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The Hungarian method is a well-known method for solving the assignment problem. This method was developed and published in 1955. It was named the Hungarian method because two theorems from two Hungarian mathematicians were used. In 1957, it was noticed that this algorithm is strongly polynomial and has a complexity of order O(n4) This is the reason why the Hungarian method is also known as the Kuhn-Munkres algorithm. Later on, in 1971 the complexity of the method was improved to order O(n3) A smallest uncovered element is selected to create a single zero at every iteration. This is a weakness and is alleviated by selecting more than one smallest uncovered element thus creating more than one zero at every iteration to come up with what we now call the Accelerating Hungarian (AH) method.
From the numerical illustration of the Hungarian method given in this paper, we require 6 iterations to reach optimality. It can also be shown that selecting a single smallest uncovered element (es) makes the Hungarian method inefficient when creating zeros. From the same numerical illustration of the proposed algorithm (AH) also given in this paper, it can be noted that only one iteration is required to reach optimality and that a total of six zeros are created in one iteration.
Assignment model and the Hungarian method have application in addressing the Weapon Target Assignment (WTA) problem. This is the problem of assigning weapons to targets while considering the maximum probability of kill. The assignment problem is also used in the scheduling problem of physicians and medical staff in the outpatient department of large hospitals with multi-branches. The mathematical modelling of this assignment problem results in complex problems. A hybrid meta-heuristic algorithm SCA-VNS combining a Sine Cosine Algorithm (SCA) and Variable Neighbourhood Search (VNS) based on the Iterated Hungarian algorithm is normally used
Keywords: Hungarian method, assignment problem, Konig theorem,
Egervary Theorem, Kuhn-Munkres algorithm -□ □-
UDC 338
[DOI: 10.15587/1729-4061.2020.209172|
DEVELOPMENT OF AN ACCELERATING HUNGARIAN METHOD FOR ASSIGNMENT PROBLEMS
Elias Munapo
PhD, Professor of Operations Research Department of Statistics and Operations Research School of Economics and Decision Sciences North-West University Mmabatho Unit 5, Mafikeng, South Africa, 2790 E-mail: emunapo@gmail.com
Received date 04.08.2020 Accepted date 17.08.2020 Published date 31.08.2020
Copyright © 2020, Ellas Munapo This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0)
1. Introduction
The Hungarian method for solving the assignment problem was developed and published in 1955 [1]. It was named the Hungarian method because two theorems by two Hungarian mathematicians [2, 3] were used. In 1957 [4], it was noticed that this algorithm was strongly polynomial and has a complexity of order O(n4). This is the reason why the Hungarian method is also known as the Kuhn-Munkres algorithm. Later on, in 1971 [5] and 1972 [6], it was improved to a complexity of order O(n3). A lot of work has been done on this algorithm after that but up to now this algorithm still has an obvious weakness. A smallest uncovered element is selected to create a single zero at every iteration. In this paper, this weakness is alleviated by selecting more than one smallest uncovered element thus creating more than one zero at every iteration to come up with what we now call the Accelerating Hungarian (AH) method.
Assignment model and the Hungarian method have application in addressing the Weapon Target Assignment (WTA) problem. This is the problem of assigning weapons to targets while considering the maximum probability of kill. The assignment problem is also used in the scheduling problem of physicians and medical staff in the outpatient department of large hospitals with multi-branches. The mathematical modelling of these assignment problems results in complex problems. A hybrid meta-heuristic algorithm SCA-VNS combining a Sine Cosine Algorithm (SCA) and Variable
Neighbourhood Search (VNS) based on the Iterated Hungarian algorithm is normally used.
2. Literature review and problem statement
A lot of work has been done on the Hungarian algorithm but up to now this algorithm still has an obvious weakness. A single smallest uncovered element is selected to create a single zero at every iteration. In this paper, this weakness is alleviated by selecting more than one smallest uncovered element thus creating more than one zero at every iteration to come up with what we now call the Accelerating Hungarian (AH) method.
