1251Specially Structured Linear Programmes I : Transportation and Transhipment Problems

A typical transportation problem is like this. Suppose that a manufacturer of refrigerators has three plants situated at places called A, B, and C. Suppose further that his buyers are located in three regions X, Y and Z where he has to supply them the goods. Also, assume that the demands in the three regions and the production in different plants per unit time period are known and equal in aggregate, and further that the cost of transporting one refrigerator from each plant to each of the requirement centers is given and constant. The manufacturer’s problem is to determine as to how he should route his product from his plants to the market places so that the total cost involved in the transportation is minimized. In other words, he has to decide as to how many refrigerators should be supplied from A to X, Y and Z; how many from B to X, Y, and Z and how many from C to X, Y, and Z, to achieve it at the least cost. This is illustrated in Figure given below:

The places where the goods originate from (the plants in our example) are called the sources or the origins and places where they are to be supplied are the destinations. In this terminology, the problem of the manufacturer is to decide as to how many units be transported from different origins to different destinations so that the total transportation cost is the minimum.

PROBLEM STATEMENT

The classical transportation problem can be stated mathematically as follows:

Let = quantity of product available at origin i = quantity of product required at destination j = the cost of transporting one unit of product from source/origin i to destination j = the quantity transported from origin i to destination j

Assume that which means that the total quantity available at the origins is

precisely equal to the total amount required at the destinations.

With these, the problem can be stated as a linear programming problem as:

Z

A X

B Y

C

Supply Demand

ROUTES

PLANT MARKET

Figure – Transportation Problem

Minimize Total Cost,

Subject to

for i = 1,2….,m

for j = 1,2….,n

and for all i =1,2…,m, and j =1,2,…,n

The transportation model can also be portrayed in a tabular form by means of a transportation tableau, shown in table given below.

Table – Transportation Tableau

Origin (i)Destination (j)

Supply, 1 2 N

1

…

2

…

… … … … … … m

…

Demand, …

This tableau can be thought of as a matrix within a matrix, of the dimension m x n. The one is the per unit cost matrix which represents the unit transportation costs for each of the possible transportation routes. Its elements are given by , indicating the cost of shipping a unit from the ith origin to the jth destination. Superimposed on this matrix is the matrix in which each cell contains a transportation variable-that is, the number of units shipped from the row-designated origin to the column designated destination. Each such variable is represented by , the amount shipped from ith source to jth destination. Right and bottom sides of the transportation tableau show, respectively, the amount of supplies available at source i and the amount demanded at each destination j. The

and represent the supply and demand constraints.

The aggregate transportation cost is determined by multiplying the various with corresponding and then adding them up all. The solution to the transportation problem calls for determining values of as would yield the minimum aggregate transportation cost. When a problem is solved, some of the would assume positive values indicating utilized routes. The cells containing such values are called occupied or filled cells and each of them represents the presence of a basic variable. For the remaining cells, called the empty cells, would be zero. These are the routes that are not utilized by the transportation pattern and their corresponding variables ( ) are regarded to be non-basic.

Remember that the transportation costs are assumed to be linear in nature. Further, it is assumed here that the aggregate supply at sources ( ) is equal to the aggregate

demand at destinations ( . We shall consider later the situation when they do not match.

As observed earlier, the number of constraints in a transportation tableau is m + n (the number of rows plus the number of columns). The number of variables required for forming a basis is one less, i.e., m + n – 1. This is so, because there are only m + n –1 independent variables in the solution basis. In other words, with values of any m + n –1 independent variables being given, the remaining would automatically be determined on the basis of those values. Also, considering the conditions of feasibility and non-negativity, the number of basic variables representing transportation routes that are utilized equals m + n –1, and all other variables be non-basic, or zero, representing the unutilized routes. It means that a basic feasible solution of the transportation problem has exactly m + n –1 positive components in comparison to the m + n positive components required for a basic feasible solution in respect of a general linear programming problem in which there are m + n structural constraints to satisfy.

Solution to the Transportation Problem

A transportation problem can be solved by two methods, using (a) simplex method, and (b) transportation method. We shall illustrate these with the help of an example.

Example

A firm owns facilities at six places. It has manufacturing plants at places. A, B and C with daily production of 50, 40 and 60 units respectively. At point D, E, and F it has three warehouses with daily demand of 20, 95, and 35 units respectively. Per unit shipping costs are given in the following table. If the firm wants to minimize its total transportation cost, how should it route its products?

