# Time Minimizing Transportation Problem

In the Time Minimizing Transportation Problem the objective is to minimize the time.

This problem is same as the transportation problem of minimizing the cost, expect that the unit transportation cost is replaced by the time tij.

"One thing you can't recycle is wasted time." -Anon.

In a cost minimization problem, the cost of transportation depends on the quantity shipped. To the contrary, in a time minimization problem, the time involved is independent of the amount of commodity shipped.

## Steps in Time Minimizing Problem

1. Determine an initial basic feasible solution using any one of the following:

2. Find Tk for this feasible plan and cross out all the unoccupied cells for which tij ≥ Tk.

3. Trace a closed path for the occupied cells corresponding to Tk. If no such closed path can be formed, the solution obtained is optimum otherwise, go to step 2.

### Example 1: Time Minimizing Transportation Problem

The following matrix gives data concerning the transportation times tij

Destination
Origin D1 D2 D3 D4 D5 D6 Supply
O1 25 30 20 40 45 37 37
O2 30 25 20 30 40 20 22
O3 40 20 40 35 45 22 32
O4 25 24 50 27 30 25 14
Demand 15 20 15 25 20 10

Solution.

We compute an initial basic feasible solution by north west corner rule which is shown in table 1.

Table 1

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Destination
Origin D1 D2 D3 D4 D5 D6 Supply
O1 40 45 37 37
O2 30 25 40 20 22
O3 40 20 40 22 32
O4 25 24 50 27 14
Demand 15 20 15 25 20 10

Here, t11 = 25, t12 = 30, t13 = 20, t23 = 20, t24 = 30, t34= 35, t35 = 45, t45 =30, t46 = 25

Choose maximum from tij, i.e., T1 = 45. Now, cross out all the unoccupied cells that are ≥ T1.
The unoccupied cell (O3D6) enters into the basis as shown in table 2.

Table 2

Choose the smallest value with a negative position on the closed path, i.e., 10. Clearly only 10 units can be shifted to the entering cell. The next feasible plan is shown in the following table.

Table 3

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Destination
Origin D1 D2 D3 D4 D5 D6 Supply
O1 40 37 37
O2 30 25 40 20 22
O3 40 20 40 32
O4 25 24 27 25 14
Demand 15 20 15 25 20 10

Here, T2 = Max(25, 30, 20, 20, 20, 35, 45, 22, 30) = 45. Now, cross out all the unoccupied cells that are ≥ T2.

Table 4

By following the same procedure as explained above, we get the following revised matrix.

Table

Destination
Origin D1 D2 D3 D4 D5 D6 Supply
O1 37 37
O2 30 25 20 22
O3 20 32
O4 25 24 27 25 14
Demand 15 20 15 25 20 10

T3 = Max(25, 30, 20, 20, 30, 40, 35, 22, 30) = 40. Now, cross out all the unoccupied cells that are ≥ T3.

Now we cannot form any other closed loop with T3.
Hence, the solution obtained at this stage is optimal.
Thus, all the shipments can be made within 40 units.