The Accelerating Hungarian method is not a new idea. This idea has been applied to the Hungarian method before in [7]. In that paper, parallel versions of two different variants which are the classical and alternating tree of the Hungarian algorithm for solving the Linear Assignment Problem (LAP) were used. The main contribution of that paper noted was the efficient parallelization of the augmenting path search phase of the Hungarian algorithm. Computational experiments on problems with up to 25 million variables reveal that the GPU-accelerated versions are extremely efficient in solving large assignment problems, as compared to their CPU counterparts. Tremendous parallel speedups were achieved for problems with up to 400 million variables.
The Hungarian method has application in addressing the Weapon Target Assignment (WTA) problem. This is the problem of assigning weapons to targets while considering the maximum probability of kill. In [8], various model formulations, exact algorithms, and heuristic algorithms for the static and dynamic WTA were considered. The formulations used in that paper were placed into a comparable form so as to provide an insight into the developments of the defence-related WTA problems. The available solution methods were compared.
The existing Hungarian method has the weakness of assigning some jobs to dummy or pseudo machines in those cases whereby the number of jobs is more than the machines. In real life, this does not make sense since those jobs assigned to dummy machines are not done. In real life, there is a need to execute all the jobs. This purpose can be served by assigning multiple jobs to a single machine. The paper [9] proposes a Modified Hungarian Method for solving unbalanced assignment problems which gives the optimal policy of assignment of several jobs to a single machine.
The paper [10] investigated the scheduling problem of physicians and medical staff in the outpatient department of large hospitals with multi-branch. The mathematical modelling of this assignment problem resulted in a complex problem. A hybrid meta-heuristic algorithm SCA-VNS combining a Sine Cosine Algorithm (SCA) and variable neighbourhood search (VNS) based on the Iterated Hungarian algorithm was used. The performance of the scheduling model obtained is better than in other compared algorithms.
The paper [11] focuses on an extension of the assignment problem with additional conflict (pair) constraints in conjunction with the assignment constraints and binary restrictions. Unlike the well-known assignment problem, this problem is NP-hard. A branch-and-bound algorithm and a Russian Doll Search algorithm using the assignment problem relaxations for lower bound computations were used. From the computational experiments done, the proposed algorithm is efficient.
In [12], The Reviewer Assignment Problem (RAP) is presented. This is one of the most important and apparent processes that involves the process of assigning a reviewer to a paper. Of the factors studied, it was observed that 'unanimity' is the most appropriate measure for the overall review process. The use of performance measures such as matching score, authority, coverage, and diversity is the most suitable for measuring the performance of phase one -matching paper and reviewer. It was also noticed that load balance, fairness and accuracy are suitable for measuring the performance of phase two that assigns a reviewer to a paper by satisfying constraints. The study concluded that a hybrid approach with the proper combination of these measures is the best for measuring the performance of the process of assigning a reviewer to a paper.
The paper [13] introduced the concept of the demand correlation pattern (DCP) to describe the correlation among items, based on which a new model is constructed to address the storage location assignment problem (SLAP). To solve the model, a heuristic and a simulated annealing method were developed. The proposed methods were examined and compared using both real data collected from an online retailer and numerical instances that are randomly generated.
In [14], a new graph-theoretic proof of the tropical Jacobi identity is presented. This identity was applied to optimal assignments with supervisions. That is, optimally assigning
multiple tasks to one team, or daily tasks to multiple teams, where each team has a supervisor task or a supervised task.
In general, the assignment problem is formulated as given in Section 4. The variants of the assignment problems presented in sources [7-14] are obtained by modifying the constraints and objective function of Table 1.
3. The aim and objectives of the study
The aim of the study is to improve the Hungarian method for linear assignment problems so that it becomes more efficient.
To achieve the aim, the following objectives have been set:
- to increase the number of the smallest uncovered elements;
- to create more than one zero elements at each iteration;
- to illustrate the proposed algorithm by an example.
