4 Transportation Problem- An Introduction

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An Introduction


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One important area of LPP is the physicaltransportation of goods and services from severalsupply origins to several demand destinations

Since this problem involves a large number of variablesand constraints, two methods have been developednamely

Stepping Stone Method Modified Distribution Method (MODI)


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Objective is To determine the number of units which should be

shipped from an origin to a destination in order tosatisfy the required quantity of goods or services at

each demand destinations , within the limitedquantity of goods or services at each supply origin, atthe minimum transportation cost and/or time.

Therefore, the target is to:-

Minimize total cost of transporting a homogeneouscommodity (product) from supply origins to demanddestinations

Maximize the profit


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General equation Minimize (Total Cost) Z = i j cij xij

wherei=1,2,3,.m; j=1,2,3n

Subject to the constraintsj xij =ai i=1,2,3,.m; j=1,2,3n (supply constraints)

i cij= bj j=1,2,3n; i=1,2,3,.m (demand constraints)

and xij 0 for all i and j


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Existence of feasible solution Total supply = Total demand

i ai = j bj (Rim conditions)

i=1,2,3,.m ; j=1,2,3ni.e. the total capacity (or supply) = total requirement (or

demand)

Constraints of a problem = m+n

Variables in a problem = m x n


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RemarksAny feasible solution for a transportation must have

exactly (m+n-1) non- negative basic decision variablesxij satisfying rim conditions

When total supply equals total demand, the problemis called a balanced transportation problem,otherwise it is called an unbalanced transportationproblem

Whenthe number of positive allocations (xij ) is lessthan the required number of independent constraintequations (m+n-1), the solution is said to bedegenerate, otherwise non- degenerate


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Cells in the transportation table having positiveallocation will be called occupied cells, otherwiseempty or non- occupied cells


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The Transportation Method Step I:- Formulate the problem and set up in the

matrix form

Step II:- Obtain an initial basic feasible solution. Forthis, various basic methods used are:-

North- West Corner Method (NWCM)

Least Cost Method (LCM)

Vogels Approximate (or Penalty) Method (VAM)


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The initial solution obtained by any of the threemethods must satisfy the following conditions:-

The solution must be feasible ie it must satisfy all the supply

and demand constraints (rim conditions) The number of positive allocations must be equal to m+n-1,

where m is the number of rows and n is the number ofcolumns

Any solution that satisfies the above conditions is calledthe non- degenerate basic feasible solution otherwisedegenerate solution


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Step III:- test the initial solution for optimality usingMODI method.

If the result is optimal, then stop. Otherwise determinea new improved solution

Step IV:- Updating the solution

Repeat step III until optimal solution is found


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North- West Corner Method

(NWCM) This method does not take into account the cost of

transportation on any route of transportation

Step I:- Start with the cell at the upper left (north-

west) corner of the transportation matrix and allocateas much as possible equal to the minimum of the rimvalues to the first row and the first column (a1, b1)

Step II:-

(a)- if allocation made in step I is equal to the supplyavailable at the first source (a1 in first row), then move

vertically down to cell (2,1) in the second row firstcolumn and apply step I again for the next allocation


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(b)- if allocation made in step I is equal to the demandavailable at the first destination (b1 in first column),then move horizontally to cell (1,2) in the first row and

second column and apply step I again for the nextallocation

(c)- If a1 =b1, allocate x11= a1 or b1 and move diagonally tothe cell (2,2)

Step III:- Continue the procedure step by step till anallocation is made in the south- east corner cell of thetransportation table


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This method takes into account the minimum unitcost of transportation for obtaining initial solution

Step I:- Select the cell with the lowest unit cost in theentire transportation table and allocate as much aspossible to this cell and eliminate that row or columnin which demand is exhausted . If both row andcolumn are satisfied simultaneously, only one may becrossed out. In case the smallest unit cost cell is notunique, then select the cell where maximumallocation can be made


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Step II:- after adjusting the supply and demand for alluncrossed- out rows and columns repeat the procedurewith the next lowest unit cost among the remaining

rows and columns of the transportation table andallocate as much a possible to this cell and eliminatethat row and column in which either supply ordemand is exhausted

Step III:- Repeat the procedure until the entireavailable supply at various sources and demand atvarious destinations is satisfied. The solution soobtained need not be non- degenerate


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Remarks.. If there is a tie between the least values, choose any

one the values arbitrarily


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It is a heuristic method and is preferred to the othertwo methods

In this method, each allocation is made on the basis ofopportunity cost that would have been incurred ifallocations in certain cells with minimum unittransportation cost were missed

Advantage is that it gives an initial solution which isnear to the optimal solution or is the optimal solutionitself


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Step I:- calculate penalties for each row (column) bytaking the difference between the smallest and thenext smallest unit transportation cost in the same row

(column). This difference indicates the penalty or extracost which has to be paid if one fails to allocate to thecell with the minimum unit transportation cost

Step II:- select the row or column with the largest

penalty and allocate as much as possible in the cellhaving the least cost in the selected row or columnsatisfying the rim conditions. If there is a tie, select thecell with the least per unit cost.


