Qadm Ppt New

Suchita Nevrekar

Aarti AsraniMartha DsouzaPallavi Karande

Urja ShahPooja Bhalja

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Hungarian Method

The Hungarian method is a combinatorial optimization algorithm which solves the assignment problem in polynomial time and which anticipated later primal-dual methods.

In every workplace there are jobs to be done and there are people available to do them, but everyone is not equally efficient at every job. Someone may be more efficient on one and less efficient on other job. The relative efficiency is reflected in terms of time taken for or cost associated with, performance of different jobs by different people and obvious problem for manager to handle is to assign jobs to various workers in manner that they can be done in the most efficient way. Such problems can be formulated as liner programming problems and this method called as “Hungarian Assignment Method”.

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Introduction

It’s allows a manager to obtain answers to the questions like following

How to deal with situation when the number of jobs do no match with number of jobs performer

How should the salesman of the company be assign to different sales zone so that total excepted sales are maximize.

How to schedule flight s or the bus routes between two cities so that the lay over times for the crew can be minimize

How to assign the given jobs by some works on one to one basic when completion times of performance is given for each combination and it is desired that the jobs are completed in the list time and list cost.

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Hungarian Assignment Method

Let m be the number of jobs as well as the operators, and tij be the processing time of the job i if it is assigned to the operator j. Here the objective is to assign the jobs to the operators such that the total processing time is minimized.

Operators

Job

1 2 … j … m1 t11 t12 t1j t1m

2.i ti1 tij tim

.m tm1 tm2 tmj tmm

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General format of assignment problem

Examples of assignment problem

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Row entity Column entity Cell entity

jobs operators Processing time

Programmer program Processing time

operators machine Processing time

Drivers Routes Travel time

Teachers Subjects Students pass percentage

Consists of two phases. First phase: row reductions and column reductions

are carried out. Second phase :the solution is optimized in iterative

basis.

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Hungarian method

Step 0: Consider the given cost matrix Step 1: Subtract the minimum value of each row

from the entries of that row, to obtain the next matrix.

Step 2: Subtract the minimum value of each column from the entries of that column , to obtain the next matrix.

Treat the resulting matrix as the input for phase 2.

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Phase 1: Row and column reductions

Step3: Draw a minimum number of lines to cover all the zeros of the matrix.

Procedure for drawing the minimum number of lines: 3.1 Row scanning 1 Starting from the first row ,if there’s only one zero in

a row mark a square round the zero entry and draw a vertical line passing through that zero. Otherwise skip the row.

2.After scanning the last row, check whether all the zeros are covered with lines. If yes go to step 4. Otherwise do column scanning. Ctd

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Phase 2: Optimization

3.2 Column scanning.1. Starting from the first column: if there’s only

one zero in a column mark a square round the zero entry and draw a horizontal line passing through that zero. otherwise skip the column.

2.After scanning the last column, check whether all the zeros are covered with lines. If yes go to step 4. Otherwise do row scanning. ctd

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Step 4: check whether the number of squares marked is equal to the number of rows/columns of the matrix.

If yes go to step 7. Otherwise go to step 5. Step 5: Identify the minimum value of the undeleted

cell values ,say ‘x’. Obtain the next matrix by the following steps.

5.1 Copy the entries covered by the lines ,but not on the intersection points.

5.2 add x to the intersection points 5.3 subtract x from the undeleted cell values.Step 6: go to step 3.Step 7: optimal solution is obtained as marked by the

squares

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If the problem is a maximization problem ,convert the problem into a minimization problem by multiplying by -1.

Then apply the usual procedure of an assignment problem.

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Maximization problem

Example : Assign 4 sales persons to four different sales regions such that the total sales is maximized.

Sales region

Sales person

1 2 3 4

1 10 22 12 14

2 16 18 22 10

3 24 20 12 18

4 16

14 24 20

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Modified data , after multiplying the cell entries by -1.

Sales region

Sales person

1 2 3 4

1 -10 -22 -12 -14

2 -16 -18 -22 -10

3 -24 -20 -12 -18

4 -16 -14 -24 -20

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After step 1

Sales region

Sales person

1 2 3 4

1 12 0 10 8

2 6 4 0 12

3 0 4 12 6

4 8 10 0 414

After step 2

Sales region

Sales person

1 2 3 4

1 12 0 10 4

2 6 4 0 8

3 0 4 12 2

4 8 10 0 015

Phase 2

Sales region

Sales person

1 2 3 4

1 12 0 10 4

2 6 4 0 8

3 0 4 12 2

4 8 10 0 0

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Note that the number of squares is equal to the number of rows of the matrix. solution is feasible and optimal.

Result: Salesman Sales region Sales

1 2 22

2 3 22

3 1 24

4 4 20

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Operator job

1 2 3 4 5

1 10 12 15 12 8

2 7 16 14 14 11

3 13 14 7 9 9

4 12 10 11 13 10

5 8 13 15 11 15

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Example : Assign the 5 operators to the 5 jobs such that the total processing time is minimized.

Example :ROW SCANNING.

Operator job

1 2 3 4 5

1 10 12 15 12 8

2 7 16 14 14 11

3 13 14 7 9 9

4 12 10 11 13 10

5 8 13 15 11 15

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Example : Assign the 5 operators to the 5 jobs such that the total processing time is minimized.

Operator job

1 2 3 4 5

1 10 12 15 12 8

2 7 16 14 14 11

3 13 14 7 9 9

4 12 10 11 13 10

5 8 13 15 11 15

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Example : Assign the 5 operators to the 5 jobs such that the total processing time is minimized.

Operator job

1 2 3 4 5

1 2 4 7 4 0

2 0 9 7 7 4

3 6 7 0 2 2

4 2 0 1 3 0

5 0 5 7 4 8

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Example : Assign the 5 operators to the 5 jobs such that the total processing time is minimized.

Operator job

1 2 3 4 5

1 2 4 7 2 0

2 0 9 7 5 4

3 6 7 0 0 2

4 2 0 1 1 0

5 0 5 7 2 8

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Example : Assign the 5 operators to the 5 jobs such that the total processing time is minimized.

Operator job

1 2 3 4 5

1 2 4 6 1 0

2 0 9 6 4 4

3 7 8 0 0 3

4 2 0 0 0 0

5 0 5 6 1 8

MA 402--assignment problem 24