1. 1 1 Slide © 2008 Thomson South-Western. All Rights Reserved Slides by JOHN LOUCKS St. Edward’s University

2. 2 2 Slide © 2008 Thomson South-Western. All Rights Reserved Chapter 6, Part A Distribution and Network Models n Transportation Problem Network Representation Network Representation General LP Formulation General LP Formulation n Assignment Problem Network Representation Network Representation General LP Formulation General LP Formulation n Transshipment Problem Network Representation Network Representation General LP Formulation General LP Formulation

3. 3 3 Slide © 2008 Thomson South-Western. All Rights Reserved 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, transshipment, shortest-route, and maximal flow problems of this chapter as well as the minimal spanning tree and PERT/CPM problems (in others chapter) are all examples of network problems.

4. 4 4 Slide © 2008 Thomson South-Western. All Rights Reserved Transportation, Assignment, and Transshipment Problems n Each of the five models of this chapter can be formulated as linear programs and solved by general purpose linear programming codes. n For each of the five 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 ) that contain separate computer codes for these models which take advantage of their network structure.

5. 5 5 Slide © 2008 Thomson South-Western. All Rights Reserved 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 © 2008 Thomson South-Western. All Rights Reserved Transportation Problem n 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

7. 7 7 Slide © 2008 Thomson South-Western. All Rights Reserved Transportation Problem n Linear Programming Formulation Using the notation: Using the notation: x ij = number of units shipped from x ij = number of units shipped from origin i to destination j origin i to destination j c ij = cost per unit of shipping from c ij = cost per unit of shipping from origin i to destination j origin i to destination j s i = supply or capacity in units at origin i s i = supply or capacity in units at origin i d j = demand in units at destination j d j = demand in units at destination j continued

8. Modeling source supply capacity xi1xi1xi1xi1 ii Total supply from source i cannot be more than the capacity at source i: x i 1 + x i 2 + x i 3 + x i 4 xi2xi2xi2xi2 xi3xi3xi3xi3 xi4xi4xi4xi4

9. Modeling destination demand x1jx1jx1jx1j jj Total arrivals to destination j have to be at least equal to the demand at destination j: x 1j + x 2j + x 3j + x 4j x2jx2jx2jx2j x3jx3jx3jx3j x4jx4jx4jx4j

10. 10 Slide © 2008 Thomson South-Western. All Rights Reserved x ij > 0 for all i and j Transportation Problem n Linear Programming Formulation (continued)

11. 11 Slide © 2008 Thomson South-Western. All Rights Reserved n LP Formulation Special Cases The objective is maximizing profit or revenue: The objective is maximizing profit or revenue: 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. Transportation Problem Solve as a maximization problem.

12. 12 Slide © 2008 Thomson South-Western. All Rights Reserved Transportation Problem: Example #1 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

13. 13 Slide © 2008 Thomson South-Western. All Rights Reserved n 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 Transportation Problem: Example #1 n Total Delivery Costs D 1 (Nw) D2 (Ww) ED3 (Ew) D 1 (Nw) D2 (Ww) ED3 (Ew) Plant 1 24x 11 30x 12 40x 13 Plant 1 24x 11 30x 12 40x 13 Plant 2 30x 21 40x 22 42x 23 Plant 2 30x 21 40x 22 42x 23

14. Network diagram for Acme 2 2 c 11 x 11 c 12 x 12 c 13 x 13 c 21 x 21 c 22 x 22 c 23 x 23 Plants (sources) Suburban locations (destinations) 3 3 2 2 1 1 1 1 Total arrivals: Total departures: x 1 1 + x 1 2 + x 1 3 x 2 1 + x 2 2 + x 2 3 x 1 1 + x 2 1 x 1 2 + x 2 2 x 1 3 + x 2 3 Total costs:

15. 15 Slide © 2008 Thomson South-Western. All Rights Reserved n Partial Spreadsheet Showing Problem Data Transportation Problem: Example #1

16. 16 Slide © 2008 Thomson South-Western. All Rights Reserved n Partial Spreadsheet Showing Optimal Solution Transportation Problem: Example #1

17. 17 Slide © 2008 Thomson South-Western. All Rights Reserved 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 Transportation Problem: Example #1

18. 18 Slide © 2008 Thomson South-Western. All Rights Reserved n Partial Sensitivity Report (first half) Transportation Problem: Example #1 If transportation cost c13 is reduced by $4, from $40 to $36, there will be some non-zero shipments from source 1 to destination 3

19. 19 Slide © 2008 Thomson South-Western. All Rights Reserved n Partial Sensitivity Report (second half) Transportation Problem: Example #1 If Westwood demand is increased by 1 ton, from 45 tons to 46 tons, the total cost will increase by $36

20. 20 Slide © 2008 Thomson South-Western. All Rights Reserved 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.

21. 21 Slide © 2008 Thomson South-Western. All Rights Reserved 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

22. 22 Slide © 2008 Thomson South-Western. All Rights Reserved n Linear Programming Formulation Using the notation: Using the notation: x ij = 1 if agent i is assigned to task j x ij = 1 if agent i is assigned to task j 0 otherwise 0 otherwise c ij = cost of assigning agent i to task j c ij = cost of assigning agent i to task j Assignment Problem continued

23. 23 Slide © 2008 Thomson South-Western. All Rights Reserved n Linear Programming Formulation (continued) Assignment Problem x ij > 0 for all i and j

24. 24 Slide © 2008 Thomson South-Western. All Rights Reserved n LP Formulation Special Cases Number of agents exceeds the number of tasks: Number of agents exceeds the number of tasks: 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. Assignment Problem Extra agents simply remain unassigned.

