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CHAPTER – V

VARIANT CONSTRAINT TRAVELLING SALESMAN

PROBLEM

5.1. Introduction:

There are many algorithms for usual one man TSP developed by

researchers from time to time. But the problem has not received

much attention in its restricted context. However, literatures which

are available in regard to the TSP with variations are discussed (Das -

1976, 78, Jaillet P - 1985, Kubo & Kasugai - 1991, Pandit - 1961, 63,

64, 65, Ramesh - 1981, Raviganesh G.S. Murthy & Das - 1998 and

Srivastava Kumar, R.C. Garg and P. Sen - 1969).

The earliest papers on the Generalized Travelling Salesman

Problem discuss the problem in the context of particular applications

by Henry-Labordere (1969), Saskena (1970), and Srivasatava et al

(1970). Laporte et al. (1985&1987) and Laporte and Nobert (1983)

began to study exact algorithm and some theoretical aspects of the

problem in the 1980s. Fischetti et al (1995) discuss facet-defining

inequalities for the GTSP polytope, and in a later paper Fischetti et al

(1997) use the polyhedral results to develop a branch and cut

algorithm. Other exact algorithms are presented by Noon (1991) (A

lagrangian-based branch and bound algorithm] and Chenstov (1997)

(A dynamic programming algorithm).

A variety of descriptions of the GTSP are presented in the

literature. Some papers assume symmetry or the triangle inequality;

other do not some require the subsets to form a strict partition of the

node set, others allow them to overlap. Some require that exactly one

node per cluster be visited. (The two formulations are equivalent if the

distance matrix satisfies the triangle inequality). Only a few authors

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Laporte et al (1983), Saskena et al (1970) handle fixed costs for

including a node on the tour, possibly because such fixed costs can be

incorporated into the distance costs via a simple transformation.

Applications of the GTSP are discussed by Ben-Arieh et al

(2003), Dimitrijevic et al(1997) and Noon(1988). Dror et al(1997)

transforms the clustered rural postman problem into the GTSP. Ben-

Arieh et al (2003), Laporte et al(1999),Lien et al (1993) and Noon(1993)

present transformations of the GTSP into standard TSP instances that

can then be solved using TSP algorithms and heuristics. Some of the

resulting TSP instances have nearly the same number of nodes as the

original GTSP instances and have nice properties like symmetry and

the triangle inequality; others have many more nodes and are more

irregular. Some transformations of the GTSP into the TSP by Noon

(1993) have the property that an optimal solution to the related TSP

can be converted to an optimal solution to the GTSP, but a (sub-

optimal) feasible solution for the TSP may not perform well for the

GTSP; for example, “generalized” versions of Christofides‟s heuristic

have worst-case bounds that are strictly greater than its worst-case

bound of 3/2 for the TSP by Dror et al (1997) and Slavik(1997).

Two approximation algorithms have been published for the

GTSP. Slavik (1997) develops a 3 /2-approximation algorithm for the

GTSP, where is the number of nodes in the largest cluster, that is,

=max {1vi1},i=1,2,….,m. As may be quite large, the worst-case

bound may be relatively weak. Garg et al (2000) present an

approximation algorithm for the group Steiner problem, which

provides as a byproduct an O (log21v1loglog1v1logm)-approximation

algorithm for GTSP. Both papers assume that the distance matrix

satisfies the triangle inequality.

94

Noon (1988) proposed several heuristics for the GTSP, the most

promising of which is an adaptation of the well-known nearest-

neighbor heuristic of the TSP. Fischetti et al (1997) propose similar

adaptations of the farthest-insertion, nearest-insertion, and cheapest-

insertion heuristics.

Renaud and Boctor (1998) develop the most sophisticated

heuristic published to date, which they call GI3 [Generalized

Initialization, Insertion, and Improvement]. GI3 is a generalization of

the I3 heuristic proposed by Renuaud (1996). It consists of three

phases. The first, Initialization, constructs an initial tour by choosing

one node from each cluster that is “close” to other clusters, then

greedily building a tour that passes through some, but necessarily all,

of the chosen nodes. The second phase, Insertion, completes the tour

by successively inserting nodes from unvisited clusters in the

cheapest possible manner between two consecutive clusters on the

tour, allowing the visited node to change for the adjacent clusters;

after each insertion, the heuristic performs a modification of the 3-opt

improvement method. The third phase, Improvement, uses

modifications of 2-opt and 3-opt to improve the tour. The

modifications, called G2-opt, G3-opt and G-opt, allow the visited

nodes from each cluster to change as the tour is being re-ordered by

the 2-opt or 3-opt procedure. But in all the above attempts the simple

combinatorial structure of the GTSP is not at all taken into

consideration.

