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- 1. Assignment and Transportation Problem Amit Kumar Bardhan Faculty of Management Studies University of Delhi
- 2. Step1. Find a bfs. Find the minimum element in each row of the mxm cost matrix. Construct a new matrix by subtracting from each cost the minimum cost in its row. For this new matrix, find the minimum cost in each column. Construct a new matrix (reduced cost matrix) by subtracting from each cost the minimum cost in its column. Step2. Draw the minimum number of lines (horizontal and/or vertical) that are needed to cover all zeros in the reduced cost matrix. If m lines are required, an optimal solution is available among the covered zeros in the matrix. If fewer than m lines are required, proceed to step 3. Step3. Find the smallest nonzero element (call its value k) in the reduced cost matrix that is uncovered by the lines drawn in step 2. Now subtract k from each uncovered element of the reduced cost matrix and add k to each element that is covered by two lines. Return to step2. Assignment Problems: Hungarian method
- 3. Example 1: Assignment Problem Personnel Assignment: Time (hours) Job 1 Job 2 Job 3 Job 4 Person 1 14 5 8 7 Person 2 2 12 6 5 Person 3 7 8 3 9 Person 4 2 4 6 10
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- 5. Swimmer Selection [contd.] 31 28 29 26 29 Freestyle 33 30 38 28 33 Butterfly 41 34 42 33 43 Breaststroke 35 37 33 32 37 Backstroke Vir Ali Vipin Ajay Mark Stroke
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- 7. Swimmer Selection [Answer] Freestyle -> Mark Butterfly -> Ajay Breaststroke -> Ali Backstroke -> Vipin
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- 9. Flight Assignment [contd.] 1.54 √ √ √ 8 0.84 √ √ 7 1.12 √ √ 6 1.32 √ √ √ 5 1.60 √ √ √ 4 1.52 √ √ √ 3 0.96 √ √ 2 1.40 √ √ 1 6 5 4 3 2 1 Cost Flight Work Pattern
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- 11. Transportation Problem
- 12. Power Co: Shipping costs, Supply, and Demand for Powerco Example Transportation Tableau 30 30 20 45 Demand (Million kwh) 40 5 16 9 14 Plant 3 50 7 13 12 9 Plant 2 35 9 10 6 8 Plant 1 Supply (Million kwh) City 4 City 3 City 2 City 1 To From
- 13. Powerco Power Plant: Formulation Define Variables: Let Xij = number of (million kwh) produced at plant i and sent to city j. Objective Function: Min z = 8*X11 + 6*X12 + 10*X13 + 9*X14 + 9*X21 + 12*X22 + 13*X23 + 7*X24 + 14*X31 + 9*X32 + 16*X33 + 5*X34 Supply Constraints: X11 + X12 + X13 + X14 < = 35 (Plant 1) X21 + X22 + X23 + X24 < = 50 (Plant 2) X31 + X32 + X33 + X34 < = 40 (Plant 3)
- 14. Powerco Power Plant: Formulation (Cont’d.) Demand Constraints: X11 + X21 +X31 > = 45 (City 1) X12 + X22 +X32 > = 20 (City 2) X13 + X23 +X33 > = 30 (City 3) X14 + X24 +X34 > = 30 (City 4) Nonnegativity Constraints: Xij > = 0 (i=1,..,3; j=1,..,4) Balanced? Total Demand = 45 + 20 + 30 + 30 = 125 Total Supply = 35 + 50 + 40 = 125 Yes, Balanced Transportation Problem! (If problem is unbalanced, add dummy supply or demand as required.)
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- 16. Example: Cost-less Corp Ltd. [contd.] Plant 1, 2, 3 and 4 make 10, 20, 20 and 10 shipments respectively. Retail outlets 1, 2, 3 and 4 need to receive 20, 10, 10 and 20 shipments per month respectively. The distribution manager wants to determine the best plan for how many shipments to send from each plant to respective retail outlets each month. The objective is to minimize the total shipping cost.
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- 18. Transshipment Problems A transportation problem allows only shipments that go directly from supply points to demand points. In many situations, shipments are allowed between supply points or between demand points. Sometimes there may also be points (called transshipment points) through which goods can be transshipped on their journey from a supply point to a demand point. Fortunately, the optimal solution to a transshipment problem can be found by solving a transportation problem.
- 19. The following steps describe how the optimal solution to a transshipment problem can be found by solving a transportation problem. Step 1 : If necessary, add a dummy demand point (with a supply of 0 and a demand equal to the problem’s excess supply) to balance the problem. Shipments to the dummy and from a point to itself will be zero. Let s= total available supply. Step 2 : Construct a transportation tableau as follows: A row in the tableau will be needed for each supply point and transshipment point, and a column will be needed for each demand point and transshipment point. Solving Transshipment Problems
- 20. Each supply point will have a supply equal to it’s original supply, and each demand point will have a demand to its original demand. Let s= total available supply. Then each transshipment point will have a supply equal to (point’s original supply)+s and a demand equal to (point’s original demand)+s. This ensures that any transshipment point that is a net supplier will have a net outflow equal to point’s original supply and a net demander will have a net inflow equal to point’s original demand. Although we don’t know how much will be shipped through each transshipment point, we can be sure that the total amount will not exceed s.
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- 24. Example - 1 - 0 121 109 89 Houston 0 - 82 117 210 Tampa 78 113 0 115 105 Atlanta 119 110 111 0 145 Detroit 225 90 100 140 0 L.A. Tampa Houston Atlanta Detroit L.A. From To ($)
- 25. Example – 1a (answer) 100 5,500 6,400 4,000 4,000 4,000 4,000 0 0 M 82 117 210 Tampa 4,000 0 M 0 121 109 89 Houston 4,000 0 78 113 0 115 105 Atlanta 6,900 0 119 110 111 0 145 Detroit 5,100 0 225 90 100 140 0 L.A Dummy Tampa Houston Atlanta Detroit L.A.
- 26. Example – 1b (answer) 100 5,500 6,400 4,000 4,000 4,000 4,000 0 0 M 82 117 210 Tampa 4,000 0 M 0 121 109 89 Houston 4,000 0 78 113 0 115 105 Atlanta 6,900 0 119 110 111 0 M Detroit 5,100 0 225 90 100 M 0 L.A Dummy Tampa Houston Atlanta Detroit L.A.
- 27. Example – 1c (answer) 100 5,500 6,400 4,000 4,000 4,000 4,000 0 0 5 82 117 210 Tampa 4,000 0 5 0 121 109 89 Houston 4,000 0 78 113 0 115 105 Atlanta 6,900 0 119 110 111 0 145 Detroit 5,100 0 225 90 100 140 0 L.A Dummy Tampa Houston Atlanta Detroit L.A.
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- 31. Question 1… Rs. 410 35 March Rs. 420 30 February Rs. 400 35 January Unit production cost Production capacity Month
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