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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)
1. Algorithm & Example-1
(Previous example)		3. Unbalanced supply and demand example
(Next example)

2. Example-2


Find Solution using Russell's Approximation 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: 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_4`Supply`bar U_i`
`S_1`11131714250`17`
`S_2`16181410300`18`
`S_3`21241310400`24`
Demand200225275250
`bar V_j``21``24``17``14`


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) = 11 - (17 +21) = color{blue}{-27} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 13 - (17 +24) = color{blue}{-28} `

`3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 17 - (17 +17) = color{blue}{-17} `

`4. Delta_14 = c_14 - (bar U_1 + bar V_4) = 14 - (17 +14) = color{blue}{-17} `

`5. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (18 +21) = color{blue}{-23} `

`6. Delta_22 = c_22 - (bar U_2 + bar V_2) = 18 - (18 +24) = color{blue}{-24} `

`7. Delta_23 = c_23 - (bar U_2 + bar V_3) = 14 - (18 +17) = color{blue}{-21} `

`8. Delta_24 = c_24 - (bar U_2 + bar V_4) = 10 - (18 +14) = color{blue}{-22} `

`9. Delta_31 = c_31 - (bar U_3 + bar V_1) = 21 - (24 +21) = color{blue}{-24} `

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

`11. Delta_33 = c_33 - (bar U_3 + bar V_3) = 13 - (24 +17) = color{blue}{-28} `

`12. Delta_34 = c_34 - (bar U_3 + bar V_4) = 10 - (24 +14) = color{blue}{-28} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`11 [-27]13 [-28]17 [-17]14 [-17]250`17`
`S_2`16 [-23]18 [-24]14 [-21]10 [-22]300`18`
`S_3`21 [-24]24 [-24]13 [-28]10 [-28]400`24`
Demand200225275250
`bar V_j``21``24``17``14`


The most negative `Delta_(ij)` is -28 in cell `S_3 D_4`

The allocation to this cell is min(400,250) = 250.
This satisfies the entire demand of `D_4` and leaves 400 - 250 = 150 units with `S_3`

Table-1: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`11 13 17 14 250
`S_2`16 18 14 10 300
`S_3`21 24 13 10 (250)150
Demand2002252750


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_4`Supply`bar U_i`
`S_1`11131714250`17`
`S_2`16181410300`18`
`S_3`21241310(250)150`24`
Demand2002252750
`bar V_j``21``24``17`--


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) = 11 - (17 +21) = color{blue}{-27} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 13 - (17 +24) = color{blue}{-28} `

`3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 17 - (17 +17) = color{blue}{-17} `

`4. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (18 +21) = color{blue}{-23} `

`5. Delta_22 = c_22 - (bar U_2 + bar V_2) = 18 - (18 +24) = color{blue}{-24} `

`6. Delta_23 = c_23 - (bar U_2 + bar V_3) = 14 - (18 +17) = color{blue}{-21} `

`7. Delta_31 = c_31 - (bar U_3 + bar V_1) = 21 - (24 +21) = color{blue}{-24} `

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

`9. Delta_33 = c_33 - (bar U_3 + bar V_3) = 13 - (24 +17) = color{blue}{-28} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`11 [-27]13 [-28]17 [-17]14250`17`
`S_2`16 [-23]18 [-24]14 [-21]10300`18`
`S_3`21 [-24]24 [-24]13 [-28]10(250)150`24`
Demand2002252750
`bar V_j``21``24``17`--


The most negative `Delta_(ij)` is -28 in cell `S_1 D_2`

The allocation to this cell is min(250,225) = 225.
This satisfies the entire demand of `D_2` and leaves 250 - 225 = 25 units with `S_1`

Table-2: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`11 13 (225)17 14 25
`S_2`16 18 14 10 300
`S_3`21 24 13 10 (250)150
Demand20002750


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_4`Supply`bar U_i`
`S_1`1113(225)171425`17`
`S_2`16181410300`16`
`S_3`21241310(250)150`21`
Demand20002750
`bar V_j``21`--`17`--


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) = 11 - (17 +21) = color{blue}{-27} `

`2. Delta_13 = c_13 - (bar U_1 + bar V_3) = 17 - (17 +17) = color{blue}{-17} `

`3. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (16 +21) = color{blue}{-21} `

