JOHANNES KEPLER UNIVERSITÄT LINZ BERICHTE DER MATHEMATISCHEN INSTITUTE

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1 JOHANNES KEPLER UNIVERSITÄT LINZ BERICHTE DER MATHEMATISCHEN INSTITUTE JOHANNES KEPLER UNIVERSITY LINZ REPORTS OF THE MATHEMATICAL INSTITUTES (PREPRINT SERIES) THE WAITING TIME ANALYSIS OF MULTI-SERVER QUEUE WITH CONSTANT RETRIAL RATE AND DIFFERENT CONTROL POLICIES by Dmtry Efrosnn and Janos Sztrk Number 564 November 007 INSTITUT FÜR STOCHASTIK A 4040 LINZ/AUSTRIA

2 THE WAITING TIME ANALYSIS OF MULTI-SERVER QUEUE WITH CONSTANT RETRIAL RATE AND DIFFERENT CONTROL POLICIES Efrosnn, Dmtry Johannes Kepler Unversty of Lnz, Altenberger Str, 69,4040 Lnz, Austra Sztrk, Janos Unversty of Debrecen, PO Box, 400 Debrecen, Hungary Abstract A controlled retral queueng system wth two heterogeneous servers, frst-come frst-served dscplne for the customers on the orbt, possble drect access for the prmary customers, Posson arrvals and fnte retral group s consdered Accordng to the optmal polcy the faster server must be actve whenever the customer enters the servce area whle the slower one can be actve only when the number of customers n the orbt exceeds a prescrbed threshold level The problem of dervng the equlbrum watng tme and soourn tme dstrbutons s formulated We ntroduce the recursve method for the calculaton of the correspondng Laplace-Stltes transforms and the nverson methods are used to get the dstrbuton functons Also the recursve method for the obtanng z-transforms s used to get some dscrete dstrbutons for the number of drectly served prmary customers and the number of retrals made by a customer untl the servce starts The methods are appled also for other heurstc control polces, namely for the Schedulng threshold polcy (STP), Fastest Free Server (FFS) or Random Server Selecton (RSS) Keywords: Retral queue, Controlled queueng system, Steady-state probabltes, Threshold control polcy, Watng tme dstrbuton, Soourn tme dstrbuton Introducton We consder controllable retral queueng system wth several exponental servers, functonng at dfferent rates The arrvng customers form a Posson process and can have a drect access to the servce area In our prevous paper 8 we mnmzed the expected soourn tme over all customers and found that there s a threshold polcy whch uses a slow server only f the orbt sze exceeds a certan threshold level Wth respect to ths polcy we have appled the matrx-geometrc method for the calculaton of steady-states probabltes and dfferent mean performance characterstcs In ths paper we formulate the problem of determnng the statonary watng and soourn tme dstrbutons For the uncontrolled classcal queueng systems wthout retrals for the dervng of the watng tme dstrbuton t s enough to consder the system state at the arrval tme of a tagged customer, eg as s presented by Klenrock In controlled case 0 t was shown that the watng and soourn tme dstrbutons correspond to the lnear combnaton of Erlang dstrbutons The watng tme n uncontrolled retral queues s more dffcult Some methods are descrbed n the monograph by Faln and Tempelton 9 For retral systems (uncontrolled or controlled) wth drect access of prmary customers to the servce area the watng tme analyss becomes more complcated because the watng tme of a customer n ths case depends also on future arrvals In systems wth classcal retral polcy, where the retral rate depends on the number of orbtng customers, t s necessary to take nto account that some later arrved customer can The nvestgaton s supported by the Austro-Hungaran Cooperaton Grant No 66o:u, 006

3 be earler served accordng to the random order polcy for the orbtng customers The recursve scheme for the computaton of the Laplace-Stltes transforms of the watng tme of a tagged customer s ntroduced by Artaleo et al 4 Ths paper motvate us to perform the watng tme analyss for the controlled queue wth constant retral polcy, where the retral rate s ndependent of the number of orbtng customers In systems wth constant retral rate wth FCFS servce dscplne for the orbtng customers for a threshold polcy future arrvals can nfluence the watng tme of the tagged customer by nfluencng the servers that can be actve n the future Therefore t s also necessary to consder the arrval process after the arrval of the tagged customer up to the tme where servce of ths customer starts For the system under consderaton the calculaton of the watng tme dstrbuton s acheved by analysng the transent Markov process wth absorpton at the tme the tagged customer starts servce Ths requres an extenson of the state representaton snce we have to know at each tme the poston of the tagged customer n the lst of orbtng customers In ths paper we express the Laplace transforms of the watng and soourn tme dstrbutons Representng the Laplace-Stltes transforms of the watng or soourn tme denstes of the tagged customer as vector we get that for threshold system t satsfes the threshold depended block-threedagonal system For uncontrolled queue wth classcal retral rate such a structure was shown n 4 But n case of FIFO dscplne for the orbtng customers ths system s recurrent wth respect to the poston of the tagged customer n the orbt The organzaton of the paper s as follows In sectons and 3, the steady-state dstrbutons are derved for a general threshold system and heurstc control polces In sectons 4 and 5, we develop recursve equatons for the calculaton of the statonary watng and soourn tme dstrbutons for a system under threshold and heurstc control polces as well as the correspondng moments of the arbtrary orders In sectons 6 and 7, we obtan, respectvely, dscrete dstrbuton functons for the number of drectly served customers and number of retrals made by a customer In secton 8 we gve the results of numercal nverson of Laplace- and z-transforms and compare them for threshold and heurstc control polces In further sectons we wll use the notatons e(n), e (n) and I n for the column-vector of dmenson n consstng of s, column vector of dmenson n wth n the -th (begnnng from 0-th) poston and 0 elsewhere, and an dentty matrx of dmenson n n The notatons wthout specfyng the dmensons wll be used for the sutably dmensoned vectors and matrces Descrpton of the mathematcal model Consder the queueng model M/M/c n whch prmary customers arrve accordng to a Posson stream wth rate λ, two heterogeneous exponental servers c = wth rates µ > µ, constant retral rate γ > 0 and the number of places n the retral orbt K Accordng to the control polcy an arrvng customer can on the orbt or have drect access to the accessble dle servers The arrval process, servce tmes and retral tmes are assumed to be mutually ndependent Let Q(t) s the number of customers n the retral orbt at tme t, D (t), D (t) descrbe the states of the servers at ths tme, { 0, f the -th server s dle at tme t and D (t) =, f the -th server s busy The observed process s a contnuous-tme Markov process wth state space defned as {X(t)} t 0 = {Q(t), D (t), D (t)} t 0 E = {x = (q, d, d ); 0 q K, d = {0, }, =, } N {0, }, where q and d, =, denote, respectvely, the number of customers on the orbt and states of the servers

