1. 1 1 © 2003 Thomson  /South-Western Slide Slides Prepared by JOHN S. LOUCKS St. Edward’s University

2. 2 2 © 2003 Thomson  /South-Western Slide Chapter 7 Transportation, Assignment, and Transshipment Problems n Transportation Problem Network Representation and LP Formulation Network Representation and LP Formulation Transportation Simplex Method Transportation Simplex Method n Assignment Problem Network Representation and LP Formulation Network Representation and LP Formulation Hungarian Method Hungarian Method n The Transshipment Problem Network Representation and LP Formulation Network Representation and LP Formulation

3. 3 3 © 2003 Thomson  /South-Western 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 the shortest route, minimal spanning tree, and maximal flow problems (Chapter 9) and PERT/CPM problems (Chapter 10) are all examples of network problems.

4. 4 4 © 2003 Thomson  /South-Western 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 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 © 2003 Thomson  /South-Western 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 © 2003 Thomson  /South-Western 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 © 2003 Thomson  /South-Western Slide Transportation Problem n 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 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 © 2003 Thomson  /South-Western 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 © 2003 Thomson  /South-Western 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. 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?

10. 10 © 2003 Thomson  /South-Western Slide Example: BBC 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

11. 11 © 2003 Thomson  /South-Western Slide Example: BBC n Partial Spreadsheet Showing Problem Data

12. 12 © 2003 Thomson  /South-Western Slide Example: BBC n Partial Spreadsheet Showing Optimal Solution

13. 13 © 2003 Thomson  /South-Western 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

14. 14 © 2003 Thomson  /South-Western Slide Example: BBC n Partial Sensitivity Report (first half)

15. 15 © 2003 Thomson  /South-Western Slide Example: BBC n Partial Sensitivity Report (second half)

16. 16 © 2003 Thomson  /South-Western Slide Transportation Simplex Method n The transportation simplex method requires that the sum of the supplies at the origins equal the sum of the demands at the destinations. n If the total supply is greater than the total demand, a dummy destination is added with demand equal to the excess supply, and shipping costs from all origins are zero. (If total supply is less than total demand, a dummy origin is added.) n When solving a transportation problem by its special purpose algorithm, unacceptable shipping routes are given a cost of + M (a large number).

17. 17 © 2003 Thomson  /South-Western Slide Transportation Simplex Method n A transportation tableau is given below. Each cell represents a shipping route (which is an arc on the network and a decision variable in the LP formulation), and the unit shipping costs are given in an upper right hand box in the cell. Supply 30 50 35 20 40 30 30 Demand104525 S1 S2 D3D2D1 15

18. 18 © 2003 Thomson  /South-Western Slide Transportation Simplex Method n The transportation problem is solved in two phases: Phase I -- Obtaining an initial feasible solution Phase I -- Obtaining an initial feasible solution Phase II -- Moving toward optimality Phase II -- Moving toward optimality n Phase I: The Minimum-Cost Procedure can be used to establish an initial basic feasible solution without doing numerous iterations of the simplex method. n Phase II: The Stepping Stone Method - using the MODI Method for evaluating the reduced costs - may be used to move from the initial feasible solution to the optimal one.

19. 19 © 2003 Thomson  /South-Western Slide Transportation Simplex Method n Phase I - Minimum-Cost Method Step 1: Select the cell with the least cost. Assign to this cell the minimum of its remaining row supply or remaining column demand. Step 1: Select the cell with the least cost. Assign to this cell the minimum of its remaining row supply or remaining column demand. Step 2: Decrease the row and column availabilities by this amount and remove from consideration all other cells in the row or column with zero availability/demand. (If both are simultaneously reduced to 0, assign an allocation of 0 to any other unoccupied cell in the row or column before deleting both.) GO TO STEP 1. Step 2: Decrease the row and column availabilities by this amount and remove from consideration all other cells in the row or column with zero availability/demand. (If both are simultaneously reduced to 0, assign an allocation of 0 to any other unoccupied cell in the row or column before deleting both.) GO TO STEP 1.

20. 20 © 2003 Thomson  /South-Western Slide Transportation Simplex Method n Phase II - Stepping Stone Method Step 1: For each unoccupied cell, calculate the reduced cost by the MODI method described on an upcoming slide. Step 1: For each unoccupied cell, calculate the reduced cost by the MODI method described on an upcoming slide. Select the unoccupied cell with the most negative reduced cost. (For maximization problems select the unoccupied cell with the largest reduced cost.) If none, STOP.

