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Transportation problem

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Transportation problem

  1. 1. Transportation Modeling Problem 10/13/13 1
  2. 2. THE PROBLEM The Epsilon Computers Co. sells desktop computers to universities along University belt, and ship them from three distribution warehouses. The firm is able to supply the following numbers of desktop computers to the universities by the beginning of the academic year: 10/13/13 2
  3. 3. Distribution warehouse Supply Sta. Mesa Taft Ave Divisoria 150 200 50 total 400 Universities have ordered desktop computers that must be delivered and installed by the beginning of the academic year: University/College Demand CEU FEU UE 100 80 220 total 400 10/13/13 3
  4. 4. The shipping cost per desktop computer from each distributor to each university are as follows: To From 1(Sta. Mesa) 2 (Taft Ave) 3 (Divisoria) A B (CEU) (FEU) 7 10 6 5 12 3 C (UE) 9 10 4 With cost minimization as criterion, Epsilon Company wants to determine how many desktop computers should be shipped from each warehouse to each university. Compare alternatives using, a.Northwest Corner Rule (NCR) b. Least Cost Method (LCM) c. Stepping Stone Method d. Vogel’s Approximation Method (VAM) 10/13/13 4
  5. 5. From To 1 (STA. MESA) 2 (TAFT AVE) 3 (DIVISORIA) Demand 10/13/13 A (CEU) B (FEU) X1a 7 X1b 5 100 9150 12 X2b 10 X3b 80 200 14 50 X2c 3 6 X3a Supply X1c 10 X2a C (UE) X3c 220 400 5
  6. 6. Objective Function: C= cost of shipment of all Xij= no. of computerd delivered i= origin ; j= destination Minimize: C= 7X1a + 5X1b + 9X1c + 10X2a + 12X2b + 10X2c + 6X3a + 3X3b + 14 X3c Constraints: 10/13/13 X1a +X1b + X1c= 150 X2a + X2b + X2c = 200 X3a + X3b + X3c = 50 X1a + X2a + X3a= 100 X1b + X2b + X3b= 80 X1c+ X2c + x3c= 220 Xij ≥ 0 6
  7. 7. From To A (CEU) B (FEU) 7 1 (STA. MESA) 100 Supply 5 9150 12 10 50 10 2 30 (TAFT AVE) 100 80 200 170 3 6 3 (DIVISORIA) Demand C (UE) 50 220 14 50 400 C= 700 + 250 + 360 + 1700 + 700 = P 3710 10/13/13 7
  8. 8. From To A (CEU) B (FEU) 100 7 30 5 20 2 (TAFT AVE) 10 12 3 (DIVISORIA) 6 1 (STA. MESA) Demand 100 LEGEND: 10/13/13 12345 50 80 C (UE) Supply 9150 200 10 14 3 220 200 50 400 C = P 3190 8
  9. 9. From To A (CEU) B (FEU) 7 1 (STA. MESA) 100 5 9150 12 10 10 30 (TAFT AVE) 100 80 200 170 3 6 3 (DIVISORIA) 10/13/13 Supply 50 2 Demand C (UE) 50 220 Table derived through Northwest Corner Rule 14 50 400 9
  10. 10. Compute for Improvement Indices: Improvement Index- the increase/decrease in total cost that would result from reallocating one unit to an unused square. Unused Square Closed Path Improvement Indices X1c +X1c-X1b+X2b-X2c +9-5+12-10= +6 X2a +X2a-X1a+X1b-X2b +10-7+5-12= -4 X3a +X3a-X1a+X1b-X2b+X2c-X3c +6-7+5-12+10-14= -12 X3b +X3b-X2b+X2c-X3c +3-12+10-14= -13 10/13/13 10
  11. 11. From To A (CEU) B (FEU) 7 1 (STA. MESA) 100 5 9150 12 10 10 30 (TAFT AVE) 100 80 200 170 3 6 3 (DIVISORIA) 10/13/13 Supply 50 2 Demand C (UE) 50 220 14 50 400 11
  12. 12. From To A (CEU) B (FEU) 7 1 (STA. MESA) 100 C (UE) 5 9150 12 10 200 14 50 50 10 2 (TAFT AVE) 6 3 (DIVISORIA) Demand Supply 100 30 80 3 200 20 220 400 Improved Transportation Tableau @ C= P 3,320 10/13/13 12
  13. 13. NOTE: Test the solution for improvement by computing again the improvement indices of the improved transportation tableau. If all indices are positive then an optimal solution has been obtained, otherwise, repeat the same steps for SSM. Unused Square Closed Path Improvement Indices X1c +X1c-X1b+X3b-X3c +9-5+3-14= -7 X2a +X2a-X1a+X1b-X3b+X3c-X2b +10-7+5-3+14-10= +9 X2b +X2b-X2c+X3c-X3b +12-10+14-3= +13 X3a +X3a-X1a+X1b-X3b +6-7+5-3= +1 10/13/13 13
