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					       UNIT II
       TOPIC
ASSIGNMENT PROBLEM



           gaurav sonkar
                                Gaurav Sonkar

          ASSIGNMENT PROBLEM
The assignment problem is a particular case of
  transportation problem in which the objective is to
  assign a number of resources to an equal number of
  activities so as to minimize total cost or maximize
  total profit.
Typical examples of decision making situations where
  assignment model can be used are:
  workers to machines
  sales personnel to different sales areas
  products to factories
  jobs to machines
  classes to rooms                                  2
A GENERAL ASSIGNMENT PROBLEM LOOKS
                           Gaurav Sonkar


LIKE:

      1      2        ……….            n
  1    C11   C12      …………..         C1n   1
  2    C21   C22      ……………          C2n   1
  :     .     .         .              .
        .     .         .              .
                      ……………
       Cn1   Cn2                     Cnn
  n                     .                  1
      1           1   ………..            1



                                               3
                                        Gaurav Sonkar


  HOW TO SOLVE AN A.P.


Step I:
  Row Operation:
Subtracting smallest element in each row from all other elements of
that row.
  Column Operation:
Subtracting smallest element in each column from all other elements
of that column.




                                                                 4
Step II: MAKING ASSIGNMENTS:
                                  Gaurav Sonkar




Search for an optimal assignment in the finally
  modified cost matrix as follows:
  Examine the first row. If there is only one zero in it,
  then enclose this zero in a box ( ) cross (X) all the
  zero of that column passing through the enclosed
  zero.
  Similarly done with other rows.
  If any row has more than one zero, then do not
  touch that row & passes on the next row.
  Repeat the same procedure with columns.
                                                       5
Step III:                              Gaurav Sonkar

 If each row and each column of the reduced matrix has
  exactly one enclosed zero, then the enclosed zero yield an
  optimal assignment.
 If not, then go to next step.
Step IV:
 Draw minimum number of horizontal and/or vertical lines
  to cover all the zeros of the reduced matrix.
 Since the assignment is not optimal, the number of lines
  will be less than n (i.e. order of matrix).
 In order to move towards optimality, generates more zeros
  as follows:
 Find the smallest of the elements of the reduced matrix not
  covered by any of the lines. Let this element be α.
 Subtract α from each of the element not covered by the
  lines and add α to the element at the intersection of these
  lines.
 Do not change the remaining elements                       6
                              Gaurav Sonkar



Step V:
Go to step II and repeat the procedure till an
  optimal assignment is achieved.




                                                 7
                    ILLUSTRATION 1       Gaurav Sonkar


A department head has four subordinates, and four tasks to be
  performed. The subordinates differ in efficiency and the tasks
  differ in their intrinsic difficulty. His estimate of the times each
  man would take to perform each task is given in the effectiveness
  matrix below. How should the task be allocated, one to a man, so
  as to minimize the total man hours?


                         I        II        III          IV
               A        8        26        17            11
               B        13       28         4            26
               C        38       19        18            15
               D        19       26        24            10




                                                                    8
                          SOLUTION       Gaurav Sonkar




Step I:
  Row Operation:
Subtracting smallest element in each row from all other elements of
  that row.
      I      II     III     IV              I            II   III   IV

 A                                   A
 B                                   B
 C                                   C
 D                                   D
   Column Operation:
 Subtracting smallest element in each column from all other
 elements of that column.                                                9
Step II: MAKING ASSIGNMENTS:         Gaurav Sonkar


                    I    II    III           IV

               A    0    14    9            3
               B    9    20    0           22
               C   23     0    3            0
               D    9    12    14           0
Since every row and every column have one assignment,
  hence the current solution is optimal.




                                                        10
, THE OPTIMAL SOLUTION OBTAINED IS
                        Gaurav Sonkar




  AS UNDER

     Task     Subordinate               Man Hour




      TOTAL MAN HOURS
                                                   11
                   ILLUSTRATION 2
                                      Gaurav Sonkar




A department has four subordinates and four tasks to be
  performed. The subordinates differ in efficiency and the
  tasks differ in their intrinsic difficulties. The estimates of
  time (in man-hours) each man would take to perform
  each task is given by:
                            SUBORDINATES
          JOBS
                     I        II     III              IV
            A       18       26      17               11
            B       13       28      14               26
            C       38       19      18               15
            D       29       26      24               10

How should the tasks to be allocated to men so as to
  optimize the total man-hours?                   12
                          SOLUTION       Gaurav Sonkar


STEP I:
Subtracting the smallest element of each row from all the
elements of that row, we obtain the reduced matrix (a) given below:
Further subtracting the smallest element of each column from all
the elements of that column, we obtain the reduced matrix (b)
given below:

        I     II    III    IV                    I       II   III   IV
 A     7     15     6      10        A           7       11   5     0
 B     0     15     1      13        B           0       11   0     13
 C     23     4     3      0         C         23        0    2     0
 D     19    16     14     0         D         19        12   13    0



                                                                        13
STEP II:                                      Gaurav Sonkar


Examine the rows of matrix (b) successively until a row with exactly
  one zero is found, enclose that zero in a box and cross all the other
  zero in its column. Thus we obtain matrix (c)

                           I     II     III          IV
                    A     7      11     5             0
                    B     0      11     0            13
                    C     23     0      2             0
                    D     19    12     13             0


Further examine the columns of matrix (c) successively until a column
  with exactly one unmarked zero is found, enclose that zero in a box
  and cross all the other zero in its row
                                                                    14
   STEP III:                              Gaurav Sonkar


Matrix (c) does not provide an optimal solution, since fourth row as
  well as third column does not have an enclosed zero.
  STEP IV:
Now we draw minimum number of horizontal and / or vertical lines to
  cover all the zeros of the reduced matrix (c)
                          I     II    III    IV
                   A      7    11      5     0
                   B      0    11      0     13
                   C     23     0      2     0
                   D     19    12     13     0



                                                                  15
The smallest element in matrix (d) not covered by any of the lines above is 5.
                                                    Gaurav Sonkar

Subtracting this element from all uncovered elements and adding the same to the
   elements lying at the intersection of these lines, we obtain matrix (e) given
   below:

                                I       II     III     IV
                        A       2       6       0      0
                        B       0      11       0      18
                        C      23       0       2       5
                        D      14       7       8      0



STEP V:
Following the procedure of enclosing and crossing the zeros as in
  (step II) in matrix (e).
                                                                              16
STEP VI:                      Gaurav Sonkar



Since each row and each column in matrix (g) has
  exactly one enclosed zero, we have attained the
  following optimal assignment schedule:
      A     III, B    I, C  II, D   IV
The minimum total time for this assignment
  schedule is:
                 Z = 17+13+19+10= 59 man-hours.




                                                    17

				
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