1. 1 1 Slide Transportation, Assignment, and Transshipment Professor Ahmadi

2. 2 2 Slide Chapter 7 Transportation, Assignment, and Transshipment Problems n The Transportation Problem: The Network Model and a Linear Programming Formulation n The Assignment Problem: The Network Model and a Linear Programming Formulation n The Transshipment Problem: The Network Model and a Linear Programming Formulation

3. 3 3 Slide Transportation, Assignment, and Transshipment Problems n A network model is one which can be represented by a set of nodes, a set of arcs, and functions (e.g. costs, supplies, demands, etc.) associated with the arcs and/or nodes. n Transportation, assignment, and transshipment problems are all examples of network problems.

4. 4 4 Slide Transportation, Assignment, and Transshipment Problems n Each of the three models of this chapter (transportation, assignment, and transshipment models) can be formulated as linear programs. n For each of the three models, if the right-hand side of the linear programming formulations are all integers, the optimal solution will be in terms of integer values for the decision variables. n These three models can also be solved using a standard computer spreadsheet package.

5. 5 5 Slide Transportation Problem n 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. n The network representation for a transportation problem with two sources and three destinations is given on the next slide.

6. 6 6 Slide Transportation Problem n Network Representation 1 1 2 2 3 3 1 1 2 2 c 11 c 12 c 13 c 21 c 22 c 23 d1d1d1d1 d2d2d2d2 d3d3d3d3 s1s1s1s1 s2s2 SOURCESDESTINATIONS

7. 7 7 Slide Transportation Problem n LP Formulation The linear programming formulation in terms of the amounts shipped from the origins to the destinations, x ij, can be written as: Min  c ij x ij Min  c ij x ij i j i j s.t.  x ij < s i for each origin i s.t.  x ij < s i for each origin i j  x ij = d j for each destination j  x ij = d j for each destination j i x ij > 0 for all i and j x ij > 0 for all i and j

8. 8 8 Slide Transportation Problem n LP Formulation Special Cases The following special-case modifications to the linear programming formulation can be made: Minimum shipping guarantees from i to j :Minimum shipping guarantees from i to j : x ij > L ij x ij > L ij Maximum route capacity from i to j :Maximum route capacity from i to j : x ij < L ij x ij < L ij Unacceptable routes:Unacceptable routes: delete the variable delete the variable

9. 9 9 Slide Example: BBC-1 Building Brick Company (BBC) has orders for 80 tons of bricks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, and Eastwood -- 10 tons. BBC has two plants. Plant 1 produces 50 and plant 2 produces 30 tons per week. How should end of week shipments be made to fill the above orders given the following 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

10. 10 Slide Example: BBC-1 n LP Formulation Decision Variables DefinedDecision Variables Defined x ij = amount shipped from plant i to suburb j x ij = amount shipped from plant i to suburb j where i = 1 (Plant 1) and 2 (Plant 2) where i = 1 (Plant 1) and 2 (Plant 2) j = 1 (Northwood), 2 (Westwood), j = 1 (Northwood), 2 (Westwood), and 3 (Eastwood) and 3 (Eastwood)

11. 11 Slide Transportation Problem n Network Representation of BBC-1 Northwood 1 Northwood 1 Westwood 2 Westwood 2 Eastwood 3 Eastwood 3 Plant 1 Plant 1 Plant 2 Plant 2 24 30 40 30 40 42 25 45 10 50 30 SOURCES DESTINATIONS

12. 12 Slide Example: BBC-1 n LP Formulation Objective FunctionObjective Function Minimize total shipping cost per week: Min 24 x 11 + 30 x 12 + 40 x 13 + 30 x 21 + 40 x 22 + 42 x 23 Min 24 x 11 + 30 x 12 + 40 x 13 + 30 x 21 + 40 x 22 + 42 x 23 ConstraintsConstraints s.t. x 11 + x 12 + x 13 < 50 (Plant 1 capacity) s.t. x 11 + x 12 + x 13 < 50 (Plant 1 capacity) x 21 + x 22 + x 23 < 30 (Plant 2 capacity) x 11 + x 21 = 25 (Northwood demand) x 11 + x 21 = 25 (Northwood demand) x 12 + x 22 = 45 (Westwood demand) x 12 + x 22 = 45 (Westwood demand) x 13 + x 23 = 10 (Eastwood demand) x 13 + x 23 = 10 (Eastwood demand) all x ij > 0 (Non-negativity) all x ij > 0 (Non-negativity)

13. 13 Slide Example: BBC-1 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

14. 14 Slide Assignment Problem n An assignment problem seeks to minimize the total cost assignment of m workers to m jobs, given that the cost of worker i performing job j is c ij. n It assumes all workers are assigned and each job is performed. n An assignment problem is a special case of a transportation problem in which all supplies and all demands are equal to 1; hence assignment problems may be solved as linear programs. n The network representation of an assignment problem with three workers and three jobs is shown on the next slide.

