PRESENTATION ON TRANSPORTATION &

ASSIGNMENT PROBLEM

TRANSPORTATION PROBLEM

Transportation Problem is one of the subclasses of LPPs.

The objective is to transport various quantities of a single homogenious commodity that are initially stored at various origins to different destinations in such a way that the transportation cost is minimum

To achieve this we must know the amount & location of available supplies and the quantities demanded.

In addition, we must know the costs that result from transporting one unit of commodity from various origins to various destinations

DEFINITIONS… Feasible Solution: Any set of non negative allocations (xij =0) which satisfies the row &

coloumn sum (RIM requirement)

Basic feasible Solution: A feasible solution is called a basic feasible solution if the number of non

negative allocations is equal to m+n-1, Where ‘m’ is the no: of rows, ‘n’ the no: of columns in a transportation table

Non-Degenerate Basic Feasible Solution

Any feasible solution to a transportation problem containing m origins & n destinations is said to be non-degenerate, if it contains m+n-1 occupied cells & each allocation is in independent positions

Degenerate Basic Feasible Solution

If a basic feasible solution contains less than m-n-1 non negetive allocations, it is said to be degenerate.

OPTIMAL SOLUTION Optimal Solution is a feasible solution which minimises the total

cost

The solution of a transportion problem can be obtained in two stages, namely initial solution & optimum solution.

Initial solutuin can be obtained by using any one of the following 3 methods : North West Corner Rule (NWCR) Last Cost Method or Matrix Minima

Method Vogel’s Approximation Method (VAM)

VAM is preferred over the other two methods since the initial basic feasible solution obtained by this method is either optimal or very close to the the optimal solution.

NORTH WEST CORNER RULE (NWCR)

Step 1: Starting with cell ath the upper left corner (North West) of the transportation matrix we allocate as much as possible so that either the capacity of the first row is exhausted or the destination requirement of the first column is satisfied ie, xn = min(a1b1)

Step 2: If b1>a1, we move down vertically to the second row & make the second allocation of magnitude x22=min (a2b1-xn) in the cell (2,1)

If b1< a1, move horizontally to the second column and ,make the second allocation of magnitude x12=min (a1,xn-b1) in the cell (1,2)

if b1=a1 there is a tie for the second allocation. We make the second allocations of magnitude

x12=min(a1-a1,b1)=0 in the cell (1,2)

or x21=min (a2,b1-b1)=0 in the cell (2,1)

Step 3 : Repeat steps 1 & 2 moving down towards the lower right corner of the transportation table untill all the rim requirements are satisfied.

LEAST COST METHOD OR MATRIX MINIMA METHOD

Step 1: Determine the smallest cost in the cost matrix of the transportation table. Let it be cij. Allocate xij=min(ai,bj) in the cell (i,j)

Step 2: xij=a, cross of the ith row of the transportation table and decrease bj by ai. Then go to step 3

xij=bj, cross of the jth column of the transportation table and decrease ai by bj. Then go to step 3

if xij=ai=bj cross off either the ith row or the jth column but not both.

Step 3: Repeat step 1 & 2 for the resulting reduced

transportation tabe untill all the rim reqts are satisfied.

Whenever the minimum cost is not unique, make an

arbitrary choice among the minima.

VOGEL’S APPROXIMATION METHOD (VAM)

Step 1: Find the penalty cost of, namely the diff between the smallest and the next smallest costs in each row & column.

Step 2: Among the penalties as found in step 1, choose the maximum penalty. If this maximum penalty is more than one, chose any one arbitarily

Step 3: In the selected row or column as by step 2, find outthe cell having the least cost. Allocate to this cell as much as possible depending on the capacity & reqts.

Step 4: Delete the row or column which is fully exhausted. Again compute the column & row penalties for the reduced transportation table & then go to step 2. Repeat the procedure untill all the rim reqts are satisfied.

ASSIGNMENT PROBLEM

The objective of assignment problem is to assign a no: of origins (jobs) to the equal no: of destinations (persons) at a minimum cost or maximum Profit.

It is concerned with assigning jth jobs to nth persons.

The assignment problem can be stated in the form of n x n cost matrix [cij] of real nos

DIFF B/W TRANSPORTATION AND ASSIGNMENT PROBLEM

Tran: No: of sources & no: of destinations need not be equal.

Assi: The no: of sources & Destinations must be equal. The cost Matrix Must be a Square Matrix Tran: xij , the quantity to be transported from ‘ith’ origin to ‘jth’ destination can take any possible positive values and satisfies the rim requirements.

Assi: xij the ‘jth’ job to be assigned to the ith’ person and can take either the value 1 0r zero

Tran: The capacity & the requirement value is equal to ai & bj for the ith’ source and jth’ destinations ( i=1, 2mj=1,2,….n)

Assi: The capacity and the requirement value is exactly one.

Tran: The problem is unbalanced if the total supply and total demand are not equal.

Assi: The problem is unbalanced if the cost matrix is not a square matrix

HUNGARIAN METHOD PROCEDURE

Step1 : Prepare a cost matrix. If the cost matrix is not a square matrix then add a dummy row (column) with zero cost element.

Step 2 : Subtract the minimum element in each row from all the elements of the respective rows.

Step 3 : Further modify the resulting matrix by subtracting the minimum element of each column from all the elements of the respective column .Thus, obtain the modified the matrix .

Step 4 :Then draw minimum number of horizontal and vertical lines to cover all zeros in the resulting matrix. Let the minimum number of lines be N. Now there are two cases.

case1 If N = n,where n is the order of matrix, then an optimal assignment can be made. So make the assignment to get the required solution.

Case 2 If N <n, then proceed to step 5

o Step 5 : Determine the smallest uncovered element in the matrix(element not covered by N lines). Subtract this minimum element from all uncovered elements and add the same element at the intersection of horizontal and vertical lines. Thus the second modified matrix is obtained.

Step 6 : Repeat step 3 and 4 until we get the case 1 of step 3.

Step 7 : (to make zero assignment) Examine the rows successively until a row-wise exactly single zero is found. Circle(o)this zero to make assignment. Then mark a cross(x) overall zeros if lying in the column of the circled zero, showing that they can be considered for future assignment. Continue in this manner until all the zeros have been examined. Repeat the same procedure for column also

Step 8 : Repeat the step 6 successively until one of the following situations arises-

(1) if no unmarked zero is left .then the process ends or

(2) if there lies more than one of the unmarked zero in any column or row then , circle one of the unmarked zero arbitrarily and mark a cross in the cells of remaining zeros in its row and column. Repeat the process until no unmarked zero is left in the matrix.

Step 9 Thus exactly one marked circled zero in each row and each column of the matrix is obtained. The assignment corresponding to these marked circled zeros will give the optimal assignment.

UNBALANCED ASSIGNMENT PROBLEM

Any assignment problem is said to be unbalanced if the number of columns and rows are not equal.

To make its balanced we add a dummy row or dummy column with all the entries as zero.

MAXIMISATION IN ASSIGNMENT PROBLEM

Objective is to maximize the profit

To solve this we first convert the given profit matrix in to the loss matrix by subtracting all the elements from the highest element of the given profit matrix.

For this converted loss matrix we apply the steps in Hungarian method to get the optimum assignment.

TRAVELLING SALESMAN PROBLEM

The objective is to select the sequence which the cities are visited in such a way that the total travelling time is minimized .

o For eg : to visit two cities (A and B) there is no choice. To visit three cities we have two possible routes. For four cities we have three possible routes. In general to visit “ n ” there are “n-1” possible routes.