the new approach to traveling salesman problem using ... · the new approach to traveling salesman...
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Advances in Fuzzy Mathematics.
ISSN 0973-533X Volume 12, Number 4 (2017), pp. 1017-1033
© Research India Publications
http://www.ripublication.com
The New Approach to Traveling Salesman Problem
using Branch and Bound Method with case study of
Domino’s Pizza Centers
Ms. Nilofer1 and Dr. Mohd. Rizwanullah2
1Research Scholar, Department of Mathematics and Statistics
Manipal University Jaipur, Jaipur, Rajasthan-303007, India.
2Associate Professor, Department of Mathematics and Statistics,
Manipal University Jaipur, Jaipur, Rajasthan-303007, India.
Abstract
The goal of this paper is to optimize delivering of packages at twelve randomly
chosen pizza centers in the city of Jaipur. This problem is also known as the
Travelling Salesman Problem. The Travelling salesman problem involves a
salesman who must make a tour of a number of cities using the shortest path. In
this problem the minimum cost tour of few pizza centers is needed, which are
connected. The cost of different paths is given. The tour should be started from
a given node and after completing the tour the Travelling Salesman has to return
to the starting node. The main ingredients are Two Optimality and Branch and
Bound methods for the TSP.
Key Words:-Branch and Bound technique, Two optimality, Cost matrix,
Hungrerian method.
1 E-mail id:[email protected]
1018 Ms. Nilofer and Dr. Mohd. Rizwanullah
INTRODUCTION
The Travelling Salesman Problem is one of the most studied problems in mathematical
optimization. The possibility to apply this problem to various human activities is what
it makes one of the most recognizable mathematical problems without an ideal solution
so far. The formulation of the problem is simple. The travelling salesman needs to pass
through several towns and return to the starting point of his travel, making the shortest
trip possible. Instead of examining the shortest route, one could examine the one that
costs the least, one that allows the greatest flow of information, etc. By its nature, the
TSP falls into the category of NP-complete problems; meaning there is no algorithm
that would provide a solution for the problem in terms of polynomial time but if the
solution appears, one could test it and conclude whether it is an optimal one. The goal
of this research is to optimize the delivery of packages to appointed addresses in Jaipur.
The delivery man has to deliver packages using the most efficient and shortest route
moving from the distribution center to which he is required return after the delivery of
the last package. This is in fact a Travelling Salesman Problem (Bosancic, V. Golemac,
A. Vojkovic T.) and it can be solved using the branch and bound method (Pielic, M).
The origins of the travelling salesperson problem are unclear. A handbook for travelling
salesmen from 1832 mentions the problem and includes example tours through
Germany and Switzerland, but contains no mathematical treatment. The travelling
salesperson problem was mathematically formulated in the 1800s by the Irish
mathematician W.R. Hamilton and by the British mathematician Thomas Kirkman.
Hamilton’s Icosian Game was a recreational puzzle based on finding a Hamiltonian
cycle. The general form of the TSP appears to have been first studied by mathematicians
during the 1930s in Vienna and at Harvard, notably by Karl Menger, who defines the
problem, considers the obvious brute-force algorithm, and observes the non-optimality
of the nearest neighbour heuristic.
We denote by messenger problem (since in practice this question should be solved by
each postman, anyway also by many travelers) the task to find, for finitely many points
whose pairwise distances are known, the shortest route connecting the points. Of
course, this problem is solvable by finitely many trials. Rules which would push the
number of trials below the number of permutations of the given points, are not known.
The rule that one first should go from the starting point to the closest point, then to the
point closest to this, etc., in general does not yield the shortest route.
