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Slide 1 of 82 Linear Programming: Transportation Problem J. Loucks, St. Edward's University (Austin, TX, USA) J. Loucks, St. Edward's University (Austin, TX, USA) J. Loucks, St. Edward's University (Austin, TX, USA) Chapter 2B

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Slide 2 of 82 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 problem (TP), as well as many other problems, are all examples of network problems. n Efficient solution algorithms exist to solve network problems. Transportation Problem

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Slide 3 of 82 n TP can be formulated as linear programs and solved by general purpose linear programming codes. n For the TP, 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, which contain separate computer codes for the TP which take advantage of its network structure. Transportation Problem

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Slide 4 of 82 Transportation Problem n The TP problem has the following characteristics: m sources and n destinations m sources and n destinations number of variables is m x nnumber of variables is m x n number of constraints is m + n (constraints are for source capacity and destination demand)number of constraints is m + n (constraints are for source capacity and destination demand) costs appear only in objective function (objective is to minimize total cost of shipping)costs appear only in objective function (objective is to minimize total cost of shipping) coefficients of decision variables in the constraints are either 0 or 1coefficients of decision variables in the constraints are either 0 or 1

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Slide 5 of 82 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. Transportation Problem

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Slide 6 of 82 Transportation Problem n Network Representation c 11 c 12 c 13 c 21 c 22 c 23 d1d1d1d1 d2d2d2d2 d3d3d3d3 s1s1s1s1 s2s2 SOURCESDESTINATIONS

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Slide 7 of 82 The Relationship of TP to LP n TP is a special case of LP n How do we formulate TP as an LP? Let x ij = quantity of product shipped from source i to destination jLet x ij = quantity of product shipped from source i to destination j Let c ij = per unit shipping cost from source i to destination jLet c ij = per unit shipping cost from source i to destination j Let s i be the row i total supplyLet s i be the row i total supply Let d j be the column j total demandLet d j be the column j total demand n The LP formulation of the TP problem is:

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Slide 8 of 82 n LP Formulation The linear programming 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 The Relationship of TP to LP

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Slide 9 of 82 Transportation Problem n LP Formulation Special Cases The following special-case modifications to the linear programming formulation can be made: Minimum shipping guarantees from i to j :Minimum shipping guarantees 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 routes:Unacceptable routes: delete the variable (or put a very high cost, called the Big-M method, on a selected pair of i and j) delete the variable (or put a very high cost, called the Big-M method, on a selected pair of i and j)

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Slide 10 of 82 n To solve the transportation problem by its special purpose algorithm, it is required that the sum of the supplies at the origins equal the sum of the demands at the destinations. 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. Similarly, 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). Transportation Problem

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Slide 11 of 82 n The transportation problem is solved in two phases: Phase I — Obtaining an initial feasible solutionPhase I — Obtaining an initial feasible solution Phase II — Moving toward optimalityPhase II — Moving toward optimality n In 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 In 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. Transportation Problem

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Slide 12 of 82 n Phase I - Minimum-Cost Method (simple and intuitive) 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. Transportation Algorithm

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Slide 13 of 82 Transportation Algorithm n Phase II - Stepping Stone Method Step 1: For each unoccupied cell, calculate the reduced cost by the MODI method described below. 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.Step 1: For each unoccupied cell, calculate the reduced cost by the MODI method described below. 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. 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.

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Slide 14 of 82 Transportation Algorithm 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. (Note: An occupied cell on the stepping stone path now becomes 0 (unoccupied).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.) GO TO STEP 1.

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Slide 15 of 82 Transportation Algorithm 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.

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Slide 16 of 82 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 Demand S1 S2D3D2D11515 Example 1: TP

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Slide 17 of 82 Example 1: TP 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. How should end of week shipments be made to fill the above orders given the following delivery cost per ton: Northwood Westwood Eastwood Northwood Westwood Eastwood Plant Plant Plant Plant

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Slide 18 of 82 Example 1: TP 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 DemandSupply Dummy Plant 1 Plant 2 EastwoodWestwoodNorthwood2424

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Slide 19 of 82 Example 1: TP n Least Cost Starting 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. 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.

