one...
To know and to be able to do that, first we need to learn the things right from the beginning... as a responsible teacher I ought to consider all types of students and also the fact that the proportion of beginners trying to learn from this platform is very large... So, i always try to explain the things from the very beginning and leave it to the learners to cut the process according to their understanding and ability... :)
thank you
Thanks sir
Sir, can you plz give us a lecture on queue theory. thank you.
There are already... Please visit and subscribe my channel; there are playlists on many chapters/topics... :)
Sir xplaination was xcellent..Bs mudde ki baat phle kijiye🙃🙃
Sir aap hindi mein samjhao toh jada aacha samajh aayega
dont take so much tym 2 explain.
.dont keep repeating the same lines again and again
Pavi dharshini
As a teacher I ought to consider the last benchers also... 😊
even a person who has a donkey in his brain can understand your lessons, you are the best teacher
Though your lecture is long but it covers all concept. I can find many lectures of this chapter in RUclips with short duration but each n every concept is not clear but in ur case u have cleared every concept. Now I am confident that I can solve all questions of transportation. Thank You so much sir. God bless u.
Thank you very much sir
thank you sir