assignment problem kauser wise

How to Solve the Assignment Problem: A Complete Guide

Table of Contents

Assignment problem is a special type of linear programming problem that deals with assigning a number of resources to an equal number of tasks in the most efficient way. The goal is to minimize the total cost of assignments while ensuring that each task is assigned to only one resource and each resource is assigned to only one task. In this blog, we will discuss the solution of the assignment problem using the Hungarian method, which is a popular algorithm for solving the problem.

Understanding the Assignment Problem

Before we dive into the solution, it is important to understand the problem itself. In the assignment problem, we have a matrix of costs, where each row represents a resource and each column represents a task. The objective is to assign each resource to a task in such a way that the total cost of assignments is minimized. However, there are certain constraints that need to be satisfied – each resource can be assigned to only one task and each task can be assigned to only one resource.

Solving the Assignment Problem

There are various methods for solving the assignment problem, including the Hungarian method, the brute force method, and the auction algorithm. Here, we will focus on the steps involved in solving the assignment problem using the Hungarian method, which is the most commonly used and efficient method.

Step 1: Set up the cost matrix

The first step in solving the assignment problem is to set up the cost matrix, which represents the cost of assigning a task to an agent. The matrix should be square and have the same number of rows and columns as the number of tasks and agents, respectively.

Step 2: Subtract the smallest element from each row and column

To simplify the calculations, we need to reduce the size of the cost matrix by subtracting the smallest element from each row and column. This step is called matrix reduction.

Step 3: Cover all zeros with the minimum number of lines

The next step is to cover all zeros in the matrix with the minimum number of horizontal and vertical lines. This step is called matrix covering.

Step 4: Test for optimality and adjust the matrix

To test for optimality, we need to calculate the minimum number of lines required to cover all zeros in the matrix. If the number of lines equals the number of rows or columns, the solution is optimal. If not, we need to adjust the matrix and repeat steps 3 and 4 until we get an optimal solution.

Step 5: Assign the tasks to the agents

The final step is to assign the tasks to the agents based on the optimal solution obtained in step 4. This will give us the most cost-effective or profit-maximizing assignment.

Solution of the Assignment Problem using the Hungarian Method

The Hungarian method is an algorithm that uses a step-by-step approach to find the optimal assignment. The algorithm consists of the following steps:

Subtract the smallest entry in each row from all the entries of the row.
Subtract the smallest entry in each column from all the entries of the column.
Draw the minimum number of lines to cover all zeros in the matrix. If the number of lines drawn is equal to the number of rows, we have an optimal solution. If not, go to step 4.
Determine the smallest entry not covered by any line. Subtract it from all uncovered entries and add it to all entries covered by two lines. Go to step 3.

The above steps are repeated until an optimal solution is obtained. The optimal solution will have all zeros covered by the minimum number of lines. The assignments can be made by selecting the rows and columns with a single zero in the final matrix.

Applications of the Assignment Problem

The assignment problem has various applications in different fields, including computer science, economics, logistics, and management. In this section, we will provide some examples of how the assignment problem is used in real-life situations.

Applications in Computer Science

The assignment problem can be used in computer science to allocate resources to different tasks, such as allocating memory to processes or assigning threads to processors.

Applications in Economics

The assignment problem can be used in economics to allocate resources to different agents, such as allocating workers to jobs or assigning projects to contractors.

Applications in Logistics

The assignment problem can be used in logistics to allocate resources to different activities, such as allocating vehicles to routes or assigning warehouses to customers.

Applications in Management

The assignment problem can be used in management to allocate resources to different projects, such as allocating employees to tasks or assigning budgets to departments.

Let’s consider the following scenario: a manager needs to assign three employees to three different tasks. Each employee has different skills, and each task requires specific skills. The manager wants to minimize the total time it takes to complete all the tasks. The skills and the time required for each task are given in the table below:

	Task 1	Task 2	Task 3
Emp 1	5	7	6
Emp 2	6	4	5
Emp 3	8	5	3

The assignment problem is to determine which employee should be assigned to which task to minimize the total time required. To solve this problem, we can use the Hungarian method, which we discussed in the previous blog.

Using the Hungarian method, we first subtract the smallest entry in each row from all the entries of the row:

	Task 1	Task 2	Task 3
Emp 1	0	2	1
Emp 2	2	0	1
Emp 3	5	2	0

Next, we subtract the smallest entry in each column from all the entries of the column:

	Task 1	Task 2	Task 3
Emp 1	0	2	1
Emp 2	2	0	1
Emp 3	5	2	0
	0	0	0

We draw the minimum number of lines to cover all the zeros in the matrix, which in this case is three:

Since the number of lines is equal to the number of rows, we have an optimal solution. The assignments can be made by selecting the rows and columns with a single zero in the final matrix. In this case, the optimal assignments are:

Emp 1 to Task 3
Emp 2 to Task 2
Emp 3 to Task 1

This assignment results in a total time of 9 units.

I hope this example helps you better understand the assignment problem and how to solve it using the Hungarian method.

Solving the assignment problem may seem daunting, but with the right approach, it can be a straightforward process. By following the steps outlined in this guide, you can confidently tackle any assignment problem that comes your way.

How useful was this post?

Click on a star to rate it!

Average rating 0 / 5. Vote count: 0

No votes so far! Be the first to rate this post.

We are sorry that this post was not useful for you! 😔

Let us improve this post!

Tell us how we can improve this post?

Operations Research

1 Operations Research-An Overview

History of O.R.
Approach, Techniques and Tools
Phases and Processes of O.R. Study
Typical Applications of O.R
Limitations of Operations Research
Models in Operations Research
O.R. in real world

2 Linear Programming: Formulation and Graphical Method

General formulation of Linear Programming Problem
Optimisation Models
Basics of Graphic Method
Important steps to draw graph
Multiple, Unbounded Solution and Infeasible Problems
Solving Linear Programming Graphically Using Computer
Application of Linear Programming in Business and Industry

3 Linear Programming-Simplex Method

Principle of Simplex Method
Computational aspect of Simplex Method
Simplex Method with several Decision Variables
Two Phase and M-method
Multiple Solution, Unbounded Solution and Infeasible Problem
Sensitivity Analysis
Dual Linear Programming Problem

4 Transportation Problem

Basic Feasible Solution of a Transportation Problem
Modified Distribution Method
Stepping Stone Method
Unbalanced Transportation Problem
Degenerate Transportation Problem
Transhipment Problem
Maximisation in a Transportation Problem

5 Assignment Problem

Solution of the Assignment Problem
Unbalanced Assignment Problem
Problem with some Infeasible Assignments
Maximisation in an Assignment Problem
Crew Assignment Problem

6 Application of Excel Solver to Solve LPP

Building Excel model for solving LP: An Illustrative Example

7 Goal Programming

Concepts of goal programming
Goal programming model formulation
Graphical method of goal programming
The simplex method of goal programming
Using Excel Solver to Solve Goal Programming Models
Application areas of goal programming

8 Integer Programming

Some Integer Programming Formulation Techniques
Binary Representation of General Integer Variables
Unimodularity
Cutting Plane Method
Branch and Bound Method
Solver Solution

9 Dynamic Programming

Dynamic Programming Methodology: An Example
Definitions and Notations
Dynamic Programming Applications

10 Non-Linear Programming

Solution of a Non-linear Programming Problem
Convex and Concave Functions
Kuhn-Tucker Conditions for Constrained Optimisation
Quadratic Programming
Separable Programming
NLP Models with Solver

11 Introduction to game theory and its Applications

Important terms in Game Theory
Saddle points
Mixed strategies: Games without saddle points
2 x n games
Exploiting an opponent’s mistakes

12 Monte Carlo Simulation

Reasons for using simulation
Monte Carlo simulation
Limitations of simulation
Steps in the simulation process
Some practical applications of simulation
Two typical examples of hand-computed simulation
Computer simulation

13 Queueing Models

Characteristics of a queueing model
Notations and Symbols
Statistical methods in queueing
The M/M/I System
The M/M/C System
The M/Ek/I System
Decision problems in queueing

[#1]Assignment Problem[Easy Steps to solve - Hungarian Method with Optimal Solution] by kauserwise

Channel | channels.

Kauser Wise

691K subscribers

Hungarian Maximum Matching Algorithm

Already have an account? Log in here.

The Hungarian matching algorithm , also called the Kuhn-Munkres algorithm, is a $O\big(|V|^3\big)$ algorithm that can be used to find maximum-weight matchings in bipartite graphs , which is sometimes called the assignment problem . A bipartite graph can easily be represented by an adjacency matrix , where the weights of edges are the entries. Thinking about the graph in terms of an adjacency matrix is useful for the Hungarian algorithm.

A matching corresponds to a choice of 1s in the adjacency matrix, with at most one 1 in each row and in each column.

The Hungarian algorithm solves the following problem:

In a complete bipartite graph $G$, find the maximum-weight matching. (Recall that a maximum-weight matching is also a perfect matching.)

This can also be adapted to find the minimum-weight matching.

Say you are having a party and you want a musician to perform, a chef to prepare food, and a cleaning service to help clean up after the party. There are three companies that provide each of these three services, but one company can only provide one service at a time (i.e. Company B cannot provide both the cleaners and the chef). You are deciding which company you should purchase each service from in order to minimize the cost of the party. You realize that is an example of the assignment problem, and set out to make a graph out of the following information: $\quad$ Company$\quad$ $\quad$ Cost for Musician$\quad$ $\quad$ Cost for Chef$\quad$ $\quad$ Cost for Cleaners$\quad$ $\quad$ Company A$\quad$ $\quad$ $108$\quad$ $\quad$ $125$\quad$ $\quad$ $150$\quad$ $\quad$ Company B$\quad$ $\quad$ $150$\quad$ $\quad$ $135$\quad$ $\quad$ $175$\quad$ $\quad$ Company C$\quad$ $\quad$ $122$\quad$ $\quad$ $148$\quad$ $\quad$ $250$\quad$ Can you model this table as a graph? What are the nodes? What are the edges? Show Answer The nodes are the companies and the services. The edges are weighted by the price.