Warehouse____________________________________D E F

A 6 4 1Plant B 3 8 7

C 4 4 2

(a) The simplex method - The given problem can be express as an LPP as follows:Let represent the number of units shipped from plant i to warehouse j. With Z representing the total cost we can state the problem as follows.

Minimize Z = Subject to

Supply constraints

Demand constraints

for i = 1, 2, 3 and j = 1, 2, 3

Now, this problem can be solved as any other problem using the simplex algorithm. But the solution is going to be very lengthy and a cumbersome process because of the involvement of a large number of decision and artificial variables. Hence, we look for an alternate solution procedure called the transportation method which is an efficient one that yields results faster and with less computational effort. A significant point of difference between the simplex and the transportation methods is regarding the determination of initial basic feasible solution. As we use simplex method to solve a minimization problem, we must add artificial variables to make the solution artificially feasible. As we progress from one tableau to another, the artificial variables are dropped one after the other as they become non-basic. Then, eventually an optimal solution is obtained where all the artificial variables are excluded. The transportation method obviates the need to use artificial variables because with this method it is fairly easy to find an initial solution that is feasible, without using the artificial variables.

(b) The transportation method - The transportation method is shown schematically in Figure 1. The method consists of the following three steps:

(i) Obtaining an initial solution, that is to say making an initial assignment in such a way that a basic feasible solution is obtained; (ii) ascertaining whether it is optimal or not, by determining opportunity costs associated with the empty cells, and if the solution is not optimal; (iii) revising the solution until an optimal solution is obtained.

Step 1 : Obtaining the initial feasible solution. The first step in using the transportation method is to obtain a feasible solution, namely, the one that satisfies the rim requirements (i.e. the requirements of demand and supply). The initial feasible solution can be obtained by several methods. The commonly used are the North-West Corner (NWC) Rule, Least Cost Method (LCM) and the vogel’s Approximation Method (VAM). We shall discuss these methods in turn.

1. North-West Corner Rule. The North-West Corner rule (N-W Corner rule) may be stated as follows. Start with the north-west corner of the transportation tableau and consider the cell in the first column and the first row. Corresponding to this cell are the values and respectively in row 1 and column 1. Proceed as follows:

If , then assign quantity in this cell. This implies that we put . If , then assign in the cell so that . Simply speaking, put the lower

of and as . If , then . To illustrate, if supply = 50 and demand =30, then assign 30 units as the quantity to be supplied from first source to the first destination. If supply, =30 and demand =50, the =30 and if =30, =30, then .

Now, if , then move horizontally to the next column in the first row, if , move vertically in the same column to the next row; and if , then move diagonally to the next column and next row. In operational terms, if , so that the supply is greater than demand, then having assigned quantity equal to demand ( ), the remaining quantity is considered alongwith demand at the next destination ( ), whereas if supply

falls short of demand ( ), then having exhausted the available supply at source 1, consider obtaining from the next source ( ). Obviously, if supply and demand match, then consider the next source and next destination ( ).

Once is the next cell, again compare the supply available at the source and demand at the destination, corresponding to the cell chosen, and assign lower of the two values. Move to the next cell appropriately as explained earlier and again assign the quantity considering demand and available supply.

Continue in the zig zag manner until the last source and last destination are covered, so that the south-east corner of the tableau is reached.

For example, the initial basic feasible solution using this method is obtained as follows. First start with the cell on the intersection of A and D. The column total corresponding to this cell is 20 while the row total is 50. So allocate 20 to this cell. Now, the destination requirement having been satisfied, move horizontally in the row to the cell AE. The column total is 95 while a total of 30 is left in the row. Thus assign 30 to this cell. With this, the supply of the row origin is exhausted. Next, move vertically to the cell BE. For this cell, the destination requirement being 65 and the source supply being 40, assign 40 to this cell and exhaust the supply at source B. Then move to the cell CE and allocate 25 units. The remaining supply of 35 at source C is sufficient to meet the demand at destination F. So assign 35 to the cell CF. These assignments are shown in Table given below.

Table 2 – Initial Basic Feasible Solution – NW Corner Method

To

From

D E F Supply

A 20 6

30 4 1 50

B 3 40 8

7 40

C 4 25 4

35 2

60

Demand 20 95 35 160

Total cost : 6 x 20 + 4 x 30 + 8 x 40 + 4 x 25 + 2 x 35 = Rs.730

This routing of the units meets all the rim requirements and entails 5(=3 + 3 –1) shipments as there are 5 occupied cells. It involves a total cost of Rs.730.

