2. Transportation, Assignment, and Transshipment Problems Pertemuan 7 Matakuliah: K0442-Metode Kuantitatif Tahun: 2009

3. Bina Nusantara University 3 Material Outline Transportation Problem Network Representation General LP Formulation Assignment Problem Network Representation General LP Formulation Transshipment Problem Network Representation General LP Formulation

4. Transportation, Assignment, and Transshipment Problems 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. Transportation, assignment, and transshipment problems of this chapter as well as the PERT/CPM problems (in another chapter) are all examples of network problems.

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

6. 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.

7. Transportation Problem 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

8. Transportation Problem 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

9. LP Formulation Special Cases The following special-case modifications to the linear programming formulation can be made: –Minimum shipping guarantee from i to j: x ij > L ij –Maximum route capacity from i to j: x ij < L ij –Unacceptable route: Remove the corresponding decision variable. Transportation Problem

10. Example: Acme Block Co. Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, and Eastwood -- 10 tons. Acme has two plants, each of which can produce 50 tons per week. 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? AcmeAcme

11. n Delivery Cost Per Ton Northwood Westwood Eastwood Plant 1 24 30 40 Plant 2 30 40 42 Example: Acme Block Co.

12. n Optimal Solution From To Amount Cost Plant 1 Northwood 5 120 Plant 1 Westwood 45 1,350 Plant 2 Northwood 20 600 Plant 2 Eastwood 10 420 Total Cost = $2,490 Example: Acme Block Co.

13. n Partial Sensitivity Report (first half) Example: Acme Block Co.

14. n Partial Sensitivity Report (second half) Example: Acme Block Co.

15. Assignment Problem 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. It assumes all workers are assigned and each job is performed. 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. The network representation of an assignment problem with three workers and three jobs is shown on the next slide.

16. Assignment Problem 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. LP Formulation Min  c ij x ij i j s.t.  x ij = 1 for each agent i j  x ij = 1 for each task j i 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 1 job. Assignment Problem

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

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

20. An electrical 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 Subcontractor A B C Westside 50 36 16 Federated 28 30 18 Goliath 35 32 20 Universal 25 25 14 How should the contractors be assigned to minimize total mileage costs? Example: Who Does What?

21. n Network Representation 50 36 16 28 30 18 35 32 20 25 25 14 West. CC BB AA Univ.Univ. Gol.Gol. Fed. Fed. Projects Subcontractors

22. Example: Who Does What? n Linear Programming Formulation Min 50x 11 +36x 12 +16x 13 +28x 21 +30x 22 +18x 23 +35x 31 +32x 32 +20x 33 +25x 41 +25x 42 +14x 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 Agents Tasks

23. Example: Who Does What? The optimal assignment is: Subcontractor Project Distance Westside C 16 Federated A 28 Goliath (unassigned) Universal B 25 Total Distance = 69 miles

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

25. Transshipment Problem Network Representation 2 2 33 44 55 66 7 7 1 1 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 Sources Destinations s2s2s2s2 Demand Supply

26. Transshipment Problem Linear Programming Formulation x ij represents the shipment from node i to node j Min  c ij x ij i j s.t.  x ij < s i for each origin i j  x ik -  x kj = 0 for each intermediate i j node k  x ij = d j for each destination j i x ij > 0 for all i and j

27. Example: Zeron Shelving The Northside and Southside facilities of Zeron Industries supply three firms (Zrox, Hewes, Rockrite) 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 Rockrite. Both Arnold and Supershelf can supply at most 75 units to its customers. Additional data is shown on the next slide.

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

29. Example: Zeron Shelving 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 ZeronN ZeronS Rock-Rite

30. Example: Zeron Shelving Linear Programming Formulation –Decision Variables Defined x ij = amount shipped from manufacturer i to supplier j x jk = amount shipped from supplier j to customer k where i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) –Objective Function Defined Minimize Overall Shipping Costs: Min 5x 13 + 8x 14 + 7x 23 + 4x 24 + 1x 35 + 5x 36 + 8x 37 + 3x 45 + 4x 46 + 4x 47

31. Example: Zeron Shelving Constraints Defined Amount Out of Arnold: x 13 + x 14 < 75 Amount Out of Supershelf: x 23 + x 24 < 75 Amount Through Zeron N: x 13 + x 23 - x 35 - x 36 - x 37 = 0 Amount Through Zeron S: 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 Rockrite: x 37 + x 47 = 40 Non-negativity of Variables: x ij > 0, for all i and j.

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

33. Example: Zeron Shelving 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 ZeronN ZeronS Rock-Rite 75 75 50 25 35 40

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

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

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