What are some ways to solve the problem above? Since the table above can be thought of as a $3 \times 3$ matrix, one could certainly solve this problem using brute force, checking every combination and seeing what yields the lowest price. However, there are $n!$ combinations to check, and for large $n$, this method becomes very inefficient very quickly.

The Hungarian Algorithm Using an Adjacency Matrix

The hungarian algorithm using a graph.

With the cost matrix from the example above in mind, the Hungarian algorithm operates on this key idea: if a number is added to or subtracted from all of the entries of any one row or column of a cost matrix, then an optimal assignment for the resulting cost matrix is also an optimal assignment for the original cost matrix.

The Hungarian Method [1] Subtract the smallest entry in each row from all the other entries in the row. This will make the smallest entry in the row now equal to 0. Subtract the smallest entry in each column from all the other entries in the column. This will make the smallest entry in the column now equal to 0. Draw lines through the row and columns that have the 0 entries such that the fewest lines possible are drawn. If there are $n$ lines drawn, an optimal assignment of zeros is possible and the algorithm is finished. If the number of lines is less than $n$, then the optimal number of zeroes is not yet reached. Go to the next step. Find the smallest entry not covered by any line. Subtract this entry from each row that isn’t crossed out, and then add it to each column that is crossed out. Then, go back to Step 3.

Solve for the optimal solution for the example in the introduction using the Hungarian algorithm described above. Here is the initial adjacency matrix: Subtract the smallest value in each row from the other values in the row: Now, subtract the smallest value in each column from all other values in the column: Draw lines through the row and columns that have the 0 entries such that the fewest possible lines are drawn: There are 2 lines drawn, and 2 is less than 3, so there is not yet the optimal number of zeroes. Find the smallest entry not covered by any line. Subtract this entry from each row that isn’t crossed out, and then add it to each column that is crossed out. Then, go back to Step 3. 2 is the smallest entry. First, subtract from the uncovered rows: Now add to the covered columns: Now go back to step 3, drawing lines through the rows and columns that have 0 entries: There are 3 lines (which is $n$), so we are done. The assignment will be where the 0's are in the matrix such that only one 0 per row and column is part of the assignment. Replace the original values: The Hungarian algorithm tells us that it is cheapest to go with the musician from company C, the chef from company B, and the cleaners from company A. We can verify this by brute force. 108 + 135 + 250 = 493 108 + 148 + 175 = 431 150 + 125 + 250 = 525 150 + 148 + 150 = 448 122 + 125 + 175 = 422 122 + 135 + 150 = 407. We can see that 407 is the lowest price and matches the assignment the Hungarian algorithm determined. $_\square$

The Hungarian algorithm can also be executed by manipulating the weights of the bipartite graph in order to find a stable, maximum (or minimum) weight matching. This can be done by finding a feasible labeling of a graph that is perfectly matched, where a perfect matching is denoted as every vertex having exactly one edge of the matching.

How do we know that this creates a maximum-weight matching?

A feasible labeling on a perfect match returns a maximum-weighted matching. Suppose each edge $e$ in the graph $G$ connects two vertices, and every vertex $v$ is covered exactly once. With this, we have the following inequality: \[w(M’) = \sum_{e\ \epsilon\ E} w(e) \leq \sum_{e\ \epsilon\ E } \big(l(e_x) + l(e_y)\big) = \sum_{v\ \epsilon\ V} l(v),\] where $M’$ is any perfect matching in $G$ created by a random assignment of vertices, and $l(x)$ is a numeric label to node $x$. This means that $\sum_{v\ \epsilon\ V}\ l(v)$ is an upper bound on the cost of any perfect matching. Now let $M$ be a perfect match in $G$, then \[w(M) = \sum_{e\ \epsilon\ E} w(e) = \sum_{v\ \epsilon\ V}\ l(v).\] So $w(M’) \leq w(M)$ and $M$ is optimal. $_\square$

Start the algorithm by assigning any weight to each individual node in order to form a feasible labeling of the graph $G$. This labeling will be improved upon by finding augmenting paths for the assignment until the optimal one is found.

A feasible labeling is a labeling such that

$l(x) + l(y) \geq w(x,y)\ \forall x \in X, y \in Y$, where $X$ is the set of nodes on one side of the bipartite graph, $Y$ is the other set of nodes, $l(x)$ is the label of $x$, etc., and $w(x,y)$ is the weight of the edge between $x$ and $y$.

A simple feasible labeling is just to label a node with the number of the largest weight from an edge going into the node. This is certain to be a feasible labeling because if $A$ is a node connected to $B$, the label of $A$ plus the label of $B$ is greater than or equal to the weight $w(x,y)$ for all $y$ and $x$.

A feasible labeling of nodes, where labels are in red [2] .

Imagine there are four soccer players and each can play a few positions in the field. The team manager has quantified their skill level playing each position to make assignments easier.

How can players be assigned to positions in order to maximize the amount of skill points they provide?

The algorithm starts by labeling all nodes on one side of the graph with the maximum weight. This can be done by finding the maximum-weighted edge and labeling the adjacent node with it. Additionally, match the graph with those edges. If a node has two maximum edges, don’t connect them.

Although Eva is the best suited to play defense, she can't play defense and mid at the same time!

If the matching is perfect, the algorithm is done as there is a perfect matching of maximum weights. Otherwise, there will be two nodes that are not connected to any other node, like Tom and Defense. If this is the case, begin iterating.

Improve the labeling by finding the non-zero label vertex without a match, and try to find the best assignment for it. Formally, the Hungarian matching algorithm can be executed as defined below:

The Hungarian Algorithm for Graphs [3] Given: the labeling $l$, an equality graph $G_l = (V, E_l)$, an initial matching $M$ in $G_l$, and an unmatched vertex $u \in V$ and $u \notin M$ Augmenting the matching A path is augmenting for $M$ in $G_l$ if it alternates between edges in the matching and edges not in the matching, and the first and last vertices are free vertices , or unmatched, in $M$. We will keep track of a candidate augmenting path starting at the vertex $u$. If the algorithm finds an unmatched vertex $v$, add on to the existing augmenting path $p$ by adding the $u$ to $v$ segment. Flip the matching by replacing the edges in $M$ with the edges in the augmenting path that are not in $M$ $($in other words, the edges in $E_l - M).$ Improving the labeling $S \subseteq X$ and $T \subseteq Y,$ where $S$ and $T$ represent the candidate augmenting alternating path between the matching and the edges not in the matching. Let $N_l(S)$ be the neighbors to each node that is in $S$ along edges in $E_l$ such that $N_l(S) = \{v|\forall u \in S: (u,v) \in E_l\}$. If $N_l(S) = T$, then we cannot increase the size of the alternating path (and therefore can't further augment), so we need to improve the labeling. Let $\delta_l$ be the minimum of $l(u) + l(v) - w(u,v)$ over all of the $u \in S$ and $v \notin T$. Improve the labeling $l$ to $l'$: If $r \in S,$ then $l'(r) = l(r) - \delta_l,$ If $r \in T,$ then $l'(r) = l(r) + \delta_l.$ If $r \notin S$ and $r \notin T,$ then $l'(r) = l(r).$ $l'$ is a valid labeling and $E_l \subset E_{l'}.$ Putting it all together: The Hungarian Algorithm Start with some matching $M$, a valid labeling $l$, where $l$ is defined as the labelling $\forall x \in X, y \in Y| l(y) = 0, l(x) = \text{ max}_{y \in Y}(w\big(x, y)\big)$. Do these steps until a perfect matching is found $($when $M$ is perfect$):$ (a) Look for an augmenting path in $M.$ (b) If an augmenting path does not exist, improve the labeling and then go back to step (a).

Each step will increase the size of the matching $M$ or it will increase the size of the set of labeled edges, $E_l$. This means that the process will eventually terminate since there are only so many edges in the graph $G$. [4]

When the process terminates, $M$ will be a perfect matching. By the Kuhn-Munkres theorem , this means that the matching is a maximum-weight matching.

The algorithm defined above can be implemented in the soccer scenario. First, the conflicting node is identified, implying that there is an alternating tree that must be reconfigured.

There is an alternating path between defense, Eva, mid, and Tom.

To find the best appropriate node, find the minimum $\delta_l$, as defined in step 4 above, where $l_u$ is the label for player $u,$ $l_v$ is the label for position $v,$ and $w_{u, v}$ is the weight on that edge.

The $\delta_l$ of each unmatched node is computed, where the minimum is found to be a value of 2, between Tom playing mid $(8 + 0 – 6 = 2).$

The labels are then augmented and the new edges are graphed in the example. Notice that defense and mid went down by 2 points, whereas Eva’s skillset got back two points. However, this is expected as Eva can't play in both positions at once.

Augmenting path leads to relabeling of nodes, which gives rise to the maximum-weighted path.

These new edges complete the perfect matching of the graph, which implies that a maximum-weighted graph has been found and the algorithm can terminate.

The complexity of the algorithm will be analyzed using the graph-based technique as a reference, yet the result is the same as for the matrix-based one.

Algorithm analysis [3] At each $a$ or $b$ step, the algorithm adds one edge to the matching and this happens $O\big(|V|\big)$ times. It takes $O\big(|V|\big)$ time to find the right vertex for the augmenting (if there is one at all), and it is $O\big(|V|\big)$ time to flip the matching. Improving the labeling takes $O\big(|V|\big)$ time to find $\delta_l$ and to update the labelling accordingly. We might have to improve the labeling up to $O\big(|V|\big)$ times if there is no augmenting path. This makes for a total of $O\big(|V|^2\big)$ time. In all, there are $O\big(|V|\big)$ iterations each taking $O\big(|V|\big)$ work, leading to a total running time of $O\big(|V|^3\big)$.