4. Assignment problem
Let any assignment problem be given by Table 1.
Table 1
Assignment problem in general
Source
C11 C12 C13 c1n 1
C21 c22 c23 c2n 1
C31 C32 C33 c3n 1
C11 cm2 cm3 cmn 1
Demand 1 1 1 1 1 D
Where Cj is the cost of the assignment. This is a balanced problem, i. e. the number of rows (n) is equal to the number of columns (m), i. e. D=n=m. The objective is to minimize the total assignment cost.
4. 1. Kónig's theorem
In a bipartite graph G=(S, T; E), the minimum number of elements of S exposed by matching is equal to the maximum of the deficit h(X) over the subsets of S where h(X):= X| -|r(X)| and r(X) denotes the set of elements of T having a neighbour in X. In particular, there is a matching covering S if and only if |r(X)| > |X| holds for every subset X c S.
Proof.
There are two main ways of proving this theorem. These are the constructive proof and using the linear programming primal-dual relationship. In this paper, we use the constructive proof. For any matching M, we orient the edges in M from T to S and all other edges from S to T. Let Rs and RT denote the set of nodes of S and of T, respectively that are exposed by M. Let Z denote the set of nodes of the resulting directed graph which can be reached from RT by a directed path. If the intersection of RT and Z is not empty (i. e. RT f| Z ^ ) then there is a path from Rs to RT that alternates in M and then the symmetric difference of M and P is a matching M 1with Jm1] = |M| +1. In addition, if R | Z ^ then L := (T | Z) U (S - Z) is a set of nodes covering all edges and Ml = LI. This theorem was published in 1931 [1].
4. 2. Egerv6ry's theorem
In the same year of 1931, Konig's theorem was extended to Egervary's theorem [2]. Let G=(S, T; E) be a complete bipartite graph with |S| = |T| and let c: E ^ Z+ be a nonnegative integer-valued weight function. The maximum weight of a perfect matching of G is equal to the minimum weight of a nonnegative, integer-valued, weighted-covering of c where a weighted-covering is a function n: S U T ^ Z+ for which n(u)+n(v)>c(uv) for every uv eE and the weight of n is defined to be (v):v e SUT].
Proof.
Let n be a nonnegative integer-valued weighted-covering of c with minimum weight. If there is a perfect matching M in the subgraph Gn of tight edges, where an edge uv is called tight if n(u) + n(v)ec(uv) then Mis a maximum weight perfect matching of G whose weight is equal to the weight of n. If there is no perfect matching in Gn then Konig's theorem implies that there is a deficient set X c S in Gn. Increase the n-value of each node in rG (X)by 1 and decrease the n-value in X by 1. Using this we obtain another weighted covering n1 of c whose weight is smaller than that of n. In that case where n1 is negative, increase n1-values on the elements of S by 1 and decrease the n1-values on the elements of T by 1. Since G is complete bipartite and c>0. the resulting n11 is a nonnegative weighted-cover of c whose weight is smaller than that of n in a contradiction with the minimum choice of n.
4. 3. Hungarian method
Hungarian algorithm was developed and published in 1955 [1]. It was named the Hungarian algorithm because it was based on the works of two Hungarian mathematicians. In 1957, it was shown that the Hungarian algorithm is strictly polynomial.
Step 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.
Step 2. 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.
Step 3. Draw lines through the row and columns that have 0 entries such that the fewest lines possible are drawn.
Step 4. 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 5.
Step 5. Find the smallest entry (es) 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 covered by two lines. Then, go back to Step 3.
4. 3. 1. Numerical illustration of the Hungarian method
Solve the following assignment problem using the Hungarian method (Table 2).
Table 2
Given assignment problem
Worker
10 8 3 9 24 13 1
14 24 2 32 18 12 2
44 16 19 22 15 19 3
2 2 3 1 1 1 4
31 32 4 43 28 41 5
25 62 2 29 46 22 6
Job 1 2 3 4 5 6
The Hungarian algorithm steps are presented from Table 3 to Table 17.
