Dagsvik, John

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STATISTICAL RESEARCH REPORT Institute of Mathematics University of Oslo ON THE WAITING TIME DISTRIBUTION OF BULK QUEUES by John Dagsvik No 8 1973 Abstract The waiting time process of the n-th arriving group is considered for the general bulk queue·ing model GIX/Gy/1 • A generalisation of Lindley's waiting time equation is established. By a generalisation of Kingman's method [3], this equation is solved for the models GIX/E, y /1 .K and ~x/GY/1 • When the service time is Erlang distributed ~' the results are applied to the case where the service- and the arrival groups are of constant size. Key words: Wendel projection, Group Waiting time, Restbatch, Waiting time equation, Erlang distributions, Hy~erexponential distribution, Stationary distri-butions. Contents page 1 • Introduction 1 2. The algebraic formalism 1 3. Group waiting time 3 4. The model GIX/E:K:y/1 12 5. Special cases 16 6. The model E1cx/GY/1 19 7. The stationary solution 21 - 1 -1. Introduction In the present paper we shall assume that customers arrive in groups en , n = 0,1,2, •••• The group size is a stochastic variable X , with probability distribution f(o) • The inter-arrival intervals ~, n = 0,1, ••• are independent and have the same distribution a(o) • The service mechanism is described as follows: At the end of a service period the server accepts Y customers from the waiting line, or a smaller number if the line is shorter. Y is called the service group capasity. ~1e length of the service time B , has the distribution b(o) • We shall assume the existence of two integers m, 1 such that X~l,Y:5m. The most general works on bulk ~ueues seems to be those of Keilson [2], Cohen [1], Le Gall [5], Lambotte and Teghem [4]. They obtain the distribution of the queue length from which the waiting time distribution is derived. However, there exists no such results for general distributions a(o), b(o), f(o) and g(o) • Earlier works are restricted to the case where a(Q) or b(o) are the exponential distribution. Even if b(o) is exponential the analysis are only limited to bulk service models (Cohen [1], Le Gall [5]). 2. The algebraic formalism. Let On denote the set of n x n matrices whose components are finite complex measures on the Borel subsets of the real line. According to Kingman [3] the product of two measures is defined as their convolution. An operator T ~ o1 ~ n1 is defined by - 2 -(2.1) where e is the measure if if and R+ = (0, oo) • 0 E E 0 t E This operator has the property that if X is a random variable with distribution v(E) = PriX E E}, then (Tv)(E) = Pr!X+ E E} • Kingman shows that 0 1 is an commutative algebra over the com-plex field r. with identity e • Furthermore; he shows that the image 0~ and the kernel 0~ of T are both disjoint sub-algebras of o1 • T is extended to 0 by T l v .. ! = {Tv .. l , n lJ lJ {vijl E On and multiplication in On is defined in the obvious way. With Ine as identity it is easy to verify that (~ has the same properties as o1 except that On is no longer commu-tative. The norm on On is defined by II vii = max