Transportation problems
- 2. WHAT IS TRANSPORTATION PROBLEM ? • The transportation problem is a special type of linear programming problem where the objective is to minimize the cost of distributing a product from a number of sources or origins to a number of destinations. • Because of its special structure, the usual simplex method is not suitable for solving transportation problems. These problems require a special method of solution.
- 3. AIM OF TRANSPORTATION PROBLEM: • To find out the optimum transportation schedule keeping in mind cost of Transportation to be minimized. • The origin of a transportation problem is the location from which shipments are despatched. • The destination of a transportation problem is the location to which shipments are transported. • The unit transportation cost is the cost of transporting one unit of the consignment from an origin to a destination
- 4. • Determination of a transportation plan of a single commodity • From a number of sources • To a number of destinations, • Such that total cost of transportation id minimized. OBJECTIVES OF TRANSPORTATION PROBLEM
- 5. • It is used to compute transportation routes in such a way as to minimize transportation cost for finding out locations of warehouses. • It is used to find out locations of transportation corporations depots where insignificant total cost difference may not matter. • Minimize shipping costs from factories to warehouses(or from warehouses to retail outlets). • Determine lowest cost location for new factory, warehouse, office ,or other outlet facility. • Find minimum cost production schedule that satisfies firms demand and production limitations. APPLICATIONS OF TRANSPORTATION
- 6. • The transportation problem is one of the most frequently encountered application in real life situations and is a special type of linear programming problem. • The transportation problem indicates the amount of consignment to be transported from various origins to different destinations so that the total transportation cost is minimized without violating the availability constraints and the requirement constraints. CONCLUSION
- 7. TWO TYPES OF TRANSPORTATION PROBLEM:- • BALANCED TRANSPORTATION PROBLEM:- Where the total supply equally total demand. • UNBALANCED TRANSPORTATION PROBLEM:- Where the total supply is not equally total demand. PHASES OF SOLUTION OF TRANSPORTATION PROBLEM:- PHASE I:- Obtains the Initial Basic Feasible Solution. PHASE II:- Obtains the Optimal Basic Solution.
- 8. INITIAL BASIC FEASIBLE SOLUTION:- NORTH-WEST CORNER RULE(NWCR) LEAST COST METHOD(LCM) VOGLE APPROXIMATION METHOD(VAM) OPTIMUM BASIC SOLUTION:- MODIFIED DISTRIBUTION METHOD (MODI METHOD) STEPPING STONE METHOD
- 9. NORTH-WEST CORNER RULE (NWCR):- • DEFINATION:- The North-West Corner Rule is a method adopted to compute the Initial Feasible Solution of the transportation problem. The name North-West Corner is given to this method because the basic variables are selected from the extreme left corner. • It is most systematic and easiest method for obtaining Initial Feasible Basic Solution.
- 10. STEPS IN NORTH-WEST CORNER METHOD STEP 1:- Select the upper left (north-west) cell of the transportation matrix and allocate minimum of supply and demand, i.e min(A1,B1) value in that cell. STEP 2:- If A1<B1, then allocation made is equal to the supply available at the first source(A1 in first row), then move vertically down to the cell (2,1). If A1>B1, then allocation made is equal to demand of the first destination (B1 in first column), then move horizontally to the cell (1,2). If A1=B1, then allocate the value of A1 or B1and then move to cell (2,2). STEP 3:- Continue the process until an allocation is made in the south- east corner cell of the transportation table.
- 11. In the table, three sources A,B and C with the Production Capacity of 50units, 40units, 60 units of product respectively is given. Every day the demand of three retailers D,E,F is to be furnished with at least 20units, 95units and 35units of product respectively. The transportation costs are also given in the matrix. Step 1:- Check whether Total Demand is equal to Total Supply. In case the demand is more then supply, then dummy origin is added to the table. The cost associated with the dummy origin will be zero. Step 2:- Select the North-West or Extreme left corner of the matrix, assign as many units as possible to cell AD, within the supply and demand constraints. Such as 20units are assigned to the first cell, that satisfies the demand of destination D while the supply is in Surplus. Now Assign 30 units to the cell AE. Since 30units are available with the source A, the supply gets fully saturated. SOURCE TO D E F SUPPLY A 5 8 4 50 B 6 6 3 40 C 3 9 6 60 DEMAND 20 95 35 150/150
- 12. STEP 3:- Now move vertically, and assign 40 units to cell BE. The supply of source B also gets fully saturated. Again move vertically, assign 25units to cell CE, demand of destination E is fulfilled. Move horizontally in the matrix and assign 35units to cell CF, both the demand and supply of origin and destination get saturated. Now Total Cost can be computed. Total Cost can be computed by multiplying the units assigned to each cell with the concerned transportation cost. INTIAL TRANSPORTATION COST TOTAL COST = (20*5)+(30*8)+(40*6)+(25*9)+(35*6)= Rs 1015. SOURCES / TO D E F SUPPLY A 5 (20) 8 (30) 4 50 B 6 6 (40) 3 40 C 3 9 (25) 6 (35) 60 DEMAND 20 95 35 150/150
- 13. TRANPORTATION PROBLEM (BALANCED) USES OF LEAST COST METHOD LEAST COST METHOD TRANPORTATION PROBLEM( UNBALANCED) LEAST COST METHOD Contents
- 14. Destination constraints:Cost As A Priority Solve the Problem Supply constraints: Constraints LEAST COST METHOD
- 15. 1 FINANCIAL ANALYSIS 2 ECONOMIC ANALYSIS 3 RISK ASSESSMENT 4 TRANPORTATION 5 ENVIRONMENT WHERE LEAST COST METHOD IS USED
- 16. TRANSPORTATION PROBLEM WITH BALANCED SITUATIONS ASSUMING THEM INDUSTRIES ASSUMING THEM STORES BALANCED COST
- 17. 1 The minimum cost in the matrix is Rs 3, but there is a tie in the cell BF, and CD, now the question arises in which cell we shall allocate. Generally, the cost where maximum quantity can be assigned should be chosen to obtain the better initial solution. Therefore, 35 units shall be assigned to the cell BF.