Matching Algorithms
Bruff, D. The Assignment Problem and the Hungarian Method . Retrieved June 26, 2016, from http://www.math.harvard.edu/archive/20_spring_05/handouts/assignment_overheads.pdf
Golin, M. Bipartite Matching &amp; the Hungarian Method . Retrieved Retrieved June 26th, 2016, from http://www.cse.ust.hk/~golin/COMP572/Notes/Matching.pdf
Grinman, A. The Hungarian Algorithm for Weighted Bipartite Graphs . Retrieved June 26, 2016, from http://math.mit.edu/~rpeng/18434/hungarianAlgorithm.pdf
Golin, M. Bipartite Matching & the Hungarian Method . Retrieved June 26, 2016, from http://www.cse.ust.hk/~golin/COMP572/Notes/Matching.pdf

Problem Loading...

Note Loading...

Set Loading...

Assignment Problem: Meaning, Methods and Variations | Operations Research

After reading this article you will learn about:- 1. Meaning of Assignment Problem 2. Definition of Assignment Problem 3. Mathematical Formulation 4. Hungarian Method 5. Variations.

Meaning of Assignment Problem:

An assignment problem is a particular case of transportation problem where the objective is to assign a number of resources to an equal number of activities so as to minimise total cost or maximize total profit of allocation.

The problem of assignment arises because available resources such as men, machines etc. have varying degrees of efficiency for performing different activities, therefore, cost, profit or loss of performing the different activities is different.

Thus, the problem is “How should the assignments be made so as to optimize the given objective”. Some of the problem where the assignment technique may be useful are assignment of workers to machines, salesman to different sales areas.

Definition of Assignment Problem:

ADVERTISEMENTS:

Suppose there are n jobs to be performed and n persons are available for doing these jobs. Assume that each person can do each job at a term, though with varying degree of efficiency, let c ij be the cost if the i-th person is assigned to the j-th job. The problem is to find an assignment (which job should be assigned to which person one on-one basis) So that the total cost of performing all jobs is minimum, problem of this kind are known as assignment problem.

The assignment problem can be stated in the form of n x n cost matrix C real members as given in the following table:

1 $\begingroup$ or.stackexchange.com $\endgroup$ – Rodrigo de Azevedo Commented Oct 31, 2019 at 21:09

A very general argument for the case with multiple optimal solutions applies here.

Consider an $n\times n$ instance of the assignment problem with multiple optimal solutions. If we pick any specific optimal solution $S$ , then decrease the cost of every edge used in $S$ by a small amount $\epsilon$ , then the total cost of $S$ decreases by $n\epsilon$ , whereas the cost of any other optimal solution decreases by $k\epsilon$ for some $k<n$ . So the same solution $S$ is the unique optimal solution for the modified problem.

If you compare what the Hungarian algorithm does in the original problem and the modified problem, then (assuming $\epsilon$ is sufficiently small) there is only one way for the algorithm to make different choices in the two cases: when it's comparing two costs that were equal in the original problem, but where one cost ended up being decreased in the modified problem.

Therefore the Hungarian algorithm, which is guaranteed to find solution $S$ for the modified problem, is also capable of finding $S$ in the original program, if it breaks ties correctly: if, whenever it compares two equal costs, it does whatever it would for the modified problem.

Since $S$ was arbitrary, it follows that for every optimal solution to the original problem, the Hungarian algorithm is capable of finding it, depending on how it breaks ties. This is a sense in which the Hungarian algorithm finds all optimal solutions.

We could actually generate all the optimal solutions by forking into two paths every time that the Hungarian algorithm has a choice between two equally good options. But if there are many ties in the cost matrix, then this could require lots of forks, and might not be efficient.

$\begingroup$ Thank you Misha or your elaboration! So performing the Hungarian method just once will not lead to all optimal solutions. This comes from the fact that the decisions that you make (covering rows and columns with lines to cover all zero elements for instance) lead to some specific optimal solution and hence taking another path in the algorithme could have given you another optimal solution. Is that correct? $\endgroup$ – Ruudferguson Commented Nov 1, 2019 at 8:33
$\begingroup$ Right, that's the idea. In cases with multiple optimal solutions, you'll have some choices about which zero elements to pick as assignments, and how to cover rows/columns. $\endgroup$ – Misha Lavrov Commented Nov 1, 2019 at 13:14
$\begingroup$ Actually, just the covering-zeroes-by-the-fewest-rows-and-columns aspect of things shouldn't be the reason for multiple solutions, since it can happen whether or not there are ties between costs. The reason for multiple outcomes is when you have more than one zero in a row or column to pick as the "assigned" zero. $\endgroup$ – Misha Lavrov Commented Nov 1, 2019 at 13:35
$\begingroup$ So if I understand correctly, you say that once the Hungarian method is performed and an optimal solution (or 0-assignment) is found in some modified matrix, all other optimal solutions (if any) should also be visible in this modified matrix? So it is not possible that there are two optimal assignments for which only one of them is visible in the final step of the Hungarian method and the other assignment does not contain all zero elements? $\endgroup$ – Ruudferguson Commented Nov 1, 2019 at 14:15
$\begingroup$ I'm not saying that either. It might be true, though. $\endgroup$ – Misha Lavrov Commented Nov 2, 2019 at 4:21

You must log in to answer this question.

Not the answer you're looking for browse other questions tagged graph-theory optimization ..

Upcoming Events
2024 Community Moderator Election ends in 5 days
Featured on Meta
We've made changes to our Terms of Service & Privacy Policy - July 2024
Bringing clarity to status tag usage on meta sites
Upcoming Moderator Election
2024 Community Moderator Election

Hot Network Questions

Guitar amplifier placement for live band
Next Bitcoin Core client version
Why would Space Colonies even want to secede?
Can a Statute of Limitations claim be rejected by the court?
How to prevent erasing buffer content when an external program used in filter fails?
Why did evolution fail to protect humans against sun?
Cover letter format in Germany
How to Increase a Length in Inches by a Point without Manual Conversion?
Repeats: Simpler at the cost of more redundant?
What does "hypothecate" mean?
Car LED circuit
Did the Space Shuttle weigh itself before deorbit?
Using relative coordinates with \draw plot
Making wobbly 6x4’ table stable
Can I cast True Strike, then cast Message to give someone else advantage?
What was the reason for not personifying God's spirit in NABRE's translation of John 14:17?
Questions about best way to raise the handlebar on my bike
Clean up verbose code for Oracle SQL
Stargate "instructional" videos
Has the application of a law ever being appealed anywhere due to the lawmakers not knowing what they were voting/ruling?
What connotation does "break out the checkbook" have?
Finding the point that a sequence of midpoints converges to
Non-linear recurrence for rational sequences with generating function with radicals?
What exactly is component based architecture and how do I get it to work?

$assignment problem kauser wise$

Procedure, Example Solved Problem | Operations Research - Solution of assignment problems (Hungarian Method) | 12th Business Maths and Statistics : Chapter 10 : Operations Research

Chapter: 12th business maths and statistics : chapter 10 : operations research.

Solution of assignment problems (Hungarian Method)

First check whether the number of rows is equal to the numbers of columns, if it is so, the assignment problem is said to be balanced.

Step :1 Choose the least element in each row and subtract it from all the elements of that row.

Step :2 Choose the least element in each column and subtract it from all the elements of that column. Step 2 has to be performed from the table obtained in step 1.

Step:3 Check whether there is atleast one zero in each row and each column and make an assignment as follows.

Step :4 If each row and each column contains exactly one assignment, then the solution is optimal.

Example 10.7

Solve the following assignment problem. Cell values represent cost of assigning job A, B, C and D to the machines I, II, III and IV.

Here the number of rows and columns are equal.

∴ The given assignment problem is balanced. Now let us find the solution.

Step 1: Select a smallest element in each row and subtract this from all the elements in its row.

Look for atleast one zero in each row and each column.Otherwise go to step 2.

Step 2: Select the smallest element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.

Step 3 (Assignment):

Thus all the four assignments have been made. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost

Example 10.8

Consider the problem of assigning five jobs to five persons. The assignment costs are given as follows. Determine the optimum assignment schedule.

∴ The given assignment problem is balanced.

Now let us find the solution.

The cost matrix of the given assignment problem is

Column 3 contains no zero. Go to Step 2.

Thus all the five assignments have been made. The Optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = ` 9

Example 10.9

Solve the following assignment problem.

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.

Step 1: is not necessary, since each row contains zero entry. Go to Step 2.

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = ₹ 35

Collectives™ on Stack Overflow

Find centralized, trusted content and collaborate around the technologies you use most.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Get early access and see previews of new features.

Hungarian Algorithm: finding minimum number of lines to cover zeroes?

I am trying to implement the Hungarian Algorithm but I am stuck on the step 5 . Basically, given a n X n matrix of numbers, how can I find minimum number of vertical+horizontal lines such that the zeroes in the matrix are covered?

Before someone marks this question as a duplicate of this , the solution mentioned there is incorrect and someone else also ran into the bug in the code posted there .

I am not looking for code but rather the concept by which I can draw these lines...

