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Transportation Problems MHA 6350

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Medical Supply Transportation Problem A Medical Supply company produces catheters in packs at three productions facilities. The company ships the packs from the production facilities to four warehouses. The packs are distributed directly to hospitals from the warehouses. The table on the next slide shows the costs per pack to ship to the four warehouses.

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Medical Supply Seattle New York Phoenix Miami FROM PLANT Juarez $19 $ 7 $ 3 $21 Seoul 1521 18 6 Tel Aviv 1114 15 22 TO WAREHOUSE Capacity Juarez100 Seoul300 Tel Aviv200 Demand Seattle150 New York100 Phoenix200 Miami150 Source: Adapted from Lapin, 1994

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JXjsXjnXjpXjm100 SNPM XssXsnXspXsm XtsXtnXtpXtm 150100200150 S300 T200 Warehouse Demand 600 TO WAREHOUSE Plant Capacity From Plant Number of constraints = number of rows + number of columns Total plant capacity must equal total warehouse demand. Although this may seem unrealistic in real world application, it is possible to construct any transportation problem using this model. Source: Adapted from Lapin, 1994

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SNPM 150100200150 J100 S300 T200 Demand 600 Capacity From To 197321 6 22 18 15 21 14 15 11 Northwest Corner Method Begin with a blank shipment schedule. Note the shipping costs in the upper right hand corner of each cell. Source: Adapted from Lapin, 1994

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SNPM 150100200150 J100 S300 T200 Demand 600 Capacity From To 197321 6 22 18 15 21 14 15 11 Northwest Corner Method 100 Start in the upper left-hand corner, “northwest corner” of the schedule and place the largest amount of capacity and demand available in that cell. Seattle demands 150 and Jaurez has a capacity of 100. Source: Adapted from Lapin, 1994

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SNPM 150100200150 J100 S300 T200 Demand 600 Capacity From To 197321 6 22 18 15 21 14 15 11 Northwest Corner Method 100 Since Juarez capacity is depleted move down to repeat the process for the Seoul to Seattle cell. Seoul has sufficient capacity but Seattle can only take another 50 packs of demand. 50 Source: Adapted from Lapin, 1994

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SNPM 150100200150 J100 S300 T200 Demand 600 Capacity From To 197321 6 22 18 15 21 14 15 11 Northwest Corner Method 100 50 Now move to the next cells to the right and assign capacity for Seoul to warehouse demand until depleted. Then move down to the Tel Aviv row and repeat the process. 100150 50150 Source: Adapted from Lapin, 1994

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SNPM 150100200150 J100 S300 T200 Demand 600 Capacity From To 197321 6 22 18 15 21 14 15 11 Northwest Corner Method The previous slides show the process of satisfying all constraints and allows us to begin with a starting feasible solution. Multiply the quantity in each cell by the cost. 1900 21007502700 7503300 1900 750 2100 2700 750 3300 C =11,500 Source: Adapted from Lapin, 1994

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r j = 0 k s = 19 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 a b For non empty cells: c ij = r i + k j Assign zero as the row number for the first row. 19 = (0) + k s Source: Adapted from Lapin, 1994

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r j = 0 r s = -4 k s = 19 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 c a b For non empty cells: c ij = r i + k j Assign zero as the row number for the first row. 15 = r s + 19 r s = -15 + 19 = -4 Note: Always use the newest r value to compute the next k. Source: Adapted from Lapin, 1994

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r j = 0 r s = -4 k s = 19k p = 22 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 * c a bd For non empty cells: c ij = r i + k j Assign zero as the row number for the first row. 18 =-4 + k p 18 + 4 = k p = 22 Skip cell SN, mark it * for later and move on to cell SP. Source: Adapted from Lapin, 1994

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r j = 0 r s = -4 r t = -7 k s = 19k p = 22 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 * c a e bd For non empty cells: c tp = r t + k p Assign zero as the row number for the first row then use the newest r value to compute the next k. 15 = r t + 22 15 - 22 = r t = -7 Source: Adapted from Lapin, 1994

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r j = 0 r s = -4 r t = -7 k s = 19k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 * c a e bdf For non empty cells: c ij = r i + k j Assign zero as the row number for the first row then use the newest r value to compute the next k. 22 = -7 + k m 22 + 7 = k m = 29 Source: Adapted from Lapin, 1994

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r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 * c a e bgdf For non empty cells: c ij = r i + k j Assign zero as the row number for the first row then use the newest r value to compute the next k. 21= -4 + k n 21 + 4 = k n = 25 Source: Adapted from Lapin, 1994