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Step III:- Adjust the supply and demand and cross outthe satisfied row or column. If a row and a column aresatisfied simultaneously, only one of them is crossedout and the remaining row (column) is assigned a zerosupply (demand). Any row or column with zero supplyor demand should not be used in computing futurepenalties

Step IV:- Repeat steps 1 to3 until the entire availablesupply at various sources and demands at variousdestinations are satisfied


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Once an initial solution is found, the next step is tocheck its optimality

The optimal solution is the one in which there is noopportunity cost, ie there is no other set oftransportation routes (allocations) that will reduce thetotal transportation cost.

For testing the solution, following two methods arewidely used

Stepping Stone Method

Modified Distribution Method (MODI)

Test for Optimality


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The two methods differ in their computationalapproach but they give exactly the same results anduse the same testing procedure

The procedure being used is to test each unoccupiedcell one at a time, by computing the cost change. If theinclusion of any unoccupied cell can decrease thetransportation cost then this unoccupied cell will beconsidered for inclusion in the improved solution

That unoccupied cell is selected for which the costchange is the most negative


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In this method, the net cost change is calculated that canbe obtained by introducing any of the unoccupied cells into the solution.

Every increase (or decrease) in the supply/ demand at oneoccupied cell must be associated by a decrease(or increase)in supply at another

Net cost change is determined by listing the unit costs

associated with each cell and then summing over the pathto find the net effect

Signs are alternate from + to depending upon whethershipments are being added or subtracted at a given point


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Step I:- Determine an initial basic feasible solution using anyone of the three methods (NWCM, LCM, VAM)

Step II:- make sure that the number of occupied cells is exactlyequal to m+n-1 where m- no. of rows, n-no. of columns

Step III:-Evaluate the cost effectiveness of the shipping goods viatransportation routes not currently in the solution. This testingof each unoccupied cell is conducted by the following steps: Select an unoccupied cell, where a shipment should be made.

Trace a closed path using the most direct route through at leastthree occupied cells used in the solution and then back to theoriginal occupied cell and moving with only horizontal and

vertical moves. Cells at the turning points are called the StepingStones on the path


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Assigning (+) or (-) signing alternatively on each cornercell of the closed path are traced, starting at a plus signat the unoccupied cell to be evaluated

Compute the net change in the cost along the closedpath by adding together the unit cost figures found ineach cell containing a plus sign and then subtractingthe unit cost in each square containing the minus sign

Repeat all the above steps until net change in thetransportation cost has been evaluated for allunoccupied cells of the transportation table


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Step IV:- Check the sign of each of the net changes. If allnet changes are greater than or equal to zero, an optimalsolution has been reached. If not, it is possible to improvethe current solution and decrease the total shipping costs

Step V:- Select the unoccupied cells having the highestnegative net cost change and determine the maximumnumber of units that can be assigned to a cell marked witha minus sign on the closed path corresponding to this cell.

Add this number to the unoccupied cell and to all other

cells on the path marked with a plus sign. Subtract thisnumber from cells on the closed path marked with a minussign

Step VI:- Repeat the procedure from Step II unless anoptimal solution is obtained


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Step I:- For an initial basic solution with m+n-1occupied cells, calculate uj and vj for rows andcolumns. The initial solution cane be obtained by

using any of the methods described earlier Any of the uj and vj values is assigned the value zero.

Complete the calculation of uj and vj by using therelation

cij = uj + vj


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Step II:- for unoccupied cells, calculate opportunitycost as

dij= cij (u j + vj ) for all i and j

Step III:- Examine sign of each dij If dij >0, current basic solution is optimal

If dij = 0, current basic solution remains unaffected but analternative solution exists

If one or more dij


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Step III:- Select the unoccupied cell with the largestnegative opportunity cost as the cell to be included inthe next solution.

Step IV:- Construct a closed path for the unoccupiedcell selected in Step III. Right angle turns in this pathare allowed only. Every closed loop must have evennumber of turns and is formed with horizontal and

vertical lines only Step V:- Assign alternate (+) and (-) signs at the

unoccupied cells on the corner points of the closedpath with a (+) sign at the cell being evaluated


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Step VI:- Determine the maximum number of unitsthat should be shipped to this unoccupied cell. Thesmallest stone with a negative position on the closed

path indicates the number of units that can be shippedto the entering cell. This quantity is added to the cellmarked plus sign and subtracted from the cell markedminus sign

Step VII:- Repeat the whole procedure until theoptimal solution is found

Step VIII:- Finally, obtain the total transportation costfor the new solution

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4 Transportation Problem- An Introduction