EDIT: Please do not post the simple (but wrong) greedy algorithm: Given this input:

I select, column 2 obviously (0-indexed):

Now I can either select row 2 or col 1 both of which have two "remaining" zeroes. If I select col2, I end up with incorrect solution down this path:

The correct solution is using 4 lines:

hungarian-algorithm

1 Check step 3 of this wikipedia article if it helps. – Mohit Jain Commented May 1, 2014 at 2:57
3 I saw the Wikipedia article before posting here and no it does not help. Its too high level when they in step 3 to "Initially assign as many tasks as possible then do the following".... – pathikrit Commented May 1, 2014 at 5:32
To replicate my now-lost comment: there's a nice way to test the output of the Hungarian algorithm (or any other primal-dual algorithm for that matter). Modify the implementation to return the potentials for each row and column. Check that the potentials are valid: each matrix entry is greater than or equal to (for a min-cost matching) the sum of the row potential and the column potential. Check that the entries corresponding to the matching are equal to the sum of row potential and column potential. – David Eisenstat Commented May 2, 2014 at 17:27
What you search for is called the zero-term rank of a matrix. – Anne van Rossum Commented May 3, 2014 at 21:59
1 I imagine that the contents of your two links were cribbed from the Wikipedia article linked from the first comment, which (IMO) isn't very good. – David Eisenstat Commented Apr 28, 2015 at 4:15

8 Answers 8

I have implemented the Hungarian Algorithm in the same steps provided by the link you posted: Hungarian algorithm

Here's the files with comments: Github

Algorithm (Improved greedy) for step 3: (This code is very detailed and good for understanding the concept of choosing line to draw: horizontal vs Vertical. But note that this step code is improved in my code in Github )

Calculate the max number of zeros vertically vs horizontally for each xy position in the input matrix and store the result in a separate array called m2 .
While calculating, if horizontal zeros > vertical zeroes, then the calculated number is converted to negative. (just to distinguish which direction we chose for later use)
Loop through all elements in the m2 array. If the value is positive, draw a vertical line in array m3 , if value is negative, draw an horizontal line in m3

Follow the below example + code to understand more the algorithm:

Create 3 arrays:

m1: First array, holds the input values
m2: Second array, holds maxZeroes(vertical,horizontal) at each x,y position
m3: Third array, holds the final lines (0 index uncovered, 1 index covered)

Create 2 functions:

hvMax(m1,row,col); returns maximum number of zeroes horizontal or vertical. (Positive number means vertical, negative number means horizontal)
clearNeighbours(m2, m3,row,col); void method, it will clear the horizontal neighbors if the value at row col indexes is negative, or clear vertical neighbors if positive. Moreover, it will set the line in the m3 array, by flipping the zero bit to 1.

PS: Your example that you pointed to, will never occur because as you can see the first loop do the calculations by taking the max(horizontal,vertical) and save them in m2. So col1 will not be selected because -3 means draw horizontal line, and -3 was calculated by taking the max between horizontal vs vertical zeros. So at the first iteration at the elements, the program has checked how to draw the lines, on the second iteration, the program draw the lines.

1 @wrick if the zeros Horizontally and vertically are equal, then it doesn't matter in which direction you draw the line. In my code, when both direction have the same number of zeros, the line will be drawn horizontally. You can see this in the above code in method maxVH() return vertical > horizontal ? vertical : horizontal * -1; . In my full implementation in github, I also considered this case to draw a horizontal line. – CMPS Commented May 6, 2014 at 1:47
6 Your implementation is wrong. Just ran it on my system, and does not solve this wikihow.com/Use-the-Hungarian-Algorithm particular example. There is a problem in the line drawing algorithm. – Rohan Sood Commented Aug 29, 2014 at 11:19
1 @RohanSood Correction: I see what you are talking about, and I saw where the problem is. Give me some time to fix it and post it here and on Github. – CMPS Commented Aug 29, 2014 at 15:33
2 This solution still wrong. The line calculation does not work fine. It looks like the recalculation of the lines in every step is wrong. – Bence Végert Commented Jun 4, 2020 at 9:11
1 Line covering logic is incorrect. It doesn't not use minimum number of lines in some cases. – Gurwinder Singh Commented Oct 3, 2020 at 12:28

Greedy algorithms may not work for some cases.

Firstly, it is possible reformulate your problem as following: given a bipartite graph, find a minimum vertex cover. In this problem there are 2n nodes, n for rows and n for columns. There is an edge between two nodes if element at the intersection of corresponding column and row is zero. Vertex cover is a set of nodes (rows and columns) such that each edge is incident to some node from that set (each zero is covered by row or column).

This is a well known problem and can be solved in O(n^3) by finding a maximum matching. Check wikipedia for details

The whole Hungarian algorithm is O(n^3). How can a substep of it be O(N^3) itself? – pathikrit Commented Apr 29, 2015 at 7:19
@wrick Saying that the Hungarian Algorithm finds a maximum matching is sort of misleading; it finds several maximum matchings, computing each from the last more efficiently than starting from scratch. – David Eisenstat Commented Apr 29, 2015 at 14:21
1 O(n^3) is just a bounding. If you have an algorithem that has a step that is O(n), a step that is O(n^3), and then another step that is O(n), your entire algorithm is O(n^3), because n + n + n^3 is less than 3n^3, and 3n^3 is O(n^3) – Psymunn Commented May 1, 2015 at 21:35

There are cases where Amir's code fails.

Consider the following m1:

The best solution is to draw vertical lines in the first two columns.

Amir's code would give the following m2:

And the result would draw the two vertical lines AS WELL AS a line in the first row.

It seems to me the problem is the tie-breaking case:

return vertical > horizontal ? vertical : horizontal * -1;

Because of the way the code is written, the very similar m1 will NOT fail:

Where the first row is moved to the bottom, because the clearing function will clear the -2 values from m2 before those cells are reached. In the first case, the -2 values are hit first, so a horizontal line is drawn through the first row.

I've been working a little through this, and this is what I have. In the case of a tie, do not set any value and do not draw a line through those cells. This covers the case of the matrix I mentioned above, we are done at this step.

Clearly, there are situations where there will remain 0s that are uncovered. Below is another example of a matrix that will fail in Amir's method (m1):

The optimal solution is four lines, e.g. the first four columns.

Amir's method gives m2:

Which will draw lines at the first four rows and the first column (an incorrect solution, giving 5 lines). Again, the tie-breaker case is the issue. We solve this by not setting a value for the ties, and iterating the procedure.

If we ignore the ties we get an m2:

This leads to covering only the first row and the first column. We then take out the 0s that are covered to give new m1:

Then we keep repeating the procedure (ignoring ties) until we reach a solution. Repeat for a new m2:

Which leads to two vertical lines through the second and third columns. All 0s are now covered, needing only four lines (this is an alternative to lining the first four columns). The above matrix only needs 2 iterations, and I imagine most of these cases will need only two iterations unless there are sets of ties nested within sets of ties. I tried to come up with one, but it became difficult to manage.

Sadly, this is not good enough, because there will be cases that will remain tied forever. Particularly, in cases where there is a 'disjoint set of tied cells'. Not sure how else to describe this except to draw the following two examples:

The upper-left 3x3 sub-matrices in these two examples are identical to my original example, I have added 1 or 2 rows/cols to that example at the bottom and right. The only newly added zeros are where the new rows and columns cross. Describing for clarity.

With the iterative method I described, these matrices will be caught in an infinite loop. The zeros will always remain tied (col-count vs row-count). At this point, it does make sense to just arbitrarily choose a direction in the case of a tie, at least from what I can imagine.

The only issue I'm running into is setting up the stopping criteria for the loop. I can't assume that 2 iterations is enough (or any n), but I also can't figure out how to detect if a matrix has only infinite loops left within it. I'm still not sure how to describe these disjoint-tied-sets computationally.

Here is the code to do what I have come up with so far (in MATLAB script):

It appears that if label ties as vertical (positive) the entire problem goes away? – nigelhenry Commented Jan 11, 2020 at 22:36

Step 5: The drawing of line in the matrix is evaluated diagonally with a maximum evaluations of the length of the matrix.

Based on http://www.wikihow.com/Use-the-Hungarian-Algorithm with Steps 1 - 8 only.

