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# Transportation Problem in Operational Research

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• 1. INITIAL BASICFEASIBLE SOLUTION
• 2. TRANSPORTATION PROBLEM Transport various quantities of a single homogeneous commodity to different destinations in such a way that total transportation cost is minimum.
• 3. TERMINOLOGY USED IN TRANSPORTATIONAL MODELFeasible solution: Non negative values of xij where i=1, 2……….m and j=1, 2,…n whichsatisfy the constraints of supply and demand is called feasible solution.Basic feasible solution: If the no of positive allocations are (m+n-1).Optimal solution: A feasible solution is said to be optimal solution if it minimizes thetotal transportation cost.Balanced transportation problem: A transportation problem in which the total supplyfrom all sources is equal to the total demand in all the destinations.Unbalanced transportation problem: Problems which are not balanced are calledunbalanced.Matrix terminology: In the matrix, the squares are called cells and form columnsvertically and rows horizontally.Degenerate basic feasible solution: If the no. of allocation in basic feasible solutions isless than (m+n-1).
• 4. OPTIMAL SOLUTION OF TRANSPORTATION PROBLEM Obtain an optimal solution by making Initial Basic successiveSTEP1 Feasible STEP2 improvements in Solution IBFS until no further decrease in transportation cost is possible
• 5. INITIAL BASIC FEASIBLE SOLUTION Methods Available NORTH WEST LOWEST COST VOGEL’S CORNER ENTRY APPROXIMATIONMETHOD(NWCM) METHOD(LCEM) METHOD(VAM)
• 6. UNBALANCED TRANSPORTATION PROBLEM Demand>SupplyAdd dummy column in matrix with zero cost Supply>Demand Add dummy row in matrix with zero cost
• 7. DEMAND>SUPPLYPRODUCT P Q SUPPLYOFFICEA 10 15 20B 20 30 50DEMAND 30 60 70 90PRODUCT P Q SUPPLYOFFICEA 10 15 20B 20 30 50DUMMY 0 0 20DEMAND 30 60 90 90
• 8. SUPPLY>DEMAND PRODUCT P Q SUPPLY OFFICE A 10 15 30 B 20 30 60 DEMAND 20 50 90 70PRODUCT P Q DUMMY SUPPLYOFFICEA 10 15 0 20B 20 30 0 50DEMAND 20 50 20 90 90
• 9. METHOD FOR OPTIMAL SOLUTION Stepping stone method Modified Distribution Method(MODI)
• 10. NORTH WEST CORNER METHODMost systematic and easiest method for obtaining initial feasible basic solution
• 11. STEPS TO SOLVE THE PROBLEMStep1: construct an empty m*n matrix, completed with rows & columns.Step2: indicate the rows and column totals at the end.Step3: starting with (1,1)cell at the north west corner of the matrix, allocate maximum possible quantitykeeping in view that allocation can neither be more than the quantity required by the respective warehousesnor more than quantity available at the each supply centre.Step 4: adjust the supply and demand nos. in the rows and columns allocations.Step5: if the supply for the first row is exhausted then move down to the first cell in the second row and firstcolumn and go to the step 4.Step 6: if the demand for the first column is satisfied, then move to the next cell in the second column and firstrow and go to step 4.Step 7: if for any cell, supply equals demand then the next allocation can be made in cell either in the nextrow or column.Step 8: continue the procedure until the total available quantity is fully allocated to the cells as required.
• 12. MINIMIZATION USING NWCM WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28P3 4 5 8 17DEMAND 21 25 19
• 13. Step1: Check whether problem is balanced WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28P3 4 5 8 17DEMAND 21 25 19 65 65
• 14. Allocate P1W1 max quantity such that itshould not exceed supply/requirement WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21/1 25 19 65P1 can supply 20 units at max so W1 can be allocated 20 units only by P1
• 15. Check whether the requirements of W1 fulfilled WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20/0 20P2 5 7 3 28P3 4 5 8 17DEMAND 21/1 25 19 65Now allocate the remaining requirement to next north west corner element i.e. P2W1
• 16. Allocate 1 to P2W1 WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/27 1 P3 4 5 8 17 DEMAND 21/1/0 25 19 65In this way repeat it for all columns until demand of all warehouses is fulfilled
• 17. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/27/0 1 25 2 P3 4 5 8 17/0 17 DEMAND 21/1/0 25/0 19/17/0 65 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
• 18. TOTAL TRANSPORTATION COSTPLANT WAREHOUSE COSTP1 W1 7*20=140P2 W2 5*1=5P2 W2 7*25=175P2 W2 3*2=6P3 W3 8*17=136TOTAL COST 462
• 19. LEAST COST ENTRY METHOD This method takes into consideration the lowest cost and therefore takes less time to solve the problem
• 20. STEPS TO SOLVE THE PROBLEM• Step1: select the cell with the lowest transportation cost among all the rows and columns of the transportation table. If the minimum cost is not unique then select arbitrarily any cell with the lowest cost.• Step2: allocate as many units as possible to the cell determined in step 1 and eliminate that row in which either capacity or requirement is exhausted.• Step3:adjust the capacity and the requirement for the next allocations.• Step4: repeat the steps1to3 for the reduced table until the entire capacities are exhausted to fill the requirements at the different destinations.
