success of Business enterprise especially in manufacturing organization. Goods manufactured by firm need to be distributed to dealers, distributers

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1 INTRODUCTION ASSIGNMENT V/S TRANSPORTATION ASSUMPTIONS INITIAL BASIC FEASIBLE SOLUTION [IBFS] 5 METHODS. DEGENERACY IN TRANSPORTATION OPTIMAL SOLUTION [MODI METHOD] HOW TO PREPARE LOOP PROHIBITED PROBLEM MAXIMIZATION

2 Transportation is a key element for the success of Business enterprise especially in manufacturing organization. Goods manufactured by firm need to be distributed to dealers, distributers etc in such a manner that overall transportation cost of the enterprise is minimize. It is Firstly Utilized during the Second world war.

3 ASSIGNMENT TRANSPORTATION 1 Assignment is on one to one Transportation is on one to many basis. basis. 2 Matrix Should be Balanced Matrix. i.e. Number of row's and Columns are same. Matrix Should be Balanced Matrix. i.e. availability and requirement should be same.

4 Number of Supply center are finite and known. Number of Demand center are finite and known. Supply from each supply centre is constant for a given problem Demand at each demand centre is constant for given problem. Cost of transporting one unit from each supply centre to each demand centre is constant for the given problem.

5 There are five methods in all for developing IBFS. *NORTH WEST CORNER RULE COLUMN MINIMA ROW MINIMA *MATRIX MINIMA OR LEAST COST METHOD *VOGEL S APPROXIMATION METHOD (VAM) Note:* methods are in syllabus

6 Here we first start from the north-west corner of the matrix and we make allocation of the quantity which is equal to lower of availability and requirement applicable to the position. After this, we cancel the row or column depending on where the quantity gets exhausted. In the revised matrix after this cancellation, we again catch the North west corner of this revised matrix and we make the allocation. We go on doing this as long as the entire allocation is not over. The IBFS on this method ignores the cost completely and therefore, it is farthest from the optimum solution.

7 Here we take up each column, one by one, beginning with the first column and make allocation of quantity in the least cost cell of the column. As long as column quantity is not exhausted, we go to the next least cost cell in the same column & make the allocation. This way when all columns are exhausted, the IBFS is obtained.

8 Here we take up each Row, one by one, beginning with the first Row and make allocation of quantity in the least cost cell of the Row. As long as Row quantity is not exhausted, we go to the next least cost cell in the same Row & make the allocation. This way when all Row are exhausted, the IBFS is obtained.

9 Here we make the allocation in the least cost cell of the entire matrix. Depending on, where the quantity gets exhausted, we cancel the row or as the case may be column. From the revised matrix, we again identify the least cost cell and make the allocation. Once all allocation are over, the IBFS is obtained.

10 This is closest to the optimum solution STEPS FOR VOGEL S APPROXIMATION METHOD 1. Calculate penalty in respect of each row and each column. And Penalty is difference between two least cost. 2. Compare all penalties and make allocation in that row or that column where penalty is maximum [ enter from highest penalty] 3. If penalty is same for 2 or more option then give preference to least cost 4. If least cost is also same for 2 or more option then give preference to maximum quantity allocation. [if qty is also same for two or more option then select anyone at random] 5. We go on doing this as long as the entire allocation is not over.

11 When RIM Condition is satisfied i.e. Actual allocation = M+N-1 then solution is not degenerate. when RIM condition is not satisfied then solution is degenerate. there for we have to introduce epsilon (є). If the є survive in final matrix then it is known as Permanent Degeneracy If the є does not survive in final matrix then it is known as a Temporary Degeneracy

12 We can start the process of optimality test only when RIM Condition are satisfied RIM Condition is Total number of allocation = Number of Row + Number of Colum 1 When the actual number of allocation is less than required allocation as per the RIM condition then we have to make dummy allocation in Least cost unallocated independent cell. The independent cell is that cell from where Loop can not be prepare. Calculate Value in respect of each row and each column. To start value calculation put zero either in front of any one row or in front of any one column. Value (u & v) of remaining row & Column is to be calculated by following: Cost price of allocated cell = value in front of that row + value in front of that column. Calculate net evaluation in respect of each unallocated cell by following method: Net evaluation of unallocated cell ( ) = Cost price of unallocated cell (value in front of row + value in front of column) When Net evaluation ( ) of any cell in the final matrix is Zero then there is Alternate optimum Solution. When None of the net evaluation ( ) is negative Optimality is reached.

13 Start from highest negative net evaluation cell. Go any where in the matrix either Horizontally or Vertically. You can take turn only if there is allocation in that cell. Comeback to same place from where we have started. When more than one Loop from same point is prepare then select Loop which can shift the highest allocation. Put + sign at starting point, and at the alternate corner put sign and adjust Θ Θ is the Lower of the all allocation at ive sign & add this qty at + and deduct this from sign allocation.

14 When it is not possible to transport the product from one place(origin) to other place(destination) then it is said to be Prohibited Problem. For solving such problem a very Large cost M or for each prohibited route. No allocation were made at that cell.

15 When the Object is Maximization of profit then following additional step is to be followed: 1. If the given matrix is unbalanced, then first make it balanced matrix by introducing dummy 2. Convert maximization matrix in to minimization (i.e. deduct all element of the matrix from the highest value) 3. Remaining all steps to solve the Problems are same.

16 THANK YOU

Fundamentals of Operations Research. Prof. G. Srinivasan. Department of Management Studies. Indian Institute of Technology, Madras. Lecture No.

Fundamentals of Operations Research Prof. G. Srinivasan Department of Management Studies Indian Institute of Technology, Madras Lecture No. # 13 Transportation Problem, Methods for Initial Basic Feasible