Transportation CHAP 7

1

CHAPTER 7

Transportation Problem

7.1 Introduction

The transportation problem is one of the subclasses of LPPs in which the objective is to transport various

quantities of a single homogeneous commodity that are initially stored at various origins to different

destinations in such a way that the transportation cost is minimum. To achieve this objective we must

know the amount and location of available supplies and the quantities demanded. In addition we must

known the costs that result from transporting one unit of commodity from various origins to various

destinations.

Let

m be the number of sources

n be the number of destinations

ai be the supply at the source i

bj be the demand at the destination j

cij be the cost of transportation per unit from source i to destination j

Xij be the number of units to be transported from the source i to the destination j. The Linear

programming model representing the transportation problem is given by

Minimize ∑∑= =

=m

1i

n

1jijijXCZ

subject to the constraints

m....3,2,1i,aXn

1jijij =≤∑

=

(Row Sum) and

n....3,2,1j,bXm

1iijij =≥∑

=

(Column Sum)

0Xij ≥ for all i and j

The objective function minimizes the total cost of transportation (z) between various sources and

destinations. The constraint i in the first set of constraints ensures that the total units transported from the

source i is less than or equal to its supply. The constraint j in the second set of constraints ensures that the

total units transported to the destination j is greater than or equal to its demand.

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Fig. 7.1 Schematic Diagram of Simple Transportation

Example 3.1 Consider the following transportation problem (Table 3.3) involving 3 sources and 3

destinations. Develop a linear programming (LP) model for this problem and solve it.

The given transportation problem is said to be balanced if

∑∑==

=n

1jj

m

1ii ba

ie. if the total supply is equal to the total demand.

This restriction causes one of the constraints to be redundant (and hence it can be deleted) so that the

problem will have (m + n – 1) constraints and (m x n) unknowns.

Note that a transportation problem will have a feasible solution only if the above restriction is satisfied.

Thus, ∑∑==

=n

1jj

m

1ii ba is necessary as well as a sufficient condition for a transportation problem to have a

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feasible solution. Problems that satisfy this condition are called balanced transportation problems.

Techniques have been developed for solving balanced or standard transportation problems only. It

follows that any non-standard problem in which the supplies and demands do not balance, must be

converted to a standard transportation problem before it can be solved. This conversion can be achieved

by the use of a dummy source/destination.

The above information can be put in the form of a general matrix shown below:

TABLE 7.1 Cost Matrix of Transportation Problem

Definitions

A few terms in connection with Transportation model are defined below

Feasible Solution Any set of non negative allocations ( 0Xij > ) which satisfies the row and column sum

(rim requirement) is called a feasible solution.

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 here m is the number of rows, n is the number of columns in a

transportation table.

Non-degenerate Basic Feasible Solution

Any feasible solution to a transportation problem containing in origins and n destinations is said to be

non-degenerate, if it contains m+n-1 occupied cells and each allocation is in independent positions.

The allocations are said to be in independent positions, if it is impossible to form a closed path.

Closed path means by allowing horizontal and vertical lines and all the corner cells are occupied.

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Optimal Solution A feasible solution that minimizes (maximizes) that the transportation cost (profit) is

called an optimal solution. The solution of a transportation problem. can be obtained in two stages,

namely initial solution and optimal solution. Initial solution can be obtained by using any one-of the

three methods

(i) North west corner rule (NWCR)

(ii) Least cost method or Matrix minima method(LCM)

(iii) 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 optimal solution.

The cells in the transportation table can be classified as occupied cells and unoccupied cells. The

allocated cells in the transportation table is called occupied cells and empty cells in a transportation table

is called unoccupied cells.

The improved solution of the initial basic feasible solution is called optimal solution which is, the

second stage of solution that can be obtained by MODI (modified distribution method).

The allocations in the following tables are not in independent positions.

The allocations in the following tables are in independent positions.

TABLE 7.2 Allocation of Different Possible Ways of Items

* *

* *

* *

* *

* *

* *

*

* *

* *

* *

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Degenerate Basic Feasible Solution If a basic feasible solution contains less than m+n-1 non negative

allocations, it is said to be degenerate.

