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Home > Operation Research calculators > Transshipment Problem example
Transshipment Problem example ( Enter your problem )
( Enter your problem )
Algorithm and examples
  1. Example-1
  2. Example-2
  3. Example-3 by using M value = 1000 and M value = M
  4. Example-4 by using M value = 1000 and M value = M
  5. Example-5 by using M value = 1000 and M value = M
 		Other related methods
  1. north-west corner method
  2. least cost method
  3. vogel's approximation method
  4. Row minima method
  5. Column minima method
  6. Russell's approximation method
  7. Heuristic method-1
  8. Heuristic method-2
  9. modi method (optimal solution)
  10. stepping stone method (optimal solution)
  11. Transshipment Problem
  12. LP Model Formulation
4. Example-4 by using M value = 1000 and M value = M
(Previous example)		12. LP Model Formulation
(Next method)

5. Example-5 by using M value = 1000 and M value = M


Example 5.1 by using M value = 1000 and Example 5.2 by using M value = M
5.1) Find Solution of Transshipment Problem using vogel's approximation method
45678Supply
110.3MMM900
20.84.3MMM1400
324.6MMM1000
400.50.24.560
5M032.11.90
Demand00110010001200


Solution:
Problem Table is
`4``5``6``7``8`Supply
`1`10.3100010001000900
`2`0.84.31000100010001400
`3`24.61000100010001000
`4`00.50.24.560
`5`1000032.11.90
Demand00110010001200


`1,2,3` are pure supply nodes

`6,7,8` are pure demand nodes

`4,5` are transshipment nodes

Add Total value `=3300` in supply and demand for transshipment nodes `4,5`

So again Problem Table is
`4``5``6``7``8`Supply
`1`10.3100010001000900
`2`0.84.31000100010001400
`3`24.61000100010001000
`4`00.50.24.563300
`5`1000032.11.93300
Demand33003300110010001200


Now, we solve this Transshipment problem
Table-1
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.7=1-0.3`
`2`0.84.31000100010001400`3.5=4.3-0.8`
`3`24.61000100010001000`2.6=4.6-2`
`4`00.50.24.563300`0.2=0.2-0`
`5`1000032.11.93300`1.9=1.9-0`
Demand33003300110010001200
Column
Penalty		`0.8=0.8-0``0.3=0.3-0``2.8=3-0.2``2.4=4.5-2.1``4.1=6-1.9`


The maximum penalty, 4.1, occurs in column `8`.

The minimum `c_(ij)` in this column is `c_55`=1.9.

The maximum allocation in this cell is min(3300,1200) = 1200.
It satisfy demand of `8` and adjust the supply of `5` from 3300 to 2100 (3300 - 1200=2100).

Table-2
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.7=1-0.3`
`2`0.84.31000100010001400`3.5=4.3-0.8`
`3`24.61000100010001000`2.6=4.6-2`
`4`00.50.24.563300`0.2=0.2-0`
`5`1000032.11.9(1200)2100`2.1=2.1-0`
Demand33003300110010000
Column
Penalty		`0.8=0.8-0``0.3=0.3-0``2.8=3-0.2``2.4=4.5-2.1`--


The maximum penalty, 3.5, occurs in row `2`.

The minimum `c_(ij)` in this row is `c_21`=0.8.

The maximum allocation in this cell is min(1400,3300) = 1400.
It satisfy supply of `2` and adjust the demand of `4` from 3300 to 1900 (3300 - 1400=1900).

Table-3
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.7=1-0.3`
`2`0.8(1400)4.31000100010000--
`3`24.61000100010001000`2.6=4.6-2`
`4`00.50.24.563300`0.2=0.2-0`
`5`1000032.11.9(1200)2100`2.1=2.1-0`
Demand19003300110010000
Column
Penalty		`1=1-0``0.3=0.3-0``2.8=3-0.2``2.4=4.5-2.1`--


The maximum penalty, 2.8, occurs in column `6`.

The minimum `c_(ij)` in this column is `c_43`=0.2.

The maximum allocation in this cell is min(3300,1100) = 1100.
It satisfy demand of `6` and adjust the supply of `4` from 3300 to 2200 (3300 - 1100=2200).

Table-4
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.7=1-0.3`
`2`0.8(1400)4.31000100010000--
`3`24.61000100010001000`2.6=4.6-2`
`4`00.50.2(1100)4.562200`0.5=0.5-0`
`5`1000032.11.9(1200)2100`2.1=2.1-0`
Demand19003300010000
Column
Penalty		`1=1-0``0.3=0.3-0`--`2.4=4.5-2.1`--


The maximum penalty, 2.6, occurs in row `3`.

