slm-unit-05-mb0048

Operations Research Unit 5

Sikkim Manipal University Page No. 93

Unit 5 Transportation Problem

Structure:

5.1 Introduction

Learning objectives

5.2 Formulation of Transportation Problem (TP)

5.3 Transportation Algorithm (MODI Method)

5.4 The Initial Basic Feasible Solution

North west corner rule

Matrix minimum method

Vogel‟s approximation method

5.5 Moving Towards Optimality

Improving the solution

Modified distribution method / MODI method / U – V Method

Degeneracy in transportation problem

5.6 Summary

5.7 Terminal Questions

5.8 Answers to SAQs and TQs

Answers to Self Assessment Questions

Answers to Terminal Questions

5.9 References

5.1 Introduction

Welcome to the unit on transportation model in Operations Research

Management. Transportation model is an important class of linear

programs. For a given supply at each source and a given demand at each

destination, the model studies the minimisation of the cost of transporting a

commodity from a number of sources to several destinations.

The transportation problem involves m sources, each of which has available

ai (i = 1, 2… m) units of homogeneous product and n destinations, each of

which requires bj (j = 1, 2…., n) units of products. Here ai and bj are positive

integers. The cost cij of transporting one unit of the product from the ith

source to the jth destination is given for each i and j. The objective is to

develop an integral transportation schedule that meets all demands from the

inventory at a minimum total transportation cost.



It is assumed that the total supply and the total demand are equal.

n

1j

m

1i

i bja (1)

The condition (1) is guaranteed by creating either a fictitious destination with

a demand equal to the surplus if total demand is less than the total supply or

a (dummy) source with a supply equal to the shortage if total demand

exceeds total supply. The cost of transportation from the fictitious

destination to all sources and from all destinations to the fictitious sources

are assumed to be zero so that total cost of transportation will remain the

same.

Learning objectives

By the end of this unit, you should be able to:

Formulate the transportation problem

Find the initial basic feasible solution

Compare the advantages of various methods of finding initial basic

feasible solution

Solve the degeneracy in the transportation problem

Apply the model to minimise the cost of transporting a commodity

5.2 Formulation of Transportation Problem

The standard mathematical model for the transportation problem is as

follows.

Let xij be number of units of the homogenous product to be transported from

source i to the destination j

Then objective is to

Minimise z = ij

m

1i

n

1j

ij xC

Subject to

n.,..........,2,1ij;bjx

m,......,2,1i,ax

m

1i

ij

n

1j

iij (2)

(2) (2)

With all xij 0 and integrals



Theorem: A necessary and sufficient condition for the existence of a

feasible solution to the transportation problem (2) is:

n

1j

m

1i

i bja

Self Assessment Questions

Fill in the blanks

1. Transportation problems are a special type of ___________.

2. The number of rows and columns need not always be ___________.

3. Transportation problem develops a schedule at _______ and ________.

5.3 Transportation Algorithm (MODI Method)

The first approximation to (2) is integral. Therefore, you always need to find

a feasible solution. Rather than determining a first approximation by a direct

application of the simplex method, it is more efficient to work with the

transportation table given below. The transportation algorithm is the simplex

method specialised to the format of table involving the following steps:

i) Finding an integral basic feasible solution

ii) Testing the solution for optimality

iii) Improving the solution, when it is not optimal

iv) Repeating steps (ii) and (iii) until the optimal solution is obtained

The solution to TP is obtained in two stages.

In the first stage, you find the basic feasible solution using any of the

following methods a) North-west corner rule b) Matrix Minima Method or

least cost method c) Vogel‟s approximation method. In the second stage,

you test the basic feasible solution for its optimality either by MODI method

or by stepping stone method.



Table 5.1: Transportation Table

D1 D2 Dn Supply ui

S1

x11

x12

x1n a1 u1

S2

x21

x22

x2n a2 u2

S3

x31

x32

x3n a3 u3

Sm

xm1

xm2

xmn am um

Demand b1 b2 bn ai = bi

vj v1 V2 vm


State Yes or No

4. In transportation problems, ai = bj is a sufficient and necessary

condition for getting a feasible solution.