Example 7.1 Consider the following transportation problem (Table 7.3) involving 3 sources and 3

destinations. Develop a linear programming (LP) model for this problem and solve it

Table 7.3 Example 7.1

Solution Let Xij be the number of units to be transported from the source i to the destination j, where i =

1, 2, 3 and j = 1, 2, 3. An LP model of this problem is:

333231232221311211 X9X7X3X8X2X6X8X5X4ZMinimize ++++++++=

subject to

.3,2,1jand3,2,1i,0X100XXX200XXX200XXX200XXX200XXX

100XXX

ij

332313

322212

312111

333231

232221

311211

==≥≥++≥++≥++≤++≤++≤++

Application of simplex method to the above LP model yields the optimal shipping plan as presented

in Table 7.4.

Table 3.4 Optimal Shipping Plan (Example 3.1)

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Source Destination Quantity shipped

1 3 100

2 2 200

3 1 200

Total minimum cost = Rs. 1800

7.2 TYPES OF TRANSPORTATION PROBLEM

The transportation problem can be classified into balanced transportation problem and unbalanced

transportation problem.

7.2.1 Balanced Transportation Problem

If the sum of the supplies of all the sources is equal to the sum of the demands of all the destinations, then

the problem is termed as balanced transportation problem. This may be represented by the relation:

∑∑==

=n

1jj

m

1ii ba

Example 3.1 represents a balanced transportation problem.

7.2.2 Unbalanced Transportation Problem

It is the sum of the supplies of all the sources is not equal to sum of the demands of all the destinations,

then the problem is termed as unbalanced transportation problem. That means, for any unbalanced

transportation problem, we have

∑∑==

≠n

1jj

m

1ii ba

Example 7.2 Convert the transportation problem shown in Table 7.5 into a balanced transportation

problem.

Table 7.5 Example 3.2

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Solution For the given problem,

500band,400an

1jj

m

1ii == ∑∑

==

Here, ∑∑==

≠n

1jj

m

1ii ba

Hence it is an unbalanced transportation problem. Under this situation, an additional source or

destination is to be included in the table as per the guidelines discussed now.

If ∑∑==

>n

1jj

m

1ii ba , then include a dummy destination to absorb the excess supply. The demand of the

dummy destination is equal to .ban

1jj

m

1ii ∑∑

==

− The cost coefficients in the dummy destination are

assumed as zeros. If ,abm

1ii

n

1jj ∑∑

==

> then include a dummy source to supply the excess demand. The

supply of the dummy source is equal to .abm

1ii

n

1jj ∑∑

==

− The cost coefficients in the dummy source are

assumed as zeros.

Table 7.5 is modified by including a dummy source with a supply of 100 units. This is shown in Table

7.6, which is now a balanced transportation table.

Table 7.6 Balanced Transportation Problem (Example 3.2)

Example 7.3 Convert the transportation problem shown in Table 7.7 into a balanced transportation

problem.

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Solution We have

400band,500an

1jj

m

1ii == ∑∑

==

Here, ∑∑==

>n

1jj

m

1ii ba

It is unbalanced transportation problem. This is converted into balanced transportation by including a

dummy destination as shown in Table 7.8

TABLE 7.7 Example 7.3

TABLE 7.8 Balanced Transportation of Problem of Example 7.3

7.3 FINDING THE INITIAL BASIC SOLUTION

There are there method available to find initial feasible basic solution of transportation problem.

7.3.1. North West Corner Method(NWCM)

Algorithm for northwest corner cell method

Step 1: Find the minimum of the supply and demand values with respect to the current northwest corner

cell of the cost matrix.

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Step 2: Allocate this minimum value to the current northwest corner cell and subtract this minimum

from the supply and demand values with respect to the current northwest corner cell.

Step 3: Check whether exactly one of the row/column corresponding to the northwest comer cell

has zero supply/demand, respectively. If so, go to step 4 otherwise, go to step 5.