Run code snippet and see results in console

Console Output

JSFiddle: http://jsfiddle.net/jjcosare/6Lpz5gt9/2/

// http://www.wikihow.com/Use-the-Hungarian-Algorithm var inputMatrix = [ [0, 1, 0, 1, 1], [1, 1, 0, 1, 1], [1, 0, 0, 0, 1], [1, 1, 0, 1, 1], [1, 0, 0, 1, 0] ]; //var inputMatrix = [ // [10, 19, 8, 15], // [10, 18, 7, 17], // [13, 16, 9, 14], // [12, 19, 8, 18], // [14, 17, 10, 19] // ]; var matrix = inputMatrix; var HungarianAlgorithm = {}; HungarianAlgorithm.step1 = function(stepNumber) { console.log("Step " + stepNumber + ": Matrix"); var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { var sb = ""; for (var j = 0; j < matrix[i].length; j++) { currentNumber = matrix[i][j]; sb += currentNumber + " "; } console.log(sb); } } HungarianAlgorithm.step2 = function() { var largestNumberInMatrix = getLargestNumberInMatrix(matrix); var rowLength = matrix.length; var columnLength = matrix[0].length; var dummyMatrixToAdd = 0; var isAddColumn = rowLength > columnLength; var isAddRow = columnLength > rowLength; if (isAddColumn) { dummyMatrixToAdd = rowLength - columnLength; for (var i = 0; i < rowLength; i++) { for (var j = columnLength; j < (columnLength + dummyMatrixToAdd); j++) { matrix[i][j] = largestNumberInMatrix; } } } else if (isAddRow) { dummyMatrixToAdd = columnLength - rowLength; for (var i = rowLength; i < (rowLength + dummyMatrixToAdd); i++) { matrix[i] = []; for (var j = 0; j < columnLength; j++) { matrix[i][j] = largestNumberInMatrix; } } } HungarianAlgorithm.step1(2); console.log("Largest number in matrix: " + largestNumberInMatrix); function getLargestNumberInMatrix(matrix) { var largestNumberInMatrix = 0; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { for (var j = 0; j < matrix[i].length; j++) { currentNumber = matrix[i][j]; largestNumberInMatrix = (largestNumberInMatrix > currentNumber) ? largestNumberInMatrix : currentNumber; } } return largestNumberInMatrix; } } HungarianAlgorithm.step3 = function() { var smallestNumberInRow = 0; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { smallestNumberInRow = getSmallestNumberInRow(matrix, i); console.log("Smallest number in row[" + i + "]: " + smallestNumberInRow); for (var j = 0; j < matrix[i].length; j++) { currentNumber = matrix[i][j]; matrix[i][j] = currentNumber - smallestNumberInRow; } } HungarianAlgorithm.step1(3); function getSmallestNumberInRow(matrix, rowIndex) { var smallestNumberInRow = matrix[rowIndex][0]; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { currentNumber = matrix[rowIndex][i]; smallestNumberInRow = (smallestNumberInRow < currentNumber) ? smallestNumberInRow : currentNumber; } return smallestNumberInRow; } } HungarianAlgorithm.step4 = function() { var smallestNumberInColumn = 0; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { smallestNumberInColumn = getSmallestNumberInColumn(matrix, i); console.log("Smallest number in column[" + i + "]: " + smallestNumberInColumn); for (var j = 0; j < matrix[i].length; j++) { currentNumber = matrix[j][i]; matrix[j][i] = currentNumber - smallestNumberInColumn; } } HungarianAlgorithm.step1(4); function getSmallestNumberInColumn(matrix, columnIndex) { var smallestNumberInColumn = matrix[0][columnIndex]; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { currentNumber = matrix[i][columnIndex]; smallestNumberInColumn = (smallestNumberInColumn < currentNumber) ? smallestNumberInColumn : currentNumber; } return smallestNumberInColumn; } } var rowLine = {}; var columnLine = {}; HungarianAlgorithm.step5 = function() { var zeroNumberCountRow = 0; var zeroNumberCountColumn = 0; rowLine = {}; columnLine = {}; for (var i = 0; i < matrix.length; i++) { zeroNumberCountRow = getZeroNumberCountInRow(matrix, i); zeroNumberCountColumn = getZeroNumberCountInColumn(matrix, i); if (zeroNumberCountRow > zeroNumberCountColumn) { rowLine[i] = i; if (zeroNumberCountColumn > 1) { columnLine[i] = i; } } else if (zeroNumberCountRow < zeroNumberCountColumn) { columnLine[i] = i; if (zeroNumberCountRow > 1) { rowLine[i] = i; } } else { if ((zeroNumberCountRow + zeroNumberCountColumn) > 2) { rowLine[i] = i; columnLine[i] = i; } } } var zeroCount = 0; for (var i in columnLine) { zeroCount = getZeroNumberCountInColumnLine(matrix, columnLine[i], rowLine); if (zeroCount == 0) { delete columnLine[i]; } } for (var i in rowLine) { zeroCount = getZeroNumberCountInRowLine(matrix, rowLine[i], columnLine); if (zeroCount == 0) { delete rowLine[i]; } } console.log("horizontal line (row): " + JSON.stringify(rowLine)); console.log("vertical line (column): " + JSON.stringify(columnLine)); HungarianAlgorithm.step1(5); //if ((Object.keys(rowLine).length + Object.keys(columnLine).length) == matrix.length) { // TODO: // HungarianAlgorithm.step9(); //} else { // HungarianAlgorithm.step6(); // HungarianAlgorithm.step7(); // HungarianAlgorithm.step8(); //} function getZeroNumberCountInColumnLine(matrix, columnIndex, rowLine) { var zeroNumberCount = 0; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { currentNumber = matrix[i][columnIndex]; if (currentNumber == 0 && !(rowLine[i] == i)) { zeroNumberCount++ } } return zeroNumberCount; } function getZeroNumberCountInRowLine(matrix, rowIndex, columnLine) { var zeroNumberCount = 0; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { currentNumber = matrix[rowIndex][i]; if (currentNumber == 0 && !(columnLine[i] == i)) { zeroNumberCount++ } } return zeroNumberCount; } function getZeroNumberCountInColumn(matrix, columnIndex) { var zeroNumberCount = 0; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { currentNumber = matrix[i][columnIndex]; if (currentNumber == 0) { zeroNumberCount++ } } return zeroNumberCount; } function getZeroNumberCountInRow(matrix, rowIndex) { var zeroNumberCount = 0; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { currentNumber = matrix[rowIndex][i]; if (currentNumber == 0) { zeroNumberCount++ } } return zeroNumberCount; } } HungarianAlgorithm.step6 = function() { var smallestNumberInUncoveredMatrix = getSmallestNumberInUncoveredMatrix(matrix, rowLine, columnLine); console.log("Smallest number in uncovered matrix: " + smallestNumberInUncoveredMatrix); var columnIndex = 0; for (var i in columnLine) { columnIndex = columnLine[i]; for (var i = 0; i < matrix.length; i++) { currentNumber = matrix[i][columnIndex]; //matrix[i][columnIndex] = currentNumber + smallestNumberInUncoveredMatrix; matrix[i][columnIndex] = "x"; } } var rowIndex = 0; for (var i in rowLine) { rowIndex = rowLine[i]; for (var i = 0; i < matrix.length; i++) { currentNumber = matrix[rowIndex][i]; //matrix[rowIndex][i] = currentNumber + smallestNumberInUncoveredMatrix; matrix[rowIndex][i] = "x"; } } HungarianAlgorithm.step1(6); function getSmallestNumberInUncoveredMatrix(matrix, rowLine, columnLine) { var smallestNumberInUncoveredMatrix = null;; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { if (rowLine[i]) { continue; } for (var j = 0; j < matrix[i].length; j++) { if (columnLine[j]) { continue; } currentNumber = matrix[i][j]; if (!smallestNumberInUncoveredMatrix) { smallestNumberInUncoveredMatrix = currentNumber; } smallestNumberInUncoveredMatrix = (smallestNumberInUncoveredMatrix < currentNumber) ? smallestNumberInUncoveredMatrix : currentNumber; } } return smallestNumberInUncoveredMatrix; } } HungarianAlgorithm.step7 = function() { var smallestNumberInMatrix = getSmallestNumberInMatrix(matrix); console.log("Smallest number in matrix: " + smallestNumberInMatrix); var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { for (var j = 0; j < matrix[i].length; j++) { currentNumber = matrix[j][i]; matrix[j][i] = currentNumber - smallestNumberInMatrix; } } HungarianAlgorithm.step1(7); function getSmallestNumberInMatrix(matrix) { var smallestNumberInMatrix = matrix[0][0]; var currentNumber = 0; for (var i = 0; i < matrix.length; i++) { for (var j = 0; j < matrix[i].length; j++) { currentNumber = matrix[i][j]; smallestNumberInMatrix = (smallestNumberInMatrix < currentNumber) ? smallestNumberInMatrix : currentNumber; } } return smallestNumberInMatrix; } } HungarianAlgorithm.step8 = function() { console.log("Step 8: Covering zeroes with Step 5 - 8 until Step 9 is reached"); HungarianAlgorithm.step5(); } HungarianAlgorithm.step9 = function(){ console.log("Step 9..."); } HungarianAlgorithm.step1(1); HungarianAlgorithm.step2(); HungarianAlgorithm.step3(); HungarianAlgorithm.step4(); HungarianAlgorithm.step5(); HungarianAlgorithm.step6();

Although your code is very clear to read, can you still explain in English, what you did for step 5? – pathikrit Commented Apr 30, 2015 at 3:28
Evaluated the lines diagonally in the matrix and kept record on both rowLine and columnLine separately. After the initial evaluation there will be extra lines in both, removed the extra line by evaluating it against its opposite line. For columnLine, we need to check if the rowLine has already covered all its zero, if yes, we delete that columLine and vice versa. – jjcosare Commented Apr 30, 2015 at 4:30
This algorithm fails for the identity matrix or any element on the diagonal that is the only element in the corresponding row and column. – George Ogden Commented Feb 24, 2022 at 12:11

Do the assignment using the steps mentioned below:

assign a row if it has only one 0, else skip the row temporarily
cross out the 0's in the assigned column
Do the same for every column

After doing the assignment using the above steps, follow the steps below to get the minimum number of lines which cover all the 0's

step 1 - Tick an unassigned row
step 2 - If a ticked row has a 0, then tick the corresponding column
step 3 - If a ticked column has an assignment, then tick the corresponding row
step 4 - Repeat steps 2 and 3, till no more ticking is possible
step 5 - Draw lines through un-ticked rows and ticked columns

For your case: (0-indexing for rows and columns)

skip row 0, as it has two 0's
assign row 1, and cross out all the 0's in column 2
skip row 2, as it has two uncrossed 0's
skip row 3, as it has no uncrossed 0
skip row 4, as it has 2 uncrossed 0's
assign column 0
skip column 1 as it has two uncrossed 0's (in row-2 and row-4)
skip column 2, as it has an already assigned 0
assign column 3,and cross out the 0 in row 2
assign column 4, and cross out the 0 in row 4
assigned 0's are shown by '_' and 'x' shows crossed out 0's

( _ 1 x 1 1 ), ( 1 1 _ 1 1 ), ( 1 x x _ 1 ), ( 1 1 x 1 1 ), ( 1 x x 1 _ )

The matrix looks like the one shown above after doing the assignments

Now follow the 5 steps mentioned above to get the minimum number of lines that cover all the 0's

Tick row 3 as it is not assigned yet
Since row 3 has a 0 in column 2, tick column 2
Since column 2 has an assignment in row 1, tick row 1
Now draw lines through un-ticked rows (i.e. row 0,2,4) and ticked columns(i.e. column 2)
These 4 lines will cover all the 0's

Hope this helps:)

PS : For cases where no initial assignment is possible due to multiple 0's in each row and column, this could be handled by taking one arbitrary assignment (For the cases where multiple 0's are present in each row and column, it is very likely that more than one possible assignment would result in an optimal solution)

1 What if there is no row with only one 0? – tuket Commented Mar 14, 2020 at 23:35
1 Can can't find the solution with the following matrix ((0, 2, 2, 2), (0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 0, 2)). Thanks for your answer – tuket Commented Mar 15, 2020 at 14:49
1 I'm viewing several pages in internet about the Hungarian method, but none of them clearly states if - and this is quite frequent - I can choose whatever combination of covered lines that cover the zeros in the matrix with minimum number of lines. So can you @Yash Plvirwal, tell me if really I can choose any such combination? – Caco Commented Mar 26, 2020 at 12:15

@CMPS answer fails on quite a few graphs. I think I have implemented a solution which solves the problem.