L:J ldv · ·I , j i lJ v = l v .. ! . lJ The set whose elements are Fourier-Stieltjes transform of ele-ments from o1 w·e denote n1, and extend the definition to On in the obvious manner. - 3 -3. Group waiting timee We shall study the group waiting time process defined as follows: Definition 3.1 By the group waiting tn1e Wn, n = 0,1,2, ••• we mean the waiting time (excludli1g service time) of the first customer from On who is taken into service. It is convenient to work with the random variable Zn which is defined as the time en spends in the queue until there is no one ahead, except the ones being served. If Ln is the time interval from the instant when Zn becomes zero and until the last service group with customers from On starts, then the sequence iznl , n = 0,1, 2,... satisfies ( 3.1) with the initial condition z = z • 0 We recognize the expression above as ru1 equation of the similar type as the waiting time equation found by Lindley [6]. However, there is an important difference: The random variables Zn and Ln are no longer independent as in the model GI/G/1 • Consider now the servicing of the n-th arrival group On • Service may be performed in one or several groups. The last service group which has customers from en will be called the n-th rest batch. The rest batch may be filled with customers from On or there may be places available for customers from 0n+1 • - 4 -We defDLe two random variables Tn and Jn , as follows: If the n-th restbatch can accept s customers and contains only t < s customers from Cn , then jJnl = s-t • If the n-th restbatch contains customers from en_1 ; Jn = -\Jnl • T is the number of customers which can n be accepted from en+1 in the first service group with customers from en+1 • We define J_1 = T_ 1 ; the capasity of the initial service group. If S(Tn_ 1) denotes service time of en , excluding the service time of the n-th rest batch, we realize that Zn > 0 implies that when (3.2) when J > 0 n-J < 0 ' n since en will have to wait an extra service period when Jn-1 = 0 ' Jn ~ 0 • When zn = 0 the first service group from en has capacity y and en must wait for the time wn before service starts, hence when when and Tn_ 1 = Y • Jn 2: 0 J < 0 ' n We define matrices of distribution functions and H ( t) ::. l Hi j} for i,j = -(m-2),-(m-1), ••• ,m-1 , {K .. } lJ - 5 -by uni(t) = Pr{Z < t, J 1 = il, n - n-~n· (t) = Pr{W < t, J 1 = i}, n - n-~ r'r\S' (T,-1) ::;t, Jn=i \Jn-1=j 'Zn > Oj when i > 0 K .. J.J Pr{O ~ t, Jn=J.!Jn-1=J} when i < 0 ' '-,..,. = ~ lir!S' (Tn_1 ) ~ t, J =i\J 1=j,Z =0} when i > 0 n n- n H .. J.J l -0 when i < 0 • If ~ is the probability distribution of a random variable X , ~* denote the distribution of (-X) • Let Considering the possible events between the n-th and the n+1-th arrival we find (3.8 b) ui+1 (t) = L:[Pr{M+ < t,Z > O,D .. }+Pr!N+< t,Z =O,D .. } ] n . -11 - n J.J n- n J.J J when i :::; 0 , + Pr{ (Z +B-.A )+ < t,Z -.A < O,Z > O,D .. } ] n n - n n - n J.J . when i < 0 • (3.9 a) can be written - 6 -+Pr{(Z -A)++B < t, n - sn > o, D .. } l.J -Pr{B < t,(Z -A)+= O, Z > 0, D .. } J - n -n n l.J = ((bo(O,i)_b)(ui -eui (o)))(t) n+1 n+1 + L Pr{(Z -A +B)+< t, Zn > 0, D1.J.