Step 4: Delete that row/column with respect to the current northwest corner cell which has the zero

supply/demand and go to step 6.

Step 5: Delete both the row and the column with respect to the current northwest corner cell. Step 6:

Check whether exactly one row or column is left out. If yes, go to step 7 otherwise go to step 1.

Step 7: Match the supply/demand of that row/column with the remaining demands/supplies of the

undeleted columns/rows.

EXAMPLE 7.4

The application of the procedure to the model of example 7.1gives the starting basic solution in Table

7.9. The arrows show that the order in which the allocated amounts are generated.

TABLE 7.9 Starting Basic Solution of Example 7.1

The starting basic solution is given as

,100X,100X,100X,100X

,100X

3332

2221

11

====

=

The associated cost of the schedule as

2800.Rs10091007100210061004Z =×+×+×+×+×=

EXAMPLE 7.5

Obtain the initial basic feasible solution of transportation problem using North West Corner Method,

whose cost and rim requirement is given in Table 7.10.

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TABLE 7.10 Cost Matrix of Example 7.5

D1 D2 D3 Supply

O1 2 7 4 5

O2 3 3 1 8

O3 5 4 7 7

O4 1 6 2 14

Demand 7 9 18 34

TABLE 7.11 Initial Basic Feasible Solution of Example 7.5


,14X,4X,2X,6X,2X

,5X

43

3332

2221

11

=====

=


.102.Rs1424734632352Z =×+×+×+×+×+×=

7.3.2 Algorithm for Least Cost Cell Method(LCM)

Step 1: Find the minimum of the (undeleted) values in the cost matrix (i.e. find the matrix minimum).

Step 2: Find the minimum of the supply and demand values (X) with respect to the cell corresponding to

the matrix minimum.

Step 3: Allocate X units to the cell with the matrix minimum. Also, subtract X units from the supply and

the demand values corresponding to the cell with the matrix minimum.

Source/Destination

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Step 4: Check whether exactly one of the row/column corresponding to the cell with the matrix

minimum has zero supply/zero demand, respectively. If yes, go to step 5 otherwise, go to step 6.

Step 5: Delete that row/column with respect to the cell with the matrix minimum which has the zero

supply/zero demand and go to step 7.

Step 6: Delete both the row and the column with respect to the cell with the matrix minimum.

Step 7: Check whether exactly one row or column is left out. If yes, go to step 8 otherwise, go to step 1.

Step 8: Match the supply/demand of that row/column with the remaining demands/supplies of the

undeleted columns/rows.

EXAMPLE 7.6

The least-cost method is applied to Example 7.1 in the following manner:

1. Cell (2, 2) has the least unit cost in the tableau (= Rs. 2). The most that can be shipped through (2, 2)

is X22 = 200 truckloads, which happens to satisfy both row 2 and column 2 simultaneously. We

arbitrarily cross out column 2 and adjust the supply in row 2 to 0.

2. Cell (3, 1) has the smallest uncrossed-out unit cost (=Rs. 3). Assign X31 = 200, cross out column 1 and

adjust the supply in row 3 to 0.

3. Continuing in the same manner, we successively assign 100 truckloads to cell (1, 3), 0 truckloads to

cell (1, 3). The resulting starting solution is summarized in Table 7.12. The arrows show the order in

which the allocations are made. The starting solution (consisting of 3 basic variables) is



,100X,200X,200X

13

31

22

===

12


1800.Rs100820022003Z =×+×+×=

The quality of the least-cost starting solution is better than that of the northwest-corner method

(Example 7.5) because it yields a smaller value of Z (Rs. 83 versus Rs. 102 in the northwest-corner

method).

EXAMPLE 7.7

Obtain the initial basic feasible solution of transportation problem using Least Cost Cell Method, whose

cost and rim requirement is given in Table 7.10.


The least-cost method is applied to Example 7.5 in the following manner:

1. Cell (4,1) has the least unit cost in the tableau (= Rs. 1). The most that can be shipped through (4,1) is

X41 = 7 truckloads, cross out column 1becuase it is satisfied, and adjust the supply in row 4 to 14 – 7=

7.

