Easy Learning - Assignment Problems

8/22/2019 Easy Learning - Assignment Problems

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Operations Research Unit 7

Sikkim Manipal University 1

ASSIGNMENT PROBLEM

Consider a small manufacturing unit and imagine you are the manager of this unit.

If there is only one machine and one worker to operate the machine, how would you employ the worker?

Your immediate answer might be, the available worker will operate the machine.

Suppose two machines and two workers are engaged at different rates to operate the machines, which worker must operate

which machine to ensure maximum profit?

Similarly, if there are n machines available and n workers are engaged at different rates to operate them, which worker

should be assigned to which machine to ensure maximum efficiency?

In such situations, while answering the above questions, you will have to find such an assignment which offers maximum

profit on minimum cost. Such problems are known as " Assignment Problems" .

The Assignment Problem is a special case of Transportation Problem. It deals with assigning equal number of resources to

equal number of activities on one to one basis so as to minimize the total cost/time or maximize total profit of allocation.

The Assignment Model is useful in solving problems such as assignment of machines to jobs, assignment of salesman to

sales territories, travelling salesman problem and many more similar situations.


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STRUCTURE OF ASSIGNMENT PROBLEM

The structure of assignment problem of assigning operators to jobs is shown below:

Operators

Jobs

1 2 j n

1 C11 C12 . C1j C1n

2 C21 C22 . C2j C2n

. . . . . . .

i Ci1 Ci2 ..... Cij Cin

. . . . . .

n Cn1 Cn2 . Cnj . Cnn

Let there be n number of jobs and n number of operators.

The Operators are to be assigned to jobs on one to one basis.

Let Cijbe the cost of assigning ith

operator to jth

job (C11be the cost of assigning 1st

operator to the 1st

job, C12be the cost of

assigning 1st operator to the 2nd job and so on).

Xij denotes the assignment of ith

operator to jth

job.

Now the problem is which job is to be assigned to which operator so that the cost of completion of work will be minimum.


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Mathematically, we can express the problem as follows:

To minimize Z (cost) = ]n...2,1j;n...2,1iijxn

1j

n

1iijC

1; if ith

operator is assigned to the jth

job

Where xij =

0; if ith

operator is not assigned to the jth

job

With the restrictions,

1. n....3,2,1j;1n

1i ijx

, ie the i

thoperator will work on only one job

2. n....3,2,1i;1n

1j ijx

, ie thej

thjob will be done only by one operator

NETWORK REPRESENTATION OF ASSIGNMENT PROBLEM

For an operator job assignment problem, the time taken (in mins) by operators to perform the jobs is given below:

Operators

Jobs

1 2 3

A 10 16 7

B 9 17 6

C 6 13 5


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An assignment problem can be viewed as a transportation problem in which the capacity from each source is 1 and the

demand at each destination is 1

Source Destination

1 1

Capacity

1

Demand

1

1

Time taken (in mins)

1

JOB1

JOB

2

JOB3

OPERATOR

B

OPERATORC

OPERATOR

A

10

16

7

9

17

6

6

13

5


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TYPES OF ASSIGNMENT PROBLEM

Balanced Assignment Problem: If in an assignment problem, number of rows is equal to number of columns or if a given

problem is a square matrix, then the problem is termed as a Balanced Assignment Problem.

Unbalanced Assignment Problem: If in an assignment problem, number of rows is not equal to number of columns or if a

given problem is not a square matrix, then the problem is termed as an Unbalanced Assignment Problem.

The fol lowing example will help you in understanding the Assignment Problem.

Example 1:

Three Operators A, Band C are to be assigned to three Jobs J1, J2 and J3. The cost of assigning each Operator to each

Job (in Rs.) is given in the following matrix. Find which Operator is to be assigned to which Job so as to minimize the

total cost.

Operators

Jobs

J1 J2 J3

A 4 2 7

B 8 5 3

C 4 5 6


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Solution:

Here the objective of this assignment problem is to assign Operators (3) to the Jobs (3) in a way that will result in total

minimum cost.

In order to find the assignment, we apply the Hungarian algorithmas follows:

Step 1: Prepare the square matrix

The given matrix is a square matrix [Number of rows (Operators) = Number of columns (Jobs)].

Step 2: Reduce the Matrix Row Wise

Prepare a row reduced matrix by selecting the smallest element in each row of the matrix and subtracting this smallest

element from all the other elements in the row. This gives an opportunity cost matrix with at least one zero in each row.


