International Journal of Research in Engineering Technology and Management
ISSN 2347 - 7539
Method of Finding an Optimal Solution of an Assignment Problem
Rupsha Roy1, Rukmani Rathore2
1
2
Research Scholar, Department of Mathematics, IIT, Indore, Madhya Pradesh, India, [email protected]
Lecturer, Department of Mathematics, LNCT, Indore, Madhya Pradesh, India, [email protected]
Abstract
In this paper a method is applied for finding an optimal solution for assignment problem. This method requires least iterations to
reach optimality, compared to the existing methods available in the literature. Here numerical examples are solved to check the
validity of the proposed method.
Keywords: Assignment Problem, Linear Programming Problem, Zero suffix method.
--------------------------------------------------------------------***----------------------------------------------------------------------
1. INTRODUCTION
In real life, we are faced with the problem of allocating
different personnel/ workers to different jobs. Not everyone
has the same ability to perform a given job. Different persons
have different abilities to execute the same task and these
different capabilities are expressed in terms of
cost/profit/time involved in executing a given job. Therefore,
we have to decide how to assign different workers to different
jobs” so that, cost of performing such job is minimized. The
assignment problem is one of the first studied combinatorial
optimization problems. It was investigated by G. Monge
[1784]. Various methods are available to solve the
assignment problem to abtain an optimal solution. Typical
well known methods include Hungarian methodIt was
developed and published by Harold Kuhn in 1955, who gave
the name "Hungarian method" because the algorithm was
largely based on the earlier works of two Hungarian
mathematicians: Dénes Kőnig and Jenő Egerváry. James
Munkres reviewed the algorithm in 1957 and observed that it
is (strongly) polynomial. Since then the algorithm has been
known also as Kuhn-Munkres or Munkres assignment
algorithm.
Step 5: Choose the maximum of S, if it has one maximum
value then assign that task to the person and if it has more
than one maximum value then also assign the tasks to their
respective persons (if the zeros don’t lie in the same column
or row).
And if the zeros lie in the same row or column then assign the
job to that person whose cost is minimum. Now create a new
assignment table by deleting that row & column which has
been assigned.
Step 6: Repeat step 2 to step 3 until all the tasks has not been
assigned to the persons.
2.1 Example:
(i) Consider the following cost minimization assignment
problem
J1
J2
J3
J4
B1
15
11
13
15
B2
17
12
12
13
B3
14
15
10
14
B4
16
18
11
17
2. ZERO SUFFIX METHOD
The working rule of finding the optimal solution is as follows:
Step 1: Construct the assignment problem.
Step2: Subtract each row entries of the assignment table from
the row minimum element.
Step 3: Subtract each column entries of the assignment table
from the column minimum element.
Step 4: In the reduced cost matrix there will be at least one
zero in each row and column, then find the suffix value of all
the zeros in the reduced cost matrix by following
simplification, the suffix value is denoted by S.
Solution: On applying row minimum operation we get
J1
J2
J3
J4
B1
4
0
2
4
B2
5
0
0
1
costs added
_______________________________________________________________________________________
Volume: 02 Issue: 01 | Jan-2014, Available @ http://www.ijretm.com | Paper id - IJRETM-2014-02-01-221
1
International Journal of Research in Engineering Technology and Management
B3
4
5
0
4
B4
5
7
0
6
J1
B1
Again on applying column minimum operation we get
J1
J2
J3
3
1
0 [1.5]
0 [2]
B3
0 [3]
5
3
0
0
2
3
B2
1
0
0
0
From all of the above suffix, 3 is the maximum so assign the
job J1 to the person B3
B3
0
5
0
3
Next delete the 3rd row and 1st column from the above table
and apply the same process.
1
J1
B4
0 [1]
J4
7
0
0
J2
J3
J4
[0.5]
0 [0.6]
2
3
1
0 [1.5]
0 [0.5]
0 [2]
0 [2.3]
1
5
7
0 [2]
0 [4]
J2
J4
B1
0 [1.5]
3
B2
0 [0]
0 [1.5]
5
S = Add the cost of nearest adjacent sides of zeros/ No. of
costs added
B3
J4
B2
Now find the suffix value of each element whose value is zero
& write the suffix value within the bracket [ ]. The suffix
value is calculated by using the formula
B2
[0.5]
J2
B1
B4
B1
0
ISSN 2347 - 7539
From the above table, it is clear that the job J2 should be
assigned to the person B1 and the job J4 should be assigned to
the person B2
Finally the assignments are as follows
B1 → J2, B2 → J4, B3 → J1, B4 → J3
& the minimum assignment cost = Rs (11 + 13 + 14 + 11)
= Rs 49
(ii) Consider the following travelling salesman problem
D1
D2
D3
D4
O1
∞
46
16
40
O2
41
∞
50
40
O3
82
32
∞
60
O4
40
40
36
∞
3
5
From all of the above suffix, 4 is the maximum so assign the
job J3 to the person B4
Next delete the 4th row and 3rd column from the above table
and apply the same process.
