7/29/2019 Assignment Problem 2

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HUNGARIAN METHOD

EXCEPTIONAL CASES

7/29/2019 Assignment Problem 2

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MAXIMIZATION

If the problem is of maximization

Hungarian method cannot be applicable

directly.

To apply the method first convert it into

minimization by subtracting all the

elements of the matrix from the largest

element of that matrix

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Unbalanced A.P.

Any assignment problem is said to be

unbalanced if the cost matrix is not a square

matrix

i.e. the number of rows and the number ofcolumns are not equal.

To make it balanced we add a dummy row or

dummy column with all the entries as zero.

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Practice problem

There are four jobs to

be assigned to the

machines. Only one job

could be assigned toone machine. The

amount of time in hours

required for the jobs in

a machine are given in

the following

matrix.Find an optimum

assignment.

A B C D E

1 4 3 6 2 7

2 10 12 11 14 16

3 4 3 2 1 5

4 8 7 6 9 6

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Hungarian method

Since the cost matrix is

unbalanced we add a

dummy job 5 with

corresponding entrieszero.

Job number 5 is a

dummy job

A B C D E

1 4 3 6 2 7

2 10 12 11 14 16

3 4 3 2 1 5

4 8 7 6 9 6

5 0 0 0 0 0

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HUNGARIAN METHOD

Step-1

Row minimization

Step-2

Column minimization

No.of lines = 4

No. of rows= 5

Go to next step

L3

2 1 4 0 5

L4 0 2 1 4 6

3 2 1 0 4

L2 2 1 0 3 0

L1 0 0 0 0 0

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HUNGARIAN METHOD

STEP-3

Subtracting smallest

element 1 from

uncovered, adding 1 incross elements and

keeping the other

elements same the

modified matrix is No of lines=rows

Assignment is possible.

L5 1 0 3 0 4

L4 0 2 1 5 6

L3 2 1 0 0 3

L2 2 1 0 4 0

L1 0 0 0 1 0

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HUNGARIAN METHOD

Different assignments are

1- B 1-B 1 - D

2- A 2- A 2 - A

3- D 3- D 3 - C

4- C 4- E 4 - E

5- E 5- C 5 - B Total minimum time = 20 min

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HUNGARIAN METHOD

Five jobs are to beprocessed and fivemachines are available.Any machine can

process any job withthe resulting profit (inrupees) as follows:

What is the maximum

profit that may beexpected if an optimumassignment is made?

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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Hungarian method

Step1

Since the problem is of

maximization type

subtracting all elementsof the matrix from the

largest element the

modified matrix will

become

9 3 1 13 1

1 17 13 20 5

0 14 8 11 4

19 3 0 5 5

12 8 1 6 2

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Hungarian method

Step 2

By row minimization the

matrix becomes 8 0 0 7 0

0 14 12 14 4

0 12 8 6 4

19 1 0 0 5

11 5 0 0 1

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Hungarian method

Step-3

By column minimization

No of lines =4

No. of rows = 5

L2

L1 8 0 0 7 0

0 14 12 14 4

0 12 8 6 4

L3 19 1 0 0 8

L4 11 5 0 0 1

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Hungarian method

Step-4

Selecting smallest

element 4 from

uncovered elementsand adding the same at

the intersection and

keeping the other

elements same themodified matrix

becomes

L4 L3 L2 L1

L5 12 0 0 7 0

0 10 8 10 0

0 8 4 2 0

23 1 0 0 8

15 5 0 0 1

7/29/2019 Assignment Problem 2

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Hungarian method

Different assignments are

1- B 1- B 1- B 1 - B

2- A 2- E 2- A 2 - E

3- E 3 A 3 - E 3 - A

4- C 4 D 4 D 4 - C

5- D 5 C 5 C 5 D

Total profit = 191 units