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Identify the smallest element of each row.

Operators Jobs Row Minimum

J1 J2 J3

A 4 2 7 2

B 8 5 3 3

C 4 5 6 4

The smallest element of each row is subtracted from the elements of the respective row to get a reduced matrix.

Operators

Jobs

J1 J2 J3

A 42= 2 2 - 2=0 7 - 2= 5

B 8 - 3= 5 53= 2 33= 0

C 44 = 0 5 - 4 = 1 64 =2

The reduced matrix is:

Operators

Jobs

J1 J2 J3A 2 0 5

B 5 2 0

C 0 1 2


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Step 3: Reduce the Matrix Column Wise

In the reduced matrix given in step 2, determine the smallest element in each column and subtract this smallest element

from all the other elements in the column. This gives an opportunity cost matrix with at least one zero in each column.

In the reduced matrix given in step 2, identify the smallest element of each Column.

OperatorsJobs

J1 J2 J3

A 2 0 5

B 5 2 0

C 0 1 2

Column M inimum 0 0 0

The smallest element of each column is subtracted from the elements of the respective column to get a reduced matrix.

Operators

Jobs

J1 J2 J3

A 20 = 2 00 = 0 50 = 5

B 5 - 0= 5 20 = 2 00 = 0

C 0 - 0 = 0 10 = 1 20 = 2


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Operators

Jobs

J1 J2 J3

A 2 0 5

B 5 2 0

C 0 1 2

Step 4: In the reduced matrix given in step 3, search for optimum solution as follows:

Make Assignment as follows:

If there is only one zero in a row, an assignment is made by enclosing this zero by a square ( ) and crossing (X) or

cancelling other zeros in its column. Continue the process till all the rows in the matrix is examined.

In the reduced matrix given in step 3, first row has only one zero (cell 1,2) and hence make assignment by enclosing

this zero by a square ( ). Similarly, second and third row has only one zero and hence make assignment by

enclosing this zero by a square ( ).

Operators

Jobs

J1 J2 J3

A 2 5

B 5 2

C 1 2


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Note:

After examining the rows, examine the columns successfully until a column with exactly one zero is found. Make

assignment by enclosing this zero by a square ( ) and cross (X) or cancel other zeros in its row.

In case of atleast two zeros in a particular row and column is found, choose arbitrarily any one of these and cross out all

other zeros of that row and column.

Repeat step 4 till all zeros are either assigned or crossed.

If the number of assignments is equal to number of rows/columns, you have arrived at an optimal solution.

In this problem, all zeros are either crossed out or assigned. Also number of assignments (3) is equal to number of

rows/columns (3). As the number of assignments is equal to the number of rows/columns, solution is optimal.

The optimal allocation is to assign Operator Ato Job J2, Operator Bto Job J3 and Operator C to Job J1. The total cost (in Rs.)

of this assignment is computed from the original table.

Operators

Jobs

J1 J2 J3

A 2 5

B 5 2

C 1 2

Original Table:

Operators

Jobs

J1 J2 J3

A 4 2 7B 8 5 3

C 4 5 6


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The optimal assignment is:

Operators Jobs Cost of assignment

(Rs.)

A J2 2

B J3 3

C J1 4

Total 9

The total number of steps in the above problem: 4 Steps

Step1: Check for Number of rows (Operators) = Number of columns

(Jobs)Result of 1st step: A square matrix

Step 2: Select the smallest element in each row of the matrix andsubtract this smallest element from all the other elements in the row.

Result of 2nd step: Row Reduced matrix

Step 3: In the reduced matrix given in step 2, select the smallest element

in each column and subtract this smallest element from all the other

elements in the column.

Result of 3rd

step: Column Reduced matrix

Step 4: In the reduced matrix given in step 3, search for optimum

solution by making assignments as follows:If there is only one zero in a row, an assignment is made by enclosing

this zero by a square ( ) and crossing (X) or cancelling other zeros in

its column. Continue the process till all the rows in the matrix isexamined.

Result of 4th step: All zeros are either crossed outor assigned and the number ofassignments is equal to the

number of rows/columns,

therefore the solution is optimal


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Example 2:

Five machines M1, M2, M3, M4 and M5 are to be assigned to five jobs J1, J2, J3, J4, J5. The cost of assigning each

machine to each job in Rs. is given in the following matrix. Find which machine is to be assigned to which job so as to

minimize the total cost of assigning machines to jobs.