Table 3
Row minima and column minima
10 8 3 9 24 13
14 24 2 32 18 12
44 16 2 22 15 19
2 2 3 1 1 1
31 32 4 43 28 41
25 62 2 29 46 22
Table 4
Subtracting the row minima
7 5 0 6 21 10
12 22 0 30 16 10
42 14 0 20 13 17
1 1 2 0 0 0
27 28 0 39 24 37
23 60 0 27 44 20
Table 5
Subtracting the column minima
6 4 0 6 21 10
11 21 0 30 16 10
41 13 0 20 13 17
0 0 2 0 0 0
26 27 0 39 24 37
22 59 0 27 44 20
Table 6
Covering zeros with minimum number of lines (iteration 1)
6 4 6 21 10
11 21 30 16 10
41 13 20 13 17
26 27 39 24 37
22 59 27 44 20
The number of lines hi=2. This implies that the solution is not optimal. The smallest uncovered element is es=4. We then add 4 to the elements covered by two lines and subtract 4 from all elements that are uncovered as given in Table 7.
Table 7
Adding 4 to the elements covered by two lines and subtract 4 from the uncovered elements
2 0 0 3 17 6
7 17 0 26 12 6
37 9 0 16 9 13
0 0 6 0 0 0
22 23 0 35 20 33
28 55 0 23 40 16
Table 8
Covering zeros with minimum number of lines (iteration 2)
7 17 26 12 6
37 9 16 9 13
22 23 35 20 33
28 55 23 40 16
The number of lines h2=3. This implies that the solution is not optimal. The smallest uncovered element is es=6. We then add 6 to the elements covered by two lines and subtract 6 from all elements that are uncovered as given in Table 9.
Table 9
Adding 6 to the elements covered by two lines and subtracting 6 from the uncovered elements
2 0 6 3 17 6
1 11 0 20 6 0
31 3 0 10 3 7
0 0 12 0 0 0
16 17 0 29 14 27
22 49 0 17 34 10
Table 10
Covering zeros with minimum number of lines (iteration 3)
2 3 17
1 20 6
31 10 3
16 29 14
22 17 34
Table 14
Covering zeros with minimum number of lines (iteration 5)
13 15 0 26 11 25
19 47 0 14 31 8
The number of lines hs=5. This implies that the solution is not optimal. The smallest uncovered element is es=8. We then add 8 to the elements covered by two lines and subtract 8 from all elements that are uncovered as given in Table 15.
Table 15
Add 9 to the elements covered by two lines and subtract 8 from the uncovered elements
1 0 16 3 17 6
0 11 10 19 5 0
28 1 8 7 0 5
0 1 23 0 0 1
5 7 0 8 3 17
11 39 0 6 23 0
Table 16
Covering zeros with minimum number of lines (iteration 6)
The number of lines h3=4. This implies that the solution is not optimal. The smallest uncovered element is es=1. We add 1 to all elements covered by two lines and subtract 1 from all elements that are uncovered as given in Table 11.
Table 11
Adding 1 to the elements covered by two lines and subtracting 1 from the uncovered elements
28 1 5
5 7 17
11 39 0
The number of lines h6=6. This implies that an optimal solution is available as given in Table 17.
Table 17
1 0 6 3 17 6 Optimal allocation
0 11 0 19 5 0 1 0 16 3 17 6
30 3 0 9 2 7 0 11 10 19 5 0
0 1 13 0 0 1 28 1 8 7 0 5
15 17 0 28 13 27 0 1 23 0 0 1
21 49 0 16 33 10 5 7 0 8 3 17
Table 12 11 39 0 6 23 0
Covering zeros with minimum number of lines (iteration 4)
30 3 9 2 7
15 17 28 13 27
21 49 16 33 10
The number of lines h4=4. This implies that the solution is not optimal. The smallest uncovered element is es=2. We add 2 to all elements covered by two lines and subtract 2 from all elements that are uncovered as given in Table 13.
