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Home > Operation Research calculators > Optimal solution using MODI method example
9. modi method (optimal solution) example ( Enter your problem )
( Enter your problem )
Algorithm and examples
  1. Algorithm & Example-1
  2. Example-2
  3. Alternate optimal solutions example
  4. Degeneracy at initial solution
  5. Degeneracy at subsequent iterations
 		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)
1. Algorithm & Example-1
(Previous example)		3. Alternate optimal solutions example
(Next example)

2. Example-2


Find Solution using Voggel's Approximation method, also find optimal solution using modi method,
D1D2D3D4Supply
S111131714250
S216181410300
S321241310400
Demand200225275250


Solution:
TOTAL number of supply constraints : 3
TOTAL number of demand constraints : 4
Problem Table is
`D_1``D_2``D_3``D_4`Supply
`S_1`11131714250
`S_2`16181410300
`S_3`21241310400
Demand200225275250


Table-1
`D_1``D_2``D_3``D_4`SupplyRow Penalty
`S_1`11131714250`2=13-11`
`S_2`16181410300`4=14-10`
`S_3`21241310400`3=13-10`
Demand200225275250
Column
Penalty		`5=16-11``5=18-13``1=14-13``0=10-10`


The maximum penalty, 5, occurs in column `D_1`.

The minimum `c_(ij)` in this column is `c_11` = 11.

The maximum allocation in this cell is min(250,200) = 200.
It satisfy demand of `D_1` and adjust the supply of `S_1` from 250 to 50 (250 - 200 = 50).

Table-2
`D_1``D_2``D_3``D_4`SupplyRow Penalty
`S_1`11(200)13171450`1=14-13`
`S_2`16181410300`4=14-10`
`S_3`21241310400`3=13-10`
Demand0225275250
Column
Penalty		--`5=18-13``1=14-13``0=10-10`


The maximum penalty, 5, occurs in column `D_2`.

The minimum `c_(ij)` in this column is `c_12` = 13.

The maximum allocation in this cell is min(50,225) = 50.
It satisfy supply of `S_1` and adjust the demand of `D_2` from 225 to 175 (225 - 50 = 175).

Table-3
`D_1``D_2``D_3``D_4`SupplyRow Penalty
`S_1`11(200)13(50)17140--
`S_2`16181410300`4=14-10`
`S_3`21241310400`3=13-10`
Demand0175275250
Column
Penalty		--`6=24-18``1=14-13``0=10-10`


The maximum penalty, 6, occurs in column `D_2`.

The minimum `c_(ij)` in this column is `c_22` = 18.

The maximum allocation in this cell is min(300,175) = 175.
It satisfy demand of `D_2` and adjust the supply of `S_2` from 300 to 125 (300 - 175 = 125).

Table-4
`D_1``D_2``D_3``D_4`SupplyRow Penalty
`S_1`11(200)13(50)17140--
`S_2`1618(175)1410125`4=14-10`
`S_3`21241310400`3=13-10`
Demand00275250
Column
Penalty		----`1=14-13``0=10-10`


The maximum penalty, 4, occurs in row `S_2`.

The minimum `c_(ij)` in this row is `c_24` = 10.

The maximum allocation in this cell is min(125,250) = 125.
It satisfy supply of `S_2` and adjust the demand of `D_4` from 250 to 125 (250 - 125 = 125).

Table-5
`D_1``D_2``D_3``D_4`SupplyRow Penalty
`S_1`11(200)13(50)17140--
`S_2`1618(175)1410(125)0--
`S_3`21241310400`3=13-10`
Demand00275125
Column
Penalty		----`13``10`


The maximum penalty, 13, occurs in column `D_3`.

The minimum `c_(ij)` in this column is `c_33` = 13.

The maximum allocation in this cell is min(400,275) = 275.
It satisfy demand of `D_3` and adjust the supply of `S_3` from 400 to 125 (400 - 275 = 125).

Table-6
`D_1``D_2``D_3``D_4`SupplyRow Penalty
`S_1`11(200)13(50)17140--
`S_2`1618(175)1410(125)0--
`S_3`212413(275)10125`10`
Demand000125
Column
Penalty		------`10`


The maximum penalty, 10, occurs in row `S_3`.

The minimum `c_(ij)` in this row is `c_34` = 10.

The maximum allocation in this cell is min(125,125) = 125.
It satisfy supply of `S_3` and demand of `D_4`.