`4. Delta_23 = c_23 - (bar U_2 + bar V_3) = 14 - (16 +17) = color{blue}{-19} `

`5. Delta_31 = c_31 - (bar U_3 + bar V_1) = 21 - (21 +21) = color{blue}{-21} `

`6. Delta_33 = c_33 - (bar U_3 + bar V_3) = 13 - (21 +17) = color{blue}{-25} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`11 [-27]13(225)17 [-17]1425`17`
`S_2`16 [-21]1814 [-19]10300`16`
`S_3`21 [-21]2413 [-25]10(250)150`21`
Demand20002750
`bar V_j``21`--`17`--


The most negative `Delta_(ij)` is -27 in cell `S_1 D_1`

The allocation to this cell is min(25,200) = 25.
This exhausts the capacity of `S_1` and leaves 200 - 25 = 175 units with `D_1`

Table-3: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`11 (25)13 (225)17 14 0
`S_2`16 18 14 10 300
`S_3`21 24 13 10 (250)150
Demand17502750


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_4`Supply`bar U_i`
`S_1`11(25)13(225)17140--
`S_2`16181410300`16`
`S_3`21241310(250)150`21`
Demand17502750
`bar V_j``21`--`14`--


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

`1. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (16 +21) = color{blue}{-21} `

`2. Delta_23 = c_23 - (bar U_2 + bar V_3) = 14 - (16 +14) = color{blue}{-16} `

`3. Delta_31 = c_31 - (bar U_3 + bar V_1) = 21 - (21 +21) = color{blue}{-21} `

`4. Delta_33 = c_33 - (bar U_3 + bar V_3) = 13 - (21 +14) = color{blue}{-22} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`11(25)13(225)17140--
`S_2`16 [-21]1814 [-16]10300`16`
`S_3`21 [-21]2413 [-22]10(250)150`21`
Demand17502750
`bar V_j``21`--`14`--


The most negative `Delta_(ij)` is -22 in cell `S_3 D_3`

The allocation to this cell is min(150,275) = 150.
This exhausts the capacity of `S_3` and leaves 275 - 150 = 125 units with `D_3`

Table-4: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`11 (25)13 (225)17 14 0
`S_2`16 18 14 10 300
`S_3`21 24 13 (150)10 (250)0
Demand17501250


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_4`Supply`bar U_i`
`S_1`11(25)13(225)17140--
`S_2`16181410300`16`
`S_3`212413(150)10(250)0--
Demand17501250
`bar V_j``16`--`14`--


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

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

`2. Delta_23 = c_23 - (bar U_2 + bar V_3) = 14 - (16 +14) = color{blue}{-16} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`11(25)13(225)17140--
`S_2`16 [-16]1814 [-16]10300`16`
`S_3`212413(150)10(250)0--
Demand17501250
`bar V_j``16`--`14`--


The most negative `Delta_(ij)` is -16 in cell `S_2 D_3`

The allocation to this cell is min(300,125) = 125.
This satisfies the entire demand of `D_3` and leaves 300 - 125 = 175 units with `S_2`

Table-5: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`11 (25)13 (225)17 14 0
`S_2`16 18 14 (125)10 175
`S_3`21 24 13 (150)10 (250)0
Demand175000


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_4`Supply`bar U_i`
`S_1`11(25)13(225)17140--
`S_2`161814(125)10175`16`
`S_3`212413(150)10(250)0--
Demand175000
`bar V_j``16`------


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

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

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`11(25)13(225)17140--
`S_2`16 [-16]1814(125)10175`16`
`S_3`212413(150)10(250)0--
Demand175000
`bar V_j``16`------


The most negative `Delta_(ij)` is -16 in cell `S_2 D_1`

The allocation to this cell is min(175,175) = 175.
Table-6: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`11 (25)13 (225)17 14 0
`S_2`16 (175)18 14 (125)10 0
`S_3`21 24 13 (150)10 (250)0
Demand0000



Initial feasible solution is
`D_1``D_2``D_3``D_4`Supply
`S_1`11 (25)13 (225)17 14 250
`S_2`16 (175)18 14 (125)10 300
`S_3`21 24 13 (150)10 (250)400
Demand200225275250


The minimum total transportation cost `= 11 xx 25 + 13 xx 225 + 16 xx 175 + 14 xx 125 + 13 xx 150 + 10 xx 250 = 12200`

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


1. Algorithm & Example-1
(Previous example)		3. Unbalanced supply and demand example
(Next example)


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