MachinesJobs

J1 J2 J3 J4 J5

M1 11 17 8 16 20

M2 9 7 12 6 15

M3 13 16 15 12 16

M4 21 24 17 28 26

M5 14 10 12 11 15

Solution:

Here the objective of this assignment problem is to assign Machines (5) to the Jobs (5) in a way that will result in total

minimum cost.

In order to find the assignment we apply the Hungarian algorithmas follows:


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The given matrix is a square matrix [Number of rows (Machines) = Number of columns (Jobs)].


Prepare row reduced matrix by selecting the smallest element in each row of the matrix and subtracting this smallest

element from all the other elements in the row. This gives an opportunity cost matrix with at least one zero in each row.

Identify the smallest element of each row.

Machines

Jobs Row Minimum

J1 J2 J3 J4 J5

M1 11 17 8 16 20 8

M2 9 7 12 6 15 6

M3 13 16 15 12 16 12

M4 21 24 17 28 26 17M5 14 10 12 11 15 10


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The smallest element of each row is subtracted from the elements of the respective row to get a reduced matrix.

Machines

Jobs

J1 J2 J3 J4 J5

M1 11-8 = 3 17-8=9 8-8=0 16-8=8 20-8=12

M2 9-6=3 7-6=1 12-6=6 6-6=0 15-6=9

M3 13-12=1 16-12=4 15-12=3 12-12=0 16-12=4M4 21-17=4 24-17=7 17-17=0 28-17=11 26-17=9

M5 14-10=4 10-10=0 12-10=2 11-10=1 15-10=5


Machines

Jobs

J1 J2 J3 J4 J5

M1 3 9 0 8 12

M2 3 1 6 0 9

M3 1 4 3 0 4

M4 4 7 0 11 9

M5 4 0 2 1 5


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from all the other elements in the column. This gives an opportunity cost matrix with at least one zero in each column.

In the reduced matrix given in step 2, identify the smallest element of each Column.

MachinesJobs

J1 J2 J3 J4 J5

M1 3 9 0 8 12

M2 3 1 6 0 9

M3 1 4 3 0 4

M4 4 7 0 11 9

M5 4 0 2 1 5

Column minimum 1 0 0 0 4


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The smallest element of each column is subtracted from the elements of the respective column to get a reduced matrix.

Machines

Jobs

J1 J2 J3 J4 J5

M1 3-1=2 9-0=9 0-0=0 8-0=8 12-4=8

M2 3-1=2 1-0=1 6-0=6 0-0=0 9-4=5

M3 1-1=0 4-0=4 3-0=3 0-0=0 4-4=0M4 4-1=3 7-0=7 0-0=0 11-0=11 9-4=5

M5 4-1=3 0-0=0 2-0=2 1-0=1 5-4=1


Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 0 8 8

M2 2 1 6 0 5

M3 0 4 3 0 0

M4 3 7 0 11 5

M5 3 0 2 1 1


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In the reduced matrix given in Step 3,a. First row has only one unmarked zero (cell 1, 3) and hence make assignment by enclosing this zero by a square ( ) and

mark (X) to the zeros in the same column (cell 4,3).

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 0 5

M3 0 4 3 0 0

M4 3 7 11 5

M5 3 0 2 1 1


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b. In the second row the cell (2,4) contains one unmarked zero and hence make assignment by enclosing this zero by a

square ( ) and mark (X) to the zeros in the same column (cell 3,4).

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 5

M3 0 4 3 0

M4 3 7 11 5

M5 3 0 2 1 1

c. In the fifth row there is only one unmarked zero in cell (5,2) and hence make assignment by enclosing this zero by a

square ( ) .

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 5

M3 0 4 3 0M4 3 7 11 5

M5 3 2 1 1


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d. After examining the rows, examine the columns successfully until a column with exactly one zero is found. Make


The first column contains one unmarked zero in cell (3,1) and hence make assignment by enclosing this zero by a square (

) and mark (X) to the zero in the same row in (cell 3,5).

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 5

M3 4 3

M4 3 7 11 5

M5 3 2 1 1

In this problem, all zeros are either crossed out or assigned. Also number of assignments (4) is not equal to number of

rows/columns(5). As the number of assignments is not equal to the number of rows/columns, solution is not optimal.