Table 13
Adding 2 to the elements covered by two lines and subtracting 2 from the uncovered elements
1 0 8 3 17 6
0 11 2 19 5 0
28 1 0 7 0 5
0 1 15 0 0 1
13 15 0 26 11 25
19 47 0 14 31 8
The Hungarian method has many versions and the original algorithm had asymptotic complexity O(n4) as given in [4]. Then later on it was shown to have a complexity of order O(n3) as given in [5, 6].
Strengths of the Hungarian method.
This method is easier to apply than the transportation simplex method. It is made up of only addition and subtraction operations and application of the optimality test. It can handle degenerate transportation problems better than the transportation simplex method.
Weakness of the Hungarian method.
The Hungarian method has the obvious disadvantage of selecting the smallest uncovered element which may happen to be only one number. In this case, it implies one zero is created. Creating a single zero at every iteration is a weakness when the problem is large.
4. 4. Accelerating Hungarian method
A way can be used to alleviate the weakness of creating a single zero at every iteration. This is done by creating more than one smallest uncovered element at every iteration. Creating more than one zero in an iteration is the same as the
accelerating Hungarian method. After covering all the zeros with the minimum number of lines it can be noticed that some rows and columns can be changed without necessarily changing the minimum number of lines. In this paper, we call such rows and columns flexible rows or columns. This important fact can be used to increase the number of the smallest uncovered elements to more than one. In order to do this, there is a need to identify the flexible rows and columns immediately after covering all the zeros. Sometimes a set of columns or rows can be flexible but cannot be changed at the same time. Such a set of rows or columns is called a flexible dependent set.
4. 4. 1. Numerical illustration 2 - flexible dependent set
Table 18 is a table of reduced cost entries with zeros covered by a minimum number of lines.
Table 18
Zeros covered with a minimum number of lines
(c4+8), respectively. The resulting cost matrix is presented in Table 21. The number of the smallest uncovered elements has increased from 1 to 3.
Table 21
Adjusted cost matrix
28 35 35
47 28 41
38 49 28
This process of increasing the number of the smallest uncovered elements makes the proposed approach (AH) more powerful than the original Hungarian version. Subtracting 28 from uncovered elements and adding 28 to all elements covered by two lines we have Table 22.
Table 22
Resulting cost matrix
12 35 27
31 28 23
22 49 20
ci c2 c4
0 7 8 7
19 0 6 13
0 0 31 0
10 21 0 0
From the columns c1 to c4, it can be noted that three of these columns are flexible and dependent. The columns are c1, c2, c4 and these cannot all be changed at the same time. Only two columns can be changed at a time without changing the minimum number of lines covering the zeros.
4. 4. 2. Changing the flexible row or columns
Once the flexible rows and columns have been identified, then these can be changed to increase the number of the smallest uncovered elements so as to make the Hungarian method efficient. As an illustration, we use Table 19 to Table 23.
In general, for any assignment cost matrix of the form given in Table 23, with columns cj, c2, c3,..., cn, integers can be used to change the columns to cj+^i, c2+^2, c3+X3,..., cn+Xn, such that the elements, (c2i+^i)= =(ci2+^2)=(c23+X3)=...=(c3n+Xn) as given in Table 23 and Table 24.
Table 23
Assignment cost matrix in general
Table 19
Identifying the smallest uncovered element
12 I 35 I 27
31__28__23
22__49__20
Ci C2 C4
cn c12 c13 c1n
c21 c22 c23 c2n
c31 c32 c33 c3n
cn1 cn2 cn3 cnn
c1 c2 c3 Cn
Table 24
Cost matrix after adding integers to columns
In Table 19, the only smallest uncovered element is 12. There is a need to change the flexible columns so as to increase the number of the smallest uncovered elements. To change the flexible columns, we select the smallest uncovered elements in the other two columns as given in Table 20. These numbers are 28 and 20.