15. 15 Slide Assignment Problem n Network Representation 22 33 11 22 33 11 c 11 c 12 c 13 c 21 c 22 c 23 c 31 c 32 c 33 WORKERSJOBS

16. 16 Slide Assignment Problem n Linear Programming Formulation Min  c ij x ij Min  c ij x ij i j i j s.t.  x ij = 1 for each worker i s.t.  x ij = 1 for each worker i j  x ij = 1 for each job j  x ij = 1 for each job j i x ij = 0 or 1 for all i and j. x ij = 0 or 1 for all i and j. Note: A modification to the right-hand side of the first constraint set can be made if a worker is permitted to work more than one job.Note: A modification to the right-hand side of the first constraint set can be made if a worker is permitted to work more than one job.

17. 17 Slide Example: Assignment A contractor pays his subcontractors a fixed fee plus mileage for work performed. On a given day the contractor is faced with three electrical jobs associated with various projects. Given below are the distances between the subcontractors and the projects. Project Project A B C A B C Westside 50 36 16 Westside 50 36 16 Subcontractors Federated 28 30 18 Subcontractors Federated 28 30 18 Goliath 35 32 20 Goliath 35 32 20 Universal 25 25 14 Universal 25 25 14 How should the contractors be assigned to minimize total distance (and total cost)?

18. 18 Slide Example: Assignment n Network Representation 50 36 16 28 30 18 35 32 20 25 25 14 West.West. CC BB AA Univ.Univ. Gol.Gol. Fed.Fed.

19. 19 Slide Example: Assignment n LP Formulation Decision Variables DefinedDecision Variables Defined x ij = 1 if subcontractor i is assigned to project j = 0 otherwise = 0 otherwise where: i = 1 (Westside), 2 (Federated), where: i = 1 (Westside), 2 (Federated), 3 (Goliath), and 4 (Universal) 3 (Goliath), and 4 (Universal) j = 1 (A), 2 (B), and 3 (C) j = 1 (A), 2 (B), and 3 (C)

20. 20 Slide Example: Assignment n LP Formulation Objective FunctionObjective Function Minimize total distance: Min 50 x 11 + 36 x 12 + 16 x 13 + 28 x 21 + 30 x 22 + 18 x 23 + 35 x 31 + 32 x 32 + 20 x 33 + 25 x 41 + 25 x 42 + 14 x 43 + 35 x 31 + 32 x 32 + 20 x 33 + 25 x 41 + 25 x 42 + 14 x 43

21. 21 Slide Example: Assignment n LP Formulation ConstraintsConstraints x 11 + x 12 + x 13 < 1 (no more than one x 11 + x 12 + x 13 < 1 (no more than one x 21 + x 22 + x 23 < 1 project assigned x 21 + x 22 + x 23 < 1 project assigned x 31 + x 32 + x 33 < 1 to any one x 41 + x 42 + x 43 < 1 subcontractor) x 41 + x 42 + x 43 < 1 subcontractor) x 11 + x 21 + x 31 + x 41 = 1 (each project must x 12 + x 22 + x 32 + x 42 = 1 be assigned to just x 13 + x 23 + x 33 + x 43 = 1 one subcontractor) all x ij > 0 (non-negativity)

22. 22 Slide Example: Assignment n Optimal Assignment Subcontractor Project Distance Subcontractor Project Distance Westside C 16 Westside C 16 Federated A 28 Federated A 28 Universal B 25 Universal B 25 Goliath (unassigned) Goliath (unassigned) Total Distance = 69 miles Total Distance = 69 miles

23. 23 Slide Variations of Assignment Problem n Total number of agents not equal to total number of tasks n Maximization objective function n Unacceptable assignments

24. 24 Slide Transshipment Problem n Transshipment problems are transportation problems in which a shipment may move through intermediate nodes (transshipment nodes)before reaching a particular destination node. n Transshipment problems can be converted to larger transportation problems and solved by a special transportation program. n Transshipment problems can also be solved as linear programs. n The network representation for a transshipment problem with two sources, three intermediate nodes, and two destinations is shown on the next slide.