The standard TSP is called „Two Dimensional TSP‟. The standard

TSP has been generalized in many directions In the present problem

we have considered the third dimension time (availing facility) which

influences the cost, implementing the Lexicographic Search Approach,

and make use of Pattern Recognition Technique which take care of the

simple combinatorial structure of the problem.

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In this chapter we study a problem called “Variant Constraint

Travelling Salesman Problem” (VCTSP).

“There are n cities and N= {1, 2, 3,... ,n}. The cost array C (i, j, k)

is the cost of a salesman visiting from city i to city j at time (availing

facility) k is known (i, j=1, 2, 3,…, n; k=1, 2, 3,…..m)”.Here the third

dimension need not be the usual time which is continuous, but a

factor which influences the cost C and can be a facility.

A variant of well – known Traveling Salesman Problem where a

tour does not necessarily visit all cities is called the Generalized

Traveling Salesman Problem. More specifically, the set of „n‟ cities are

divided into r sets such that the N = 𝑁1, 𝑁2, 𝑁3, … 𝑁𝑟 and 𝑁𝑖 ∩ 𝑁𝑗 = ∅.

A subset with m < n cities has to be traveled by the salesman. The

number of cities travelled by as salesman is m. The Travelling

Salesman has to visit 𝑛𝑝 cities in the 𝑁𝑝 sets. The problem is to find a

minimum cost tour by visiting „m‟ cities with given number of 𝑛𝑝 cities,

where 𝑛𝑝𝑟𝑝=1 ≤ 𝑚.

In the sequel we will develop a lexi-search algorithm based on

the “Pattern Recognition Technique” to solve this problem which takes

care of the simple combinatorial structure of the problem.

Lawrence V. Snyder and Mark S. Daskin (2006), have also

attempted the above problem with two dimensional problem and the

travelling salesman has to visit only one city in each cluster, which

was published in European Journal of Operational Research.

5.2. Mathematical Formulation:

𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 𝐶 𝑖, 𝑗, 𝑘 𝑋(𝑖, 𝑗, 𝑘)𝑚𝑘=1

𝑛𝑗=1

𝑛𝑖=1 ----------- (5.1)

Subject to

𝑋 𝑖, 𝑗, 𝑘 = 1,𝑛𝑗=1

𝑛𝑖=1 k = 1,2, …, m ---------(5.2)

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𝑋 𝑖, 𝑗, 𝑘 = 𝑚,𝑚𝑘=1

𝑛𝑗=1

𝑛𝑖=1 -------- (5.3)

𝑋 𝑖, 𝑗, 𝑘 = 0 𝑜𝑟 1 ------- (5.4)

Constraints (5.2) and (5.3) and the restrictions

are 𝑋 𝑖, 𝑗, 𝑘 = 𝜉𝑖𝑚𝑘=1

𝑛𝑗=1 , 𝜉𝑖 = 1, 𝑖 𝜖 𝑁𝑝 & 𝜉𝑖 = 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 and 𝜉𝑖 =𝑖𝜖 𝑁𝑝

𝑛𝑝 , 𝑋 𝑖, 𝑗, 𝑘 = 𝜉𝑗𝑚𝑘=1

𝑛𝑖=1 , 𝜉𝑗 = 1, 𝑗 𝜖 𝑁𝑝 & 𝜉𝑗 = 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 and

𝜉𝑗 = 𝑛𝑝𝑗𝜖 𝑁𝑝 define the constraint set of the generalized TSP, whose

objective function (5.1) is minimum. Constraint (5.4.) represents that

the Salesman visits the city j from city i at time or facility k is 1

otherwise 0. X is a feasible tour is it satisfies all the constraints and

restrictions.

5.3. Numerical Illustration:

The concepts and the algorithm developed will be illustrated by

a numerical example for which n = 6, m = 4 and let N = [1, 2, 3, 4, 5,

6]. The costs of the numerical example are given below the table – 5.1.