The minimum `c_(ij)` in this row is `c_31`=2.

The maximum allocation in this cell is min(1000,1900) = 1000.
It satisfy supply of `3` and adjust the demand of `4` from 1900 to 900 (1900 - 1000=900).

Table-5
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.7=1-0.3`
`2`0.8(1400)4.31000100010000--
`3`2(1000)4.61000100010000--
`4`00.50.2(1100)4.562200`0.5=0.5-0`
`5`1000032.11.9(1200)2100`2.1=2.1-0`
Demand9003300010000
Column
Penalty		`1=1-0``0.3=0.3-0`--`2.4=4.5-2.1`--


The maximum penalty, 2.4, occurs in column `7`.

The minimum `c_(ij)` in this column is `c_54`=2.1.

The maximum allocation in this cell is min(2100,1000) = 1000.
It satisfy demand of `7` and adjust the supply of `5` from 2100 to 1100 (2100 - 1000=1100).

Table-6
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.7=1-0.3`
`2`0.8(1400)4.31000100010000--
`3`2(1000)4.61000100010000--
`4`00.50.2(1100)4.562200`0.5=0.5-0`
`5`1000032.1(1000)1.9(1200)1100`1000=1000-0`
Demand9003300000
Column
Penalty		`1=1-0``0.3=0.3-0`------


The maximum penalty, 1000, occurs in row `5`.

The minimum `c_(ij)` in this row is `c_52`=0.

The maximum allocation in this cell is min(1100,3300) = 1100.
It satisfy supply of `5` and adjust the demand of `5` from 3300 to 2200 (3300 - 1100=2200).

Table-7
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.7=1-0.3`
`2`0.8(1400)4.31000100010000--
`3`2(1000)4.61000100010000--
`4`00.50.2(1100)4.562200`0.5=0.5-0`
`5`10000(1100)32.1(1000)1.9(1200)0--
Demand9002200000
Column
Penalty		`1=1-0``0.2=0.5-0.3`------


The maximum penalty, 1, occurs in column `4`.

The minimum `c_(ij)` in this column is `c_41`=0.

The maximum allocation in this cell is min(2200,900) = 900.
It satisfy demand of `4` and adjust the supply of `4` from 2200 to 1300 (2200 - 900=1300).

Table-8
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.3`
`2`0.8(1400)4.31000100010000--
`3`2(1000)4.61000100010000--
`4`0(900)0.50.2(1100)4.561300`0.5`
`5`10000(1100)32.1(1000)1.9(1200)0--
Demand02200000
Column
Penalty		--`0.2=0.5-0.3`------


The maximum penalty, 0.5, occurs in row `4`.

The minimum `c_(ij)` in this row is `c_42`=0.5.

The maximum allocation in this cell is min(1300,2200) = 1300.
It satisfy supply of `4` and adjust the demand of `5` from 2200 to 900 (2200 - 1300=900).

Table-9
`4``5``6``7``8`SupplyRow Penalty
`1`10.3100010001000900`0.3`
`2`0.8(1400)4.31000100010000--
`3`2(1000)4.61000100010000--
`4`0(900)0.5(1300)0.2(1100)4.560--
`5`10000(1100)32.1(1000)1.9(1200)0--
Demand0900000
Column
Penalty		--`0.3`------


The maximum penalty, 0.3, occurs in row `1`.

The minimum `c_(ij)` in this row is `c_12`=0.3.

The maximum allocation in this cell is min(900,900) = 900.
It satisfy supply of `1` and demand of `5`.


Initial feasible solution is
`4``5``6``7``8`SupplyRow Penalty
`1`10.3(900)100010001000900 0.7 |  0.7 |  0.7 |  0.7 |  0.7 |  0.7 |  0.7 |  0.3 |  0.3 |
`2`0.8(1400)4.31000100010001400 3.5 |  3.5 | -- | -- | -- | -- | -- | -- | -- |
`3`2(1000)4.61000100010001000 2.6 |  2.6 |  2.6 |  2.6 | -- | -- | -- | -- | -- |
`4`0(900)0.5(1300)0.2(1100)4.563300 0.2 |  0.2 |  0.2 |  0.5 |  0.5 |  0.5 |  0.5 |  0.5 | -- |
`5`10000(1100)32.1(1000)1.9(1200)3300 1.9 |  2.1 |  2.1 |  2.1 |  2.1 | 1000 | -- | -- | -- |
Demand33003300110010001200
Column
Penalty		0.8
0.8
1
1
1
1
1
--
--
0.3
0.3
0.3
0.3
0.3
0.3
0.2
0.2
0.3
2.8
2.8
2.8
--
--
--
--
--
--
2.4
2.4
2.4
2.4
2.4
--
--
--
--
4.1
--
--
--
--
--
--
--
--