21. 21 © 2003 Thomson  /South-Western Slide Transportation Simplex Method n Phase II - Stepping Stone Method (continued) Step 2: For this unoccupied cell generate a stepping stone path by forming a closed loop with this cell and occupied cells by drawing connecting alternating horizontal and vertical lines between them. Step 2: For this unoccupied cell generate a stepping stone path by forming a closed loop with this cell and occupied cells by drawing connecting alternating horizontal and vertical lines between them. Determine the minimum allocation where a subtraction is to be made along this path.

22. 22 © 2003 Thomson  /South-Western Slide Transportation Simplex Method n Phase II - Stepping Stone Method (continued) Step 3: Add this allocation to all cells where additions are to be made, and subtract this allocation to all cells where subtractions are to be made along the stepping stone path. Step 3: Add this allocation to all cells where additions are to be made, and subtract this allocation to all cells where subtractions are to be made along the stepping stone path. (Note: An occupied cell on the stepping stone path now becomes 0 (unoccupied). If more than one cell becomes 0, make only one unoccupied; make the others occupied with 0's.) (Note: An occupied cell on the stepping stone path now becomes 0 (unoccupied). If more than one cell becomes 0, make only one unoccupied; make the others occupied with 0's.) GO TO STEP 1.

23. 23 © 2003 Thomson  /South-Western Slide Transportation Simplex Method n MODI Method (for obtaining reduced costs) Associate a number, u i, with each row and v j with each column. Step 1: Set u 1 = 0. Step 1: Set u 1 = 0. Step 2: Calculate the remaining u i 's and v j 's by solving the relationship c ij = u i + v j for occupied cells. Step 2: Calculate the remaining u i 's and v j 's by solving the relationship c ij = u i + v j for occupied cells. Step 3: For unoccupied cells ( i, j ), the reduced cost = c ij - u i - v j. Step 3: For unoccupied cells ( i, j ), the reduced cost = c ij - u i - v j.

24. 24 © 2003 Thomson  /South-Western Slide Example: BBC n Initial Transportation Tableau Since total supply = 100 and total demand = 80, a dummy destination is created with demand of 20 and 0 unit costs. 4242 404000 004040 3030 3030 Demand Supply 50 50 20104525 Dummy Plant 1 Plant 2 EastwoodWestwoodNorthwood 2424

25. 25 © 2003 Thomson  /South-Western Slide Example: BBC n Phase I: Minimum-Cost Procedure Iteration 1: Tie for least cost (0), arbitrarily select x 14. Allocate 20. Reduce s 1 by 20 to 30 and delete the Dummy column. Iteration 1: Tie for least cost (0), arbitrarily select x 14. Allocate 20. Reduce s 1 by 20 to 30 and delete the Dummy column. Iteration 2: Of the remaining cells the least cost is 24 for x 11. Allocate 25. Reduce s 1 by 25 to 5 and eliminate the Northwood column. Iteration 2: Of the remaining cells the least cost is 24 for x 11. Allocate 25. Reduce s 1 by 25 to 5 and eliminate the Northwood column.

26. 26 © 2003 Thomson  /South-Western Slide Example: BBC n Phase I: Minimum-Cost Procedure (continued) Iteration 3: Of the remaining cells the least cost is 30 for x 12. Allocate 5. Reduce the Westwood column to 40 and eliminate the Plant 1 row. Iteration 3: Of the remaining cells the least cost is 30 for x 12. Allocate 5. Reduce the Westwood column to 40 and eliminate the Plant 1 row. Iteration 4: Since there is only one row with two cells left, make the final allocations of 40 and 10 to x 22 and x 23, respectively. Iteration 4: Since there is only one row with two cells left, make the final allocations of 40 and 10 to x 22 and x 23, respectively.

27. 27 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II – Iteration 1 MODI Method MODI Method 1. Set u 1 = 0 2. Since u 1 + v j = c 1 j for occupied cells in row 1, then v 1 = 24, v 2 = 30, v 4 = 0. v 1 = 24, v 2 = 30, v 4 = 0. 3. Since u i + v 2 = c i 2 for occupied cells in column 2, then u 2 + 30 = 40, hence u 2 = 10. 4. Since u 2 + v j = c 2 j for occupied cells in row 2, then 10 + v 3 = 42, hence v 3 = 32. 10 + v 3 = 42, hence v 3 = 32.