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Slide 20 of 82 Example 1: TP n Iteration 1 MODI MethodMODI 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 = Since u i + v 2 = c i 2 for occupied cells in column 2, then u = 40, hence u 2 = Since u 2 + v j = c 2 j for occupied cells in row 2, then 10 + v 3 = 42, hence v 3 = v 3 = 42, hence v 3 = 32.

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Slide 21 of 82 Example 1: TP n 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) = 8 (1,3) = 8 (2,1) = -4 (2,1) = -4 (2,4) = -10 (2,4) = -10

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Slide 22 of 82 Example 1: TP n Iteration 1 Tableau vjvjvjvj uiuiuiui Dummy Plant 1 Plant 2 EastwoodWestwoodNorthwood 2424

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Slide 23 of 82 Example 1: TP n 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 = = 20 (0 is its current allocation) x 14 = = 0 (blank for the next tableau) x 14 = = 0 (blank for the next tableau) x 12 = = 25 x 12 = = 25 x 22 = = 20 x 22 = = 20 The other occupied cells remain the same. The other occupied cells remain the same.

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Slide 24 of 82 Example 1: TP n Iteration 2 MODI MethodMODI Method The reduced costs are found by calculating the u i 's and v j 's for this tableau. 1. Set u 1 = 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 = Since u i + v 2 = c i 2 for occupied cells in column 2, then u = 40, or u 2 = 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 = v 3 = 42 or v 3 = 32; and, 10 + v 4 = 0 or v 4 = -10.

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Slide 25 of 82 Example 1: TP n 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) = 8 (1,3) = 8 (1,4) (-10) = 10 (1,4) (-10) = 10 (2,1) = -4 (2,1) = -4

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Slide 26 of 82 Example 1: TP n Iteration 2 Tableau vjvjvjvj uiuiuiui Dummy Plant 1 Plant 2 EastwoodWestwoodNorthwood 2424

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Slide 27 of 82 Example 1: TP n Iteration 2 Stepping Stone MethodStepping 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. Thus for the next tableau: x 21 = = 20 (0 is its current allocation) x 21 = = 20 (0 is its current allocation) x 11 = = 5 x 11 = = 5 x 12 = = 45 x 12 = = 45 x 22 = = 0 (blank for the next tableau) x 22 = = 0 (blank for the next tableau) The other occupied cells remain the same. The other occupied cells remain the same.

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Slide 28 of 82 Example 1: TP n Iteration 3 MODI MethodMODI 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 = Since u i + v 1 = c i 1 for occupied cells in column 2, then u = 30 or u 2 = 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 = v 3 = 42 or v 3 = 36, and 6 + v 4 = 0 or v 4 = -6.

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Slide 29 of 82 Example 1: TP n 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) = 4 (1,3) = 4 (1,4) (-6) = 6 (1,4) (-6) = 6 (2,2) = 4 (2,2) = 4

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Slide 30 of 82 Example 1: TP n Iteration 3 Tableau Since all the reduced costs are non-negative, this is the optimal tableau vjvjvjvj uiuiuiui Dummy Plant 1 Plant 2 EastwoodWestwoodNorthwood 2424

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Slide 31 of 82 Example 1: TP n Optimal Solution From To Amount Cost From To Amount Cost Plant 1 Northwood Plant 1 Westwood 45 1,350 Plant 1 Westwood 45 1,350 Plant 2 Northwood Plant 2 Northwood Plant 2 Eastwood Plant 2 Eastwood Total Cost = $2,490 Total Cost = $2,490

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Slide 32 of 82 Example 2: TP Des Moines (100 units capacity) Cleveland (200 units Req.) Boston (200 units req.) Fort Lauderdale Fort Lauderdale (300 units capacity) Albuquerque (300 units Req.) Evansville (300 units capacity)

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Slide 33 of 82 Example 2: TP

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Slide 34 of 82 Example 2: TP

