assignment problem algorithm r

Solve linear assignment problem using LAPJV

Description.

Use the algorithm of Jonker and Volgenant (1987) to solve the Linear Sum Assignment Problem (LSAP).

The Linear Assignment Problem seeks to match each row of a matrix with a column, such that the cost of the matching is minimized.

The Jonker & Volgenant approach is a faster alternative to the Hungarian algorithm (Munkres 1957), which is implemented in clue::solve_LSAP() .

Note: the JV algorithm expects integers. In order to apply the function to a non-integer n , as in the tree distance calculations in this package, each n is multiplied by the largest available integer before applying the JV algorithm. If two values of n exhibit a trivial difference – e.g. due to floating point errors – then this can lead to interminable run times. (If numbers of the magnitude of billions differ only in their last significant digit, then the JV algorithm may undergo billions of iterations.) To avoid this, integers over 2^22 that differ by a value of 8 or less are treated as equal.

LAPJV() returns a list with two entries: score , the score of the optimal matching; and matching , the columns matched to each row of the matrix in turn.

C++ code by Roy Jonker, MagicLogic Optimization Inc. [email protected] , with contributions from Yong Yang [email protected] , after Yi Cao

Jonker R, Volgenant A (1987). “A shortest augmenting path algorithm for dense and sparse linear assignment problems.” Computing , 38 , 325–340. doi:10.1007/BF02278710 . Munkres J (1957). “Algorithms for the assignment and transportation problems.” Journal of the Society for Industrial and Applied Mathematics , 5 (1), 32–38. doi:10.1137/0105003 .

Implementations of the Hungarian algorithm exist in adagio , RcppHungarian , and clue and lpSolve ; for larger matrices, these are substantially slower. (See discussion at Stack Overflow .)

The JV algorithm is implemented for square matrices in the Bioconductor package GraphAlignment::LinearAssignment() .

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Very simple implementation of the Hungarian Method for solving an asignation problem coded in R language.

sergioreyblanco/hungarian_method

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Implementation of the hungarian method in r language.

Here is a very simple implementation of the Hungarian Method in R language. It is used to solve assignment problems where a set of elements have to be asigned to an amount of other elements. The key of the problem is to find an optimal group of assignments of elements of the two set avoiding repetion. The word optimal is used because each assignment has a cost/profit. An example of this type of problem can be found in the following image (which corresponds to the example1 , in the terminology of the code developed):

The whole algorithm is implementated in the function assignmentSolver . It is divided in four well diferenced steps: caculation of the reduced matrix, the strikethrough crossed out of zeros, the update of the reduced matrix and the final gathering of the solution found. Moreover, this main function named assignmentSolver uses one auxiliary function called solutionChecking . Its usefulness is that of checking that the solution gathered is valid.

An instance of this function execution with example1 (using the terminology of the code) as input data is shown here:

As you can see, the cost/profit is given with its sign changed. The same happens with the values of each cell of the matrix used as input: the numbers have their sign changed. This is because the function minimizes a given matrix, but does not maximizes them.

The function takes as an input parameter a representation of a graph in the way of a matrix where each row is one element of the first set and each column represents one element of the second set. Thus, no assignment between elements of the same set or one with itself can be made. The introduced matrix must be a square one, so the method of the big M is used when there are less elements in the first set than in the second one, or viceversa. Each cell represents the profit (or cost) of assigning the corresponding element of the first set with other in the second one. Again, if one assingment between elements is not possible because of circumstances of the problem itself, an M is put in the corresponding cell. The function gives a return value consisting of a list with two components in which the first is the optimal assignment found and the other is the profit/cost of that assignment.

Example matrices are given (located in the last part of the code after the functions). Thus, you can test the function and create new ones. The first example matrix will result in a graph like this one:

The code does not need any additional libraries to run, besides the classical R environment.

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Matching algorithms in R (bipartite matching, Hungarian algorithm)

I wonder how to set up some example some fundamental matching procedures in R. There are many examples in various programming languages, but I have not yet found a good example for R.

Let’s say I want to match students to projects and I would consider 3 alternative approaches which I came across when googling on this issue:

1) Bipartite matching case: I ask each student to name 3 projects to work on (without stating any preference ranking among those 3).

2) Hungarian algorithm: I ask each student name 3 projects to work on WITH stating a preference ranking among those 3. As far as I understood the reasoning when applying the algorithm in this case would be something like: the better the rank the lower the “costs” to the student.

