the transportation models

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The Transportation Models

Transportation model is one of the class of linear programming models, so named because of the linear relationships among variables in transportation model, transportation cost are treated as a direct linear function of the number of units shipped.

Transportation models can be used to determine how to allocate the supplies available from various factories to the warehouses that stock or demand those goods, in such a way that total shipping cost (time, distance) is minimized. The shipping (supply) points can be factories, warehouses, department or any other place from which goods are sent. The destination (demand) can be factories, warehouses, departments or other place that receive goods.

Major assumptions for the transportation model

1. The items to be shipped are homogeneous. 2. Shipping cost per unit is the same regardless of the number of units shipped. 3. There is only one route or mode of transportation being used between each source and each destination.

The information needed to use the model

1. A list of the origins and each ones capacity or supply quantity per period 2. A list of the destinations and each ones demand per period 3. The unit cost of shipping items from each origin to each destination.

The transportation model starts with the development of a feasible solution, which is then sequentially tested and improved until an optimal solution is obtained.

Major Steps 1.Obtaining an initial solution 2.Testing the solution for optimality 3.Improving sub optimal solution

Approaches

1. Intuitive Lowest cost approach 2. North West Corner stepping stone method 3. Modified Transportation Approach

Intuitive Lowest-Cost MethodWith the intuitive approach, cell allocations are made according to cell cost, beginning with the lowest cost. The procedure involves these steps:

Intuitive Lowest-Cost Method1. Identify the cell with the lowest cost 2. Allocate as many units as possible to that cell without exceeding supply or demand; then cross out the row or column (or both) that is exhausted by this assignment

3. Find the cell with the lowest cost from the remaining cells4. Repeat steps 2 and 3 until all units have been allocated

Intuitive Lowest-Cost MethodTo From Des Moines Evansville Fort Lauderdale Albuquerque \$5 \$8 \$9 Boston \$4 \$4 \$7 Cleveland \$3 \$3 \$5

Intuitive Lowest-Cost MethodDes Moines (100 units capacity) Cleveland (200 units required) Boston (200 units required)

Albuquerque (300 units required)

Evansville (300 units capacity) Fort Lauderdale (300 units capacity)

Figure C.1

Transportation MatrixFigure C.2 To From Des Moines Albuquerque Boston Cleveland Factory capacity Des Moines capacity constraint Cell representing a possible source-todestination shipping assignment (Evansville to Cleveland)

\$5 \$8 \$9

\$4 \$4 \$7

\$3 \$3 \$5

100 300

Evansville

Fort Lauderdale Warehouse requirement

300 700

300

200

200

Cost of shipping 1 unit from Fort Lauderdale factory to Boston warehouse

Cleveland warehouse demand

Total demand and total supply

Intuitive Lowest-Cost MethodTo From (D) Des Moines (A) Albuquerque (B) Boston (C) Cleveland

Factory capacity

\$5 \$8 \$9

\$4 \$4 \$7

100

\$3 \$3 \$5

100300 300 700

(E) Evansville

(F) Fort Lauderdale Warehouse requirement

300

200

200

First, \$3 is the lowest cost cell so ship 100 units from Des Moines to Cleveland and cross off the first row as Des Moines is satisfiedFigure C.4

Intuitive Lowest-Cost MethodTo From (D) Des Moines (A) Albuquerque (B) Boston (C) Cleveland

Factory capacity

\$5 \$8 \$9

\$4 \$4 \$7

100 100

\$3 \$3 \$5

100300 300 700

(E) Evansville

(F) Fort Lauderdale Warehouse requirement

300

200

200

Second, \$3 is again the lowest cost cell so ship 100 units from Evansville to Cleveland and cross off column C as Cleveland is satisfiedFigure C.4

Intuitive Lowest-Cost MethodTo From (D) Des Moines (A) Albuquerque (B) Boston (C) Cleveland

Factory capacity

\$5 \$8 \$9

\$4 \$4 \$7

100 100

\$3 \$3 \$5

100300 300 700

(E) Evansville

200

(F) Fort Lauderdale Warehouse requirement

300

200

200

Third, \$4 is the lowest cost cell so ship 200 units from Evansville to Boston and cross off column B and row E as Evansville and Boston are satisfiedFigure C.4

Intuitive Lowest-Cost MethodTo From (D) Des Moines (A) Albuquerque (B) Boston (C) Cleveland

Factory capacity

\$5 \$8 \$9

\$4 \$4 \$7

100 100

\$3 \$3 \$5

100300 300 700

(E) Evansville

200

(F) Fort Lauderdale Warehouse requirement

300 300

200

200

Finally, ship 300 units from Albuquerque to Fort Lauderdale as this is the only remaining cell to complete the allocationsFigure C.4

Intuitive Lowest-Cost MethodTo From (D) Des Moines (A) Albuquerque (B) Boston (C) Cleveland

Factory capacity

\$5 \$8 \$9

\$4 \$4 \$7

100 100

\$3 \$3 \$5

100300 300 700

(E) Evansville

200

(F) Fort Lauderdale Warehouse requirement

300 300

200

200

Total Cost

= \$3(100) + \$3(100) + \$4(200) + \$9(300) = \$4,100Figure C.4

Intuitive Lowest-Cost MethodTo From (A) Albuquerque (B) Boston (C) Cleveland

Factory capacity

(D) Des Moines improvement over the previous solution, (E) Evansville

\$5 This is a feasible solution, and an

\$4 \$4 \$7

100 100

\$3 \$3 \$5

100300 300 700

but not necessarily the lowest cost \$8 alternative \$9

200

(F) Fort Lauderdale Warehouse requirement

300 300

200

200

Total Cost

= \$3(100) + \$3(100) + \$4(200) + \$9(300) = \$4,100Figure C.4

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