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Slide 35 of 82 Example 2: TP n What is the initial solution? Since we want to minimize the cost, it is reasonable to “be greedy” and start at the cheapest cell ($3/unit) while shipping as much as possible. This method of getting the initial feasible solution is called the least- cost ruleSince we want to minimize the cost, it is reasonable to “be greedy” and start at the cheapest cell ($3/unit) while shipping as much as possible. This method of getting the initial feasible solution is called the least- cost rule Any ties are broken arbitrarily. (Des Moines to Cleveland - $3) (Evansville to Cleveland - $3) Let’s start ad cell 1-3 (Des Moines to Cleveland)Any ties are broken arbitrarily. (Des Moines to Cleveland - $3) (Evansville to Cleveland - $3) Let’s start ad cell 1-3 (Des Moines to Cleveland) In making the flow assignments make sure not to exceed the capacities and demands on the margins!In making the flow assignments make sure not to exceed the capacities and demands on the margins!

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Slide 36 of 82 Example 2: TP Start Start

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Slide 37 of 82 Example 2: TP

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Slide 38 of 82 Example 2: TP

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Slide 39 of 82 Example 2: TP

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Slide 40 of 82 Example 2: TP n The initial feasible solution via the least-cost rule is: Ship 100 units from Des Moines to ClevelandShip 100 units from Des Moines to Cleveland Ship 100 units from Evansville to ClevelandShip 100 units from Evansville to Cleveland Ship 200 units from Evansville to BostonShip 200 units from Evansville to Boston Ship 300 units from Fort Lauderdale to AlbuquerqueShip 300 units from Fort Lauderdale to Albuquerque n The total cost of this solution is: 100(3) + 100(3) + 200(4) + 300(9) = $4,100100(3) + 100(3) + 200(4) + 300(9) = $4,100 n Is the current solution optimal? Perhaps, but we are not sure. Hence, some more work is needed.Perhaps, but we are not sure. Hence, some more work is needed.

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Slide 41 of 82 Example 2: TP n The cells that have a number in them are referred to as basic cells (or basic variables) In general, all basic variables are strictly positiveIn general, all basic variables are strictly positive When at least one basic variable is equal to zero, the corresponding solution is called degenerateWhen at least one basic variable is equal to zero, the corresponding solution is called degenerate When a solution is degenerate, exercise caution in using sensitivity reportsWhen a solution is degenerate, exercise caution in using sensitivity reports

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Slide 42 of 82 Example 2: TP n How many basic variables (cells) should there be in the transportation tableau? Answer: (R+C-1), where R = number of rows and C = number of columns. Here, (R+C-1) = = 5Answer: (R+C-1), where R = number of rows and C = number of columns. Here, (R+C-1) = = 5 n What are they (there are 4 of them)? Ship 100 units from Des Moines to ClevelandShip 100 units from Des Moines to Cleveland Ship 100 units from Evansville to ClevelandShip 100 units from Evansville to Cleveland Ship 200 units from Evansville to BostonShip 200 units from Evansville to Boston Ship 300 units from Fort Lauderdale to AlbuquerqueShip 300 units from Fort Lauderdale to Albuquerque

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Slide 43 of 82 Example 2: TP n Is this solution degenerate? Answer: Yes, because (R+C-1) = = 5 > 4Answer: Yes, because (R+C-1) = = 5 > 4 Hence, we can arbitrarily make (5-4) = 1 variable which is at level zero basic.Hence, we can arbitrarily make (5-4) = 1 variable which is at level zero basic. Let us make Des Moines to Albuquerque a basic variable at level zeroLet us make Des Moines to Albuquerque a basic variable at level zero n How do we determine if the current solution is optimal? Let us price currently nonbasic variables (cells), which are all at level zero using the stepping stone algorithmLet us price currently nonbasic variables (cells), which are all at level zero using the stepping stone algorithm

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Slide 44 of 82 Example 2: TP

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Slide 45 of 82 Example 2: TP n Evansville – Albuquerque: 8 – – 5 = 3 > 08 – – 5 = 3 > 0

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Slide 46 of 82 Example 2: TP

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Slide 47 of 82 Example 2: TP n Des Moines - Boston: 4 – = 04 – = 0

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Slide 48 of 82 Example 2: TP

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Slide 49 of 82 Example 2: TP n Fort Lauderdale - Boston: 7 – – 4 = -1 < 07 – – 4 = -1 < 0

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Slide 50 of 82 Example 2: TP