4 Theorem The optmal polcy for the retral queueng system M/M/ wth heterogeneous servers and constant retral rate s of threshold and monotone type, e the fastest dle server must be swtched on whenever a prmary or retral customer arrves and another one must be swtched on f and only f the fastest server s busy and the orbt length reaches the threshold level q q The analytcal representaton of the threshold level s qute complcated, but by means of the value teraton algorthm t can be done numercally If future arrvals are not taken nto account (schedulng problem), e when the obectve s to mnmze the soourn tme for an ndvdual customer, the explct soluton for the threshold level exsts Theorem If λ = 0 then there exsts a threshold level ( ) q γ µ = µ + γ µ +, () such that f q q n state x = (q,, 0) then upon retral arrval t s optmal to dspatch a customer to the slower server, f q q then the slower server must be dle 3 Steady-state dstrbuton of the system under optmal polcy In ths secton we derve the equlbrum state dstrbuton under the optmal threshold polcy (OTP) The dervaton works va a standard matrx-geometrc approach 3, takng nto account the specal structure of the boundary states where not all servers are actve To dstngush the system under OTP from other control polces whch wll be dscussed later we wll use the upper ndex for the concerned values Let q be the threshold level for actvaton of the second server As t was mentoned above they can be found numercally (eg usng the value teraton algorthm 0) Consder a Markov process {X(t)} t 0 defned by () wth a state space E Ths process s a QBD process wth block - three-dagonal nfntesmal matrx Note that the blocks have dfferent szes dependng on the queue length 3 Fnte retral group Frst consder the system wth fnte retral group, e K < In ths case the states are parttoned as follows: block 0 ncludes the sngle state: (0, 0, 0); block ncludes the states: (0, 0, ), (0,, 0), (, 0, 0); block ncludes the states: (0,, ), (, 0, ), (,, 0), (, 0, 0); blocks, 3 K nclude the states: (,, ), (, 0, ), (,, 0), (,0, 0); block K + ncludes the states: (K,, ), (K, 0, ), (K,, 0); block K + ncludes the sngle state: (K,, ); 3

5 Λ = Denote by Λ the nfntesmal matrx of dmenson 4(K + ) for the system under OTP, λ A D 0 (C B ) A D (C B ) A D (C B ) A D (C 3 B 3 ) A D 3 (C 3 B 3 ) A D 3 (C 4 B 4 ) A D 4 M where λ + A e = D 0 e (C D e (C D 3 e (C B 4 B 4 B )e + A e = D e (C )e + A 3 e = (D (C 3 B )e + A 6 e = D 4 e M = 0, B 3 ) + A 4 )e + A 3 e = )e = D 3 e (C 3 B 3 (3), )e + A 5 e = M = µ + µ Matrces A and B represent prmary and retral arrvals, respectvely, dependng on whether the queue length are above or below threshold level: A = ( 0 λ 0 ), A A 6 = λ λ λ = λ λ λ 0, A 3 = λ λ λ λ 0, A 4 = λ λ λ λ 0, A 5 = λ 0 0 λ 0 0 λ λ, and B = γ 0, B = γ γ 0, B 3 = γ γ γ 0, B 4 = γ 0 0 γ 0 0 Matrces C C = represent cases when the system stays at certan states: C 3 = Matrces D D 0 = µ µ 0 λ + µ λ + µ λ + γ, C = λ + M λ + µ + γ λ + µ + γ λ + γ λ + M λ + µ + γ λ + µ λ + γ, C 4 = represent departures wth elements dependng on actve servers:, D = µ µ µ 0 0 µ 0 0 0, D = 0 µ µ µ µ , λ + M λ + µ + γ λ + µ + γ, D 3 = 0 µ µ µ µ, D 4 = ( 0 µ µ ) 4

6 Denote by π = (π 0, π, ) the row-vector of the steady-state probabltes, π = {π x = π (q,d,d ) : x E} = lm P{X(t) = x}, t by {π k : k 0} the subvectors that specfy the states wth k obs n the system Obvously the row-vector π of the steady-state probabltes of the system under optmal polcy satsfes the equatons π Λ = 0, π e = (4) The probabltes for the system states can be represented n the form of recursve relaton wth some matrces M k, π k = π k+ M k, k = 0,,,K +, namely π 0 = π D 0 λ = π M 0 ; π = π D (C B D 0 λ A ) = π M ; π = π 3 D (C B M A ) = π 3 M ; π 3 = π 4 D (C B M A 3 ) = π 4 M 3 ; π = π + D (C B M A 3 ) = π + M, 4 q; π = π + D (C 3 B 3 M A 4 ) = π + M, q + K ; π K = π K+ D 3 (C 3 B 3 M K A 4 ) = π K+ M N ; π K+ = π K+ D 4 (C 4 B 4 M N A 5 ) = π K+ M K+ (5) To calculate the values π k Step Evaluate the matrces M k t s necessary: Step Evaluate the value π K+ = π (K,,) = K+ k=0 Step 3 Substtute π K+ nto π K+ k e = π K+ + π K+ π = π, k = 0,,K + accordng to relaton (5) from the normalzaton condton K+ =0 =K+ K+ K+ =K+ M e = π K+ + K+ M, = 0,,K + K+ =0 =K+ M e (6) Note that ths method works effcently as long as K < s not too large But for large K the matrx geometrc soluton correspondng to K = s a good approxmaton 3 Infnte retral group In case of nfnte retral group K = the nfntesmal matrx Λ has nfnte sze and s obtaned from the matrx (3) by removng the last three rows We consder frst the matrx-geometrc part of the equatons, above the threshold level q : π q +A 4 + π q ++D = π q ++(C 3 B 3 ), 0 5

7 Conecturng the matrx-geometrc form π q + = π q (R ) we substtute ths guess nto the last equaton, then we get the followng equaton for matrx R (R ) D R (C 3 B 3 ) + A 4 = 0 (7) Ths s a quadratc equaton n the matrx R, whch s typcally solved numercally usng the followng teraton procedure: R (0) = 0, R (n + ) = A 4 (C 3 B 3 ) + (R ) (n)d (C 3 B 3 ), where the teraton halts when entres n R (n+) and R (n) dffer n absolute value by less that a gven small constant The general theory 3 states that the necessary and suffcent condton for stablty s p D e > p A 4 e, where p = (p 0, p, p, p 3 ) s a statonary probablty vector gven by p (A 4 (C 3 B 3 ) + D ) = 0, p e = Theorem 3 For the system under optmal polcy, the statonary vector p of A 4 (C 3 B 3 ) + D s gven by p 0 = p = (λ + γ) (λ + µ + γ) (λ + µ + γ)((λ + γ)(λ + µ + γ) + µ M), µ λ + µ + γ p 0, p = µ (λ + M + γ) (λ + γ)(λ + µ + γ) p 0, p 3 = µ µ ((λ + γ) + M) (λ + γ) (λ + µ + γ) p 0 The system s stable f and only f the load factor ρ satsfes ρ = λ(λ + γ) (λ + µ + γ) Mγ(λ + γ) + γµ µ (3(λ + γ) + µ ) + µ γ(λ + µ < (9) + γ) Proof: By elementary calculatons Equatons for the boundary states below the threshold level are stll to be solved, namely: π 0 = π D 0 λ = π M 0 ; π = π D (C B D 0 λ A ) = π M ; π = π 3 D (C B M A ) = π 3 M ; π 3 = π 4 D (C B M A 3 ) = π 4 M 3 ; π = π + D (C B M A 3 ) = π + M, 4 q ; π q = π q (M q A 3 + R D )(C B ) (0) 6 (8)

8 The followng algorthm s used to n the calculatons: Step Solve (7) for matrx R, usng teratons (8) Step Evaluate the Matrces M for = 0,,q Step 3 Evaluate the value π q by solvng the normalsaton condton = k=0 q = π q q π k e = π q =0 q =0 M =q q M e + π q =q e + (I R ) e (R ) e () =0 wth the equaton Step 4 Substtute π q n π q (M q A 3 (C B ) + R D ) = 0 for the values = 0,,q and calculate for > 0 π = π q q π q + = π q =q M () (R ) (3) 4 Steady-state dstrbutons of the system under heurstc polces To measure the advantages of optmal threshold polcy (OTP) three more servers selecton dscplnes wll be consdered, namely Schedulng threshold polcy (STP) Fastest free server selecton (FFS) Random server selecton (RSS) The polces STP, FFS and RSS wll be denoted by ndces m = {, 3, 4}, respectvely The formulas for the calculaton of the steady-state dstrbuton of the system under the STP are exactly the same as for the optmal polcy wth only one excepton that the threshold level may dffer from the optmal one In the next subsectons calculaton procedures are treated for systems under the FFS and RSS polces 4 Fnte retral group In STP case the fastest server must be busy whenever the arrval occurs whereas the slower server can be swtched on only f upon arrval of a prmary or retral customer the orbt has length defned by () The polcy FSS means that the fastest free server must be occuped upon an arrval of a prmary or retral customer Under the polcy RSS arrvals choose any free server wth equal probablty Snce the system under FFS and RSS control polces s descrbed by the same Markov process {X(t)} t 0 wth the state space E as under the optmal control polcy, we have a smlar states parttonng 7