5. Transportation problems can also be solved by simplex method.

6. Matrix-minima method gives optimum solution.

5.4 The Initial Basic Feasible Solution

Let us consider a TP involving m-origins and n-destinations. Since the sum

of origin capacities equals the sum of destination requirements, a feasible

solution always exists. Any feasible solution satisfying m + n – 1 of the

m + n constraints is a redundant one and hence it can be deleted. This also

means that a feasible solution to a TP can have only m + n – 1 positive

component; otherwise the solution will degenerate.

It is always possible to assign an initial feasible solution to a TP, satisfying

all the rim requirements. This can be achieved either by inspection or by

following some simple rules. You can begin by imagining that the

transportation table is blank that is initial xij = 0. The simplest procedures for

initial allocation are discussed in the following section.

C11

C21

C31

Cm

1

C12

C22

C32

Cm

2

C1n

C2n

C3n

Cm

n



5.4.1 North West Corner rule

Step1: The first assignment is made in the cell occupying the upper left

hand (north-west) corner of the transportation table. The maximum feasible

amount is allocated here is

x11 = min (a1, b1)

Either the capacity of origin O1 is used up or the requirement at destination

D1 is satisfied or both. This value of x11 is entered in the upper left hand

corner (small square) of cell (1, 1) in the transportation table.

Step 2: If b1 > a1, the capacity of origin O is exhausted and the requirement

at destination D1 is still not satisfied. Then at least one variable in the first

column will have to take on a positive value. Move down vertically to the

second row and make the second allocation of magnitude:

x21 = min (a2, b1 – x21) in the cell (2, 1)

This either exhausts the capacity of origin O2 or satisfies the remaining

demand at destination D1.

If a1 > b1 ,the requirement at destination D1 is satisfied, but the capacity of

origin O1 is not completely exhausted. Move to the right in a horizontal

position to the second column to make the second allocation of magnitude:

x12 = min (a1 – x11, b2) in the cell (1, 2)

This either exhausts the remaining capacity of origin O1 or satisfies the

demand at destination D2.

If b1 = a1, the origin capacity of O1 is completely exhausted as well as the

requirement at destination is completely satisfied, then there is a tie at the

second allocation. An arbitrary tie breaking choice is made. Make the

second allocation of magnitude

x12 = min (a1 – a1, b2) = 0 in the cell (1, 2)

OR

x21 = min (a2, b1 – b2) = 0 in the cell (2, 1)

Step 3: Start from the new north-west corner of the transportation table

satisfying the destination requirements and exhausting the origin capacities

one at a time, moving down towards the lower right corner of the

transportation table until all the rim requirements are satisfied.



Solved Problem 1

Determine an initial basic feasible solution to the following transportation

problem using the north west corner rule:

D1 D2 D3 D4

Availability

01 6 4 1 5 14

02 8 9 2 7 16

03 4 3 6 2 5

6 10 15 4 35

Requirements

Where Oi and Dj represent the ith origin and the jth destination respectively.

Solution: The transportation table of the given T.P. has 12 cells.

14

16

5

6 10 15 4

Following north west corner rule, the first allocation is made in the cell

(1,1), the magnitude being x11 = min (14, 6) = 6

The second allocation is made in the 6 10 15 4 cell (1, 2) and the

magnitude of allocation is given by x12 = min (14 – 6, 10) = 8

The third allocation is made in the cell (2, 2), the magnitude being

x22 = min (16, 10 – 8) = 2.

The magnitude of fourth allocation, in the cell (2, 3) is given by x23 = min

(16 – 2, 15) = 14.

The fifth allocation is made in the cell (3, 3), the magnitude being x33 = min

(5, 15 –14) =1.

The sixth allocation in the cell (3, 4) is given by x34 = min (5 –1, 4) = 4.