3) ??? approach: This should be pretty much related to (2). However, I think it is probably a better/ fairer approach (at least in the setting of the example). The students cannot pick projects, they even don’t know about the projects, but they have rate their qualifications (1 “not existent” to 10 “professional level”) with regards to a certain skillset. Further, the lecturer has rated the required skillset for every project. In addition to (2), a first step would be to calculate a similarity matrix and then to run the optimization routine from above.

I would appreciate any help in implementing those 3 approaches in R. Thank you.

UPDATE: The following questions seem to be related, but none show how to solve it in R: https://math.stackexchange.com/questions/132829/group-membership-assignment-by-preferences-optimization-problem https://superuser.com/questions/467577/using-optimization-to-assign-by-preference

• optimization
• linear-programming

• The R language is designed with statistical vector processing in mind. I would not expect it to be ideal for this sort of thing, or for many others. Because of this a quick google search will find you lots of information about calling other languages from R. A very simple way is to have R call other programs via system() as described for example in darrenjw.wordpress.com/2010/12/30/calling-c-code-from-r - although for the purposes of this method it doesn't matter much what the other program is written in so C could be almost anything. –  mcdowella Commented May 23, 2013 at 4:11
• As those seem to be very fundamental techniques, I was wondering whether R does not also provide this functionality through e.g. the package optmatch or the package clue (i.e. solve_LSAP() ). –  majom Commented May 23, 2013 at 20:17
• You can solve these using solve_LSAP(), you need to get the constraints and the cost functions right. You might even want to look at the package optimx. –  jackStinger Commented Dec 9, 2013 at 5:44

Here are possible solutions using bipartite matching and the Hungarian algorithm.

My proposed solution using bipartite matching might not be what you have in mind. All the code below does is sample randomly for a specified number of iterations, after which at least one solution hopefully will have been identified. This might require a large number of iterations and a long time with large problems. The code below found three solutions to your example problem within 200 iterations.

Here is code for the Hungarian algorithm using package clue and solve_LSAP() as @jackStinger suggested. For this to work I had to replace the missing observations and I arbitrarily replaced them with 4. Person 5 did not get their first choice and Person 7 did not get any of their three choices.

Here is a link to a site showing how the Hungarian algorithm works: http://www.wikihow.com/Use-the-Hungarian-Algorithm

EDIT: June 5, 2014

Here is my first stab at optimizing the third scenario. I randomly assign each student to a project, then calculate the cost for that set of assignments. Cost is calculated by finding the difference between a student's skill set and the project's required skills. The absolute values of those differences are summed to give a total cost for the seven assignments.

Below I repeat the process 10,000 times and identify which of those 10,000 assignments results in the lowest cost.

An alternative approach would be to do an exhaustive search of all possible student-project assignments.

Neither the random search nor the exhaustive search is likely what you had in mind. However, the former will give an approximate optimal solution and the latter would give an exact optimal solution.

I might return to this problem later.

EDIT: June 6, 2014

Here is the exhaustive search. There are only 5040 possible ways to assign projects to the seven students. This search returns four optimal solutions:

• The third case perhaps can be solved with Munkres' Assignment Algorithm. I might look into that. However, Munkres' Assignment Algorithm looks like it might be a little complex to program and the exhaustive search approach I have already posted identifies the optimal solution quickly and easily. Although, the exhaustive search might not be feasible when the number of students and projects is large. –  Mark Miller Commented Jun 6, 2014 at 17:10

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Hungarian Maximum Matching Algorithm

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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 & 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

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Constrained assignment problem (Linear Programming, Genetic Algorithm, etc...)

I'm looking for advice on how I should approach a specific problem. I have about 1000 shops that I have to assign to about 20 different supply centers out of a possible 28, and I'm trying to pick the best 20 (or less) centers that minimizes the overall distance between shops and supply centers.

I looked into the 'assignment problem' and how linear programming could help, but it seems this can only work if I have an equal number of shops and centers. I also looked into Genetic Algorithms and tried to code the algorithm to 1) ensure each shop is assigned to only one center and the center count must be no greater than 20, but results aren't that great (I can provide my R code if requested)

Should I continue trying to work with a Genetic Algorithm or can you folks suggest some other tool I should use or resource to look over? Thank you.

EDIT: Some nice user who's comment is now deleted suggested researching the Facility Location Problem . This seems more in line with what I'm looking for but most of the examples online cover single-facility problems, not multi-facility problems. Can anyone recommend a good resource for R or Python?

• optimization
• genetic-algorithms
• operations-research

EDIT: Apparently, I should have posted my original recommendation about the facility location problem as a reply to your question, not as an answer. I am new to the forum, but I get it now.

Any way, I believe your problem can be formulated as follows.

