The New Approach to Traveling Salesman Problem using ... · The New Approach to Traveling Salesman...

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]

2 [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

https://www.dominos.co.in/store-location/jaipur/alankar-plaza,-vidyadhar-nagar,-jaipur-(rajasthan)/785

https://www.dominos.co.in/store-location/jaipur/d.p-metro-d.p.-colony,-jaipur,-rajasthan/1086

https://www.dominos.co.in/store-location/jaipur/gt-central-mall,-jaipur-(rajasthan)/760

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

https://www.dominos.co.in/store-location/jaipur/krishna-enclave,-jaipur,-rajasthan/998

https://www.dominos.co.in/store-location/jaipur/malviya-nagar/151

https://www.dominos.co.in/store-location/jaipur/mansarovar-plaza,-jaipur-(rajasthan)/691

https://www.dominos.co.in/store-location/jaipur/metropolitan-mall/293

https://www.dominos.co.in/store-location/jaipur/pink-square/443

https://www.dominos.co.in/store-location/jaipur/raj-mandir/139

https://www.dominos.co.in/store-location/jaipur/triton-mall,-jaipur,-rajasthan/1127

https://www.dominos.co.in/store-location/jaipur/vaibhav-cine-multiplex-vaishali-nagar/378

https://www.dominos.co.in/store-location/jaipur/world-trade-park,-jaipur,-rajasthan/1066















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 ∞


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


(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


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


(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.


(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


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


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.


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


Step5:-j=5


This has length is 49km.

Step5:-j=6


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


This has length is 55km.

Step5:- j=8

Step4:- consider the tour 1-8-7-5-6-4-3-210-9-11-12-1


Step5:- j=9



Step5:- j=10



Step5:- j=11



Step5:- j=12



Step5:- j=13, set i=2

Step3:- j=i+2=4


This has length is 45.2km.

Step5:- j=5



Step5:- j=6

Step4:- consider the tour1-2-5-6-4-3-7-8-10-9-11-12-1


This has length is 48.2

Step5:-j=7



Step5:-j=8



Step5:-j=9



Step5:- j=10


This has length is 59.1km

Step5:-j=11



Step5:-j=12


This has length is 53.1km

Step5:-j=13, set i=3

Step3:-j=i+2=5



Step5:-j=6




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.

REFERENCES

[1] A.K. Rastogi, A.K. Shrivastava, N. Payal, R. Singh, A Proposed solution to

Travelling Salesman Problem using Branch and Bound, International Journal

of Computer Applications, Vol.65, 2013, No.5, (0975-8887).

[2] C. Ding, Cheng, Y., He, M.: Two-Level Genetic algorithm for Clustered

Traveling Salesman Problem with Application in Large Scale TSPs, Tsinghua

Science and Technology, Vol.12.No.4 (2007) pp. 459-465.

[3] D.L. Applegate, Bixby, R.E., Chvatal, V., Cook, W.J.: The Traveling Salesman

Problem, A Computational Study, Princeton University Press, Princeton and

Oxford (2006) pp. 10-11.

[4] E. Balas, Branch and Bound methods for the Travelling Salesman Problem,

Management Science Research Report no. MSRR 488, March 1983.

[5] E.J. Teoh, K.C., Tang, H.J., Xiang, C., Goh, C.K.: An asynchronous recurrent

linear threshold network approach to solving the traveling salesman problem,

Neuro-computing 71 (2008) pp. 1359137.(12).

[6] J. Botzheim, Drobics M., Kóczy L. T.: Feature selection using bacterial

optimization, in Proceedings of the International Conference on Information

Processing and Management of Uncertainty in Knowledge-based Systems,

IPMU 2004, Perugia, Italy, (July 2004) pp. 797–804.

[7] H. Bashirzadeh, An Approach for solving Fuzzy Transportation Problem,

Applied Mathematical sciences, Vol.5, 2011, no.32, (1549-1566).(7)

[8] L. Yu-Hsin, A hybrid scatter search for the probabilistic traveling salesman

problem, Computers & Operations Research 34 (2007) pp. 2949-2963.

[9] N.E. Nawa, Furuhashi, Fuzzy System Parameters Discovery by Bacterial

Evolutionary Algorithm, IEEE Tr. Fuzzy Systems 7 (1999) pp. 608-616.


[10] R. Weiner, Branch and Bound Implementation for the Travelling salesperson

Problem, Journal of Object Technology, Vol.2, 2003, no.2.

[11] S. Dhanasekar, S.Hariharan, P.Sekar,“Classical Travelling Salesman Problem

based approach to solve Fuzzy TSP using Yagers Ranking, International

Journal of Computer Applications, Vol.74, 2013, no.13, (0975-8887).

[12] S. Chandrasekaran, G. Kokila, J.Saju, A New Approach to solve Fuzzy

Travelling salesman Problems by using Ranking Functions, International

Journal of science and Research, Vol.4, 5 may 2015.(11)

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