- 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 MODEL Feasible solution: Non negative values of xij where i=1, 2โฆโฆโฆ.m and j=1, 2,โฆn which satisfy 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 the total transportation cost. Balanced transportation problem: A transportation problem in which the total supply from all sources is equal to the total demand in all the destinations. Unbalanced transportation problem: Problems which are not balanced are called unbalanced. Matrix terminology: In the matrix, the squares are called cells and form columns vertically and rows horizontally. Degenerate basic feasible solution: If the no. of allocation in basic feasible solutions is less than (m+n-1).
- 4. OPTIMAL SOLUTION OF TRANSPORTATION PROBLEM Obtain an optimal solution by making Initial Basic successive STEP1 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 VOGELโS CORNER ENTRY APPROXIMATION METHOD(NWCM) METHOD(LCEM) METHOD(VAM)
- 6. UNBALANCED TRANSPORTATION PROBLEM Demand>Supply Add dummy column in matrix with zero cost Supply>Demand Add dummy row in matrix with zero cost
- 7. DEMAND>SUPPLY PRODUCT P Q SUPPLY OFFICE A 10 15 20 B 20 30 50 DEMAND 30 60 70 90 PRODUCT P Q SUPPLY OFFICE A 10 15 20 B 20 30 50 DUMMY 0 0 20 DEMAND 30 60 90 90
- 8. SUPPLY>DEMAND PRODUCT P Q SUPPLY OFFICE A 10 15 30 B 20 30 60 DEMAND 20 50 90 70 PRODUCT P Q DUMMY SUPPLY OFFICE A 10 15 0 20 B 20 30 0 50 DEMAND 20 50 20 90 90
- 9. METHOD FOR OPTIMAL SOLUTION Stepping stone method Modified Distribution Method(MODI)
- 10. NORTH WEST CORNER METHOD Most systematic and easiest method for obtaining initial feasible basic solution
- 11. STEPS TO SOLVE THE PROBLEM Step1: 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 quantity keeping in view that allocation can neither be more than the quantity required by the respective warehouses nor 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 first column 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 first row 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 next row 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 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19
- 13. Step1: Check whether problem is balanced 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 65
- 14. Allocate P1W1 max quantity such that it should 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 65 P1 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 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21/1 25 19 65 Now 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 65 In 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 COST PLANT WAREHOUSE COST P1 W1 7*20=140 P2 W2 5*1=5 P2 W2 7*25=175 P2 W2 3*2=6 P3 W3 8*17=136 TOTAL 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 โข 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. โข 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. โข Step3:adjust the capacity and the requirement for the next allocations. โข 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 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19 65 Step1 :This problem is balanced problem
- 22. Find out the cell having least cost 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 65 P2W3 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 be allocated because in P3W2 P3 cant supply anymore units
- 24. Final matrix with all allocations 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
- 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 COST PLANT WAREHOUSE COST P1 W2 6*20=120 P2 W1 5*4=20 P2 W2 7*5=35 P2 W3 3*19=57 P3 W1 4*17=68 TOTAL COST 300
- 27. MAXIMIZATION USING LCEM INVESTMENT P Q R S AVAILABLE 1 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 60 210 Convert maximization into minimization
- 28. Subtract all the elements from the largest element 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 60 210 Here 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 60 210 This 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. VOGELโS APPROXIMATION METHOD โข BASIS OF ALLOCATION IS UNIT COST PENALTY โข THE SUBSEQUENT ALLOCATIONS IN CELLS ARE DONE KEEPING IN VIEW THE HIGHEST UNIT COST โข IBFS OBATINED BY THIS METHOD IS EITHER OPTIMAL OR VERY NEAR TO OPTIMAL SOLUTION โข SO AMOUNT OF TIME REQUIRED TO CALCULATE THE OPTIMUM SOLUTION IS REDUCED
- 34. STEPS TO SOLVE THE PROBLEM Step1: 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 order Maximum difference least cost cell. Maximum difference tie least cost cell Maximum unitโs allocations tie arbitrary Step 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 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 Step1 :This problem is balanced problem
- 36. Find out โPenaltiesโ(P) for each row & column by subtracting the smallest element from next smallest element in the row/column WAREHOUSE W1 W2 W3 SUPPLY P1 PLANT P1 7 6 9 20 7-6=1 P2 5 7 3 28 5-3=2 In W3 3 is min cost 19 P3 4 5 8 17 5-4=1 DEMAND 21 25 19 65 P1 5-4=1 6-5=1 9-3=6 6 is max penalty Find 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 65 W3 is having zero demand now so it will be cut and wont be considered
- 38. Now find next penalty P2 in same way WAREHOUSE W1 W2 W3 SUPPLY P2 PLANT P1 7 6 9 20 7-6=1 P2 5 9 7 3 28/9/0 7-5=2 2 is max penalty P3 4 5 8 17 5-4=1 19 DEMAND 21/12 25 19/0 P2 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 done WAREHOUSE W1 W2 W3 SUPPLY P3 PLANT P1 7 6 9 20 7-6=1 P2 5 9 7 3 28/9/0 ---------- P3 4 12 5 8 17/5 5-4=1 19 DEMAND 21/12/0 25 19/0 In P3 4 is min cost P3 7-4=3 6-5=1 ------------ 3 is max penalty
- 40. Final matrix with all allocations 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
- 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 COST PLANT WAREHOUSE COST P1 W2 6*20=120 P2 W1 5*9=45 P2 W3 3*19=57 P3 W1 12*4=48 P3 W2 5*5=25 TOTAL 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 CAPACITY A 40 25 22 33 100 B 44 35 30 30 30 C 38 38 28 30 70 REQUIREMENT 40 20 60 30 200 150 This is an unbalanced maximization problem for finding out maximum profit involved
- 45. To solve this first we will convert this in minimization problem by subtracting the largest element from all elements I II III IV CAPACITY A 4 19 22 11 100 B 0 9 14 14 30 C 6 6 16 14 70 REQUIREMENT 40 20 60 30 200 150 44 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 CAPACITY A 4 19 22 11 0 100 B 0 9 14 14 0 30 C 6 6 16 14 0 70 REQUIREMENT 40 20 60 30 50 200 200 Now this can be solved by any of the IBFS methods
- 47. CHOCOLATE WINING QUESTION FOR AUDIENCE
- 48. Obtain IBFS for the following Maximization Problem using VAM: I II III SUPPLY A 6 4 4 100 B 5 6 7 25 C 3 4 6 75 DEMAND 60 80 85
- 49. SOLUTION: I II III SUPPLY A 6 60 4 40 4 100 B 5 6 15 7 10 25 C 3 4 6 75 75 DUMMY 0 0 25 0 25 DEMAND 60 80 85 MAXIMUM PROFIT= Rs. 1130