Modified Distribution Method

Added: 2026-05-06

[00:00:00]ify distribution method. So when giving a transportation problem, you first use the northwest corner method or the VA method or that's the approximation method to draw your initial solution.

[00:00:15]After that then use the modified distribution or the stepping stone to calculate or to get your uh optimal tableau. So in this method what we are going to do is that we are going to assign UI values for each row and then I'm going to assign UI values for each row and then VJ values for columns of 0 to new one that's the initial value or the first row is the first row. So that's going to be zero always remember this.

[00:00:58]So always u1 is equal to z. Then we compute the net cost change for each non-basic variable. So after we use the allocated cell for the values of u the ui values and the values and then we use the same values using the expression the net cost is equal to the c i value minus ui minus. So this will give us the net cost change value.

[00:01:29]After that we compare the values. If you have a negative value it means that the solution is not optimal. So we take the most negative value and then according to the the stepping stone process using the step one we find a close loop and then do the allocations again and then we draw another table and then calculate for the new UI and values. Then we continue the process until our all our net costing values are positive. Until all of them are positive, our solution is not optimal.

[00:02:03]So let's get started.

[00:02:06]All right. So we are going to assign our UI and VJ to the rows and columns respectively. So here is going to be U1 and here is going to be U2 U3 and here is going to be V1 V2 V3 but remember always U1 is equal to zero. So for so we going to use the relationship this equation UI plus VJ is equal to CIG.

[00:02:41]All right. So for cell X11 we are going to get what's the UI value here? Zero. What's the VI value?

[00:02:52]VI here we don't know. So is equal to the C value here is eight.

[00:02:58]So from here can say V is equal to 8. So we do same for this one we doing for only the allocated cells. So you move to the next one that's X12 sorry X21 X21 and then that will be U2 V1. So U2 we don't know the value. So U2 plus we know our value for V1 which is what we had here + 8. So + 8 is equal to this value equal to 15. So when 8 cross to the other side we going to have u2 to be equal to 15 - 8 and that's going to give us 7.

[00:03:49]So for 31 sorry this is the 31 is not an allocated cell. You don't have it's not an allocated cell. So the next allocated cell is here and that's 22 2. So for X U2 we are going to have what the value for U2 is 7 plus you don't know the value for V2. So V2 is equal to the value 10. So therefore V2 is equal to 10 - 7 and that's going to give us 3.

[00:04:28]So move to the next one.

[00:04:32]move to the this allocated cell. So x 23 sorry x32 this gives us what the value for u3 we don't know we've not computed it yet3 plus v2 and it is so and then v2 is so u3 plus the v2 is is equal to this value 9. So therefore we have our U3 to be equal to 6 9 - 3 that will be six. We move to the next one that's the next allocated cell that's X33. This the next allocated cell. We already know our U3 to be 6. So 6 + V3 is equal to this value 10. So therefore V3 is equal to N - 6 that's four.

[00:05:41]All right. So we know our VI and VJ values we know of. So what we are going to do is to check for optimality that we are going to calculate for the value of each non-basic variable. The net cost change for each basic variable as I said earlier the net cost change is equal to C I J minus UI minus VG. So let's do that here.

[00:06:15]All right. So for let's look at the unallocated cell. So let's start from here. That's going to be let's start from here.

[00:06:25]That's going to be C12. So C12 is equal what's the C value? That's 5 minus the U value that's 0 minus V3.

[00:06:35]So that's going to give us two. Then we move to the next one. C13 C13 this one because the C value the C value is three.

[00:06:48]So 3 minus this the U value here is six.

[00:07:02]6 this supposed to be 31 6 - this one 8.

[00:07:09]So 3 - 6 - 8 that's going to be - 11. So we are done with this one. Done with this one. For me to the next one that's the C3.

[00:07:21]So the value here is 6. We have 6 - here is 0 6 - 0us this one the v value we add v3 to be 4.

[00:07:34]So 6 - 0 - 4 and that's going to give us positive2.

[00:07:41]So our next this one we calculate for 23. this one that's going to give us 12 because you have this one 12 - 7 - 4. You know how the u value here is seven. So seven and this four. So 12 - 7 - 4 that's 1. All right. So this our net costing values and then we have a negative number here which means that our table is not optimal. So - 11 and that's in which cell that's the 31. This is C 31. So where is C31?

[00:08:28]We find our C31.

[00:08:39]Let me draw a close loop.

[00:08:42]I want to let this mark.

[00:08:57]All right. So this is 63 sorry C31. So we start here. We form a close loop. Start positive here. Which direction are we going? We go this way.

