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Chapter 6
TRAVELLING SALESMAN PROBLEM
6.1 Introduction
The Traveling Salesman Problem (TSP) is a problem in combinatorial
optimization studied in operations research and theoretical computer science.
Given a list of cities and their pair wise distances, the task is to find a shortest
possible tour that visits each city exactly once.
The problem was first formulated as a mathematical problem by Menger
(1930) and is one of the most intensively studied problems in optimization.
However it was unnoticed till Menger (1994) published a book where he narrated
the foundation of mathematical problem for the travelling salesman problem. It is
used as a benchmark for many optimization methods. Even though the problem
is computationally difficult, a large number of heuristics and exact methods are
known, so that some instances with tens of thousands of cities can be solved.
The TSP has several applications even in its purest formulation, such as
planning, logistics, and the manufacture of microchips. If a slight modification is
made in the problem, it appears as a sub-problem in many areas, such as
genome sequencing. In these applications, the concept city represents, for
example, customers, soldering points, or DNA fragments, and the concept
distance represents traveling times or cost, or a similarity measure between DNA
fragments. In many applications, additional constraints such as limited resources
or time windows make the problem considerably harder. In the theory of
computational complexity, the decision version of TSP belongs to the class of
NP-complete problems. Thus, it is assumed that there is no efficient algorithm for
solving TSP problems. In other words, it is likely that the worst case running time
for any algorithm for TSP increases exponentially with the number of cities, so
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even some instances with only hundreds of cities will take many CPU years to
solve exactly.
The origins of the traveling salesman problem are unclear. A handbook for
traveling salesmen from 1832 mentions the problem and includes example tours
through Germany and Switzerland, but contains no mathematical treatment.
Hamilton (1800) and Kirkman (1800) expressed the concept of Mathematical
problems related to the travelling salesman problem. The general form of the
TSP appears to have been first studied by mathematicians notably by Menger
(1930). Further Menger (1930) also defines the problem related with salesman
ship based on brute-force algorithm, and observes the non-optimality of the
nearest neighbor heuristic. However Whitney (1930) introduced the name
travelling salesman problem.
During the period 1950 to 1960, the travelling salesman problem started
getting popularity in scientific circle is especially in Europe and the USA. Many
researchers like Dantzig, Fulkerson and Johnson (1954) at the RAND
Corporation in Santa Monica expressed the problem as an integer linear program
and developed the cutting plane method for its solution. With these new methods
they solved an instance with 49 cities to optimality by constructing a tour and
proving that no other tour could be shorter. In the following decades, the problem
was studied by many researchers from mathematics, science, chemistry,
physics, and other sciences. Karp (1972) showed that the Hamiltonian cycle
problem was NP-complete, which implies the NP-hardness of TSP. This supplied
a scientific explanation for the apparent computational difficulty of finding optimal
tours.
6.2 Application of New Alternate Method of Assignment Problem in TSP
We have so far already discussed the algorithm and examples for solving an
assignment problem using a new alternate method in Chapter 5. Now in this
chapter we discuss how the new alternate method for solving an assignment
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problem can be applied for Travelling Salesman Problem. For this we have
considered an example related with travelling salesman problem and explain in
detail how to find optimal solution using new alternate method of assignment
problem.
Example 6.2.1 A salesman has to visit five cities A.B, C, D and E. The distances
(in hundred kilometers) between the five cities are shown in Table 6.1.
Table 6.1
From
City
A
B
C
D
E
A
7
6
8
4
B
1
8
5
6
To city
C
6
8
9
7
D
8
5
9
8
E
4
6
7
8
-
If the salesman starts from city A and has to come back to city A, which route
should he select so that total distance traveled become minimum?
Solution
Consider the effective matrix. This is shown in Table 6.2.
Table 6.2
From
City
A
B
C
D
E
A
7
6
8
4
B
1
8
5
6
To city
C
6
8
9
7
D
8
5
9
8
E
4
6
7
8
-
In this matrix first, we will take first row which is referred a city. We select that
column (assignment) for which it contains minimum distance. For this example,
incase of first row, column B (assignment) has the minimum value. In the similar
way, we select all the rows and find the minimum value for the respective
columns. These are given in Table 6.3.
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Table 6.3
Column 1(City)
Colum 2(Assignment)
A
B
B
D
C
A
D
B
E
A
In this table, we observed that assignment D occur only once with city B. That is
city B is unique for city D and hence we assign city B to D. This is shown in Table
6.4.
Table 6.4
A
B
A
7
B
1
-
C
6
8
C
D
E
6
8
4
8
5
6
9
7
D
8
E
4
6
9
8
7
8
-
5
However, for other job assignment occur more than once. Hence they are not
unique. So how other job will be assigned further we discuss below.
Next delete row B and column D. Again select minimum cost value for the
remaining cities which is shown below Table 6.5.
Table 6.5
Column 1(City)
Colum 2(Assignment)
A
B
C
A
D
B
E
A
Since assignment B occur with city A and D. Hence first we take the difference
between the value of B and next minimum value (here tie is happens). Here
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maximum difference is 2 for A and hence we assign B to city A. This is shown in
Table 6.6.
Table 6.6
A
A
B
-
1
C
D
E
6
8
4
8
5
6
C
6
E
9
7
7
8
-
4
Next delete row A and column B. Again select minimum cost value for the
remaining cities which is shown below Table 6.7.
Table 6.7
Column 1(City)
Colum 2(Assignment)
C
A
D
A, E
E
A
Since assignment A occur with city C, D and E. Hence we take the difference
between the value of A and next minimum value, here the maximum difference is
3 for Job E. And hence we assign A to city E. This is shown in Table 6.8.
Table 6.8
A
6
8
4
C
D
E
C
9
7
E
7
8
-
Next delete row E and column A. Again select minimum cost value for the
remaining cities which is shown below Table 6.9.
Table 6.9
Column 1(City)
Column 2(Assignment)
C
E
D
E
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Here, we cannot assign C to city C. Therefore we only assign E to city C. Then
obviously, we have no other choice rather to assign C for City D.
Finally, we can assign all the cities along with distance which is shown in Table
6.10.
Table 6.10
Column 1(City)
A
Colum 2(Assignment) Distance
B
1
B
D
5
C
E
7
D
C
9
E
A
4
Total
26
This solution is happened to be same as that of Hungarian method. Hence we
can say that the minimum value is still 26 in both the methods. So this solution is
optimal. However our method seems to be very simple, easy and takes very few
steps in solving the method.
6.3 Conclusion
In this chapter, we have applied new alternate method of an assignment problem
for solving Travelling salesman problem where it is shown that this method also
gives optimal solution. Moreover the optimal solution obtained using this method
is same as that of optimal solution obtained by Hungarian method. So we
conclude that the Hungarian method and our method give same optimal solution.
However the technique for solving Travelling Salesman problem using our
method is more simple and easy as it takes few steps for the optimal solution.