2. Least Cost Method (LCM). The NW corner rule described earlier considers only the availability and supply requirements in making assignments. It takes no account of the shipping costs given in the tableau. It is therefore not a sound method as it ignores the very factor (cost) which is sought to be minimized. The least cost method and the Vogel’s Approximation method consider the shipping costs while making allocations. The former of these is discussed below.

As the name suggests, of all the routes (that is, combinations of sources and destinations) select the one where shipping cost is the least. Now, consider the supply available at the corresponding source and demand at the corresponding destination and put the lower of the two as the quantity to be transported through that route. After this, delete the source/destination, whichever is satisfied. Consider the remaining routes and again choose the one with the smallest cost and make assignments. Continue in this manner until all the units are assigned.

It may be mentioned that at any stage, if there is a tie in the minimum cost, so that two or more routes have the same least cost of shipping, then, conceptually, either of them may be selected. However, a better initial solution is obtained if the route chosen is the one where largest quantity can be assigned. Thus, if there are three cells for which the (least) cost value is equal, then consider each one of these one by one and determine the quantity (by reference to the demand and supply quantities given) which can be dispatched, and choose the cell with the largest quantity. If there is still a tie, then either of them may be selected.

For example table given above, the initial solution by using least cost method is obtained in Table given below. To begin with, the lowest of all cost elements is 1, for the cell AF, with corresponding supply and demand being 50 and 35. Accordingly, assign 35 to this cell, delete the column and reduce the quantity available of 15. Of the remaining cost elements (excluding the column values of 1,7 and 2), the least is 3 corresponding to the cell BD. The quantity for this cell is 20, being lower of 20 (demand) and 40 (supply). Delete the column headed D as well and adjust the available supply at B to 20. Now, since only the one market remains, its requirement is met by transferring the available supply at different sources. Accordingly, it gets the 15 units remaining at A, 20 remaining at B and 60 from C.

Table – 3 - Initial Feasible Solution – Least Cost Method

To

From

D E F Supply

A 6

15 4

35 1 50 15

B 20 3

20 8 7 40 20

C 4

60 4

2 60

Demand 20

95 35 150

The transportation schedule obtained in table given below. It involves a total cost of Rs.555.

Table 4 – Initial Feasible Solution

To

From

D E F Supply Iteration

I II

A 6

15 4

35 1

50 15 3

B 20 3

20 8 7 40 20 1

C 4

60 4

2 60 2 2

Demand 20

95 35 150

I 1 0 1

II -- 0 1

Total Cost: 4 x 15 + 1 x 35 + 3 x 20 + 8 x 20 + 4 x 60 = Rs.555

3. Vogel’s Approximation Method (VAM). Like the Least Cost Method, the Vogel’s Approximation Method (VAM) also consider the shipping costs, but in a relative sense, when making allocations. This method is given here.

First, consider each row of the cost matrix individually and find the difference between two least cost cells in it. Then repeat the exercise for each of the columns. Identify the column or row with the largest difference value. Now, consider the cell with the minimum cost in that column (or row, as the case may be) and assign the maximum units possible, looking at the demand and availability corresponding to that cell. In case of a tie in the largest cost difference, although either of them may be chosen but it is preferable to choose the cost difference corresponding to which the largest number of units may be assigned or corresponding to which the cell chosen has minimum cost. To illustrate, suppose the largest cost difference is found to be tied for a row and a column. In applying first of the rules, determine the quantity (considering supply availability and demand) which can be assigned in each case and select the one where larger quantity can be assigned. In the other case, compare unit cost values of the two least cells, in the row and the column, and choose the one which has a lower value.

Delete the column/row which has been satisfied. Again, find out the differences and proceed in the same manner as discussed earlier. Continue until all units have been assigned. The VAM is shown schematically in figure 2.

For example Table 1, the initial feasible solution by using this method is given in table 5.

The differences between the two least-cost cells are calculated for each row and column. The largest of these being 4(=7-3), the row designated as B is selected. In the lower cost cell of the row, BD, a value 20 is assigned and the column D is deleted as the demand is satisfied. In the second iteration, again the cost differences are calculated and the first row is selected as it shows the greatest difference value of 3. In the cell AF, 35 is assigned and column F is deleted. Now, only the column is left and therefore no differences need be calculated. The assignments can be easily made having reference to the supply at various origins. The assignment made by VAM involves a total cost of Rs.555-a solution same as that by LCM, and better than the one obtained by NW corner rule involving a total cost of Rs.730.