25. 25 Slide © 2008 Thomson South-Western. All Rights Reserved 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 t tasks: An agent is permitted to work t tasks:

26. 26 Slide © 2008 Thomson South-Western. All Rights Reserved 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 Projects Subcontractor A B C Westside 50 36 16 Federated 28 30 18 Federated 28 30 18 Goliath 35 32 20 Universal 25 25 14 Universal 25 25 14 How should the contractors be assigned to minimize total mileage costs? Assignment Problem: Example

27. 27 Slide © 2008 Thomson South-Western. All Rights Reserved 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 Assignment Problem: Example

28. 28 Slide © 2008 Thomson South-Western. All Rights Reserved n Linear Programming Formulation Min 50 x 11 +36 x 12 +16 x 13 +28 x 21 +30 x 22 +18 x 23 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 s.t. x 11 + x 12 + x 13 < 1 x 21 + x 22 + x 23 < 1 x 21 + x 22 + x 23 < 1 x 31 + x 32 + x 33 < 1 x 31 + x 32 + x 33 < 1 x 41 + x 42 + x 43 < 1 x 41 + x 42 + x 43 < 1 x 11 + x 21 + x 31 + x 41 = 1 x 11 + x 21 + x 31 + x 41 = 1 x 12 + x 22 + x 32 + x 42 = 1 x 12 + x 22 + x 32 + x 42 = 1 x 13 + x 23 + x 33 + x 43 = 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 Assignment Problem: Example

29. 29 Slide © 2008 Thomson South-Western. All Rights Reserved n The optimal assignment is: 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 Assignment Problem: Example

30. 30 Slide © 2008 Thomson South-Western. All Rights Reserved End of Chapter 6 Essentials – Part A

31. 31 Slide © 2008 Thomson South-Western. All Rights Reserved 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.

32. 32 Slide © 2008 Thomson South-Western. All Rights Reserved Transshipment Problem n 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

33. 33 Slide © 2008 Thomson South-Western. All Rights Reserved Transshipment Problem n Linear Programming Formulation Using the notation: Using the notation: x ij = number of units shipped from node i to node j x ij = number of units shipped from node i to node j c ij = cost per unit of shipping from node i to node j c ij = cost per unit of shipping from node i to node j s i = supply at origin node i s i = supply at origin node i d j = demand at destination node j d j = demand at destination node j continued

34. Intermediate node convention x1jx1jx1jx1j jj Subtract arrivals (sinks) to intermediate node j -x 1j - x 2j - x 3j - x 4j x2jx2jx2jx2j x3jx3jx3jx3j x4jx4jx4jx4j xj5xj5xj5xj5 xj6xj6xj6xj6 xj7xj7xj7xj7 xj8xj8xj8xj8 Add departures (sources) from intermediate node j + x j5 + x j6 + x j7 + x j8 = 0 Total quantity generated at node j is zero

35. 35 Slide © 2008 Thomson South-Western. All Rights Reserved Transshipment Problem x ij > 0 for all i and j n Linear Programming Formulation (continued) continued

36. 36 Slide © 2008 Thomson South-Western. All Rights Reserved Transshipment Problem n LP Formulation Special Cases Total supply not equal to total demand Total supply not equal to total demand Maximization objective function Maximization objective function Route capacities or route minimums Route capacities or route minimums Unacceptable routes Unacceptable routes The LP model modifications required here are identical to those required for the special cases in the transportation problem.

37. 37 Slide © 2008 Thomson South-Western. All Rights Reserved 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. Transshipment Problem: Example

38. 38 Slide © 2008 Thomson South-Western. All Rights Reserved Because of long standing contracts based on past orders, unit costs from the manufacturers to the suppliers are: Zeron N Zeron S Zeron N Zeron S Arnold 5 8 Arnold 5 8 Supershelf 7 4 Supershelf 7 4 The costs to install the shelving at the various locations are: Zrox Hewes Rockrite Zrox Hewes Rockrite Thomas 1 5 8 Thomas 1 5 8 Washburn 3 4 4 Washburn 3 4 4 Transshipment Problem: Example

39. 39 Slide © 2008 Thomson South-Western. All Rights Reserved 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 ZeronN ZeronS Rock-Rite Transshipment Problem: Example

40. 40 Slide © 2008 Thomson South-Western. All Rights Reserved n Linear Programming Formulation Decision Variables Defined Decision 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 (Zeron N), 4 (Zeron S) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) Objective Function Defined Objective 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 Transshipment Problem: Example

41. 41 Slide © 2008 Thomson South-Western. All Rights Reserved n 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. Transshipment Problem: Example

42. 42 Slide © 2008 Thomson South-Western. All Rights Reserved n The Management Scientist Solution 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 Transshipment Problem: Example

43. 43 Slide © 2008 Thomson South-Western. All Rights Reserved n 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 Transshipment Problem: Example

44. 44 Slide © 2008 Thomson South-Western. All Rights Reserved n The Management Scientist 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 Transshipment Problem: Example

45. 45 Slide © 2008 Thomson South-Western. All Rights Reserved n The Management Scientist 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 Transshipment Problem: Example

46. 46 Slide © 2008 Thomson South-Western. All Rights Reserved n The Management Scientist 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 Transshipment Problem: Example

47. 47 Slide © 2008 Thomson South-Western. All Rights Reserved End of Chapter 6, Part A