2. Cell (3, 1) has the smallest uncrossed-out unit cost (=Rs. 2). Assign X43 = 7, cross out row 4 and

adjust the supply in column 3 to 18-7= 11.

3. Continuing in the same manner, we successively assign 7 truckloads to cell (3,2), 3 truckloads to cell

(1, 3) and 2 truckloads to cell (1, 2). The resulting starting solution is summarized in Table 7.13. The

arrows show the order in which the allocations are made. The starting solution is


13

,7X,7X,7X,8X

,3X,2X

4341

32

23

1312

====

==

The associated cost of the schedule as 83.Rs727174813427Z =×+×+×+×+×+×=

The quality of the least-cost starting solution is better than that of the northwest-corner method

(Example 7.5) because it yields a smaller value of Z (Rs. 1800 versus Rs. 2800 in the northwest-corner

method).

7.3.3 Vogel's Approximation Method (VAM)

The steps involved in this method for finding the initial solution are as follows.

Step 1 Find the penalty cost, naively the difference between the smallest and next smallest costs in each

row and column.

Step 2 Among the penalties as found in step(l) choose the maximum penalty. If this maximum penalty is

more than one (i.e if there is a tie) choose any one arbitrarily.

Step 3 In the selected row or column as by step(2) find out the cell having the least cost. Allocate to this

cell as much as possible depending on the capacity and requirements.

Step 4 Delete the row or column which is fully exhausted. Again compute the column and row penalties

for the reduced transportation table and then go to step (2). Repeat the procedure until all the rim

requirements are satisfied.

Note If the column is exhausted, then there is a change in row penalty and vice versa.

EXAMPLE 7.8

Vogel’s Approximation Method is applied to Example 7.1, Table 7.3


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Row 2 and 3 have largest penalty (=4) and cell (2,2) has the smallest unit cost in that row, the amount

200 is assigned to X22. Column 2 and row 2 are satisfied and must be cross out. Next new penalties are

calculated as in Table 7.15.

TABLE 7.15 First Iteration of VAM for Example 7.1

Table 7.15 shows the row 3 has the highest penalty (= 6) and cell (3,1) has the smallest unit cost in that

row, the amount 200 is assigned to X31. Column 2 and row 3 are satisfied and must be cross out. Next

new penalties are calculated as in Table 7.16.

TABLE 7.16 Second Iteration of VAM for Example 7.1

Table 7.16 shows only one is remaining in the cell (1,3), the amount 100 units is assigned. Now column

3 also satisfied as shown in Table 7.17.

TABLE 7.17 Final Iteration of VAM for Example 7.1

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The Associated cost of the schedule as 1800.Rs100820022003Z =×+×+×=

EXAMPLE 7.9

Vogel’s Approximation Method is applied to Example 7.5, Table 7.10

TABLE 7.18 First Iteration of VAM for Example 7.5

In Table 7.18, row penalties and column penalties are computed. The maximum of these penalties is 2

which occur in rows 1 and 2. Hence, the cell with the least cost in these rows are to be identified. This

occurs at the cell (2, 3). The supply and the demand values corresponding to the cell (2, 3) are 8 and 18,

respectively. The minimum of these values is 8. Thus 8 units are allocated to the cell (2, 3) and the same

is subtracted from the supply and demand values of the cell (2, 3).

In this process, the demand at the source 2 is fully satisfied. Hence, this row is highlighted and the

resultant data is shown in Table 7.19.

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Since a row has been deleted, it is important to revise the row penalties as shown in Table 7.19. The

maximum of these penalties is 5 which occur in rows 1 and 4. Therefore, the cell with the least cost is

identified in row 4. This occurs at the cell (4,1). The supply and the demand values corresponding to the

cell (4,1) are 14 and 7 respectively. The minimum of these values is 7. The supply at the source 4 is

fully satisfied. Hence, this row is highlighted and the resultant data is shown in Table 7.20.