I followed the Wikipedia article on the Hungarian algorithm and I made an implementation that seems to work all the time. From Wikipedia, here is a the method to draw the minimum number of lines:

First, assign as many tasks as possible. Mark all rows having no assignments. Mark all (unmarked) columns having zeros in newly marked row(s). Mark all rows having assignments in newly marked columns. Repeat for all non-assigned rows.

Here is my Ruby implementation:

Best Option

I found a performant and optimized algorithm. I specifically created a Github repository about this. You can find Info, Codes, Documentations, Benchmarks and there. I implemented it in C++, JavaScript and going to implement other versions.

It's rather complex yet. but I can elaborate a little.

As I said in readme of the repository:

This algorithm doesn't have anything to do about Hungarian Algorithm . It's just an algorithm to find the minimum amount of lines in a 2-Dimensional matrix!

Loop through input matrix and count the zeros and collect necessary information about it.

The power matrix is a helper which indicates the number of zeros in each column or row.
The line matrix is a helper which indicates where lines are and where not.
The crosses matrix is a helper which contains every cross lines on a specific line.
The sides is a pair of numbers that indicates the count of rows and columns which contain a zero!

In the next step we loop through matrix but from both directions (horizontal and vertical). Using a for main axis can be 0: row and 1:column .

first we need to find two weakest cross lines. Weak means containing less zeros!

Now we have to choose!

If the weakest cross power was 1 or sum of weakest pair cross is less than current line: Draw Current Line
If power of current line is 1 so probably the cross line is better: Draw Cross Line
If we are in second axis which means we don't get back and should not leave any zeros without line:

Draw the line in direction which less lines are needed!

In every step just check and end the progress if:

The lines count reach the shortest direction power!
Any of sides shrinks to zero!

Notice: When a column crosses a row on a zero the power would be at least 1 because of the zero in the cross!

Note: matrix is the variable which contains your matrix!

I tested the algorithm against an optimized brute force and find amazing results. You can check benchmark results in repository .

You can find a step-by-step algorithm, complete with proof that it works, in Munkres's paper Algorithms for the assignment and transportation problems . In particular look at page 33, steps 1-3. Each zero of the matrix can be starred or primed, as well as being covered by a horizontal or vertical line.

To initialize, go through each zero in some order. If a zero has no other starred zero in its row or column, star it. In your example, going through the zeroes in the natural order we get

(0*, 1, 0, 1, 1) (1, 1, 0*, 1, 1) (1, 0*, 0, 0, 1) (1, 1, 0, 1, 1) (1, 0, 0, 1, 0*)

Then cover every column containing a starred zero. This will cover columns 1,2,3 and 5 in the above matrix (indexing from 1). Next comes the main loop.

Step 1: Choose a non-covered zero and prime it. If there is no starred zero in its row, go to step 2. Otherwise, let Z be the starred zero in its row. Cover the row and uncover the column of Z. [In the above example, this would remove the vertical cover of the second row and add a horizontal cover to the fourth row.] Repeat for all uncovered zeroes. [In our example, we are done. The algorithm produced vertical lines covering columns 1,3,5 and a horizontal line covering row 3.]
Step 2: This is the most complicated step, but it's actually not that hard. Let Z0 be the unique uncovered primed zero. There is a sequence of starred and primed zeroes (possibly with only one element) starting from Z0 defined as follows. Z1 is a starred 0 in Z0's column (if there isn't one, the sequence stops here.) Then Z2 is a primed 0 in Z1's row, etc. Continue until the sequence stops at a primed zero Z_2k. Now unstar each starred zero of this sequence, and star each primed zero of the sequence. Erase all the primes, uncover every row, and cover every column containing a starred zero. If all columns are covered, we are done. Otherwise return to step 1.

Your Answer

Reminder: Answers generated by artificial intelligence tools are not allowed on Stack Overflow. Learn more

Sign up or log in

Post as a guest.

Required, but never shown

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .

Not the answer you're looking for? Browse other questions tagged algorithm matrix matching greedy hungarian-algorithm or ask your own question .

The Overflow Blog
Scaling systems to manage all the metadata ABOUT the data
Navigating cities of code with Norris Numbers
Featured on Meta
We've made changes to our Terms of Service & Privacy Policy - July 2024
Bringing clarity to status tag usage on meta sites
Feedback requested: How do you use tag hover descriptions for curating and do...

Hot Network Questions

Is there any point "clean-installing" on a brand-new MacBook?
Many and Many of - a subtle difference in meaning?
What is a word/phrase that best describes a "blatant disregard or neglect" for something, but with the connotation of that they should have known?
How predictable are the voting records of members of the US legislative branch?
Repeats: Simpler at the cost of more redundant?
Can I cast True Strike, then cast Message to give someone else advantage?
Is an invalid date considered the same as a NULL value?
Guitar amplifier placement for live band
What is the purpose of toroidal magnetic field in tokamak fusion device?
Violation of the Law of Total Expectation in the Unit Square?
If there is no free will, doesn't that provide a framework for an ethical model?
Is there a French noun equivalent for "twofer"?
Using relative coordinates with \draw plot
On Schengen Visa - Venice to Zagreb - by bus
Why was I was allowed to bring 1.5 liters of liquid through security at Frankfurt Airport?
Does the First Amendment protect deliberately publicizing the incorrect date for an election?
What does "hypothecate" mean?
Cover letter format in Germany
How do you "stealth" a relativistic superweapon?
Has the application of a law ever being appealed anywhere due to the lawmakers not knowing what they were voting/ruling?
How did Jason Bourne know the garbage man isn't CIA?
Erase the loops
Power line crossing data lines via the ground plane
What's the airplane with the smallest wingspan to fuselage ratio?

Algorithms: The Assignment Problem

One of the interesting things about studying optimization is that the techniques show up in a lot of different areas. The “assignment problem” is one that can be solved using simple techniques, at least for small problem sizes, and is easy to see how it could be applied to the real world.

Assignment Problem

Pretend for a moment that you are writing software for a famous ride sharing application. In a crowded environment, you might have multiple prospective customers that are requesting service at the same time, and nearby you have multiple drivers that can take them where they need to go. You want to assign the drivers to the customers in a way that minimizes customer wait time (so you keep the customers happy) and driver empty time (so you keep the drivers happy).

The assignment problem is designed for exactly this purpose. We start with m agents and n tasks. We make the rule that every agent has to be assigned to a task. For each agent-task pair, we figure out a cost associated to have that agent perform that task. We then figure out which assignment of agents to tasks minimizes the total cost.

Of course, it may be true that m != n , but that’s OK. If there are too many tasks, we can make up a “dummy” agent that is more expensive than any of the others. This will ensure that the least desirable task will be left to the dummy agent, and we can remove that from the solution. Or, if there are too many agents, we can make up a “dummy” task that is free for any agent. This will ensure that the agent with the highest true cost will get the dummy task, and will be idle.

If that last paragraph was a little dense, don’t worry; there’s an example coming that will help show how it works.

There are special algorithms for solving assignment problems, but one thing that’s nice about them is that a general-purpose solver can handle them too. Below is an example, but first it will help to cover a few concepts that we’ll be using.

Optimization Problems

Up above, we talked about making “rules” and minimizing costs. The usual name for this is optimization. An optimization problem is one where we have an “objective function” (which tells us what our goals are) and one or more “constraint functions” (which tell us what the rules are). The classic example is a factory that can make both “widgets” and “gadgets”. Each “widget” and “gadget” earns a certain amount of profit, but it also uses up raw material and time on the factory’s machines. The optimization problem is to determine exactly how many “widgets” and how many “gadgets” to make to maximize profit (the objective) while fitting within the material and time available (the constraints).

If we were to write this simple optimization problem out, it might look like this:

In this case, we have two variables: g for the number of gadgets we make and w for the number of widgets we make. We also have three constraints that we have to meet. Note that they are inequalities; we might not use all the available material or time in our optimal solution.

Just to unpack this a little: in English, the above is saying that we make 45 dollars / euros / quatloos per gadget we make. However, to make a gadget needs 120 lbs of raw material 1, 80 lbs of raw material 2, and 3.8 hours of machine time. So there is a limit on how many gadgets we can make, and it might be a better use of resources to balance gadgets with widgets.

Of course, real optimization problems have many more than two variables and many constraint functions, making them much harder to solve. The easiest kind of optimization problem to solve is linear, and fortunately, the assignment problem is linear.

Linear Programming

A linear program is a kind of optimization problem where both the objective function and the constraint functions are linear. (OK, that definition was a little self-referential.) We can have as many variables as we want, and as many constraint functions as we want, but none of the variables can have exponents in any of the functions. This limitation allows us to apply very efficient mathematical approaches to solve the problem, even for very large problems.

