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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)
5. Column minima method
(Previous method)		2. Example-2
(Next example)

1. Algorithm & Example-1


Algorithm
Russell's Approximation Method (RAM):
Step-1: For each source row still under consideration, determine its `bar U_i` (largest cost in row i).
Step-2: For each destination column still under consideration, determine its `bar V_j` (largest cost in column j).
Step-3: For each variable, calculate `Delta_(ij) = c_(ij) - (bar U_i + bar V_j)`.
Step-4: Select the variable having the most negative `Delta` value, break ties arbitrarily.
Step-5: Allocate as much as possible. Eliminate necessary cells from consideration. Return to Step-1.
Example-1
1. Find Solution using Russell's Approximation method
D1D2D3D4Supply
S1193050107
S2703040609
S3408702018
Demand58714


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`193050107
`S_2`703040609
`S_3`408702018
Demand58714


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`193050107`50`
`S_2`703040609`70`
`S_3`408702018`70`
Demand58714
`bar V_j``70``30``70``60`


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) = 19 - (50 +70) = color{blue}{-101} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 30 - (50 +30) = color{blue}{-50} `

`3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 50 - (50 +70) = color{blue}{-70} `

`4. Delta_14 = c_14 - (bar U_1 + bar V_4) = 10 - (50 +60) = color{blue}{-100} `

`5. Delta_21 = c_21 - (bar U_2 + bar V_1) = 70 - (70 +70) = color{blue}{-70} `

`6. Delta_22 = c_22 - (bar U_2 + bar V_2) = 30 - (70 +30) = color{blue}{-70} `

`7. Delta_23 = c_23 - (bar U_2 + bar V_3) = 40 - (70 +70) = color{blue}{-100} `

`8. Delta_24 = c_24 - (bar U_2 + bar V_4) = 60 - (70 +60) = color{blue}{-70} `

`9. Delta_31 = c_31 - (bar U_3 + bar V_1) = 40 - (70 +70) = color{blue}{-100} `

`10. Delta_32 = c_32 - (bar U_3 + bar V_2) = 8 - (70 +30) = color{blue}{-92} `

`11. Delta_33 = c_33 - (bar U_3 + bar V_3) = 70 - (70 +70) = color{blue}{-70} `

`12. Delta_34 = c_34 - (bar U_3 + bar V_4) = 20 - (70 +60) = color{blue}{-110} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`19 [-101]30 [-50]50 [-70]10 [-100]7`50`
`S_2`70 [-70]30 [-70]40 [-100]60 [-70]9`70`
`S_3`40 [-100]8 [-92]70 [-70]20 [-110]18`70`
Demand58714
`bar V_j``70``30``70``60`


The most negative `Delta_(ij)` is -110 in cell `S_3 D_4`

The allocation to this cell is min(18,14) = 14.
This satisfies the entire demand of `D_4` and leaves 18 - 14 = 4 units with `S_3`

Table-1: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`19 30 50 10 7
`S_2`70 30 40 60 9
`S_3`40 8 70 20 (14)4
Demand5870


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`193050107`50`
`S_2`703040609`70`
`S_3`4087020(14)4`70`
Demand5870
`bar V_j``70``30``70`--


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) = 19 - (50 +70) = color{blue}{-101} `

`2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 30 - (50 +30) = color{blue}{-50} `

`3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 50 - (50 +70) = color{blue}{-70} `

`4. Delta_21 = c_21 - (bar U_2 + bar V_1) = 70 - (70 +70) = color{blue}{-70} `

`5. Delta_22 = c_22 - (bar U_2 + bar V_2) = 30 - (70 +30) = color{blue}{-70} `

`6. Delta_23 = c_23 - (bar U_2 + bar V_3) = 40 - (70 +70) = color{blue}{-100} `

`7. Delta_31 = c_31 - (bar U_3 + bar V_1) = 40 - (70 +70) = color{blue}{-100} `

`8. Delta_32 = c_32 - (bar U_3 + bar V_2) = 8 - (70 +30) = color{blue}{-92} `

`9. Delta_33 = c_33 - (bar U_3 + bar V_3) = 70 - (70 +70) = color{blue}{-70} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`19 [-101]30 [-50]50 [-70]107`50`
`S_2`70 [-70]30 [-70]40 [-100]609`70`
`S_3`40 [-100]8 [-92]70 [-70]20(14)4`70`
Demand5870
`bar V_j``70``30``70`--


The most negative `Delta_(ij)` is -101 in cell `S_1 D_1`

The allocation to this cell is min(7,5) = 5.
This satisfies the entire demand of `D_1` and leaves 7 - 5 = 2 units with `S_1`

Table-2: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`19 (5)30 50 10 2
`S_2`70 30 40 60 9
`S_3`40 8 70 20 (14)4
Demand0870


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`19(5)3050102`50`
`S_2`703040609`40`
`S_3`4087020(14)4`70`
Demand0870
`bar V_j`--`30``70`--