- 18. 5 Again the minimum cost in the matrix is Rs 3. Therefore, 20 units shall be assigned to the cell CD. With this, the demand of retailer D gets fulfilled. Only 40 units are left with the source C.
- 19. 40 The next minimum cost is 8, assign 50 units to the cell AE. The supply of source A gets saturated. The next minimum cost is Rs 9; we shall assign 40 units to the cell CE. With his both the demand and supply of all the sources and origins gets saturated.
- 20. Total Cost = 50*8 + 5*6 + 35*3 +20*3 +40*9 = Rs 955. The total cost can be calculated by multiplying the assigned quantity with the concerned cost of the cell. Therefore,
- 21. Example of Unbalanced Transportation Problem Plant Warehouse Supply W1 W2 W3 A 28 17 26 500 B 19 12 16 300 Demand 250 250 500 The total demand is 1000, whereas the total supply is 800. Si<Dj Total supply < total demand.
- 22. Plant Warehouse Supply W1 W2 W3 A 28 17 26 500 B 19 12 16 300 Unsatisfied demand 0 0 0 200 Demand 250 250 500 1000
- 23. Plant Warehouse Supply W1 W2 W3 A 28 17 26 500 B 19 12 16 300 Unsatisfied demand 0 0 0 200 Demand 250 250 500 1000 200 250 50 50 450
- 24. VOGEL APPROXIMATION METHOD (VAM METHOD) The Vogel Approximation Method is an improved version of the Minimum Cell Cost Method and the Northwest Corner Method that in general produces better initial basic feasible solution, that report a smaller value in the objective (minimization) function of a balanced Transportation Problem. (sum of the supply = sum of the demand). Applying the Vogel Approximation Method requires the following STEPS: • Step 1: Determine a penalty cost for each row (column) by subtracting the lowest unit cell cost in the row (column) from the next lowest unit cell cost in the same row (column).
- 25. • Step 2: Identify the row or column with the greatest penalty cost. Break the ties arbitrarily (if there are any). Allocate as much as possible to the variable with the lowest unit cost in the selected row or column. Adjust the supply and demand and cross out the row or column that is already satisfied. If a row and column are satisfied simultaneously, only cross out one of the two and allocate a supply or demand of zero to the one that remains. • Step 3: • If there is exactly one row or column left with a supply or demand of zero, stop. • If there is one row (column) left with a positive supply (demand), determine the basic variables in the row (column) using the Minimum Cell Cost Method. Stop. • If all of the rows and columns that were not crossed out have zero supply and demand (remaining), determine the basic zero variables using the Minimum Cell Cost Method. Stop. • In any other case, continue with Step 1.
- 26. The concept of Vogel’s Approximation Method can be well understood through an ILLUSTRATION given below: • First of all the difference between two least cost cells are calculated for each row and column, which can be seen in the iteration given for each row and column. Then the largest difference is selected, which is 4 in this case. So, allocate 20 units to cell BD, since the minimum cost is to be chosen for the allocation. Now, only 20 units are left with the source B.
- 27. • Column D is deleted, again the difference between the least cost cells is calculated for each row and column, as seen in the iteration below. The largest difference value comes to be 3, so allocate 35 units to cell AF and 15 units to the cell AE. With this, the Supply and demand of source A and origin F gets saturated, so delete both the row A and Column F.
- 28. • Now, single column E is left, since no difference can be found out, so allocate 60 units to the cell CE and 20 units to cell BE, as only 20 units are left with source B. Hence the demand and supply are completely met.
- 29. • Now the total cost can be computed, by multiplying the units assigned to each cell with the cost concerned. Therefore, • Total Cost = 20*3 + 35*1 + 15*4 + 60*4 + 20*8 = Rs 555 • Note: Vogel’s Approximation Method is also called as Penalty Method because the difference costs chosen are nothing but the penalties of not choosing the least cost routes.
