1. Lecture 13: Transportation Introduction AGEC 352 Fall 2012 – October 24 R. Keeney

2. Commodity Properties Based on final use ◦ Form  Products are converted from original to one or more consumable types. ◦ Time  Products are inventoried converting them from current to future consumption possibilities. ◦ Place  Products are moved converting them to consumption possibilities at another location.

3. Classes of problems Production type model: Basic resources are converted to consumable or saleable products. ◦ Ex. Labor and lumber to make chairs & tables. Blending type model: Basic consumables are blended together to meet requirements. ◦ Ex. Combine fertilizers together to make a new product with different composition.

4. Models to date have been about form, now we deal with place Company has two plants and three warehouses (all in different locations) ◦ Must transport the output of the plants to the warehouses ◦ Production capacity is limited at each plant ◦ Demand at each warehouse is limited and each warehouse location faces a different price

5. Transportation Problem Source 1 Source 2 Destination 1 Destination 2 Destination 3 All material must be moved from a source to a destination. Decision variables have two dimensions (from, to) = (source, dest.) Objective coefficients have two dimensions (from, to) = (s,d). Notation P(1,2) = profit per unit from shipping from S1 to D2. X(1,2) = amount moved from shipping from S1 to D2. P(1,2)*X(1,2) = total profit from shipping from S1 to D2. Summing all P*X’s gives total profit for firm.

6. Matrix Formulation Dest. 1Dest. 2Dest. 3 Source 1X(1,1)X(1,2)X(1,3) Source 2X(2,1)X(2,2)X(2,3) Dest. 1Dest. 2Dest. 3 Source 1P(1,1)P(1,2)P(1,3) Source 2P(2,1)P(2,2)P(2,3) Activities Matrix Objective Coefficient Matrix

7. Costs and Objective Values Warehouse 1 (12 $ SP) Warehouse 2 (14 $ SP) Warehouse 3 (15$ SP) Plant 1Cost = 8 Profit = 4 Cost = 10 Profit = 4 Cost = 12 Profit = 3 Plant 2Cost = 7 Profit = 5 Cost = 9 Profit = 5 Cost = 11 Profit = 4

8. Lab 7 Problem Sources Destinations LeipzigNancyLiegeTillburg Amsterdam120.0130.041.059.5 Antwerp61.040.0100.0110.0 Le Havre102.590.0122.042.0 *Could compare these routes or compare sources and destinations *Statistician might average costs from a source or to a destination *What should we do?

9. Information for a Model Location Port CitiesQuantity of Motors to Ship (<=) Amsterdam500 Antwerp700 Le Havre800 Factory CitiesQuantity of Motors to Receive (>=) Leipzig400 Nancy900 Liege200 Tilburg500 *All of the locations are not the same, they have different capacities and requirements. Simple averaging would be incorrect…

10. Problem Size Transportation Problem ◦ S = # of sources ◦ D = # of destinations Then ◦ SxD = # of decision variables ◦ S+D = # of constraints (not counting non- negativity constraints) Problems can get big quickly…

11. Algebraic Simplification *We use subscripts to keep track. We use s to indicate a source and d a destination. *X 23 is a shipment from source 2 to destination 3

12. Spreadsheet Setup Three matrix approach First ◦ Unit cost coefficients (from the data) Second ◦ Decision variables (including consraints) Third ◦ Cost contributions (links the first two and determines the total cost)