Table – 5.1

𝐶 𝑖, 𝑗, 1 =

999 40 7 45 1 960 999 30 51 39 74 7 999 7 21 20

36 6 15 999 46 391 29 3 21 999 405 29 2 40 7 999

𝐶 𝑖, 𝑗, 2 =

999 14 30 2 10 2617 999 15 29 4 822 19 999 16 39 501 7 17 999 5 29

47 30 36 5 999 2041 39 14 1 7 999

97

𝐶 𝑖, 𝑗, 3 =

999 3 1 40 20 3042 999 30 40 17 75 35 999 29 1 42 16 29 999 45 38

41 30 35 16 999 117 21 38 4 2 999

𝐶 𝑖, 𝑗, 4 =

999 7 41 1 17 2139 999 4 5 17 289 21 999 35 39 15

34 2 1 999 16 401 54 14 32 999 6

26 6 34 7 1 999

In the numerical example given in Table-1, C (i, j, k), i=j, i,

j=1,2,3,4,5,6, k=1,2,3,4 are taken to be very high number 9999

because they are irrelevant for finding tours of the salesman. Though

the entire C (i, j, k) are taken as non – negative integers it can be

easily seen that this is not a necessary condition and the cost as well

be negative quantities. Similarly n1, n2, n3 need not be 1 and can take

any integer values where in m n . In this example we consider the

first two cities i.e. 1 and 2 are considered as a first group and the

salesman has compulsory visited one city from this group, 3 and 4 are

considered as second group and the salesman has compulsory visited

one city from this group, the remaining cities 5 and 6 are considered

as third group and the salesman has compulsory visited one city from

this group. A tour is a feasible trip-schedule for the salesman. Trip-

schedule of the Salesman can be represented by an appropriate n*n*m

indicator array X=[X (i, j, k), X (i, j, k) = 0 or 1 in which X (i, j, k) =1

indicates that the salesman visits city j from city i at time (availing

facility) k and if there is no such trip it is indicated by X (i, j, k) =0]. X

is called a „solution‟. In this problem, the sum of cities in the tours

must be equal to number of truncated cities m. The indicator X given

in Table-5.2 is represented as four matrices for different values of k

i.e., 1,2,3,4 is a solution to the numerical example and is represented

as follows:

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Table – 5.2

𝑋 𝑖, 𝑗, 1 =

0 0 0 0 0 00 0 0 0 0 00 0 0 0 1 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

𝑋 𝑖, 𝑗, 2 =

0 0 1 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

𝑋 𝑖, 𝑗, 3 =

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 1 0 0 0 00 0 0 0 0 0

𝑋 𝑖, 𝑗, 4 =

0 0 0 0 0 01 0 0 0 1 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

The representation of the solution X to the problem is that the

salesman visits city 3 from city 1 that is from group 1 to group 2 at

time 2, salesman visits city 5 from city 3 that is from group 2 to group

3 at time 1, salesman visits city 2 from city 5 that is from group 3 to

group 1 at time 3 and visits city 1 from city 2 that is group 1 to group

1 at time 4. This solution is a feasible solution.

5.4. Concepts and Definitions:

5.4.1. Definition of a Pattern:

An indicator three-dimensional array which is associated with a

tour is called a pattern. A pattern is said to be feasible if X is a feasible

solution. The pattern represented in the table-5.2 is a feasible pattern.

Now V(x) the value of the pattern X is defined as

𝑉 𝑋 = 𝐶 𝑖, 𝑗, 𝑘 𝑋(𝑖, 𝑗, 𝑘)𝑚𝑘=1

𝑛𝑗=1

𝑛𝑖=1

The value V(X) gives the total cost of the tour for the solution

represented by X. Thus the value of the feasible pattern gives the total

cost represented by it. In the algorithm, which is developed in the

sequel, a search is made for a feasible pattern with the least value.

99

Each pattern of the solution X is represented by the set of ordered

triples [(i, j, k)] for which X (i, j, k) =1, with the understanding that the

other X (i, j, k)‟s zeros.

There are M= n n m ordered triples in the three-dimensional

array X. For convenience these are arranged in ascending order of

their corresponding costs and are indexed from 1 to M (Sundara

Murthy-1979). Let SN= [1, 2, 3…M] be the set of M indicies. Let V be

the corresponding array of costs. If a, bSN and a < b then D(a) D

(b). Also let the arrays R, C, T be the array of row, column and time

indicies of the ordered triples represented by SN and DC be the array

of cumulative sums of the elements of D. The arrays SN, D, DC, R, C,

T for the numerical example are given in the table-5.3. If P SN then

(R(P), C(P),T(P)) is the ordered triple and D(a)= C(R(a),C(a),T(a)) is the

value of the ordered triple and DC(a)= 𝐷(𝑖)𝑎𝑖=1 .