28. 28 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II – Iteration 1 MODI Method (continued) MODI Method (continued) Calculate the reduced costs (circled numbers on the next slide) by c ij - u i + v j. Unoccupied Cell Reduced Cost Unoccupied Cell Reduced Cost (1,3) 40 - 0 - 32 = 8 (1,3) 40 - 0 - 32 = 8 (2,1) 30 - 24 -10 = -4 (2,1) 30 - 24 -10 = -4 (2,4) 0 - 10 - 0 = -10 (2,4) 0 - 10 - 0 = -10

29. 29 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II – Iteration 1 MODI Method (continued) MODI Method (continued) 252555 -4-4 +8+82020 40401010-10-104242 404000 004040 3030 3030 vjvjvjvj uiuiuiui 10 0 0323024 Dummy Plant 1 Plant 2 EastwoodWestwoodNorthwood 2424

30. 30 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II – Iteration 1 Stepping Stone Method Stepping Stone Method The stepping stone path for cell (2,4) is (2,4), (1,4), (1,2), (2,2). The allocations in the subtraction cells are 20 and 40, respectively. The minimum is 20, and hence reallocate 20 along this path. Thus for the next tableau: x 24 = 0 + 20 = 20 (0 is its current allocation) x 14 = 20 - 20 = 0 (blank for the next tableau) x 14 = 20 - 20 = 0 (blank for the next tableau) x 12 = 5 + 20 = 25 x 12 = 5 + 20 = 25 x 22 = 40 - 20 = 20 x 22 = 40 - 20 = 20 The other occupied cells remain the same. The other occupied cells remain the same.

31. 31 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II - Iteration 2 MODI Method MODI Method The reduced costs are found by calculating the u i 's and v j 's for this tableau. 1. Set u 1 = 0. 2. Since u 1 + v j = c ij for occupied cells in row 1, then v 1 = 24, v 2 = 30. v 1 = 24, v 2 = 30. 3. Since u i + v 2 = c i 2 for occupied cells in column 2, then u 2 + 30 = 40, or u 2 = 10. 4. Since u 2 + v j = c 2 j for occupied cells in row 2, then 10 + v 3 = 42 or v 3 = 32; and, 10 + v 4 = 0 or v 4 = -10. 10 + v 3 = 42 or v 3 = 32; and, 10 + v 4 = 0 or v 4 = -10.

32. 32 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II - Iteration 2 MODI Method (continued) MODI Method (continued) Calculate the reduced costs (circled numbers on the next slide) by c ij - u i + v j. Unoccupied Cell Reduced Cost Unoccupied Cell Reduced Cost (1,3) 40 - 0 - 32 = 8 (1,3) 40 - 0 - 32 = 8 (1,4) 0 - 0 - (-10) = 10 (1,4) 0 - 0 - (-10) = 10 (2,1) 30 - 10 - 24 = -4 (2,1) 30 - 10 - 24 = -4

33. 33 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II - Iteration 2 MODI Method (continued) MODI Method (continued) 25252525 -4-4 +8+8+10+10 2020101020204242 404000 004040 3030 3030 vjvjvjvj uiuiuiui 10 0 -6363024 Dummy Plant 1 Plant 2 EastwoodWestwoodNorthwood 2424

34. 34 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II - Iteration 2 Stepping Stone Method Stepping Stone Method The most negative reduced cost is = -4 determined by x 21. The stepping stone path for this cell is (2,1),(1,1),(1,2),(2,2). The allocations in the subtraction cells are 25 and 20 respectively. Thus, the new solution is obtained by reallocating 20 on the stepping stone path.

35. 35 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II - Iteration 2 Stepping Stone Method (continued) Stepping Stone Method (continued) Thus, for the next tableau: x 21 = 0 + 20 = 20 (0 is its current allocation) x 21 = 0 + 20 = 20 (0 is its current allocation) x 11 = 25 - 20 = 5 x 11 = 25 - 20 = 5 x 12 = 25 + 20 = 45 x 12 = 25 + 20 = 45 x 22 = 20 - 20 = 0 (blank for the next tableau) x 22 = 20 - 20 = 0 (blank for the next tableau) The other occupied cells remain the same. The other occupied cells remain the same.

36. 36 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II - Iteration 3 MODI Method MODI Method The reduced costs are found by calculating the u i 's and v j 's for this tableau. 1. Set u 1 = 0 2. Since u 1 + v j = c 1 j for occupied cells in row 1, then v 1 = 24 and v 2 = 30. v 1 = 24 and v 2 = 30. 3. Since u i + v 1 = c i 1 for occupied cells in column 2, then u 2 + 24 = 30 or u 2 = 6. 4. Since u 2 + v j = c 2 j for occupied cells in row 2, then 6 + v 3 = 42 or v 3 = 36, and 6 + v 4 = 0 or v 4 = -6. 6 + v 3 = 42 or v 3 = 36, and 6 + v 4 = 0 or v 4 = -6.