Domino’s Pizza Centers in Jaipur (Rajasthan):
Alankar plaza, Vidyadhar nagar
D.P. metro D.P. colony
GT central mall
The New Approach to Traveling Salesman Problem using Branch and Bound.. 1019
krishna enclave
Malviya Nagar
Mansarovar plaza
Metropolitan mall
Pink square
Raj Mandir
Triton mall
Vaibhav cine multiplex, Vaishali Nagar
World trade park
Distance in Kilo Meters (Distance Matrix) by Road
mal
viy
a n
agar
gt
cen
tral
mal
l
wo
rld
tra
de
par
k
man
saro
var
pla
za
d.p
met
ro,
viv
ek v
ihar
met
rop
oli
tan
mal
l (m
gf)
raj
man
dir
pin
k s
qu
are
vai
bh
av c
ine
mu
ltip
lex
vai
shal
i n
agar
alan
kar
pla
za, v
idy
adh
ar
nag
ar
trit
on
mal
l
kri
shn
a en
clav
e, j
ho
twar
a
malviya nagar 0 .2 .3 5 7 8 8 10 12 12 14 18
gt central mall .2 0 .1 5.1 7 8 8 10 12 12 14 17
world trade park .3 .1 0 5 7 8 8 10 11 11 14 18
mansarovar plaza 5 5.1 5 0 3 2 2 3 5 5 7 11
d.p metro, vivek
vihar 7 7 7 3 0 1.5 1 2 4 4 6 10
metropolitan mall
(mgf) 8 8 8 2 1.5 0 1 2 4 4 6 12
raj mandir 8 8 8 2 1 1 0 2 4 4 10 10
pink square 10 10 10 3 2 2 2 0 2 1 4 8
vaibhav cine
multiplex vaishali
nagar
12 12 11 5 4 4 4 2 0 1 2 6
alankar plaza,
vidyadhar nagar 12 12 11 5 4 4 4 1 1 0 2 4
triton mall 14 14 14 7 6 6 10 4 2 2 0 2.5
krishna enclave,
jhotwara 18 17 18 11 10 12 10 8 6 4 2.5 0
The New Approach to Traveling Salesman Problem using Branch and Bound.. 1021
Solution to the problem using branch and bound method:
The input to the method is the cost matrix, which is prepared using the convention:
CIJ={∞, If there is no direct path from vi to vI}
={WIJ, If there is a direct path from vi to vj}
While solving the problem, we first prepare the state space tree, which represents all
possible solution. Here in this problem V=12.which is the number of total nodes on the
graph or the domino’s pizza centers in the map.
The input array for the method is given by:
Step1: Reduce each row and each column in such a way that there must be at least one
zero in each row and each column for doing this, we need to reduce the minimum value
from each element in each row and column.
Malviya
nagar GT WTP Mansarover DPmetro Mgf rajmandir
Pink
square vaishali Vdn triton Jhotwra
Malviya
nagar ∞ 0 0.1 4.8 6.8 7.8 7.8 9.8 11.8 11.8 13.8 17.8
GT 0.1 ∞ 0 5 6.9 7.9 7.9 9.9 11.9 11.9 13.8 16.9
WTP 0.1 0 ∞ 4.9 6.9 7.9 7.9 9.9 10.9 10.9 13.9 17.9
Mansarover 3 3.1 3 ∞ 1 0 0 1 3 3 5 9
DPmetro 6 6 6 2 ∞ 0.5 0 1 3 3 5 9
Mgf 7 7 7 1 0.5 ∞ 0 1 3 3 5 11
Rajmandir 7 7 7 1 0 0 ∞ 1 3 3 9 9
Pink squ. 9 9 9 2 1 1 1 ∞ 1 0 3 7
Vdn 11 11 10 4 3 3 3 1 ∞ 0 1 5
Vaishali 11 11 10 4 3 3 3 0 0 ∞ 1 3
Triton 12 12 12 5 4 4 8 2 0 0 ∞ .5
Jhotwra 15.5 14.5 15.5 8.5 7.5 9.5 7.5 5.5 3.5 1.5 0 ∞
1022 Ms. Nilofer and Dr. Mohd. Rizwanullah
After reducing column
Malviya
nagar