9 Analogously as n prevous sectons we can wrte down the three-dagonal nfntesmal block matrces Λ, m = {, 3, 4} of dmenson 4(K + ) It s obvous that for STP matrx has the same form as for the optmal polcy For polces FFS and RSS polces n case K < the nfntesmal matrces are of the form, Λ = λ A D 0 (C B ) A D (C 3 B 3 ) A D (C 3 B 3 ) A D 3 (C 3 B 3 ) A D 3 (C 4 B 4 ) A D 4 M where A (3) = A, A(3) C (3) = C (4) = C, C(3) A (4) = ( λ A (4) 6 = A 6, B(4) = λ λ λ 0 = C, A (3) = A, = 4, 5,6, B (3) = B, B(3) 3 = B 3, B(3) 4 = B (4) 4 = B 4, = C (4), = 3,4, D (3) = D (4) = D, 0 4, λ λ 0 ) λ 0 0 0, A (4) = λ 0 0 0, A (4) 4 = λ λ λ λ λ λ = 0 0 0, B (4) γ γ 3 = γ γ γ γ 0 0, A(4) 5 = 0 The steady-state probablty vector, m = {, 3, 4} satsfes the system Λ = 0, e = To solve the system for FFS and RSS polces we represent the equatons n the form 0 = D 0 λ = π M 0 ; = D (C B D 0 λ A ) = M ; = 3 D (C 3 B 3 M A ) = 3 M ; K λ 0 0 λ 0 0 λ 0 0 = + D (C 3 B 3 M A 4 ) = + M, 3 K ; = K+ D 3 (C 3 B 3 M K A 4 ) = K+ M K ; K+ = K+ D 4 (C 4 B 4 M K A 5 ) = To calculate the values k, m = {3, 4} t s necessary: Step Evaluate matrces M k, k = 0,,K + by relaton (4) Step Evaluate the value K+ from the normalzaton condton = K+ k=0 k e = K+ K+ + π K+ K+ =0 =K+ M e = K+ K+ M K+ + K+ λ λ,, m = {3, 4} (4) K+ =0 =K+ M e (5) Step 3 Substtute K+ nto for = 0,,K + = K+ K+ =K+ M 8

10 4 Infnte retral group In case of nfnte retral group, K =, the nfntesmal matrces Λ, m = {, 3, 4} are obtaned from the above matrces by removng the last three rows In ths case we conecture the matrx-geometrc form + = π (R ), 0 where the matrx R s to be found by solvng the followng quadratc equaton (R ) D R (C 3 B 3 ) + A 4 = 0 (6) As before the necessary and suffcent condton for stablty s p D e > p A 4 e, where p s a statonary probablty vector gven by p (A 4 (C 3 B 3 ) + D ) = 0, p e = Theorem 4 For the system under STP and FFS control polces, the statonary probablty vector p of A 4 (C 3 B 3 ) + D, m = {, 3} s gven by formulas 3 The system s stable f and only f the load factor ρ, defned by (9) satsfes ρ <, m = {, 3} Theorem 5 For the system under RSS control polcy, the statonary probablty vector p (4) of A (4) 4 (C(4) 3 B (4) 3 ) + D(4) s gven by p (4) 0 = (λ + γ) (λ + γ)(λ + M + γ) + µ µ, p (4) = µ λ + γ p(4) 0, p (4) = µ λ + γ p(4) 0, p (4) 3 = µ µ (λ + γ) p(4) 0 The system s stable f and only f the load factor ρ (4) satsfes ρ (4) = Proof: By elementary calculatons Probabltes 0 and satsfy relatons 0 = D 0 λ(λ + γ) Mγ(λ + γ) + γµ µ < (7) λ = π M = D (C B D 0 0, (8) λ A ) = M In case m = {3, 4} the followng algorthm s used for the calculatons: Step Solve equatons (6) for matrx R, usng teratons startng from R (0) = 0 Step Evaluate matrces M 0 and M Step 3 Evaluate value by solvng the normalsaton condton = k=0 = k e = M M 0 e + M e + M M 0 e + M e + (I R ) e (R ) e (9) =0 9

11 wth the equaton Step 4 Substtute nto and calculate (M A (C 3 B 3 ) + R D ) = 0 0 = M M 0, = M (0) 5 Statonary dstrbuton of the watng tme + = π (R ), > 0 () In ths secton we fnd the dstrbuton of the watng tme For the controllable queue wth heterogeneous servers the watng tme of some customer can depend on the future arrval because n ths case the number of actve servers can be changed Therefore, the watng tme of the tagged customer n the system under OTP strongly depends not only on ts poston n the queue, but also on the queue length durng ts watng tme In the alternatve models we have also consder not only the system state at the arrval tme of the tagged customer, but also the possblty that the customer who comes later wll be serverd by free server Thus, to calculate the watng tme dstrbuton we wll consder the process ust after the arrval of the tagged customer wth absorpton at the tme when he starts the servce Let us ntroduce the transent Markov process X(t) = (Q(t), D (t), D (t), J(t)) () wth the same genenerators as the models of the prevous secton The state spaces E = {x = (q, d, d, ); 0 q K, d = {0, }, =,, 0 q}, where the last component J(t) denotes the poston of the fxed ob n the lst of watng obs at tme t Ths component can take the values {0,,, }, and decreases for the system under threshold polcy at tme of a retral arrval when the frst server s dle, at tme of a retral arrval when at least one of the servers s dle and the queue length s greater than q, and for other systems - at tme of a retral arrval when at least one server s dle The process s absorbed when the component J(t) become equal to zero Note that Q(t) J(t) at any tme t when the targed customer has to wat n the orbt We denote the state of the process X(t) by x = (q, d, d, ) At the pont of tme of a new arrval t + (the ntal tme for the transent Markov process) t s obvous that J(t + ) = Q(t + ) f tagged customer has to wat n the orbt and J(t + ) = 0 f upon arrval the customer can be served mmedately Defne W rv of the watng tme n the system under polcy m, W x rv of the resdual watng tme of the tagged customer gven that the system state s x, w x w x (t) the dencty functon of the resdual watng tme, (s) = Ee sw x = 0 e st w x (t)dt, Res 0 correspondng Laplace-Steltes transform 0

