Variant Multi Objective Tsp Model

Transcription

1 Volume Issue 06 Pages June-06 ISSN e): Varant Mult Objectve Tsp Model K.Vjaya Kumar, P.Madhu Mohan Reddy, C. Suresh Babu, M.Sundara Murthy Department of Mathematcs, Sr Venkateswara Unversty, Trupat Andhra Pradesh, Inda-5750 Dept of Mathematcs, Sddharth Insttute of Engg &Technology, Puttur, Andhra Pradesh, Inda Correspondng Author: Abstract: Let there be a set of 'n' ctes. Par of dstances between the ctes are known and dstance th th between ctes to j cty s gven. The cost of travel between par of ctes s gven. The cost of C and Dstance D need not be related. These two matrces C and D can be smply wrte as C and D respectvely. The travellng salesman starts hs busness from head quarter cty and he wants to travel n 0 ctes less than n.the travellng salesman n hs tour vsts n 0 ctes wth mnmum dstances then the same route need not be wth mnmum costs and same way a route wth mnmum costs need not be wth a mnmum dstance. The cost and dstance are dfferent unrelated factors but the salesman wants to mnmze n both cases as much as possble. The objectve s he wants to vst n 0 ctes nearly wth mnmum cost and mnmum dstance. Both the factors dstance and cost beng ndependent, absolutely n both cases mnmum s not possble. The salesman may be nterested n one factor when compared wth the other factor or vce versa. We want to suggest tours where both factors are consdered and consderable mnmum tours are planned, wth dfferent constrants under consderatons. Keywords: travellng salesman, tour, dstance, cost, head quarter INTRODUCTION: Travellng salesman problem TSP) s a very popular model. The travelng salesman problem nvolves of a salesman and a set of ctes. The salesman has to vst each one of the ctes startng from a hometown and returnng to the same cty. The challenge of the problem s that the travelng salesman wants to mnmze the total length of the trp. So many researchers developed dfferent algorthms. The earlest papers on the Generalzed Travellng Salesman Problem dscuss the problem n the context of partcular applcatons by Henry-Labordere [], Saskena [6], and Srvasatava et al [0]. Fschett et al [9] use the polyhedral results to develop a branch and cut algorthm A lagrangan-based branch and bound algorthm] FathTasgetren et al.[] and Dmtrjevc [5] dscussed the varant travellng salesman problem. Only a few authors Laporte et al [7] handle fxed costs for ncludng a node on the tour, possbly because such fxed costs can be ncorporated nto the dstance costs va a smple transformaton.laporte[8] descrbes An overvew of Exact and Approprate algorthm. Madhu Mohan Reddy et al.[] descrbes aproblem called an alternate travellng salesman problem. The objectve of ths problem s to fnd a Tour such that total length of the tour mnmum. An exact algorthm s proposed for ths TSP. The algorthm solves the problem on dentfy the key patterns whch optmze the objectve of thecost/dstance. Bhavan and Sundara Murthy [3] explaned Tme-Dependent Travelng Salesman Problem.Ben-Areh et al. [4] explaned Process Plannng for rotatonal parts usng the generalzed TSP. Problem Descrpton: Letset of th ctes to 'n' ctes be N {,,3,..., n}. Par of dstance between the ctes s known and let t be D th j cty s gven. The cost of travel between par of ctes C s gven. The cost of,.e. dstance between C and Dstance K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