7.4.2 Optimizing the Basic Feasible Solution Applying U-V Method

Step 1: Row 1, row 2,..., row m of the cost matrix are assigned with variables U1, L,1 ... I U1111

respectively and the column 1, column 2,..., column n are assigned with variables V1, V,,.... V,

respectively.

Sri p 2 Check whether the number of basic cells in the set of initial basic feasible solution is equal to m

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+ n – 1. If yes, go to step 4, otherwise, go to step 3.

Step 3 Convert the necessary number of non-basic cells into basic cells to satisfy the condition stated in

step 2 (while doing this, sufficient care should be taken such that there is no closed loop formation with

the inclusion of the new basic cell(s)). The concept of the closed loop is explained in step 8.

Step 4: Compute the values for U,, U,_,..., U,,,, and V1, I V,, by applying the following formula

2

to all the basic cells only.

U1 + Vi Cii (assume U, = 0)

Step 5: Compute penalties Pij for the non-basic cells by using the formula: P'j = Uj -+- Vi – cjj

Step 6: Check whether all P;j values are less than or equal to zero. If yes, go to step 12, otherwise, go to

step 7.

Step T Identify the non-basic cell which has the maximum positive penalty, and term that cell as the new

basic cell.

Step 8: Starting from the new cell, draw a closed loop consisting of only horizontal and vertical lines

passing through some basic cells. (Note: Change of direction of the loop should be with 90 degrees only

at some basic cell.)

Step 9: Starting from the new basic cell, alternatively assign positive (+) and negative (–) signs at the

corners of the closed loop.

Step 10: Find the minimum of the allocations made amongst the negatively signed cells.

Step 11: Obtain the table for the next iteration by doing the following steps and then go to step 2.

(i) Add the minimum allocation obtained in the previous step to all the positively signed cells and

subtract minimum allocation from all the negatively signed cells and then treat the net allocations as the

allocations in the corresponding cells of the next iteration.

(ii) Copy the allocations which are on the closed loop but not at the corner points of the closed loop, as

well as the allocations which are not on the loop as such without any modifications to the corresponding

cells of the next iteration.

Step 12. The optimality is reached. Treat the present allocations to the set of basic cells as the optimum

allocations.

Step 13. Stop.

Step 1 Starting with the cell at 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

18

of the first column is satisfied i.e., ).b,amin(X 111i =

Step 2 If ,ab 11 > we move down vertically to the second row and make the second allocation of

magnitude ),Xb,amin(X 1i1222 −= in the cell (2, 1)

If ,ab 11 > move right horizontally to the second column and make the second allocation of magnitude

),bX,amin(X 11i112 −= in the cell (1, 2)

If 11 ab = b, there is a tie for the second allocation. We make the second allocations of magnitude

0)b,aamin(X 11112 =−= in cell (1,2)

or 0)bb,amin(X 11221 =−= in the cell (2,1)

Step 3 Repeat steps I and 2 moving down towards the lower right corner of the transportation table until

all the rim requirements are satisfied.

Example Obtain the initial basic feasible solution of a transportation problem whose cost and rim

requirement table is given below.

Example Determine an initial basic Feasible solution to the following transportation problem using

N.W.C.R

Origin/Ds D1 D2 D3 D4 Supply

O1 6 4 1 5 14

O2 8 9 2 7 16

O3 4 3 6 2 5

Required 6 10 15 4 35

Least Cost-or Matrix Minima Method

Step 1 Determine the smallest cost in the cost matrix of the transportation table.

Let it be ijC . Allocate ).b,amin(X 11ij = in the cell (i, j)

Step 2 If iij aX = cross off the ith row of the transportation table and decrease bj by ai. Then go to step3.

Origin\Destination D1 D2 D3 Supply

O1 2 7 4 5

O2 3 3 1 8

O3 5 4 7 7

O4 1 6 2 14

Demand 7 9 18 34

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If jij bX = cross off the jth column of the transportation table and decrease ai by bj. Go to step3.

If jiij baX == cross off either the ith row or the jth column but not both.

Step 3 Repeat steps 1 and 2 for the resulting reduced transportation table untill all the rim requirements

are satisfied. Whenever the minimum

cost is not unique, make an arbitrary choice among the minima.