12 Because of the markovty of the process X(t), the resdual watng tme n state x conssts of the tme the system spend n state x untl the next transton wth densty λ x e λxt plus the resdual tme n a new state y after possble transton from the state x, whch take place wth probablty λxy λ x Thus from the low of total probablty for the densty w x (t) we get w x (t) = λ xy λ x e λxt w y (t), x E (3) λ x y x where * denotes convoluton Applyng the Laplace-Steltes transforms to the relaton (3) we get w x (s) = λ xy w y (s), x E (4) s + λ x y x We partton the above Laplace-Steltes transforms accordng to the partton of the system states: defne the column-vectors w, (s) n whch denotes the number of customers n the system and the poston of the tagged customer: w, (s) = w (,0,0,)(s), K (5) w +, (s) = ( w (,0,,)(s), w (,,0,)(s), w (+,0,0,) (s))t, K, w, (s) = ( w (,,,)(s), w (,0,,)(s), w (,,0,)(s), w (,0,0,) (s))t, K, w K+, (s) = ( w (K,,,)(s), w (K,0,,)(s), w (K,,0,) (s))t, K, w K+,K (s) = ( w (K,0,,K)(s), w (K,,0,K) (s))t, w K+, (s) = w (K,,,)(s), K, w (s) = ( w, (s), w +, (s),, w K+, (s))t, K, w (s) = ( w (s), w (s),, w K (s))t The followng theorem gves recurrent relaton for the vectors w (s), K of the Laplace- Steltes transforms w x, x E of the condtonal watng tme denctes w x (t) Theorem 6 The vectors of the Laplace-Steltes transforms w (s), K of the condtonal watng tme denctes under the control polcy m = {,, 3, 4} are related by the followng recurrent block threedagonal system Λ W, (s) w (s) = Γ e, (6) Λ W, The matrces Λ W, (s) = (Φ (s) w (s) = Γ w (s), K si 4(K +) ) and Γ, are of the dmenson 4(K + ) All

13 matrces have K + 3 block-columns and K + 3 block-rows The matrces Φ for are of the form: (λ + γ) A D 0 Ĉ A D C A D C A Φ = D C 3 A D 3 C 3 A D 3 C 4 A D 4 M (λ + γ) A Φ = (λ + γ) Φ = D 0 Ĉ Â D C 3 A D C 3 A D 3 C 3 A D 3 C 4 A D 4 M D 0 Ĉ 3 (λ + γ) Φ K = where A Â D C 3 A D C 3 A D 3 C 3 A D 3 C 4 A D 4 M D 0 Ĉ 3 Â = ( 0 λ ), Â 5 = A 0 0 Â ˆD C 4 A D 4 M Â = A, m = {3, 4}, Â(4) 5 = Ĉ = λ 0 0 λ λ λ + µ + γ λ + µ λ + γ ( Ĉ λ + µ + γ 0 4 = 0 λ + µ + γ λ 0 0 λ 0 0 λ λ 0 (λ + γ), Φ K =, m = {,, 3}, Â (4) = ( λ q, q, m = {, } K q K, = q, m = {, } K, Â 0 ˆD 0 Ĉ 4 Â 6 0 ˆD 4 M λ, Â 6 = ( λ λ ),m = {,,3, 4},m = {, }, Ĉ 3 = ), ˆD 0 = ( ) µ, µ ˆD = ), Â =, λ + µ + γ λ + µ + γ λ + γ µ µ µ 0 0 µ λ λ λ 0, q K, m = {, } K, m = {3, 4}, m = {, },, ˆD 4 = ( µ µ ),m = {,, 3,4}

14 The matrces Γ are of the form Γ = B ˆB B B B q, q, m = {, } K q Γ = Γ = B ˆB B B B ˆB B B K, = q, m = {, } K, q K, m = {, } K, m = {3, 4} Γ K = where B ˆB B B 0 = ( 0 γ 0 ), ˆB 3 = γ {, }, ˆB(4) 3 = γ γ γ 0 0, Γ K = γ γ γ 0, ˆB 4 = B ˆB , Γ (4), m = {,,3}, B (4) 0 = ( γ ( γ 0 0 γ 0 0 ), m = {,,3, 4} ( (4) K+ = B ), γ 0 ), ˆB = γ γ 0, m = Proof: Consder the model under optmal polcy m = Note that f J(t) = 0 then the fxed ob should be mmedately served, so the watng tme s 0, e w x (s) =, x = (q, d, d, 0) E For other states and postons of the tagged customer accordng to the equaton (4) the problem reduces to the above case by backward nducton Ths s always possble snce the poston J(t + ) s a decreasng 3

15 nteger w (q,,0,) (s) = s + λ + µ + γ for q q K, w (q,0,,) (s) = s + λ + µ + γ for q K, w (q,0,0,) (s) = s + λ + γ λ w (q,,,) (s) + µ w (q,0,0,) (s) + γ λ w (q,,,) (s) + µ w (q,0,0,) (s) + γ λ w (q,,0,) (s) + γ for q K, w (q,,0,) (s) = λ w s + λ + µ (q+,,0,) (s) + µ w (q,0,0,) (s) for q, q q, w (q,,0,) (s) = λ w s + λ + µ (q,,,) (s) + µ w (q,0,0,) (s) for q, q = q, w (q,,0,) (s) = s + λ + µ + γ for q K, q K, w (q,0,,) (s) = s + λ + µ + γ w (q,0,0,) (s) = s + λ + γ for K, q K, λ w (q,,,) (s) + µ w (q,0,0,)(s) + γ w (q,,, ) (s) λ w (q,,,) (s) + µ w (q,0,0,)(s) + γ w (q,,, ) (s) λ w (q,,0,)(s) + γ w (q,,0, ) (s) for K, q K, w (q,,,) (s) = λ w s + λ + µ + µ (q+,,,) (s) + µ w (q,0,,) (s) + µ w (q,,0,) (s) for K, q K, w (K,,,) (s) = µ w s + µ + µ (K,0,,) (s) + µ w (K,,0,) (s) for K (7) Now, after routne block dentfcaton, we may express the system for m = n (7) as gven n (6) For the STP the recurrent expressons are the same wth correspondng threshold level q For the polcy FFS t s enough to set q = To get the expressons for RSS polcy we take the system for FSS polcy and the equaton for the state q, 0, 0, rewrte as follows w (4) (q,0,0,) (s) = λ s + λ + γ ( w(4) (q,,0,)(s) + w(4) (q,0,,) (s)) + γ ( w(4) (q,,0, )(s) + w(4) (q,0,, ) (s)) for K, q K The tagged customer must wat n orbt f upon arrval he fnds the system n some state of the subset E W = {(q,, 0); 0 q q } {(q,, ); 0 q K }, m = {, } E W = {(q,, ); 0 q K }, m = {3, 4} 4

16 Denote by W, m = {,, 3, 4} the row-vectors of the dmenson K(K+) whch nclude the subvectors of steady-state probabltes n set E W : W W = (π e (3)e t (4K) + e 0 (4)e t 4(4K), e (4)e t (4(K )) + e 0 (4)e t 4(4(K )),, q +e 0(4)e t 4(4(K q + )),, K+ e 0(3)e t 3(4)), m = {, }, = (π e 0 (4)e t 4(4K), 3 e 0 (4)e t 4(4(K )),, K+ e 0(3)e t 3(4)), m = {3, 4} Accordng to the PASTA property the condtonal probablty of the state x upon arrval consdes wth the uncondtonal one Hence for the Laplace transform of the uncondtonal watng tme dstrbuton wth respect to all possble ntal states x of the process X(t) and the correspondng states x before an arrval we have The formula (8) ncludes two contrbutons: q W e = K W e = W (s) = ( π W s e + π W w (s)) (8) K (q,,0) + q=0 q=0 q=0 (q,,) = K (q,,) = (q,0,) + π q=0 (q,0,) + π (q,,0) + π (q,0,0) q=0 (q,0,0) +, m = {3, 4} K (q,,0) q=q, m = {, } s a steady-state probablty that the tagged customer does not have to wat for the servce; the transform q W w (s) = q=0 (q,,0) w K (q+,,0,q+) (s) + q=0 K W w (s) = (q,,) w (q+,,,q+)(s), m = {3, 4} q=0 (q,,) w (q+,,,q+)(s), m = {, } of the contrbuton of the watng tme wth dencty functon w c (t) gven W > 0 Note that 0 w c (u)du = W e The lmt property of the Laplace-Stltes transform allows us to get the value of the functon w c (t) at pont t = 0: γ, f q q K, d =, d = 0, = lm s s w (q,d,d,) (s) = γ, f q K, d = 0, d =, =, m = {, } γ, f q K, d = 0, d = 0, = lm s s w (q,d,d,) (s) = 0, { otherwse γ, f q K, 0 d + d, 0, otherwse, m = {3, 4} Therefore we have lm t 0 w c (t) = lm s s W w (s) = 0 5