}. j n n -When i ~ 0 the expression is more complicated; (3.9 b) v~ + 1 ( t ) = ~I Pr { ~ + B c ( o , J n) ~ t, ~ > o , z > o , D. . l J n l.J +Pr{ (M +B)+_< t, M < 0, Z > 0, D .. } -"'h -"'h - n l.J +Pr{(N +B)+< t, N < 0, Z = 0, D1.J.}]. n - n- n If (3.9 b) is revvritten in the same way as (3.9 a) we obtain vi (t) = ((bc(O,i)_b)(ui -€Ui (o)))(t) n+1 n+1 n+1 + ~[Pr{ (Mn+B)+ :5 t, Z > O, D .. } J n l.J + Pr{ (N +B)+ < t, Z = 0, D .. }], i > 0 • n - n l.J - 7 -Since Zn > 0 implies !(Nn+B)+ ~ t, Zn = 0} W = Z +B6(0,J 1), the probability of n n n-can be written Pr{N +B)+< t, Z = 0} = Pr{(N +B)+< tt n - n n - ~ -Pr{(Z +S(Y)+(1+&(0,J 1 ))B-A )+ < t, Z > 0}, n n- -11 - n whence (3.10 a) ui = 2:: T(a*K .. (Uj -e:Uj(O)) when i < 0 , n+1 j lJ n n (3.10 b) ui = L:[T(a*(K .. -bo(O,j)H .. )(Uj -e:Uj(O))) n+1 j lJ lJ n n and (3.10 c) + T(a*H .. vj) J when i > 0 lJ n + L: T(a*bK .. (Uj -e:Uj(O))) when i < 0 , j lJ n n Hence we have established a set of equation for Un and Vn • Lemma 3.1 (i) - 8 -~ f f ( I j I +i) , when i < 0 l ~ br ~ (f(g*)r-1)(p+lj\)g(i+p)+o(O,i)f( \j j)e ~ r,2:1 p>O when i > 0 , · r r when i > 0 (ii) \( ~ b ~ (f(g*) )(p)g(i+p) ' H .. = r>O p>O 1.J 0 when i < 0 , (iii) bH. . = K. when i > 0 • lJ lO Proof: Let R(Jn_1 )+1 denote the number of service groups with customers from en • R(Jn_ 1 ) must satisfy· vJhen J > 0 ' J 1 = j ~ 0 ' z > 0 ' n- n- n (3.11 a) R(j) R(j)-1 J +X= ~ Y +\j\ , X> ~ Y +\j\ n s=1 s s=1 s because customers are served in R(j)+1 groups and the rest of en is served in the n-th restbatch. \'Jhen is complete and Tn_ 1 = Y , so that R(O) R(0)-1 J 1 = 0 , the restbatch n-(3.11 b) = ~ y +Y s=1 s X > ~ Y +Y • s s=1 - 9 -By an elementary argument i ~ 0 , j ~ 0 , r 2 1 , When Jn <0, zn > o, the relation must be valid. Furthermore S(Y) is seen to have the same distribution as S(O) • The theorem now follows easily. It is convenient to introduce some further matrix notations. Let F = { F . . }, G = { G . . } , E = { E . . } , K 1 = { K' . .j and ~J ~J ~J ~J ~ = l~ij} be the matrices with entries r r F .. = f(j-i), K' .. = L: br L: (f(g*)r)(p+j)g(i+p), ~J ~J r20 p>O for i,j = 0,1,2, ••• m-1 and F .. = K' .. = 0 ~J ~J for G .. = g(i-j), E .. = L: g(i+p)f(p+j), for i,j = 0,1, ••• m+(l-m)+-1 , ~J ~J p>O c(i,O) for i,j = 0,1, ••• r-1 - 10 -Lemma 3.2 (i) K' = E(ei -bG)-1I m m (ii) Proof: K .. ~J I 1eF .. , when = -~J bK . . + 6 ( 0 , i ) F . . e ~J ~J (ii) is seen immediately. i < 0, j 2: 0 when i ~ 0, j > 0 • (i): Since E~ = f(EGr).J.}={ ~ (f(g*)r)(p+j)g(i+p)}, ~ p>O i, j = 0, 1, ••• m-1, the theorem follov, s. Observe that Gr = 0 when r > m+(l-m)+ , hence there is no convergence problem. Remark: Even if the matrices involved are defined m dimensional they are understood to be m+(n-m)+ dimensional with the undefined entries equal to zero. By the substitutions ~ = !Q~kl, Pn = {P~k}, where and (3.12 b) pik = n (u~i+u* -o(O,i)~(u~r+U~))6(k,O) r i,k = 0,1, ••• ,m-1 , it is possible to write (3.10) on the form .. 