1. 1 1 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Slides by JOHN LOUCKS St. Edward’s University INTRODUCTION TO MANAGEMENT SCIENCE, 13e Anderson Sweeney Williams Martin © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2. 2 2 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 6, Part A Distribution and Network Models n Transportation Problem n Assignment Problem n Transshipment Problem n Shortest Route Problem n Maximum Flow Problem

3. 3 3 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Transportation Problems (p.262) n Origin or Source node n Destination node n arcs n Minimize the total transportation cost meeting the supply availability and demand requirements. n Example (p.261) Production capacity of 3 plants Production capacity of 3 plants Demand forecasts for 4 distribution centers Demand forecasts for 4 distribution centers Network representation (p.262) Network representation (p.262) Transportation costs for each route. Transportation costs for each route.

4. 4 4 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 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

5. 5 5 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Transportation Problem n Linear Programming Formulation (p.267) x ij > 0 for all i and j

6. 6 6 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Transportation Problems n LP formulation (p.264) n Output (p.264) n What is the solution? n How much is the total transportation cost n Is it dual solution or degenerate solution? Row Slack or Surplus Dual Price Row Slack or Surplus Dual Price 1 39500.00 -1.000000 1 39500.00 -1.000000 2 0.000000 3.000000 2 0.000000 3.000000 3 0.000000 0.000000 3 0.000000 0.000000 4 0.000000 4.000000 4 0.000000 4.000000 5 0.000000 -6.000000 5 0.000000 -6.000000 6 0.000000 -5.000000 6 0.000000 -5.000000 7 0.000000 -2.000000 7 0.000000 -2.000000 8 0.000000 -3.000000 8 0.000000 -3.000000 Variable Value Reduced Cost X11 3500.000 0.000000 X11 3500.000 0.000000 X12 1500.000 0.000000 X12 1500.000 0.000000 X13 0.000000 8.000000 X13 0.000000 8.000000 X14 0.000000 6.000000 X14 0.000000 6.000000 X21 0.000000 1.000000 X21 0.000000 1.000000 X22 2500.000 0.000000 X22 2500.000 0.000000 X23 2000.000 0.000000 X23 2000.000 0.000000 X24 1500.000 0.000000 X24 1500.000 0.000000 X31 2500.000 0.000000 X31 2500.000 0.000000 X32 0.000000 4.000000 X32 0.000000 4.000000 X33 0.000000 6.000000 X33 0.000000 6.000000 X34 0.000000 6.000000 X34 0.000000 6.000000

7. 7 7 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Transportation Problems n If total supply is greater than the total demand? No change No change n If total supply is less than the total demand? Use dummy origin with 0 cost Use dummy origin with 0 cost example : Lexington demand is 2500

8. 8 8 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 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 : (p.266) Minimum shipping guarantee from i to j : (p.266) 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.

9. 9 9 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Assignment Problem (p.268) n Assign m workers to m jobs. n Minimize the total assignment cost n Cost info (p.268) n Assignment problem is a special case of transportation problem with all RHS are 1. n Network representation (p.269)

10. 10 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 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

11. 11 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. n Linear Programming Formulation (continued) Assignment Problem x ij > 0 for all i and j

12. 12 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Assignment Problem n LP formulation (p.270) n Output (p.270) n What is the solution? n What is the total completion time? n Is the solution dual solutions or degenerate? Variable Value Reduced Cost X11 0.000000 2.000000 X11 0.000000 2.000000 X12 1.000000 0.000000 X12 1.000000 0.000000 X13 0.000000 4.000000 X13 0.000000 4.000000 X21 0.000000 1.000000 X21 0.000000 1.000000 X22 0.000000 3.000000 X22 0.000000 3.000000 X23 1.000000 0.000000 X23 1.000000 0.000000 X31 1.000000 0.000000 X31 1.000000 0.000000 X32 0.000000 1.000000 X32 0.000000 1.000000 X33 0.000000 0.000000 X33 0.000000 0.000000 Row Slack or Surplus Dual Price Row Slack or Surplus Dual Price 1 26.00000 -1.000000 1 26.00000 -1.000000 2 0.000000 0.000000 2 0.000000 0.000000 3 0.000000 0.000000 3 0.000000 0.000000 4 0.000000 2.000000 4 0.000000 2.000000 5 0.000000 -8.000000 5 0.000000 -8.000000 6 0.000000 -15.00000 6 0.000000 -15.00000 7 0.000000 -5.000000 7 0.000000 -5.000000

13. 13 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 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.