Initial feasible solution is
`D_1``D_2``D_3``D_4`SupplyRow Penalty
`S_1`11(200)13(50)1714250 2 |  1 | -- | -- | -- | -- |
`S_2`1618(175)1410(125)300 4 |  4 |  4 |  4 | -- | -- |
`S_3`212413(275)10(125)400 3 |  3 |  3 |  3 |  3 | 10 |
Demand200225275250
Column
Penalty		5
--
--
--
--
--
5
5
6
--
--
--
1
1
1
1
13
--
0
0
0
0
10
10


The minimum total transportation cost `= 11 xx 200 + 13 xx 50 + 18 xx 175 + 10 xx 125 + 13 xx 275 + 10 xx 125 = 12075`

Here, the number of allocated cells = 6 is equal to m + n - 1 = 3 + 4 - 1 = 6
`:.` This solution is non-degenerate


Optimality test using modi method...
Allocation Table is
`D_1``D_2``D_3``D_4`Supply
`S_1`11 (200)13 (50)17 14 250
`S_2`16 18 (175)14 10 (125)300
`S_3`21 24 13 (275)10 (125)400
Demand200225275250


Iteration-1 of optimality test
1. Find `u_i` and `v_j` for all occupied cells(i,j), where `c_(ij) = u_i + v_j`

`1.` Substituting, `u_1 = 0`, we get

`2. c_11 = u_1 + v_1=>v_1 = c_11 - u_1=>v_1 = 11 -0=>v_1 = 11`

`3. c_12 = u_1 + v_2=>v_2 = c_12 - u_1=>v_2 = 13 -0=>v_2 = 13`

`4. c_22 = u_2 + v_2=>u_2 = c_22 - v_2=>u_2 = 18 -13=>u_2 = 5`

`5. c_24 = u_2 + v_4=>v_4 = c_24 - u_2=>v_4 = 10 -5=>v_4 = 5`

`6. c_34 = u_3 + v_4=>u_3 = c_34 - v_4=>u_3 = 10 -5=>u_3 = 5`

`7. c_33 = u_3 + v_3=>v_3 = c_33 - u_3=>v_3 = 13 -5=>v_3 = 8`

`D_1``D_2``D_3``D_4`Supply`u_i`
`S_1`11 (200)13 (50)17 14 250`u_1=0`
`S_2`16 18 (175)14 10 (125)300`u_2=5`
`S_3`21 24 13 (275)10 (125)400`u_3=5`
Demand200225275250
`v_j``v_1=11``v_2=13``v_3=8``v_4=5`


2. Find `d_(ij)` for all unoccupied cells(i,j), where `d_(ij) = c_(ij) - (u_i + v_j)`

`1. d_13 = c_13 - (u_1 + v_3) = 17 - (0 +8) = color{blue}{9} `

`2. d_14 = c_14 - (u_1 + v_4) = 14 - (0 +5) = color{blue}{9} `

`3. d_21 = c_21 - (u_2 + v_1) = 16 - (5 +11) = color{blue}{0} `

`4. d_23 = c_23 - (u_2 + v_3) = 14 - (5 +8) = color{blue}{1} `

`5. d_31 = c_31 - (u_3 + v_1) = 21 - (5 +11) = color{blue}{5} `

`6. d_32 = c_32 - (u_3 + v_2) = 24 - (5 +13) = color{blue}{6} `

`D_1``D_2``D_3``D_4`Supply`u_i`
`S_1`11 (200)13 (50)17 [9]14 [9]250`u_1=0`
`S_2`16 [0]18 (175)14 [1]10 (125)300`u_2=5`
`S_3`21 [5]24 [6]13 (275)10 (125)400`u_3=5`
Demand200225275250
`v_j``v_1=11``v_2=13``v_3=8``v_4=5`


Since all `d_(ij)>=0`.

So final optimal solution is arrived.
`D_1``D_2``D_3``D_4`Supply
`S_1`11 (200)13 (50)17 14 250
`S_2`16 18 (175)14 10 (125)300
`S_3`21 24 13 (275)10 (125)400
Demand200225275250


The minimum total transportation cost `= 11 xx 200 + 13 xx 50 + 18 xx 175 + 10 xx 125 + 13 xx 275 + 10 xx 125 = 12075`

Notice alternate solution is available with unoccupied cell `S_2D_1 : d_(21` = [0], but with the same optimal value.

This material is intended as a summary. Use your textbook for detail explanation.
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1. Algorithm & Example-1
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