_______________________________________________________________________________________
Volume: 02 Issue: 01 | Jan-2014, Available @ http://www.ijretm.com | Paper id - IJRETM-2014-02-01-221
2
International Journal of Research in Engineering Technology and Management
Solution: On applying row minimum operation we get
D1
D2
D3
ISSN 2347 - 7539
From all of the above suffix, 21.3 is the maximum (take the
finite value only) so assign the origin O1 to the destination D3
D4
Next delete the 1st row and 3rd column from the above table
and apply the same process.
∞
O1
O2
30
∞
1
O3
50
O4
0
4
4
0
10
∞
0
24
D1
D2
D4
O2
0
∞
0
O3
49
0
28
O4
3
4
∞
0
28
∞
Again on applying column minimum operation we get
D1
D2
D3
D4
On applying the row minimum operation we get
O1
∞
30
0
24
D1
D2
D4
O2
0
∞
10
0
O2
0
∞
0
O3
49
0
∞
28
O3
49
0
28
O4
3
4
0
∞
O4
0
1
∞
Now find out the suffix value of each element whose value is
zero & write the suffix value within the bracket [ ]. The suffix
value is calculated by using the formula
Now find out the suffix value of each element whose value is
zero
S = Add the cost of nearest adjacent sides of zeros/ No. of
costs added
D1
D2
D3
D4
O1
∞
30
0[21.3]
24
O2
0[∞]
∞
10
0[20.6]
O3
49
0[∞]
∞
28
O4
3
4
36[∞]
D1
D2
D4
O2
0[∞]
∞
0[∞]
O3
49
0[∞]
28
O4
0[25]
1
∞
∞
_______________________________________________________________________________________
Volume: 02 Issue: 01 | Jan-2014, Available @ http://www.ijretm.com | Paper id - IJRETM-2014-02-01-221
3
International Journal of Research in Engineering Technology and Management
ISSN 2347 - 7539
From all of the above suffix, 25 is the maximum (take the
finite value only) so assign the origin O4 to the destination D1
Next delete the 3rd row and 1st column from the above table
and apply the same process.
D2
D4
O2
∞
0
O3
0
28
From the above table it is clear that O2 → D4 , O3 → D2
The optimum assignment is
O1 → D3 , O2 → D4 , O3 → D2, O4 → D1
& the minimum cost = Rs (16 + 40 + 32 + 40 )
= Rs 128
3. LIMITATION
This method is not applicable for unbalanced transportation
problem & it is also not necessarily applicable in that kind of
problem where in any iteration if we get dummy row or
column.
4. REFERENCES
[1] Abdul Quddoos, Shakeel Javaid , M. M. Khalid , A New
Method for Finding an Optimal Solution for Transportation
Problems, International Journal on Computer Science and
Engineering (IJCSE), ISSN: 0975-3397, Vol. 4 No. 07 July
2012.
[2] V. J .Sudhakar , N. Arunsankar, T. Karpagam , A New
Approach for Finding an Optimal Solution for Transportation
Problems, European Journal of Scientific Research ,ISSN
1450-216X Vol.68 No.2 (2012), pp. 254-257.
[3] Kasana Dr. H. S. and Dr. K. D. Kumar,
INTRODUCTORY OPERATIONS RESEARCH, Theory
and Applications, Springer Pvt. Ltd., India, 2011.
[4] Sharma. J.K., Operations Research-Theory and
applications, Macmillan India (LTD), New Delhi, 2005.
[5] Zimmermann H.J. “Fuzzy sets theory and its
applications”, Kluwer - Boston.1996.
_______________________________________________________________________________________
Volume: 02 Issue: 01 | Jan-2014, Available @ http://www.ijretm.com | Paper id - IJRETM-2014-02-01-221
4