2 Volume Issue 06 Pages June-06 ISSN e): D need not be related. These two matrces C and D can be smply wrte as C and D respectvely. The travellng salesman starts hs busness n head quarter cty and he wants to travel n 0 ctes less than n.the travellng salesman n hs tour vsts n 0 ctes wth mnmum dstancesthen the route need not be wth mnmum costs and same way a route wth mnmum costs need not be wth a mnmum dstance when t s wth mnmum cost. The cost and dstance are dfferent unrelated factors but the salesman wants to mnmze n both cases as much as possble. The objectve s he wants to vst n 0 ctes mostly wth mnmum cost and mnmum dstance. Both the factors dstance and cost beng ndependent, absolutely n both cases mnmum s not possble. The salesman may be nterested n one factor when compared wth the other factor or vce versa. We want to suggest tours where both factors wth consderable mnmum tour are planned. The travellng salesman wants a tour of n0 ctes wth the followng plans. A tour where the average of dstance and cost be mnmum. The values of the tour dstance and costs be DV, CV ) N n. and the set of ctes s N, 0 A tour wth mnmum dstance rrespectve of cost. The values of the tour dstance and costs be N n. set of ctes s N, 0 DV, CV ) and the 3 A tour wth mnmum cost rrespectve of dstance. The values of the tour dstance and costs be DV3, CV3) and the set of ctes s N 3, 3 n0 N. 4 A tour where the dstance s 0% say) more than the mnmum dstance and the cost s correspondngly mnmum. The values of the tour dstance and costs be DV4, CV4) and the set of ctes s 4 N n. N, 0 5 A tour where the cost s 0% more than the mnmum cost and the dstance s correspondngly mnmum. The values of the tour dstance and costs be DV5, CV5) and the set of ctes s 5 N, 5 n0 N. 4 The set of 5 values nvolvng n total dstance and total costs for the above 5 tours s gven to the customer for hs choce. So, the objectve of the problem s to dentfy the above 5 sets of tour values whch nvolves the tour of n 0 ctes. 3 MATHEMATICAL FORMULATION: Let the tour of n0 ctes be n 0, N Let X be a 0, ) ndcator matrx such that x, ) and0, for other par of values. Ths X also represents the above tour of n0 ctes. N, N 0,,... n0, 0 K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

3 Volume Issue 06 Pages June-06 ISSN e): Let E [ D C ] ) Let X represents tour of n0 ctes. Then Z E X, j N and N...) j The objectve s to fnd X such that 0 0 N Mn Z X X ) Z X ) Z D D X j...3)...4) Z C C X. j...5) Consder the matrx D alone. Z X ) D X j Mn Z X ) Z X ) Z X ) ZC C X j D...6)...6.)...7) Smlarlyconsder the matrx C alone. Z X ) C X j Mn Z X ) Z 3 X 3) Z 3 X ) Z3D D x3 j Z4 X 4) Mn C X Z4c X j c... ) Z X 5) Mn D X Z5 5 D X j...8)...9)... 0)... ) In Equaton ), E represents that average values of correspondng dstance matrx D and cost matrx C respectvely. K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

4 Volume Issue 06 Pages June-06 ISSN e): Equaton ) represents that the problem to fnd total value n a tour wth respect to the matrx E,. In equaton 3), Z X ) s mnmum and let t be X, where X gves the tour wth mnmum average correspondng values of both dstance and cost matrces. In Equaton 4), In Equaton 5), Z D s the total dstance of the tour represented by tour X wth respect to the matrx D. Z C s the total cost of the tour represented by tour X wth respect to the matrx C. matrx D. Equaton 6) represents that the problem to fnd total dstance value n a tour wth respect to the In equaton 6.), Z X ) s mnmum and let t be X, where X gves the tour wth mnmum dstance wth respect to the matrx D. In Equaton 7) Z c s the total cost n a tour represented byx wth respect to the matrx C. matrx C. Equaton 8) represents that the problem to fnd total cost value n a tour wth respect to the Equaton 9) represents matrx C. Z X ) s mnmum and let t be X 3, where X 3 gves the tour wth mnmum cost wth respect to the In Equaton 0),Z 3c s the total cost n a tour represented by X 3 wth respect to the matrx D. Equaton ) represents 4 X s the tour wth Z D) Z ) Z 4 D D Equaton) represents X 5 s the tour wth Z 5 C Z 3 C) Z 3 C) 4 NUMERICAL EXAMPLE: The algorthm and concepts are developed and llustrated by a numercal example. For whch the total number of ctes N = {,, 3, 4, 5, 6, 7}. For the followng example, frst we fnd the optmal soluton n Lex-Search approach usng the Pattern Recognton Technque. Here cty s taken as the head quarter. In the table the dstance ) s taken as - ndcates dsconnectvty to the correspondng ctes, whch can also be taken as. Here the entres D C and E taken as non-negatve ntegers, t can be easly seen that ths s not a necessary condton and the dstance, cost be any postve quantty. Table- Table- Dstance matrx: [ ] D Cost matrx: [ C ] K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