11 -whence theorem 3.3 follows. Theorem 3.3 Let and assume that U0 (t) = I 2m_ 1e(z+t), V0 (t) = r 2m_ 1e(w+t) • Then Un and Vn are uniquely determined by (3.13) and where - 0' ( Im -vm )K' -1 • 0, bv K' m Equation (3.14) is called the waiting time equation. Corollary 3. Llr _! When the trafic intensity p = E(X)E(B)/E(Y)E(A) is less than unity the stationary distribution "" "" P = lim P n~oo n exists and is determined by - 12 -A proof of the existence of the stationary distribution is given in [7] and [2]. We will assume that the customers in an arrival group are ordered and the queuedisiplin is first come first served. Let wP n be the waiting time of customer nr. p that en contains at least p customers. Let rP = PrjwP < tl . n n-in given The problem of finding rP can be solved by the following argument; n If en contains exactly p customer then obviously r P n i = ~ un+1 • ~ Hence we must have where K' p and is replaced by KP are 0 (p' 0) • 4. The model GIX/Eky /1. K' and K respectively, when f(.) Within this model it is possible to obtain solutions of the waiting time equation (by using the approach suggested in [3] p.p. 312-313). Let e8; 8 E W , denote the complex exponential measure defined by e8 (E) = J eexp(-ex)dx , if Re e > 0 , EnR+ = J eexp(-ex)dx , En-R+ if Re e < 0 , - 13 -The following le~na is proved in the same way as the analogous results in [3] p.p. 312-313. Lemma 4.1 (i) where n - .1:!:... ~-" - p-e . (ii) where c::o -py ( )m Y m = J e m! l:lY a ( dy ) • 0 By the introduction of the generating functions (3.14) is transformed to the equivalent equation - 14 -By the transformation we get (4. 2 i) ¢ = x (I - v ) T (a* ( F ( ¢ - € $ ( 0 ) ) + Ew ) ) + P , X mm X X X 0 (4.2ii) wx-b (Gwx +¢x-€1jix (0) )=xvmT(a*b (F( ¢x-e¢x (0) )+Ewx)+vmQo • By assumption b k = ell • Assume that (4.2) has a solution of hyperexponential type; i.e, (\ qqo~ ) e + ~ ( q~ \) e • >1 q. e. J_ J I J I ' If uo = e:U0 (o) a."ld V = eV (0) , 0 0 it follows from the definit-rv rv r~-vm)Uo I ion of pn that p = 0 vmvo I I • Inserting (4.3) in (4. 2) gives (e:I -bG)q 11 + I: (e q'!-e b(Gq'!+qq) m o j>O ej J ej J J = xv I: T(a*be 6 (Fq~+Eql!))+xv T(a*bEq")+v V • mj>O j J J m o m o • From Lemma 4.1 follows - 15 --x(T -v )Eq 11 e+(I -v )U = 0 , -m m o m m o (4.4-ii) k -xv 2.:: (Fq~+Eq 1!)((1-~.) 2.:: ~~fJ .(er-e)+(1-~~)e) mj>O J J J r=1 J r J 1-l J k r -xv Eq 11 ( 2.:: (e -e )yk +€ )-v V = 0 , m o r=1 1-1 -r m o where k fJ . = 2.:: ~-J.rYs-r' ~·=~<e.) • rJ s>r J J The equations above imply that the coefficients of e, (e~-e), r = 1 ,2, •• ,k, and e 8 .' j = 1 ,2,... are zero. Hence, J we are led to the equations (4.5 i) 1-a. . q' = 2.:: ...:__:..;:]_ q ~+x(I -v )Eq"+(I -v )U , o j>O aj J m m o m m o (4.5ii) - 16 -(4.5iii) (4.5iv) +(f3.-1)(n .a:: 1 -~~J.t~)v (I-f3~G)]q'!, r = 1,2, ••• ,k, J rJ J J m · m J J (4. 5 v) where ~j'j == 1,2, ••• hk , are the roots of' where h = m+(l-m)+. Observe that (4.5 iv-v) constitutes a set of' 2kh equations. Hence it is sufficient that (4.5 vi) has kh roots. 5. Special cases. Uni'ortunately, the assumptions (4 .