37. 37 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II - Iteration 3 MODI Method (continued) MODI Method (continued) Calculate the reduced costs (circled numbers on the next slide) by c ij - u i + v j. Unoccupied Cell Reduced Cost Unoccupied Cell Reduced Cost (1,3) 40 - 0 - 36 = 4 (1,3) 40 - 0 - 36 = 4 (1,4) 0 - 0 - (-6) = 6 (1,4) 0 - 0 - (-6) = 6 (2,2) 40 - 6 - 30 = 4 (2,2) 40 - 6 - 30 = 4

38. 38 © 2003 Thomson  /South-Western Slide Example: BBC n Phase II - Iteration 3 MODI Method (continued) MODI Method (continued) Since all the reduced costs are non-negative, this is the optimal tableau. 554545 2020 +4+4+6+6 +4+4101020204242 404000 004040 3030 3030 vjvjvjvj uiuiuiui 6 0 -6363024 Dummy Plant 1 Plant 2 EastwoodWestwoodNorthwood 2424

39. 39 © 2003 Thomson  /South-Western Slide Example: BBC n Optimal Solution From To Amount Cost From To Amount Cost Plant 1 Northwood 5 120 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

40. 40 © 2003 Thomson  /South-Western 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.

41. 41 © 2003 Thomson  /South-Western 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

42. 42 © 2003 Thomson  /South-Western Slide Assignment Problem n LP 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 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. 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.

43. 43 © 2003 Thomson  /South-Western Slide 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. Assignment Problem

44. 44 © 2003 Thomson  /South-Western 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

45. 45 © 2003 Thomson  /South-Western Slide 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 Subcontractor A B C Subcontractor A B C Westside 50 36 16 Westside 50 36 16 Federated 28 30 18 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? Example: Hungry Owner

46. 46 © 2003 Thomson  /South-Western Slide Example: Hungry Owner 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

47. 47 © 2003 Thomson  /South-Western 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 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

48. 48 © 2003 Thomson  /South-Western Slide Hungarian Method n The Hungarian method solves minimization assignment problems with m workers and m jobs. n Special considerations can include: number of workers does not equal the number of jobs -- add dummy workers or jobs with 0 assignment costs as needed number of workers does not equal the number of jobs -- add dummy workers or jobs with 0 assignment costs as needed worker i cannot do job j -- assign c ij = + M worker i cannot do job j -- assign c ij = + M maximization objective -- create an opportunity loss matrix subtracting all profits for each job from the maximum profit for that job before beginning the Hungarian method maximization objective -- create an opportunity loss matrix subtracting all profits for each job from the maximum profit for that job before beginning the Hungarian method

49. 49 © 2003 Thomson  /South-Western Slide Hungarian Method n Step 1: For each row, subtract the minimum number in that row from all numbers in that row. n Step 2: For each column, subtract the minimum number in that column from all numbers in that column. n Step 3: Draw the minimum number of lines to cover all zeroes. If this number = m, STOP -- an assignment can be made. n Step 4: Subtract d (the minimum uncovered number) from uncovered numbers. Add d to numbers covered by two lines. Numbers covered by one line remain the same. Then, GO TO STEP 3.

50. 50 © 2003 Thomson  /South-Western Slide Hungarian Method n Finding the Minimum Number of Lines and Determining the Optimal Solution Step 1: Find a row or column with only one unlined zero and circle it. (If all rows/columns have two or more unlined zeroes choose an arbitrary zero.) Step 1: Find a row or column with only one unlined zero and circle it. (If all rows/columns have two or more unlined zeroes choose an arbitrary zero.) Step 2: If the circle is in a row with one zero, draw a line through its column. If the circle is in a column with one zero, draw a line through its row. One approach, when all rows and columns have two or more zeroes, is to draw a line through one with the most zeroes, breaking ties arbitrarily. Step 2: If the circle is in a row with one zero, draw a line through its column. If the circle is in a column with one zero, draw a line through its row. One approach, when all rows and columns have two or more zeroes, is to draw a line through one with the most zeroes, breaking ties arbitrarily. Step 3: Repeat step 2 until all circles are lined. If this minimum number of lines equals m, the circles provide the optimal assignment. Step 3: Repeat step 2 until all circles are lined. If this minimum number of lines equals m, the circles provide the optimal assignment.

