1. 1 1 Slide © 2001 South-Western College Publishing/Thomson Learning Anderson Sweeney Williams Anderson Sweeney Williams Slides Prepared by JOHN LOUCKS QUANTITATIVE METHODS FOR BUSINESS 8e QUANTITATIVE METHODS FOR BUSINESS 8e

2. 2 2 Slide Chapter 10 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 of this chapter, as well as PERT/CPM problems (Chapter 12), are all examples of network problems.

4. 4 4 Slide Transportation, Assignment, and Transshipment Problems n Each of the three models of this chapter can be formulated as linear programs and solved by general purpose linear programming codes. 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 However, there are many computer packages (including The Management Scientist ) which contain separate computer codes for these models which take advantage of their network structure.

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 guarantee from i to j :Minimum shipping guarantee 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 route:Unacceptable route: Remove the corresponding decision variable. Remove the corresponding decision variable.

9. 9 9 Slide Example: BBC 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, each of which can produce 50 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 n Partial Spreadsheet Showing Problem Data

11. 11 Slide Example: BBC n Partial Spreadsheet Showing Optimal Solution

12. 12 Slide 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 Example: BBC

13. 13 Slide Example: BBC n Partial Sensitivity Report (first half)

14. 14 Slide Example: BBC n Partial Sensitivity Report (second half)

15. 15 Slide Assignment Problem n An assignment problem seeks to minimize the total cost assignment of m agents to m tasks, given that the cost of agent i performing task j is c ij. n It assumes all agents are assigned and each task 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 agents and three tasks is shown on the next slide.

16. 16 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 AGENTSTASKS

17. 17 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 agent i s.t.  x ij = 1 for each agent i j  x ij = 1 for each task j  x ij = 1 for each task j i x ij = 0 or 1 for all i and j. x ij = 0 or 1 for all i and j.

18. 18 Slide Assignment Problem n LP Formulation Special Cases Number of agents exceeds the number of tasks:Number of agents exceeds the number of tasks:  x ij < 1 for each agent i j Number of tasks exceeds the number of agents:Number of tasks exceeds the number of agents: Add enough dummy agents to equalize the Add enough dummy agents to equalize the number of agents and the number of tasks. number of agents and the number of tasks. The objective function coefficients for these The objective function coefficients for these new variable would be zero. new variable would be zero.

19. 19 Slide Assignment Problem n LP Formulation Special Cases (continued) The assignment alternatives are evaluated in terms of revenue or profit:The assignment alternatives are evaluated in terms of revenue or profit: Solve as a maximization problem. Solve as a maximization problem. An assignment is unacceptable:An assignment is unacceptable: Remove the corresponding decision variable. Remove the corresponding decision variable. An agent is permitted to work a tasks:An agent is permitted to work a tasks:  x ij < a for each agent i j

20. 20 Slide Example: Hungry Owner 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. Projects Projects 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 costs?

21. 21 Slide Example: Hungry Owner 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. Projects Subcontractors

22. 22 Slide Example: Hungry Owner n Linear Programming Formulation 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 s.t. x 11 + x 12 + x 13 < 1 x 21 + x 22 + x 23 < 1 x 31 + x 32 + x 33 < 1 x 41 + x 42 + x 43 < 1 x 11 + x 21 + x 31 + x 41 = 1 x 12 + x 22 + x 32 + x 42 = 1 x 13 + x 23 + x 33 + x 43 = 1 x ij = 0 or 1 for all i and j x ij = 0 or 1 for all i and j Agents Tasks

23. 23 Slide Example: Hungry Owner n Optimal Assignment Subcontractor Project Distance Subcontractor Project Distance Westside C 16 Westside C 16 Federated A 28 Federated A 28 Goliath (unassigned) Universal B 25 Total Distance = 69 miles Total Distance = 69 miles

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 by general purpose linear programming codes. 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 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 ik -  x kj = 0 for each intermediate  x ik -  x kj = 0 for each intermediate i j node k i j node k  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

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 Arnold SuperShelf Hewes Zrox Thomas Wash-Burn Rock-Wright

30. 30 Slide Example: Transshipping n Linear Programming Formulation Decision Variables DefinedDecision Variables Defined 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 Supershelf: x 23 + x 24 < 75 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 Into Zrox: x 35 + x 45 = 50 Amount Into Hewes: x 36 + x 46 = 60 Amount Into Rockwright: x 37 + x 47 = 40 Non-negativity of Variables: x ij > 0, for all i and j.

32. 32 Slide Example: Transshipping n Optimal Solution (from Management Scientist ) Objective Function Value = 1150.000 Objective Function Value = 1150.000 Variable Value Reduced Costs Variable Value Reduced Costs X13 75.000 0.000 X13 75.000 0.000 X14 0.000 2.000 X14 0.000 2.000 X23 0.000 4.000 X23 0.000 4.000 X24 75.000 0.000 X24 75.000 0.000 X35 50.000 0.000 X35 50.000 0.000 X36 25.000 0.000 X36 25.000 0.000 X37 0.000 3.000 X37 0.000 3.000 X45 0.000 3.000 X45 0.000 3.000 X46 35.000 0.000 X46 35.000 0.000 X47 40.000 0.000 X47 40.000 0.000

33. 33 Slide Example: Transshipping n Optimal Solution ARNOLD WASH BURN ZROX HEWES 75 75 50 60 40 5 8 7 4 1 5 8 3 4 4 Arnold SuperShelf Hewes Zrox Thomas Wash-Burn Rock-Wright 75 75 50 25 35 40

34. 34 Slide Example: Transshipping n Optimal Solution (continued) Constraint Slack/Surplus Dual Prices Constraint Slack/Surplus Dual Prices 1 0.000 0.000 1 0.000 0.000 2 0.000 2.000 2 0.000 2.000 3 0.000 -5.000 3 0.000 -5.000 4 0.000 -6.000 4 0.000 -6.000 5 0.000 -6.000 5 0.000 -6.000 6 0.000 -10.000 6 0.000 -10.000 7 0.000 -10.000 7 0.000 -10.000

35. 35 Slide Example: Transshipping n Optimal Solution (continued) OBJECTIVE COEFFICIENT RANGES OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit Variable Lower Limit Current Value Upper Limit X13 3.000 5.000 7.000 X13 3.000 5.000 7.000 X14 6.000 8.000 No Limit X14 6.000 8.000 No Limit X23 3.000 7.000 No Limit X23 3.000 7.000 No Limit X24 No Limit 4.000 6.000 X24 No Limit 4.000 6.000 X35 No Limit 1.000 4.000 X35 No Limit 1.000 4.000 X36 3.000 5.000 7.000 X36 3.000 5.000 7.000 X37 5.000 8.000 No Limit X37 5.000 8.000 No Limit X45 0.000 3.000 No Limit X45 0.000 3.000 No Limit X46 2.000 4.000 6.000 X46 2.000 4.000 6.000 X47 No Limit 4.000 7.000 X47 No Limit 4.000 7.000

36. 36 Slide Example: Transshipping n Optimal Solution (continued) RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit Constraint Lower Limit Current Value Upper Limit 1 75.000 75.000 No Limit 1 75.000 75.000 No Limit 2 75.000 75.000 100.000 2 75.000 75.000 100.000 3 -75.000 0.000 0.000 3 -75.000 0.000 0.000 4 -25.000 0.000 0.000 4 -25.000 0.000 0.000 5 0.000 50.000 50.000 5 0.000 50.000 50.000 6 35.000 60.000 60.000 6 35.000 60.000 60.000 7 15.000 40.000 40.000 7 15.000 40.000 40.000

37. 37 Slide The End of Chapter 10