TABLE-5.3

ALPHABET - TABLE

S.NO. D DC R C T

1 1 1 1 5 1

2 1 2 5 1 1

3 1 3 4 1 2

4 1 4 6 4 2

5 1 5 1 3 3

6 1 6 3 5 3

7 1 7 1 4 4

8 1 8 4 3 4

9 1 9 5 1 4

10 1 10 6 5 4

100

11 2 12 6 3 1

12 2 14 1 4 2

13 2 16 4 1 3

14 2 18 6 5 3

15 2 20 4 2 4

16 3 23 5 3 1

17 3 26 1 2 3

18 4 30 3 1 1

19 4 34 2 5 2

20 4 38 3 6 3

21 4 42 6 4 3

22 4 46 2 3 4

23 5 51 6 1 1

24 5 56 4 5 2

25 5 61 5 4 2

26 5 66 3 1 3

27 5 71 2 4 4

28 6 77 4 2 1

29 6 83 5 6 4

30 6 89 6 2 4

31 7 96 1 3 1

32 7 103 2 6 1

33 7 110 3 2 1

34 7 117 3 4 1

35 7 124 6 5 1

36 7 131 4 2 2

101

37 7 138 6 5 2

38 7 145 2 6 3

39 7 152 6 1 3

40 7 159 1 2 4

41 7 166 6 4 4

42 8 174 2 6 2

43 9 183 1 6 1

44 9 192 3 1 4

45 10 202 1 5 2

-- -- -- -- -- --

-- -- -- -- -- --

-- -- -- -- -- --

Let us consider 12 SN. It represents the ordered triple

(R(12),C(12),T(12))=(1, 4, 2). Then D (12) =C (1, 4, 2) = 2 and DC

(12) = 14

5.4.2. Definition of an Alphabet-Table and a word:

Let LK= {a1, a2,……,aK}, ai SN be an ordered sequence of k

indicies from SN. The pattern represented by the ordered triples

whose indicies are given by Lk is independent of the order of ai in the

sequence. Hence for uniqueness the indicies are arranged in the

increasing order such that ai < ai+1, i=1, 2… n-1. The set SN is defined

as the “Alphabet-Table” with alphabetic order as (1, 2, 3,…, M) and

the ordered sequence LK is defined as a “word” of length k. A word Lk

is called a “Sensible word” if ai<ai+1, for i=1, 2, 3…k-1 and if this

condition is not met it is called a “insensible word”. A word LK is said

to be feasible if the corresponding pattern X is feasible and same is

with the case of infeasible and partial feasible. Therefore a partial

102

feasible word is said to be feasible word if k=m. A partial word Lk is

said to be feasible if the block of words represented by LK has at least

one feasible word or, equivalently the partial pattern represented by Lk

should not have any inconsistency. Any of the letters in SN can

occupy the first place in the partial word L k . Consider LK-1 =

(a1,a2,……,ak-1). The alphabet table for the kth position is SNak-1=(ak-

1+1,ak-1+2…M), where SNP is defined as SNP =(p+1,p+2,…….,M). Thus

for example consider a word with two letters as (a1, a2) = (1, 3). Then

SNa2=SN3= (4, 5, 6…120) is the alphabet for the third position. We

concentrate on the set of words of length at most m (for the numerical

example it is 4). A leader Lk (k<m) is said to be feasible, if the block of

words defined by it contains at least one feasible word or equivalently

there should not be inconsistency in the partial pattern defined by the

partial word.

The value of the LK , V (LK) is defined recursively as V (LK)=V (LK-1)+ D

(aK) with V (L0)=0 obviously this V(LK) and V(X) the values of the

pattern X represented by L k will be same (Sundara Murthy-1979).

5.4.3. Lower Bound of a partial word LB (LK):

A lower bound LB (L k ) for the values of the block of words

represented by L k can be defined as follows:

LB (LK) = V (LK) + j)D(akm

1j

k

= V (LK) +DC (ak + m-K)-DC (ak)

5.4.4. Feasibility Criterion of a Partial word:

A recursive algorithm is developed for checking the feasibility of

a partial word LK+1= (a1, a2,…..,ak, ak+1) given that LK is a feasible

partial word. We will introduce some more notations which will be

useful in the sequel.

103

IR be an array where IR (i) =1, i N represents that the salesman

is visiting some city from city i, otherwise zero.

IC be an array where IC (i) =1, i N represents that the salesman

is coming to city i from Some city; otherwise zero.

IT be an array where IT (i) =1, i N represents that the salesman

at time (facility) „i‟ travels one pair of cities.

SW be an array where SW (i) is the city that the salesman is visiting

from city i,Sw(i) = 0 indicates that the salesman is not visiting any

city from city i.

L be an array where L (i) is the letter in the ith position of a

word.

NL be an array where NL (i) =q indicates that q cities are covered up

to ith Position of a Word.

GS be an array where GS (i) represents ith group number of cities.

Then for a given partial word LK = (a1,a2,…..,aK) the values of the

arrays RI, CI, TI, SW, L ,NL and GS as follows.

IR(R (ai)) =1, i=1, 2, 3……..K

IC(C (ai)) =1, i=1, 2, 3……..K

IT (T (ai)) =1, i=1, 2, 3……. K

SW(R (ai)) =C (ai), i=1, 2, 3……..K

L (i) =ai, i=1, 2, 3……..K

NL(i)=NL(i-1)+1, if IC(R(ai))=0 and NL(i)=NL(i-1)+1, if IR(C(ai))=0

i=1,2,3….K

GS (GN(R (ai)) =GS (GN(R (ai)) +1 if IC(R (ai)) =0 and

GS (GN(C (ai)) =GS (GN(C (ai)) +1 if IR(C (ai)) =0 i=1,2,3….K

The recursive algorithm for checking the feasibility of a partial

word LP is given as follows: In the algorithm first we equate IX=0. At

the end if IX=1 then the partial word is feasible, otherwise it is

104

infeasible. For this algorithm we have TR=R (ap+1), TC=C (aP+1) and

TT=T (aP+1).