14. 14 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 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:

15. 15 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Assignment Problem n If there are two more tasks with completion times (10, 18, 3) and (9, 14, 9) Min=10*x11+15*x12+9*x13+9*x21+18*x22+5*x23+6*x31+14*x32+3*x33 +10*x14+9*x15+18*x24+14*x25+3*x34+9*x35; +10*x14+9*x15+18*x24+14*x25+3*x34+9*x35; x11+x12+x13 +x14 +x15 <= 1; x21+x22+x23 +x24 +x25 <= 1; x21+x22+x23 +x24 +x25 <= 1; x31+x32+x33 +x34 +x35 <= 1; x31+x32+x33 +x34 +x35 <= 1; x11 +x21 +x31 +xd1 = 1; x12 +x22 +x32 +xd2 = 1; x12 +x22 +x32 +xd2 = 1; x13 +x23 +x33 +xd3 = 1; x13 +x23 +x33 +xd3 = 1; x14 +x24 +x34 +xd4 = 1; x14 +x24 +x34 +xd4 = 1; x15 +x25 +x35 +xd5 = 1; x15 +x25 +x35 +xd5 = 1; xd1 +xd2 +xd3 +xd4 +xd5 <= 2; n If task 1, 2 should be assigned remove xd1, xd2 in constraints 4 and 5.

16. 16 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Assignment Problem n If there are two more tasks with completion times (10, 18, 3) and (9, 14, 9) Objective value = 17 Variable Value Reduced Cost Variable Value Reduced Cost X11 0.000000 1.000000 X11 0.000000 1.000000 X12 0.000000 6.000000 X12 0.000000 6.000000 X13 0.000000 4.000000 X13 0.000000 4.000000 X21 0.000000 0.000000 X21 0.000000 0.000000 X22 0.000000 9.000000 X22 0.000000 9.000000 X23 1.000000 0.000000 X23 1.000000 0.000000 X31 0.000000 0.000000 X31 0.000000 0.000000 X32 0.000000 8.000000 X32 0.000000 8.000000 X33 0.000000 1.000000 X33 0.000000 1.000000 X14 0.000000 4.000000 X14 0.000000 4.000000 X15 1.000000 0.000000 X15 1.000000 0.000000 X24 0.000000 12.00000 X24 0.000000 12.00000 X25 0.000000 5.000000 X25 0.000000 5.000000 X34 1.000000 0.000000 X34 1.000000 0.000000 X35 0.000000 3.000000 X35 0.000000 3.000000 XD1 1.000000 0.000000 XD1 1.000000 0.000000 XD2 1.000000 0.000000 XD2 1.000000 0.000000 XD3 0.000000 4.000000 XD3 0.000000 4.000000 XD4 0.000000 3.000000 XD4 0.000000 3.000000 XD5 0.000000 0.000000 XD5 0.000000 0.000000

17. 17 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Transshipment Problem n A special case of transportation problem with transhipement nodes. n For transhipment nodes Flow out = Flow in  Flow out – Flow in = 0  Flow out – Flow in = 0 n Network representation (p.274) n Unit transportation cost (p.274, Table 6.5)

18. 18 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 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

19. 19 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Transshipment Problem x ij > 0 for all i and j n Linear Programming Formulation (p.279) continued

20. 20 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Transshipment Problem n LP formulation (p.276) n Output (p.276) n What is the solution? n How much is the total transportation cost? n Modified Example (p.277, Figure 6.9) n Formulation for modified example (p.278) n Output (p.278)

21. 21 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 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.

