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Home > Operation Research calculators > Russell's approximation method example
6. russell's approximation method example ( Enter your problem )
( Enter your problem )
Algorithm and examples
  1. Algorithm & Example-1
  2. Example-2
  3. Unbalanced supply and demand example
 		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)
2. Example-2
(Previous example)		7. Heuristic method-1
(Next method)

3. Unbalanced supply and demand example


Unbalanced supply and demand
If the total supply is not equal to the total demand then the problem is called unbalanced transportation problem.

It's solution :
1. If the total supply is more than the total demand, then we add a new column, with transportation cost 0
2. If the total demand is more than the total supply, then we add a new row, with transportation cost 0
Example
Find Solution using Russell's Approximation method
D1D2D3Supply
S148876
S216241682
S38162477
Demand7210241


Solution:
TOTAL number of supply constraints : 3
TOTAL number of demand constraints : 3
Problem Table is
`D_1``D_2``D_3`Supply
`S_1`48876
`S_2`16241682
`S_3`8162477
Demand7210241


Here Total Demand = 215 is less than Total Supply = 235. So We add a dummy demand constraint with 0 unit cost and with allocation 20.
Now, The modified table is
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`488076
`S_2`162416082
`S_3`81624077
Demand721024120


Table-1: Calculate `bar U_i` and `bar V_j` (where `bar U_i` is the largest cost in row and `bar V_j` is the largest cost in column)

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`488076`8`
`S_2`162416082`24`
`S_3`81624077`24`
Demand721024120
`bar V_j``16``24``24``0`


2. Compute reduced cost of each cell `Delta_(ij)`, where `Delta_(ij) = c_(ij) - (bar U_i + bar V_j)`

`1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +16) = color{blue}{-20} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +24) = color{blue}{-24} `

`3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 8 - (8 +24) = color{blue}{-24} `

`4. Delta_14 = c_14 - (bar U_1 + bar V_4) = 0 - (8 +0) = color{blue}{-8} `

`5. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (24 +16) = color{blue}{-24} `

`6. Delta_22 = c_22 - (bar U_2 + bar V_2) = 24 - (24 +24) = color{blue}{-24} `

`7. Delta_23 = c_23 - (bar U_2 + bar V_3) = 16 - (24 +24) = color{blue}{-32} `

`8. Delta_24 = c_24 - (bar U_2 + bar V_4) = 0 - (24 +0) = color{blue}{-24} `

`9. Delta_31 = c_31 - (bar U_3 + bar V_1) = 8 - (24 +16) = color{blue}{-32} `

`10. Delta_32 = c_32 - (bar U_3 + bar V_2) = 16 - (24 +24) = color{blue}{-32} `

`11. Delta_33 = c_33 - (bar U_3 + bar V_3) = 24 - (24 +24) = color{blue}{-24} `

`12. Delta_34 = c_34 - (bar U_3 + bar V_4) = 0 - (24 +0) = color{blue}{-24} `

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`4 [-20]8 [-24]8 [-24]0 [-8]76`8`
`S_2`16 [-24]24 [-24]16 [-32]0 [-24]82`24`
`S_3`8 [-32]16 [-32]24 [-24]0 [-24]77`24`
Demand721024120
`bar V_j``16``24``24``0`


The most negative `Delta_(ij)` is -32 in cell `S_3 D_2`

The allocation to this cell is min(77,102) = 77.
This exhausts the capacity of `S_3` and leaves 102 - 77 = 25 units with `D_2`

Table-1: This leads to the following table
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4 8 8 0 76
`S_2`16 24 16 0 82
`S_3`8 16 (77)24 0 0
Demand72254120


Table-2: Calculate `bar U_i` and `bar V_j` (where `bar U_i` is the largest cost in row and `bar V_j` is the largest cost in column)

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`488076`8`
`S_2`162416082`24`
`S_3`816(77)2400--
Demand72254120
`bar V_j``16``24``16``0`


2. Compute reduced cost of each cell `Delta_(ij)`, where `Delta_(ij) = c_(ij) - (bar U_i + bar V_j)`

`1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +16) = color{blue}{-20} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +24) = color{blue}{-24} `

`3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 8 - (8 +16) = color{blue}{-16} `

`4. Delta_14 = c_14 - (bar U_1 + bar V_4) = 0 - (8 +0) = color{blue}{-8} `

`5. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (24 +16) = color{blue}{-24} `

`6. Delta_22 = c_22 - (bar U_2 + bar V_2) = 24 - (24 +24) = color{blue}{-24} `

`7. Delta_23 = c_23 - (bar U_2 + bar V_3) = 16 - (24 +16) = color{blue}{-24} `

`8. Delta_24 = c_24 - (bar U_2 + bar V_4) = 0 - (24 +0) = color{blue}{-24} `

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`4 [-20]8 [-24]8 [-16]0 [-8]76`8`
`S_2`16 [-24]24 [-24]16 [-24]0 [-24]82`24`
`S_3`816(77)2400--
Demand72254120
`bar V_j``16``24``16``0`


The most negative `Delta_(ij)` is -24 in cell `S_2 D_(dummy)`

The allocation to this cell is min(82,20) = 20.
This satisfies the entire demand of `D_(dummy)` and leaves 82 - 20 = 62 units with `S_2`

Table-2: This leads to the following table
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4 8 8 0 76
`S_2`16 24 16 0 (20)62
`S_3`8 16 (77)24 0 0
Demand7225410


