Definition
The values of P_{1}, P_{2}, P_{3}, represent the supply sources
The values of T_{1}, T_{2},…, T_{13 }represent the destinations
x_{ij} represent the amount of to be transported from the i-th supply source to the j-th demand destination ( where i = 1, …, 3; j = 1, …, 13);
c_{ij} represent the transport cost transportation per unit of goods transported.
a_{i} represent the available quantity at the i-th origin (i = 1 ,…, 3) for supply.
b_{j} – amount of goods demanded at the j-th demand destination) (j = 1 ,…, 13)
we wish to find the optimal values of the variables x_{ij} (i=1,…,3; j=1,…,13) this refers to the optimum amount transported at minimum cost.
Optimize the equation
m | n | |
^{z} ^{=} ∑ ∑^{c}ij^{x}ij | ||
i=1 | j=1 |
Obtain the basic initial solution and optimizing
Using the Vogel’s method
Step 1
We obtain the row and column difference
That is the values of u_{i} to represent the row difference and the values of v_{j} to represent the column difference.
The differences can be labelled as 1^{st}, 2^{nd} …
The firs difference is obtained by finding the difference between the least 2 column or row values.
The table above shows the values of u’ and v’ calculated by finding the difference between the minimum values of cost.
Step 2
u’ = min values between the least two C_{ij }along the rows
v’ = min values between the least two C_{ij }across the columns
The u’ and v’ are then ranked to obtain the maximum
u’+ v’ = c’_{ij}
u’’ + v’’ = c’’_{ij}
The min value between the demand and supply capacities is then allocated to the cell with the least cost.
I our case, 28 is the highest penalty, we allocate 3211 (value of demand since demand is less than supply) to column Minna, and row Sureja since 20 is the least cost in that column.
We then subtract the demand (the allocated) from the supply.
We then look for u’’ …… and v’’ …. Consecutively until the allocation inside the cells is complete.
The end result of the process is
The initial basic solution using this method is also the optimal solution.
The optimal transport cost is obtained by the formula.
3 | 13 | |
^{z} ^{=} ∑ ∑^{c}ij^{x}ij | ||
i=1 | j=1 |
Step 3
The calculations of the u’ and v’ leads to costs c’_{ij . }This values are creates linear equations.
C_{ij }= c’_{ij }+λc’’_{ij}
Y_{y}= α_{ij}+ λ β_{ij}
Where
When β_{ij < 0}
_{–}
On the other hand, when
When β_{ij} < 0
When β_{ij < 0}
This can be used to estimate the limits of coefficients
Lower | Objective | Upper | Objective | |
Limit | Result | Limit | Result | |
2067 | 546884.4 | 2067 | 546884.4 | |
1515 | 546884.4 | 1515 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 817 | 590593.9 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
672 | 546884.4 | 672 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
208 | 546884.4 | 208 | 546884.4 | |
1318 | 546884.4 | 1318 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
219 | 546884.4 | 219 | 546884.4 | |
641 | 546884.4 | 641 | 546884.4 | |
0 | 546884.4 | 70 | 552029.4 | |
1264 | 546884.4 | 1264 | 546884.4 | |
1236 | 546884.4 | 1236 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
564 | 546884.4 | 564 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
480 | 546884.4 | 480 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
3211 | 546884.4 | 3211 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
1 | 546884.4 | 1 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 | |
0 | 546884.4 | 0 | 546884.4 |
Leave a Reply