5 Volume Issue 06 Pages June-06 ISSN e): Table-3Average of dstance and cost matrx: [ E ] From the above table- D,5), means the dstance of the connectng the cty to 5 s 3. From the above table-, 3,7) C =4, means the cost of the connectng the cty 3 to 7 s 4. From the above table-3 E 6,3) =, means the average of D 6,3) + C6,3). The objectve s of the problem s to dentfy the above 5 sets of values whch nvolves the tour of n 0 5 ctes, salesman wants to vst wth consderable mnmum dstance and cost s gven below.. A tour where the average of dstance and cost be mnmum. The values of the tour dstance and costs be DV, CV ) and the set s n N, N 5. DV, CV ). A tour wth mnmum dstance rrespectve of cost. The values of the tour dstance and costs be set s n N, 5 N. and the 3. A tour wth mnmum cost rrespectve of dstance. The values of the tour dstance and costs be DV3, CV3) and the set s n 3 N, 3 5 N. 4. A tour where the dstance s 0% say) more than the mnmum dstance and the cost s correspondngly mnmum. The values of the tour dstance and costs be DV4, CV4) and the set s n N 4, 4 5 N. 5. A tour where the cost s 0% more than the mnmum cost and the dstance s correspondngly mnmum. The values of the tour dstance and costs be DV5, CV5) and the set s n N 5, 5 5 N. 5 CONCEPTS AND DEFINITIONS 5. Defnton of a pattern An ndcator two-dmensonal array whch s assocated wth a tour s called a pattern. A pattern s sad to be feasble f X s a feasble tour. V X) = D X 5. Feasble soluton of the problem K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

6 Volume Issue 06 Pages June-06 ISSN e): Consder an ordered par set{,4),4,5),5,6),6,3),3,)}and t represents the tour , whch s a feasble soluton for average of dstance and cost matrx E.The Fgure-, represents the feasble soluton of the above ordered pars. In the graphs fgure-, fgure-, fgure-3, fgure-4)hexagons represents ctes and the value n hexagons ndcates the name of the respectve ctes also values at each arc representcorrespondng average value of costand dstance matrx E. Fgure- From the Fgure-, the cty connected to 4, cty 4 connected to 5, cty 5 connected to 6, cty 6 connected to 3 and cty 3 connected to. Here, the feasble soluton s 4.Average of total cost and dstance =E,4)+E4,5)+E5,6)+E6,3)+E3,) =++4++5=4. From the soluton of average of dstance and cost matrx EFgure-), we can splt the correspondng solutons from the dstance matrxtable-) and cost matrxtable-)s gven n fgure-. fgure-. Average of total cost and dstance =E,4)+E4,5)+E5,6)+E6,3)+E3,)=++4++5=4 Total cost=c,4)+c4,5)+c5,6)+c6,3)+c3,)= =. Total dstance=d,4)+d4,5)+d5,6)+d6,3)+d3,)=++5++7=6. K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June 06 66

7 Volume Issue 06 Pages June-06 ISSN e): From the Fgure-, the cty connected to 4, cty 4 connected to 5, cty 5 connected to 6, cty 6 connected to 3 and cty 3 connected to. Values on arcs represent correspondng cost and dstances betweennodes respectvely. Total dstance=d,4)+d4,5)+d5,6)+d6,3)+d3,)=++5++7=6from table-). cost=c,4)+c4,5)+c5,6)+c6,3)+c3,)= = from table-). Total Let the total feasble dstance be DV 6 and the total feasble cost s CV. Therefore, the values of the tour dstance and costs be, CV ) 6,) DV and the number ctes 5 N. 6Soluton Approach:In the above fgure-, for the feasble soluton we observe that there are 5 ordered pars taken along wth the value from the average of dstance and cost matrx [ E ] dstance matrx )], [ D j, cost matrx [ C j )] of the numercal example n table-,,3 respectvely. The 5 ordered pars are selected such that they represent a feasble soluton n fgure-,,3. So the problem s that we have to select 5 ordered pars from the dstance matrx along wth values such that the total value s mnmum and represents a feasble soluton. For ths selecton of 5 ordered pars we arrange the 7 7=49 ordered pars n the ncreasng order and call ths formaton as alphabet table and we wll develop an algorthm for the selecton along wth the checkng for the feasblty. 6.Alphabet table partal) for average of dstance and cost matrx E: There s m= n x n ordered pars n the two-dmensonal array X. Let E be the correspondng array of average of dstance and cost matrx. For convenence these are arranged n ascendng order of ther correspondngmatrx E and are ndexed from to k SundaraMurthy-979). Let SN= [,, 3 k] be the set of k ndces. If a, bsn and a < b then E a) E b). Also let the arrays R and C be the array of row and column ndces of the ordered pars are represented by SN and CE be the array of cumulatve sum of the elements of E. The arrays SN, E, CE, R, C, for the numercal example are gven n the Table-3. If psn then Rp),Cp)) s the ordered pars and Ea)=ERa),Ca)) s the value of the ordered pars and CE a). Table-4 S.N E CE R C The frst column represents seral number, second column represents the ncreasng order of the values of E,thrd column represents the cumulatve of E, fourth and ffth columnsrepresent the correspondng row and column of E. Let us consder 7ЄSN. It represents the ordered par R 7), C 7)) = 3, 5). Then E 7) =E3, 5) =4 andce 3, 5) =7. K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June 06 66