Now all the rim requirements have been satisfied and hence an initial

feasible solution to the TP has been obtained. The solution is displayed as



Table 5.2: Initial feasible solution to the TP

D1 D2 D3 D4

01

02

03

6

6

8

4

4

5 14

8

2

9

14

2

16

4

3

1

6

4

2 5

6 10 15 4

Clearly, this feasible solution is non-degenerate basic feasible solution as

the allocated cells do not form a loop. The transportation cost according to

the above loop is given by.

Z = x11 c11 + x12 c12 + x22 c22 + x23 c23 + x33 c33 + x34 c34

= 66 + 48 + 92 + 214 + 2461

= 128

5.4.2 Matrix minimum 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: If xij = ai cross the ith row of the transportation table, decrease bj by

ai and proceed to step 3.

If xij = bj cross the ith column of the transportation table, decrease ai by bj and

proceed to step 3.

If xij = ai= bj cross either the ith row or the ith column, but not both.

Step 3: Repeat steps 1 and 2 to reduce transportation table until all the rim

requirements are satisfied. Whenever the minimum cost is not unique, make

an arbitrary choice among the minima.



Solved Problem 2

Obtain an initial basic feasible solution to the following TP using the

matrix minima method.

Table 5.3: Initial table

D1 D2 D3 D4

01 1 2 3 4 6 Capacity

02 4 3 2 0 8

03 0 6 8 6 10

4 6 8 5 24

Demand

Where 0i and Di denote ith origin and jth destination respectively.

Solution: The transportation table of the given TP has 12 cells. Following

the matrix minima method,

The first allocation is made in the cells (3, 1), the magnitude being x21 = 4.

This satisfies the requirement at destination D1 and thus we cross the first

column from the table. The second allocation is made in the cell (2, 4),

the magnitude being x24 = min (6, 8) = 6. Cross the fourth column of the

table. This yields the table (i) shown below

There is again a tie for the third allocation. Arbitrarily choose the cell

(1, 2) and allocate x12 = min (6, 6) = 6. Cross the second column of the

first row. Next, choose to cross off the first row of the table. The next

allocation of magnitude x22 = 0 is made in the cell (3, 2), cross the

second column getting table(ii), as shown above



Again you choose arbitrarily to make the next allocation in cell (2, 3) of

magnitude x23 = min (2, 8) = 2, cross the second row to get table (iii) as

show above. The last allocation of magnitude x23 = min (6, 6) = 6 is made

in the cell (3, 3).

Now that all the rim requirements have been satisfied, an initial feasible

solution has been determined. This solution is shown above in table (iv).

Since the cells do not form a loop, the solution is basic and degenerate.

The transportation cost according to the above route is given by

Z=6×2+2×2+6×0+4×0+0×2+6×2=28

5.4.3 Vogel’s approximation method

The Vogel‟s approximation method (VAM) takes into account the least cost

cij, but also the cost that just exceeds cij. The steps of the method are given

below.

Step 1: For each row of the transportation table, identify the smallest and

the next to smallest costs. Determine the difference between them for each

row. Display them alongside the transportation table by enclosing them in

parenthesis against the respective rows. Similarly, compute the differences

for each column.

Step 2: Identify the row or column with the largest difference among all the

rows and columns. If a tie occurs, use any arbitrary tie breaking choice. Let

the greatest difference correspond to the ith row and let Cij be the smallest

cost in the ith row. Allocate the maximum feasible amount xij = min (ai, bj) in

the (i, j)th cell and cross off the ith row or the jth column in the usual manner.

Step 3: Recompute the column and row differences for the reduced

transportation table and go to step 2. Repeat the procedure until all the rim

requirements are satisfied.



Remarks:

1. A row or column “difference” indicates the minimum unit penalty incurred

by failing to make an allocation to the last smallest cell in that row or

column.

2. It is clear that VAM determines an initial basic feasible solution, which is

very close to the optimum solution. But the number of iterations required

to reach the optimal solution is small.

