Assignment Problems

Compiled By: Group cSandeep Amin

The Hungarian method is a combinatorial optimization algorithm which solves the assignment problem in polynomial time and which anticipated later primal-dual methods. It was developed and published by Harold Kuhn in 1955, who gave the name "Hungarian method" because the algorithm was largely based on the earlier works of two Hungarian mathematicians: Dénes Kőnig and Jenő Egerváry.

Suppose there are two machines in the press and twooperators are engaged at different rates to operate them. Which operator should operate which machine for maximizing profit?

Similarly, if there are n machines available and n persons are engaged at different rates to operate them. Which operator should be assigned to which machine to ensure maximum efficiency?

While answering the above questions we have to think about the interest of the press, so we have to find such an assignment by which the press gets maximum profit on minimum investment.Such problems are known as "assignment problems"

Phase 1: Row and column reductions

Step 0: Consider the given cost matrixStep 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.

The Hungarian Method

Consider the assignment problem:

8 6 5 7

6 5 3 4

7 8 4 6

6 7 5 6

Row Min

p1 = 5

p2 = 3

p3 = 4

p4 = 5

Worker

Job

1

1

2

2

3

3

4

4

Step 1: From each entry of a row, we subtract the minimum value in that row and get the following reduced cost matrix:

3 1 0 2

3 2 0 1

3 4 0 2

1 2 0 1

Column Minimum

q1=1 q2=1 q3=0 q4=1

Step 2: From each entry of a column, we subtract the minimum value in that column and get the following reduced cost matrix:

2 0 0 1

2 1 0 0

2 3 0 1

0 1 0 0

Step 3: Now we test whether an assignment can be made as follows. If such an assignment is possible, it is the optimal assignment.

a.Examine the first row. If there is only one zero in that row, then make an ( ) and cross ( ) all the other zeros in the column passing through the surrounded zero and draw a vertical line on that column

b.Then starting with the first column if there is one zero then make an ( ) cross all the zero in that row & draw horizontal line on that row cont till all zero are crossed or even assignment

Step 3(a) gives the following table.

2 0 0 1

2 1 0 0

2 3 0 1

0 1 0 0

Step 3(b): Now repeat the above procedure for columns. (Remember to interchange row and column in that step.)

Step 3(b) gives the following table.

2 0 0 1

2 1 0 0

2 3 0 1

0 1 0 0

If there is now a surrounded zero in each row and each column, the optimal assignment is obtained.

Worker 1 is assigned to Job 2 = 6

Worker 2 is assigned to Job 4 = 4

Worker 3 is assigned to Job 3 = 4

Worker 4 is assigned to Job 1 = 6 Minimum total time = 20 hrs

The optimal solution is unique

In our example, there is a surrounded zero in each row and each column and so the optimal assignment is: Hrs

If the final stage is reached (that is all the zeros are either surrounded or crossed) and if there is no surrounded zero in each row and column, it is not possible to get the optimal solution at this stage. We have to do some more work. Again we illustrate with a numerical example.Solve the following unbalanced assignment problem (Only one job to one man and only one man to one job): 7 5 8 4

5 6 7 4

8 7 9 8

Since the problem is unbalanced, we add a dummy worker 4 with cost 0 and get the following starting cost matrix:

7 5 8 4

5 6 7 4

8 7 9 8

0 0 0 0

Applying Step 1, we get the reduced cost matrix

p1=4

p2=4

p3=7

p4=0 Dummy

Row Min

Worker

Job

3 1 4 0

1 2 3 0

1 0 2 1

0 0 0 0

Now Step 2 is Not needed. We now apply Step 3(a) and get the following table.

3 1 4 0

1 2 3 0

1 0 2 1

0 0 0 0

Now all the zeros are either surrounded or crossed but there is no surrounded zero in Row 2. Hence assignment is NOT possible. We go to Step 4.

3 1 4 0

1 2 3 0

1 0 2 1

0 0 0 0

Step 4(b) Select the smallest element, say, u, from among all elements uncovered by all the lines.

In our example, u = 1

Step 4(c) Now subtract this u from all uncovered elements but add this to all elements that lie at the intersection of two lines

2 1 3 0

0 2 2 0

0 0 1 1

0 1 0 1

Doing this, we get the table:

Step 5: Reapply Step 3.

We thus get the table2 1 3 0

0 2 2 0

0 0 1 1

0 1 0 1

Thus the optimum allocation is:

W1 → J4 W2 → J1 W3 → J2 W4 → J3

And the optimal cost = 4+5+7+0 = 16Hence Job 3 is not done by any (real) worker.

The optimal assignment is unique

MAXIMIZATION TYPE

•Hungarian method is valid for balanced & minimization type

•The assignment problem can be converted to minimization by finding the opportunity loss•The opportunity loss matrix is found by subtracting all the element of the matrix from the largest element

•Efficiency of each professor to teach each subject as follow :

Consider the assignment problem

ABC

Professor

SUBJECT

1 2 3 4

Find which professor to be assigned to which subject so that total efficiency can be maximize . (-) indicates that professor b cannot be assigned to sub 2 also find sub for which we do not have professor

10 5 9 15

6 - 3 12

16 8 5 9

10 5 9 15

6 - 3 12

16 8 5 9

0 0 0 0

Since the problem is unbalanced, we add a dummy professor D with cost 0 and get the following starting cost matrix:

ABC D

Professor

1 2 3 4

SUBJECT

Dummy

OPPORTUNITY LOSS MATRIX•Subtracting all the element of the matrix from the largest element that is 16 •We get this table

6 11 7 1

10 - 13 4

0 8 11 7

16 16 16 16

ABC D

Professor

1 2 3 4

SUBJECT

APPLY HUNGARIAN METHOD

5 10 6 0

6 - 6 0

0 8 11 7

0 0 0 0

ABC D

Professor

1 2 3 4

SUBJECT

Complete assignment is not formed

Apply Step 1 ,step 2 is not needed

Doing this, we get the table:

NOW SUBTRACT MINIMUM ELEMENT FROM ALL UNCOVERED ELEMENTS BUT ADD THIS TO ALL ELEMENTS THAT LIE AT THE INTERSECTION OF

TWO LINES

5 10 6 0

6 - 6 0

0 8 11 7

0 0 0 0

ABC D

Professor

1 2 3 4

SUBJECT

Doing this, we get the table:

5 4 0 0

6 - 3 0

0 2 5 7

6 0 0 6

Complete assignment is formed

ABC D

Professor

1 2 3 4

SUBJECT

•The Optimal assignment is • Professor Subject Efficiency• A 3 9• B 4 12• C 1 16 • D 2 0 •Maximum total efficiency 37•The optimal assignment is unique•Subject 2 is not assigned to any professor

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