  14. 14. From To A (CEU) B (FEU) 7 1 (STA. MESA) 100 C (UE) 5 9150 12 10 200 14 50 50 10 2 (TAFT AVE) 6 3 (DIVISORIA) Demand Supply 100 30 80 3 200 20 220 400 Previously improved transportation tableau 10/13/13 14
  15. 15. From To A (CEU) B (FEU) 7 1 (STA. MESA) 100 C (UE) 5 30 12 10 200 14 50 200 (TAFT AVE) 6 3 (DIVISORIA) Demand 9150 20 10 2 Supply 100 50 80 3 220 400 New improved Transportation Tableau @ C= P 3180 10/13/13 15
  16. 16. Again, computing for improvement indices: Unused Square X2a +X2a-X1a+X1c-X2c Improvement Indices +10-7+9-10= +2 X2b +X2b-X1b+X1c-X2c +12-5+9-10= +6 X3a +X3a-X1a+X1b-X3b +6-7+5-3= +1 X3c +X3c-X3b+X1b-X1c +14-3+5-9= +7 Closed Path Since all improvement indices are all positive, optimal solution is obtained @ C= P 3,180. 10/13/13 16
  17. 17. This method is an algorithm that finds an initial feasible solution to transportation problem by considering the “penalty cost” of not choosing cheapest available route. Opportunity Cost- the cost of opportunities that are sacrificed in order to take a specific action. 10/13/13 17
  18. 18. 1. For each row with an available supply and each column with an unfilled demand, calculate an opportunity/penalty cost by subtracting the smallest cost per unit entry from the second smallest entry for a minimization problem. While, for maximization problem, opportunity cost is calculated by getting the difference between the highest and the second highest entry. 2. Identify the row or column with largest opportunity(difference) cost. 3. Allocate maximum amount possible to the available route with the lowest cost for minimization or highest revenue for maximization in the row/column selected from step 2. 4.Reduce appropriate supply and demand by the amount allocated in step 3. 5. Remove any rows with zero available supply and columns with unfilled demand for further consideration. 6.Return to step 1. 10/13/13 18
  19. 19. From To A (CEU) 7 1 (STA. MESA) X1a 2 (TAFT AVE) 3 (DIVISORIA) Demand B (FEU) C (UE) 5 X1b 12 X2b 100 10 X3b 80 200 14 50 X2c 3 6 X3a 9 150 X1c 10 X2a Supply X3c 220 400 Original Transportation Tableau 10/13/13 19
  20. 20. OPPORTUNITY COST FOR THE FIRST ALLOCATION Row/ Column Lowest Cost Opportunity Cost 1 7 5 2 2 10 10 0 3 6 3 3 A 7 6 1 B 5 3 2 C 10/13/13 Second Lowest Cost 10 9 1 20
  21. 21. From To A (CEU) B (FEU) C (UE) Supply 7 5 9150 2 10 12 10 200 (TAFT AVE) 3 (DIVISORIA) 6 14 50 1 (STA. MESA) Demand 100 50 3 80 220 400 First allocation 10/13/13 21
  22. 22. OPPORTUNITY COST FOR THE SECOND ALLOCATION Row/ Column Lowest Cost Opportunity Cost 1 7 5 2 2 10 10 0 A 10 7 1 B 12 5 7 C 10/13/13 Second Lowest Cost 10 9 1 22
  23. 23. From To A (CEU) B (FEU) 7 1 (STA. MESA) Supply 5 9150 12 10 200 14 50 30 2 10 (TAFT AVE) 3 (DIVISORIA) 6 Demand C (UE) 100 50 3 80 220 400 Second allocation 10/13/13 23
  24. 24. OPPORTUNITY COST FOR THE THIRD ALLOCATION Row/ Column Lowest Cost Opportunity Cost 1 9 7 2 2 10 10 0 A 10 7 1 C 10/13/13 Second Lowest Cost 10 9 1 24
  25. 25. From To 1 (STA. MESA) A (CEU) B (FEU) 7 100 30 (TAFT AVE) 3 (DIVISORIA) 6 12 80 220 10 200 14 200 50 3 Third allocation 10/13/13 9150 20 10 100 Supply 5 2 Demand C (UE) 50 400 C= P 3180 25
  26. 26. • DEGENERATE PROBLEMS >> (column + row) – 1 , SSM not applicable >> put zero (0) to any unused square and proceed with SSM steps • UNBALANCED PROBLEMS >> demand ≠ supply >> add dummy column/row 10/13/13 26
  27. 27. Otsukaresamadeshita!!! Thank You for listening!!!  10/13/13 27

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