MODI UV METHOD Optimal Solution IN TRANSPORTATION PROBLEM

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First Watch Initial Basic Feasible Solution by Vogel's Approximation Method (VAM) • • Initial basic feasible solution by VA... • If you like and understand our video then SUBSCRIBE our YouTube channel and Share our video. • #MODIMETHOD #TRANSPORTATIONPROBLEM • Transportation Algorithm for Minimization Problem (MODI Method) • Step 1 • Construct the transportation table entering the origin capacities (supply) ai • , the destination • requirement (demand) bj • and the cost cij. • Step 2 • Find an initial basic feasible solution by Vogel’s method or by any of the given method. • Step 3 • Test for optimality: • 1. If number of positive independent allocations xij = m+n-1 • Then IBFS is non-degenerate and optimality test can be performed. • 2. If number of positive independent allocations xij less than m+n-1 • Then IBFS is degenerate and optimality test cannot be performed. And first we • have to convert degenerate solution into non degenerate. • Step 4 • Apply MODI Method • 1. Set up cost matrix for allocated (occupied) cell only. • 2. Enter the set of number ui for row and vj • for column such that for occupied cell • cij = ui • vj • . • 3. Set any ui • or vj • = 0 which have maximum allocation to calculate the values of ui • , vj. • 4. Compute ui • vj for unoccupied cells. • 5. Compute the opportunity cost • dij = cij – ( ui • vj • ) for unoccupied cells. • Step 5 • Apply optimality test by examining the sign of each dij • 1. If all dij greater than or equal to 0, the current initial basic feasible solution is optimal. • 2. If at least one dij less than 0, the current initial basic feasible solution is not optimal and we have to improve the solution. • Step 6 • Iterative towards an optimal solution: • 1. Identify the most negative cell in dij matrix and called this cell as identified cell. • (The most negative value is the rate by which total transportation cost is reduced if • one unit is transported to this cell) • 2. Write IBFS again which needs to be improved. • 3.Marked (√) the identified cell and make a closed loop starting and ending in the • identified cell. All the corner of the loop should be on occupied cell except identified • cell. • 4. Mark the identified cell as +ive and other corner of the loop with –ive, +ive, ive sign • alternatively. • 5. Find minimum allocated value on the loop where we have –ive sign. • 6. Add these minimum allocations in the loop where we have +ive sign and subtract • where we have–ive sign. ( by these most negative cell gets an allocation and • minimum allocated cell with –ive sign gets zero allocation) • 7. By doing all the above steps we get second basic feasible solution. • Step 7 • Now, return to step 4 and repeat the process until an optimal solution is obtained. • ~-~~-~~~-~~-~ • Please watch: UNBALANCED ASSIGNMENT PROBLEM IN OPERATION RESEARCH | USING HUNGARIAN METHOD | Lecture 03 • • Unbalanced Assignment Problem | Using... • ~-~~-~~~-~~-~

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