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1 1 Slides by John Loucks St. Edward’s University Modifications by A. Asef-Vaziri

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2 2 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

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3 3 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.

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4 4 Transportation, Assignment, and Transshipment Problems n Each of the five problems of this chapter can be formulated as linear programs and solved by general purpose linear programming codes. n For each of the five problems, 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 that contain separate computer codes for these problems which take advantage of their network structure.

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5 5 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.

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6 6 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

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7 7 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?

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8 8 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 Decision Variables. the tons of concrete blocks, x ij, to be shipped from source i to destination j. Northwood Westwood Eastwood Northwood Westwood Eastwood Plant 1 x 11 x 12 x 13 Plant 1 x 11 x 12 x 13 Plant 1 x 21 x 22 x 23 Plant 1 x 21 x 22 x 23

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9 9 n Define the Objective Function Minimize the total shipping cost. Minimize the total shipping cost. Min: (shipping cost per ton for each origin to destination) × (number of pounds shipped from each origin to each destination). Min: (shipping cost per ton for each origin to destination) × (number of pounds shipped from each origin to each destination). Min: 24 x 11 + 30 x 12 + 40 x 13 + 30 x 21 + 40 x 22 + 42 x 23 Min: 24 x 11 + 30 x 12 + 40 x 13 + 30 x 21 + 40 x 22 + 42 x 23 Transportation Problem: Example #2

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10 n Define the Constraints Supply constraints: Supply constraints: (1) x 11 + x 12 + x 13 = 50 (1) x 11 + x 12 + x 13 = 50 (2) x 21 + x 22 + x 23 = 50 (2) x 21 + x 22 + x 23 = 50 Demand constraints: Demand constraints: (4) x 11 + x 21 = 25 (4) x 11 + x 21 = 25 (5) x 12 + x 22 = 45 (5) x 12 + x 22 = 45 (6) x 13 + x 23 = 10 (6) x 13 + x 23 = 10 Non-negativity of variables: Non-negativity of variables: x ij > 0, i = 1, 2 and j = 1, 2, 3 x ij > 0, i = 1, 2 and j = 1, 2, 3 Transportation Problem: Example #2 = Constraints

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11 n Define the Constraints Supply constraints: Supply constraints: (1) x 11 + x 12 + x 13 ≤ 50 (1) x 11 + x 12 + x 13 ≤ 50 (2) x 21 + x 22 + x 23 ≤ 50 (2) x 21 + x 22 + x 23 ≤ 50 Demand constraints: Demand constraints: (4) x 11 + x 21 ≥ 25 (4) x 11 + x 21 ≥ 25 (5) x 12 + x 22 ≥ 45 (5) x 12 + x 22 ≥ 45 (6) x 13 + x 23 ≥ 10 (6) x 13 + x 23 ≥ 10 Non-negativity of variables: Non-negativity of variables: x ij > 0, i = 1, 2 and j = 1, 2, 3 x ij > 0, i = 1, 2 and j = 1, 2, 3 Transportation Problem: Example #2 ≤ and ≥ Constraints

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12 n Partial Spreadsheet Showing Problem Data Transportation Problem: Example #1

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13 Transportation Problem: Example #1

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14 Transportation Problem: Example #1 Sensitivity Report

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15 Transportation Problem: Example #2 The Navy has 9,000 pounds of material in Albany, Georgia that it wishes to ship to three installations: San Diego, Norfolk, and Pensacola. They require 4,000, 2,500, and 2,500 pounds, respectively. Government regulations require equal distribution of shipping among the three carriers. The shipping costs per pound for truck, railroad, and airplane transit are shown on the next slide. Formulate and solve a linear program to determine the shipping arrangements (mode, destination, and quantity) that will minimize the total shipping cost.