Let $y_i$ = 1 if supply center i is opened and 0 otherwise

$x_i,_j$ = 1 if supply center i feeds shop j and 0 otherwise

$c_i,_j$ = distance from supply center i to shop j

Then you want to solve the following linear programming problem.

    minimize  $\sum_i\sum_jc_i,_jx_i,_j$

    subject to $\sum_ix_i,_j = 1$                j = 1,...,1000 (or the number of shops you have)

                    $\sum_jx_i,_j \le 1000y_i$      i = 1,...,28 (If you have a max capacity for each supply center, use that number instead of 1000; 1000 assumes there is no max capacity so one center could feasibly supply all shops)

                    $\sum_iy_i \le 20$                 (max of 20 supply centers chosen)

                    $x_i,_j$ = 0 or 1                  i=1,...,28; j =1,...,1000

                    $y_i$ = 0 or 1                     i=1,...,14

The Rsymphony package in R can solve this problem for you. You will need a vector of the distances $c_i,_j$ and 0's for each $y_i$ as well as a matrix to take care of the constraints (the three summations starting after "subject to"). If you aren't familiar with the constraint matrix, each column contains the coefficients for that variable. The help file for the Rsymphony package has a couple examples that should help with that. Don't forget to include columns for the $y_i$ in the matrix as well. You will also need to set the $x_i,_j$ and $y_i$ to be binary by using

     types = "B"

As an example, suppose we have 3 shops and 2 supply centers with distances between them as $x_1,_1$ = 2.8, $x_1,_2$ = 5.4, $x_1,_3$ = 1.4, $x_2,_1$ = 4.2, $x_2,_2$ = 3.0, $x_2,_3$ = 6.3. Then

tells us that we should open both supply centers and supply center 1 will supply shops 1 and 3 and supply center 2 will supply shop 2, for a total distance of 7.2.

I hope this helps.

• 1 $\begingroup$ Actually, this is a very direct answer to the question that suggests the use of the RSymphony package for integer linear programming to solve the problem. These types of integer programming problems are actually quite easy to solve exactly, so there's no need to use an heuristic approach such as genetic algorithms. $\endgroup$ –  Brian Borchers Commented Feb 7, 2015 at 21:30
• $\begingroup$ @TAS thanks! Sorry for the late reply but I think this is exactly what I'm looking for! I'm going to test it out on my own data set and let you guys know soon. $\endgroup$ –  gtnbz2nyt Commented Feb 11, 2015 at 15:12
• $\begingroup$ Glad to hear. I think the most difficult part will be coming up with your constraint matrix. $\endgroup$ –  TAS Commented Feb 11, 2015 at 18:24
• $\begingroup$ @TAS Thanks a freaking million dude! It worked, and building the constraint matrix wasn't that hard once I took your example constraint matrix and broke it into different 'blocks' then binded them all together! $\endgroup$ –  gtnbz2nyt Commented Feb 12, 2015 at 18:21

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Hungarian Algorithm for Assignment Problem | Set 1 (Introduction)

• For each row of the matrix, find the smallest element and subtract it from every element in its row.
• Do the same (as step 1) for all columns.
• Cover all zeros in the matrix using minimum number of horizontal and vertical lines.
• Test for Optimality: If the minimum number of covering lines is n, an optimal assignment is possible and we are finished. Else if lines are lesser than n, we haven’t found the optimal assignment, and must proceed to step 5.
• Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3.

Try it before moving to see the solution

Explanation for above simple example:

  An example that doesn’t lead to optimal value in first attempt: In the above example, the first check for optimality did give us solution. What if we the number covering lines is less than n.

Time complexity : O(n^3), where n is the number of workers and jobs. This is because the algorithm implements the Hungarian algorithm, which is known to have a time complexity of O(n^3).

Space complexity :   O(n^2), where n is the number of workers and jobs. This is because the algorithm uses a 2D cost matrix of size n x n to store the costs of assigning each worker to a job, and additional arrays of size n to store the labels, matches, and auxiliary information needed for the algorithm.

In the next post, we will be discussing implementation of the above algorithm. The implementation requires more steps as we need to find minimum number of lines to cover all 0’s using a program. References: http://www.math.harvard.edu/archive/20_spring_05/handouts/assignment_overheads.pdf https://www.youtube.com/watch?v=dQDZNHwuuOY

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EngineersTutor

Solved assignment problems – algorithms and flowcharts.