The minimum total transportation cost `=0.3 xx 900+0.8 xx 1400+2 xx 1000+0 xx 900+0.5 xx 1300+0.2 xx 1100+0 xx 1100+2.1 xx 1000+1.9 xx 1200=8640`

Here, the number of allocated cells = 9 is equal to m + n - 1 = 5 + 5 - 1 = 9
`:.` This solution is non-degenerate
5.2) Find Solution of Transshipment Problem using vogel's approximation method
45678Supply
110.3MMM900
20.84.3MMM1400
324.6MMM1000
400.50.24.560
5M032.11.90
Demand00110010001200


Solution:
Problem Table is
`4``5``6``7``8`Supply
`1`10.3MMM900
`2`0.84.3MMM1400
`3`24.6MMM1000
`4`00.50.24.560
`5`M032.11.90
Demand00110010001200


`1,2,3` are pure supply nodes

`6,7,8` are pure demand nodes

`4,5` are transshipment nodes

Add Total value `=3300` in supply and demand for transshipment nodes `4,5`

So again Problem Table is
`4``5``6``7``8`Supply
`1`10.3MMM900
`2`0.84.3MMM1400
`3`24.6MMM1000
`4`00.50.24.563300
`5`M032.11.93300
Demand33003300110010001200


Now, we solve this Transshipment problem
Table-1
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.7=1-0.3`
`2`0.84.3MMM1400`3.5=4.3-0.8`
`3`24.6MMM1000`2.6=4.6-2`
`4`00.50.24.563300`0.2=0.2-0`
`5`M032.11.93300`1.9=1.9-0`
Demand33003300110010001200
Column
Penalty		`0.8=0.8-0``0.3=0.3-0``2.8=3-0.2``2.4=4.5-2.1``4.1=6-1.9`


The maximum penalty, 4.1, occurs in column `8`.

The minimum `c_(ij)` in this column is `c_55`=1.9.

The maximum allocation in this cell is min(3300,1200) = 1200.
It satisfy demand of `8` and adjust the supply of `5` from 3300 to 2100 (3300 - 1200=2100).

Table-2
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.7=1-0.3`
`2`0.84.3MMM1400`3.5=4.3-0.8`
`3`24.6MMM1000`2.6=4.6-2`
`4`00.50.24.563300`0.2=0.2-0`
`5`M032.11.9(1200)2100`2.1=2.1-0`
Demand33003300110010000
Column
Penalty		`0.8=0.8-0``0.3=0.3-0``2.8=3-0.2``2.4=4.5-2.1`--


The maximum penalty, 3.5, occurs in row `2`.

The minimum `c_(ij)` in this row is `c_21`=0.8.

The maximum allocation in this cell is min(1400,3300) = 1400.
It satisfy supply of `2` and adjust the demand of `4` from 3300 to 1900 (3300 - 1400=1900).

Table-3
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.7=1-0.3`
`2`0.8(1400)4.3MMM0--
`3`24.6MMM1000`2.6=4.6-2`
`4`00.50.24.563300`0.2=0.2-0`
`5`M032.11.9(1200)2100`2.1=2.1-0`
Demand19003300110010000
Column
Penalty		`1=1-0``0.3=0.3-0``2.8=3-0.2``2.4=4.5-2.1`--


The maximum penalty, 2.8, occurs in column `6`.

The minimum `c_(ij)` in this column is `c_43`=0.2.

The maximum allocation in this cell is min(3300,1100) = 1100.
It satisfy demand of `6` and adjust the supply of `4` from 3300 to 2200 (3300 - 1100=2200).

Table-4
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.7=1-0.3`
`2`0.8(1400)4.3MMM0--
`3`24.6MMM1000`2.6=4.6-2`
`4`00.50.2(1100)4.562200`0.5=0.5-0`
`5`M032.11.9(1200)2100`2.1=2.1-0`
Demand19003300010000
Column
Penalty		`1=1-0``0.3=0.3-0`--`2.4=4.5-2.1`--


The maximum penalty, 2.6, occurs in row `3`.

The minimum `c_(ij)` in this row is `c_31`=2.

The maximum allocation in this cell is min(1000,1900) = 1000.
It satisfy supply of `3` and adjust the demand of `4` from 1900 to 900 (1900 - 1000=900).

Table-5
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.7=1-0.3`
`2`0.8(1400)4.3MMM0--
`3`2(1000)4.6MMM0--
`4`00.50.2(1100)4.562200`0.5=0.5-0`
`5`M032.11.9(1200)2100`2.1=2.1-0`
Demand9003300010000
Column
Penalty		`1=1-0``0.3=0.3-0`--`2.4=4.5-2.1`--


The maximum penalty, 2.4, occurs in column `7`.