Since one row and one column are not assigned, optimal assignment is not possible in the current matrix and hence find

minimum number of lines crossing all zeros.

Step 5: Find minimum number of lines required to cross all zeros.

Draw minimum number of horizontal/vertical lines to cover all the zeroes in the resulting matrix:


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This consists of the following sub steps:

a. Mark () the row that do not have assignment. In this case it is the 4th row.

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 5

M3 4 3

M4 3 7 11 5

M5 3 2 1 1

b. Mark () the column which contains zeros in marked row. In this case it is the 3rd

column.

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 5

M3 4 3

M4 3 7 11 5

M5 3 2 1 1


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c. Mark () the row which contains assignment in the marked column. In this case it is the 1st

row.

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 5

M3 4 3

M4 3 7 11 5

M5 3 2 1 1

d. Draw the lines through unmarked rows and marked columns.

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 5

M3 4 3

M4 3 7 11 5

M5 3 2 1 1


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e. Revise the matrix as follows:

Examine the uncovered elements (on which there are no lines). Find the smallest element among the uncovered elements of

the table. Here the smallest element is 2.

Subtract this element (2) from the uncovered elements.

Add this element (2) to all the values at the point of intersection of any two of the lines.

Other elements crossed by the lines remain unchanged.Given matrix:

Machines

Jobs

J1 J2 J3 J4 J5

M1 2 9 8 8

M2 2 1 6 5

M3 4 3

M4 3 7 11 5

M5 3 2 1 1

The smallest element is 2.


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Revised matrix:

Machines

Jobs

J1 J2 J3 J4 J5

M1 0 7 0 6 6

M2 2 1 8 0 5

M3 0 4 5 0 0

M4 1 5 0 9 3

M5 3 0 4 1 1

Repeat Step 4 to make assignments.

Machines

Jobs

J1 J2 J3 J4 J5

M1 7 6 6

M2 2 1 8 5

M3 4 5

M4 1 5 9 3

M5 3 4 1 1

Since all the rows and columns are assigned or number of rows/columns is equal to number of assignments, this is an

optimal assignment.


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The optimal allocation is to assign Machine M1 to J 1, M2 to J4, M3 to J5, M4 to J3, M5 to J2 . The total cost (in Rs.) of this

assignment is computed from the original table.

Machines

Jobs

J1 J2 J3 J4 J5

M1 7 6 6

M2 2 1 8 5

M3 4 5

M4 1 5 9 3

M5 3 4 1 1

Original Table:

Machines

Jobs

J1 J2 J3 J4 J5

M1 11 17 8 16 20

M2 9 7 12 6 15

M3 13 16 15 12 16M4 21 24 17 28 26

M5 14 10 12 11 15


Machine Jobs Cost of assignment

(Rs.)

M1 J1 11

M2 J4 06

M3 J5 16

M4 J3 17

M5 J2 10

Total 60


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SPECIAL CASES IN ASSIGNMENT PROBLEM

1. Unbalanced Assignment Problem

Unbalanced assignment problem is a problem where the number of rows is not equal to the number of columns and vice

versa. For example, the number of machines may be more than the number of jobs or the number of jobs may be more than

the number of machines. In such a situation, you will have to introduce dummy rows or columns in the matrix. The dummy

rows or columns will contain all cost elements as zero. This balances the problem and then you can use Hungarian methodto find the optimal assignment.

Example 3:

Suppose six operators are to be assigned to five jobs and the cost of assignment in Rs. is given in the matrix below.

Operators Jobs

J1 J2 J3 J4 J5

1 6 2 5 2 6

2 2 5 8 7 7

3 8 9 5 4 7

4 2 5 6 8 9

5 3 7 6 5 96 5 4 7 9 5

The given problem is an Unbalanced Assignment Problem since the number of Operators is not equal to the number of

Jobs. The given matrix is not a square matrix and hence introduce a dummy column (fictitious Job 6) to make it a square

matrix with zero cost for the dummy column.


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Therefore the Balanced Assignment Problem is:

Operators Jobs

J1 J2 J3 J4 J5 J6

1 6 2 5 2 6 0

2 2 5 8 7 7 0

3 8 9 5 4 7 0

4 2 5 6 8 9 0

5 3 7 6 5 9 0

6 5 4 7 9 5 0

Example 4:

Suppose five operators are to be assigned to six jobs and the cost of assignment in Rs. is given in the matrix below.

