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A
PRESENTATION
ON
Assignment problem Unbalance
• Guided by :
• Prof. Amita garg
Group No:- 1
Presented by :
Jignesh Anaghan Roll no:- 01
Enrollment No:- 127110592006
Vishal Bhadiyadra Roll no:- 02
Enrollment No:- 127110592018
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
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.
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.
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
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 (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.
Step 4(b) Select the smallest element, say, u, from
among all elements uncovered by all the lines.
2 1 3 0
0 2 2 0
0 0 1 1
0 1 0 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:
In our example, u = 1.
Step 5: Reapply Step 3.
We thus get the table
2 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 = 16 (= p1+p2+p3+p4+u)
Hence Job 3 is not done by any (real) worker.
Qa 2

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Qa 2

  • 1. A PRESENTATION ON Assignment problem Unbalance • Guided by : • Prof. Amita garg Group No:- 1 Presented by : Jignesh Anaghan Roll no:- 01 Enrollment No:- 127110592006 Vishal Bhadiyadra Roll no:- 02 Enrollment No:- 127110592018
  • 2. 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
  • 3. 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.
  • 4. 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.
  • 5. 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
  • 6. 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
  • 7. 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.
  • 8. 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.
  • 9. 3 1 4 0 1 2 3 0 1 0 2 1 0 0 0 0 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.
  • 10. Step 4(b) Select the smallest element, say, u, from among all elements uncovered by all the lines. 2 1 3 0 0 2 2 0 0 0 1 1 0 1 0 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: In our example, u = 1.
  • 11. Step 5: Reapply Step 3. We thus get the table 2 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 = 16 (= p1+p2+p3+p4+u) Hence Job 3 is not done by any (real) worker.