Caselet

A company has four warehouses. Each warehouse supplies inventory to

five stores.

The company needs to develop an integral transportation schedule that

meets all demands from the inventory at a minimum total transportation

cost.

Assuming that the total supply and the total demand are equal, the

company can use the transportation model of a LPP to arrive at a basic

feasible solution.

Solved Problem 3

Obtain an initial basic feasible solution to the following TP using the

Vogel‟s approximation method:

Table 5.4: Transportation Table

Ware houses Stores

Availability I II III IV

A 5 1 3 3 34

B 3 3 5 4 15

C 6 4 4 3 12

D 4 -1 4 2 19

Requirement 21 25 17 17 80

Solution: The transportation table of the given TP has 16 cells. The

differences between the smallest and next to smallest costs in each row

and each column are computed and displayed inside the parenthesis

against the respective columns and rows. The largest of these differences

is (3) and is associated with the fourth row of the transportation table.



The minimum cost in the fourth row is C42 = –1. Accordingly, you allocate

x42 = min (19, 25) = 19 in the cell (4, 2). This exhausts the availability at

warehouse D. Cross the fourth row. The row and column differences are

now computed for the resulting reduced transportation table (ii) as shown

below:

Table 5.5: Resulting reduced transportation table

5 1 3 3 34(2)

3 3 5 5 15(0)

6 4 4 3 12(1)

4 19-1 4 2 19(3)

21 (1)

25

(2)

17 (1)

17 (1)

(i)

Table 5.6: Resulting reduced transportation table

5 1 3 3 34 (2)

3 3 5 4 15 (0)

6 4 4 1 12 (1)

21 (2) 6 (2) 17 (1) 17 (0)

(ii)

The largest of this is (2) and is associated with the first row as well as the

first and second column. Arbitrarily, select the first row whose minimum

cost is C12 = 1. Thus the second allocation of magnitude x12 = min (34, 6)

= 6 is made in the cell (1, 2). Cross the second column from the table.

Continuing in this way, the subsequent reduced transportation tables and

the differences for the surviving rows and columns are shown below in

figure 5.1.

Figure 5.1: Surviving rows and columns

6

19



Eventually, the basic feasible solution is obtained as shown in table 5.6

below:

Table 5.7

The transportation cost according to the above route is given by:

Z = 6 x 5 + 6 x 1 + 17 x 3 + 5 x 3 + 15 x 3 + 12 x 3 + 19 x (– 1) = 164


State True or False

7. In matrix-minima method, you start allocating from the left-top cell of

the table.

8. In Vogel‟s approximation method, you first construct penalty and then

start allocating.

9. North-west corner rule gives optimum solution.

10. Vogel‟s approximation method gives solution near to the optimum

solution.

5.5 Moving Towards Optimality

After evaluating an initial basic feasible solution to a transportation problem,

the next question is how to get the optimum solution. The basic techniques

are illustrated below.

1. Determine the net evaluations for the non–basic variables (empty cells)

2. Determine the entering variable

3. Determine the leaving variable

4. Compute a better basic feasible solution

5. Repeat steps (1) to (4) until an optimum solution has been obtained



5.5.1 Improving the solution

Definition: A loop is the sequence of cells in the transportation table such

that:

i) Each pair of consecutive cells lie either in the same row or same

column

ii) No three consecutive cells lies in the same row or same column

iii) The first and the last cells of the sequence lies in the same row or

column

iv) No cell appears more than once in the sequence

Consider the non-basic variable corresponding to the most negative of the

quantities cij – ui – vj, calculated in the test for optimality; it is made the

incoming variable. Construct a loop consisting exclusively of this incoming

variable (cell) and current basic variables (cells). Then allocate the incoming

cell to as many units as possible after appropriate adjustments have been

made to the other cells in the loop. Avoid violating the supply and demand

constraints, allow all allocations to remain non-negative and reduce one of

the old basic variables to zero (where upon it ceases to be basic).