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

`1. Delta_12 = c_12 - (bar U_1 + bar V_2) = 30 - (50 +30) = color{blue}{-50} `

`2. Delta_13 = c_13 - (bar U_1 + bar V_3) = 50 - (50 +70) = color{blue}{-70} `

`3. Delta_22 = c_22 - (bar U_2 + bar V_2) = 30 - (40 +30) = color{blue}{-40} `

`4. Delta_23 = c_23 - (bar U_2 + bar V_3) = 40 - (40 +70) = color{blue}{-70} `

`5. Delta_32 = c_32 - (bar U_3 + bar V_2) = 8 - (70 +30) = color{blue}{-92} `

`6. Delta_33 = c_33 - (bar U_3 + bar V_3) = 70 - (70 +70) = color{blue}{-70} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`19(5)30 [-50]50 [-70]102`50`
`S_2`7030 [-40]40 [-70]609`40`
`S_3`408 [-92]70 [-70]20(14)4`70`
Demand0870
`bar V_j`--`30``70`--


The most negative `Delta_(ij)` is -92 in cell `S_3 D_2`

The allocation to this cell is min(4,8) = 4.
This exhausts the capacity of `S_3` and leaves 8 - 4 = 4 units with `D_2`

Table-3: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`19 (5)30 50 10 2
`S_2`70 30 40 60 9
`S_3`40 8 (4)70 20 (14)0
Demand0470


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`19(5)3050102`50`
`S_2`703040609`40`
`S_3`408(4)7020(14)0--
Demand0470
`bar V_j`--`30``50`--


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

`1. Delta_12 = c_12 - (bar U_1 + bar V_2) = 30 - (50 +30) = color{blue}{-50} `

`2. Delta_13 = c_13 - (bar U_1 + bar V_3) = 50 - (50 +50) = color{blue}{-50} `

`3. Delta_22 = c_22 - (bar U_2 + bar V_2) = 30 - (40 +30) = color{blue}{-40} `

`4. Delta_23 = c_23 - (bar U_2 + bar V_3) = 40 - (40 +50) = color{blue}{-50} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`19(5)30 [-50]50 [-50]102`50`
`S_2`7030 [-40]40 [-50]609`40`
`S_3`408(4)7020(14)0--
Demand0470
`bar V_j`--`30``50`--


The most negative `Delta_(ij)` is -50 in cell `S_2 D_3`

The allocation to this cell is min(9,7) = 7.
This satisfies the entire demand of `D_3` and leaves 9 - 7 = 2 units with `S_2`

Table-4: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`19 (5)30 50 10 2
`S_2`70 30 40 (7)60 2
`S_3`40 8 (4)70 20 (14)0
Demand0400


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`19(5)3050102`30`
`S_2`703040(7)602`30`
`S_3`408(4)7020(14)0--
Demand0400
`bar V_j`--`30`----


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

`1. Delta_12 = c_12 - (bar U_1 + bar V_2) = 30 - (30 +30) = color{blue}{-30} `

`2. Delta_22 = c_22 - (bar U_2 + bar V_2) = 30 - (30 +30) = color{blue}{-30} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`19(5)30 [-30]50102`30`
`S_2`7030 [-30]40(7)602`30`
`S_3`408(4)7020(14)0--
Demand0400
`bar V_j`--`30`----


The most negative `Delta_(ij)` is -30 in cell `S_1 D_2`

The allocation to this cell is min(2,4) = 2.
This exhausts the capacity of `S_1` and leaves 4 - 2 = 2 units with `D_2`

Table-5: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`19 (5)30 (2)50 10 0
`S_2`70 30 40 (7)60 2
`S_3`40 8 (4)70 20 (14)0
Demand0200


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`19(5)30(2)50100--
`S_2`703040(7)602`30`
`S_3`408(4)7020(14)0--
Demand0200
`bar V_j`--`30`----


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

`1. Delta_22 = c_22 - (bar U_2 + bar V_2) = 30 - (30 +30) = color{blue}{-30} `

`D_1``D_2``D_3``D_4`Supply`bar U_i`
`S_1`19(5)30(2)50100--
`S_2`7030 [-30]40(7)602`30`
`S_3`408(4)7020(14)0--
Demand0200
`bar V_j`--`30`----


The most negative `Delta_(ij)` is -30 in cell `S_2 D_2`

The allocation to this cell is min(2,2) = 2.
Table-6: This leads to the following table
`D_1``D_2``D_3``D_4`Supply
`S_1`19 (5)30 (2)50 10 0
`S_2`70 30 (2)40 (7)60 0
`S_3`40 8 (4)70 20 (14)0
Demand0000



Initial feasible solution is
`D_1``D_2``D_3``D_4`Supply
`S_1`19 (5)30 (2)50 10 7
`S_2`70 30 (2)40 (7)60 9
`S_3`40 8 (4)70 20 (14)18
Demand58714


The minimum total transportation cost `= 19 xx 5 + 30 xx 2 + 30 xx 2 + 40 xx 7 + 8 xx 4 + 20 xx 14 = 807`

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


5. Column minima method
(Previous method)		2. Example-2
(Next example)


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