Transportation Problem

Transportation Method

A transportation tableau is given below. Each cell represents a shipping route (which is an arc on the network and a decision variable in the LP formulation), and the unit shipping costs are given in an upper right hand box in the cell.

To solve the transportation problem by its special purpose algorithm, the sum of the supplies at the origins must equal the sum of the demands at the destinations (Balanced transportation problem).

• If the total supply is greater than the total demand, a dummy destination is added with demand equal to the excess supply, and shipping costs from all origins are zero.

• Similarly, if total supply is less than total demand, a dummy origin is added. The supply at the dummy origin is equal to the difference of the total supply and the total demand. The costs associated with the dummy origin are equal to zero.

When solving a transportation problem by its special purpose algorithm, unacceptable shipping routes are given a cost of +M (a large number).

Develop an Initial Solution

Two methods to get the initial solution:

• Northwest Corner Rule

• Minimum Cell-Cost Method

Northwest Corner Rule

1. Starting from the northwest corner of the transportation tableau, allocate as much quantity as possible to cell (1,1) from Origin 1 to Destination 1, within the supply constraint of source 1 and the demand constraint of destination 1.

2. The first allocation will satisfy either the supply capacity of Source 1 or the destination requirement of Destination 1.

ƒ If the demand requirement for destination 1 is satisfied but the supply capacity for Source 1 is not exhausted, move on to cell (1,2) for next allocation.

ƒ If the demand requirement for destination 1 is not satisfied but the supply capacity for Source 1 is exhausted, move to cell (2,1)

ƒ If the demand requirement for Destination 1 is satisfied and the supply capacity for Source 1 is also exhausted, move on to cell (2,2).

3. Continue the allocation in the same manner toward the southeast corner of the transportation tableau until the supply capacities of all sources are exhausted and the demands of all destinations are satisfied.

Initial tableau developed using Northwest Corner Method

Total Cost = 12(400)+13(100)+4(700)+9(100)+12(200)+4(500)= 142,000

Minimum Cell-Cost Method

Although the North-west Corner Rule is the easiest, it is not the most attractive because our objective is not included in the process.

Steps of Minimum Cell-Cost Method

1. Select the cell with the minimum cell cost in the tableau and allocate as much to this cell as possible, but within the supply and demand constraints.

2. Select the cell with the next minimum cell-cost and allocate as much to this cell as possible within the demand and supply constraints.

3. Continue the procedure until all of the supply and demand requirements are satisfied. In a case of tied minimum cell-costs between two or more cells, the tie can be broken by selecting the cell that can accommodate the greater quantity.

Initial tableau developed using Minimum Cell-Cost Method

Total Cost = 12(300)+4(200)+4(700)+10(100)+9(200)+4(500)= 120,000

MODI Method (for obtaining reduced costs)

Associate a number, ui, with each row and vj with each column.

• Step 1: Set u1 = 0.

• Step 2: Calculate the remaining ui's and vj's by solving the relationship cij

= ui + vj for occupied cells.

• Step 3: For unoccupied cells (i,j), the reduced cost = cij - ui - vj.

Step 1: For each unoccupied cell, calculate the reduced cost by the MODI method. Select the unoccupied cell with the most negative reduced cost. (For maximization problems select the unoccupied cell with the largest reduced cost.) If none, STOP.

Step 2: For this unoccupied cell, generate a stepping stone path by forming a closed loop with this cell and occupied cells by drawing connecting alternating horizontal and vertical lines between them. Determine the minimum allocation where a subtraction is to be made along this path.

Step 3: Add this allocation to all cells where additions are to be made, and subtract this allocation to all cells where subtractions are to be made along the stepping stone path. (Note: An occupied cell on the stepping stone path now

becomes 0 (unoccupied). If more than one cell becomes 0, make only one unoccupied; make the others occupied with 0's.)

GO TO STEP 1.

Example: Acme Block Co. (ABC)

Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, and Eastwood -- 10 tons. Acme has two plants, each of which can produce 50 tons per week. Delivery cost per ton from each plant to each suburban location is shown below.