5.5.2 Modified distribution method / Modi method / U–V method

Step 1: Under this method, you construct penalties for rows and columns

by subtracting the least value of row / column from the next least value.

Step 2: Then select the highest penalty constructed for both row and

column. Enter that row/column and select the minimum cost and allocate

min (ai, bj)

Step 3: Delete the row or column or both if the rim availability/ requirements

are met.

Step 4: You repeat steps 1 to 2 to till all allocations are over.

Step 5: For allocating all forms of equations ui + vj = cj, set one of the dual

variable ui / vj to zero and solve for others.

Step 6: Use this value to find ij = cij - ui - vj. If all ij 0, then it is the optimal

solution.

Step 7: If any ij 0, select the most negative cell and form loop. Starting

point of the loop is positive and alternative corners of the loop are negative

and positive. Examine the quantities allocated at negative places. Select the



minimum, add it to the positive places and subtract from the negative

places.

Step 8: Form new table and repeat steps 5 to 7 till ij 0

Balanced TP

Solved Problem 4

Solve the following transportation problem with cost coefficients demands

and supplies as given in the following table:

Destinations

A B C Supply

Sources I

II

III

6 8 4 14

12

5

4 9 8

1 2 6

Demand 6 10 15

Solution: Since total demand = 31 = Total supply, the problem is

balanced. The initial basic feasible solution is obtained by Vogel‟s

approximation method. The following table gives the initial solution:

Supply

6

8

4

14 14

12

5

4

6

9

5

8

1

1 2

5

6

Demand 6 10 15

Destination



D1 D2 D3

1 6 8 4 14

14

2 4 6

9 5

8 1

12

3 1 2 5

6 5

6 10 15

The optimum allocations are

x13 = 14, x21 = 6, x22 = 5, x23 = 1, x32 = 5

The minimum transportation cost is

144+64+59+18+52 = 143

For allocated cells

u1 + v3 = 4 u1 = -4

Set u2 = 0

u1 + v3 = 4 u1 = -4

u2 + v2 = 9 v1 = 4

u2 + v3 = 8 v2 = 9

u3 + v2 = 5 v3 = 8

Note: Select that variable ui / vi is repeated very often for easy calculation.

Here u2 is repeated often.

For unallocated cells

ij = cij – ui – vj

11 = 6 – (-4) – 4 = 6

12 = 8 – (-4) – 9 = 3

31 = 1 – (-4) – 4 = 1

33 = 6 – (-4) – 8 = 2

Since all ij 0, the optimum solution is

X13 = 14 x 4 = 56

X21 = 6 x 4 = 24

X22 = 5 x 9 = 45

X23 = 1 x 8 = 8

X32 = 5 x 2 = 10

Total cost = 143

Supply Sources



Unbalanced T.P

A car company is faced with an allocation problem resulting from rental

agreement that allow cars to be returned to locations other than those

from where they were originally rented. At the present time there are two

cars with 15 and 13 simplex cars respectively and 4 locations requiring 9,

6, 7 and 9 cars respectively. The unit transportation costs (in dollars)

between the locations an given below:

Destinations

D1 D2 D3 D4

Sources S1 45 17 21 30

S2 14 18 19 31

Obtain a minimum cost schedule.

Solution: Since the supply and requirements are not equal it is called an

unbalanced TP. In general, if ai bj then it is called an unbalanced TP.

We introduce either a dummy row or a column with cost zero quantities

and bi aj respectively. Applying Vogel‟s approximation method we find

the basic feasible solution.