GT WTP Mansarover DPmetro mgf Rajmandir Pinksquare vaishali Vdn triton Jhotwara
Malviya
nagar
∞ 0 0.1 3.8 6.8 7.8 7.8 9.8 11.8 11.8 13.8 17.8
GT 0 ∞ 0 4 6.9 7.9 7.9 9.9 11.9 11.9 13.9 16.4
WTP 0.1 0 ∞ 3.9 6.9 7.9 7.9 9.9 10.9 10.9 13.9 17.4
Mansarover 2.9 3.1 3 ∞ 1 0 0 1 3 3 5 8.5
DPmetro 5.9 6 6 1 ∞ 0.5 0 1 3 3 5 8.5
Mgf 6.9 7 7 0 0.5 ∞ 0 1 3 3 5 10.5
Rajmandir 6.9 7 7 0 0 0 ∞ 1 3 3 9 8.5
Pinks quare 8.9 9 9 1 1 1 1 ∞ 1 0 3 6.5
Vaishali 10.9 11 10 3 3 3 3 2 ∞ 0 1 4.5
Vdn 10.9 11 10 3 3 3 3 0 0 ∞ 1 2.5
Triton 11.9 12 12 4 4 4 8 2 0 0 ∞ 0
Jhotwara 15.4 14.5 15.5 7.5 7.5 9.5 7.5 5.5 3.5 1.5 0 ∞
Total expected cost of expanding root node L(A)=12.8
Because we have to plan the path starting 1, so 1 will be the root of the tree and would
be the first node to be expanded.
Step2.We have discovered the node 1 so the next node to be expanded will be any node
from GT, WTP, Mansarover, DPmetro, Mgf, Rajmandir, Pink square, Vaishali, Vdn,
Triton, Jhotwara. The formula for finding the cost is
L(node) = L(Parent node) + Parent(i, j) + Total cost of reduction
The New Approach to Traveling Salesman Problem using Branch and Bound.. 1023
(a) Obtain cost of expanding using cost matrix for node 2 in the tree;
Change all the elements in 1st row and 2nd column andm1 (2, 1) to ∞.
Malviya
nagar
GT WTP Mansarover DP
metro
Mgf Rajmandir Pink
square
vaishali Vdn Triton Jhotwara
Malviya nagar ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞
GT 0 ∞ 0 4 6.9 7.9 7.9 9.9 11.9 11.9 13.9 16.9
WTP 0.1 ∞ ∞ 3.9 6.9 7.9 0 9.9 10.9 10.9 13.9 17.9
Mansarover 2.9 ∞ 3 ∞ 1 0 0 1 3 3 5 9
DPmetro 5.9 ∞ 6 1 ∞ 0.5 0 1 3 3 5 9
Mgf 6.9 ∞ 7 0 0.5 ∞ 0 1 3 3 5 11
Rajmandir 6.9 ∞ 7 0 0 0 ∞ 1 3 3 9 9
Pinksquare 8.9 ∞ 9 1 1 1 1 ∞ 1 0 3 7
Vaishali 10.9 ∞ 10 3 3 3 3 2 ∞ 0 1 5
Vdn 10.9 ∞ 10 3 3 3 3 0 0 ∞ 1 3
Triton 11.9 ∞ 12 4 4 4 8 2 0 0 ∞ .5
Jhotwara 15.4 ∞ 15.5 7.5 7.5 9.5 7.5 5.5 3.5 1.5 0 ∞
Each row and column already has a zero.so reduction cost=0 so the total cost of
expanding node 2,
L(2)=L(1) +M1(1,2) +0
L(2)=12.8+0+0=12.8
L(3)=L(1)+M1(1,3)+0
L(3)=12.8+.1+0=12.9
L(4)=L(1)+M1(1,4)+0
L(4)=12.8+3.8+0=16.6
L(5)=L(1)+M1(1,5)+0
L(5)=12.8+6.8+0=19.6
L(6)=L(1)+M1(1,6)+0
L(6)=12.8+7.8+0=20.6
1024 Ms. Nilofer and Dr. Mohd. Rizwanullah
L(7)=L(1)+M1(1,7)+1
L(7)=12.8+7.8+1=21.6
L(8)=L(1)+M1(1,8)+0
L(8)=12.8+9.8+0=22.6
L(9)=L(1)+M1(1,9)+0
L(9)=12.8+11.8+0=24.6
L(10)=L(1)+M1(1,10)+2
L(10)=12.8+11.8+2=26.6
L(11)=L(1)+M1(1,11)+1.5
L(11)=12.8+13.8+1.5=28.1
L(12)=L(1)+M1(1,12)+0
L(12)=12.8+17.3+0=30.1
Now to GT is promising. We choose GT. We have 10 nodes still to be traversed. So we
can expand WTP, Mansarover, DPmetro, Mgf, Rajmandir, Pink square, Vaishali, Vdn,
Triton, Jhotwara as the next node. So using same method. We will find the expansion
cost of each of those nodes.