8 Volume Issue 06 Pages June-06 ISSN e): Defnton of a word Let SN =,, ) be the set of ndces, E be an array of correspondng dstances of the ordered pars and Cumulatve sums of elements n E s represented as an array CE. Let arrays R, C be respectvely the row, column ndces of the ordered pars. Let L k = {a, a, , a k }, a SN be an ordered sequence of k ndces from SN. The pattern s represented by the ordered pars whose ndces are gven by L k s ndependent of the order of a n the sequence. Hence for unqueness the ndces are arranged n the ncreasng order such that a < a +, =,, , k Value of the word The value of the partal word L k, V L k ) s defned recursvely as V L k ) = VL k- ) +E a k ) wth V L o ) = 0 where E a k ) s the dstance/cost array arranged such that E a k ) < E a k+ ). V L k ) and Vx) the values of the pattern X wll be the same. X s the partal) pattern represented by L k, Sandara Murthy 979). 6.4 Feasblty crteron of a partal word An algorthm was developed, n order to check the feasblty of a partal word L k+ = a, a, a k, a k+ ) gven that L k s a feasble partal word. We wll ntroduce some more notatons whch wll be useful n the sequel. 6.5Algorthm- Feasble checkng) STEP : IX=0 STEP : IS IR RA) = ) IF YES GOTO 8 IF NO GOTO 3 STEP3 : W = CA GOTO 4 STEP 4 : IS W = RA IF YES GOTO 8 IF NO GOTO 5 STEP5 : IS SW W) = 0 IF YES GOTO 7 IF NO GOTO 6 STEP 6 : W = SW W) GOTO 4 STEP7 : IX= IF YES GOTO 8 IF NO GOTO STEP 8 : STOP We start wth the partal word L = a ) = ). A partal word L p s constructed as L p = L p- * p ). Where * ndcates chan formulaton. We wll calculate the values of V L p ) and LB L p ) smultaneously. Then two stuatons arses one for branchng and other for contnung the search Algorthm- : Lex - Search algorthm) STEP : Intalzaton) The arrays SN, D, DC, R, C and the values of n, Hgh, MAX are made avalable IR,SW L, V, LB are ntalzed to zero. The values I=, J=0, IR RA) =0, VT = Hgh K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

9 Volume Issue 06 Pages June-06 ISSN e): STEP : J=J+ IS J>MAX) IF YES GOTO IF NO GOTO 3 STEP 3 : L I) = J RA=R J) CA=C J) GOTO 4 STEP 4 : V I) =V I-) +D J) LB I) =V I) +CE J+N-I)-CE J) GOTO 5 STEP 5 : IS LB I)VT) IF YES GOTO 0 IF NO GOTO 6 STEP 6 : CHECK THE FEASIBILITY OF L USING ALGORITHM-) IS IX=0) IF YES GOTO 6.7 Search table for average of dstance and cost matrx partal): The workng detals of gettng an optmal word usng the above algorthm for the llustratve numercal example s gven n the Table-5. The columns named ), ), 3),, gves the letters n the frst, second, thrd and so on places respectvely. The columns R, C gve the row, column ndces of the letter. The last column gves the remarks regardng the acceptablty of the partal words. In the followng table 8 ndcates ACCEPT and R ndcates REJECT. Table-5 S.N 4 5 V LB T R C Remarks 8 7 A 8 4 R A RTR) RTR) A * R A * 7 R A, VT= * R =VT) 7 7 5* R=VT) A R R RTR) R * R,=VT) K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

