Transportation and Distribution Planning Matthew J. Liberatore John F. Connelly Chair in Management Professor, Decision and Information Techologies

TRANSPORTATION PROBLEM Mathematical programming has been successfully applied to important supply chain problems. These problems address the movement of products across links of the supply chain (supplier, manufacturers, and customers). We now focus on supply chain applications in transportation and distribution planning.

TRANSPORTATION PROBLEM A manufacturer ships TV sets from three warehouses to four retail stores each week. Warehouse capacities (in hundreds) and demand (in hundreds) at the retail stores are as follows: Capacity Demand Capacity Demand Warehouse 1200Store 1100 Warehouse 2 150Store Warehouse 3 300Store Store

The shipping cost per hundred TV sets for each route is given below: To To FromStore 1 Store 2 Store 3 Store 4 FromStore 1 Store 2 Store 3 Store 4 warehouse 1$ warehouse warehouse TRANSPORTATION PROBLEM

What are the decision variables? XIJ=number of TV sets (in cases) shipped from warehouse I to store J I is the index for warehouses (1,2,3) J is the index for stores (1,2,3,4) TRANSPORTATION PROBLEM

What is the objective? Minimize the total cost of transportation which is obtained by multiplying the shipping cost by the amount of TV sets shipped over a given route and then summing over all routes OBJECTIVE FUNCTION ; MIN = 10*X11+5*X12+12*X13+ 3*X14+ 4* X21+9*X22+15*X23+ 6*X24+ 4* X21+9*X22+15*X23+ 6*X24+ 15*X31+8*X32+ 6*X33+11*X34 15*X31+8*X32+ 6*X33+11*X34 TRANSPORTATION PROBLEM

How are the supply constraints expressed? For each warehouse the amount of TV sets shipped to all stores must equal the capacity at the warehouse X11+X12+X13+X14=200; SUPPLY CONSTRAINT FOR WAREHOUSE 1 X21+X22+X23+X24=150; SUPPLY CONSTRAINT FOR WAREHOUSE 2 X31+X32+X33+X34=300; SUPPLY CONSTRAINT FOR WAREHOUSE 3 TRANSPORTATION PROBLEM

How are the demand constraints expressed? For each store the amount of TV sets shipped from all warehouses must equal the demand of the store X11+X21+X31=100; DEMAND CONSTRAINT FOR STORE 1 X12+X22+X32=200; DEMAND CONSTRAINT FOR STORE 2 X13+X23+X33=125; DEMAND CONSTRAINT FOR STORE 3 X14+X24+X34=225; DEMAND CONSTRAINT FOR STORE 4 TRANSPORTATION PROBLEM

Since partial shipment cannot be made, the decision variables must be integer –valued However, if all supplies and demands are integer-valued, the values of our decision variables will be integer valued TRANSPORTATION PROBLEM

After solution in Solver: The total shipment cost is $3500, and the optimal shipments are: warehouse 1 ships 25 cases to store 2 and 175 to store 4; warehouse 2 ships 100 to store 1 and 50 to store 4, and warehouse 3 ships 175 to store 2 and 125 to store 3. The reduced cost of X11 is 9, so the cost of shipping from warehouse 1 to store 1 would have to be reduced by $9 before this route would be used TRANSPORTATION PROBLEM

UNBALANCED PROBLEMS Suppose warehouse 2 actually has 175 TV sets. How should the original problem be modified? Since total supply across all warehouses is now greater than total demand, all supply constraints are now “<=“ Since total supply across all warehouses is now greater than total demand, all supply constraints are now “<=“ Referring to the original problem, suppose store 3 needs 150 TV sets. How should the original problem be modified? The demand constraints are now “<=“

RESTRICTED ROUTE Referring to the original problem, suppose there is a strike by the shipping company such that the route from warehouse 3 to store 2 cannot be used. How can the original problem be modified to account for this change? Add the constraint: X32=0 X32=0

WAREHOUSE LOCATION Suppose that the warehouses are currently not open, but are potential locations. The fixed cost to construct warehouses and their capacity values are given as: WAREHOUSES FIXED COST CAPACITY Warehouse 1125, Warehouse 2185, Warehouse 3100,000325

How do we model the fact that the warehouses may or may not be open? Define a set of binary decision variables YI, I =1,2,3, where warehouse I is open if YI = 1 and warehouse I is closed if YI = 0 WAREHOUSE LOCATION

How must the objective function change? Additional terms are added to the objective function which multiply the fixed costs of operating the warehouse by YI and summing over all warehouses I: *Y *Y *Y3 Why can’t we use the current capacity constraints? Product cannot be shipped from a warehouse if it is not open. Since the capacity is available only if the warehouse is open, we multiply warehouse 1’s capacity by Y1. WAREHOUSE LOCATION

Also, we must make the YI variables binary integer Total fixed and shipping costs are $289,100; warehouses 2 and 3 are open; warehouse 2 ships 100 to store 1, 225 to store 4; and warehouse 3 ships 200 to store 2 and 125 to store 4 Total fixed and shipping costs are $289,100; warehouses 2 and 3 are open; warehouse 2 ships 100 to store 1, 225 to store 4; and warehouse 3 ships 200 to store 2 and 125 to store 4 WAREHOUSE LOCATION