The minimum `c_(ij)` in this column is `c_54`=2.1.

The maximum allocation in this cell is min(2100,1000) = 1000.
It satisfy demand of `7` and adjust the supply of `5` from 2100 to 1100 (2100 - 1000=1100).

Table-6
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.7=1-0.3`
`2`0.8(1400)4.3MMM0--
`3`2(1000)4.6MMM0--
`4`00.50.2(1100)4.562200`0.5=0.5-0`
`5`M032.1(1000)1.9(1200)1100`M=M-0`
Demand9003300000
Column
Penalty		`1=1-0``0.3=0.3-0`------


The maximum penalty, M, occurs in row `5`.

The minimum `c_(ij)` in this row is `c_52`=0.

The maximum allocation in this cell is min(1100,3300) = 1100.
It satisfy supply of `5` and adjust the demand of `5` from 3300 to 2200 (3300 - 1100=2200).

Table-7
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.7=1-0.3`
`2`0.8(1400)4.3MMM0--
`3`2(1000)4.6MMM0--
`4`00.50.2(1100)4.562200`0.5=0.5-0`
`5`M0(1100)32.1(1000)1.9(1200)0--
Demand9002200000
Column
Penalty		`1=1-0``0.2=0.5-0.3`------


The maximum penalty, 1, occurs in column `4`.

The minimum `c_(ij)` in this column is `c_41`=0.

The maximum allocation in this cell is min(2200,900) = 900.
It satisfy demand of `4` and adjust the supply of `4` from 2200 to 1300 (2200 - 900=1300).

Table-8
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.3`
`2`0.8(1400)4.3MMM0--
`3`2(1000)4.6MMM0--
`4`0(900)0.50.2(1100)4.561300`0.5`
`5`M0(1100)32.1(1000)1.9(1200)0--
Demand02200000
Column
Penalty		--`0.2=0.5-0.3`------


The maximum penalty, 0.5, occurs in row `4`.

The minimum `c_(ij)` in this row is `c_42`=0.5.

The maximum allocation in this cell is min(1300,2200) = 1300.
It satisfy supply of `4` and adjust the demand of `5` from 2200 to 900 (2200 - 1300=900).

Table-9
`4``5``6``7``8`SupplyRow Penalty
`1`10.3MMM900`0.3`
`2`0.8(1400)4.3MMM0--
`3`2(1000)4.6MMM0--
`4`0(900)0.5(1300)0.2(1100)4.560--
`5`M0(1100)32.1(1000)1.9(1200)0--
Demand0900000
Column
Penalty		--`0.3`------


The maximum penalty, 0.3, occurs in row `1`.

The minimum `c_(ij)` in this row is `c_12`=0.3.

The maximum allocation in this cell is min(900,900) = 900.
It satisfy supply of `1` and demand of `5`.


Initial feasible solution is
`4``5``6``7``8`SupplyRow Penalty
`1`10.3(900)MMM900 0.7 |  0.7 |  0.7 |  0.7 |  0.7 |  0.7 |  0.7 |  0.3 |  0.3 |
`2`0.8(1400)4.3MMM1400 3.5 |  3.5 | -- | -- | -- | -- | -- | -- | -- |
`3`2(1000)4.6MMM1000 2.6 |  2.6 |  2.6 |  2.6 | -- | -- | -- | -- | -- |
`4`0(900)0.5(1300)0.2(1100)4.563300 0.2 |  0.2 |  0.2 |  0.5 |  0.5 |  0.5 |  0.5 |  0.5 | -- |
`5`M0(1100)32.1(1000)1.9(1200)3300 1.9 |  2.1 |  2.1 |  2.1 |  2.1 | M | -- | -- | -- |
Demand33003300110010001200
Column
Penalty		0.8
0.8
1
1
1
1
1
--
--
0.3
0.3
0.3
0.3
0.3
0.3
0.2
0.2
0.3
2.8
2.8
2.8
--
--
--
--
--
--
2.4
2.4
2.4
2.4
2.4
--
--
--
--
4.1
--
--
--
--
--
--
--
--


The minimum total transportation cost `=0.3 xx 900+0.8 xx 1400+2 xx 1000+0 xx 900+0.5 xx 1300+0.2 xx 1100+0 xx 1100+2.1 xx 1000+1.9 xx 1200=8640`

Here, the number of allocated cells = 9 is equal to m + n - 1 = 5 + 5 - 1 = 9
`:.` This solution is non-degenerate


This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here


4. Example-4 by using M value = 1000 and M value = M
(Previous example)		12. LP Model Formulation
(Next method)


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