Operators Jobs

J1 J2 J3 J4 J5 J6

1 6 2 5 2 6 5

2 2 5 8 7 7 2

3 8 9 5 4 7 9

4 2 5 6 8 9 6

5 3 7 6 5 9 7


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The given problem is an Unbalanced Assignment Problem since the number of Operators is not equal to the number of

Jobs. The given matrix is not a square matrix and hence introduce a dummy row (fictitious Operator 6) to make it a square

matrix with zero cost for the dummy row.

Therefore the Balanced Assignment Problem is:

Operators Jobs

J1 J2 J3 J4 J5 J6

1 6 2 5 2 6 5

2 2 5 8 7 7 2

3 8 9 5 4 7 9

4 2 5 6 8 9 6

5 3 7 6 5 9 7

6 0 0 0 0 0 0

2. Restrictions on Assignment or Impossible Assignments

It is sometimes possible that a particular person is incapable of doing certain job or a specific job cannot be performed on a

particular machine. The solution of the assignment problem should take into account these restrictions so that the infeasible

assignment can be avoided. This can be achieved by assigning a very high cost (say or M) to the cells where assignments

are prohibited, thereby, restricting the entry of this pair of job machine or resourceactivity into the final solution. After

inserting a high value at the cell we need to apply Hungarian method to solve the problem.


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Example 5:

The cost of performing different tasks by different persons is given below. The task-person marking (x) indicates that the

individual involved cannot perform the particular task. Using this information, determine the optimal assignment.

Tasks

Persons

P1 P2 P3 P4

T1 20 x 32 27

T2 15 20 17 18

T3 16 18 x 20

T4 x 20 18 24

Solution: Assign a cost of to the cells wherever there is a restriction on assignment.

The resulting matrix is:

Tasks

Persons

P1 P2 P3 P4

T1 20 32 27

T2 15 20 17 18

T3 16 18 20

T4 20 18 24


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Here the objective of this assignment problem is to assign Tasks (4) to the Persons (3) in a way that will result in total

minimum cost.



The given matrix is a square matrix [Number of rows (Tasks) = Number of columns (Persons)].



element from all the other elements in the row.


Tasks

Persons

P1 P2 P3 P4

T1 0 12 7

T2 0 5 2 3

T3 0 2 4

T4 2 0 6


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from all the other elements in the column.


Tasks

Persons

P1 P2 P3 P4

T1 0 12 4

T2 0 3 2 0T3 0 0 1

T4 0 0 3





Tasks

Persons

P1 P2 P3 P4

T1 12 4

T2 3 2

T3 1

T4 3


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In this problem, all zeros are either crossed out or assigned. Also number of assignments (4) is equal to number of

rows/columns (4). As the number of assignments is equal to the number of rows/columns, solution is optimal.

The optimal allocation is to assign Tasks T1 to Person P1, T2 to P4, T3 to P2, and T4 to P3 . The total cost (in Rs.) of this

assignment is computed from the original table.


Tasks Persons Cost of assignmentT1 P1 20

T2 P4 18

T3 P2 18

T4 P3 18

Total 74

3. Multiple Optimal Solutions

While making assignment in the reduced assignment matrix, it is possible to have two or more ways to strike off certain

number of zeroes. In such cases, assignments are to be made arbitrarily and hence assignments can be made in many ways.Such a situation indicates multiple optimum solutions with the same total cost (or profit). If the problem has only one

solution then the solution is said to be Unique solution.


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4. Maximization Problems

In Maximization Problems, the objective is to maximize the profit, revenue etc. In such problems, the given maximization

problems may be converted into minimization problem by subtracting all profits from the highest element of the profit

matrix. This transformed matrix or the opportunity cost matrix can be solved by using the Hungarian method.

Example 6:A marketing manager has 5 salesmen and 5 sales districts. Considering the capabilities of the salesman and the nature of

districts, the marketing manager estimates that the sales per month (in hundred rupees) for each salesman in each district

would be as follows. Find the assignment of salesmen to districts that will result in maximum sales.