Destinations

S1 D1 D2 D3 D4 P1 P2 P3

P4

Sources

S2

6

3

6

15/9/6

4 4 4 11

S3

9

4

13/4

4 4 4 12

1st

Cancel

3

3

0 - - -

Demand 9/0 6 7/3 9/6

P1 14 17 19 30

P2 31 1 2 1

P3 – 1 2 1

P4 – – 2 1

45

14

4

0

17

18

0

21

19

0

30

31

0



X34 = Min [3, 9] = 3 x 10 = 0

X21 = Min [13, 9] = 9 x 14 = 126

X12 = Min [15, 6] = 6 x 17 = 102

X23 = Min [4, 7] = 4 x 9 = 76

X14 = Min [6, 6] = 6 x 30 =

180

Total Cost 547

Testing for optimality

u1 + v2 = 17 Set u1 = 0

u1 + v3 = 21 u1 = 2

u1 + v4 = 30 u3 = 30

u2 + v1 = 14 v1 = 16

u2 + v3 = 19 v2 = 17

u3 + v4 = 0 v3 = 21

v4 = 30

For unallocated cells

ij = cij - ui - vj

11 = 45 – 0 – 16 = 29

22 = 8 + 2 – 17 = 3

24 = 31 +2 – 30 = 3

31 = 0 + 30 – 26 = 4

32 = 0 + 30 – 17 = 13

33 = 0 + 30 – 21 = 9

For non-allocated cells, determine cij – uI – vi. Since all then quantities are

non-negative, the current solution is optimal. The minimum transportation

cost is equal to

617+321+630+914+419+3.0 = 470.

This is achieved by transporting x12 = 6 cars from source 1 to destination

2, x13 = 3, x14 = 6 cars from sources 1 to destinations 3 and 4

respectively; x21 = 9 and x33 = 4 cars from sources 2 to destinations 1

and 3 respectively.



5.5.3 Degeneracy in transportation problem

It is shown that a basic solution to an m-origin, n destination; transportation

problem can have at the most m+n-1 positive basic variables (non-zero),

otherwise the basic solution degenerates. It follows that whenever the

number of basic cells is less than m + n – 1, the transportation problem is a

degenerate one. The degeneracy can develop in two ways:

Case 1: The degeneracy develops while determining an initial assignment

via any one of the initial assignment methods discussed earlier.

To resolve degeneracy, you must augment the positive variables by as

many zero-valued variables as is necessary to complete the required

m + n – 1 basic variable. These zero-valued variables are selected in such a

manner that the resulting m + n – 1 variable constitutes a basic solution.

The selected zero valued variables are designated by allocating an

extremely small positive value ε to each one of them. The cells containing

these extremely small allocations are then treated like any other basic cells.

The ε‟s are kept in the transportation table until temporary degeneracy is

removed or until the optimum solution is attained, whichever occurs first. At

that point, we set each ε = 0.

Case 2: The degeneracy develops at the iteration stage. This happens

when the selection of the entering variable results in the simultaneous drive

to zero of two or more current (pre-iteration) basic variables.

To resolve degeneracy, the positive variables are augmented by as many

zero-valued variables as it is necessary to complete m+n-1 basic variables.

These zero-valued variables are selected from among those current basic

variables, which are simultaneously driven to zero. The rest of the

procedure is exactly the same as discussed above in case 1.

Note: The extremely small value ε is infinitely small and it never affects the

value it is added to or subtracted from. Introduce „‟ in unallocated minimum

cost cell to avoid forming a loop.



Solved Problem 5

Obtain an optimum basic feasible solution to the following degenerate T.P.

Table 5.7: Initial table

From

7 3 4 2 Available 2 1 3 3

3 4 6 5

4 1 5 10

Demand

Solution: Following the North West Corner rule, an initial assignment is

made as shown in the table 5.7. Since the basic cells do not form a loop,

the solution is basic. However, since the number of basic cells is 4, which

is less than 5. (= m + n - 1) the basic solution degenerates.

Table 5.8: Initial assignment

7

3 4 2

2 1 3 3

3 4 6 5

In order to complete the basis and remove degeneracy, you require only

one more positive basic variable. Select the variable x23 and allocate a

negligibly small positive quantity in the cells (2, 3) as shown in the table

5.8.

Table 5.9: Augmented solution

7

3 4 2

2 1 3 3 + = 3

3 4 6 5

4 1 5 + = 5

2

2

1

5

1 2

1

5



Notice that even after the inclusion of cell (2, 3) in the basis; the basic

cells do not form a loop, that is, the augmented solution remains basic.

