Inenaional Mahemaical Foum, Vol 8, 03, no 0, 463-47 HIKARI Ld, wwwm-hikaicom Combinaoial Appoach o M/M/ Queues Using Hypegeomeic Funcions Jagdish Saan and Kamal Nain Depamen of Saisics, Univesiy of Delhi, Delhi-0007, India Copyigh 03 Jagdish Saan and Kamal Nain This is an open access aicle disibued unde he Ceaive Commons Aibuion License, which pemis unesiced use, disibuion, and epoducion in any medium, povided he oiginal wok is popely cied Absac In his pape we use a well known eflecion pinciple fo laice pah couning o apply fo he analysis of he M/M/ ueuing sysem The join disibuion of i aivals and j depaues ove a ime ineval of lengh is obained, when hee ae k cusomes in he beginning of he sysem The deivaion uses he laice pah appoach beween wo poins, in wo dimensional x-y plane wih ceain esicions Finally a known esul of ueue lengh is veified fom Say (96) Keywods: M/M/ ueues, Combinaoial Mehod, Laice Pah, Idle peiod, Bessel s funcion, Hypegeomeic Funcions INTRODUCTION Mohany and Panny (990) obained a ansien soluion by fis disceizing he coninuous ime model and hen epesening he same by a laice pah Sen, Jain and Gupa (993) used he same appoach fo solving he M/M/ model In his pape we use he laice pah appoach fo solving M/M/ ueuing model and obain an expession fo he pobabiliy ( Pi, j ; k() ) of i aivals, j depaues in he ime ineval of lengh, unde he assumpion ha he numbe of cusomes in he sysem a ime = 0 is k We hen use his esul o veify a known esul of Say (96) Fo some elaed wok based on laice pah combinaoics and applicaions,
464 J Saan and K Nain one may efe o Mohany (979), Csaki, Mohany and Saan (990), Saan and Rani (994) and Saan (994, 996) Combinaoial Appoach Fo deemining he ansien soluion, we fis disceize he model by dividing he ime ineval (0,) ino sub inevals(slos) of lengh h In each subineval, one of he following hee muually exclusive evens occus, viz Aival of a cusome Depaue of a cusome 3 Neihe aival no depaue of a cusome The evens in diffeen ime slos ae independen and each can be epesened on gaph by aking a uni hoizonal sep fo an aival, a uni veical sep fo each depaue and say a he posiion if neihe aival no depaue of an even occus In ohe wods, we can say ha if a any poin, he sysem is a sae (u,v), i akes one of he posiions (u+,v) o (u, v +) o (u, v) in he nex subineval of lengh h Le he aivals occu accoding o Poisson pocess wih ae λ and sevice imes, independen of aivals, ae iid exponenially disibued wih mean /µ Duing busy peiod, in a ime slo of lengh h Pobabiliy of an aival is: λh+ o(h) Pobabiliy of a depaue is: μh +o(h) and Pobabiliy of neihe aival no depaue is: - (λ + h + o(h) Wheneve he sysem is empy, seve sis idle, ill a fis cusome aives Seve sas poviding sevice immediaely afe he fis aival Duing idle peiod Pobabiliy of an aival is: λh + o(h) Pobabiliy of a depaue is: o(h) and Pobabiliy of neihe aival no depaue is: - λh + o(h) COMPUTATION OF P, ;() i j k Le he sysem become empy imes in he peiod of duaion