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16 Destination Destination Mode San Diego Norfolk Pensacola Truck $12 $ 6 $ 5 Railroad 20 11 9 Airplane 30 26 28 Transportation Problem: Example #2

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17 n Define the Decision Variables We want to determine the pounds of material, x ij, to be shipped by mode i to destination j. The following table summarizes the decision variables: San Diego Norfolk Pensacola San Diego Norfolk Pensacola Truck x 11 x 12 x 13 Truck x 11 x 12 x 13 Railroad x 21 x 22 x 23 Railroad x 21 x 22 x 23 Airplane x 31 x 32 x 33 Airplane x 31 x 32 x 33 Transportation Problem: Example #2

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18 n Define the Objective Function Minimize the total shipping cost. Minimize the total shipping cost. Min: (shipping cost per pound for each mode per destination pairing) x (number of pounds shipped by mode per destination pairing). Min: (shipping cost per pound for each mode per destination pairing) x (number of pounds shipped by mode per destination pairing). Min: 12 x 11 + 6 x 12 + 5 x 13 + 20 x 21 + 11 x 22 + 9 x 23 Min: 12 x 11 + 6 x 12 + 5 x 13 + 20 x 21 + 11 x 22 + 9 x 23 + 30 x 31 + 26 x 32 + 28 x 33 + 30 x 31 + 26 x 32 + 28 x 33 Transportation Problem: Example #2

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19 n Define the Constraints Equal use of transportation modes: Equal use of transportation modes: (1) x 11 + x 12 + x 13 = 3000 (1) x 11 + x 12 + x 13 = 3000 (2) x 21 + x 22 + x 23 = 3000 (2) x 21 + x 22 + x 23 = 3000 (3) x 31 + x 32 + x 33 = 3000 (3) x 31 + x 32 + x 33 = 3000 Destination material requirements: Destination material requirements: (4) x 11 + x 21 + x 31 = 4000 (4) x 11 + x 21 + x 31 = 4000 (5) x 12 + x 22 + x 32 = 2500 (5) x 12 + x 22 + x 32 = 2500 (6) x 13 + x 23 + x 33 = 2500 (6) x 13 + x 23 + x 33 = 2500 Non-negativity of variables: Non-negativity of variables: x ij > 0, i = 1, 2, 3 and j = 1, 2, 3 x ij > 0, i = 1, 2, 3 and j = 1, 2, 3 Transportation Problem: Example #2

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20 Transportation Problem: Example #2

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21 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

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22 Transportation Problem n Linear Programming Formulation (continued) x ij > 0 for all i and j >

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23 n LP Formulation Special Cases Total supply exceeds total demand: Total supply exceeds total demand: Total demand exceeds total supply: Total demand exceeds total supply: Add a dummy origin with supply equal to the shortage amount. Assign a zero shipping cost per unit. The amount “shipped” from the dummy origin (in the solution) will not actually be shipped. Transportation Problem No modification of LP formulation is necessary.

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24 n LP Formulation Special Cases (continued) 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.

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25 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.

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26 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

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27 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 so that total mileage is minimized? Assignment Problem: Example

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28 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

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29 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

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30 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

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31 Transportation Problem: Example #2

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32 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

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33 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

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34 n Linear Programming Formulation (continued) Assignment Problem x ij > 0 for all i and j

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35 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.

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36 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:

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37 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.

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38 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

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39 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

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40 Transshipment Problem x ij > 0 for all i and j n Linear Programming Formulation (continued) continued

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41 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.

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42 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

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43 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 Zeron N 1 5 8 Zeron N 1 5 8 Zeron S 3 4 4 Zeron S 3 4 4 Transshipment Problem: Example

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44 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

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45 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

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46 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

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47 Transshipment Problem: Example

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48 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

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49 Transshipment Problem: Example

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50 n Computer Output (continued) Constraint Slack/Surplus Dual Values Constraint Slack/Surplus Dual Values 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

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51 n Computer Output (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

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52 n Computer Output (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

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53 Transshipment Transformed into Transportation Problem

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54 End of Chapter 6, Part A

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