Transportation Model
Source: YT – Kauser Wise
Transportation Problem – is a special kind of linear programming problem in which goods are
transported from a set of sources to a set of destinations subject to supply and demand of the sources
and destination, respectively, such that the total cost of transportation is minimized.
Types:
1. Balanced Transportation Problem (S=D)
2. Unbalanced Transportation Problem (S=/=D)
Methods:
1. Find the initial feasible solution.
General Steps:
✔ Plot the given information using the table - Sources, Destination, Demand, Supply and Cost
of Transportation for given destination.
✔ Compare the demand and supply if it is balanced.
✔ If not, proceed to finding the difference and add another row/column as a dummy account
to convert the unbalanced market to a balanced market.
✔ Apply any of the following method available to solve the initial feasible solution:
a. Northwest Corner Cell Method
● In this method, you need to start the allocation of demand and supply
whichever is lower to the northwest corner of the table.
● Once the row/column was consumed, marked the remaining column to
determine that demand/supply was already provided.
● Repeat steps until demand and supply becomes zero.
● Add all the products of the allocated cell to compute the transportation value of
the initial feasible solution.
b. Least Cost Cell Method
● In this method, you need to start the allocation of demand and supply
whichever is lower to the lowest transportation cost. If the lowest cost has the
same amount, randomly choose between the available value.
● Once the row/column was consumed, marked the remaining column to
determine that demand/supply was already provided.
● Repeat steps until demand and supply becomes zero.
● Add all the products of the allocated cell to compute the transportation value of
the initial feasible solution.
c. Vogel’s Approximation Method
● In this method, you need to get the difference between the row and the column
by extending the column and row right side and downside.
● Select the least value among rows and columns then subtract them with
absolute value result and plot to the extended column and row.
● Select the maximum value from row and column difference.
● Start the allocation of demand and supply whichever is lower to the least cost
cell to column/row of which the maximum value of difference is situated.
● Once the row/column was consumed, marked the remaining column to
determine that demand/supply was already provided.
● Repeat steps until demand and supply becomes zero.
● Add all the products of the allocated cell to compute the transportation value of
the initial feasible solution.
Note: The method that will result in the least cost of initial feasible solution is Vogel Approximation
Method.
2. Optimization of the initial feasible solution.
a. U-V (Modified Distribution) Method
● Using this method, select the table of initial feasible solution as base.
● Identify the u-value and v-value in the table using the formula:
Formula:
ui + vj = cij
Whereas:
u - row
i - row value
v - column
j - column value
c - cell value
Note: u1 should always be equal to zero as starting point of finding u and v
Test the allocated cell which must be equal to the result of the formula:
Formula:
m + n - cij
● After getting the u and v value, compute the penalties of the values of
non-allocated cells using the formula:
Formula:
Pij = ui + vj - cij
Whereas:
P - penalties
u - row
i - row value
v - column
j - column value
c - cell value
● To conclude that the solution is optimal, penalties of all the non-allocated cells
must be equal to or less than zero.
● If some of the penalties are above zero, you need to redistribute the allocation
starting from the highest penalty cell above zero. This cell will be the new basic
cell.
● From the new basic cell draw horizontal and vertical lines passing through
allocated cells only to make a close loop on where to redistribute the allocation.
● After getting the close loop, identify the starting point as positive, second point
as negative, third point as positive and fourth point as negative following the
destination where lines pass through. Then from the two negative points,
choose the lowest cost, and deduct it to the negative cell and add it to the
positive cells, making the lowest negative points to zero which is the new
allocated distribution.
● After the new distribution, go back to the first step until the value of the
penalties of all non-allocated cells will become equal or less than zero. This
means that the solution is optimal.
● Add all the products of allocated cells to the computed optimal solution.