How should end of week shipments be made to fill the above orders?

Since total supply = 100 and total demand = 80, a dummy destination is created with demand of 20 and 0 unit costs.

Iteration 1: Tie for least cost (0), arbitrarily select x14. Allocate 20. Reduce s1 by 20 to 30 and delete the Dummy column.

Iteration 2: Of the remaining cells the least cost is 24 for x11. Allocate 25. Reduce s1 by 25 to 5 and eliminate the Northwood column.

Iteration 3: Of the remaining cells the least cost is 30 for x12. Allocate 5. Reduce the Westwood column to 40 and eliminate the Plant 1 row. Iteration 4: Since there is only one row with two cells left, make the final allocations of 40 and 10 to x22 and x23, respectively.

1. Set u1 = 0

2. Since u1 + vj = c1j for occupied cells in row 1, then v1 = 24, v2 = 30, v4 = 0.

3. Since ui + v2 = ci2 for occupied cells in column 2, then u2 + 30 = 40, hence u2 = 10.

4. Since u2 + vj = c2j for occupied cells in row 2, then 10 + v3 = 42, hence v3 = 32.

Calculate the reduced costs (circled numbers on the previous slide) by cij - ui - vj.

Unoccupied Cell Reduced Cost (1,3) 40 - 0 - 32 = 8 (2,1) 30 - 24 -10 = -4 (2,4) 0 - 10 - 0 = -10

Iteration 1:

The stepping stone path for cell (2,4) is (2,4), (1,4), (1,2), (2,2). The allocations in the subtraction cells are 20 and 40, respectively. The minimum is 20, and hence reallocate 20 along this path. Thus for the next tableau:

x24 = 0 + 20 = 20 (0 is its current allocation) x14 = 20 - 20 = 0 (blank for the next tableau) x12 = 5 + 20 = 25

x22 = 40 - 20 = 20

The other occupied cells remain the same.

1. Set u1 = 0.

2. Since u1 + vj = cij for occupied cells in row 1, then v1 = 24, v2 = 30.

3. Since ui + v2 = ci2 for occupied cells in column 2, then u2 + 30 = 40, or u2 =

10.

4. Since u2 + vj = c2j for occupied cells in row 2, then 10 + v3 = 42 or v3 = 32;

and, 10 + v4 = 0 or v4 = -10.

Iteration 2

Calculate the reduced costs (circled numbers on the previous slide) by cij - ui - vj.

Unoccupied Cell Reduced Cost (1,3) 40 - 0 - 32 = 8 (1,4) 0 - 0 - (-10) = 10 (2,1) 30 - 10 - 24 = -4

The most negative reduced cost is = -4 determined by x21. The stepping stone path for this cell is (2,1),(1,1),(1,2),(2,2). The allocations in the subtraction cells are 25 and 20 respectively. Thus the new solution is obtained by reallocating 20 on the stepping stone path. Thus for the next tableau:

x21 = 0 + 20 = 20 (0 is its current allocation)

x11 = 25 - 20 = 5

x12 = 25 + 20 = 45

x22 = 20 - 20 = 0 (blank for the next tableau)

The other occupied cells remain the same.

1. Set u1 = 0

2. Since u1 + vj = c1j for occupied cells in row 1, then v1 = 24 and v2 = 30.

3. Since ui + v1 = ci1 for occupied cells in column 2, then u2 + 24 = 30 or u2 =6.

4. Since u2 + vj = c2j for occupied cells in row 2, then 6 + v3 = 42 or v3 = 36, and 6 + v4 = 0 or v4 = -6.

Iteration 3

Calculate the reduced costs (circled numbers on the previous slide) by cij - ui - vj.

Unoccupied Cell Reduced Cost

(1,3)	40 - 0 - 36 =	4
(1,4)	0 - 0 - (-6) =	6
(2,2)	40 - 6 - 30 =	4

Since all the reduced costs are non-negative, this is the optimal tableau.