Algorithm – 1:

STEP1: IX=0

NXA=NL [I-1]

TCX=TC

GOTO 2

STEP 2: IS (IR (TR) = 1) IF YES GOTO 10

IF NO GOTO 2A

STEP2A: IS (IC (TC) = 1) IF YES GOTO 10

IF NO GOTO 2B

STEP2B: IS (IT (TT) = 1) IF YES GOTO 10

IF NO GOTO 3

STEP3: IS IC (TR) =0 IF YES NXA=NXA+1 GOTO 3A

IF NO GOTO 3A

STEP3A: IS IR (TC) =0 IF YES NXA=NXA+1 GOTO 3B

IF NO GOTO 3B

STEP3B: IS (NXA>M) IF YES GOTO 10

IF NO GOTO 4

105

STEP4: IS IC (TR) =0 IF YES GS (GN (TR)) = GS (GN

(TR)) +1

GOTO 4A

IF NO GOTO 4A

STEP4A: IS IR (TC) =0 IF YES GS (GN (TC)) = GS (GN

(TC)) +1

GOTO 4B

IF NO GOTO 4B

STEP4B : IS GS (GN (TR)) <= GP (TR) IF YES GNZ1=0

GOTO 4C

IF NO GNZ1=1 GOTO

4C

STEP4C : IS GS (GN (TC)) <= GP (TC) IF YES GNZ2=0

GOTO 4D

IF NO GNZ2=1 GOTO

4D

STEP4D: NPA=NP-(GNZ1+GNZ2) GOTO 5

STEP5: IS M – NXA <= NPA IF YES GOTO 6

IF NO GOTO 7

STEP6: IS (SW (TCX) = 0) IF YES IX=1 GOTO 10

IF NO IK=SW (TCX) GOTO 6A

STEP6A: IS (IK=TR) IF YES GOTO 6B

106

IF NO TCX=IK GOTO 6

STEP6B: IS (I=M) IF YES IX=1 GOTO 10

IF NO GOTO 10

STEP7: IS IC (TR) =0 IF YES GS (GN (TR)) = GS (GN

(TR)) – 1

GOTO 7A

IF NO GOTO 7A

STEP7A: IF IR (TC) =0 IF YES GS (GN (TC)) = GS (GN

(TC)) – 1

GOTO 10

IF NO GOTO 10

STEP10: STOP.

This recursive algorithm will be used as a subroutine in the lexi-

search algorithm. We start the algorithm with a very large value, say,

9999 as a trial value of VT. If the value of a feasible word is known, we

can as well start with that value as VT. During the search the value of

VT is improved. At the end of the search the current value of VT gives

the optimal feasible word. We start with the partial word L1= (a1) = (1).

A partial word Lp=Lp-1 (ap) where indicates chain form or

concatenation. We will calculate the values of V (Lp) and LB (Lp)

simultaneously. Then two cases arises (one for branching and other

for continuing the search).

107

1. LB (Lp) < VT. Then we check whether Lp is feasible or not. If it is

feasible we proceed to consider a partial word of order (p+1),

which represents a sub block of the block of words represented

by Lp. If Lp is not feasible then consider the next partial word of

order p by taking another letter which succeeds ap in the pth

position. If all the words of order p are exhausted then we

consider the next partial word of order (p-1).

2. LB (LP) VT. In this case we reject the partial word meaning that

the block of words with Lp as leader is rejected for not having an

optimal word and we also reject all partial words of order p that

succeeds Lp.

Now we are in a position to develop lexi search algorithm to

find an optimal feasible word.

ALGORITHM 2: (LEXI-SEARCH ALGORITHM)

The following algorithm gives an optimal feasible word.