Table 5 – Initial Basic Feasible Solution: VAM

To

From

D E F Supply Iteration

I II

A 6

15 4

35 1

50 15 3

B 20 3

20 8 7 40 20 1

C 4

60 4

2 60 2 2

Demand 20

95 35 150

I 1 0 1

II -- 0 1

Total Cost: 4 x 15 + 1 x 35 + 3 x 20 + 8 x 20 + 4 x 60 = Rs.555Start

For each row and column, find the difference between the two least cost cells that have not been allocated

Select the largest of the differences of rows and columns, and, in case of a tie, choose the one corresponding to which largest number of units can be assigned or the cost value is the lowest

Assign the largest quantity permissible by the rim requirements to the cell in that row/column with the smallest cost

Eliminate the row/column that has been satisfied

Are all rim conditions satisfied?

STOP

Figure – Schematic Presentation of VAM

The Vogel’s approximation method is also called the Penalty Method because the cost differences that it uses are nothing but the penalties of not using the least-cost routes. Since the objective function is the minimization of the transportation cost, in each iteration that route is selected which involves the maximum penalty of not being used.

Step 2: Testing the optimality. After obtaining the initial basic feasible solution, the next step is to test whether it is optimal or not. There are two methods of testing the optimality of a basic feasible solution. The first of these is called the stepping-stone method in which the optimality test is applied by calculating the opportunity cost of each empty cell. The second method employed for testing optimality is called the modified distribution method (MODI). This method is easier and more efficient than the stepping-stone method. It is based on the concept of the dual variables that are used to evaluate the empty cells. Using these dual variables the opportunity cost of each of the empty cells is determined. The opportunity cost values in both the methods indicate the optimality or otherwise of a given solution.

Step 3: Improving the solution. By applying either of these method, if the solution is found to be optimal, then the problem is solved. If the solution is not optimal, then a new and better basic feasible solution is obtained. It is done by exchanging a non-basic variable for one basic variable. In simple terms, rearrangement is made by transferring units from an occupied cell to an empty cell that has the largest opportunity cost, and then

shifting the units from other related cells so that all the rim requirements are satisfied. This is achieved by first tracing a closed loop.

The solution is obtained is again tested for optimality (step 2) and revised if necessary. We continue in this manner until an optimal solution is finally obtained.

We shall now discuss the stepping stone and the MODI methods in turn.

1. Stepping-stone method. This is a procedure of determining the potential, if any, for improving each of the non-basic variables in terms of the objective function. To determine this potential, each of the non-basic variables (empty cells) is considered one-by-one. For each such cell we find out as to what effect on the total cost would be if one unit is assigned to this cell. With this information, then, we come to know whether the solution is optimal or not. If not, we improve that solution. To understand it, we refer to the Table 6 which is a reproduction of the Table 2 containing the initial basic feasible solution obtained by NW corner rule.

In this table, there are four empty cells: BD, CD, AF and BF, with corresponding non-basic variables as . To start with, let us assess the potential for improvement of the non-basic variable . A shipment of one unit of the item on this route would mean an increase of Rs 3 but also it will mean a reduction of one unit from

and, thereby, a reduction of Rs 8 in the cost. However, to maintain feasibility, we should subtract one unit from and add one unit to (notice that otherwise the column totals are disturbed). This has the effect of decreasing he cost of Rs 6 in respect of the former and increasing Rs 4 for the latter. The net effect of this operation would be a reduction of Rs 7 in the cost. It is shown as follows:

Increase in cost by increasing by 1 unit = +3Decrease in cost by decreasing by 1 unit = -8Increase in cost by increasing by 1 unit= +4Decrease in cost by decreasing by 1 unit = -6

____Net effect on the cost = -7

____

A reduction in the cost of Rs 7 per unit can be effected by adopting the route BD, implies that the opportunity cost of this route is +7. Since the opportunity cost here is positive, it means that it is worth considering making variables a basic variable.

Table 6 – Initial Basic Feasible Solution : Testing for Optimality

To

From

D E F Supply

A 20 6

30 4 +

1 50

B 3 +

40 8 7 40

C 4

25 4

35 2 60

Demand 20

95 35 150

It is significant to note that the net cost effect that has been calculated in an evaluation of the marginal (or per unit) effect on the total cost function of using a transportation route which is not adopted hitherto. Such a change in the routes involves tracing the effects from one utilized route, which constitutes a stepping stone, to another in such a way that the rim requirements are satisfied. Because of the dependency relationships involved, we start with an empty cell and then proceed through the succession of stepping stone to reach back to the empty cell, the pattern is called the close loop. We shall digress on the drawing of a closed loop a little later.