Table-3: Calculate `bar U_i` and `bar V_j` (where `bar U_i` is the largest cost in row and `bar V_j` is the largest cost in column)

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`488076`8`
`S_2`1624160(20)62`24`
`S_3`816(77)2400--
Demand7225410
`bar V_j``16``24``16`--


2. Compute reduced cost of each cell `Delta_(ij)`, where `Delta_(ij) = c_(ij) - (bar U_i + bar V_j)`

`1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +16) = color{blue}{-20} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +24) = color{blue}{-24} `

`3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 8 - (8 +16) = color{blue}{-16} `

`4. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (24 +16) = color{blue}{-24} `

`5. Delta_22 = c_22 - (bar U_2 + bar V_2) = 24 - (24 +24) = color{blue}{-24} `

`6. Delta_23 = c_23 - (bar U_2 + bar V_3) = 16 - (24 +16) = color{blue}{-24} `

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`4 [-20]8 [-24]8 [-16]076`8`
`S_2`16 [-24]24 [-24]16 [-24]0(20)62`24`
`S_3`816(77)2400--
Demand7225410
`bar V_j``16``24``16`--


The most negative `Delta_(ij)` is -24 in cell `S_2 D_1`

The allocation to this cell is min(62,72) = 62.
This exhausts the capacity of `S_2` and leaves 72 - 62 = 10 units with `D_1`

Table-3: This leads to the following table
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4 8 8 0 76
`S_2`16 (62)24 16 0 (20)0
`S_3`8 16 (77)24 0 0
Demand1025410


Table-4: Calculate `bar U_i` and `bar V_j` (where `bar U_i` is the largest cost in row and `bar V_j` is the largest cost in column)

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`488076`8`
`S_2`16(62)24160(20)0--
`S_3`816(77)2400--
Demand1025410
`bar V_j``4``8``8`--


2. Compute reduced cost of each cell `Delta_(ij)`, where `Delta_(ij) = c_(ij) - (bar U_i + bar V_j)`

`1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +4) = color{blue}{-8} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +8) = color{blue}{-8} `

`3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 8 - (8 +8) = color{blue}{-8} `

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`4 [-8]8 [-8]8 [-8]076`8`
`S_2`16(62)24160(20)0--
`S_3`816(77)2400--
Demand1025410
`bar V_j``4``8``8`--


The most negative `Delta_(ij)` is -8 in cell `S_1 D_3`

The allocation to this cell is min(76,41) = 41.
This satisfies the entire demand of `D_3` and leaves 76 - 41 = 35 units with `S_1`

Table-4: This leads to the following table
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4 8 8 (41)0 35
`S_2`16 (62)24 16 0 (20)0
`S_3`8 16 (77)24 0 0
Demand102500


Table-5: Calculate `bar U_i` and `bar V_j` (where `bar U_i` is the largest cost in row and `bar V_j` is the largest cost in column)

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`488(41)035`8`
`S_2`16(62)24160(20)0--
`S_3`816(77)2400--
Demand102500
`bar V_j``4``8`----


2. Compute reduced cost of each cell `Delta_(ij)`, where `Delta_(ij) = c_(ij) - (bar U_i + bar V_j)`

`1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +4) = color{blue}{-8} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +8) = color{blue}{-8} `

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`4 [-8]8 [-8]8(41)035`8`
`S_2`16(62)24160(20)0--
`S_3`816(77)2400--
Demand102500
`bar V_j``4``8`----


The most negative `Delta_(ij)` is -8 in cell `S_1 D_2`

The allocation to this cell is min(35,25) = 25.
This satisfies the entire demand of `D_2` and leaves 35 - 25 = 10 units with `S_1`

Table-5: This leads to the following table
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4 8 (25)8 (41)0 10
`S_2`16 (62)24 16 0 (20)0
`S_3`8 16 (77)24 0 0
Demand10000


Table-6: Calculate `bar U_i` and `bar V_j` (where `bar U_i` is the largest cost in row and `bar V_j` is the largest cost in column)

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`48(25)8(41)010`4`
`S_2`16(62)24160(20)0--
`S_3`816(77)2400--
Demand10000
`bar V_j``4`------


2. Compute reduced cost of each cell `Delta_(ij)`, where `Delta_(ij) = c_(ij) - (bar U_i + bar V_j)`

`1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (4 +4) = color{blue}{-4} `

`D_1``D_2``D_3``D_(dummy)`Supply`bar U_i`
`S_1`4 [-4]8(25)8(41)010`4`
`S_2`16(62)24160(20)0--
`S_3`816(77)2400--
Demand10000
`bar V_j``4`------


The most negative `Delta_(ij)` is -4 in cell `S_1 D_1`

The allocation to this cell is min(10,10) = 10.
Table-6: This leads to the following table
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4 (10)8 (25)8 (41)0 0
`S_2`16 (62)24 16 0 (20)0
`S_3`8 16 (77)24 0 0
Demand0000



Initial feasible solution is
`D_1``D_2``D_3``D_(dummy)`Supply
`S_1`4 (10)8 (25)8 (41)0 76
`S_2`16 (62)24 16 0 (20)82
`S_3`8 16 (77)24 0 77
Demand721024120


The minimum total transportation cost `= 4 xx 10 + 8 xx 25 + 8 xx 41 + 16 xx 62 + 0 xx 20 + 16 xx 77 = 2792`

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


2. Example-2
(Previous example)		7. Heuristic method-1
(Next method)


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