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r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 c a e bgdf Next calculate empty cells using: c ij - r i - k j JN = 7 – 0 – 25 = -18 -18 Improvement Difference >>

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 c a e bgdf Next calculate empty cells using: c ij - r i - k j JP = 3 – 0 – 22 = -19 -18 -19 Improvement Difference >>

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 c a e bgdf Next calculate empty cells using: c ij - r i - k j JM = 21 – 0 – 29 = -8 -18 -19-8 Improvement Difference >>

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 c a e bgdf Next calculate empty cells using: c ij - r i - k j SM = 6 – (-4) – 29 = -19 -18 -19-8 -19 Improvement Difference >>

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 c a e bgdf Next calculate empty cells using: c ij - r i - k j TS = 11 – (-7) – 19 = -1 -18 -19 -8 Improvement Difference >> -19

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 c a e bgdf Next calculate empty cells using: c ij - r i - k j TN = 14 – (-7) – 25 = -4 -18 -19 -8 -4 Improvement Difference >> -19

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 c a e bgdf Next calculate empty cells using: c ij - r i - k j -18 -19 -8 -4 Improvement Difference >> -19

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 100 50100150 50150 Next calculate the entering cell by finding the empty cell with the greatest absolute negative improvement difference. -18-19-8 -4 -19 Cells JP and SM are tied for the greatest improvement at $19 per pack. Break the tie and arbitrarily choose JP. JP becomes the entering cell. Place a + sign in cell JP (+) (-) (+)(-) Note: Except for the entering cell all changes must involve nonempty cells. 100

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = -7 k s = 19k n = 25k p = 22k m = 29 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 50 100 50150 -18-19-8 -4 -19 Continue around the closed loop until all tradeoffs are completed. (+) (-) (+)(-) Note: Except for the entering cell all changes must involve nonempty cells. 100 150 50 Previous cost was $11,500 and the new is: 300 2250 2100 900 750 3300 C = $9,600

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Source: Adapted from Lapin, 1994 r j = 0 r s = 15 r t = 12 k s = 0k n = 6k p = 3k m = 10 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 50 100 50 150 11911 -4 -19 (+) (-)(+) (-) Note: The r and k values and the improvement difference values have changed. 100 150 50 Begin another iteration choosing the empty cell with the greatest absolute negative improvement difference. >>>>>SM

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Source: Adapted from Lapin, 1994 r j = 0 r s = 15 r t = 12 k s = 0k n = 6k p = 3k m = 10 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 50 100 50 150 11911 -4 -19 (+) (-)(+) (-) Note: The r and k values and the improvement difference values have changed. 100 150 50 Previous cost was $9,600, now the new is: 300 2250 2100 300 1500 2200 C = $8,650 Begin another iteration choosing the empty cell with the greatest absolute negative improvement difference. SM 50 100

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Source: Adapted from Lapin, 1994 r j = 0 r s = -4 r t = 12 k s = 19k n = 25k p = 3k m = 10 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 50 100 -18011 -20-23 -19 (+) (-)(+) (-) Note: The r and k values and the improvement difference values have changed. 100 150 Previous cost was $8,650, now the new is: 300 2250 900 1400 1500 C = $6,350 Begin another iteration choosing the empty cell with the greatest absolute negative improvement difference. SM 50 100 150 0 k n = 2

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Source: Adapted from Lapin, 1994 r j = 0 r s = 16 r t = 12 k s = -1k p = 3k m = 10 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 52031 20 3 (+) (-)(+) (-) Note: The r and k values and the improvement difference values have changed. 100 150 $6,350 300 2250 900 1400 1500 C = $6,350 Begin another iteration choosing the empty cell with the greatest absolute negative improvement difference. SM 100 150 0 k n = 2

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Source: Adapted from Lapin, 1994 r j = 0 r s = 15 r t = 11 k s = 0k p = 3k m = -9 SNPM 150100200150 J100 S300 T200 Demand600 Capacity From To 197321 6 22 18 15 21 14 15 11 41930 20 3 1 (+) (-)(+) (-) Note: The r and k values and the improvement difference values have changed. 100 50 $6,250 300 750 1800 900 1100 1400 C = $6,250 Optimal Solution 100 150 100 k n = 3 In five iterations the shipping cost has moved from $11,500 to $6,250. There are no remaining empty cells with a negative value.

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