For the problem under investigation, we would evaluate each of the other empty cells in the manner we did for cell BD. This is done here.

Cell Closed Loop Net Cost Change Opportunity CostBD -7 +7AF AF-CF-CE-AE +1 – 2 + 4 – 4 = -1 +1BF BF-CF-CE-BE +7 – 2 + 4 – 8 = +1 -1CD CD-AD-AE-CE +4 – 6 + 4 – 4 = -2 +2

The rule for testing the optimality is: if none of the empty cells has a positive opportunity cost, the solution is optimal. And, if one (or more) of the empty cells have a positive opportunity cost, the solution is not optimal and calls for revision. For the purpose of revising a given solution, the cells for which the opportunity cost is negative are not considered at all because their inclusion would increase the transportation cost. If a cell has a zero opportunity cost, it has the implication that inclusion of that particular route would leave the objective function value unaffected. We concentrate, therefore, only on the cells that have positive opportunity cost and would select the one with the highest value.

In our illustration, the most favoured cell is BD for it has the largest opportunity cost. We decide, therefore, to include as a basic variable in the solution. Now, since each unit shifted has the effect of cost-saving of Rs 7, we should transfer the maximum number of units to the route BD. It would be equal to the least of the values of the cells on the closed loop involving a minus sign. For the loop under consideration, the cells AD and BE involve minus signs with quantities equal to 20 and 40 respectively. Twenty being the lower of the two, we shall transfer 20 units along this path. Note that the variable replaces the variable . The resulting solution in Table 7 given below.

Table 7 – Improved Solution: Non optimal To

From

D E F Supply

Yes

A 6

50 4

1 50

B 20 3

20 8 7 40

C 4

25 4

35 2 60

Demand 20

95 35 150

Total Cost: 4 x 50 + 3 x 20 + 8 x 20 + 4 x 25 + 2 x 35 = Rs 590

This plan also has 5 (= 3 + 3 –1) allocations. It involves a total cost of Rs 590. We will again apply step 2 to determine whether this solution is optimal or not, and then apply step 3 if it is found to be non-optimal. Application of step 2 yields the following results.

Cell Closed Loop Net Cost Change Opportunity Cost

AD AD-AE-BE-BD +6 – 4 + 8 – 3 = +7 -7AF AF-CF-CE-BE +1 – 2 + 4 – 4 = -1 +1BF BF-CF-CE-BE +7 – 2 + 4 – 8 = +1 -1CD CD-BD-BE-CE +4 – 3 + 8 – 4 = +5 -5

Since only the cell AF has a positive opportunity cost, we shall include it in the transportation schedule. The maximum number of units that can be routed through AF is 35. Accordingly, the revised solution is given in Table 8.

Table 8 – Improved Solution: optimal

To

From

D E F Supply

A 6

15 4

351 50

B 20 3

20 8 7 40

C 4

60 4

2 60

Demand 20

95 35 150

Total Cost: 4 x 15 + 1 x 35 + 3 x 20 + 8 x 20 + 4 x 60 = Rs 555This solution also involves 5 assignments at a total cost of Rs.555. This solution would now be tested for optimality. This is done here:

Cell Closed Loop Net Cost Change Opportunity Cost

AD AD-AE-BE-BD +6 – 4 + 8 – 3 = +7 -7BF BF-BF-AE-AF +7 – 8 + 4 – 1 = +2 -2CD CD-BD-BE-CE +4 – 3 + 8 – 4 = +5 -5

CF CF-CE-AE-AF +2 – 4 + 4 – 1 = +1 -1

Since the opportunity costs of all the empty cells are negative, the solution obtained is the optimal one.

In a similar fashion, we can test the initial basic feasible solution obtained by LCM or VAM. If we compare it with the solution contained in Table 8 we find that the two are identical. Clearly, therefore, the solution we obtained by VAM is optimal.

Before we discuss the MODI method for testing the optimality of a transportation solution, a few words on the tracing of a closed loop follow.

Tracing a loop. When a closed loop is to be traced, start with the empty cell which is to be evaluated (or to be included in the solution). Then, moving clockwise, draw an arrow from this cell to an occupied cell in the same row or column, as the case may be, after that, move vertically (or horizontally) to another occupied cell and draw an arrow. Follow the same procedure to other occupied cells before returning to the original empty cell. In the process of moving from one occupied cell to another, (a) move only horizontally or vertically, but never diagonally; and (b) step over empty or unoccupied cells, if the need be, without changing them. Thus, a loop would always have right angled turns with corners only on the occupied cells.