MN
GT
WTP
MS
DP MGF
RM PS
VS
VDN
TRI
JTH
The New Approach to Traveling Salesman Problem using Branch and Bound.. 1025
(b) change all the elements in 1st row and 3rd column and M3(3,1) to ∞
L(3) =13.0
L(4)=17.6
L(5)=19
L(6)=20
L(7)=21
L(8)=22
L(9)=24
L(10)=26
L(11)=28
L(12)=29
Now to WTP is promising. We choose WTP. We have 9 nodes still to be traversed. So
we can expand Mansarover, DPmetro, Mgf, Rajmandir, Pink square, Vaishali, Vdn,
Triton, Jhotwara as the next nodes. So using same method. We will find the expansion
cost of each of these nodes.
(c) Change all the elements of 1st row and 4th column and M4 to ∞
Node 4,
L(4)=17
L(5)=20
L(6)=20
L(7)=21
L(8)=23
L(9)=24
L(10)=26
L(11)=28.5
L(12)=30.5
Now Mansarover is promising. We choose Mansarover. We have 8 nodes still to be
traversed. So we can expand DPmetro, Mgf, Rajmandir, Pink square, Vaishali, Vdn,
Triton, Jhotwara as the next nodes. So using same method. We will find the expansion
cost of each of these nodes.
1026 Ms. Nilofer and Dr. Mohd. Rizwanullah
(d) change all the elements of 1st row and 6th column and M6 to ∞
Node 6,
L(5)=18
L(6)=17
L(7)=17.5
L(8)=18
L(9)=20
L(10)=22
L(11)=23.5
L(12)=25.5
Now Mgf is promising. We choose Mgf. We have 7 nodes still to be traversed. So we
can expand DPmetro, Rajmandir, Pink square, Vaishali, Vdn, Triton, Jhotwara as the
next nodes. So using same method. We will find the expansion cost of each of these
nodes.
(e) Change all the elements of 1st row and 7th column and M5 to ∞
Node 7,
L(5)=17.5
L(7)=17.5
L(8)=18
L(9)=20
L(10)=22
L(11)=23.5
L(12)=30
Now DPmetro is promising. We choose DPmetro. We have 6 nodes still to be traversed.
So we can expand Rajmandir, Pink square, Vaishali, Vdn, Triton, Jhotwara as the next
nodes. So using same method. We will find the expansion cost of each of these nodes.
(f) change all the elements of 1st row and 5th column and M5 to ∞
L(7)=17.5
L(8)=18.5
L(9)=20.5
The New Approach to Traveling Salesman Problem using Branch and Bound.. 1027
L(10)=20.5
L(11)=22.5
L(12)=25.5
Now Rajmandir is promising. We choose Rajmandir. We have 5 nodes still to be
traversed. So we can expand Pink square, Vaishali, Vdn, Ttriton, Jhotwara as the next
nodes. So using same method.
