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# Modi method

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• 1. MODI METHOD
• 2. WHAT IS MODI METHOD
• It is used to save the time over stepping stone method
• It provides a new means of finding the unused route with the largest negative improvement index.
• Once largest index identified ,we are required to trace only one path, just as with the stepping stone approach, this helps to determine the maximum number of unit that can be shipped by the best unused route.
• 3. STEPS
• 1.Construct a transportation table with the given cost of transportation and rim requirement.
• 2.Determine IBFS.
• 3.For current basic feasible solution check degeneracy and non-degeneracy.
• rim requirement=stone square(non-degeneracy)
• rim requirement != stone square(degeneracy)
• 4.Find occupied matrix.
• 5.Find unoccupied matrix.
• 4. Steps (contd…)
• 6.Find opportunity cost of unoccupied cells using formula:
• opportunity cost =actual cost-implied cost
• dij= cij - (ri+kj)
• 7.Unoccupied cell evaluation:
• (a) if dij>0 then cost of transportation unchanged.
• (b) if dij=0 then cost of transportation unchanged.
• (c) if dij<0 then improved solution can be obtain and go to next step.
• 5. STEPS(contd…)
• 8.Select an unoccupied cell with largest –ve opportunity cost among all unoccupied cell.
• 9.Construct closed path for the occupied cells determined in step 8.
• 10.Assign as many as units as possible to the unoccupied cell satisfying rim conditions.
• 11.Go to step 4 and repeat procedure until
• All dij>=0 i.e reached to the optimal solution.
• 6. SPECIAL CASES
• Balanced problem
• Unbalanced problem
• Non -degeneracy
• Degeneracy :occurs in two cases
• Degeneracy occurs in initial basic solution.
• Degeneracy occurs in during the test of optimality.
• Profit maximization
• 7. . PROBLEM1:Shipping costs are Rs. 10 per kilometer.. What shipping schedule should be used. if the matrix given below the kilometers from source to destination. Requirement 3 3 2 8 x 50 30 220 1 y 90 45 170 3 z 50 200 50 4 destination a b c availability
• 8. A B C X Y Z 30 220 50 90 45 170 50 200 50 30 220 1 2 3 2 2 2 2 3 P1 P2 P3 P4 20 20 20 20 45 45 45 _ 150 150 _ _ P1 40 15 120 P2 40 15 _ P3 40 15 _ IBFS BY USING VAM METHOD
• 9. 30 30 220 90 45 170 50 200 50 1 E 3 2 2 STONE SEQUARE =4 RIM REQUIREMENT =M+N-1=3+3-1=5 DEGENERACY NOW STONE SEQUARE =5 RIM REQUIREMENT =5 NON –DEGENERACY
• 10. 50 30 45 50 50 50 30 50 0 15 0 OCCUPIED MATRIX UNOCCUPIED MATRIX 105 25 170 50 65 30 65 170
• 11. OPTIMUL SOLUTION X A 50*1=50 X B 30*E=_ Y B 45*3=135 Z A 50*2=100 Z C 50*2=100 385 *10=3850
• 12. PROBLEM2:DETERMINE THE OPTIMUM SOLUTION FOR THE COMPANY OF TRASPOTATION PROBLEM(USING NWCM AND MODI METHOD) 8 8 15 15 10 17 3 9 10 REQUIREMENT 150 80 50 120 80 80 CAPACITY F1 F2 F3 W1 W2 W3 WAREHOUSE FACTORY
• 13. W1 W2 W3 F1 F2 F3 8 8 15 15 3 10 17 9 10 120 30 50 30 50 150 80 50 120 80 80 IBFS WITH NWCM
• 14. OCCUPIED MATRIX UNOCCUPIED MATRIX 8 15 10 9 10 15 10 11 -7 0 -1 15 10 11 -7 0 -1 5 11 6 -11 3 4 11 14
• 15. 15 8 8 15 10 17 3 9 10 10 120 30 50 30 50 + + _ _ 8 15 3 8 15 10 17 9 10 120 E 80 30 50 STONE SEQUARE=RIM REQUIREMENT DEGENERACY OCCUAR LOOP CONSTRUCT
• 16. OCCUPIED MATRIX UNOCCUPIED MATRIX 8 8 10 3 10 3 3 10 3 3 10 5 7 0 5 7 0 5 6 0 0 15 17 10 3
• 17. OPTIMUM SOLUTION F1 W1 8*120 =960 F1 W2 8*E = _ F2 W2 10*80 =800 F3 W1 3*30= 90 F3 W3 10*50 =500 2,350 RS
• 18.