Having traced the path, place plus and minus signs alternatively in the cells on each turn of the loop, beginning with a plus (+) sign in the empty cell. An important restriction is that there must be exactly one cell with a plus sign and one cell with a minus sign in any row or column in which the loop takes a turn. This restriction ensures that the rim requirements would not be violated when units are shifted among cells.

The following points may also be noted in connection with the closed loops:

(a) An even number of at least four cells must participate in a closed loop and an occupied cell can be considered only once and not more.

(b) If there exists a basic feasible solution with m + n –1 positive variables, then there would be one and only one closed loop for each cell. This is irrespective of the size of the matrix given.

(c) All cells that receive a plus or a minus sign, except the starting empty cell, must be the occupied cells.

(d) Closed loops may or may not be square or rectangular in shape. In larger transportation tables, the closed loops may have peculiar configurations and a loop may cross over itself.

(e) Although, as mentioned earlier, movement on the path set by the loop is generally clockwise, even if the progression on the path is anticlockwise, it would not affect the result.

The Modified distribution method (MODI)

As mentioned previously, the MODI method is an efficient method of testing the optimality of a transportation solution. It may be recalled that in the application of the stepping stone method, each of the empty cells is evaluated for the opportunity cost by drawing a closed loop. In situations where a large number of sources and destinations are involved, this would be a very time consuming and intricate exercise. The MODI method avoids this kind of extensive scanning and reduces the number of steps required in the evaluation of the empty cells. This method gives

a straightforward computational scheme whereby we can determine the opportunity cost of each of the empty cells. The method is shown schematically in Figure given below and its operation is discussed and illustrated here.

Step 1: Add to the transportation table a column on the RHS titled and a row in the bottom of it labeled .

Step 2:(a) Assign any value arbitrarily to a row or column variable . Generally, a value 0 (zero) is assigned to the first row i.e.

(b) Consider every occupied cell in the first row individually and assign the column value (when the occupied cell is in the jth column of the row) which is such that the sum of the row and the column values is equal to the unit cost value in the occupied cell. With the help of these values, consider other occupied cells one by one and determine the appropriate values of and , taking in each case . Thus, if is the row value of the ith row and is the column of the jth column and is the unit cost of the cell in

START

STWrite the initial solution obtained by NWC rule, LCM, or VAM etc.

Is it degenerate?eee

Remove degeneracy: assign to required number of cells

Assign an arbitrary index value (may be 0) to first row on the right margin

Using cost co-efficient of occupied cells, calculate index values for all rows and columns For column j ,

For row I,

Using and calculate opportunity cost for each empty cell

Are all Solution is optimal

STOP

Yes

Yes

Select the cell with largest opportunity cost

Trace a closed loop which(i) begins and ends at the cell chosen(ii) has alternating signs in the cells on its route starting with `+’ in the cell chosen(iii) changes direction only at the occupied cells(iv) may pass over both, occupied and empty

Consider cells with `-‘ mark and determine the least of

the amounts ( ) held.

Add this amount to the cells bearing `+’ sign and subtract from cells with `-‘ sign. A revised solution is obtained.

the ith row and jth column, then the row and column values are obtained using the following equation:

The initial basic feasible solution to Example is reproduced in Table given below.

Table – Initial Basic Feasible Solution: NW Corner Rule

For this solution, let . Using the equation given earlier, we have, for the occupied cell (1,1)

or or

Similarly, or or . With , we get and

From , we get This is how values of 0,4 and 0, and values of 6, 4 and 2 are determined.

It may be mentioned that this method of assigning row and column values is workable only when solution is non-degenerate, that is to say, for a matrix of the order m x n , there are exactly m + n -1 non-zero (occupied) cells.

Step 3 Having determined all and values, calculate for each unoccupied cell

. The represent the opportunity costs of various cells. After obtaining the opportunity costs, proceed in the same way as in the stepping stone method. If all the empty cells have negative opportunity cost, the solution is optimal and unique. If some empty cell(s) has a zero opportunity cost but if none of the other empty cells have positive opportunity cost, then it implies that the given solution is optimal but that it is not unique-there exists other solution(s) that would be as good as this solution.

However, if the solution contains a positive opportunity cost for one or more of the empty cells, the solution is not optimal. In such a case, the cell with the largest

opportunity cost value is selected, a closed loop traced and transfers of units along the route are made in accordance with the method. Then the resulting solution is again tested for optimality and improved if necessary. The process is repeated until an optimal solution is obtained.