We can state the assignment problem as a linear programming problem. First, we choose to make “i” represent each of our agents (drivers) and “j” to represent each of our tasks (customers). Now, to write a problem like this, we need variables. The best approach is to use “indicator” variables, where xij = 1 means “driver i picks up customer j” and xij = 0 means “driver i does not pick up customer j”.

We wind up with:

This is a compact mathematical way to describe the problem, so again let me put it in English.

First, we need to figure out the cost of having each driver pick up each customer. Then, we can calculate the total cost for any scenario by just adding up the costs for the assignments we pick. For any assignment we don’t pick, xij will equal zero, so that term will just drop out of the sum.

Of course, the way we set up the objective function, the cheapest solution is for no drivers to pick up any customers. That’s not a very good business model. So we need a constraint to show that we want to have a driver assigned to every customer. At the same time, we can’t have a driver assigned to mutiple customers. So we need a constraint for that too. That leads us to the two constraints in the problem. The first just says, if you add up all the assignments for a given driver, you want the total number of assignments for that driver to be exactly one. The second constraint says, if you add up all the assignments to a given customer, you want the total number of drivers assigned to the customer to be one. If you have both of these, then each driver is assigned to exactly one customer, and the customers and drivers are happy. If you do it in a way that minimizes costs, then the business is happy too.

Solving with Octave and GLPK

The GNU Linear Programming Kit is a library that solves exactly these kinds of problems. It’s easy to set up the objective and constraints using GNU Octave and pass these over to GLPK for a solution.

Given some made-up sample data, the program looks like this:

Start with the definition of “c”, the cost information. For this example, I chose to have four drivers and three customers. There are sixteen numbers there; the first four are the cost of each driver to get the first customer, the next four are for the second customer, and the next four are for the third customer. Because we have an extra driver, we add a “dummy” customer at the end that is zero cost. This represents one of the drivers being idle.

The next definition is “b”, the right-hand side of our constraints. There are eight constraints, one for each of the drivers, and one for each of the customers (including the dummy). For each constraint, the right-hand side is 1.

The big block in the middle defines our constraint matrix “a”. This is the most challenging part of taking the mathematical definition and putting it into a form that is usable by GLPK; we have to expand out each constraint. Fortunately, in these kinds of cases, we tend to get pretty patterns that help us know we’re on the right track.

The first line in “a” says that the first customer needs a driver. To see why, remember that in our cost information, the first four numbers are the cost for each driver to get the first customer. With this constraint, we are requiring that one of those four costs be included and therefore that a driver is “selected” for the first customer. The other lines in “a” work similarly; the last four ensure that each driver has an assignment.

Note that the number of rows in “a” matches the number of items in “b”, and the number of columns in “a” matches the number of items in “c”. This is important; GLPK won’t run if this is not true (and our problem isn’t stated right in any case).

Compared to the above, the last few lines are easy.

“lb” gives the lower bound for each variable.
“ub” gives the upper bound.
“ctype” tells GLPK that each constraint is an equality (“strict” as opposed to providing a lower or upper bound).
“vartype” tells GLPK that these variables are all integers (can’t have half a driver showing up).
“s” tells GLPK that we want to minimize our costs, not maximize them.

We push all that through a function call to GLPK, and what comes back are two values (along with some other stuff I’ll exclude for clarity):

The first item tells us that our best solution takes 27 minutes, or dollars, or whatever unit we used for cost. The second item tells us the assignments we got. (Note for pedants: I transposed this output to save space.)

This output tells us that customer 1 gets driver 2, customer 2 gets driver 3, customer 3 gets driver 4, and driver 1 is idle. If you look back at the cost data, you can see this makes sense, because driver 1 had some of the most expensive times to the three customers. You can also see that it managed to pick the least expensive pairing for each customer. (Of course, if I had done a better job making up cost data, it might not have picked the least expensive pairing in all cases, because a suboptimal individual pairing might still lead to an overall optimal solution. But this is a toy example.)

Of course, for a real application, we would have to take into consideration many other factors, such as the passage of time. Rather than knowing all of our customers and drivers up front, we would have customers and drivers continually showing up and being assigned. But I hope this simple example has revealed some of the concepts behind optimization and linear programming and the kinds of real-world problems that can be solved.

No category

Transportation Model (Quick Notes)

Add this document to collection(s)

You can add this document to your study collection(s)

Add this document to saved

You can add this document to your saved list

Suggest us how to improve StudyLib

(For complaints, use another form )

Input it if you want to receive answer

Question asked by Filo student

Some problems in A maths , assignment tommorow pls hurry

Views: 5,760 students

Updated on: Aug 14, 2024

Video solutions ( 1 )

Learn from their 1-to-1 discussion with Filo tutors.

Uploaded on: 8/14/2024

Connect instantly with this tutor

Connect now

Total classes on Filo by this tutor - 36,211

Teaches : Mathematics

Notes from this class ( 1 pages)

Students who ask this question also asked

Mathematics

View solution

Stuck on the question or explanation?

Connect with our Mathematics tutors online and get step by step solution of this question.

Question Text
Updated On	Aug 14, 2024
Topic	All topics
Subject	Mathematics
Class	Class 10
Answer Type	Video solution:
Upvotes	65
Avg. Video Duration	6

Are you ready to take control of your learning?

Download Filo and start learning with your favourite tutors right away!

SI SWIMSUIT
SI SPORTSBOOK

Watch: Padres Manager Slams Reporter For Asking a Painfully Obvious Question

Maren angus-coombs | aug 15, 2024.

San Diego Padres

It began with a series of questions from Kevin Acee of the San Diego Union-Tribune regarding the usage of the San Diego Padres' bullpen in recent games.

Manager Mike Shildt answered the questions in depth until it became too ridiculous for him. Acee followed up with a question about the offense not providing a bigger cushion.

"How much would it help to have a five-run lead?" asked Acee.

For all the aspiring sports journalists out there, there is such thing as a dumb question. This was one of them.

Shildt looked at Acee in disbelief saying, "Of course. I don't know what you want me to say. Of course it would help."

Acee continued to press Shildt on how he uses his bullpen and Shildt just let the Padres beat reporter have it.

"I'm pitching them according to my experience and communication with them, if you have exception with that ... I don't come here to argue. I've already answered your question very clearly for a minute. I don't know what more you want me to tell you that you're not hearing from me. I really don't."

#ICYMI , Kevin Acee ( @sdutKevinAcee ) getting schooled by Padres' manager Mike Shildt. https://t.co/EU6g6fwCdH pic.twitter.com/Aeo4KOmnR9 — Mickey Koke (@mickeykoke) August 14, 2024

The game that sparked this postgame exchange was Tuesday night, when the Padres defeated the Pittsburgh Pirates 3-0.

Michael King threw six strong innings and the Padres clinched their eighth straight series win heading into the series finale on Wednesday that they also went on to win.

Reliever Jason Adam and Tanner Scott stranded runners at third base in the seventh and eighth innings before Robert Suarez pitched a scoreless ninth for his 27th save.

Padres third baseman Manny Machado heard the criticism of the Padres not scoring more runs but they were still winning games.

“It’s been 50-50,” Machado said. “You’re just talking about one week worth of baseball. It’s been tight. I mean, baseball’s up and down. Honestly in that week we’ve been winning games. That’s all that really matters at the end of the day. Two weeks ago, we were scoring runs and pitching wasn’t giving up any runs. Now it’s flipped. Now it’s been we’ve been hitting and we’ve been giving up some runs and it’s been tight games. …

“It’s the beauty of baseball. Every week it’s something different. You just gotta try to come out on the opposite side of winning ballgames and that’s what we’ve been doing.”

The Padres won, the bullpen feels great and the exchange between Acee and Shildt was completely unnecessary. Perhaps, it can even be considered a lesson for journalism teachers.

MAREN ANGUS-COOMBS

Maren Angus-Coombs was born in Los Angeles and raised in Nashville, Tenn. She is a graduate of Middle Tennessee State University and has been a sports writer since 2008. Despite being raised in the South, her sports obsession has always been in Los Angeles. She is currently a staff writer for the LA Sports Report Network.

DSA with JS - Self Paced
JS Tutorial
JS Exercise
JS Interview Questions
JS Operator
JS Projects
JS Examples
JS Free JS Course
JS A to Z Guide
JS Formatter

How to fix SyntaxError – ‘invalid assignment left-hand side in JavaScript’?

In JavaScript, a SyntaxError : Invalid Assignment Left-Hand Side occurs when the interpreter encounters an invalid expression or structure on the left side of an assignment operator ( = ).

This error typically arises when trying to assign a value to something that cannot be assigned, such as literals, expressions, or the result of function calls. Let’s explore this error in more detail and see how to resolve it with some examples.

Understanding an error

An invalid assignment left-hand side error occurs when the syntax of the assignment statement violates JavaScript’s rules for valid left-hand side expressions. The left-hand side should be a target that can receive an assignment like a variable, object property or any array element, let’s see some cases where this error can occur and also how to resolve this error.

Case 1: Error Cause: Assigning to Literals

When you attempt to assign a value to a literal like a number, string or boolean it will result in SyntaxError: Invalid Assignment Left-Hand Side .

Resolution of error

In this case values should be assigned to variables or expressions which will be on the left side of an equation and avoid assigning values directly to literals.

Case 2: Error Cause: Assigning to Function Calls

Assigning a value directly to the result of function call will give an invalid assignment left-hand side error.

Explanation : In this example, getX() returns a value but is not a valid target for assignment. Assignments should be made to variables, object properties, or array elements, not to the result of a function call.

Therefore, store it into a variable or at least place it on the left-hand side that is valid for assignment.

Please Login to comment...