The net evaluations can now be computed and the solution is tested for

optimality.

Starting table:

Since all the net evaluations for the non-basic variables are not non-

positive, the initial solution is not optimum. The non basic cell (1, 3) must

enter the basis. The exit criterion removes the basic cell (2, 3) from the

basis max .

Table 5.10: Starting table

–

7

(3)

3

(4)

4

5

+

2

1 -

3 0

(2)

3

(0)

4

6 8

-

7

(3)

3 +

4

0

2

1

(– 4)

3

-

(6)

(4)

4 –

6

2

First iteration: Introduce the cell (1, 3) into the basis and drop the cell (2,

3) from it. Determine the current net evaluations. Since all of them are

not non-positive the current solution can be improved.

Second iteration: Introduce the cell (3, 1) and drop the cell (1, 1) from the

basis. Since some of the current net evaluations are still positive, the

current solution can further be improved.

2 1

2 2

2

1

6

Vi 2 1 3 Vj 7 6 4

Starting table First iterated table



Table 5.11: Second iterated table

Ui

(-6)

7

(-3)

3

4 –1

–

2

1

(2)

3

0

+

3

(–2)

4 –

6 1

vj 2 1 5

Table 5.12: Optimum table

ui

(– 6)

7

(–1)

3

4 4

(–2)

2

1

3 3

3

(0)

4

6 6

Vj – 3 – 2 0

Third iteration: Introduce the cell (2, 3) and drop the cell (2, 1) from the

basis. Since all the current net evaluations are non positive, the current

solution is an optimum one. The transportation cost according to the

above route is given by

6134321142z = 33


Fill in the blanks

11. All the values of Cij - ui - vj should be __________ or _________ for the

solution to be optimum.

12. In unbalanced transportation problem ai is _________ _______ to bj.

13. If the number of allocation is less than _________ then it is said to be a

degenerate transportation problem.

1

1

2

2

4

2 1

3

2

2



5.6 Summary

The transportation problem is a special type of linear programming problem

in which the objective is to transport a homogeneous product manufactured

at several plants (origins) to a number of different destinations at a minimum

total cost. In this unit, you have learnt several different techniques for

computing an initial basic feasible solution to a transportation problem, such

as north-west corner rule, matrix minimum method and Vogel‟s

approximation method. Further, you studied the degeneracy in

transportation problem with examples on obtaining an optimum basic

feasible solution.

5.7 Terminal Questions

1. Solve the following transportation problem.

2. A company has three cement factories located in cities 1,2,3 which

supply cement to four projects located in towns 1,2,3,4. Each plant can

supply daily 6,1,10 truck loads of cement respectively and the daily

cement requirements of the projects are respectively 7,5,3,2 truck loads.

The transportation cost per truck load of cement (in hundreds of rupees)

from each plant to each project site is as follows.

Factories

Determine the optimal distribution for the company so as to minimise the

total transportation cost.

1 2 3 4

1 2 3 11 7

2 1 0 6 1

3 5 8 15 9



3. Solve the following transportation problem.

9 12 9 6 9 10 5

7 3 7 7 5 5 6

6 5 9 11 3 11 2

6 8 11 2 2 10 9

4 4 6 2 4 2 22

5.8 Answers to SAQs and TQs

Answers to Self Assessment Questions

1. LPP

2. Equal

3. Minimum cost

4. Yes

5. Yes

6. No

7. False

8. True

9. False

10. True

11. zero

12. Not equal to

13. m + n – 1

Answers to Terminal Questions

1. The optimal transportation cost is Rs. 796

2 Optimal transportation cost is Rs. 10, 000

3 The minimum transportation cost is Rs. 112 as 0

5.9 References

No external sources have been referred for this unit.

slm-unit-05-mb0048

Documents

vogels approximation

vogels approximation

west corner

current net

matrix minima

initial feasible

basic feasible

basic feasible