17 The nverson of the Laplace transform s π W w (s) of the functon W c (t) = t get the dstrbuton functon of the watng tme W (t) = PW t = W e + W c (t), t 0 W x 0 w c (u)du s used to Now we obtan the n-th moments of W x whch we denote by Accordng to the ndroduced parttonng of the condtonal Laplace transforms we denote by W (n) the vector of the correspondng n-th moments: (n) = E(W x ) n, m = {,, 3, 4} W, (n) = ( W (q,d,d,) (n) d = {0, }, q + d + d = ) t, K +, mn{, K}, W (n) = ( W, (n), W +, (n),, W K+, (n))t, K, W (n) = ( W (n), W (n),, W K (n))t By dfferentaton the expressons (6) over the parameter s we get Snce Λ dn W, (s) ds nw (s) n dn ds n w (s) = 0, Λ dn W, (s) ds nw W (n) = ( ) n dn ds w n (s) n dn ds n w (s) we get s=0 (s) = Γ d n ds nw (s), K Φ Φ W (n) = n W (n) = n W (n ), n, W (n ) Γ W (0) = e(4(k + )), K Therefore for the uncondtonal moments of the watng tme we have W (n), K, n, E(W ) n = W W (n), n 0, m = {,, 3, 4} 6 Statonary dstrbuton of the soourn tme To carry out the calculaton of a soourn tme dstrbuton we apply the same Markov process () wth only excepton that the absorpton takes place at the tme when the tagged customer completes the servce that wll correspond to the case when J(t) = At the pont of tme of a new arrval t + to some of the state from the set E W the customer has to wat for the servce and The arrval to the state from the set E\E W J(t + ) = Q(t + ) mples J(t + ) = 0 that means that the customer must be served mmedately on the frst or second server accordng to the control polcy Let the process s absorbed when the component J(t) become equal to It occurs when the tagged customer leaves the system Defne T rv of the soourn tme n the system, 6

18 T t x x rv of the resdual soourn tme of the tagged customer gven that the system state s x, (τ) the condtonal densty functon of the resdual soourn tme, t x (s) = Ee st x, Res 0 correspondng Laplace-Stltes transform As above we partton the condtonal Laplace-Stltes transforms of the soourn tme denstes n the followng way: t, (s) = t (,0,0,)(s), K, (9) t,0 (s) = ( t,0,,0 (s), t,,0,0 (s)), K +, t +, (s) = ( t (,0,,) (s), t (,,0,) (s), t (+,0,0,)(s)), K, t, (s) = ( t (,,,) (s), t (,0,,) (s), t (,,0,) (s), t (,0,0,)(s)), K, t K+, (s) = ( t (K,,,) (s), t (K,0,,) (s), t (K,,0,)(s)), K, t K+,K (s) = ( t (K,0,,K) (s), t (K,,0,K) (s)), t K+, (s) = t (K,,,)(s), K, t 0 (s) = ( t,0 (s), t,0 (s),, t K+,0 (s)), t (s) = ( t, (s), t +, (s),, t K+, (s)), K, t (s) = ( t 0 (s), t (s), t (s),, t K (s)) Theorem 7 The vectors of the Laplace-Stltes transforms t (s), 0 K of the condtonal soourn tme denctes under the control polcy m = {,, 3, 4} are related by the followng recurrent block threedagonal system Λ T,0 (s) t 0 (s) = Γ 0 e, (30) Λ T, (s) t (s) = ˆΓ t 0 (s), Λ T, (s) t (s) = Γ t (s), K where Λ T,0 (s) = Φ 0 si (K+) and Γ 0 are the block dagonal matrces of the dmenson (K + ) These matrces have K + block-rows and block-columns The matrces Λ T, (s) = Λ W, (s) for ˆΓ are dagonal block matrces of the dmenson (K + ) 4K These matrces have K + block-rows and K + block-columns The matrces Φ 0 have the blocks and the matrces Γ 0 have ( ) µ 0 0 µ 7

19 the blocks ˆD 0 The matrces ˆΓ are of the form: ˆΓ = ˆΓ = G G G G G G G G G where G 0 = ( 0 γ ), m = {,, 3}, G (4) 0 = ( γ 0 γ γ 0 γ γ m = {,,3, 4}, G = γ γ, m = {, }, G 3 = γ, m = {3, 4} ), G = γ γ 0 0 γ 0 γ γ q, m = {, } N q, m = {,, 3}, G(4) 3 =, m = {, }, G (3) = γ γ 0 γ γ 0 γ γ 0 0 γ, G 4 =, G (4) = γ γ 0 Proof: Consder agan the system under OTP m = {} If J(t) = then the served tagged customer leaves the system and the soourn tme s 0, e t x (s) =, x = (q, d, d, ) E For the states where the tagged customer s n servce area the servce tme does not depend on the future arrvals or servce tme on the other server f t s busy Thus we have t (q,,0,0) (s) = µ s + µ, 0 q K t (q,0,,0) (s) = µ s + µ, 0 q K For the states where the tagged customer stayes ust before the servers J(t) = then we have the followng recursve expressons: t (q,,0,) (s) = λ t s + λ + µ + γ (q,,,) (s) + µ t (q,0,0,)(s) + γ t (q,0,,0) (s) for q q K, t (q,0,,) (s) = s + λ + µ + γ for q K, t (q,0,0,) (s) = s + λ + γ λ t (q,,,) (s) + µ t (q,0,0,)(s) + γ t (q,,0,0) (s) λ t (q,,0,)(s) + γ t (q,,0,0) (s) for q K, (3) In all other states the transforms t x (s) satsfy the equatons (3) By expressong these equatons n matrx form we obtan the expressons (30) for optmal polcy Analogously we obtan the expressons for the systems under other control polces, 8

20 The tagged customer ons the system f upon arrval he fnds the system n one of the state n the set E T = {(q, d, d ); d + d = {0, }, 0 q K} {(q, d, d ); d + d =, 0 q K }, m = {,, 3, 4} Denote by T the row-vectors of the dmenson (N + ) whch nclude the steady-state probabltes of the states n the set E ( K T = + =q T : e (4) + K+ e (3) 0 + (e 0 (3) + e (3)) + π (3) T = ( π (3) e (3) + = e t 0((K + )) m = {, } = π (3) e (4) + π (3) K+ e (3) e t 0((K + )) + π (3) 0 + π (3) (e 0(3) + e (3)) + π (4) T = ( π (4) e (3) + = + π (4) 0 + π (4) e 0(3) + + π (4) e (3) + = π (4) e (4) + π (4) K+ e (3) e t 0((K + )) = = π (4) e 3 (4) ) (e (4) + e 3 (4)) + K+ e (3) e t ((K + )), W, ) π (3) (e (4) + e 3 (4)) + π (3) K+ e (3) e t ((K + )), π (3) W, π (4) e (4) + π (4) K+ e (3) e t ((N + )) (e t 0((K + )) + e t ((K + ))), π (4) W Then for the uncondtonal Laplace transform of the soourn tme dstrbuton wth respect to all possble ntal states x of the Process X (t) and correspondng states before an arrval x we get and componentwse T t (s) = π (3) T t (3) (s) = π (4) T t (4) (s) = (q,0,) + π q=0 q + π (q,,0) t q=0 π (3) (q,0,) + π(3) q=0 π (4) (q,0,) q=0 K + q=0 T (s) = s π T t (s), (3) (q,0,0) µ + s + µ K (q+,,0,q+) (s) + µ (q,,0) q=q q=0 (q,0,0) µ + s + µ s + µ + π (4) (q,,0) q=0 π (4) (q,,) t (4) (q+,,,q+) (s) µ s + µ ) π (q,,) t (q+,,,q+)(s), m = {, } π (3) (q,,0) q=0 µ s + µ + µ s + µ + q=0 K q=0 π (3) (q,,) t (3) (q+,,,q+) (s), ( π (4) µ (q,0,0) + µ ) s + µ s + µ 9