22. 22 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Shortest-Route Problem (p.281) n Finding the shortest path in a network from one node (or set of nodes) to another node (or set of nodes). n All arcs are two-way (Figure 6.12) n Some arcs are two-way (Figure 6.13)

23. 23 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. n Linear Programming Formulation Using the notation: Using the notation: x ij = 1 if the arc from node i to node j x ij = 1 if the arc from node i to node j is on the shortest route is on the shortest route 0 otherwise 0 otherwise c ij = distance, time, or cost associated c ij = distance, time, or cost associated with the arc from node i to node j with the arc from node i to node j continued Shortest-Route Problem

24. 24 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. n Linear Programming Formulation (p.283) Shortest-Route Problem

25. 25 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Shortest-Route Problem (p.281) n LP Formulation for Figure 6.13 (p.282) n Output (p.283) n What is the solution? 1 – 3 – 2 – 4 – 6 n What is the total distance? n For a single source shortest-route problem, there is Dijkstra’s algorithm.

26. 26 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Shortest-Route Problem (p.281) n Dijkstra’s algorithm Start from the source Fro each node [min dist, previous node] 1. {1} 2 [25, 1] 3 [20, 1]*** 3 [20, 1]*** 2. {1, 3} 2 [23, 3]*** 5 [26, 3] 5 [26, 3] 3. {1, 3, 2} 4 [28, 2], 5 [26, 3]***, 6 [37, 2] 4. {1, 3, 2, 5} 4 [28, 2]***, 6 [32, 4] 5. {1, 3, 2, 5, 4} 6 [32, 4]*** 6 – 4 – 2 – 3 – 1 6 – 4 – 2 – 3 – 1

27. 27 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Minimum spanning tree (p.281) n Connect all nodes with minimum total distance n Prim’s algorithm Start with any node Connect the closest node to connected set 1. {2} 1 [25, 2], 3 [3, 2]**, 4 [5, 2], 6 [14, 2] 2. {2, 3} 1 [20, 3], 4 [5, 2]**, 5 [6, 3], 6 [14, 2] 3. {2, 3, 4} 1 [20, 3], 5 [4, 4]**, 6 [4, 4]** 4. {2, 3, 4, 5, 6} 1 [20, 3]** 5. {1, 2, 3, 4, 5, 6} 1 – 3, 5 – 4 – 6, 4 – 2, 3 – 2

28. 28 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Minimum spanning tree (p.281) n LP formulation Min = 25*x12 +20*x13 +3*x23 +5*x24 +14*x26 +6*x35 +4*x45 +4*x46 +7*x56; x12 +x13 +x23 +x24 +x26 +x35 +x45 +x46 +x56 >= 5; ! maximum arcs is 5; x13 +x23 +x24 +x26 +x35 +x45 +x46 +x56 >= 4; ! connect 1-2; x13 +x23 +x24 +x26 +x35 +x45 +x46 +x56 >= 4; ! connect 1-2; x12 +x23 +x24 +x26 +x35 +x45 +x46 +x56 >= 4; ! connect 1-3; x12 +x13 +x24 +x26 +x35 +x45 +x46 +x56 >= 4; ! connect 2-3; x12 +x13 +x23 +x26 +x35 +x45 +x46 +x56 >= 4; ! connect 2-4; x12 +x13 +x23 +x24 +x35 +x45 +x46 +x56 >= 4; ! connect 2-6; x12 +x13 +x23 +x24 +x26 +x45 +x46 +x56 >= 4; ! connect 3-5; x12 +x13 +x23 +x24 +x26 +x35 +x46 +x56 >= 4; ! connect 4-5; x12 +x13 +x23 +x24 +x26 +x35 +x45 +x56 >= 4; ! connect 4-6; x12 +x13 +x23 +x24 +x26 +x35 +x45 +x46 >= 4; ! connect 5-6; x24 +x26 +x35 +x45 +x46 +x56 >= 3; ! connect (123, 4, 5, 6); x24 +x26 +x35 +x45 +x46 +x56 >= 3; ! connect (123, 4, 5, 6); x13 +x23 +x26 +x35 +x45 +x46 +x56 >= 3; ! connect (1-2, 2-4), (124, 3, 5, 6); x13 +x23 +x26 +x35 +x45 +x46 +x56 >= 3; ! connect (1-2, 2-4), (124, 3, 5, 6); x13 +x23 +x24 +x35 +x45 +x46 +x56 >= 3; ! connect (1-2, 2-6), (126, 3, 4, 5); x13 +x23 +x24 +x35 +x45 +x46 +x56 >= 3; ! connect (1-2, 2-6), (126, 3, 4, 5); x12 +x23 +x24 +x26 +x45 +x46 +x56 >= 3; ! connect (1-3, 3-5), (135, 2, 4, 6; x12 +x13 +x26 +x35 +x45 +x46 +x56 >= 3; ! connect (2-3, 2-4), (234, 1, 5, 6); x12 +x13 +x24 +x26 +x45 +x46 +x56 >= 3; ! connect (2-3, 3-5), (235, 1, 4, 6); x12 +x13 +x24 +x35 +x45 +x46 +x56 >= 3; ! connect (2-3, 2-6), (236, 1, 4, 5); x12 +x13 +x23 +x26 +x35 +x46 +x56 >= 3; ! connect (2-4, 4-5), (245, 1, 3, 6); x12 +x13 +x23 +x35 +x45 +x56 >= 3; ! connect (246, 1, 3, 5); x12 +x13 +x23 +x24 +x35 +x45 +x46 >= 3; ! connect (2-6, 5-6), (256, 1, 3, 4); x12 +x13 +x23 +x24 +x26 +x46 +x56 >= 3; ! connect (3-5, 4-5), (345, 1, 2, 6); x12 +x13 +x23 +x24 +x26 +x45 +x46 >= 3; ! connect (3-5, 5-6), (356, 1, 2, 4); x12 +x13 +x23 +x24 +x26 +x35 >= 3; ! connect (456, 1, 2, 3);

