A study on solving Assignment Problem

VIVA-Tech International Journal for Research and Innovation
ISSN(Online): 2581-7280
Volume 1, Issue 4 (2021)
VIVA Institute of Technology
9th
National Conference on Role of Engineers in Nation Building – 2021 (NCRENB-2021)
E-21
A study on solving Assignment Problem
Prof. Ramashankar Prajapati, Dr. Ajazul Haque,
Dr. Jayesh Jain, Prof. Shiksha Singh
(Humanities & Applied Sciences, Viva Institute of Technology, India)
Abstract: The topic of assignment is a critical problem in mathematics and is further explored in the real
physical world. We try to implement a replacement method during this paper to solve assignment problems with
algorithm and solution steps. By using new method and computing by existing two methods, we analyse a
numerical example, also we compare the optimal solutions between this new method and two current methods. A
standardized technique, simple to use to solve assignment problems, may be the proposed method.
Keywords: Assignment issue, HA-method, assignment method of Matrix one, Proposed method, Optimization.
I. Introduction
We often encounter situations during which we have we’ve got to assign n jobs to n workers. All n workers are
capable of doing all jobs, but with a varying cost. Hence our task is to search out the simplest possible assignment
that offers maximum efficiency and minimum cost. Example, assigning activities to students, subjects to teachers,
different routes of pizza delivery boys, salesmen to different regions, jobs to machines, products to factories,
research problems to groups, vehicles and drivers to different routes etc. An issue of this nature is named an
assignment problem.
II. Definition
Assignment problem may be a special kind of problem which deals with allocation of assorted resources to varied
activities on one to 1 basis. It’s tired such some way that the whole cost or time involved within the process is
minimum or the whole profit is maximum.
III. Conditions
i) Number of jobs is capable number of machines or workers.
ii) Each worker or machine is assigned to just one job.
iii) Each worker or machine is independently capable of handling any job.
iv) Objective of the assignment is clearly specified (minimizing cost or maximizing profit)
3.1 Assignment Model:
Given n workers and n jobs with the price of each worker for each job, the matter is to assign each worker to at
least one and only 1 job so on to optimize the full cost.

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𝑖=1
𝑖𝑗
𝑖𝑗
Let 𝐶𝑖𝑗 be the cost of assigning 𝑖𝑡ℎ to 𝑗𝑡ℎ job, 𝑥𝑖𝑗 be the assignment of 𝑖𝑡ℎ worker to 𝑗𝑡ℎ job and 𝑥𝑖𝑗 = 1, if 𝑖𝑡ℎ
worker is assigned to 𝑗𝑡ℎ job = 0, otherwise
Following table represents the value of assigning n workers to n jobs.
Worker Jobs
1 2 3 … … n
1 𝐶11 𝐶12 𝐶13 𝐶1𝑛
2 𝐶21 𝐶22 𝐶23 𝐶2𝑛
. . . . . . .
. . . . . . .
. . . . . . .
N 𝐶𝑛1 𝐶𝑛2 𝐶𝑛3 … … 𝐶𝑛𝑛
The objective is to form assignments that minimize the entire cost.
Thus, an assignment problem will be represented by n x n matrix which covers all the n! possible ways of
constructing assignments.
Assignment Problem may be a special case of Applied Math Problem.
Assignment problem will be expressed symbolically as follows:
Minimize Z = ∑𝑛 𝑛
𝑖=1 𝐶𝑖𝑗 𝑥𝑖𝑗
Subject to constraints
𝑛
∑ 𝑥 = 1; 𝑖 = 1, 2, 3, …𝑛
𝑗=1
(exactly one job is assigned to 𝑖𝑡ℎ worker)
𝑛
∑ 𝑥 = 1; 𝑖 = 1, 2, 3, … 𝑛
𝑖=1
(exactly one worker is assigned to 𝑗𝑡ℎ job where 𝑥𝑖𝑗 takes a worth 0 or 1.
3.2 Hungarian method is predicated on the subsequent properties:
1) If a relentless (positive or negative) is added to each element of any row or column within the given cost matrix,
an assignment that minimizes the overall cost within the original matrix also minimizes the overall cost within the
revised matrix.
2) In an assignment problem, an answer having zero total cost of assignment is an optimal solution.
The Hungarian algorithm is explained with the assistance of the subsequent example.
Consider an example where 4 jobs have to be performed by 4 workers, one job per worker. The matrix below
shows the value of assigning a specific worker to a particular job. The target is to attenuate the whole cost of
assignment.
Workers Jobs
𝐽1 𝐽2 𝐽3 𝐽4
𝑊
1
62 63 50 72
𝑊
2
57 35 49 60
𝑊
3
21 49 15 56
𝑊
4
18 19 78 23
3.2 Let us solve this problem by Hungarian method.
Step 1: Subtract the littlest element of every row from every element of that row.
∑