10 Volume Issue 06 Pages June-06 ISSN e): * 3 A A * R * RTR) * R=VT) * R=VT) A * R * 5 3 5* R=VT) * R>T) A A R A R * R A R A, VT= * R=VT) * R=VT) * R The partal word s L 5 = 9, 30, 3, 35, 37) s a feasble partal word. At the end of the search the current value of the VT s 4 and t s the value of optmal feasble word. L 5 = 9, 30, 3, 35, 37) gven n the 37 th row of the search table. Therefore, value of the optmal soluton of the Lex search algorthm usng pattern recognton technque s 4. The Fgure- represents the feasble optmal soluton..e. Average of total cost and dstance = E,4)+ E4,5) + E5,6) + E6,3) + E3,) =++4++5=4 From the soluton of average of dstance and cost matrx E Fgure-), we can splt the correspondng solutons from the dstance matrx table-) and cost matrx table-) s gven n fgure-..e. Total dstance= D,4)+D4,5)+D5,6)+D6,3) +D3,)=++5++7=6 from table-).total cost= C,4)+C4,5)+C5,6)+C6,3)+C3,)= = from table-). Let the total feasble dstance be DV 6 and the total feasble cost s Therefore, the values of the tour dstance and costs be, CV ) 6,) 7. Alphabet tablepartal) for dstance matrx D: CV. DV and the number ctes 5 Alphabet table s ncreasng order of the dstance matrx. It s explaned n secton7. N. K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

11 Volume Issue 06 Pages June-06 ISSN e): Alphabet Table-6 S.N D CD R C Ct Searcch Table: The workng detals of gettng an optmal word usng the above algorthm for the llustratve numercal example s gven n the Table-7. The columns named ), ), 3),, gves the letters n the frst, second, thrd and so on places respectvely. The columns R, C gve the row, column ndces of the letter. The last column gves the remarks regardng the acceptablty of the partal words. In the followng table 8 ndcates ACCEPT and R ndcates REJECT. Searcch Table-7 t V LB R C Ct Remarks 5 4 A - 5 * 7 R * - R A * 3 - R * 5 - R A * 5* - R A A,VT=0 0 7 * R>VT) 8 4 0* R=VT) A A * - RT) * - R The partal word s L 5 =,4,7,9,0) s a feasble partal word. K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

12 Volume Issue 06 Pages June-06 ISSN e): At the end of the search the current value of the VT s 0 and t s the value of optmal feasble word. L 5 =,4,7,9,0) gven n the 0 th row of the search table. Therefore, value of the optmal soluton of the Lex search algorthm usng pattern recognton technque s 0. The Fgure-3 represents the feasble optmal soluton. The Fgure-3, represents the feasble soluton. The hexagons represent ctes and the value n hexagons ndcates the name of the ctes also values at each arc represents dstance and cost between the respectve two nodes from table- and table- respectvely. Fgure-3 From the above Fgure-3, the cty connected to 4, cty 4 connected to, cty connected to 5, cty 5 connected to 7 and cty 7 connected to. Here, the feasble soluton wth respect to cost matrx s 0 and wth respect to dstance matrx s 8. Total dstance =D,4)+D4,5)+D5,6)+D6,3)+D3,)=++3+3+=0and correspondng cost= C,4)+C4,)+C,5)+C5,7)+C7,)=+++5+0=8. Let the total feasble dstance be DV 0 and the total feasble cost s CV 8. Therefore, the values of the tour dstance and costs be, CV ) 0,8) DV and the number ctes 5 N. Let β=0% ncrease n a optmal dstance soluton we get 0+=. By usng the algorthm we get total cost=c,4)+c4,)+c,7)+c7,3)+c3,)=+++4+3= and correspondng dstance= D,4)+D4,)+D,7)+D7,3)+D 3,)= =5. Therefore, the values of the tour dstances be, CV ),5) DV and the number ctes N ALPHABET TABLE FOR COST ARRAYPARTIAL): Alphabet table s ncreasng order of the dstance matrx. K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

13 Volume Issue 06 Pages June-06 ISSN e): Table-8 S.N C CC R C D Search table: The workng detals of gettng an optmal word usng the above algorthm for the llustratve numercal example s gven n table-9. The columns named ), ), 3),, gves the letters n the frst, second, thrd and so on places respectvely. The columns R, C gve the row, column ndces of the letter. The last column gves the remarks regardng the acceptablty of the partal words. In the followng table 8 ndcates ACCEPT and R ndcates REJECT. Table-9 t V LB R C Ct Remarks 7 4 A 7 * 7 - R A A * 5 - R * 4 - R A * 6 - RT) * 3 - RT) AVT=) * 6 - RT) * R=VT) * 5 - R 6 5 * 4 - R=VT) * 7 - R=VT) A K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

