The Assignment Model

1
The Assignment Model " The best person for job"
is an apt description of the assignment
model. The general assignment model with n
workers and n jobs is presented below
Jobs
1 2 . n
1 c11 c12 c1n
Workers 2 c21 c22 c2n
n cn1 cn2 cnn
2
The element cij is the cost of assigning the
worker i to the job j. More meaningfully, it may
be thought of as the time taken by the worker i
to complete the job j. There is no loss of
generality in assuming that the number of workers
the number of jobs. If there are more workers,
we may introduce dummy jobs and if there are less
workers, we may introduce dummy workers. One
important assumption we make is that each worker
is assigned to one and only one job. And each job
is done by one and only one worker.
3
The assignment model is actually a special case
of the transportation model in which the sources
are workers, the destinations are jobs, and the
number of sources the number of destinations.
Also from the last assumption we get that the
availabilities at each source equals 1 and the
demand at each destination equals 1. xij 1 if
worker i assigned to job j and 0 otherwise.
Though we can apply the transportation algorithm
to solve the assignment model, a special simple
algorithm, called the Hungarian algorithm, is
used to solve such models.
4
Consider the assignment problem
The solution is obvious
W1 ? J2, W2 ? J4, W3 ? J1, W4 ? J3
5
We note that in any assignment problem, if a
constant c is subtracted from the costs of any
row ( or column), the optimum solution does not
change, but the assignment cost decreases by c.
This is because in each row we have to assign 1
to one and only one cell.
The Hungarian algorithm exploits this fact and
tries to get a zero in each row (column) by
subtracting the minimum value in that row
(column) to get at least one zero.
6
The Hungarian algorithm Consider the assignment
problem
Row Min
p1 5
p2 3
p3 4
p4 5
7
Step 1 From each entry of a row, we subtract the
minimum value in that row and get the following
reduced cost matrix
Column Minimum
q11 q21 q30 q41
8
Step 2 From each entry of a column, we subtract
the minimum value in that column and get the
following reduced cost matrix
9
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, surround it by a square ( 0 ) and
cross ( ) all the other zeros in the column
passing through the surrounded zero.
Next examine the other rows and repeat the above
procedure for each row having only one zero.
If a row has more than one zero, do nothing to
that row and pass on to the next row.
10
Step 3(a) gives the following table.
Step 3(b) Now repeat the above procedure for
columns. (Remember to interchange row and column
in that step.)
11
Step 3(b) gives the following table.
12
If there is now a surrounded zero in each row and
each column, the optimal assignment is obtained.
In our example, there is a surrounded zero in
each row and each column and so the optimal
assignment is
Worker 1 is assigned to Job 2
Worker 2 is assigned to Job 4
Worker 3 is assigned to Job 3
Worker 4 is assigned to Job 1
And the optimal cost p1p2p3p4q1q2q3q4
20
13
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
14
Since the problem is unbalanced, we add a dummy
worker 4 with cost 0 and get the following
starting cost matrix
p14
p24
p37
p40
Applying Step 1, we get the reduced cost matrix
15
Now Step 2 is Not needed. We now apply Step 3(a)
and get the following table.
16
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.
17
Step 4 (a) We now draw minimum number m of
vertical and horizontal lines to cover all the
zeros.
Thus we draw 3 (2 vertical and one horizontal)
lines to cover all the zeros and get the
following table.
18
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. Doing this, we get
the table
19
Step 5 Reapply Step 3. We thus get the table
Thus the optimum allocation is
W1 ? J4 W2 ? J1 W3 ? J2 W4 ? J3
Hence Job 3 is not done by any (real) worker.
And the optimal cost
16 ( p1p2p3p4u)
20
Problem 1(i) Problem Set 5.4A Page 200
Solve the Assignment Model
p1 2 p2 2 p3 2 p4 2 p5 6
Applying Step 1, we get the reduced cost matrix
21
q1 1 q2 2 q3 0 q4 1 q5 1
Applying Step 2, we get the reduced cost matrix
22
Applying Step 3, we get the table
23
Now all the zeros are either surrounded or
crossed but there is no surrounded zero in Row 5.
Hence assignment is NOT possible. We go to Step
4.
24
Step 4(a) Thus we draw 4 (3 vertical and one
horizontal) lines to cover all the zeros and get
the above table.
25
Step 4(b) The smallest element, say, u, from
among all elements uncovered by all the lines is
2
Step 4(c) Now subtract this u(2) from all
uncovered elements but add this to all elements
that lie at the intersection of two lines. Doing
this, we get the table
26
We reapply Step 3.
Thus the optimum allocation is
W1 ? J5, W2 ? J3, W3 ? J2, W4 ? J4, W5 ?J1
And the optimal cost
21
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Problem 1(ii) Problem Set 5.4A Page 200
Solve the Assignment Model
p1 2 p2 1 p3 3 p4 1 p5 2
Applying Step 1, we get the reduced cost matrix
28
q1 1 q2 0 q3 0 q4 1 q5 0
Applying Step 2, we get the reduced cost matrix
29
Applying Step 3, we get the table
30
Remark Now in Rows 1 and 4 (and also in columns
1 and 4) there is no single zero. Now we modify
the step 3 for this situation as follows.
31
In row 1 we arbitrarily surround a zero and cross
all the remaining zeros in that row and in column
1. Thus we get the table
32
W J 1 ? 1 2 ? 3 3 ? 5 4 ? 4 5 ? 3
Opt. Cost 11
Now we are left with only one zero in row 4 and
surround it with a square and thus get the
optimum allocation