• 21. MINIMIZATION USING LCEM WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28P3 4 5 8 17DEMAND 21 25 19 65 Step1 :This problem is balanced problem
• 22. Find out the cell having least cost WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28/9 19P3 4 5 8 17DEMAND 21 25 19/0 65P2W3 is having least cost i.e. 3 so it is allocated 19 units as per the requirement of W3
• 23. In this way allocate next minimum costs until all requirements are fulfilled WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/5/0 4 5 19 P3 4 5 8 17/0 17 DEMAND 21/4/0 25/5/0 19/0 65 Now P2W1 and P3W2 are having same cost, we can allocate any of them but in this case P2W1 will beallocated because in P3W2 P3 cant supply anymore units
• 24. Final matrix with all allocations WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20/0 20P2 5 7 3 28/9/5/0 4 5 19P3 4 5 8 17/0 17DEMAND 21/4/0 25/5/0 19/0 65
• 25. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/5/0 4 5 19 P3 4 5 8 17/0 17 DEMAND 21/4/0 25/5/0 19/0 65 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
• 26. TOTAL TRANSPORTATION COSTPLANT WAREHOUSE COSTP1 W2 6*20=120P2 W1 5*4=20P2 W2 7*5=35P2 W3 3*19=57P3 W1 4*17=68TOTAL COST 300
• 27. MAXIMIZATION USING LCEMINVESTMENT P Q R S AVAILABLE1 95 80 70 60 702 75 65 60 50 403 70 45 50 40 904 60 40 40 30 10DEMAND 40 50 60 60 210Convert maximization into minimization
• 28. Subtract all the elements from the largestelement thereby giving minimization case. INVESTMENT P Q R S AVAILABLE 1 95 95 80 70 60 70 2 75 65 60 50 40 3 70 45 50 40 90 4 60 40 40 30 10 DEMAND 40 50 60 60 210Here 95 is maximum element so subtract all elements from 95
• 29. This is the matrix after subtracting all the elements from 95 INVESTMENT P Q R S AVAILABLE 1 0 15 25 35 70 2 20 30 35 45 40 3 25 50 45 55 90 4 35 55 55 65 10 DEMAND 40 50 60 60 210This is now minimization case which can be further solved as discussed before
• 30. TRANSPORTATION PROBLEM Categorized into two types: Minimization Maximization problem problem
• 31. MINIMIZATION PROBLEM In this transportation cost is given which is to be minimized.
• 32. MAXIMIZATION PROBLEM In this profit is given which is to be maximized.To solve this problem we convert the problem into minimization.Conversion is done by selecting the largest element from Profit Pay off matrix and then subtracting all elements from largest element including itself. Reduced matrix obtain becomes minimization case and then same steps are taken to solve it as is done in minimization problem
• 33. VOGEL’S APPROXIMATION METHOD• BASIS OF ALLOCATION IS UNIT COST PENALTY• THE SUBSEQUENT ALLOCATIONS IN CELLS ARE DONE KEEPING IN VIEW THE HIGHEST UNIT COST• IBFS OBATINED BY THIS METHOD IS EITHER OPTIMAL OR VERY NEAR TO OPTIMAL SOLUTION• SO AMOUNT OF TIME REQUIRED TO CALCULATE THE OPTIMUM SOLUTION IS REDUCED
• 34. STEPS TO SOLVE THE PROBLEMStep1: for each row of the table identify the lowest and the next lowest cost cell. Find their least cost than the difference shall be zero.Step 2: similarly find the difference of each column and place it below each column. These differences found in the steps 1 and 2 are also called penalties.Step 3: looking at all the penalties. Identify the highest of them and the row or column relative to that penalty. Allocate the maximum possible units to the least cost cell in the selected row or column. Ties should be broken in this orderMaximum difference least cost cell.Maximum difference tie least cost cellMaximum unit’s allocations tie arbitraryStep 4: adjust the supply and demand and cross the satisfied row or column.Step 5:Recompute the column and row differences ignoring deleted rows/columns and go to the step3. Repeat the procedure until the entire column and row totals are satisfied.