Salesman

Sales Districts

A B C D E

1 32 38 40 28 40

2 40 24 28 21 36

3 41 27 33 30 37

4 22 38 41 36 36

5 29 33 40 35 39


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Solution:

The given maximization problem is converted into minimization problem by subtracting from the highest sales value (i.e.,

41) with all the elements of the given table. The cost data obtained is given in the table below:

Salesmen

Sales Districts

A B C D E

1 9 3 1 13 1

2 1 17 13 20 5

3 0 14 8 11 4

4 19 3 0 5 5

5 12 8 1 6 2



The given matrix is a square matrix [Number of rows (Salesmen) = Number of columns (Sales Districts)].

Step 2: Reduce the Matrix Row WisePrepare row reduced matrix by selecting the smallest element in each row of the matrix and subtracting this smallest

element from all the other elements in the row.


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Salesmen

Sales District

A B C D E

1 8 2 0 12 0

2 0 16 12 19 4

3 0 14 8 11 4

4 19 3 0 5 5

5 11 7 0 5 1





Salesmen

Sales District

A B C D E

1 8 0 0 7 0

2 0 14 12 14 4

3 0 12 8 6 44 19 1 0 0 5

5 11 5 0 0 1


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After examining the rows, examine the columns successfully until a column with exactly one zero is found. Make


In case of atleast two zeros in a particular row and column is found, choose arbitrarily any one of these and cross out allother zeros of that row and column.

Repeat step 4 till all zeros are either assigned or crossed.

If the number of assignments is equal to number of rows/columns, you have arrived at an optimal solution.

Salesmen

Sales District

A B C D E

1 8 7

2 14 12 14 4

3 12 8 6 4

4 19 1 5

5 11 5 1


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From the last table we see that the total number of assignment i.e., 4 < 5 (Number of rows or columns). If the number of

assignment is not equal to number of rows/columns then the solution is not optimal.

Since one row and one column are not assigned, optimal assignment is not possible in the current matrix and hence find

minimum number of lines crossing all zeros.

Step 5: Find minimum number of lines required to cross all zeros.

Draw minimum number of horizontal/vertical lines to cover all the zeroes in the resulting matrix:This consists of the following sub steps:

a. Mark () the row that do not have assignment. In this case it is the 3rd

row.

Salesmen

Sales District

A B C D E

1 8 7

2 14 12 14 4

3 12 8 6 4

4 19 1 5

5 11 5 1


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b. Mark () the column which contains zeros in marked row. In this case it is the 1st

column.

Salesmen

Sales District

A B C D E

1 8 7

2 14 12 14 4

3 12 8 6 4

4 19 1 5

5 11 5 1

c. Mark () the row which contains assignment in the marked column. In this case it is the 2nd

row.

Salesmen

Sales District

A B C D E

1 8 7

2 14 12 14 4

3 12 8 6 4

4 19 1 5

5 11 5 1


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d. Draw the lines through unmarked rows and marked columns.

Salesmen

Sales District

A B C D E

1 8 7

2 14 12 14 4

3 12 8 6 4

4 19 1 5

5 11 5 1

e. Revise the matrix as follows:

Examine the uncovered elements (on which there are no lines). Find the smallest element among the uncovered elements of

the table. Here the smallest element is 4.

Subtract this element (4) from the uncovered elements.

Add this element (4) to all the values at the point of intersection of any two of the lines.

Other elements crossed by the lines remain unchanged.


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The revised matrix is :

Salesmen

Sales District

A B C D E

1 12 0 0 7 0

2 0 10 8 10 0

3 0 8 4 2 0

4 23 1 0 0 55 15 5 0 0 1

Repeat Step 4 to make assignment.

There are two alternative assignments due to presence of zero elements in cells (4, C), (4, D), (5, C) and (5, D).


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Two Alternative Assignments

Salesmen

Sales District

A B C D E

1 12 7

2 10 8 10

3 8 4 2

4 23 1 55 15 5 1

Salesmen

Sales District

A B C D E

1 12 7

2 10 8 10

3 8 4 2

4 23 1 55 15 5 1

Since all the rows and columns are assigned or number of rows/columns (5) is equal to number of assignments (5), this is an

optimal assignment. The maximum sales (in Rs.) of this assignment is computed from the original table.

The two optimal assignments are as follows:

Assignment Set 1

Salesmen District Sales (00) Rs

1 B 38

2 A 40

3 E 374 C 41

5 D 35

Total Rs.= 191.00

Assignment Set II

Salesmen District Sales (00) Rs

1 B 38

2 E 36

3 A 414 C 41

5 D 35

Total Rs.= 191.00

Therefore the Maximum sales is Rs. 19,100.