STEP 1 : (Initialization)

The arrays SN, D, DC, R, C, T and values of

N, M, GN, GP are made available IR, IC, IT, SW,

L, NL, GS, V, LB are initialized to zero. The values

I=1, J=0, VT=9999, NZ=NNM –N, MAX=NZ-1

STEP 2 : J=J+1

IS (J>MAX) IF YES GOTO 11

IF NO GOTO 3

STEP 3 : L (I) = J

JA = J + M - I

IS (I = 1) IF YES NXA = 0, V (I) = D (J) GOTO

3B

108

IF NO NXA=NL (I-1) GOTO 3A

STEP 3A : V (I) = V (I -1) + D (J)

GOTO 3B

STEP 3B : LB (I) = V (I) + DC (JA) – DC (J)

GOTO 4

STEP 4 : IS (LB (I) VT) IF YES GOTO 11

IF NO GOTO 5

STEP 5 : TR=R (J)

TC=C (J)

TT=T (J)

GOTO 6

STEP 6 : CHECK THE FEASIBILITY OF L (USING

ALGORITHM-1)

IS (IX=0) IF YES GOTO 2

IF NO GOTO 7

STEP 7 : IS (I=M) IF YES GOTO 10

IF NO GOTO 8

STEP 8 : L (I) = J

IR (TR) = 1

IC (TC) = 1

109

IT (TT) = 1

SW (TR) = TC

NL (I) = NXA

GOTO 9

STEP 9 : I=I+1

MAX=MAX+1

GOTO 2

STEP10 : L (I) =J

L (I) IS FULL LENGTH WORD AND IS FEASIBLE.

VT=V (I), record L (I), VT,

GOTO 13

STEP11 : IS (I=1) IF YES GOTO 14

IF NO GOTO 12

STEP12 : I=I-1

MAX=MAX+1

GO TO 13

STEP13 : J=L (I)

NL (I) = 0

TR = R (J)

TC = C (J)

TT = T (J)

IR (TR) = 0

IC (TC) = 0

110

IT (TT) = 0

SW (TR) = 0

GS (GN (TR)) = GS (GN (TR)) -1

GS (GN (TC)) = GS (GN (TC)) -1

GOTO 2

STEP14 : STOP

END

The current value of VT at the end of the search is the value of

the optimal feasible word. At the end if VT = 9999 it indicates that

there is no feasible solution.

5.5. Search Table:

The working details of getting an optimal word, using the above

algorithm for the illustrative numerical example are given in the

Table-5. The columns (1), (2), (3) and (4) gives the letters in the first,

second, third and fourth places respectively. The corresponding V (I)

and L B (I) are indicated in the next two columns. The rows R, C and T

give the row, column and time indices of the letter. The last column

gives the remarks regarding the acceptability of the partial words. In

the following table A indicates ACCEPT and R indicates REJECT.