21 Let t (τ), m = {,, 3, 4} denotes the uncondtonal densty assocated wth the Laplace-Stltes transform t (s) Its value at pont τ = 0 satsfes µ, f 0 q K, d = 0, d =, = 0 lm s t s (q,d,d,) (s) = µ, f 0 q K, d =, d = 0, = 0 0, otherwse Therefore, lm τ 0 t (τ) = lm s s T t (s) = µ K + µ 0 + (e 0 (3) + e (3)) + = =q lm τ 0 t(3) (τ) = lm sπ (3) t s T (3) (s) = µ π (3) e (3) + + µ π (3) 0 + π (3) (e 0(3) + e (3)) + = lm τ 0 t(4) (τ) = lm sπ (4) t s T (4) (s) = µ π (4) e (3) + + µ π (4) 0 + π (4) e 0(3) + = e (4) + K+ e (3) (e (4) + e 3 (4)) + K+ e (3), m = {, } = π (3) e (4) + π (3) K+ e (3) π (3) (e (4) + e 3 (4)) + π (3) K+ e (3), = π (4) e (4) + π (4) K+ e (3) π (4) e (4) + π (4) K+ e (3) + M π (4) e (3) + = π (4) e 3 (4) Now t suffces to nvert the Laplace transform T (s) to get the dstrbuton functon of the soourn tme T (τ) = PT τ = τ 0 t (u)du, τ 0 We now fnd the n-th moment of T x, m = {,, 3, 4} whch s denoted by n 0 Let T (n) denote the vector contanng the moments parttoned as a correspondng Laplace transforms: T x (n) = E(T x ) n for T,0 (n) = ( T (,0,,0)(n), T (,,0,0)(n)), N +, T, (n) = ( T (q,d,d,) (n) d = {0, }, q + d + d = ) t, K +, mn{, K}, T 0 (n) = ( T,0 T (n) = ( T, (n), T (n) = ( T 0 (n), (n), T,0 (n),, T T +, (n),, T T (n),, K+,0 (n)), K+, (n))t, K, T K (n))t By dfferentaton the Laplace transforms (30) over the parameter s snce T (n) = ( ) n dn ds t (s) n we get ( T n! 0 (n) =, n!,, n!, n! ) t, n, µ µ µ µ Φ T (n) = n T (n ) T (0) = e(4(k + )), K T (n), K, n s=0 0

22 For the uncondtonal moments of the soourn tme we have E(T ) n = T T (n), n 0 7 The number of customers served by drect access In ths secton we calculate the dstrbuton for the dscrete random value Θ of the number of customers that have been served by drect access to some dle server untl the targed customer on the orbt reaches the servce area Denote by Θ rv of the number of prmary customers served by drect access to the dle server, rv of the number of prmary customers that wll be served untl the tagged customer reaches the servce area, gven that the system s n state x, θ x (k) = PΘ x = k the condtonal dencty functon of the rv Θ x, (z) = Ez Θ x = (k)z k, z correspondng z transform Θ x θ x k=0 θ x By the low of total probablty for the Markov process X(t) the condtonal dencty functon θ x (k) has the form θ x (k) = a xy θ a y (k ) + θ y (k) (33) x a x y x,y a xy where the frst term represents the arrval of a prmary customer that can be mmedately served and the second one corresponds to all other possble transtons that do not change the event under consderaton Applyng z transforms to the relaton (33) we get θ x (z) = za xy a x θ y (z) + a xy y x,y a θ y x We partton the above z transforms accordng to the partton of the system states: defne the columnvectors θ, (z) n whch denotes the number of customers n the system and the poston of the tagged customer: θ, (z) = θ (,0,0,)(z), K (34) θ +, (z) = ( θ (,0,,)(z), θ (,,0,)(z), θ (+,0,0,) (z))t, K, θ, (z) = ( θ (,,,)(z), θ (,0,,)(z), θ (,,0,)(z), θ (,0,0,) (z))t, K, θ K+, (z) = ( w (K,,,)(z), θ (K,0,,)(z), θ (K,,0,) (z))t, K, θ K+,K (z) = ( θ (K,0,,K)(z), θ (K,,0,K) (z))t, θ K+, (z) = θ (K,,,)(z), K, θ (z) = ( θ (z), θ, +, (z),, θ K+, (z))t, K, θ (z) = ( θ (z), θ (z),, θ K (z)) t Theorem 8 The vectors of z-transforms θ (z), K of the condtonal denctes under polcy m = {,, 3, 4} are related by the followng recurrent block three-dagonal system (z) Λ Θ, (z) θ (z) = Γ e, (35) Λ Θ, (z) θ (z) = Γ θ (z), K

23 The matrces Λ Θ, (z) = (Φ + ( z)q ) and Γ, are of the dmenson 4(K + ) All are of the form: matrces have K + 3 block-columns and K + 3 block-rows The matrces Q Q Q = = Q K = where H = A, H H 6 = 0 H H H H H H H Ĥ H H H H Ĥ H λ λ λ λ Proof: λ λ 0 = λ λ 0, Q K =, H 3 = q, q, m = {, } K q K, q K, m = {, } 0 Ĥ Ĥ λ λ 0,, H 4 = K, m = {3, 4} λ λ λ 0, H 5 = λ 0 0 λ λ, m = {, }, H (3) = H, H(3) = A (3), H(3) = H, = 4,5, 6, H (4) = A (4), H(4) = A (4), H(4) 4 =, H(4) 5 = λ 0 0 λ 0 0 λ λ, Ĥ = Â, =,, 5, 6, m = {,, 3} θ (q,,0,) (z) = λ θ λ + µ (q+,,0,) (z) + µ θ (q,0,0,) (z) for q, q q, (36) θ (q,,0,) (z) = zλ θ λ + µ (q,,,) (z) + µ θ (q,0,0,) (z) θ (q,,0,) (z) = λ + µ + γ for q, q = q for q q N, θ (q,0,,) (z) = λ + µ + γ for q K, zλ θ (q,,,) (z) + µ θ (q,0,0,) (z) + γ zλ θ (q,,,) (z) + µ θ (q,0,0,) (z) + γ,