25. 25 Slide Transshipment Problem n Network Representation 22 33 44 55 66 77 11 c 13 c 14 c 23 c 24 c 25 c 15 s1s1s1s1 c 36 c 37 c 46 c 47 c 56 c 57 d1d1d1d1 d2d2d2d2 INTERMEDIATE NODES NODESSOURCESDESTINATIONS s2s2s2s2

26. 26 Slide Transshipment Problem n Linear Programming Formulation x ij represents the shipment from node i to node j x ij represents the shipment from node i to node j Min  c ij x ij Min  c ij x ij all arcs all arcs s.t.  x ij -  x ij < s i for each origin node i s.t.  x ij -  x ij < s i for each origin node i arcs out arcs in arcs out arcs in  x ij -  x ij = 0 for each intermediate  x ij -  x ij = 0 for each intermediate arcs out arcs in node arcs out arcs in node  x ij -  x ij = - d j for each destination arcs out arcs in node j (Note the order)  x ij -  x ij = - d j for each destination arcs out arcs in node j (Note the order) x ij > 0 for all i and j x ij > 0 for all i and j

27. 27 Slide Example: Transshipping Thomas Industries and Washburn Corporation supply three firms (Zrox, Hewes, Rockwright) with customized shelving for its offices. They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc. Currently weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockwright. Both Arnold and Supershelf can supply at most 75 units to its customers. Additional data is shown on the next slide.

28. 28 Slide Example: Transshipping Because of long standing contracts based on past orders, unit costs from the manufacturers to the suppliers are: Thomas Washburn Thomas Washburn Arnold 5 8 Arnold 5 8 Supershelf 7 4 Supershelf 7 4 The cost to install the shelving at the various locations are: Zrox Hewes Rockwright Zrox Hewes Rockwright Thomas 1 5 8 Thomas 1 5 8 Washburn 3 4 4 Washburn 3 4 4

29. 29 Slide Example: Transshipping n Network Representation ARNOLD WASH BURN ZROX HEWES 75 75 50 60 40 5 8 7 4 1 5 8 3 4 4 Arnold1 Supershelf2 Hewes6 Zrox5 Thomas3 Washburn4 Rockwright7

30. 30 Slide Example: Transshipping n LP Formulation Decision Variables DefinedDecision Variables Defined x ij = amount shipped from manufacturer i to supplier j x ij = amount shipped from manufacturer i to supplier j x jk = amount shipped from supplier j to customer k x jk = amount shipped from supplier j to customer k where i = 1 (Arnold), 2 (Supershelf) where i = 1 (Arnold), 2 (Supershelf) j = 3 (Thomas), 4 (Washburn) j = 3 (Thomas), 4 (Washburn) k = 5 (Zrox), 6 (Hewes), 7 (Rockwright) k = 5 (Zrox), 6 (Hewes), 7 (Rockwright) Objective Function DefinedObjective Function Defined Minimize Overall Shipping Costs: Min 5 x 13 + 8 x 14 + 7 x 23 + 4 x 24 + 1 x 35 + 5 x 36 + 8 x 37 Min 5 x 13 + 8 x 14 + 7 x 23 + 4 x 24 + 1 x 35 + 5 x 36 + 8 x 37 + 3 x 45 + 4 x 46 + 4 x 47 + 3 x 45 + 4 x 46 + 4 x 47

31. 31 Slide Example: Transshipping n Constraints Defined Amount out of Arnold: x 13 + x 14 < 75 Amount out of Arnold: x 13 + x 14 < 75 Amount out of Supershelf: x 23 + x 24 < 75 Amount out of Supershelf: x 23 + x 24 < 75 Amount through Thomas: x 13 + x 23 - x 35 - x 36 - x 37 = 0 Amount through Thomas: x 13 + x 23 - x 35 - x 36 - x 37 = 0 Amount through Washburn: x 14 + x 24 - x 45 - x 46 - x 47 = 0 Amount through Washburn: x 14 + x 24 - x 45 - x 46 - x 47 = 0 Amount into Zrox: x 35 + x 45 = 50 Amount into Zrox: x 35 + x 45 = 50 Amount into Hewes: x 36 + x 46 = 60 Amount into Hewes: x 36 + x 46 = 60 Amount into Rockwright: x 37 + x 47 = 40 Amount into Rockwright: x 37 + x 47 = 40 Non-negativity of variables: x ij > 0, for all i and j. Non-negativity of variables: x ij > 0, for all i and j.

32. 32 Slide Variations of Transshipment Problem n Total supply not equal to total demand n Maximization objective function n Route capacities or route minimums n Unacceptable routes

33. 33 Slide The End of Chapter