- 30. MODI METHOD(Modified Distribution Method) • 1. Determine an initial basic feasible solution using any one of the three methods given below: • North West Corner Method • Matrix Minimum Method • Vogel Approximation Method • 2. Determine the values of dual variables, ui and vj, using ui + vj = cij • 3. Compute the opportunity cost using cij – ( ui + vj ). • 4. Check the sign of each opportunity cost. If the opportunity costs of all the unoccupied cells are either positive or zero, the given solution is the optimal solution. On the other hand, if one or more unoccupied cell has negative opportunity cost, the given solution is not an optimal solution and further savings in transportation cost are possible. • 5. Select the unoccupied cell with the smallest negative opportunity cost as the cell to be included in the next solution. • 6. Draw a closed path or loop for the unoccupied cell selected in the previous step. Please note that the right angle turn in this path is permitted only at occupied cells and at the original unoccupied cell. • 7. Assign alternate plus and minus signs at the unoccupied cells on the corner points of the closed path with a plus sign at the cell being evaluated. • 8. Determine the maximum number of units that should be shipped to this unoccupied cell. The smallest value with a negative position on the closed path indicates the number of units that can be shipped to the entering cell. Now, add this quantity to all the cells on the corner points of the closed path marked with plus signs, and subtract it from those cells marked with minus signs. In this way, an unoccupied cell becomes an occupied cell.
- 31. Distribution centre D1 D2 D3 D4 Supply Plant P1 19 30 50 12 7 P2 70 30 40 60 10 P3 40 10 60 20 18 Requireme nt 5 8 7 15 onsider the transportation problem presented in the following table.
- 32. Distribution centre D1 D2 D3 D4 Supply Plant P1 19 30 50 7 P2 30 60 10 P3 60 18 Require ment 5 8 7 15 An initial basic feasible solution is obtained by Matrix Minimum Method and is shown in table 1. Table 1 = 2 Initial basic feasible solution 12 X 7 + 70 X 3 + 40 X 7 + 40 X 2 + 10 X 8 + 20 X 8 = Rs. 894.
- 33. Calculating ui and vj using ui + vj = cij Substituting u1 = 0, we get u1 + v4 = c14 ⇒ 0 + v4 = 12 or v4 = 12 u3 + v4 = c34 ⇒ u3 + 12 = 20 or u3 = 8 u3 + v2 = c32 ⇒ 8 + v2 = 10 or v2 = 2 u3 + v1 = c31 ⇒ 8 + v1 = 40 or v1 = 32 u2 + v1 = c21 ⇒ u2 + 32 = 70 or u2 = 38 u2 + v3 = c23 ⇒ 38 + v3 = 40 or v3 = 2 Distribution centre D1 D2 D3 D4 Supply ui Plant P1 19 30 50 7 0 P2 30 60 10 38 P3 60 18 8 Requi remen t 5 8 7 15 vj 32 2 2 12
- 34. Unoccupied cells Opportunity cost (P1, D1) c11 – ( u1 + v1 ) = 19 – (0 + 32) = –13 (P1, D2) c12 – ( u1 + v2 ) = 30 – (0 + 2) = 28 (P1, D3) c13 – ( u1 + v3 ) = 50 – (0 + 2) = 48 (P2, D2) c22 – ( u2 + v2 ) = 30 – (38 + 2) = –10 (P2, D4) c14 – ( u2 + v4 ) = 60 – (38 + 12) = 10 (P3, D3) c33 – ( u3 + v3 ) = 60 – (8 + 2) = 50 Calculating opportunity cost using cij – ( ui + vj )
- 35. Distribution centre D1 D2 D3 D4 Supply ui Plant P1 7 0 P2 10 38 P3 18 8 Requiremen t 5 8 7 15 vj 32 2 2 12 Now choose the smallest (most) negative value from opportunity cost (i.e., –13) and draw a closed path from P1D1. The following table shows the closed path. Table 4
- 36. Choose the smallest value with a negative position on the closed path(i.e., 2), it indicates the number of units that can be shipped to the entering cell. Now add this quantity to all the cells on the corner points of the closed path marked with plus signs and subtract it from those cells marked with minus signs. In this way, an unoccupied cell becomes an occupied cell. Now again calculate the values for ui & vj and opportunity cost. The resulting matrix is shown below. Table 5
- 37. Distribution centre D1 D2 D3 D4 Supply ui Plant P1 7 0 P2 10 51 P3 18 8 Requir ement 5 8 7 15 vj 19 2 –11 12 Choose the smallest (most) negative value from opportunity cost (i.e., –23). Now draw a closed path from P2D2 .
- 38. Now again calculate the values for ui & vj and opportunity cost
- 39. Distribution centre D1 D2 D3 D4 Supply ui Plant P1 7 0 P2 10 28 P3 18 8 Requi remen t 5 8 7 15 vj 19 2 12 12 Since all the current opportunity costs are non–negative, this is the optimal solution. The minimum transportation cost is: 19 X 5 + 12 X 2 + 30 X 3 + 40 X 7 + 10 X 5 + 20 X 13 = Rs. 799