24 θ (q,0,0,) (z) = λ + γ zλ θ (q,,0,) (z) + γ for q K, θ (q,,0,) (z) = λ + µ + γ θ (q,0,,) (z) = λ + µ + γ θ (q,0,0,) (z) = λ + γ zλ θ (q,,,) (z) + µ θ (q,0,0,)(z) + γ θ (q,,, ) (z) for q K, q K, zλ θ (q,,,) (z) + µ θ (q,0,0,)(z) + γ θ (q,,, ) (z) for K, q K, zλ θ (q,,0,)(z) + γ θ (q,,0, ) (z) for K, q K, θ (q,,,) (z) = λ θ λ + µ + µ (q+,,,) (z) + µ θ (q,0,,) (z) + µ θ (q,,0,) (z) for K, q K, θ (K,,,) (z) = µ µ + µ θ (K,0,,) (z) + µ θ (K,,0,) (z) for K For the uncondtonal z-transform of the number of drectly served customers we get θ (z) = W e + π W θ (z), (37) where W e as before means the probablty that the customer upon an arrval goes mmedately to the servce area The nverson of the contrbuton θ W (z) leads to the densty functon θ c (k), that together wth Θ c (n) = n c (k) mples the relaton for the dstrbuton functon k=0 θ Θ (n) = W e + Θ c (n), n 0 The probablty that no prmary customers wll be drectly served whle another are watng on the orbt can be expressed as follows Θ (0) = W e + π θ W (0) Θ x The n-th factoral moment of the random value Θ x denote by (n) = EΘ x x n + ), n We partton the condtonal factoral moments n the same way as before: (Θ Θ, (n) = ( Θ (q,d,d,) (n) q + d + d = ) t, K +, mn{, K}, Θ (n) = ( Θ, (n), Θ +, (n),, Θ K+, (n))t, K, Θ (n) = ( Θ (n), Θ (n),, By dfferentatng the relaton (35) we get Λ Θ, Λ Θ, dn (z) θ dzn (z) dn dz (z) nq dz θ n (z) nq dz d n θ n d n θ n Θ K (n))t (z) = 0, d n (z) = Γ dz (z), K n θ )(Θ x 3

25 Snce Θ (n) = dn dz Φ Φ n θ (z) we get z= Θ (n) = nq Θ Θ (n) = nq (n ), n, Θ (n ) Γ Θ (0) = e(4(k + )), K The moments for the uncondtonal random values have the form Θ (n), K, n, EΘ (Θ )(Θ n + ) = W Θ (n), n 0 8 The number of retrals made by a customer Denote by Ψ rv of the number of retrals made by a tagged customer untl the servce starts, Ψ x rv of the number of retrals made by a tagged customer gven that the system s n state x, ψ x ψ x (k) = PΨ = k the condtonal dencty functon of the rv Ψ x (z) = Ez Ψ x = k=0 ψ x (k)z k, z correspondng z transform By the low of total probablty for the Markov process X(t) the condtonal dencty functon ψ x (k) has the form ψ x (k) = a xy ξ a y (k ) + ψ y (k) (38) x a x where the frst term represents the retrla of a tagged customer Applyng z transforms to the relaton (38) we get ψ x (z) = za xy a x y x,y a xy ψ y (z) + a xy y x,y x, a ψ y x (z) ψ, (z) = ψ (,0,0,)(z), K (39) ψ +, (z) = ( ψ (,0,,)(z), ψ (,,0,)(z), ψ (+,0,0,) (z))t, K, ψ, (z) = ( ψ (,,,)(z), ψ (,0,,)(z), ψ (,,0,)(z), ψ (,0,0,) (z))t, K, ψ K+, (z) = ( ψ (K,,,)(z), ψ (K,0,,)(z), ψ (K,,0,) (z))t, K, ψ K+,K (z) = ( ψ (K,0,,K)(z), ψ (K,,0,K) (z))t, ψ K+, (z) = ψ (K,,,)(z), K, ψ (z) = ( ψ, (z), ψ +, (z),, ψ K+, (z))t, K, ψ (z) = ( ψ (z), ψ (z),, ψ K (z)) t Theorem 9 The vectors of z-transforms ψ (z), K of the condtonal denctes under the control polcy m = {,, 3, 4} are related by the followng recurrent block three-dagonal system Λ Ψ, (z) ψ (z) = Γ (z)e, (40) Λ Ψ, ψ (z) = Γ ψ (z), K 4

26 (z) = (Φ + ( z)v ), Λ Ψ, = Φ are of the dmenson 4(N + ) All matrces have N + 3 block-columns and N + 3 block-rows The matrx V s of the form The matrces Λ Ψ,, and Γ (z) = zγ, Γ, (e 0 + e )(e 0 + e ) t (e 0 + e )(e 0 + e ) t V = γ e 0 e t e 0 e t q, m = {, } N q V = γ Proof: e 0 e t e 0 e t e 0 e t ψ (q,,0,) (z) = λ + µ + γ, m = {3, 4} λ ψ (q+,,0,) (z) + µ ψ (q,0,0,)(z) + zγ ψ (q,,0,) (z) for q q, (4) ψ (q,,0,) (z) = λ ψ λ + µ + γ (q,,,) (z) + µ ψ (q,0,0,)(z) + zγ ψ (q,,0,) (z) for q = q ψ (q,,0,) (z) = λ + µ + γ for q q K, ψ (q,0,,) (z) = λ + µ + γ ψ (q,0,0,) (z) = λ + γ for q K, λ ψ (q,,,) (z) + µ ψ (q,0,0,) (z) + zγ λ ψ (q,,,) (z) + µ ψ (q,0,0,) (z) + zγ λ ψ (q,,0,) (z) + zγ for q K, ψ (q,,,) (z) = λ + µ + µ + γ for q K, ψ (K,,,) (z) = µ + µ + γ λ ψ (q+,,,) (z) + µ ψ (q,0,,) (z) + µ ψ (q,,0,)(z) + zγψ (q,,,) (z) µ ψ (K,0,,) (z) + µ ψ (K,,0,)(z) + zγψ (K,,,) (z) ψ (q,,0,) (z) = λ ψ λ + µ (q+,,0,) (z) + µ ψ (q,0,0,) (z) for q, q q, ψ (q,,0,) (z) = λ ψ λ + µ (q,,,) (z) + µ ψ (q,0,0,) (z) ψ (q,,0,) (z) = λ + µ + γ for q, q = q λ ψ (q,,,) (z) + µ ψ (q,0,0,)(z) + γ ψ (q,,, ) (z) for q K, q K, 5

27 ψ (q,0,,) (z) = λ + µ + γ ψ (q,0,0,) (z) = λ + γ λ ψ (q,,,) (z) + µ ψ (q,0,0,)(z) + γ ψ (q,,, ) (z) for K, q K, λ ψ (q,,0,)(z) + γ ψ (q,,0, ) (z) for K, q K, ψ (q,,,) (z) = λ ψ λ + µ + µ (q+,,,) (z) + µ ψ (q,0,,) (z) + µ ψ (q,,0,) (z) for K, q K, ψ (K,,,) (z) = µ µ + µ ψ (K,0,,) (z) + µ ψ (K,,0,) (z) for K For the uncondtonal z transform we get where W area By nverson of the contrbuton W relaton Ψ c (n) = n ψ (z) = W e + π W ψ (z), (4) e denotes the probablty that the customer upon an arrval goes mmedately to the servce ψ (z) we get the densty functon ψ c (k) and and by the k=0 ψ c (k) we derve the dstrbuton functon Ψ (n) = W e + Ψ c (n), n 0 The probablty that the customer makes no retrals concdes wth the probablty that t wll be drectly served, e Ψ (0) = W e Ψ x (n) = EΨ The n-th factoral moment of the random value Ψ x denote by x x n + ), n We partton the condtonal factoral moments n the same way as before: (Ψ Ψ, (n) = ( Ψ (q,d,d,) (n) q + d + d = ) t, K +, mn{, K}, Ψ (n) = ( Ψ, (n), Ψ +, (n),, Ψ K+, (n))t, K, Ψ (n) = ( Ψ (n), Ψ (n),, By dfferentatng the relaton (35) we get Ψ K (n))t )(Ψ x Snce Ψ (n) = dn dz ψ n Λ Ψ, Λ Ψ, dn (z) dz d n n ψ (z) nv (z) = Γ dz dz (z) we get z= n ψ dn ψ n d n dz n ψ (z) = 0, (z), K Φ Φ Ψ (n) = nv Ψ (n) = Γ Ψ (n ), n, Ψ (n), K, n, Ψ (0) = e(4(k + )), K The moments for the uncondtonal random values have the form EΨ (Ψ )(Ψ n + ) = W Ψ (n), n 0 6