Node 8,
L(8)=19
L(9)=21
L(10)=23
L(11)=28.5
L(12)=26.5
Now Pink square is promising. We choose Pink square. We have 4 nodes still to be
traversed. So we can expand Vaishali, Vdn, Triton, Jhotwara as the next nodes. So using
same method.
Node 9,
L(9)=21
L(10)=20.5
L(11)=23.5
L(12)=29
Now VDN is promising. We choose Vdn. We have 3 nodes still to be traversed. So we
can expand Vaishali, Triton, Jhotwara as the next nodes. So using same methods.
Node 10,
L(9)=22.5
L(11)=23.5
L(12)=25
Now Vaishali is promising. We choose Vaishali. We have 2 nodes still to be traversed.
So we can expand triton and Jhotwara as the next node. Again using same method.
Node 9
L(11)=23.5
1028 Ms. Nilofer and Dr. Mohd. Rizwanullah
L(12)=27
Now Triton is promising. We choose triton. We have 1 node still to be traversed.so we
can expand Jhotwara as the next node.so using same method.
Node 11
L(12)=23.5
Step5. Here Jhotwara is the most promising node so next we are going to expand this
node further. Now we are left with only one node not yet traversed. Then the tour is
completed so we will return back to the node malviya nagar. Now the sequence of
traversal is:
Malviya nagar G.T. WTP Mansarover MGF DP
metro Rajmandir Pinksquare Vdn Vaishali Triton
Jhotwara Malviya nagar.
Therefore, the total cost of traversing the graph is
i=.2+.1+5+2+1.5+1+2+1+1+2+2.5+18=36.3km
Approximate methods (By two optimality method):-A formal algorithm may be
presented as:
Step0. (Create an initial tour) select a starting point s. choose t, so that d(s, t) ≤ d(s, j)
for all j ≠ t. Set l=t, and make nodes s and t visited.
Step1. Select t from the unvisited nodes so that d(i, t) is least. Add t to the end of the
tour and set l=t. if there are further unvisited nodes, repeat step1, otherwise add s to the
tour and do step 2.
Step2. The visited tour will be an ordered set of nodes:
x1, x2, x3 ……..,xn, x1, with total length L.
Set i=1
Step3. Set j=i+2.
Step4. Consider the tour
x1, x2, ……xi, xj, xj-1, ……xi+1, xj+1, xj+2,…..x1
Create by exchanging the links (xi, xi+1) and (xj, xj+1). If this has length less than L, then
make this the new tour, and do step2.
Step5. Set j=j+1. If j ≤ n, do step 4. Otherwise, set i=i+1. If i≤ n-2, do step 3. Otherwise
stop.
The New Approach to Traveling Salesman Problem using Branch and Bound.. 1029
Solved by Two Optimal Method.
Step0: Let s=1, then t=3 and the partial tour is 1-2
Step1:- set t=3, then tour is 1-2-3
Step1:- set t=4, then tour is 1-2-3-4
Step1:- set t=6, then tour is1-2-3-4-6
Step1:- set t=5, then tour is1-2-3-4-6-5
Step1:- set t=7, then tour is 1-2-3-4-6-5-7
Step1:- set t=8, then tour is 1-2-3-4-6-5-7-8
Step1:- set t=10, then tour is 1-2-3-4-6-5-7-8-10
Step1:- set t=9, then tour is 1-2-3-4-6-5-7-8-10-9
Step1:- set t=11, then tour is 1-2-3-4-6-5-7-8-10-9-11
Step1:- set t=12, then tour is 1-2-3-4-6-5-7-8-10-9-11-12-1
There is no unvisited nodes, so the tour is established as
1-2-3-4-6-5-7-8-10-9-11-12-1
Step2:- i=1,
L=d(1,2)+d(2,3)+d(3,4)+d(4,6)+d(6,5)+d(5,7)+d(7,8)+d(8,10)+d(10,9)+d(9,11)+d(11,
12)+d(12,1)
L=.2+.1+5+2+1+1+2+1+1+2+1+18
L=34.3km.