(c11+^1) (c12+^2) (c13+^3) (c1n+^n)
(c21+>4) (c22+^2) (c23+^3) (c2n+^n)
(c31+^1) (c32+^2) (c33+^3) (c3n+^n)
(cn1+^1) (cn2+^2) (cn3+^3) (cnn+^n)
(c1 + ^1) (c2+^2) fe+^3) (cn+^n)
A new cost matrix can be created by adding constants X1, X2,..., Xn as given in Table 24 and the optimal solution does not change.
Columns of any balanced cost matrix can be changed by adding constants and this does not change the optimal solution.
Table 20
Smallest uncovered elements in the other 2 columns
12 35 27
31 28 23
22 49 20
c1 c2 c4
In this case, these two numbers cannot all be changed at the same time. We have to fix the column of 28 and change the columns of 12 and 20. The smallest uncovered element is 12 and columns cj and c4 are adjusted as (cj+16) and
4. 4. 3. Creating a column or row of only zeros
The costs in any column or any row of any balanced assignment problem can be made the same by adding constants. Suppose we select column 1 and making all entries in this column the same we have Table 25.
Table 25
First column of the same numbers
c11 C12 C1 3 C1 n
c11 C22 C23 C2n
c11 C32 C33 C3n
c11 Cn2 Cn3 c nn
Where c' = Cj + kj, kij>0 and constant. Subtracting c'11 from the first column of Table 25 we have Table 26.
Table 26
Subtracting c[x from the first column elements
0 C12 C1 3 C1 n
0 C22 C23 C2n
0 C32 C33 C3n
0 Cn2 Cn3 C nn
4. 4. 4. Subtracting or adding a constant to a row or column
In general, subtracting or adding a constant to a row or column does not change the optimal solution of any assignment problem. The assignment problem can be represented as given in (1).
Minimize
n m
Z =yY c-x.., i=1 j=1
m
I Xj =1 (1)
j
n
I x~ = 1.
¿—i j
i
Xj > 0.
Let pi be a constant subtracted from row i and qj be a constant subtracted from column j. Thus, the constant element Cj changes to Cj as given in (2).
(2) (3)
n m n m
EE]*=XX(cj- pi- Pj )>
i=1 j=1 i=1 j=1
n m n m i \
EE]*=EECx-Epi Ex] -EqIE] (4)
i=1 j=1 i=1 j=1 i y j J j
(5)
EECx=EECiX-| Epi K)- E
i=1 j=1
i=1 j=1
EE c x = EE c-x- - constant.
I] I]
i=1 j=1 i=1 j=1
(6)
This important property can be exploited to solve the assignment problem. In solving assignment problems, zeros are created in rows and columns such that the total cost of the allocated cells is zero. Such a solution is optimal since all costs are nonnegative and there is no way the total cost can be less than zero.
4. 5. Accelerating Hungarian Method
Assuming a balanced assignment problem, the proposed Accelerating Hungarian (AH) method is summarized as follows:
Step 1. Create all zero row or column. Ensure that all other rows and columns have at least a zero.
Step 2. Cover all zeros with a minimum number of lines. If the number of lines is equal to n then an optimal solution is available. Else go to Step 3.
Step 3. Select the smallest uncovered element (es). Identify flexible rows and columns and their minima.
Step 4. From the flexible rows, select the largest smallest minimum (e/s). Adjust all other flexible rows and columns so that their smallest elements also become e/s.
Step 5. Subtract e/s from all uncovered adjusted elements. Add e/s to all elements covered by two lines and return to Step 2.
4. 5. 1. Illustration 3 - proposed Accelerating Hungarian method
Apply the AH method to the illustration given in Table 27.
Step 1.