For our example, the values are given in circles in the empty cells in the Table given below. They are:

and

Here the cell (2,1) has the largest positive opportunity cost and therefore we select for inclusion as a basic variable. The closed loop starting with this cell has been shown in the table. The revised solution is shown in table given below.

Table – Improved Solution: Non-optimal

This solution is tested for optimality and is found to be non-optimal. Here the cell (1,3) has a positive opportunity cost and therefore a closed loop is traced starting with this. The solution resulting is shown in table given below. This, when tested, is found to be optimal, involving a total transportation cost of Rs 555.

Table – Improved Solution: Optimal

SOME SPECIAL TOPICS

A. Unbalanced Transportation Problems

The methods we have discussed for solving a transportation problem require that the aggregate supply is equal to the aggregate demand . In practice, however, situations may arise when the two are unequal. The two such possibilities are, first, when the aggregate supply exceeds the aggregate demand (i.e. and, second, when

aggregate supply falls short of the aggregate demand (i.e. . Such problems are called unbalanced transportation problems. Balancing must be done before they can be solved.

When the aggregate supply exceeds the aggregate demand, the excess supply is assumed to go to inventory and costs nothing for shipping. A column of slack variables is added to the transportation tableau which represents a dummy destination with a requirement equal to the amount of excess supply and the transportation costs equal to zero. On the other hand, when the aggregate demand exceeds the aggregate supply in a transportation problem, balance is restored by adding a dummy origin. The row representing it is added with an assumed total availability equal to the difference between the total demand and supply, and with each of the cells having a zero unit cost. In some cases, however, when the penalty of not satisfying the demand at a particular destination(s) is given, then such penalty value should be considered and not zero.

B. Prohibited Routes

Sometimes in a given transportation problem some route(s) may not be available. This could be due to a variety of reasons like the unfavorable weather conditions on a particular route, strike on a particular route etc. In such situations, there is a restriction on the routes available for transportation. To handle a situation of this type, we assign a

very large cost represented by M to each of such routes, which are not available. Then the problem is solved in the usual way. The effect of adding a large cost element would be that such routes would automatically be eliminated in the final solution. C. Unique vs Multiple Optimal Solutions

The optimal solution to a given problem may, or may not, be unique. Recall that the solution to a transportation problem is optimal if all the values (in terms of the MODI method) are less than, or equal to zero. For a given solution, if all the values are negative, then it is unique. If, however, some cell (or cells) has =0, then multiple optimal solutions are indicated so that there exist transportation pattern(s) other than the one obtained which can satisfy all the rim requirements for the same cost.READTo obtain an alternate optimal solution, trace a closed loop beginning with a cell having =0, and get the revised solution in the same way as a solution is improved. It may be observed that this revised solution would also entail the same total cost as before. It goes without saying that for every `zero’ value of value in the optimal solution tableau, a revised solution can be obtained.

Example

Given the following transportation problem:__________________________________________________________

MarketWarehouse _____________________________________ Supply

A B C__________________________________________________________

1 10 12 7 1802 14 11 6 1003 9 5 13 1604 11 7 9 120

__________________________________________________________Demand 240 200 220__________________________________________________________

It is known that currently nothing can be sent from warehouse 1 to market A and from warehouse 3 to market C. Solve the problem and determine the least cost transportation schedule. Is the optimal solution obtained by you unique? If not, what is/are the other optimal solution/s?

The given transportation problem is unbalanced since aggregate demand = 240 + 200 + 220 = 660 units while the aggregate supply = 180 + 100 + 160 + 120 = 560 units. Thus, a dummy warehouse (5) is introduced with supply of 660 – 560 = 100 units, and zero transportation cost since no penalties are provided for not satisfying the demand at each market. Further, since routes 1-A and 3-C are given as prohibited, the cost element for each of these is replaced by M. The information is presented in Table 9 given below.

Table 9 – Transportation Problem : Unbalanced with Prohibited Routes__________________________________________________________

Market

Warehouse __________________________________________ SupplyA B C

________________________________________________________________1 M 12 7 1802 14 11 6 1003 9 5 M 1604 11 7 9 1205 0 0 0 100

_____________________________________________________________Demand 240 200 220 660__________________________________________________________

The initial solution is contained in Table 10. It is obtained using VAM.

