The following table all the necessary information on the availability of supply to each warehouse, the requirement of each market, and unit transportation cost (in Rs.) from each warehouse to each market:

The shipping clerk has worked out the following schedule from experience: 12 units from A to Q, 1 unit from A to R, 9 units from A to S, 15 units from B to R, 7 units from C to P, 1 unit from C to R

i. Check and see if the clerk has the optimal schedule

ii. Find the optimal schedule and the minimum total transport cost

The shipping clerk’s schedule is as follows:

(Note: The small numbers in the top left corner of each box indicate the cost, while the numbers in the middle of the box indicate the allocation.)

i. Checking for optimality (MODI rule/ UV rule):

Putting $u_1$=0, and calculating other u and v values:

Note:

• UV values are calculated as follows:

  $U_1$ is assigned an initial value as zero.

  Only cost of cells that have allocations are to be considered. So using $u_1=0$:

  Cost of cell that has allocation = u + v (corresponding to the respective cell)

  For example, consider the cell which has allocation 12, and cost 3.

  Since $u_1=0, 3 = 0 + v_2$; so we get the value of $v_2$ as 3.

• The values in the brackets (of the cells that have no allocations) are calculated by:

  Value in the bracket = Cost of that particular cell – (u + v values corresponding to that cell)

  For example, consider the second row, first column:

  Cost of that particular cell – (u + v values corresponding to that cell)

  = 5 – (-3 + 2)

  = 5 – (-1) = 6, which is the value in the bracket

Now, among all the values in the bracket, there is a value calculated which is less than zero, namely the (-1) in the bottom right corner.

This shows that the solution is not optimal, i.e. the clerk does not have the optimal schedule.

We proceed further by finding out the optimal schedule.

ii. Finding optimal schedule and minimum total transport cost

Draw a loop starting from the (-1), and ending back at it, remembering that an allocation has to be present at the corners of the loop, i.e. the corners of the loop cannot be in an un-allocated cell (besides the starting point i.e. -1)

Assigning alternate (+θ) & (-θ) to the corners of the loop, starting from the (-1):

The value of θ is the minimum allocation from among of all the (-θ) in the table: so θ = 1. Creating the new allocations:

(The figures in bold are the new allocations. The zero can be ignored.)

Again checking for optimality using UV rule (Putting $u_1=0$):w

In the brackets, there are no values present which are zero or less than zero.

So this is the optimal schedule.

Associated minimum cost = 7×5 + 12×3 + 2×5 + 15×2 + 8×4 + 1×6 = Rs. 149