Combinaoial appoach o M/M/ ueues 465 Case (i) Le = 0 B(k+i, j) O A(k,0) Fig In his case he seve emains busy in (0,) and he coesponding laice pah fom A(k,0) o B(i+ k, j), as shown in Fig, does no ouch he line y = x Theefoe he numbe of pahs fom (k,0) o ( k + i, j) no ouching he line y=x is i+ j i+ j given as and hence j i+ k Lim i + j i + j ( i+ j) i j h Pi, j:() k = h ( λh)( μh)( h) h 0 j i+ k i+ j i j i+ j i+ j ( λ) ( μ) = e () j i+ k ( i+ j)! In ode o find an expession fo ueue lengh a ime, we subsiue k +i j = p o i=p + j-k and sum ove j (fom 0 o ), we ge he same
466 J Saan and K Nain p k p+ j k p+ jk λ ( λμ ) ( λμ ) = e μ j!( p + j k)! ( p + j)!( j k)! j = 0 n k λ = I I e μ n+ j x whee In ( x) = is modified Bessel s funcion of fis kind j!( n + j)! p k( λμ ) p+ k( λμ ) () j= 0 Case (ii) Le 0 P 3 B(k + i, j) P P O A(k, 0) Fig In his case, he seve sis idle imes in duaion of ime ineval (0,) and he has o wai imes fo single seve Le T denoe h waiing ime fo single seve,,,, In he coesponding laice pah (see Fig ) fom A(k,0) o B(k+i,j), we
Combinaoial appoach o M/M/ ueues 467 show his by conac poins wih he line y=x and afe each conac poin he seve sis idle fo ime T,,,, fo single hoizonal sep (single aival) The above Fig demonsaes he aival depaue pocess fo single seve, who sis idle hee imes (=3) The numbe of pahs fom (k,0) o (k+i, j) which ouches he line y=x, imes is = i + j i+ j k+ i i+ k In his case j( k) T Lim,:() i + j i + j Pijk = h h 0 = 0 T+ T+ + T + + k i k i + i j whee, = T+ T+ + T T T ( i + j ) i j h h ( λh) ( μh) ( h) ( λh)( λh) j( k) i+ j i+ j = k+ i k+ i i j ( λ( T)) ( μ( T)) ( T) λt λ ( i+ j)! e e dt dt dt e = e λt ( T) λμ( T) β λ ( μ( T)) = e β = 0 β! and, heefoe, j( k) i j β i j i j + + λμ ( λ i+ j + β Pi, j; k( ) = e ( T ) dtdt dt = k+ i i+ k β = 0 ( i+ j)! β! T + T + + T j( k) + + ( ) ( + + )! + = i j β i j i j λμ μ λ i j β i j β e = + + = 0( + )!! ( + )! k i i k β i j β i j j( k) i j β i+ j i+ j ( λ) ( μ) ( i+ j + β)! ( μ) λ = = + + e k i i k ( i+ j )! β = 0 ( i+ j+ β)! β! j( k) i j i+ j i+ j ( λ) ( μ) λ = F( i+ j + ; i+ j+ ; μ ) e = k+ i i+ k ( i+ j)!
468 J Saan and K Nain x Using he esul F ( αγ ; ; x) = e F ( γαγ ; ; x), we ge F( i+ j + ; i+ j+ ; μ ) = e F( ; i+ j+ ; μ ) μ and, heefoe, j( k) i j i+ j i+ j ( λ) ( μ) Pi, j; k() = e F(; i+ j+ ; μ) = k + i k + i ( i+ j)! Now using he esul we ge μ θ θ θ Bi (, + j + ) i+ j μθ F( : i+ j+ ; ) = ( ) e d 0 j( k) i ( λ) ( μ ) Pi, j; k() = e = ( k+ i)!( j ( k+ ) + )! ( k+ i)!( jk)! ( )! i+ j μθ ( θ ) θ e dθ j( k) i j = e ( λ) ( μ ) = ( k+ i)!( j ( k+ ) + )!( )! ( k+ i)!( jk)!( )! 0 0 θ θ i+ j μθ ( θ ) e dθ = e ( λ) ( μ) ( k i )!( j k)! 0 = θ ( k i)!( j k )! = + + θ j( k) j ( k ) ( ) i j j k θ j k θ λ + μ j θ ( ) i + j μθ e dθ ( ) ( ) θ jk j k i j θ + λ + μ = e ( λ) ( μ) + + ( k+ i)!( jk)! θ ( k+ i)!( jk)! θ 0 θ ( ) i + j μθ e dθ i+ k i+ k i j ( θ) ( θ) μθ = e ( λ) ( μ) e dθ ( k+ i)!( j k)! ( k+ i)!( jk)! 0