[00:09:11]minus plus minus. So we take the negatives this one and this one. Which one is the minimum? 20 is the minimum.

[00:09:18]So we add 20 here.

[00:09:21]We subtract 20 here. So we have 10 left.

[00:09:26]And then we add 20 here making it 70. So here we go to zero. So let's draw our table again.

[00:09:36]Let me draw it here.

[00:09:38]I can see all.

[00:09:59]All right. So yeah, write down my values. Here is eight and then 120. This one will now go to 10 and then 15. Here is now 20 and this one now go to zero.

[00:10:24]And here we go to 70.

[00:10:28]105.

[00:10:39]This is demand 70 60 280 80 here 80.

[00:10:52]So the rest remains the same. remains on teams.

[00:10:58]Yes. So this is our next table.

[00:11:02]This our next table. This our new table.

[00:11:04]So we have to also repeat the same process. Check if this our new table is also optimal. You have to repeat the process and check if it is optimal. So what do we do? We repeat the same process. We calculate for we assign the U values and the V values then we calculate for them then we check for optimality calculate the net cost all right so let's do that let's do that so let's look at the allocated cells that's what we are going to use to calculate for the uh U and V value. So from cell one one you're going to have this eight this eight or we going to have U1 plus V1 to V8. So u1 + v1 = 8. Normally this the expression ui plus vj = c minus i.

[00:12:10]Yes. So this this eight. But we know that our ui value is always zero. So therefore v1 is equal to 8. So we move to the next allocated cell that's c21. C21. So C21 that's U2 plus V1 that's equal to sorry that now that will be equal to 15.

[00:12:39]So therefore we know that V1 is 8. So let's write it.

[00:12:52]So 8. So therefore U2 is equal 50 - 8 and that that will give us 7. So we move to the next allocated cell that's C22 C22 this one. So we are going to have U2 we already know the value of U2 to be seven. So 7 + what's the value of 2? We don't know is equal to 10. So therefore our 2 will be equal to 3. 10 - 7 that's three. So move to the next allocated cell. That's C31. This cell we already know our we know our U3. No, we've noted that. But we know V1. All right. So U3 plus what is the value for V1? 8 is equal to 3. So therefore our U3 is equal to 3 - 8 that will give us -5. Move to our next allocated cell that's uh C33.

[00:14:03]All right. So we know value of U3 to minus 5 but we don't have value for V3.

[00:14:07]So - 5 + V3 is = 10.

[00:14:15]So therefore v3 is equal to 10 - - 5. This minus - 5 is crossing to the other side. So 10 + 5 that will give us 15. So now we know all our v and u values. We know all our v and u values. So what do we do? We go ahead and then calculate the net cost and then see if our table is optimal.

[00:14:45]If our table is optimal and that we use the expression C that's the net costing is equal to C minus UI minus VG. So this is what we are going to uh this formula is what we going to use to calculate for the net cost change for the optimality.

[00:15:04]So let's start it. Now we are going to do that for unallocated cells. Use the allocated cells to calculate for your V and U values and then use the you use the allocated cells to check for optimality. So let's start. Let's start with this cell that's C12.

[00:15:23]So C12 is equal what's the C value there? Five.

[00:15:29]So 5 minus the U value 0 minus the J value uh sorry V value that's 3 so - 3 is equal to 5 - 3 is 2 then the next unallocated cell is here that's U1 3 C13 sorry C13 that's equal to the C value here is 6 so 6 minus what the U value 0 and then the V value is 15 6 - 15 that's going to give us - 9. The next unallocated cell here is this.

[00:16:06]That's C 2 3 C 23 and the C value there is 12. So 12 what's the U value? 7. So - 7 minus the V value that's 15.

[00:16:26]12 - 7 - 15 that's going to give us - 10. And then our next unallocated cell is this cell that's C 32. So C 32 the next question will be the C value minus U b value here is 9 9us the U value that's - 5 minus the V value that's 3.

[00:16:55]So 9 - - 5 - 3. So the - 5. So 9 + 5 + 14. 14 14 - 3 that's 11. So this our net change.

[00:17:10]Since we have a negative value, we have negative values. This is negative values. It means that our solution is not optimal. So we we take the most negative value and the most negative value is minus 10. So our most negative value is - 10 and that's cell C 2 3 and this is 2 3 this is 2 three. So you start with plus minus and you move to this side plus and then minus. So we look at where we have so this a complete closed loop you can skip. So from here from here it goes this way it goes this way. So I don't use a different color so that can see the directions.

[00:18:02]So it goes this way then to this way.