TABLE- 5.4

SEARCH TABLE

S.N

O 1 2 3 4 V LB R C T REM

1 1 1 4 1 5 1 A

2 2 1 4 5 1 1 R

3 3 1 4 4 1 2 A

4 4 1 4 6 4 2 R

5 5 1 4 1 3 3 R

111

6 6 1 4 3 5 3 R

7 7 1 4 1 4 4 R

8 8 1 4 4 3 4 R

9 9 1 4 5 1 4 R

10 10 1 5 6 5 4 R

11 11 2 6 6 3 1 R

12 12 2 6 1 4 2 R

13 13 2 6 4 1 3 R

14 14 2 6 6 5 3 R

15 15 2 7 4 2 4 R

16 16 3 8 5 3 1 R

17 17 3 9 1 2 3 R

18 18 4 10 3 1 1 R

19 19 4 10 2 5 2 R

20 20 4 10 3 6 3 R

21 21 4 10 6 4 3 A

22 22 4 10 2 3 4 R

23 23 5 11 6 1 1 R

24 24 5 11 4 5 2 R

25 25 5 11 5 4 2 R

26 26 5 11 3 1 3 R

27 27 5 11 2 4 4 R

28 28 6 12 4 2 1 R

29 29 6 12 = VT 5 6 4 A

30 22 4 11 2 3 4 R

31 23 5 12 = VT 6 1 1 R

32 4 1 4 6 4 2 A

33 5 1 4 1 3 3 R

34 6 1 4 3 5 3 R

35 7 1 4 1 4 4 R

36 8 1 4 4 3 4 R

37 9 1 4 5 1 4 R

112

38 10 1 5 6 5 4 R

39 11 2 6 6 3 1 R

40 12 2 6 1 4 2 R

41 13 2 6 4 1 3 A

42 14 2 6 6 5 3 R

43 15 2 6 4 2 4 R

44 16 3 7 5 3 1 R

45 17 3 7 1 2 3 R

46 18 4 8 3 1 1 R

47 19 4 8 2 5 2 R

48 20 4 8 3 6 3 R

49 21 4 8 6 4 3 R

50 22 4 8 2 3 4 R

51 23 5 9 6 1 1 R

52 24 5 9 4 5 2 R

53 25 5 9 5 4 2 R

54 26 5 9 3 1 3 R

55 27 5 9 2 4 4 R

56 28 6 10 4 2 1 R

57 29 6 10 = VT 5 6 4 A

58 14 2 6 6 5 3 R

59 15 2 7 4 2 4 R

60 16 3 8 5 3 1 R

61 17 3 9 1 2 3 R

62 18 4 10 = VT 3 1 1 R

63 5 1 4 1 3 3 R

64 6 1 4 3 5 3 R

65 7 1 4 1 4 4 R

66 8 1 4 4 3 4 A

67 9 1 4 5 1 4 R

68 10 1 5 6 5 4 R

69 11 2 6 6 3 1 R

113

70 12 2 6 1 4 2 R

71 13 2 6 4 1 3 R

72 14 2 6 6 5 3 R

73 15 2 7 4 2 4 R

74 16 3 8 5 3 1 R

75 17 3 9 1 2 3 R

76 18 4 10 = VT 3 1 1 R

77 9 1 5 5 1 4 R

78 10 1 6 6 5 4 R

79 11 2 7 6 3 1 R

80 12 2 7 1 4 2 R

81 13 2 7 4 1 3 A

82 14 2 7 6 5 3 R

83 15 2 8 4 2 4 R

84 16 3 9 5 3 1 R

85 17 3 10 = VT 1 2 3 R

86 14 2 8 6 5 3 R

87 15 2 9 4 2 4 A

88 16 3 9 5 3 1 R

89 17 3 10 = VT 1 2 3 R

90 16 3 11 > VT 5 3 1 R

91 2 1 4 5 1 1 A

92 3 1 4 4 1 2 R

93 4 1 4 6 4 2 A

94 5 1 4 1 3 3 R

95 6 1 4 3 5 3 R

96 7 1 4 1 4 4 R

97 8 1 4 4 3 4 R

98 9 1 4 5 1 4 R

99 10 1 5 6 5 4 R

100 11 2 6 6 3 1 R

101 12 2 6 1 4 2 R

114

102 13 2 6 4 1 3 R

103 14 2 6 6 5 3 R

104 15 2 7 4 2 4 R

105 16 3 8 5 3 1 R

106 17 3 9 1 2 3 R

107 18 4 10 = VT 3 1 1 R

108 5 1 4 1 3 3 A

109 6 1 4 3 5 3 R

110 7 1 4 1 4 4 R

111 8 1 4 4 3 4 R

112 9 1 4 5 1 4 R

113 10 1 5 6 5 4 A

114 11 2 5 6 3 1 R

115 12 2 5 1 4 2 R

116 13 2 5 4 1 3 R

117 14 2 5 6 5 3 R

118 15 2 5 4 2 4 R

119 16 3 6 5 3 1 R

120 17 3 6 1 2 3 R

121 18 4 7 3 1 1 R

122 19 4 7 2 5 2 R

123 20 4 7 3 6 3 R

124 21 4 7 6 4 3 R

125 22 4 7 2 3 4 R

126 23 5 8 6 1 1 R

127 24 5 8 4 5 2 R

128 25 5 8 5 4 2 R

129 26 5 8 3 1 3 R

130 27 5 8 2 4 4 R

131 28 6 9 4 2 1 R

132 29 6 9 5 6 4 R

133 30 6 9 6 2 4 R

115

134 31 7 10 = VT 1 3 1 R

135 11 2 6 6 3 1 R

136 12 2 6 1 4 2 R

137 13 2 6 4 1 3 R

138 14 2 6 6 5 3 R

139 15 2 7 4 2 4 R

140 16 3 8 5 3 1 R

141 17 3 9 1 2 3 R

142 18 4 10 = VT 3 1 1 R

143 6 1 4 3 5 3 A

144 7 1 4 1 4 4 R

145 8 1 4 4 3 4 A

146 9 1 4 5 1 4 R

147 10 1 4 6 5 4 R

148 11 2 5 6 3 1 R

149 12 2 5 = VT 1 4 2 A

150 9 1 4 5 1 4 R

151 10 1 5 = VT 6 5 4 R

152 7 1 4 1 4 4 A

153 8 1 4 4 3 4 R

154 9 1 4 5 1 4 R

155 10 1 5 = VT 6 5 4 R

156 8 1 4 4 3 4 A

157 9 1 4 5 1 4 R

158 10 1 5 = VT 6 5 4 R

159 9 1 5 = VT 5 1 4 R

160 3 1 4 4 1 2 A

161 4 1 4 6 4 2 R

162 5 1 4 1 3 3 A

163 6 1 4 3 5 3 R

164 7 1 4 1 4 4 R

165 8 1 4 4 3 4 R

116

166 9 1 4 5 1 4 R

167 10 1 5 = VT 6 5 4 R

168 6 1 4 3 5 3 A

169 7 1 4 1 4 4 R

170 8 1 4 4 3 4 R

171 9 1 4 5 1 4 R

172 10 1 5 = VT 6 5 4 R

173 7 1 4 1 4 4 R

174 8 1 5 = VT 4 3 4 R

