Maths project

G.N.Khalsa CollegeBMM/BMS Department

A Project Report onOperations Research

{Assignment Problem}

Group # 07Under Prof Guide- Nimalan

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ACKNOWLEDGEMENT

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“The essence of mathematics is not to make simple things complicated, but to make complicated things simple.

”

~Albert Einstein

“The essence of mathematics is not to make simple things complicated, but to make complicated things simple.

”

~Albert Einstein

It was an immensely pleasurable experience in working on this project on

the topic of,

“Assignment Problem”

However, our efforts alone could not have been sufficient for completion of this

project.

Invaluable advice & suggestion from large number of people have gone into this

project.

The foremost among them in our guide and our Prof in charge BMS/BMM Nimalan,

Guru Nanak Khalsa College (GNKC), whose constant corrective guidance &

motivation enabled a focused effort towards completion of the project. We are also very grateful to our parents for being

extremely supportive.

Group Members

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Bird’s Eye View

Sr. # Topic Pg. #

1. Assignment Problem – Theory 5-6

2. Introduction 7

3. Hungarian Method 8

4. Minimization Case studyExample # 01

9-10

5 Example # 02 11-13

6. Alternate Solution Case studyExample # 03

14-16

ASSIGNMENT PROBLEMS

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What is an Assignment Problem? The assignment problem can be stated as a problem where

different jobs are to be assigned to different machines on the basis of the cost of doing these jobs. The objective is to minimize the total cost of doing all the jobs on different machines.

The peculiarity of the assignment problem is only one job can be assigned to one Machine i.e., it should be a one-to-one assignment

The cost data is given as a matrix where rows correspond to jobs and columns to machines and there are as many rows as the number of columns i.e. the number of jobs and number of Machines should be equal

This can be compared to demand equals supply condition in a balanced transportation problem. In the optimal solution there

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should be only one assignment in each row and columns of the given assignment table. one can observe various situations where assignment problem can exist e.g., assignment of workers to jobs like assigning clerks to different counters in a bank or sales man to different areas for sales, different contracts to bidders.

Assignment becomes a problem because each job requires different skills and the capacity or efficiency of each person with respect to these jobs can be different. This gives rise to cost differences. If each person is able to do all jobs equally efficiently then all costs will be the same and each job can be assigned to any person.

When assignment is a problem it becomes a typical optimization problem it can therefore be compared to a transportation problem. The cost elements are given and are a square matrix and requirement at each destination is one and availability at each origin is also one.

INTRODUCTIONThe assignment problem is one of the fundamental combinatorial optimization problems in the branch of optimization or operations research in mathematics. It consists of finding a maximum weight matching in a weighted bipartite graph.In its most general form, the problem is as follows:

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There are a number of agents and a number of tasks. Any agent can be assigned to perform any task, incurring some cost that may vary depending on the agent-task assignment. It is required to perform all tasks by assigning exactly one agent to each task in such a way that the total cost of the assignment is minimized.An assignment is a set of n entry positions in the cost matrix, no two of which lie in the same row or column. The sum of the n entries of an assignment is its cost. An assignment with the smallest possible cost is called an optimal assignment.For Example,Imagine, if in a printing press there is one machine and one operator is there to operate. How would you employ the worker? Your immediate answer will be, the available operator will operate the machine. Again suppose there are two machines in the press and two operators are engaged at different rates to operate them. Which operator should operate which machine for maximizing profit? Similarly, if there are “n” machines available and “n” persons are engaged at different rates to operate them. Which operator should be assigned to which machine to ensure maximum efficiency? While answering the above questions we have to think about the interest of the press, so we have to find such an assignment by which the press gets maximum profit on minimum investment. Such problems are known as "assignment problems".

Initial Stage: - Balancing the Problem

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The Hungarian Method: The following algorithm applies to a given n × n cost matrix to find an optimal assignment. Step1. Subtract the smallest entry in each row from all the entries of its row. Step2. Subtract the smallest entry in each column from all the entries of its column. Step3. Draw lines through appropriate rows and columns so that all the zero entries of the cost matrix are covered and the minimum number of such lines is used. Step4. Test for Optimality: (i) If the minimum number of covering lines is n, an optimal assignment of zeros is possible and we are finished. (ii) If the minimum number of covering lines is less than n, an optimal assignment of zeros is not yet possible. In that case, proceed to Step 5. Step5. Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Then Return to Step 3.

Minimization Case StudyExample 1: You work as a sales manager for a toy manufacturer, and you currently have three salespeople on the road meeting

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buyers. Your salespeople are in Austin, Boston, and Chicago. You want them to fly to three other cities: Denver, Edmonton, and Fargo. The table below shows the cost of airplane tickets in dollars between these cities.

From\ To D E F

A 250 400 350

B 400 600 350

C 200 400 250

Where should you send each of your salespeople in order to minimize airfare?

Solution:-Row MinimizationStep1. Subtract 250 from Row 1, 350 from Row 2, and 200 from Row 3.

