We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more   

AtoZmath.com
Support us
(New) All problem can be solved using search box
I want to sell my website www.AtoZmath.com with complete code
Home What's new College Algebra Games Feedback About us
Algebra Matrix & Vector Numerical Methods Statistical Methods Operation Research Word Problems Calculus Geometry Pre-Algebra


Home > Operation Research calculators > Transportation Problem 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
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)


Share this solution or page with your friends.


Home What's new College Algebra Games Feedback About us
 
Copyright © 2023. All rights reserved. Terms, Privacy
 
 

.