TRANSPORTATION PROBLEM

1. TRANSPORTATION PROBLEM

2. TRANSPORTATION MODEL • Special case of LP. • Deals with shipping from sources to destinations. • Objective is to determine amount shipped to each destination from sources at minimum cost. • Shipping cost are assumed to be constant and directly proportional to amount shipped. TRANSPORTATION

3. MODEL c11:x11 1 1 2 2 units of supply aj units of demand bi m n cmn:xmn destinations sources TRANSPORTATION

4. TRANSPORTATION MODEL • There are m sources and n destinations. • The lines join source (i) to destination (j). • The lines carry information about the cost of transportation cij and the amount shipped xij. • The amount of supply at source (i) is ai and the amount of demand at (j) is bj. • Objective is to determine the unknowns xij that will minimize transportation cost. TRANSPORTATION

5. Example The above table shows three sources S1, S2 & S3 and four destinations D1 to D4. The nos in the matrix are the per unit transportation costs. TRANSPORTATION

6. MODEL FORMULATION • Let xij= no of units from source i to j. • The model will be as follows • Minimize z= 19x11+30x12+50x13+10x14+70x21+30x22+40x23+ 60x24+40x31+8x32+70x33+20x34 • Capacity constraints • x11+x12+x13+x14 = 7 • x21+x22+x23+x24 = 9 • x31+x32+x33+x34 = 18 • Demand Constraints • x11+x21+x31 = 5 • x12+x22+x32 = 8 • x13+x23+x33 = 7 • x14+x24+x34 = 14 • For all xij > 0 for i and j TRANSPORTATION

7. GENERAL LP MODEL subject to constraints TRANSPORTATION

8. Conditions of TP Problem • Supply = Demand. Also called as RIM conditions. • No of constraints are m+n. • No of variables are m x n. • Supply = demand is one of the constraints and hence we have one extra redundant constraint. • Thus for any feasible TP problem we have exactly m+n-1 non-negative decision variables xij allocations satisfying all conditions. • Above is balanced model (S=D) • In unbalanced S not equal to D. • If the no of non-negative allocations is less than m+n-1 then the solution is said to degenerate. TRANSPORTATION

9. Solution to TP • Set up in Initial Matrix format. • Obtain an Initial solution by any of the following methods • North West Corner Method. • Least Count Method. • Vogel’s Approximation Method. • Test the solution for optimality. • Update the solution. TRANSPORTATION

10. North West Corner Method • Simple & Efficient Method. • Start with cell at upper left (NW). • Allocate as much as feasible without affecting RIM conditions (a1,b1) • Following things can happen • If a1=b1 move diagonally. • If a1 (supply) > b1 (demand) move horizontally. • If b1 (demand) > a1 (supply) then move vertically. TRANSPORTATION

11. Example TRANSPORTATION

12. Least Cost Method • Select the least cost cell. Allocate as much as RIM conditions are met. • Eliminate (cross) row or column if it gets exhausted. • After adjusting the S & D for all uncrossed-out rows and columns repeat the procedure with next least cost with remaining column & rows. • Repeat till all supply and demand equations are satisfied. TRANSPORTATION

13. LCM - Example TRANSPORTATION

14. LCM EXAMPLE TRANSPORTATION

15. LCM Example TRANSPORTATION

16. Vogel Approximation Method • Improved version of LCM. • Step 1 – For each row (column) with positive demand, calculate the penalty by subtracting the smallest unit cost element in the row (column) from the next smallest unit cost element in the same row (column). • Step 2 • Identify the row or column with largest penalty • Break ties arbitrarily. • Allocate as much as possible to the variable with the least unit cost in the selected row or column. • Cross out the satisfied row or column. • If a row and column are satisfied simultaneously, only one of the two is crossed out. TRANSPORTATION

17. VAM • Step 3 – • If exactly one row or column with zero supply or demand remains uncrossed out then stop. • If one row (column) with positive supply (demand) remains uncrossed out, determine the basic variables in the row (column) by the LCM and stop. • If all the uncrossed out row and columns have remaining zero supply or demand, determine the zero basic variables by the LCM and stop. • Otherwise go to step I. TRANSPORTATION

18. VAM EXAMPLE TRANSPORTATION

19. TRANSPORTATION

20. TRANSPORTATION

21. Optimality Test (MODI) • An optimal solution is one where there is no other set of route allocations that will reduce the transportation cost. • We have to review the unoccupied cells of the Initial solution. • An unoccupied cell with largest negative opportunity cost is selected for inclusion. • Concept is similar to Simplex for entering and leaving variable. • Iterations end when no more negative opportunity costs exists. TRANSPORTATION