Example Obtain an initial feasible solution to the following TP using Matrix Minima Method.

Origin/Des D1 D2 D3 D4 Supply

O1 1 2 3 4 6

O2 4 3 2 0 8

O3 0 2 2 1 10

Demand 4 6 8 6 24

Example Determine an initial basic feasible solution for the following TP, using least cost method

Origin/des D1 D2 D3 D4 Supply

01 1 6 5 3 14

02 8 9 2 7 16

03 4 3 6 2 5

Demand . 6 10 15 4 35

Example Find the initial basic feasible solution for the following transportation problem by VAM.

D1 D2 D3 D4 Supply

O1 11 13 17 14 250

O2 16 18 14 10 300

O3 21 24 13 10 400

Demand 200 225 275 250 950

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Transportation Problems 1. Obtain the initial solution for the following TP using (i) NWCR (ii) Least cost method (iii) VAM Destination

341897142614774538133254721

Demand

SupplyCBA

2. Solve the following TP where the cell entries denote the unit transportation costs. Destination

10030204010506495307538206245

DemandRQP

SupplyDCBA

3. Solve the following transportation problem. Destination

301520401313152142103221

321

Demand

Capacity

4. Find the minimum transportation cost. Warehouse

14785183070840960403070710503019

3

2

1

4321

DemandFFF

SupplyDDDD

Sour

ce

Orig

in

Sour

ce

Fact

ory

21

5. Solve the following T.P. using Vogel’s Method. Warehouse

242644minRe91022118642113119563655773725109691291

tquire

AvailableFEDCBA

6. Solve the following T.P. Destination

68641012203802342643211

Demand

SupplyDCBA

7. Solve the following T.P. Destination

403525307098128340122016212508720111

Demand

SupplyDCBA

8. Solve the following T.P to maximize the profit. Destination

306020407030283838330303035442

100332225401

Demand

SupplyDCBA

Fact

ory

Sour

ce

Sour

ce

Sour

ce

22

9.Four factories A, B, C and D are required to manufacture and supply a specific product for four stores

P, Q, R and S. Production capacities in each factory and demand at each store are given in table 1 below,

while Unit transportation cost in Rupees from each factory to each store is given in table 2.

Table 1

Table 2

(i) Construct this as a standard Transportation problem in tabular form.

(ii)Using any appropriate method, determine the optimum transportation schedule and the

corresponding transportation cost.

10. The following table gives the cost of transporting material from supply points A, B, C and D to

demand points E, F, G, H and J.

E F G H J Supply

Factory A B C D

Production capacity (units) 50 70 30 50

Store P Q R S

Demand (units) 25 35 105 35 Store P Q R S

Factory

A

B

C

D

2 4 8 11

10 8 7 5

13 3 9 12

4 6 8 3

23

A

B

C

D

8 10 12 17 15

15 13 18 11 9

14 20 6 10 13

13 19 7 6 12

100

150

180

280

Demand 90 170 50 210 190

The present allocation is as follows:

A to E 90, A to F 10, B to F 150, C to F 10, C to G 50, C to J 120, D to H 210, D to J

70.

(a) Check the allocation is optimum, if not find an optimum schedule.

(b) If in the above problem the transportation cost from A to G is reduced to 10, what will be the new

optimum schedule?.

(c) If the availability of supply point A is reduced by 10 units, use each of the following criteria to

obtain a initial basic feasible solution:

(i) Northwest corner rule

(ii) Least cost method

(d) Starting with best initial solution is found in part (c), obtain an optimal solution, and hence produce

transportation schedule.

11.Consider a transportation problem involving 3 sources and 3 destinations. Find minimum transportation cost.

400400200

183025

91210

151020

Supply

200

300500

1000

Destination1 2 3

Source

1

2

3

Demand

24

12. Find minimum transport cost of the following

302020

10752

20431

15565

25323

Supply321

Sources S1

S2

S3

S4

Demand

Destinations

Documents

Transportation CHAP 7