29. 29 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Minimum spanning tree (p.281) n LP formulation (continued) x26 +x35 +x45 +x46 +x56 >= 2; ! connect (1234, 5, 6); x26 +x35 +x45 +x46 +x56 >= 2; ! connect (1234, 5, 6); x24 +x26 +x45 +x46 +x56 >= 2; ! connect (1235, 4, 6); x24 +x26 +x45 +x46 +x56 >= 2; ! connect (1235, 4, 6); x24 +x35 +x45 +x46 +x56 >= 2; ! connect (1236, 4, 5); x24 +x35 +x45 +x46 +x56 >= 2; ! connect (1236, 4, 5); x13 +x23 +x26 +x35 +x46 +x56 >= 2; ! connect (1245, 3, 6); x13 +x23 +x26 +x35 +x46 +x56 >= 2; ! connect (1245, 3, 6); x13 +x23 +x35 +x45 +x56 >= 2; ! connect (1246, 3, 5); x13 +x23 +x35 +x45 +x56 >= 2; ! connect (1246, 3, 5); x13 +x23 +x24 +x35 +x45 +x46 >= 2; ! connect (1256, 3, 4); x13 +x23 +x24 +x35 +x45 +x46 >= 2; ! connect (1256, 3, 4); x12 +x23 +x24 +x26 +x46 +x56 >= 2; ! connect (1345, 2, 6); x12 +x23 +x24 +x26 +x45 +x46 >= 2; ! connect (1356, 2, 4); x12 +x13 +x26 +x46 +x56 >= 2; ! connect (2345, 1, 6); x12 +x13 +x35 +x45 +x56 >= 2; ! connect (2346, 1, 5); x12 +x13 +x24 +x45 +x46 >= 2; ! connect (2356, 1, 4); x12 +x13 +x23 +x35 >= 2; ! connect (2456, 1, 3); x12 +x13 +x23 +x24 +x26 >= 2; ! connect (3456, 1, 2); x26 +x46 +x56 >= 1; ! connect (12345, 6); x26 +x46 +x56 >= 1; ! connect (12345, 6); x35 +x45 +x56 >= 1; ! connect (12346, 5); x35 +x45 +x56 >= 1; ! connect (12346, 5); x24 +x45 +x46 >= 1; ! connect (12356, 4); x24 +x45 +x46 >= 1; ! connect (12356, 4); x13 +x23 +x35 >= 1; ! connect (12456, 3); x13 +x23 +x35 >= 1; ! connect (12456, 3); x12 +x23 +x24 +x26 >= 1; ! connect (13456, 2); x12 +x13 >= 1; ! connect (23456, 1); ! choose arcs from unselected nodes to any of selected nodes ! for nodes that can not be connected to any of selected nodes, choose all the outgoing arcs.