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TRAVELLING SALESMAN PROBLEM

The Travelling Salesman Problem is very similar to an Assignment problem.

Suppose a salesman wants to visit n cities. He wishes to start from a particularcity and wants to visit each city only once

and return to his starting point. The objective is to select the sequence to visit the cities in such a way that his total distance

(cost or time) is minimized. Such a problem is known as travelling salesman problem.

Example 7:

A salesman has to visit five cities A, B, C, D and E. He wishes to start from a particular city, visit each city once and thenreturn to his starting point. The travelling cost (in thousands of Rs.) of each city from a particular city is given below.

To City

From City

A B C D E

A 2 5 7 1

B 6 3 8 2

C 8 7 4 7

D 12 4 6 5

E 1 3 2 8

What should be the sequence of the salesman's visit, so that the cost is minimum?

Solution:In order to find an optimum assignment to the given travelling salesman problem, first you should follow the

same procedure as followed in the assignment problem.


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The given matrix is a square matrix [Number of rows = Number of columns].



element from all the other elements in the row. The resultant matrix is;To City

From City

A B C D E

A 1 4 6 0

B 4 1 6 0

C 4 3 0 3

D 8 0 2 1

E 0 2 1 7





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The resultant matrix is

To City

From City

A B C D E

A 1 3 6 0

B 4 0 6 0

C 4 3 0 3

D 8 0 1 1E 0 2 0 7






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To City

From City

A B C D E

A 1 3 6

B 4 6

C 4 3 3

D 8 1 1

E 2 7

Optimum Assignment Table

A E 1

B C 3

C D 4

D B 4

E A 1

Total Cost 13


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The optimum assignment table shows the solution to the assignment problem but not to the travelling salesman problem.

The following sequence can be identified from the above optimum solution table.

Sequence of the salesman:

A E A, B C D B

This represents that the travelling salesman will start from city A, then go to city E and return to city A without visiting the

other cities. This violates the additional restriction that the salesman can visit each city only once. The cycle is not

complete. Hence the problem can be improved.Case I: Now we examine the optimum assignment table to find the next best solution that satisfies the additional

restrictions. First identify the next minimum element in the assignment table. Here the next minimum element is 1. So we

make the unit assignment in the cell (A,B) instead of zero assignment in the cell (A,E) and delete the row A and column B

to avoid further assignment. Henceforth we proceed in the unit assignment as a routine assignment problem.

To City

From City

A B C D E

A 3 6

B 4 6

C 4 3 3

D 8 1

E 2 7


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The sequence of the salesman is:

A BC D E A

Case 2: If the assignment is made for cell (D, C) instead of (D, E), the feasible solution cannot be obtained. The route for

the assignment will be A B E, C D C, E A.

To City

From City

A B C D EA 3 6

B 4 0 6

C 4 3 3

D 8 1

E 2 7

This violates the additional restriction and produces no feasible solution.


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OptimumSequence

To City

From City

A B C D E

A 3 6

B 4 6

C 4 3 3

D 8 1

E 2 7

The sequence feasible for this assignment is

A B C D E A with the travelling cost of (2000 + 3000 + 4000 + 5000 + 1000) = Rs.15,000.00


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FLOW CHART FOR ASSIGNMENT PROBLEM

Yes

No

Yes

No

Yes

Arrange the data in a matrix form

Is it a

maximization

problem ?

Convert it into a minimization problem by

subtracting all the elements in the matrix from

the largest element in the matrix

Is it a

balancedproblem?

Add dummy row(s) and column(s)

nd opportunity costIdentify the smallest element in each row and subtract it from every element in that row

b. In the reduced matrix, identify the smallest element in each column and subtract it from every element

in that column


there is only one zero in a row, an assignment is made by enclosing this zero by a square ( ) andossing (X) or cancelling other zeros in its column. Continue the process till all the rows in the matrix is

xamined. Repeat same process column-wise.

Is the number of

assignments equalto the number of

rows and columns ?

OPTIMAL SOLUTION

Revise opportunity cost tablea. Draw minimum possible lines on

columns and/or rows such that all zeros

are covered.b. Choose the smallest number not

covered by lines. Subtract it from itself

and every other uncovered numberc. Add this number at the intersection of

any two lines. Other elements crossed by

the lines remain unchanged.

No

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