# Excel

Pages: 62 (16371 words) Published: February 16, 2013
Module

B

Transportation and Assignment Solution Methods

B-1

B-2

Module B

Transportation and Assignment Solution Methods

Solution of the Transportation Model
The following example was used in Chapter 6 of the text to demonstrate the formulation of the transportation model. Wheat is harvested in the Midwest and stored in grain elevators in three different cities—Kansas City, Omaha, and Des Moines. These grain elevators supply three flour mills, located in Chicago, St. Louis, and Cincinnati. Grain is shipped to the mills in railroad cars, each of which is capable of holding one ton of wheat. Each grain elevator is able to supply the following number of tons (i.e., railroad cars) of wheat to the mills on a monthly basis: Grain Elevator 1. Kansas City 2. Omaha 3. Des Moines Total Supply 150 175 275 600 tons

Each mill demands the following number of tons of wheat per month. Mill A. Chicago B. St. Louis C. Cincinnati Total Demand 200 100 300 600 tons

The cost of transporting one ton of wheat from each grain elevator (source) to each mill (destination) differs according to the distance and rail system. These costs are shown in the following table. For example, the cost of shipping one ton of wheat from the grain elevator at Omaha to the mill at Chicago is \$7. Mill Grain Elevator 1. Kansas City 2. Omaha 3. Des Moines A. Chicago \$6 7 4 B. St. Louis \$ 8 11 5 C. Cincinnati \$10 11 12

The problem is to determine how many tons of wheat to transport from each grain elevator to each mill on a monthly basis in order to minimize the total cost of transportation. The linear programming model for this problem is formulated in the equations that follow: minimize Z = \$6x1A + 8x1B + 10x1C + 7x2A + 11x2B + 11x2C + 4x3A + 5x3B + 12x3C subject to x1A + x1B x2A + x2B x3A + x3B x1A + x2A x1B + x2B x1C + x2C + + + + + + x1C = 150 x2C = 175 x3C = 275 x3A = 200 x3B = 100 x3C = 300 xij Ú 0

Solution of the Transportation Model

B-3

Transportation problems are solved manually within a tableau format.

In this model the decision variables, xij, represent the number of tons of wheat transported from each grain elevator, i (where i = 1, 2, 3), to each mill, j (where j = A, B, C). The objective function represents the total transportation cost for each route. Each term in the objective function reflects the cost of the tonnage transported for one route. For example, if 20 tons are transported from elevator 1 to mill A, the cost of \$6 is multiplied by x 1A(=20), which equals \$120. The first three constraints in the linear programming model represent the supply at each elevator; the last three constraints represent the demand at each mill. As an example, consider the first supply constraint, x1A + x1B + x1C = 150. This constraint represents the tons of wheat transported from Kansas City to all three mills: Chicago (x1A), St. Louis (x1B), and Cincinnati (x1C). The amount transported from Kansas City is limited to the 150 tons available. Note that this constraint (as well as all others) is an equation (=) rather than a … inequality because all the tons of wheat available will be needed to meet the total demand of 600 tons. In other words, the three mills demand 600 total tons, which is the exact amount that can be supplied by the three grain elevators. Thus, all that can be supplied will be, in order to meet demand. This type of model, in which supply exactly equals demand, is referred to as a balanced transportation model. The balanced model will be used to demonstrate the solution of a transportation problem. Transportation models are solved manually within the context of a tableau, as in the simplex method. The tableau for our wheat transportation model is shown in Table B-1.

Table B-1 The Transportation Tableau

To From A
6 1 7 2 4 3 Demand 200 100 300 5 12 275 600 11 11 175

B
8

C
10

Supply

150

Each cell in a transportation tableau is analogous to a decision variable that indicates the amount...