An algorithm is defined as sequence of steps to solve a problem (task) . The steps must be finite, well defined and unambiguous. Writing algorithm requires some thinking. Algorithm can also be defined as a plan to solve a problem and represents its logic. Note that an algorithm is of no use if it does not help us arrive at the desired solution

Algorithm characteristics

• It should have finite number of steps . No one can be expected to execute infinite number of steps.
• The steps must be in order and simple
• Each step should be defined clearly i.e. without un-ambiguity (without doubtfulness)
• Must include all required information
• Should exhibit at least one output

A flowchart is a pictorial (graphical) representation of an algorithm . A flowchart is drawn using different kinds of symbols. A symbol is used for a specific purpose. Each symbol has name.

Different algorithms have different performance characteristics to solve the same problem. Some algorithms are fast. Some are slow. Some occupy more memory space. Some occupy less memory space. Some are complex and some algorithms are simple.

Logically algorithm, flowchart and program are the same.

Q1 . Create a program to compute the volume of a sphere. Use the formula: V = (4/3) *pi*r 3 where pi is equal to 3.1416 approximately. The r is the radius of sphere.  Display the result.

Q2 . Write a program the converts the input Celsius degree into its equivalent Fahrenheit degree. Use the formula: F = (9/5) *C+32.

Q3 . Write a program that converts the input dollar to its peso exchange rate equivalent.  Assume that the present exchange rate is 51.50 pesos against the dollar. Then display the peso equivalent exchange rate.

Q4 . Write a program that converts an input inch(es) into its equivalent centimeters. Take note that one inch is equivalent to 2.54cms.

Q5 . Write a program that exchanges the value of two variables: x and y.  The output must be: the value of variable y will become the value of variable x, and vice versa.

Q6 . Design a program to find the circumference of a circle. Use the formula: C=2πr, where π is approximately equivalent 3.1416.

Q7 . Write a program that takes as input the purchase price of an item (P), its expected number of years of service (Y) and its expected salvage value (S). Then outputs the yearly depreciation for the item (D). Use the formula: D = (P – S) Y.

Q8 . Swapping of 2 variables without using temporary (or 3 rd variable).

Q9 . Determine the most economical quantity to be stocked for each product that a manufacturing company has in its inventory: This quantity, called economic order quantity (EOQ) is calculated as follows: EOQ=2rs/1 where: R= total yearly production requirement S=set up cost per order I=inventory carrying cost per unit.

Q10 . Write a program to compute the radius of a circle. Derive your formula from the given equation: A=πr², then display the output.

• ← Solved Assignment Problems in Java (with Algorithm and Flowchart)
• Simple if statement in C →

Gopal Krishna

Hey Engineers, welcome to the award-winning blog,Engineers Tutor. I'm Gopal Krishna. a professional engineer & blogger from Andhra Pradesh, India. Notes and Video Materials for Engineering in Electronics, Communications and Computer Science subjects are added. "A blog to support Electronics, Electrical communication and computer students".

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Hungarian Method

The Hungarian method is a computational optimization technique that addresses the assignment problem in polynomial time and foreshadows following primal-dual alternatives. In 1955, Harold Kuhn used the term “Hungarian method” to honour two Hungarian mathematicians, Dénes Kőnig and Jenő Egerváry. Let’s go through the steps of the Hungarian method with the help of a solved example.

Hungarian Method to Solve Assignment Problems

The Hungarian method is a simple way to solve assignment problems. Let us first discuss the assignment problems before moving on to learning the Hungarian method.

What is an Assignment Problem?

A transportation problem is a type of assignment problem. The goal is to allocate an equal amount of resources to the same number of activities. As a result, the overall cost of allocation is minimised or the total profit is maximised.

Because available resources such as workers, machines, and other resources have varying degrees of efficiency for executing different activities, and hence the cost, profit, or loss of conducting such activities varies.

Assume we have ‘n’ jobs to do on ‘m’ machines (i.e., one job to one machine). Our goal is to assign jobs to machines for the least amount of money possible (or maximum profit). Based on the notion that each machine can accomplish each task, but at variable levels of efficiency.

Hungarian Method Steps

Check to see if the number of rows and columns are equal; if they are, the assignment problem is considered to be balanced. Then go to step 1. If it is not balanced, it should be balanced before the algorithm is applied.

Step 1 – In the given cost matrix, subtract the least cost element of each row from all the entries in that row. Make sure that each row has at least one zero.

Step 2 – In the resultant cost matrix produced in step 1, subtract the least cost element in each column from all the components in that column, ensuring that each column contains at least one zero.