14 Volume Issue 06 Pages June-06 ISSN e): A R * 5 - R * 4 - RVT) * 7 - R=VT) 0 3 * 4 - RVT) From the above search table, the optml soluton wth respect to cost matrx s and correspondng soluton wth respect to dstance matrx s 5. The Fgure-4 represents the above feasble optmal soluton. The hexagons represent ctes and the value n crcles ndcates the name of the ctes also values at each arc represents cost and dstance between the respectve two nodes. Fgure-4 From the above Fgure-4, the cty connected to 4, cty 4 connected to, cty connected to 7, cty 7 connected to 3 and cty 3 connected to. Here, the feasble soluton wth respect to cost matrx s and correspondng soluton wth respect to dstance matrx s 5. Total cost =C,4)+C4,)+C,7)+C7,3)+C3,)=+++4+3=and Correspondng dstance = D,4)+D4,)+D,7)+D7,3)+D3,)= =5. Let the total feasble dstance be DV 3 and the total feasble cost s CV 3 5. Therefore, the values of the tour dstance and costs be DV, CV ) 0,5) and the number of ctes N 5. 9 Concluson: 3 3 In ths paper, we have studed a model namely A Varant Mult Objectve TSP model.. Many researchers studed ths travellng salesman problem. We also tred varant travellng salesman problem. In ths paper, we take random values of dstances and costs whch are n table- and table-. The costs and dstances need not be related. The correspondng average values of table- and table- are gven n table-3.to fnd the optmal soluton we developed an algorthm. By usng ths algorthm we fnd optmal solutons for table-, table- and table-3 are DV, CV ) 0,8) DV, ) 6,), CV DV, ) 0,5) and 3 CV3 If β% ncrease n optmal soluton of dstance matrx and cost matrx we get another two par of values. We wll gve respectvely. choce to customer to select any one par out of 5 solutons. K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June

15 Volume Issue 06 Pages June-06 ISSN e): REFERENCES: [] Henry-Labordere, A. L. 969), The record balancng problem: A dynamc programmng soluton of a generalzed travelng salesman problem, RAIRO,B, [] FathTasgetren, M. Suganthan, P.N. and Yun-Cha Lang007). A genetc algorthm for the generalzed travelng salesman problem, IEEE. [3] Bhavan V. andsundara Murthy, M. 005). Tme-Dependent Travelng Salesman Problem, OPSEARCH 4,pp [4] Ben-Areh,D, Gutn,G, Yeo,A and Zverovth, A003), Process Plannng for rotatonal parts usng the generalzed TSP, Internatonal Journal of Producton Research 4), [5] Dmtrjevc, V. &Sarc, Z. 997) An effcent transformaton of the generalzed travelng salesman problem nto the travelng salesman problem on dgraphs. Informaton Scences 0-4) [6] Saskena, J.P. 970) Mathematcal model of schedulng clents through welfare agences, Journal of the Canadan Operatonal Research Socety 8, [7] Laporte, G 99), The TSP: An overvew of Exact and Approprate algorthm. Euro. J. ofopns. Res. Vol. 59, p. 3. [8]Len, Y.N andwah, B.W.S. 993), Transformaton of the generalzed travelng salesman problem nto the standard travelng salesman problem. Informaton Scence, 74: [9]Fschett M, Salazar- Gonzalez, J.J, Toth, P 997), A branch-and-cut algorthm for the symmetrc generalzed travelng salesman problem. Operatons Research 45 3) [0]Srvastava, S.S, Kumar, S, Garg, R.C and Sen 970) Generalzed travelng salesman problem through n sets of nodes. Journal of the Canadan Operatonal Research Socety 7, [] Madhu Mohan Reddy, P., Sudhsksr, E.,Sreenadh,S.and Vjay Kumar Varma,S03). An alternate travellng salesman problem-journal of Mathematcal Archve, Vol.- 4, ssue-0, October-03,pp6-8. K.Vjaya Kumar, PhD, IJMEI Volume ssue 06 June