Assignment problems

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• Assignment problems.

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What is an assignment problem?

• Suppose there are n jobs to be performed and n
  persons
• are available for doing these jobs. Assume that
  each
• person can do each job at a time , though with
  varying
• degree of efficiency. Let cij be the
  cost(payment) if the
• ith person is assigned the jth job , the problem
  is to find
• an assignment (which job should be assigned to
  which
• person ) so that the total cost for performing
  all jobs is
• minimum.

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The Assignment Model

• Each assignment problem has associated with it a
  table, or matrix.
• Generally, the rows contain the objects or people
  we wish to assign, and the columns comprise the
  tasks or things we want them assigned to. Note-
  The number of persons and number of jobs are
  equal.
• The numbers in the table are the costs associated
  with each particular assignment.

Jobs? Persons 1 2 3 4
A 8 26 17 11
B 13 28 4 26
C 38 19 18 15
D 19 26 24 10
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The Assignment Model

• An assignment problem can be viewed as a
  transportation problem in which
• the capacity from each source (or person to be
  assigned) is 1 and
• the demand at each destination (or job to be
  done) is 1.
• Such a formulation could be solved using the
  transportation algorithm, but it would have a
  severe degeneracy problem.
• However, this type of problem is very easy to
  solve using the Hungarian method.

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Assignment algorithm (Hungarian assignment method)

• Find the opportunity cost table by
• Subtracting the smallest number in each row of
  the original cost table or matrix from every
  number in that row.
• Then subtracting the smallest number in each
  column of the table obtained in part (a) from
  every number in that column.

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Continue..

• Test the table resulting from step 1 to see
  whether an optimal assignment can be made.
• The procedure is to draw the minimum number of
  vertical and horizontal straight lines necessary
  to cover all zeros in the table.
• If the number of lines equals either the number
  of rows or columns in the table, an optimal
  assignment can be made.
• If the number of lines is less than the number of
  rows or columns, then proceed to step 3.

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Continue.

• Revise the present opportunity cost table.
• This is done by subtracting the smallest number
  not covered by a line from every other uncovered
  number.
• This same smallest number is also added to any
  number(s) lying at the intersection of horizontal
  and vertical lines.
• We then return to step 2 and continue the cycle
  until an optimal assignment is possible.

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Steps in the Assignment algorithm.
Not optimal
Step 1
Step 2
Optimal
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Example-

• A car hire company has one car at each of five
  depots a,b,c,d and e. A customer requires a car
  in each town , namely A,B,C,D and E. Distance (in
  km s) between depots (origins) and towns
  (destinations ) are given in the following
  matrix.

a b c d e
A 160 130 175 190 200
B 135 120 130 160 175
C 140 110 155 170 185
D 50 50 80 80 110
E 55 35 70 80 105
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Solution-

• Now we will find the opportunity table as follow.

a b c d e
A 160 130 175 190 200
B 135 120 130 160 175
C 140 110 155 170 185
D 50 50 80 80 110
E 55 35 70 80 105
a b c d e
A 30 0 45 60 70
B 15 0 10 40 55
C 30 0 45 60 75
D 0 0 30 30 60
E 20 0 35 45 70
a b c d e
A 30 0 35 30 15
B 15 0 0 10 0
C 30 0 35 30 20
D 0 0 20 0 5
E 20 0 25 15 15
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Continue
To test for optimality , we will draw the minimum
no of vertical and horizontal lines to cover
all the zeros. So we have
Now the opportunity table is as follow.
a b c d e
A 30 0 35 30 15
B 15 0 0 10 0
C 30 0 35 30 20
D 0 0 20 0 5
E 20 0 25 15 15
a b c d e
A 30 0 35 30 15
B 15 0 0 10 0
C 30 0 35 30 20
D 0 0 20 0 5
E 20 0 25 15 15
Here the number of lines are 3. which is not
equal to rows ( or columns)
So it is not an optimal solution.
Then we do the following operations to get
improved table as follow.
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a b c d e
A 30 0 35 30 15
B 15 0 0 10 0
C 30 0 35 30 20
D 0 0 20 0 5
E 20 0 25 15 15
Choose the minimum element among all uncovered
elements .
Here the minimum element is 15.
Then subtract the minimum element (15) from all
uncovered elements by the lines and add that
element to intersection of these lines.
a b c d e
A 15 0 20 15 0
B 15 15 0 10 0
C 15 0 20 15 5
D 0 15 20 0 5
E 5 0 10 0 0
So the matrix becomes
Now draw the minimum no of lines to cover all the
zeros.
Here we saw that number of lines are 5.i.e. which
is equal to number of rows (or columns).
Hence it is the optimal solution table.
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Now optimal solution is obtained as follow.
Here column a has single zero in row D so
make a zero assignment to this by square and put
on the zeros in that row . (if any).
a b c d e
A 15 0 20 15 0
B 15 15 0 10 0
C 15 0 20 15 5
D 0 15 20 0 5
E 5 0 10 0 0
0
0
Similarly we can make remaining zero assignments
in a such a way that each row and each column as
a single zero assignment.
0
0
0
Now we will take the values at zero assignment in
the table form original table.
a b c d e
A 160 130 175 190 200
B 135 120 130 160 175
C 140 110 155 170 185
D 50 50 80 80 110
E 55 35 70 80 105
So we have.
Route A-e B-c C-b D-a E-d total
Distance 200 130 110 50 80 570kms
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• Thank You ALL