Step 3 – Assign zeros

• Analyse the rows one by one until you find a row with precisely one unmarked zero. Encircle this lonely unmarked zero and assign it a task. All other zeros in the column of this circular zero should be crossed out because they will not be used in any future assignments. Continue in this manner until you’ve gone through all of the rows.
• Examine the columns one by one until you find one with precisely one unmarked zero. Encircle this single unmarked zero and cross any other zero in its row to make an assignment to it. Continue until you’ve gone through all of the columns.

Step 4 – Perform the Optimal Test

• The present assignment is optimal if each row and column has exactly one encircled zero.
• The present assignment is not optimal if at least one row or column is missing an assignment (i.e., if at least one row or column is missing one encircled zero). Continue to step 5. Subtract the least cost element from all the entries in each column of the final cost matrix created in step 1 and ensure that each column has at least one zero.

Step 5 – Draw the least number of straight lines to cover all of the zeros as follows:

(a) Highlight the rows that aren’t assigned.

(b) Label the columns with zeros in marked rows (if they haven’t already been marked).

(c) Highlight the rows that have assignments in indicated columns (if they haven’t previously been marked).

(d) Continue with (b) and (c) until no further marking is needed.

(f) Simply draw the lines through all rows and columns that are not marked. If the number of these lines equals the order of the matrix, then the solution is optimal; otherwise, it is not.

Step 6 – Find the lowest cost factor that is not covered by the straight lines. Subtract this least-cost component from all the uncovered elements and add it to all the elements that are at the intersection of these straight lines, but leave the rest of the elements alone.

Step 7 – Continue with steps 1 – 6 until you’ve found the highest suitable assignment.

Hungarian Method Example

Use the Hungarian method to solve the given assignment problem stated in the table. The entries in the matrix represent each man’s processing time in hours.

\(\begin{array}{l}\begin{bmatrix} & I & II & III & IV & V \\1 & 20 & 15 & 18 & 20 & 25 \\2 & 18 & 20 & 12 & 14 & 15 \\3 & 21 & 23 & 25 & 27 & 25 \\4 & 17 & 18 & 21 & 23 & 20 \\5 & 18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array} \)

With 5 jobs and 5 men, the stated problem is balanced.

\(\begin{array}{l}A = \begin{bmatrix}20 & 15 & 18 & 20 & 25 \\18 & 20 & 12 & 14 & 15 \\21 & 23 & 25 & 27 & 25 \\17 & 18 & 21 & 23 & 20 \\18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array} \)

Subtract the lowest cost element in each row from all of the elements in the given cost matrix’s row. Make sure that each row has at least one zero.

\(\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 5 & 10 \\6 & 8 & 0 & 2 & 3 \\0 & 2 & 4 & 6 & 4 \\0 & 1 & 4 & 6 & 3 \\2 & 2 & 0 & 3 & 4 \\\end{bmatrix}\end{array} \)

Subtract the least cost element in each Column from all of the components in the given cost matrix’s Column. Check to see if each column has at least one zero.

\(\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 3 & 7 \\6 & 8 & 0 & 0 & 0 \\0 & 2 & 4 & 4 & 1 \\0 & 1 & 4 & 4 & 0 \\2 & 2 & 0 & 1 & 1 \\\end{bmatrix}\end{array} \)

When the zeros are assigned, we get the following:

The present assignment is optimal because each row and column contain precisely one encircled zero.

Where 1 to II, 2 to IV, 3 to I, 4 to V, and 5 to III are the best assignments.

Hence, z = 15 + 14 + 21 + 20 + 16 = 86 hours is the optimal time.

Practice Question on Hungarian Method

Use the Hungarian method to solve the following assignment problem shown in table. The matrix entries represent the time it takes for each job to be processed by each machine in hours.

\(\begin{array}{l}\begin{bmatrix}J/M & I & II & III & IV & V \\1 & 9 & 22 & 58 & 11 & 19 \\2 & 43 & 78 & 72 & 50 & 63 \\3 & 41 & 28 & 91 & 37 & 45 \\4 & 74 & 42 & 27 & 49 & 39 \\5 & 36 & 11 & 57 & 22 & 25 \\\end{bmatrix}\end{array} \)

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Frequently Asked Questions on Hungarian Method

What is hungarian method.

The Hungarian method is defined as a combinatorial optimization technique that solves the assignment problems in polynomial time and foreshadowed subsequent primal–dual approaches.

What are the steps involved in Hungarian method?

The following is a quick overview of the Hungarian method: Step 1: Subtract the row minima. Step 2: Subtract the column minimums. Step 3: Use a limited number of lines to cover all zeros. Step 4: Add some more zeros to the equation.

What is the purpose of the Hungarian method?

When workers are assigned to certain activities based on cost, the Hungarian method is beneficial for identifying minimum costs.

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