ASSIGNMENT PROBLEMPROBLEM

Anggy Herny Anggraeni

13030214016

Mathematics 2013-B

Standard Assignment Problems

Consider a machine shop with M machine and J job where M=N

Xji = if job j is assigned to machine i otherwise

Since each machine is assigned exactly to one

1

0

Since each machine is assigned exactly to one job, we have

for i = 1,2,...,n

Similiary each job is assigned exactly to one machine

for j = 1,2,...,n

Nonstandard Assignment Problem

Consider a machine shop with M machines and J jobs where M ≠ J.

If more machines than jobs If more machines than jobs

(M > J)

or

If more jobs than machines

(M < J)

STEPS OF HUNGARIAN METHOD

STEP 1 : reduce all element of rows with the smallest element of each row.

STEP 2 : if there is zero more than one on the row, then do step one of the coloumn that don’ t have zero.coloumn that don’ t have zero.

STEP 3 : if there is job that doesn’t have pair, then we search it with close the row and coloumn which have zero. After that do step one and add the minimal value on the intersection of row and coloumn.

Find the optimal assignment of four jobs and four machines when the cost of assignment is given by the following table

J1 J2 J3 J4

M1 10 9 8 7M1 10 9 8 7

M2 3 4 5 6

M3 2 1 1 2

M4 4 3 5 6

Find less costs from the rows

J1 J2 J3 J4

M1 10 9 8 7M1 10 9 8 7

M2 3 4 5 6

M3 2 1 1 2

M4 4 3 5 6

This produces the following cost matrix

J1 J2 J3 J4

M1 3 2 1 0M1 3 2 1 0

M2 0 1 2 3

M3 1 0 0 1

M4 1 0 2 3

J1 J2 J3 J4

M1 3 2 1 0

M2 0 1 2 3

M3 1 0 0 1

M4 1 0 2 3

From the above table a feasible assignment using only the cells with zero costs is M1 → J4, M2→ J1, M3 → J3, and M4 → J2. Hence, this is an optimal assignment. The total cost is given by 7+3+1+3 = 14

M4 1 0 2 3

ExampleFind an optimal solution to an assignment problem with the following cost matrix:

J1 J2 J3 J4

M1 10 9 7 8M1 10 9 7 8

M2 5 8 7 7

M3 5 4 6 5

M4 2 3 4 5

First the minimum element in each row is subtracted from all the moments in that row. This gives the followin reduced-cost matrix.

J1 J2 J3 J4

M1 10 9 7 8M1 10 9 7 8

M2 5 8 7 7

M3 5 4 6 5

M4 2 3 4 5

First the minimum element in each row is subtracted from all the moments in that row. This gives the followin reduced-cost matrix.

J1 J2 J3 J4

M1 3 2 0 1M1 3 2 0 1

M2 0 3 2 2

M3 1 0 2 1

M4 0 1 2 3

Since both the machines M2 and M4

have zero cost corresponding to job J1

only, a feasible assignment using only cells with zero costs is not possible.

J1 J2 J3 J4

M1 3 2 0 1

M2 0 3 2 2

M3 1 0 2 1

M4 0 1 2 3

To get additional zeros subtract the minimum element in the fourth column from all the elements in that column.

J1 J2 J3 J4

M1 3 2 0 1M1 3 2 0 1

M2 0 3 2 2

M3 1 0 2 1

M4 0 1 2 3

To get additional zeros subtract the minimum element in the fourth column from all the elements in that column.

J1 J2 J3 J4

M1 3 2 0 1

M2 0 3 2 2

M3 1 0 2 1

M4 0 1 2 3

To get additional zeros subtract the minimum element in the fourth column from all the elements in that column.

J1 J2 J3 J4

M1 3 2 0 0

M2 0 3 2 1

M3 1 0 2 0

M4 0 1 2 2

The machines M2 and M4 have zero cost corresponding to job J1 only

J1 J2 J3 J4

M1 3 2 0 0M1 3 2 0 0

M2 0 3 2 1

M3 1 0 2 0

M4 0 1 2 2

The procedure draws a minimum number of lines through some selected rows and columns in such a way that all the cells with zero costs are covered by the lines.