22. DUAL OF TP PROBLEM • Refer General TP formulation. • Let ui and vj be the dual variables. • The dual equations will be as follows TRANSPORTATION

23. DUAL OF ORIGINAL PROBLEM • Max Z = 7u1+9u2+18u3+5v1+8v2+7v3+14v4. • subject to • u1+v1 <19 , u1+v2<30, u1+v3<50, u1+v4<10 • u2 +v1<70, u2+v2< 30, u2+v3<40, u2+v4<60 • u3+v1<40, u3+v2<8, u3+v3<70, u3+v4<20 • all ui and vj are unrestricted. TRANSPORTATION

24. Economic Interpretation of Dual TRANSPORTATION

25. Economic Interpretation • Dual variables represent the shadow prices of the supply and demand constraints. • The value ui measures the implicit contribution (location advantage) – termed as location rent. • The vi measures the implicit contribution of the demand centre if one unit is transported to jth location – termed as market price. TRANSPORTATION

26. Steps of MODI • Step 1 – Form an initial solution using any of the three methods. • Step 2 • For an initial basic solution with m+n-1 occupied cells, calculate ui & vj for rows and columns. • To start with any of the ui’s or vj’s can be assigned zero for a particular ui & vj. • Complete all ui’s and vj’s • Use the relation cij = ui + vj for all occupied cells. TRANSPORTATION

27. Step of MODI • Step 3 • For unoccupied cells calculate the opportunity cost using dij = cij – (ui + vj) • Step 4 • Examine each dij • If dij > 0 then current solution is optimal. • If dij=0, then current BFS will remain unaffected but alternate optima exists • If one or more cells have dij<0, then an improved solution can be obtained. TRANSPORTATION

28. Steps MODI • Step 5 • Construct a closed loop for the unoccupied cell with largest negative opportunity costs. • Mark the first cell as +ve. • Trace a path along rows (columns) to an occupied cell and mark corner with –ve sign and + ve sign alternatively. • Continue till you reach back the unoccupied cell. • Step 6 • Select the smallest qty amongst the cells marked with negative sign. • Allocate this value to the selected unoccupied cell and add or subtract from + ve and –ve sign respectively. TRANSPORTATION

29. Steps MODI • Step 7 • Obtain new improved solution by allocating units to the unoccupied cells according to earlier steps. • Step 8 • Test the revised solution for optimality TRANSPORTATION

30. BASIC RULES • Loop starts and ends on the unoccupied cell. • It contains successive horizontal and vertical lines whose end points must be occupied cells. • Clock wise or anti-clock wise does not matter. • Any basic feasible solution must contain m+n-1 occupied cells. • An ordered set of at least four cells in a transportation table form a loop provided • any two adjacent cells of the ordered set lie either in the same row or column. • no three or more adjacent cells in the ordered set lie in the same row or column. TRANSPORTATION

31. EXAMPLE TRANSPORTATION

32. EXAMPLE TRANSPORTATION

33. TRANSPORTATION

34. TRANSPORTATION

35. VARIATIONS IN TP PROBLEM • Unbalanced Supply & Demand • Degeneracy • At the initial solution • During subsequent iteration • Alternate Optimal solution • Prohibited transportation routes – Assign a very high cost to that route for solution. • TP as Maximization problem TRANSPORTATION

36. Supply & Demand Imbalance • If total supply exceeds total demand, an additional column (dummy demand centre) is introduced. The unit transportation costs for all the cells in this column are assigned as zero. • If the total demand exceeds total supply, an additional row (dummy supply centre) is added to account for excess demand. The transportation costs for this column are also assigned zero. • After doing so solve the problem in normal TP method. TRANSPORTATION

37. Degeneracy • A BFS solution for TP problem must consist of m+n-1 decision variables. (positive allocations) • A solution is called degenerate if the number of positive allocations are less than m+n-1. • We cannot improve the solution in such case and also we cannot evaluate ui & vj. • Hence we need to remove the degeneracy to improve the solution. • This is done by allocating a small quantity close to zero to one or more unoccupied cells. TRANSPORTATION

38. Degeneracy • For a minimization problem this is added to a cell with lowest transportation costs. • For maximization this is added to a cell with high payoff value. • The quantity added is such that is does not affect the allocation. • It is similar to artificial variables in simplex problem. TRANSPORTATION

39. Degeneracy Example • Refer excel sheet TRANSPORTATION

40. TP – Maximization Problem The criteria for maximization is opposite of min, i.e. solution is optimal if all opportunity costs dij for the unoccupied cells are zero or negative. For finding the initial solution by VAM, the differences between largest and next largest are taken. Allocations are made in those rows (columns) where payoff’s are the highest. TRANSPORTATION

41. Assignment TRANSPORTATION