0 150 10050 250 00 200 50

Column MinimizationStep2. Subtract 0 from Column 1, 150 from Column 2, 0 from column 30 0 100

50 100 0

0 50 50

Step3. Cover all the zeros of the matrix with the minimum number of horizontal or vertical lines.

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0 0 10050 100 00 50 50

Here Total number of lines =3Order of matrix= 3Hence, Optimality is achievedStep4. Since the minimal number of lines is 3, an optimal assignment of zeros is possible.

0 0 -

- - 0

0 - -

Step 5 Final step of Assignment which is made to the original matrix

From To

A B - 400

B C - 350

C A - 200

Total 950

Optimal Cost= 950

Example # 02

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A team of 5 horses and 5 riders has entered a jumping show contest. The number of penalty points to be expected when each rider rides any horse is shown below.

R1 R2 R3 R4 R5H1 5 3 4 7 1H2 2 3 7 6 5H3 4 1 5 2 4H4 6 8 1 2 3H5 4 2 5 7 1How should the horses be allotted to the riders so as to minimize the expected loss of the team?

Solution:-

Step 1:-Row Minimization

R1 R2 R3 R4 R5H1 4 2 3 6 0H2 0 1 5 4 3H3 3 0 4 1 3H4 5 7 0 1 2H5 3 1 4 6 0

Step 2:-Column Minimization

R1 R2 R3 R4 R5H1 4 2 3 5 0H2 0 1 5 3 3H3 3 0 4 0 3H4 5 7 0 0 2H5 3 1 4 5 0

Step 3:-Row scanning and Column scanning

R1 R2 R3 R4 R5

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H1 4 2 3 5 0H2 0 1 5 3 3H3 3 0 4 0 3H4 5 7 0 0 2H5 3 1 4 5 0

Here Lines is not equal to the size of the Matrix.

Lines=4 and Matrix Size = 5

Thus we have to make revised matrix

Step 4:-Revised Matrix; where the Covered numbers will remain the same; the Intersection numbers will be Minimum+ Number and the uncovered number will be Number –Minimum.

R1 R2 R3 R4 R5H1 3 1 2 4 0H2 0 1 5 3 4H3 3 0 4 0 4H4 5 7 0 0 3H5 2 0 3 4 0

Step 5:-Row scanning and Column scanning in the Revised Matrix.

Now the Lines is equal to the Matrix Size i.e. 5

Step 6:-Assignment

R1 R2 R3 R4 R5H1 3 1 2 4 1H2 0 1 5 3 4

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H3 3 0 4 4 4H4 5 7 0 0 3H5 2 0 3 4 0

Assignment Schedule:-

Horse RiderH1 R5 - 1H2 R1 - 2H3 R2 - 1H4 R4 - 2H5 R3 - 5Total 11

Optimal Cost: - 11

Alternate Solution Case study

Example # 3: A construction company has four large bulldozers located at four different garages. The bulldozers are to be moved to four different construction sites. The distances in miles between the bulldozers and the construction sites are given below.

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Bulldozer\Site A B C D1 90 75 75 802 35 85 55 653 125 95 90 1054 45 110 95 115How should the bulldozers be moved to the construction sites in order to minimize the total distance traveled?

Solution:-Step 1:- Row Minimization Subtract 75 from Row 1, 35 from Row 2, 90 from Row 3, and 45 from Row 4.

15 0 0 50 50 20 3035 5 0 150 65 50 70

Step 2:- Column Minimization Subtract 0 from Column 1, 0 from Colum 2, 0 from Column 3, and 5 from Column 4.

15 0 0 00 50 20 2535 5 0 100 65 50 65Step 3:- Cover all the zeros of the matrix with the minimum number of horizontal or vertical lines.

15 0 0 00 50 20 2535 5 0 100 65 50 65

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Here, number of lines=3Order of matrix =4So lines are not equal to given matrix

Since the minimal number of lines are less than 4, We have to proceed to Step 4.

Step 4:- Revised Matrix Add 5 that is the minimum value from the uncovered matrix to the intersection element, less from uncovered, and the rest covered elements will remain same.

20 0 5 00 45 20 2035 0 0 50 60 50 60

Here again lines=3Order of matrix = 4So lines are not equal to given matrix

Since the minimal number of lines are less than 4, We again have to proceed to Step 5.

Step 5:- Revised Matrix

40 0 5 00 25 0 055 0 0 50 40 30 40Now lines=4

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Order of matrix= 4Hence, Solution is Optimal

Step 6:- Doing Assignment (Alternate Solution) Solution 1 - 0 - 00 - 0 0- 0 0 -0 - - -

Solution 2- 0 - 00 - 0 0 - 0 0 -0 - - -

Assignment Schedule

Bulldozer Site(solution 1)

Site(solution 2)

1 B – 75 D - 80

2 D – 65 C – 55

3 C - 90 B – 95

4 A - 45 A - 45

Total 275 275

ConclusionThe following is the flow chart of all the explanation being mentioned in the project. In short this chart explains all how to start up and where to end up the given matrix in order to gain an optimal solution.

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Thank U

Sir!!!

Thank U

Sir!!!