J1 J2 J3 J4

M1 3 2 0 0M1 3 2 0 0

M2 0 3 2 1

M3 1 0 2 0

M4 0 1 2 2

Reduce with the minimum element not covered with lines on matrix

J1 J2 J3 J4

M1 3 2 0 0

M2 0 3 2 1

M3 1 0 2 0

M4 0 1 2 2

Reduce with the minimum element not covered with lines on matrix

J1 J2 J3 J4

M1 3 2 0 0M1 3 2 0 0

M2 0 3 2 1

M3 1 0 2 0

M4 0 1 2 2

The procedure draws a minimum number of lines through some selected rows and columns in such a way that all the cells with zero costs are covered by the lines.

J1 J2 J3 J4

M1 3 2 0 0M1 3 2 0 0

M2 0 2 1 0

M3 1 0 2 0

M4 0 0 1 1

Then add this number to all those covered elements that are at the intersection of two lines.

J1 J2 J3 J4

M1 3 2 0 0M1 3 2 0 0

M2 0 2 1 0

M3 1 0 2 0

M4 0 0 1 1

J1 J2 J3 J4

M1 4 2 0 0

Then add this number to all those covered elements that are at the intersection of two lines.

M1 4 2 0 0

M2 0 2 1 0

M3 2 0 2 0

M4 0 0 1 1

J1 J2 J3 J4

M1 4 2 0 0

M2 0 2 1 0

M3 2 0 2 0

M4 0 0 1 1

A feasible assignment is now possible and an optimal solution is to assign M1→J3, M2→J1, M3→J4, and M4→J2. The total cost is given by 7 + 5 + 5 + 3 = 20.

To solve Maximization Problems

Step I.Convert the problem to a minimization problem by multiplying all the elements (cij) of the assignment matrix by – 1

Step II.If some of the elements of tthe cost matrix are negative, add a sufficiently large positive are negative, add a sufficiently large positive number to the corresponding rows and columns so that all the cost elements would become nonnegative.

Step III.We now have an assignment problem with a minimization objective and all cost elements nonnegative. The Hungarian Method can now be applied directly.

ExerciseA batch of four jobs can be assigned to five different

machines. The setup time for each job on various machines is given by the following table and

J1 J2 J3 J4 J5

M1 10 11 4 2 8

find an optimal assignment of jobs to machines

which will minimize the total setup time.

M1 10 11 4 2 8

M2 7 11 10 14 12

M3 5 6 9 12 14

M4 13 15 11 10 7

Make 1 row for M5 with zero costs

J1 J2 J3 J4 J5

M1 10 11 4 2 8

M2 7 11 10 14 12

M3 5 6 9 12 14

M4 13 15 11 10 7

Make 1 row for M5 with zero costs

J1 J2 J3 J4 J5

M1 10 11 4 2 8

M2 7 11 10 14 12

M3 5 6 9 12 14

M4 13 15 11 10 7

M5 0 0 0 0 0

Find less costs from every rows

J1 J2 J3 J4 J5

M1 10 11 4 2 8

M2 7 11 10 14 12M2 7 11 10 14 12

M3 5 6 9 12 14

M4 13 15 11 10 7

M5 0 0 0 0 0

The machines M2 and M4 have zero cost corresponding to job J1 only

J1 J2 J3 J4 J5

M1 8 9 2 0 6

M2 0 4 3 7 5

M3 0 1 4 7 9

M4 6 8 4 3 0

M5 0 0 0 0 0

To get additional zeros subtract the minimum element in the second and third column from all the elements in that column.

J1 J2 J3 J4 J5

M1 8 9 2 0 6

M2 0 4 3 7 5

M3 0 1 4 7 9

M4 6 8 4 3 0

M5 0 0 0 0 0

To get additional zeros subtract the minimum element in the second and third column from all the elements in that column.

J1 J2 J3 J4 J5

M1 8 9 2 0 6

M2 0 4 3 7 5

M3 0 1 4 7 9

M4 6 8 4 3 0

M5 0 0 0 0 0

To get additional zeros subtract the minimum element in the second and third column from all the elements in that column.

J1 J2 J3 J4 J5

M1 8 8 0 0 6

M2 0 3 1 7 5

M3 0 0 2 7 9

M4 6 7 2 3 0

M5 0 0 0 0 0

J1 J2 J3 J4 J5

M1 8 8 0 0 6

M2 0 3 1 7 5

M3 0 0 2 7 9

M4 6 7 2 3 0

A feasible assignment is now possible and an optimal solution is to assign M1→J4, M2→J1, M3→J2, M4→J5, and M5→J3. The total cost is given by 2+7+6+7+0=22.

M4 6 7 2 3 0

M5 0 0 0 0 0

Thanks For Your Attention Thanks For Your Attention