175 4 1 4 6 4 2 A

176 5 1 4 1 3 3 A

177 6 1 4 3 5 3 R

178 7 1 4 1 4 4 R

179 8 1 4 4 3 4 R

180 9 1 4 5 1 4 R

181 10 1 5 = VT 6 5 4 R

182 6 1 4 3 5 3 A

183 7 1 4 1 4 4 R

184 8 1 4 4 3 4 A

185 9 1 4 5 1 4 R

186 10 1 4 6 5 4 R

187 11 2 5 = VT 6 3 1 R

188 9 1 4 5 1 4 R

189 10 1 5 = VT 6 5 4 R

190 7 1 4 1 4 4 A

191 8 1 4 4 3 4 R

192 9 1 4 5 1 4 R

193 10 1 5 = VT 6 5 4 R

194 8 1 4 4 3 4 A

195 9 1 4 5 1 4 R

196 10 1 5 =VT 6 5 4 R

197 9 1 5 = VT 5 1 4 R

117

198 5 1 4 1 3 3 A

199 6 1 4 3 5 3 R

200 7 1 4 1 4 4 R

201 8 1 4 4 3 4 R

202 9 1 5 = VT 5 1 4 R

203 6 1 4 3 5 3 A

204 7 1 4 1 4 4 A

205 8 1 4 4 3 4 R

206 9 1 4 5 1 4 R

207 10 1 5 = VT 6 5 4 R

208 8 1 4 4 3 4 A

209 9 1 4 5 1 4 R

210 10 1 5 = VT 6 5 4 R

211 9 1 5 = VT 5 1 4 R

212 7 1 4 1 4 4 A

213 8 1 4 4 3 4 R

214 9 1 5 = VT 5 1 4 R

215 8 1 5 = VT 4 3 4 R

At the end of the search the current value of VT is 5 and it is the

value of the optimal feasible word L4 = (2, 6, 8, 12). It is given in the

149th row of the search table. The array IR, IC, IT, SW takes the values

represented in the Table-5.5 given below. The Pattern represented by

the above optimal feasible word is represented in the following table-

5.6

Table – 5.5

1 2 3 4 5 6

IR 1 -- 1 1 1 --

IC 1 -- 1 1 1 --

IT 1 1 1 1 -- --

SW 4 -- 5 3 1 --

118

Table – 5.6

𝑋 𝑖, 𝑗, 1 =

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 01 0 0 0 0 00 0 0 0 0 0

𝑋 𝑖, 𝑗, 2 =

0 0 0 1 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

𝑋 𝑖, 𝑗, 3 =

0 0 0 0 0 00 0 0 0 0 00 0 0 0 1 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

𝑋 𝑖, 𝑗, 4 =

0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 1 0 0 00 0 0 0 0 00 0 0 0 0 0

From the table -5.6, the representation of the solution X to the

problem is that the salesman visits city 1 from city 5 at time 1,

salesman visits city 5 from city 3 at time 3, salesman visits city 3 from

city 4 at time 4 and visits city 4 from city 1 at time 2. This solution is

a feasible solution.

5.6. Computational Experience:

A Computer program for the above algorithm is written in C

language and is tested on the system COMPAQ. Random numbers are

used to construct the cost matrix. The following table- 5.7 gives the

list of the problems tried along with the average CPU time in seconds

required for solving them.

119

TABLE-5.7

5.7. Conclusion

The problems are solved by using the Lexi-Search algorithm

based on the Pattern Recognition Technique. In table-5.7, „m‟

represents that the salesmen has visited the number of cities. The

cost matrix was generated randomly in the interval [0,100]. For each

type of instance we considered six trials. Our algorithms have been

implemented in C. The computational experiments were performed on

a personal computer with AMD Sempron™ Processor LE-1200, 2.10

GHz, 896 RAM and OS Windows XP Professional. In table-5.7 we have

presented the computational results for solving the problem using the

Lexi-Search algorithm based on the Pattern Recognition Technique.

Problem

Dimensions

No. of

prob‟

s

Alpha

bet

Table

Total time taken(ET)

TYPE 1 TYPE 2 TYPE 3

n M MIN MAX AVG MIN MAX AVG MIN MAX AVG

10 5 6 .0010 .021 .040 .035 .026 .04 .003 .002 .004 .003

20 12 6 .0027 .002 .005 .0035 .004 .006 .0050 .002 .005 .0035

25 16 6 .0037 .002 .004 .003 .002 .004 .003 .002 .004 .003

30 19 6 .003 .002 .005 .0035 .004 .006 .0050 .002 .005 .0035

35 24 6 .0036 .002 .004 .003 .002 .004 .003 .002 .004 .003

40 28 6 .0038 .002 .005 .0035 .004 .006 .0050 .002 .005 .0035