Step3:- j=i+2, 1+2=3
Step4:- consider the tour 1-3-2-4-6-5-7-8-10-9-11-12-1
This has length is 36.5 km.
Step5:- j=4,
Step4:-consider the tour 1-4-3-2-6-5-7-8-10-9-11-12-1
This has length is 43.1 km.
Step5:-j=5
Step4:-consider the tour 1-6-4-3-2-5-7-8-10-9-11-12-1
This has length is 49km.
Step5:-j=6
1030 Ms. Nilofer and Dr. Mohd. Rizwanullah
Step4:-consider the tour1-5-6-4-3-2-7-8-10-9-11-12-1
This has length is 48 km.
Step5:-j=7
Step4:-consider the tour 1-7-5-6-4-3-2-8-10-9-11-12-1
This has length is 55km.
Step5:- j=8
Step4:- consider the tour 1-8-7-5-6-4-3-210-9-11-12-1
This has length is 56 km.
Step5:- j=9
Step4:- consider the tour 1-10-8-7-5-6-4-3-2-9-11-12-1
This has length is 58 km.
Step5:- j=10
Step4:- consider the tour 1-9-10-8-7-5-6-4-3-2-11-12-1
This has length is 59 km.
Step5:- j=11
Step4:- consider the tour 1-11-9-10-8-7-5-6-4-3-2-12-1
This has length is 65 km.
Step5:- j=12
Step4:- consider the tour 1-12-11-9-10-8-7-5-6-4-3-2-1
This has length is 36.3 km.
Step5:- j=13, set i=2
Step3:- j=i+2=4
Step4:- consider the tour 1-2-4-3-6-5-7-8-10-9-11-12-1
This has length is 45.2km.
Step5:- j=5
Step4:- consider the tour 1-2-6-4-3-5-7-8-10-9-11-12-1
This has length is 47.2km.
Step5:- j=6
Step4:- consider the tour1-2-5-6-4-3-7-8-10-9-11-12-1
The New Approach to Traveling Salesman Problem using Branch and Bound.. 1031
This has length is 48.2
Step5:-j=7
Step4:- consider the tour 1-2-7-5-6-4-3-8-10-9-11-12-1
This has length is 51.1km.
Step5:-j=8
Step4:-consider the tour 1-2-8-7-5-6-4-3-10-9-11-12-1
This has length is 54.1km.
Step5:-j=9
Step4:- consider the tour 1-2-9-8-7-5-6-4-3-10-11-12-1
This has length is 57.1km.
Step5:- j=10
Step4:- consider the tour 1-2-10-9-8-7-5-6-4-3-11-12-1
This has length is 59.1km
Step5:-j=11
Step4:-consider the tour 1-2-11-10-9-8-7-5-6-4-3-12-1
This has length is 64.1 km.
Step5:-j=12
Step4:- consider the tour 1-2-12-10-9-8-7-5-6-4-3-11-1
This has length is 53.1km
Step5:-j=13, set i=3
Step3:-j=i+2=5
Step4:-consider the tour 1-2-3-6-4-5-7-8-10-9-11-12-1
This has length is 38.2km.
Step5:-j=6
Step4:-consider the tour 1-2-3-5-6-4-7-8-10-9-11-12-1
This has length is 37.3km.
1032 Ms. Nilofer and Dr. Mohd. Rizwanullah
CONCLUSION:
The Travelling Salesman Problem, which is solving by using Two optimality and
Branch & Bound method and Branch and Bound method is better because it prepares
the matrices in different steps. At each step the cost matrix is calculated. This method
is very easy to apply and can be utilized for the Travelling Salesman Problem. Therefore
Branch and Bound method is efficient as compare to Two Optimality Method. The
problem further be extended to higher order of nodes and arcs. The distance in the initial
stages is not exact distance but it gives some idea because it is the approximate distance.
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