Table 27
Creating the first column of same numbers
44 42 37 43 58 47
44 54 32 62 48 42
44 16 2 22 15 19
44 44 45 43 43 43
44 45 17 56 41 54
44 81 21 48 65 41
The first column of same numbers is created by (7). (1 + 34), (r2 + 30), (r + 0),
(r + 42), (r5 +11), (16 +19). (7)
Creating the first column of zeros we have Table 28.
Table 28
Creating the first column of zeros
0 42 37 43 58 47
0 54 32 62 48 42
0 16 2 22 15 19
0 44 45 43 43 43
0 45 17 56 41 54
0 81 21 48 65 41
The first column of zeros is created by (cj-44). Ensuring that we have a zero in every column, we have Table 29.
Table 29
Ensuring at least a zero in all other columns
0 26 35 21 43 28
0 38 30 40 48 23
0 0 0 0 0 0
0 28 43 21 28 24
0 29 15 34 26 35
0 65 19 26 50 22
The other zeros in the other columns are created by (10).
(c2 -16), (ca - 2), (c< - 22), (c5 -15), (r5 -15),
(c6 -19). (10)
Step 2.
Covering zeros with a minimum number of lines we have Table 30.
Table 30
Covering zeros with a minimum number of lines
26 35 21 43 28
38 30 40 48 23
28 43 21 28 24
29 15 34 26 35
65 19 26 50 22
The number of lines is h1=2 and is less than 6 which implies an optimal solution is not available.
Step 3.
Columns c2 to c6 are flexible and dependent. The smallest uncovered element es=15 and the largest smallest minimum (els=26). The smallest uncovered number in the flexible columns is given in bold in Table 31.
Table 31
Smallest uncovered element and flexible columns
Step 4
The flexible columns are adjusted as (c2+0), (c3+11), (cs+0) and (c6+4). The adjusted cost matrix is given in Table 32.
Table 32
Adjusted cost matrix
26 46 26 43 32
38 41 45 48 27
28 54 26 28 28
29 26 39 26 39
65 30 31 50 26
Step 5
Subtract (els=26) from all uncovered adjusted elements. Add 26 to all elements covered by two lines as given in Table 33.
Table 33
Subtracting 26 from uncovered elements and adding 26 to c31
0 0 20 0 17 6
0 12 15 19 7 1
26 0 11 5 0 4
0 2 28 0 2 2
0 3 0 13 0 13
0 39 4 5 24 0
The minimum number of lines h2=6, implying an optimal solution is available which is given in yellow.
6. Discussion of experimental results
From the numerical illustration of the known version of the Hungarian method given in Section 4. 3. 1, a single uncovered element or cell is selected to create a single new zero element. As shown in this numerical illustration, 6 iterations are required to create new zeros so as to reach optimality.
The Accelerating Hungarian method is illustrated in Section 4. 5. 1. In this illustration, the uncovered elements are slightly modified so that the number of the smallest elements increases to 6. As a result of this, 6 new zeros are created in just one iteration to reach optimality. The Accelerating Hungarian method has the obvious superiority of creating more zeros in one iteration than the conventional Hungarian method.
The limitation of the study is that there are no computational comparisons with other exact methods in terms of processing time to optimality. In this paper, this is considered as an area for further research.
The only disadvantage of the proposed Accelerating Hungarian method is that it requires more time in handling the many uncovered elements. This challenge can be alleviated by adding additional parallel processors for those additional smallest uncovered elements.
26 35 21 43 28
38 30 40 48 23
28 43 21 28 24
29 15 34 26 35
65 19 26 50 22
c2 c3 c4 c5 c6
7. Conclusions
1. After covering the zeros with the minimum number of lines, some flexible uncovered columns can be identified. These flexible columns can be used to generate more than one smallest uncovered element.
2. More than one smallest uncovered elements results in the creation of more zeros in one iteration. More zeros mean faster Hungarian steps toward the optimal solution.
3. The proposed Accelerating Hungarian method is illustrated in Section 4.5.1. In this numerical illustration, it is shown that iterations are decreased from 6 to only one.
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