[00:18:07]So that's a closed loop. So what do we do? We look at the negative the cells with the negative in the closed loop. So this is having negative polarity. This one is having negative. Then we take the minimum number. So the minimum number is 10. So we add 10 here because we have plus here. We subtract 10 from here. So we going to get 50. We add 10 to this one.

[00:18:28]We are going to get 30. And here is -2.

[00:18:31]10 - 10. We going to get zero. Yes. So this what we are going to use to draw our new table. New table.

[00:18:45]All right.

[00:18:51]So two from so 1 2 3 1 2 3 and we have 120 here. This one is now zero. So unallocated.

[00:19:07]Then that one is is now 30 here. This one is then Now it's still 70. So now this is not having 10.

[00:19:31]We are going to have 10 here. Now here is 50.

[00:19:39]So I think that is it.

[00:19:42]So we are going to proceed and then find our new UI and then VJ values. I'm going to proceed and find our new VJ values.

[00:19:57]So this is U1. You know that one is always equal to Z. V2 V3 V1 V2 V3. So you have our supply here. You have demand. So here it's still 150 70 60 280 80 120. All right. So let's go ahead and calcate our new V and U values. So let's start with the allocated cells. Sorry, let's start with C11. So we have here to be we have our value to be zero. We already know the expression CI is equal to UI plus V= C. So this U1 value is Z. So 01 value we don't know. So B1 is equal to our C value 8. So therefore V1 is equal. So move to the next allocated cell. Next allocated cell in this row we don't have any that's C2.

[00:21:19]Okay. So we move we move to the next allocated cell that's this C2. So C22 you have U2 plus V2 to be equal to N. I see that you don't know the value for U2 and you don't know the value for V2 as well. So you can just write it like that and go ahead for the rest. By the time we finish, we get the values and then bring it back to get our to get that of the other values. So we just leave it this way. Let me move to the next allocated cell. The next allocated cell is C23.

[00:21:57]this one. So that one involves U2 + V3 which is equal to 12 and that one same we don't know any of the values so we just leave it like that and move to the next allocated cell that's this cell C31 for C31 we already know the value for V1 that's 8 to U3 plus V1 So v1 is 8 is equal to this value 3.

[00:22:32]So therefore u3 is equal to 8 3 - 8 that's going to give us - 5. And then we move to the next allocated. So that's the last one. C 33. So that's going to give us u3 plus v3. But we already know our value for u3 to be - 5.

[00:22:54]- 5 + v3 = 10. So therefore v3 is equal to positive 15 10 + 5. So we now bring the value for v3 here to get our value for u2. So we have this to be 15.

[00:23:17]This means that if it cross to the other side, we going to have four - 5.

[00:23:28]That's going to give us - 3. So therefore, u2 is = -3. And then we have u2 tous 3. So we will take it back to this equation. So u2 is -3. So we have -3 + v2 and that's going to give us 10 + 3. So we have our v2 to be equal to 13. Yes. So that's it.

[00:23:59]Now we have our vi and sorry ui and vj value. So let's check for optimality.

[00:24:12]We are going to use we are going to do that by calcating the net cost that's cig minus ui minus what vj so let's write our values to simple v2 was 13 and then uh v3 was and then u3 was -2 was -3 so we going to that for unallocated cell so let's start with here that's C1 2 that's the C value what's the C value there that's five so 5 - 0 the U value 0 - 13 that's going to give us minus 8 and then the next unallocated cell is here C13 that's going to value there is 6 so 6us the U value 0 - 15 that's the That's going to give us - 9. And the next unallocated cell is 61 here. C21 here.

[00:25:19]The C value is 15. So 15 minus the new value minus 3us value 8. And that's going to give us 10.

[00:25:32]I'm going to do that for the next unallocated cell. That's here. So I didn't write the C value and check above 70 that's N.

[00:25:46]So our C value is sorry 6 is 9US the U value that's - 5 minus the V value that's 13 - 13 this year 13 rather sorry 13 and this going to give us one.

[00:26:20][snorts] All right. So we still have negative numbers here which means that our solution is still not optimal. So we have to do another iteration or we have to draw another table. So you have to take the most negative number and that's -9 and -9 is in which cell that's C13.

[00:26:42]So where is one thing? So this let me bring my writing table so that you can clearly see how to do it.

[00:27:01]All right. So C13.

[00:27:03]So that's this cell. So we are going to assign the polarity. So here is positive. You have to get a close loop on the allocated cells only. So we come to this side and go and move to this side. Move to this side. Move to this side. You see we can't move the diagonal to that be difficult. No here. So that that part is not good for us. So the best part is to just move to this side here. So negative here we bring positive here, negative here. So our path is now this way. So this way that's our closed loop.

