## Minimissing Transportation Costs

The table contains information about the supply from three factories,
$(F_1, F_2. F_3)$
the capacities of three warehouses,
$(W_1. W_2. W_3)$
and the cost of sending stock to each warehouse from each factory.
 Source $F_1$ $F_2$ $F_3$ Demand $W_1$ 0.90 1.00 1.00 5 Destination $W_1$ 1.00 1.40 0.80 20 $W_1$ 1.00 1.30 0.80 20 Supply 20 15 10 45
The problem is to find the minimum cost of supplying the wavehaousing from the factories. Thie sort of problem is called a 'transportation problem'. We solve it by iteration, starting from any solution, then trying to find a lower cost solition.
An initial solution may be found by taking the box with the lowest entry (representing cost) in both row and column. Place in that box the lower of the demand and supply requirements, representing the quantity to be supplied from the corresponding factory in the second row to the corresponding warehouse in the second column. Repeat for the next lowest cost, and ensure the leat value pf demand and supply is met. Repeat until all demand and supply requirements are met.
For the table above we have two cost entries of 0.80. We can choose either of these. If we choose the one in the fifth row, , then the next lowest cost option with unmet requirements is from
$F_1$
to
$W_1$
, and we can send 5 units via this option. The next lowest cost options with unmet requirements are 1.00, and then 1.30. We obtain the table below, using the notation cost/units.
 Source Factory $F_1$ $F_2$ $F_3$ Demand $W_1$ 0.90/5 1.00/0 1.00/0 5 Destination $W_1$ 1.00/10 1.40/0 0.80/10 20 $W_1$ 1.00/5 1.30/15 0.80/0 20 Supply 20 15 10 45
The total cost of this solution is
$0.90 \times 5 +1.00 \times 10 +0.80 \times 10 1.30 \times 5+1.00 \times 5=46.00$
.
The route from
$F_3$
to
$W_3$
is the joint lowest cost route and is not yet used. Suppose we were to divert5 units from route
$F_1 - W_3$
to the lower cost route
$F_1 - W_3$
. Then we have to transer 5 units from route
$F_3 - W_2$
to route
$F_1 - W_2$
so that all demand and supply constraints are satisfied. The table becomes:
 Source Factory $F_1$ $F_2$ $F_3$ Demand $W_1$ 0.90/5 1.00/0 1.00/0 5 Destination $W_1$ 1.00/15 1.40/0 0.80/5 20 $W_1$ 1.00/0 1.30/15 0.80/5 20 Supply 20 15 10 45
The total cost of this solution is
$0.90 \times 5 +1.00 \times 15 +0.80 \times 5 1.30 \times 5+0.90 \times 5=47.00$
.
This is a higher cos solution.