1. Linear Programming Example 5 Transportation Problem

2. Transportation Problem The transportation problem seeks to minimize the total shipping costs of transporting goods from m origins (each with a supply s i ) to n destinations (each with a demand d j ), when the unit shipping cost from an origin, i, to a destination, j, is c ij. The network representation for a transportation problem with two sources and three destinations is given on the next slide.

3. Network Representation 2 2 c 11 c 12 c 13 c 21 c 22 c 23 d1d1d1d1 d2d2d2d2 d3d3d3d3 s1s1s1s1 s2s2 SourcesDestinations 3 3 2 2 1 1 1 1

4. LP Formulation The LP formulation in terms of the amounts shipped from the origins to the destinations, x ij, can be written as: Min  c ij x ij i j s.t.  x ij < s i for each origin i j  x ij = d j for each destination j i x ij > 0 for all i and j

5. Example: Acme Block Co. n Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, and Westwood -- 45 tons, and Eastwood -- 10 tons. Eastwood -- 10 tons. n Acme has two plants, each of which can produce 50 tons per week. n Delivery cost per ton from each plant to each suburban location is shown on the next slide. How should end of week shipments be made to fill the above orders?

6. n Delivery Cost Per Ton Northwood Westwood Eastwood Northwood Westwood Eastwood Plant 1 24 30 40 Plant 1 24 30 40 Plant 2 30 40 42 Plant 2 30 40 42 Delivery Cost

7. Decision Variable X 11 : Tons of Concrete shipped from Plan 1 to Northwood X 12 : Tons of Concrete shipped from Plan 1 to Westwood X 13 : Tons of Concrete shipped from Plan 1 to Eastwood X 21 : Tons of Concrete shipped from Plan 2 to Northwood X 22 : Tons of Concrete shipped from Plan 2 to Westwood X 13 : Tons of Concrete shipped from Plan 2 to Eastwood Objective Function Min. 24X 11 +30X 12 +40X 13 +30X 21 +40X 22 +42X 23 LP Model

8. Constraints X 11 + X 12 + X 13 <= 50Plant 1 X 21 + X 22 + X 23 <= 50Plant 2 X 11 +X 21 = 25Northwood X 12 +X 22 = 45Westwood X 13 +X 23 = 10Eastwood Non-Negativity X 11, X 12, X 13, X 21, X 22, X 23 >=0 LP Model

9. Excel Input

10. Solver Solution

11. n Optimal Solution From To Amount Cost From To Amount Cost Plant 1 Northwood 5 120 Plant 1 Westwood 45 1,350 Plant 1 Westwood 45 1,350 Plant 2 Northwood 20 600 Plant 2 Northwood 20 600 Plant 2 Eastwood 10 420 Plant 2 Eastwood 10 420 Total Cost = $2,490 Total Cost = $2,490 Solution

12. n Partial Sensitivity Report (first half) Sensitivity Report

13. n Partial Sensitivity Report (second half) Sensitivity Report