[00:27:46]All right. So what do we do? We check where we have the So what do we do? We look at where we have the negative polarities. So this cell and then this cell.

[00:28:00]Then we take the minimum number. So the minimum here is 50. We have 120. We have 50. So the minimum is 50. So we add 50 here. Let me add 50 here. So 50 + 30 that's going to give us 80. That be the new value for this cell. We have 120 here. So 120 minus 50 and that's going to give us 70. Then we come to this cell we add 50. There's nothing there. It's not allocated. So we are going to get 50. And here 50 - 50 we are going to get zero. So that will be the new allocation. So let's redraw our table again.

[00:28:45]Let's draw our table again and check for optimality.

[00:28:50]All right.

[00:28:53]So, we have one, two, three.

[00:29:07]We have demand here. We have supply 80 80 280 60 70 180. So let's look at our allocations. We have 70 here and the value here is eight and here we don't have any allocation. Our new allocation for this is 50. Then sorry five here is six.

[00:29:40]We have 10 here and 12. You have 17.

[00:29:47]We have 15 here.

[00:29:52]We have 80.

[00:29:57]And here we have nine. And then here we don't have any allocation here. All right. So this is our new table. This our new table. So I'm going to repeat the same process.

[00:30:12]How we do it? are going to assign our U and then J U and V values UI and VJ values and then uh calculate them using the allocated cells and then check for optimality. So let's do that.

[00:30:32]So always our U1 is zero U3 V1 V2 V3 All right. So let's compute for the U's and V's. So starting from here that's C11. You already know the expression. We use the expression UI plus VJ= C. That's what we use. So our UI value here, U1 value here is zero. Don't know our V value. So V1= 8. So therefore V1= 8. Then move to the next allocated cell in this row. That's here C13.

[00:31:19]C13. So you know U1 to be zero. And then you don't know our V3 that's equal to 6.

[00:31:27]So therefore V3 is equal to 6. What's the next allocated cell? That's Q3. So C2 U2.

[00:31:41]We don't know our U2 and then we don't know our V2 as well. We just write down the equation. That one should be equal to 10. Leave it. Let's go. We come back to it. And then our next allocated cell is 23. So 23.

[00:31:57]We have our we don't know our U2 value but then we know our Z3 to be six and that's equal to 12. So six will go to the other side. Then we're going to have our U2= to 6 12 - 6. So now we know our U2. So we bring it back here. So U2 is 6. So when 6 cross to the other side, you're going to have V2= 10 - 6. We are going to have our V2 positive 4. Then our next allocated cell is this one. 31. So 31.

[00:32:34]Do you know our U3 value? No. But we know our V1 value. So u3 + 8 is equal to 3. So therefore our u3 value is equal to 3 - 8 that's - 5. So let's write down all our u and v values in our table. So u2 is 4, u3 is - 5 and v1 is 8, v2 is 4.

[00:33:03]And then we have V3 to be equal to six.

[00:33:08]All right. So let's say for optimality optimality.

[00:33:15]All right. And we already know the formula to be the net cost change is equal to CI minus UI minus VJ.

[00:33:27]All right. So starting from we are going to do that on only the unallocated cells. So starting from here.

[00:33:34]So that's C131 that's going to value five. So 5 minus our U minus our value there that's four.

[00:33:46]So 5 - 4 that's positive one. And our next unallocated cell is this one that's C1.

[00:33:55]And our UI is 15 C sorry 15 minus our U value that's 4.

[00:34:03]And then our sorry think writing down our values here. Our U2 is six here. Six here. So that should be six.

[00:34:17]That should be six not four.

[00:34:22]So six.

[00:34:26]So this supposed to be six. So 15 - 6 minus our this our V value there is 8.

[00:34:37]So 15 - 6 - 8 that's also one. And then our next allocated cell is here that's C 32.

[00:34:48]The next question is C value the U value minus 5.

[00:34:54]All right, this night.

[00:35:14]So on our next unallocated cell is C 32 that is this cell. Our C value is N.

[00:35:22]And then our u value is -5 - our v value that's four.

[00:35:31]So 9 + 5 - 4 that's going to give us 10. All right. So our next unallocated cell is here. So C3 is equal to value there is 10 minus the U value - 5 minus the V value that's 6. So 10 + 5 - 6 is 9. So we can conclude that our table is optimal or we now have our optimal solution since all our values or the net cost are greater than or equal to zero.

[00:36:15]It implies that it is not possible to reduce cost further. Therefore our current solution is optimal. So that is it for modified distribution. method. Thank you for watching this video. In case you have a question, you can type it in the comment section and then we'll get back to you. Thanks.

[00:36:50][music]