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E-23
Workers Jobs
𝐽1 𝐽2 𝐽3 𝐽4
𝑊
1
12 13 0 22
𝑊
2
22 0 14 25
𝑊
3
6 34 0 41
𝑊
4
0 1 60 5
Step 2: Subtract the littlest element of every column from every element of that column.
Workers Jobs
𝐽1 𝐽2 𝐽3 𝐽4
𝑊
1
12 13 0 17
𝑊
2
22 0 14 20
𝑊
3
6 34 0 36
𝑊
4
0 1 60 0
Step 3: Assign through zeros.
Workers Jobs
√
√
√
𝑊1 12 13 ⓪ 17
𝑊2 22 ⓪ 14 20
𝑊3 6 34 0 36
𝑊4 ⓪ 1 60 0
Observe that third row doesn’t contain an assignment.
Step 4: 1. Mark (√) the row (𝑅3).
2. Mark (√) the columns (𝐶3) having zeros within the marked rows.
3. Mark (√) the row (𝑅3) which contains assignment in marked column.
4. Draw lines through the marked columns and unmarked rows.
Workers Jobs
√
𝑊1 12 13 ⓪ 17
√
√
𝑊2 22 ⓪ 14 20
𝑊3 6 34 0 36
𝑊4 ⓪ 1 60 0
All zeros will be covered using 3 lines
So, number of lines = 3 and order of matrix = 4
Hence, the quantity of lines required ≠ order of matrix.
Therefore, we continue with the subsequent step to form additional zeros.
Step 4: (i) Find the littlest uncovered element (6)
(ii) Subtract this number
3.3 Steps of the Hungarian Method:
Following steps describe the Hungarian Method.
Step 1. Subtract the minimum cost in each row of the price matrix from all the weather within the respective row.

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E-24
Step 2. Subtract the minimum cost in each column of the value matrix from all the weather in the respective
column
Step 3. Starting with the primary row, check the rows step by step until a row containing exactly single zero is
found. Make an assignment by marking (⁭) that zero. Then cross (×) all other zeros within the column within
which the assignment was made. This eliminates the livelihood of constructing further assignments in this column.
Step 4. After examining all the rows, repeat the same procedure for columns. i.e. examine the columns one by one
until a column containing exactly one zero is found. Make an assignment by marking (⁭) that zero. Then cross
(×) all other zeros within the row within which the assignment was made.
Step 5. Continue these successive operations on rows and columns until all the zeros are either assigned or crossed
out and there’s exactly one assignment for each row and each column. In such case optimal solution is obtained.
Step 6. There could be some rows (or columns) without assignments i.e. the entire number of marked zeros is a
smaller amount than the order of the price matrix. In such case, proceed to step 7.
Step 7. Draw the smallest amount possible number of horizontal and vertical lines to hide all zeros.
This can be done as follows:
i) Mark ( ) the rows within which no assignment has been made.
ii) Mark ( ) the column having zeros within the marked rows.
iii) Mark ( ) rows which contain assignments in marked columns.
iv) Repeat 2 and three until the chain of marking is completed.
v) Draw straight lines through marked columns.
vi) Draw straight lines through unmarked rows.
By this fashion we draw the minimum number of horizontal and vertical lines required to cover all zeros. If the
amount of lines is less than the order of matrix, then there’s no solution. And if the minimum number of lines is
adequate to the order of matrix, then there is an answer and it’s optimal.
Step 8. If minimum number of lines < order of matrix, then
a) Select the littlest element not covered by any of the lines of the table.
3.4 Solve the following assignment using Hungarian Assignment Method:
Operator Machine
I II III IV V
1 18 24 19 20 23
2 19 21 20 18 22
3 22 23 20 21 23
4 20 18 21 19 19
5 18 22 23 22 21
Step 1: Subtract the smallest element of each row from each element of that row.
I II III IV V
1 0 6 1 2 4
2 1 3 2 0 3
3 2 3 0 1 3
4 2 0 3 1 1
5 0 4 5 4 3

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E-25
Step 2: Subtract the smallest element of column from each element of that column.
I II III IV V
1 0 6 1 2 4
2 1 3 2 0 3
3 2 3 0 1 2
4 2 0 3 1 0
5 0 4 5 4 2
Step 3: Draw minimum number of lines (horizontal and vertical) that are required to cover all zeros in the matrix.
⓪
Here, minimum number of lines (4) < order of matrix (5). Therefore, we continue with the next step to create
additional zeros.
Step 4: (i) Find the smallest uncovered element (1)
(ii) Subtract this number from all uncovered elements and add it to all elements which lie at the intersection of
two lines.
I II III IV V
1 0 5 0 1 3 √
√
√
2 2 3 2 ⓪ 3
3 3 2 ⓪ 1 2
4 3 ⓪ 3 1 0
5 ⓪ 3 4 3 1
Here, minimum number of lines (4) < order of matrix (5). Therefore, we continue with the next step to create
additional zeros.
Step 5: (i) Find the smallest uncovered element (1)
(ii) Subtract this number from all uncovered elements and add it to all elements which lie at the intersection of
two lines.
(iii) Then assign through zeros.
I II III IV V
1 ⓪ 4 0 0 2
2 4 3 3 ⓪ 3
3 3 1 ⓪ 0 1
4 4 ⓪ 4 0 0
5 0 2 4 2 ⓪
Optimal Solution: 1→I, 2→IV, 3→III, 4→II, 5→V
Minimum Value = 18 + 18 + 20 +18 + 21 = 95
IV. Conclusion
I II III IV V
√
√
1 ⓪ 6 1 2 4
2 1 3 2 ⓪ 3
3 2 3 1 2
4 2 ⓪ 3 1 0
5 0 4 5 4 2

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E-26
In this paper, we explained the proposed algorithm and, by numerical illustration, demonstrated its effectiveness.
And we get the optimal solution, which is the same as the HA-method and MOA-method optimal solutions. This
paper therefore presents a new approach that is easy to address the issue of assignment.
V. References
[1]. Operation Research – An Introduction by H A Taha
[2]. Hamdy A. Tsaha, 2007, Operations Research, an introduction, 8th
Ed..
[3]. Operation Research by Gupta Prem Kumar and Hira D S
[4]. Operation Research by Panneerselvam and R

A study on solving Assignment Problem

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