• 35. MINIMIZATION USING VAMWAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28P3 4 5 8 17DEMAND 21 25 19 65 Step1 :This problem is balanced problem
• 36. Find out ‘Penalties’(P) for each row & column by subtracting the smallest element from next smallest element in the row/columnWAREHOUSE W1 W2 W3 SUPPLY P1PLANTP1 7 6 9 20 7-6=1P2 5 7 3 28 5-3=2 In W3 3 is min cost 19P3 4 5 8 17 5-4=1DEMAND 21 25 19 65P1 5-4=1 6-5=1 9-3=6 6 is max penaltyFind out the maximum penalty and in that row/column find minimum element and allocate it.
• 37. The row/column having zero demand/supply left will be cut by a line WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28/9 19 P3 4 5 8 17 DEMAND 21 25 19/0 65W3 is having zero demand now so it will be cut and wont be considered
• 38. Now find next penalty P2 in same wayWAREHOUSE W1 W2 W3 SUPPLY P2PLANTP1 7 6 9 20 7-6=1P2 5 9 7 3 28/9/0 7-5=2 2 is max penaltyP3 4 5 8 17 5-4=1 19DEMAND 21/12 25 19/0P2 5-4=1 6-5=1 ------------ In W1 5 is min cost
• 39. In this way find out all the penalties until all allocations are not doneWAREHOUSE W1 W2 W3 SUPPLY P3PLANTP1 7 6 9 20 7-6=1P2 5 9 7 3 28/9/0 ----------P3 4 12 5 8 17/5 5-4=1 19DEMAND 21/12/0 25 19/0 In P3 4 is min costP3 7-4=3 6-5=1 ------------ 3 is max penalty
• 40. Final matrix with all allocationsWAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20/0 20P2 5 7 3 28/9/0 9 19P3 4 5 8 17/5/0 12 5DEMAND 21/12/0 25/5/0 19/0
• 41. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/0 9 19 P3 4 5 8 17/5/0 12 5 DEMAND 21/12/0 25/5/0 19/0 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
• 42. TOTAL TRANSPORTATION COSTPLANT WAREHOUSE COSTP1 W2 6*20=120P2 W1 5*9=45P2 W3 3*19=57P3 W1 12*4=48P3 W2 5*5=25TOTAL COST 295
• 43. COMPARISON OF NWCM,LCEM,VAM WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19 65 Minimum Transportation Cost NWCM=462 LCEM=300 VAM=295
• 44. MAXIMIZATION CASE I II III IV CAPACITYA 40 25 22 33 100B 44 35 30 30 30C 38 38 28 30 70REQUIREMENT 40 20 60 30 200 150This is an unbalanced maximization problem for finding out maximum profit involved
• 45. To solve this first we will convert this inminimization problem by subtracting the largest element from all elements I II III IV CAPACITYA 4 19 22 11 100B 0 9 14 14 30C 6 6 16 14 70REQUIREMENT 40 20 60 30 200 15044 is maximum element in the matrix so subtract all the elements from 44 and re-write in matrix form
• 46. Now we will balance this matrix Requirement<Capacity Dummy column will be added I II III IV Dummy CAPACITYA 4 19 22 11 0 100B 0 9 14 14 0 30C 6 6 16 14 0 70REQUIREMENT 40 20 60 30 50 200 200 Now this can be solved by any of the IBFS methods
• 47. CHOCOLATE WININGQUESTION FOR AUDIENCE
• 48. Obtain IBFS for the followingMaximization Problem using VAM: I II III SUPPLYA 6 4 4 100B 5 6 7 25C 3 4 6 75DEMAND 60 80 85
• 49. SOLUTION: I II III SUPPLYA 6 60 4 40 4 100B 5 6 15 7 10 25C 3 4 6 75 75DUMMY 0 0 25 0 25DEMAND 60 80 85 MAXIMUM PROFIT= Rs. 1130