In the first step, cost differences are calculated and the largest one, 9, is selected and 100 units are placed in the cell 5-A. The row 5 is deleted and differences are recalculated. Now, a tie is observed in the largest value of the differences, that is 5. If the row 1 is selected then the appropriate cell would be 1-C, where, considering demand and supply, 180 units can be sent, while if row 2 is chosen, then in the least cost cell 2-C, 100 units can be placed. Thus, we may preferably choose 1-C. After this, we proceed in the usual manner-cost differences are recalculated and quantities assigned to the least cost cells of the rows/columns selected. The solution is reproduced in Table 12 and tested for optimality. Since all 0, the solution is seen to be optimal. Note that is not calculated for the prohibited cells 1-A and 3-C. Even if is calculated for a prohibited route, it is bound to be extremely negative since it would involve `-M’.

The solution given in Table 11 is not unique because in cell 3-A, = 0. To obtain an alternative optimal solution, we draw a closed loop as shown in this table. The revised solution is given in Table 12. It also involves a total cost of Rs.4300.

Table 10 – Initial Basic Feasible Solution by VAM

Table 11 – Initial Feasible Solution: Optimal

Table 12 – Alternate Optimal Solution

D. Degeneracy

We have already seen that a basic feasible solution of a transportation problem has m + n –1 basic variables, which means that the number of occupied cells in such is one less than the number of rows plus the number of columns. It may happen sometimes that the number of occupied cells is less than m + n – 1. Such a solution is called a degenerate solution.

Degeneracy in a transportation problem can figure in two ways. The problem may become degenerate in the first instance when an initial feasible solution is obtained. Secondly, the problem may become degenerate when two or more cells are vacated simultaneously in the process of transferring units along the closed path.

The problem, when a solution is degenerate, is that it cannot be tested for optimality. Both, the stepping stone method and the modified distribution method (MODI) are inoperative in such a case. The former cannot be applied because for some of the empty cells the closed loops cannot be traced, while the latter fails because all and values cannot be determined.

To overcome the problem, we proceed by assigning an infinitesimally small amount, close to zero, to one (or more if the need be) empty cell and treat the cell as an occupied cell. This amount is represented by a Greek letter (epsilon) and is taken to be such an insignificant value that would not affect the total cost. Thus, it is big enough to cause the particular route to which it is assigned to be considered as a basic variable but not large enough to cause a change in the total cost and the other non-zero amounts. Although is, theoretically, non-zero, the operations with it in the context of problem at hand are given here:

k + = k; + = ;k - = k; - = 0; and0 + = ; k x = 0

When the initial basic feasible solution is degenerate, we assign to an independent empty cell. An independent cell is the one (originating) from which a closed loop cannot be traced. An may be assigned to any of the independent cells but preferably to one with the minimum per unit cost. After introducing epsilon(s) in the requisite and appropriate cell(s), we solve the problem in the usual manner.

Example

Determine optimal solution to the problem given below. Obtain the initial solution by VAM.

To Market________________________________________

Supply________________________________________6 4 9 11 40

From 20 6 11 3 40Plant 7 1 0 14 50

7 1 12 6 90_______________________________________

Demand 90 30 50 30_______________________________________

Since the aggregate supply is 220 units and the aggregate demand is 200 units, we shall introduce a dummy market, , for an amount equal to 20 (the difference between the aggregate supply and demand), with all cost elements equal to zero. The solution is given in Table 13.

Table 13 – Initial Feasible Solution by VAM

The initial solution obtained by VAM is degenerate because it contains only seven basic variables since there are only 7 cells occupied and not 8(= 4+5-1) required for non-degeneracy. Here empty cells are independent while others are not. For removing degeneracy, let us place in the cell

and then test it for optimality. This is done in Table 14. This solution is found to be non-optimal. The improved solution is given in Table 15.

Table 14 – Initial Feasible Solution: Non-optimal

Total cost: 6 x 30 + 1 x 10 + 3 x 20 + 0 x 20 + 0 x 50 + 7 x 60 + 1 x 30 = Rs.700.

The optimality test, MODI, suggests that the solution given in Table 15 is an optimal one. The solution can be stated as:

all other = 0; Total cost = Rs.700.

Table 15 – Revised Solution: Optimal

Total cost:: 6 x 30 + 1 x 10 + 3 x 20 + 0 x 50 + 7 x 60 + 1 x 30 = Rs.700.

When the problem becomes degenerate at the solution-revision stage, epsilon () is placed in one (or more, if necessary) of the recently vacated cells, with the minimum cost. And then we proceed with the problem in the usual manner.