51. 51 © 2003 Thomson  /South-Western Slide Example: Hungry Owner n Initial Tableau Setup Since the Hungarian algorithm requires that there be the same number of rows as columns, add a Dummy column so that the first tableau is: A B C Dummy A B C Dummy Westside 50 36 16 0 Westside 50 36 16 0 Federated 28 30 18 0 Federated 28 30 18 0 Goliath 35 32 20 0 Goliath 35 32 20 0 Universal 25 25 14 0 Universal 25 25 14 0

52. 52 © 2003 Thomson  /South-Western Slide Example: Hungry Owner n Step 1: Subtract minimum number in each row from all numbers in that row. Since each row has a zero, we would simply generate the same matrix above. n Step 2: Subtract the minimum number in each column from all numbers in the column. For A it is 25, for B it is 25, for C it is 14, for Dummy it is 0. This yields: A B C Dummy A B C Dummy Westside 25 11 2 0 Westside 25 11 2 0 Federated 3 5 4 0 Federated 3 5 4 0 Goliath 10 7 6 0 Goliath 10 7 6 0 Universal 0 0 0 0 Universal 0 0 0 0

53. 53 © 2003 Thomson  /South-Western Slide Example: Hungry Owner n Step 3: Draw the minimum number of lines to cover all zeroes. Although one can "eyeball" this minimum, use the following algorithm. If a "remaining" row has only one zero, draw a line through the column. If a remaining column has only one zero in it, draw a line through the row. A B C Dummy A B C Dummy Westside25 11 2 0 Westside25 11 2 0 Federated 3 5 4 0 Federated 3 5 4 0 Goliath 10 7 6 0 Goliath 10 7 6 0 Universal 0 0 0 0 Universal 0 0 0 0

54. 54 © 2003 Thomson  /South-Western Slide n Step 4: The minimum uncovered number is 2. Subtract 2 from uncovered numbers; add 2 to all numbers covered by two lines. This gives: A B C Dummy A B C Dummy Westside23 9 0 0 Westside23 9 0 0 Federated 1 3 2 0 Federated 1 3 2 0 Goliath 8 5 4 0 Goliath 8 5 4 0 Universal 0 0 0 2 Universal 0 0 0 2 Example: Hungry Owner

55. 55 © 2003 Thomson  /South-Western Slide Example: Hungry Owner n Step 3: Draw the minimum number of lines to cover all zeroes. A B C Dummy A B C Dummy Westside 23 9 0 0 Westside 23 9 0 0 Federated 1 3 2 0 Federated 1 3 2 0 Goliath 8 5 4 0 Goliath 8 5 4 0 Universal 0 0 0 2 Universal 0 0 0 2

56. 56 © 2003 Thomson  /South-Western Slide Example: Hungry Owner n Step 4: The minimum uncovered number is 1. Subtract 1 from uncovered numbers. Add 1 to numbers covered by two lines. This gives: A B C Dummy A B C Dummy Westside 23 9 0 1 Westside 23 9 0 1 Federated 0 2 1 0 Federated 0 2 1 0 Goliath 7 4 3 0 Goliath 7 4 3 0 Universal 0 0 0 3 Universal 0 0 0 3

57. 57 © 2003 Thomson  /South-Western Slide Example: Hungry Owner n Step 3: The minimum number of lines to cover all 0's is four. Thus, there is a minimum-cost assignment of 0's with this tableau. 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

58. 58 © 2003 Thomson  /South-Western 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.

59. 59 © 2003 Thomson  /South-Western Slide 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 NODESSOURCESDESTINATIONS s2s2s2s2 Demand Supply

60. 60 © 2003 Thomson  /South-Western 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

61. 61 © 2003 Thomson  /South-Western 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.

62. 62 © 2003 Thomson  /South-Western 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 costs 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

63. 63 © 2003 Thomson  /South-Western 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

64. 64 © 2003 Thomson  /South-Western Slide Example: Transshipping 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 (Thomas), 4 (Washburn) j = 3 (Thomas), 4 (Washburn) k = 5 (Zrox), 6 (Hewes), 7 (Rockwright) k = 5 (Zrox), 6 (Hewes), 7 (Rockwright) 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

65. 65 © 2003 Thomson  /South-Western 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.

66. 66 © 2003 Thomson  /South-Western Slide Example: Transshipping n Optimal Solution (from The 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

67. 67 © 2003 Thomson  /South-Western 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

68. 68 © 2003 Thomson  /South-Western 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

69. 69 © 2003 Thomson  /South-Western 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

70. 70 © 2003 Thomson  /South-Western 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

71. 71 © 2003 Thomson  /South-Western Slide The End of Chapter 7