30. 30 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Minimum spanning tree (p.281) n LP output Objective value: 36.00000 Objective value: 36.00000 Variable Value Reduced Cost Variable Value Reduced Cost X12 0.000000 5.000000 X12 0.000000 5.000000 X13 1.000000 0.000000 X13 1.000000 0.000000 X23 1.000000 0.000000 X23 1.000000 0.000000 X24 1.000000 0.000000 X24 1.000000 0.000000 X26 0.000000 9.000000 X26 0.000000 9.000000 X35 0.000000 1.000000 X35 0.000000 1.000000 X45 1.000000 0.000000 X45 1.000000 0.000000 X46 1.000000 0.000000 X46 1.000000 0.000000 X56 0.000000 3.000000 X56 0.000000 3.000000 Row Slack or Surplus Dual Price Row Slack or Surplus Dual Price 1 36.00000 -1.000000 1 36.00000 -1.000000 2 0.000000 -2.000000 2 0.000000 -2.000000 3 1.000000 0.000000 3 1.000000 0.000000 4 0.000000 0.000000 4 0.000000 0.000000 5 0.000000 -2.000000 5 0.000000 -2.000000 6 0.000000 0.000000 6 0.000000 0.000000 7 1.000000 0.000000 7 1.000000 0.000000 8 1.000000 0.000000 8 1.000000 0.000000

31. 31 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Maximal Flow Problem n Maximal Flow Problem (p.284) n The maximal flow problem is concerned with determining the maximal volume of flow from one node (called the source) to another node (called the sink). n In the maximal flow problem, each arc has a maximum arc flow capacity which limits the flow through the arc.

32. 32 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Example: Maximal Flow n A capacitated transshipment model can be developed for the maximal flow problem. n We will add an arc from the sink node back to the source node to represent the total flow through the network. n There is no capacity on the newly added sink-to- source arc. n We want to maximize the flow over the sink-to-source arc.

33. 33 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Maximal Flow Problem n LP Formulation (as Capacitated Transshipment Problem) (as Capacitated Transshipment Problem) There is a variable for every arc. There is a variable for every arc. There is a constraint for every node; the flow out must equal the flow in. There is a constraint for every node; the flow out must equal the flow in. There is a constraint for every arc (except the added sink-to-source arc); arc capacity cannot be exceeded. There is a constraint for every arc (except the added sink-to-source arc); arc capacity cannot be exceeded. The objective is to maximize the flow over the added, sink-to-source arc. The objective is to maximize the flow over the added, sink-to-source arc.

34. 34 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Maximal Flow Problem n LP Formulation (as Capacitated Transshipment Problem) (as Capacitated Transshipment Problem) Max x k 1 ( k is sink node, 1 is source node) Max x k 1 ( k is sink node, 1 is source node) s.t.  x ij -  x ji = 0 (conservation of flow) i j s.t.  x ij -  x ji = 0 (conservation of flow) i j x ij < c ij ( c ij is capacity of ij arc) x ij > 0, for all i and j (non-negativity) x ij > 0, for all i and j (non-negativity) (x ij represents the flow from node i to node j) (x ij represents the flow from node i to node j)

35. 35 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Maximal Flow Problem n LP Formulation (pp.284–285) n Computer output (p.286) n What is the solution? n What is the maximum flow?

36. 36 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. A production and Inventory application

37. 37 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. A production and Inventory application n LP formulation (p.289) n Computer output (p.290) n What is the solution? n What is the total production and inventory cost?

38. 38 Slide © 2008 Thomson South-Western. All Rights Reserved © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. End of Chapter 6, Part A