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Slide 51 of 82 Example 2: TP n Fort Lauderdale - Cleveland: 5 – = -2 < 05 – = -2 < 0

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Slide 52 of 82 Example 1: TP n The objective function improvement potentials for the nonbasic variables are: Evansville – Albuquerque: 3 > 0Evansville – Albuquerque: 3 > 0 Des Moines - Boston: 0 = 0Des Moines - Boston: 0 = 0 Fort Lauderdale - Boston: -1 < 0Fort Lauderdale - Boston: -1 < 0 Fort Lauderdale - Cleveland: -2 < 0Fort Lauderdale - Cleveland: -2 < 0 n The largest improvement (here: reduction in cost) in the objective function can be achieved by increasing shipments from Fort Lauderdale to Cleveland n What i2 the limit on the increase?

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Slide 53 of 82 Example 2 TP 0 +K K 0 +K K K +K

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Slide 54 of 82 Example 2: TP n If we increase shipments from Fort Lauderdale to Cleveland by K, we must reduce shipments in the basic cells in the same row and column so that the totals on the margins remain the same n Clearly, because negative quantities cannot be shipped, we must impose that: K >= 0 (no problem)K >= 0 (no problem) 300 – K >= 0 (K = 0 (K <= 300) 0 + K >= 0 (no problem)0 + K >= 0 (no problem) 100 – K >=0 (K =0 (K <=100) Hence, let K = 100 n The new basis becomes:

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Slide 55 of 82 Example 2: TP

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Slide 56 of 82 Example 2: TP n Observe that the number of basic variables (cells) is equal to (R+C-1) = 5. Hence, the current basic solution is feasible and not degenerate n Also observe that Des Moines – Cleveland left the basis while Fort Lauderdale – Cleveland entered the basis (one for one interchange) n Is the current solution optimal? Let us price the non-basic variables (cells):Let us price the non-basic variables (cells):

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Slide 57 of 82 Example 2: TP

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Slide 58 of 82 Example 2: TP n Evansville - Albuquerque: 8 – – 9 = 1 > 08 – – 9 = 1 > 0

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Slide 59 of 82 Example 2: TP

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Slide 60 of 82 Example 2: TP n Des Moines - Boston: 4 – – – 4 = 2 > 04 – – – 4 = 2 > 0

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Slide 61 of 82 Example 2: TP

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Slide 62 of 82 Example 2: TP n Fort Lauderdale - Boston: 7 – = 1 > 07 – = 1 > 0

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Slide 63 of 82 Example 2: TP

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Slide 64 of 82 Example 2: TP n Des Moines - Cleveland: 3 – = 2 > 03 – = 2 > 0

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Slide 65 of 82 Example 2: TP n The objective function improvement potentials for the nonbasic variables are: Evansville - Albuquerque: 1 > 0Evansville - Albuquerque: 1 > 0 Des Moines - Boston: 2 > 0Des Moines - Boston: 2 > 0 Fort Lauderdale - Boston: 1 > 0Fort Lauderdale - Boston: 1 > 0 Des Moines - Cleveland: 2 > 0Des Moines - Cleveland: 2 > 0 n Since all improvement potentials are positive, it is not possible to achieve further improvement (here: reduction in cost) in the objective function n Hence, the current solution is optimal

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Slide 66 of 82 Example 2: TP n The final optimal (and not degenerate) solution is: Ship 100 units from Des Moines to AlbuquerqueShip 100 units from Des Moines to Albuquerque Ship 200 units from Fort Lauderdale to AlbuquerqueShip 200 units from Fort Lauderdale to Albuquerque Ship 200 units from Evansville to BostonShip 200 units from Evansville to Boston Ship 100 units from Evansville to ClevelandShip 100 units from Evansville to Cleveland Ship 100 units from Fort Lauderdale to ClevelandShip 100 units from Fort Lauderdale to Cleveland n The total cost of this solution is: 100(5) + 200(9) + 200(4) + 100(3) + 100(5)= $3,900100(5) + 200(9) + 200(4) + 100(3) + 100(5)= $3,900