28 9 Numercal results and comparson analyss Consder the system M/M/ wth prmary arrval rate λ, retral rate γ and servce rates µ and µ By nverson of the derved Laplace transforms (8) and (3) t s possble to evaluate the watng and soourn tme dstrbutons There are two practcal algorthms for the nverson of Laplace transform: The conventonal algorthm and the Fourer-seres algorthm Mathematcal packages such as Mathematca, Mathlab, Mathcad, etc nclude the standard functons that apply the conventonal one It s based on the partal functon expanson method and works effcently only n case of small order ratonal functons So we can obtan on the bass of ths method an accurate representaton of the functons W (t) and T (t) only for small t but n symbolc form The algorthms based on Furer-seres represent the numercal nverson methods, eg Euler and Post-Wdder, that can be used for large t as well The mentoned mathematcal software allow also to nvert z-transforms (37) and (4) to get the dstrbuton functons Θ(n) and Ψ(n) However there are problems by nverson of the functons of hgher order Therefore we mplement the numercal nverson of z-transforms usng the Lattce-Posson algorthm By means of Mathemaca package we have created the procedures: for the calculaton of steady-state probabltes under optmal and heurstc servce dscplnes, formulas (0 3) and (8) (), for the numercal nverson of the Laplace transforms W(s) and T(s) usng the Euler and Post-Wdder algorthms, for the numercal nverson of the z-transforms Θ(z) and Ψ(z) usng the Lattce-Posson algorthm In Fgures 4 we have ndcated the watng tme (the fgures labelled by letter "a") and the soourn tme (the fgures labelled by letter "b") for dfferent values of the prmary customer arrval rate λ and retral customer rate γ In our examples we fx the servce rates µ =, µ = 03 The followng observaton can be notced from these fgures: The curves of the watng tme dstrbutons W (t) for the system under threshold control polces (OTP, STP) le below the other curves (FFS, RSS) that specfes that the watng tme of a customer n the orbt s larger for the threshold systems Ths does not contradct the optmalty of the threshold polcy snce t mnmzes the soourn tme On the fgures wth soourn tme dstrbutons T (t) one can notce that for some small values of argument t the curves for the OTP can le below other graphs but startng from some pont of tme t they are above the others Nevertheless the mean soourn tme for the optmal polcy turns out to be the least The curves of the watng tme dstrbuton for the system under FFS control polcy le above the other curves that llustrates that ths polcy mnmzes the watng tme of a customer n the orbt At the same tme the largest soourn tme belongs to the system under RSS polcy It can be explaned by the fact that ths polcy wth equal probablty assgns a customer to the faster or slower server that n turn makes a sgnfcant contrbuton to the soourn tme ncreasng In Fgures and the prmary customer arrval rate λ s vared and the correspondng load factors are ρ = 096, m = {,, 3}, ρ (4) = 004 and ρ = 040, m = {,, 3}, ρ (4) = 044 In Fgure 3 and 4 the retral customer rate γ s vared and the load factors are, respectvely, ρ = 099, m = {,, 3}, ρ (4) = 000 and ρ = 0375, m = {,, 3}, ρ (4) = 0377 As λ or γ ncreases then the load factor ρ also ncreases that leads to the dstrbutons wth heaver tals Whle n Fgure and 3 the curves for the threshold systems (OTP, STP) look very smlar, n Fgure and 4 the suffcent dfference can be notced Thus we can assume that f the load factor s suffcently small, e the system has a so-called "lght traffc", 7

29 then the schedulng threshold polcy may be a good approxmaton for the optmal one (a) (b) W m t OTP STP FFS RSS T m t OTP STP FFS RSS t t Fgure : Dstrbuton functons (a) W (t) (b) T (t) for λ=05, µ =, µ =03, γ=5 (a) (b) W m t OTP STP FFS RSS T m t OTP STP FFS RSS t t Fgure : Dstrbuton functons (a) W (t) (b) T (t) for λ=09, µ =, µ =03, γ=5 (a) (b) W m t OTP STP FFS RSS t T m t OTP STP FFS RSS t Fgure 3: Dstrbuton functons (a) W (t) (b) T (t) for λ=05, µ =, µ =03, γ=85 8

30 (a) (b) W m t OTP STP FFS RSS T m t OTP STP FFS RSS t t Fgure 4: Dstrbuton functons (a) W (t) (b) T (t) for λ=09, µ =, µ =03, γ=85 The followng Fgures 5 (λ = 05 and λ = 09 n fgures labelled, respectvely, by letter "a" and "b") and 6 (γ = 5 and γ = 85 n fgures labelled, respectvely, by letter "a" and "b") represent the dscrete dstrbuton functons Θ(n) as a stepped curves for the number of prmary arrvals that wll be drectly served before an orbtng customer reaches the servce area The presented dagrams reveals the followng observatons At pont t = 0 the presented functons equal to the probablty that there are no customers that wll be served drectly As t was mentoned above, ths probablty equals to the probablty W (0) plus some probablty that the orbtng customer wll be served earler then a prmary arrval reaches some server As t to be expected n the system under FFS polcy the orbtng customer passes fewer prmary customers It concdes wth the observaton that the watng tme dstrbuton of an orbtng customer under ths polcy le above other graphs The polces (OTP, STP) turn out to be the worst wth respect to the value of nterest that can be also explaned by the suffcently larger watng tme of a customer under threshold polces n comparson wth others 3 For small load factor ρ (the concrete values are gven n prevous example) the number of drectly served prmary arrvals s qute small, the concrete examples show that almost surely not more then, e Θ (3) > 099 (Fgure 5(a)) and not more then, Θ () > 099 (Fgure 6(a)) prmary arrvals wll be served drectly whle the orbtng customer s watng for the servce As the load factor ncreases, more customers are served drectly, that can be confrmed by the fgures observaton, namely, Θ (7) > 099 (Fgure 5(b)) and Θ (3) > 099 (Fgure 6(b)) 9

31 (a) (b) m n OTP STP FFS RSS n m n OTP STP FFS RSS n Fgure 5: The dstrbuton functon Θ(n) (a) λ = 05 (b) λ = 09 for µ =, µ =03, γ = 5 (a) (b) m n m n OTP STP FFS RSS n 09 OTP STP FFS RSS n Fgure 6: The dstrbuton functon Θ(n) (a) λ = 05 (b) λ = 09 for µ =, µ =03, γ = 85 The next Fgures 7 (λ = 05 and λ = 09 n fgures labelled, respectvely, by letter "a" and "b") and 8 (γ = 5 and γ = 85 n fgures labelled, respectvely, by letter "a" and "b") llustrate the dscrete dstrbuton functons Ψ(n) for the number of retrals made by an orbtng customer untl t reaches the servce area The followng conclusons cam be done by observng the graphs At pont t = 0 the ump of the functons corresponds to the case when no retrals wll be made by a customer and equals to the probablty that the customer wll be served drectly, e Ψ (0) = W (0) As λ and γ ncrease the dstrbutons reveal the heaver tals Under the FFS and RSS control polces the customer makes less retrals as under threshold polces (OTP, STP) because frst two dscplnes mply the shorter watng tme 3 The number of retrals strongly depends on the retral rate In case when γ s small, Fgure 7(a,b), then the number of retrals wth large probablty wll be not very large, eg n ths example Ψ (6) > 099 and Ψ (7) > 099 Otherwse, when γ s large, Fgure 8(a,b), then the number of retral sgnfcantly ncreases, eg Ψ (0) > 096 and Ψ (0) >