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

IMAGES

TRICK to SOLVE
[#1] LPP
Lecture 19 Assignment problem : Unbalanced and maximal Assignment Problems
Easy Methods of Assignment Problems, Operations Research
ASSIGNMENT PROBLEM
Assignment problem ppt

COMMENTS

[#1]Assignment Problem[Easy Steps to solve
Here is the video about assignment problem - Hungarian method with algorithm.NOTE: After row and column scanning, If you stuck with more than one zero in th...
Unbalanced Assignment Problem
Here is the video about unbalanced Assignment problem using Hungarian method,In this video we have seen how to solve unbalanced assignment problem using step...
Kauser Wise
Dr. J. Kauser, M.Com, Mphil, MBA (Finance), Ph.D, (NET & JRF) Welcome to "Kauserwise" Channel, you're in the right place for mastering the fundamentals and advanced concepts of Financial ...
[2] Travelling salesman Problem
Here is the video for travelling salesman problem, phase 2. In that we have seen how to modify the solution by inspection method from assignment problem and ...
How to Solve the Assignment Problem: A Complete Guide
Step 1: Set up the cost matrix. The first step in solving the assignment problem is to set up the cost matrix, which represents the cost of assigning a task to an agent. The matrix should be square and have the same number of rows and columns as the number of tasks and agents, respectively.
[#1]Assignment Problem[Easy Steps to solve
Here is the video about assignment problem - Hungarian method with algorithm. NOTE: After row and column scanning, If you stuck with more than one zero in the matrix, please do the row scanning and column scanning (REPEATEDLY) as much as possible to cover that zeros with lines as per algorithm, If you still find some zeros without covered by lines, then we need to go for [DIAGONAL selection ...
Hungarian Maximum Matching Algorithm
The Hungarian matching algorithm, also called the Kuhn-Munkres algorithm, is a $O\big(|V|^3\big)$ algorithm that can be used to find maximum-weight matchings in bipartite graphs, which is sometimes called the assignment problem.A bipartite graph can easily be represented by an adjacency matrix, where the weights of edges are the entries.Thinking about the graph in terms of an adjacency ...
Assignment problem
The formal definition of the assignment problem (or linear assignment problem) is . Given two sets, A and T, together with a weight function C : A × T → R.Find a bijection f : A → T such that the cost function: (, ())is minimized. Usually the weight function is viewed as a square real-valued matrix C, so that the cost function is written down as: , The problem is "linear" because the cost ...
PDF Unit 4: ASSIGNMENT PROBLEM
Problem 4. Job shop needs to assign 4 jobs to 4 workers. The cost of performing a job is a function of the skills of the workers. Table summarizes the cost of the assignments. Worker1 cannot do job3, and worker 3 cannot do job 4. Determine the optimal assignment using the Hungarian method. Job.
Decision Science MGT3050
Youtube channel name: Kauser wise. Transportation problem [ MODI method - U V method with Optimal Solution ] : https://www.youtube.com/watch?v=-w2z3MVTcQA Assignment ...
Assignment Problem: Meaning, Methods and Variations
After reading this article you will learn about:- 1. Meaning of Assignment Problem 2. Definition of Assignment Problem 3. Mathematical Formulation 4. Hungarian Method 5. Variations. Meaning of Assignment Problem: An assignment problem is a particular case of transportation problem where the objective is to assign a number of resources to an equal number of activities so as to minimise total ...
Does the Hungarian Method find all optimal assignments?
Since S S was arbitrary, it follows that for every optimal solution to the original problem, the Hungarian algorithm is capable of finding it, depending on how it breaks ties. This is a sense in which the Hungarian algorithm finds all optimal solutions. We could actually generate all the optimal solutions by forking into two paths every time ...
PDF Chapter8 ASSIGNMENT PROBLEM
8.1 Introduction. An assignment problem is a particular case of transportation problem in which a number of operations are to be assigned to an equal number of operators, where each operator performs only one operation. The objective is to minimize overall cost or to maximize the overall profit for a given assignment schedule.
Solution of assignment problems (Hungarian Method)
Solution of assignment problems (Hungarian Method) First check whether the number of rows is equal to the numbers of columns, if it is so, the assignment problem is said to be balanced. ... Row E contains more than one zero, now proceed column wise. In column 1, there is an assignment. Go to column 2. There is exactly one zero. Mark that zero ...
Hungarian Algorithm: finding minimum number of lines to cover zeroes?
cross out the 0's in the assigned column. Do the same for every column. After doing the assignment using the above steps, follow the steps below to get the minimum number of lines which cover all the 0's. step 1 - Tick an unassigned row. step 2 - If a ticked row has a 0, then tick the corresponding column.
Algorithms: The Assignment Problem
The "assignment problem" is one that can be solved using simple techniques, at least for small problem sizes, and is easy to see how it could be applied to the real world. Assignment Problem Pretend for a moment that you are writing software for a famous ride sharing application. In a crowded environment, you might have multiple prospective ...
Kauser Wise
Dr. J. Kauser, M.Com, Mphil, MBA (Finance), Ph.D, (NET & JRF) Welcome to "Kauserwise" Channel, you're in the right place for mastering the fundamentals and advanced concepts of Financial Accounting, Financial Management, Cost & Management Accounting, Corporate Accounting, Operations Research, & Business Statistics, etc. Our channel is dedicated to providing clear, concise, and comprehensive ...
PDF UNIT 5 ASSIGNMENT PROBLEMS
e minimisation problem.3. The assignment problem wherein the number of rows is not equal to the number of columns is said t. be an unbalanced problem. Such a problem is handled by introducing dummy row(s) if the number of rows is less than the number of columns and dummy column(s) if the number of columns is le.
Transportation Model (Quick Notes)
Transportation Model Source: YT - Kauser Wise Transportation Problem - is a special kind of linear programming problem in which goods are transported from a set of sources to a set of destinations subject to supply and demand of the sources and destination, respectively, such that the total cost of transportation is minimized. Types: 1 ...
Some problems in A maths , assignment tommorow pls hurry
Some problems in A maths , assignment tommorow pls hurry. Views: 5,760 students. Updated on: Aug 14, 2024. Video solutions (1) Learn from their 1-to-1 discussion with Filo tutors. 6 mins. Uploaded on: 8/14/2024. Ask your question, on a video call with tutor. Connect now Install app. Connect instantly with this tutor.
Watch: Padres Manager Slams Reporter For Asking a Painfully Obvious
It started with a series of questions from Kevin Acee of the San Diego Union-Tribune about the recent usage of the San Diego Padres bullpen. It ended with a dumb question for Mike Shildt and an ...
Transportation Problem
An introduction to the transportation problem has been discussed in this article. In this article, the method to solve the unbalanced transportation problem will be discussed. Below transportation problem is an unbalanced transportation problem. The problem is unbalanced because the sum of all the supplies i.e. O1 , O2 , O3 and O4 is not equal to t
[#2] Assignment problem DIAGONAL RULE Hungarian method in ...
Here is the video for advanced problem for Assignment problem -by using Hungarian method with diagonal selection rule ( when there is more than one zero afte...
Kauser Wise
Kauser Wise — YouTube. ... Mrs. Kauser, M.Com, Mphil, MBA & NET & JRF In this channel you can learn Operations Research, Statistics and various accounting subjects with logically as well as quickly in the concise manner. Gizle.
How to fix SyntaxError
This JavaScript exception invalid assignment to const occurs if a user tries to change a constant value. Const declarations in JavaScript can not be re-assigned or re-declared. Message: TypeError: invalid assignment to const "x" (Firefox) TypeError: Assignment to constant variable. (Chrome) TypeError: Assignment to const (Edge) TypeError: Redeclara
Playlist Assignment Problem in Operations research Video ...
See this PlayList to watch Assignment problem in various method, List of videos: Assignment problem using hungarian Method (Minimization) Assignment problem ...

How to Solve the Assignment Problem: A Complete Guide

Understanding the Assignment Problem

Solving the Assignment Problem

Step 1: Set up the cost matrix

Step 2: Subtract the smallest element from each row and column

Step 3: Cover all zeros with the minimum number of lines

Step 4: Test for optimality and adjust the matrix

Step 5: Assign the tasks to the agents

Solution of the Assignment Problem using the Hungarian Method

Applications of the Assignment Problem

Applications in Computer Science

Applications in Economics

Applications in Logistics

Applications in Management

Operations Research

[#1]Assignment Problem[Easy Steps to solve - Hungarian Method with Optimal Solution] by kauserwise

Kauser Wise

Popular video in this channel

Hungarian Maximum Matching Algorithm

The Hungarian Algorithm Using an Adjacency Matrix

Assignment Problem: Meaning, Methods and Variations | Operations Research

Meaning of Assignment Problem:

Definition of Assignment Problem:

You must log in to answer this question.

Hot Network Questions

Procedure, Example Solved Problem | Operations Research - Solution of assignment problems (Hungarian Method) | 12th Business Maths and Statistics : Chapter 10 : Operations Research

Collectives™ on Stack Overflow

Hungarian Algorithm: finding minimum number of lines to cover zeroes?

8 Answers 8

Best Option

Your Answer

Sign up or log in

Not the answer you're looking for? Browse other questions tagged algorithm matrix matching greedy hungarian-algorithm or ask your own question .

Hot Network Questions

Algorithms: The Assignment Problem

Assignment Problem

Optimization Problems

Linear Programming

Solving with Octave and GLPK

Transportation Model (Quick Notes)

Add this document to collection(s)

Add this document to saved

Suggest us how to improve StudyLib

Some problems in A maths , assignment tommorow pls hurry

Video solutions ( 1 )

Students who ask this question also asked

Watch: Padres Manager Slams Reporter For Asking a Painfully Obvious Question

How to fix SyntaxError – ‘invalid assignment left-hand side in JavaScript’?

Understanding an error

Case 1: Error Cause: Assigning to Literals

Resolution of error

Case 2: Error Cause: Assigning to Function Calls

Please Login to comment...

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

IMAGES

COMMENTS