Combinaoial appoach o M/M/ ueues 469 i j ( μ) i k ( μ) + i+ k = e ( λ) ( μ) ( θ) θ dθ ( θ) θ dθ 0 ( k+ i)!( j k)!! ( k+ i)!( j ( k+ ))!! 0 0 ( ) i j ( ) ( ) e λ μ ( ) ( ) μ B(, i k) μ + = λ μ + + B( +, i+ k+ ) 0 ( k+ i)!( j k)!! ( k+ i)!( j ( k+ ))!! = e ( λ) ( μ) ( μ) 0 ( k+ i+ )! ( j k)! ( k+ i+ + )! ( j ( k+ ))! i j i j k+ i j+ + = e ( λ) ( μ) ( μ) 0 ( k+ i+ + )!( jk)! (3) which is he join disibuion fo i aivals and j depaues in ime ineval of lengh 3 COMPUTATION OF AN EXPRESSION FOR QUEUE LENGTH To find an expession fo ueue lengh puing k+i-j=p (o i=p+j-k ) and summing ove j ( fom 0 o ), we ge he same ( ) ( ) λ+ μ + k+ p+ kk j+ + = e ( λ) p j k ( μ) j ( μ ) j= 0 0 ( k+ ( p+ j k) + + )!( jk)! p+ j k j+ p+ j k j+ ( λ) ( μ) ( λ) ( μ) = e 0 j= 0 ( j k)!( + j+ p)! ( jk )!( + j+ p+ )! j+ + p k j+ + pk p+ k ( ) ( ) ( + ) λ μ λ λμ λμ = e 0 μ j= 0 ( j k)!( + j+ p)! j= 0 ( jk )!( + j+ p+ )! I may be noed ha wihin he inne culy backes, he limis in he fis summaion ae j k, and hose in he second summaion ae j k + Theefoe, puing j-k=s in he fis summaion and j-k-=s in he second summaion and summing ove s ( fom 0 o ), we ge he ueue lengh
470 J Saan and K Nain ( λμ ) ( λμ ) s+ + p+ k s+ + p+ k+ p+ k λ = e 0 μ s= 0 ()!( s + p+ k+ s)! s= 0 ()!( s + p+ k+ s+ )! p+ k λ = e { I+ p+ k( λμ ) I+ p+ k+ ( λμ ) } 0 μ p+ pk λ λ = e { ( ) ( )} μ I+ p+ k λμ I+ p+ k+ λμ 0 μ p+ λ μ = e { I ( λμ ) I+ ( λμ ) } (4) μ = p+ k+ λ Since evens epesened in Case (i) and Case (ii) ae muually exclusive, heefoe, he join disibuion of i aivals and j depaues in ime ineval of lengh is given by he sum of euaions () and (3) and he expession fo ueue lengh a ime is given by he addiion of () and (4), ie, = + p k p+ λ λ e I I I I μ μ = p+ k+ { p k( λμ ) p+ k( λμ )} { ( λμ ) + ( λμ )} pk pk λ λ e I I I I { ( ) ( ) } { ( ) ( ) p k λμ p+ k λμ p+ k λμ p+ k+ λμ } = + μ μ p+ { I ( λμ ) I+ ( λμ ) } λ + μ = p+ k+ = + p k p+ λ λ e I I I I μ μ = p+ k+ { p k( λμ ) p+ k+ ( λμ )} { ( λμ ) + ( λμ )} (5) Now ewiing λ μ μ = p+ k+ λ p+ { I ( λμ ) I+ ( λμ ) }
Combinaoial appoach o M/M/ ueues 47 p+ p+ k+ p+ k+ λ μ μ μ μ = Ip+ k+ ( λμ ) + I( λμ ) + Ip+ k+ ( λμ ) μ λ λ = p+ k+ λ λ λ μ I ( λμ ) μ = p+ k+ λ pk pk p λ λ λ λ μ = I p+ k+ ( λμ ) + Ip+ k+ ( λμ ) + I( λμ ) μ μ μ μ = p+ k+ λ and puing i in (5), we ge an expession fo ueue lengh as : pk pk λ μ λ λ e I ( ) I ( λμ ) μ μ ( + ) = p k λμ + p+ k+ p λ λ μ + I ( λμ ) μ μ = p+ k+ λ which is same as euaion (44) given in Say (96) ACKNOWLEDGEMENTS The auhos ae gaeful o he efeee fo giving valuable commens REFERENCES [] DG Champenowne, An elemenay mehod of soluion of he ueueing poblem wih a single seve and consan paamee, J Roy Sais Soc, Se B, 8 (956), 5-8 [] E Csaki, SG Mohany and J Saan, On andom walks in a plane, As Combinaoia, 9 (990), 309-38 [3] SG Mohany, Laice Pah Couning and Applicaions, Academic Pess, 979
47 J Saan and K Nain [4] SG Mohany, and W Panny, A discee ime analogue of he M/M/ ueue and he ansien soluion: A geomeic appoach, Sankhya, Se A, 5 (990), 364-370 [5] J Saan, A noe on he composiions of a posiive inege, As Combinaoia, 38 (994), 3-4 [6] J Saan, Some disibuions elaed o a andom walk in a plane, As Combinaoia, 44 (996), 9-7 [7] J Saan and S Rani, Some join disibuions concening andom walk in a plane, As Combinaoia, 38 (994), 33-45 [8] T LSay, Elemens of Queueing Theoy wih Applicaions, McGaw Hill, New Yok, 96 [9] K Sen, JL Jain and J M Gupa, Laice pah appoach o ansien soluion of M/M/ wih (0,k) conol policy, J Sais Plann Infe, 34, No (993), 59-67 Received: Novembe, 0