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Slide 67 of 82 Example 2: TP via LP n In our example the LP formulation is: Min z = 5 x x x x x x x x x 33 s.t. x 11 + x 12 + x 13 <= 100(row 1) x 21 + x 22 + x 23 <= 300(row 2) x 31 + x 32 + x 33 <= 300(row 3) x 11 + x 21 + x 31 >= 300(column 1) x 12 + x 22 + x 32 >= 200(column 2) x 31 + x 32 + x 33 >= 200(column 3) x ij >= 0 for i = 1, 2, 3 and j = 1, 2, 3 (nonnegativity)

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Slide 68 of 82 Example 2: TP via LP n The solver formulation and solution is:

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Slide 69 of 82 Example 2: TP via LP n The solver answer report is:

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Slide 70 of 82 Example 2: TP via LP n The solver sensitivity report is:

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Slide 71 of 82 Example 2: TP via LP n The solver limits report is:

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Slide 72 of 82 Example 3: Tropicsun (TP via LP) Tropicsun is a grower of oranges with locations in the cities of Mt. Dora, Eustis and Clermont. Tropicsun currently has 275,000 bushels of citrus at the grove in Mt. Dora; 400,000 bushels in Eustis; and 300,000 bushels in Clermont. The citrus processing plants are located in Ocala (capacity of 200,000 bushels), Orlando (600,000 bushels) and Leesburg (225,000 bushels). Tropicsun contracts with a trucking company to transport its fruit, which charges a flat rate for every mile that each bushel of fruit must be transported. Observe that the problem is not balanced, since the total of the fruit supplies (275, , ,000 = 975,000) is not equal to the total of the capacities (200, , ,000 = 1,025,000). The difference of 50,000 bushels will be assigned to a dummy source, which represents unutilized capacity. The distances (in miles) between the groves and processing plants are as follows:

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Slide 73 of 82 Example 3: Tropicsun (TP via LP) Distances (in miles) between groves and plants GroveOcalaOrlandoLeesburg Mt. Dora Eustis Clermont552025

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Slide 74 of 82 Example 3: TP via LP : Tropicsun Mt. Dora 1 Eustis 2 Clermont 3 Ocala 4 Orlando 5 Leesburg 6 Distances (in miles) Capacity Supply 275, , , , , ,000 Groves Processing Plants

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Slide 75 of 82 Example 3: Defining the Decision Variables X ij = # of bushels shipped from node i to node j Specifically, the nine decision variables are: X 14 = # of bushels shipped from Mt. Dora (node 1) to Ocala (node 4) X 15 = # of bushels shipped from Mt. Dora (node 1) to Orlando (node 5) X 16 = # of bushels shipped from Mt. Dora (node 1) to Leesburg (node 6) X 24 = # of bushels shipped from Eustis (node 2) to Ocala (node 4) X 25 = # of bushels shipped from Eustis (node 2) to Orlando (node 5) X 26 = # of bushels shipped from Eustis (node 2) to Leesburg (node 6) X 34 = # of bushels shipped from Clermont (node 3) to Ocala (node 4) X 35 = # of bushels shipped from Clermont (node 3) to Orlando (node 5) X 36 = # of bushels shipped from Clermont (node 3) to Leesburg (node 6)

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Slide 76 of 82 Example 3: Defining the Objective Function Minimize the total number of bushel-miles. MIN:21X X X X X X X X X 36

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Slide 77 of 82 Example 3: Defining the Constraints n Capacity constraints X 14 + X 24 + X 34 <= 200,000} Ocala X 15 + X 25 + X 35 <= 600,000} Orlando X 16 + X 26 + X 36 <= 225,000} Leesburg Supply constraints Supply constraints X 14 + X 15 + X 16 = 275,000} Mt. Dora X 24 + X 25 + X 26 = 400,000} Eustis X 34 + X 35 + X 36 = 300,000} Clermont n Nonnegativity conditions X ij >= 0 for all i and j

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Slide 78 of 82 Example 3: Solver Formulation

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Slide 79 of 82 Example 3: Solver Solution

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Slide 80 of 82 Example 4: The Sentry Lock Corporation

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Slide 81 of 82 Example 4: Solver Formulation

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Slide 82 of 82 The End of Chapter 2B

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