• 3) are not always full-filled. For instance when Y = m > 1 • However, in this case we are able to slighten the assumptions (4.3). Observe that Y = m > l implies G = 0 • Let whence - 17 -and (5.3) We now suppose that (5.3) has a solution of the form s = p e+ L: e8 P. 1 X 0 j>O j J that is; p0 ,pj,j > 0 , must satisfy p e+ L: e8 p. = x(F+bE)(Im-~ ) L: T(a*e 8 )p.+xE~ L: T(a*be 8 )P~ o j>O j J m j>O j J mj>O j J Exactly the same calculations as in the preceeding section give (5.4 i) p. = xo. . (F+~ ~E) p. , J J J J j = 1,2, ••• ' L: {(p.-1)P~-r~.-o(r,k)(1-o..))E(I -~) j>O J J J J m m ( 5 • 4 iii ) ( I - xEv ) p = I: { (a. -:- 1 -1 ) ( I - ~ ) + xEv + :xa . E ( I - v ) m m o j J m m m J m m k l -xa. -~. E p.+E~ V • J J J m o - 18 -We shall now consider the case when both the arrival and the service group capacity are of constant size; i.e. X o 1 Y = m • Assume first that m < 1 • Then F = 0 and It follows. that r -~ 0 ,xcc:sn l . 11-m' 0 I L ..; From Lemma 1 in Takacs [8] page 82 we conclude that (5.5) has kl distinctroots 13-=P(e.), j=1,2, ••• ,kl for Ree>O., J J Hence equations (4.5) have a solution. Ass~~e that 1 < m • Then G = 0 , and When p0 is eliminated, (5.4ii) is a set of kl equations because -- -1 ~1 o , o I E = • l I 1 , 0 L ; Accordingly, (5.4) has a solution since (5.6) has kl distinct roots. The stationary solution is obtained by multiplying (4.5) and (5.4) by 1-x and let x tend to 1- • - 19 -The "key" to the solution is the fact thatihe assumption a enables us to express the operator T in a special form Lemma 6.1 Suppose Then k k k k r T((e*) ~) = (e*) ~+(e-e*) ~a (e*) - , r=1 r where Proof: For k = 1 the Lemma reduces to equation (72) in Kingman [3]. Assume (6.1) valid for k = 1,2, ••• p • Since T(e*)r = e we get Now v = T((e*)P~) E o; implies T((e*)P+1\--L) == T(et v) = e*'v+(€-e*)a' , where cr.' = ( e*v) (,.,.. IR +) • Obviously T(e*v)(~ R+) = (e*v)(~ R+) which yields a' = ap+1 • Hence, by the assumption T(e*v) = (e*)P+ 1 ~--L+(e-e*) ~ a (e*)P+1-r+(e-e*)a r=1 r p+1 and the Lemma follows by induction. - 20 -With ~ = K1~x ru1d e = eA application of (6.1) on (4.1) gives (with I= I 2m_ 1) (6.2) k (~I-xa*K1 )$x = x(e-e*) ~ ~ (e*)k-r+xT(a*(K2-K1 ))~ (0)+~ • r=1 r x o Let t:.x = e:I-xa*K1 • Inserting t = 0 gives ru1 expression for ~k, viz., Since !Ja*K111 ::;: 1 , L'. - 1 = ~ (xa*K1 )r exists when !xI < 1 • x r> 0 Thus (6.3) and (6.2) are equivalent to If k = 1 , t = 0 determines ~x(O) by When k > 1 let - 21 -Thus, (6.4) can be written (6.6) Furthermore (E.~ )(0) = aJ. leads to J X j = 1 '2' ••• 'k-1 ' which determines aj' j = 1,2, ••• k-1, and $x(O) 1. The stationary solution In this section we shall demonstrate how the stationary solution of (6.2) can be obtained by use of the Fourier-Stieltjes trans-form. In the stationary case we have k (7.1) (~I-a*K1 )~ = (~-eA*) ~ ar(eA*)k-r+T(a*(K2-K1 ))P(O) , r=1 where now After the introduction of the Fourier transform an equation ana-logous to (6.6) is obtained (7.2) .... L~ 1 (z)(P(z)-P(o)) ... k-1 .... = - c 1 ( z )P ( o ) + ~ c ( z )o. _, • r=1 r I - 22 -Let x. be defined by AA A x.A 1 = det ll 1 , whence Suppose that det tl 1 (z) has k roots z 1 ,z2 , ••• ,zk. Then P(O),a 1 , ••• ,ak_1 , are determined except for a constant by (7 .4) A A k-1A A ~(z1.)C 1 (z 1.)P(O) = ~ R(z.)c (z.)a r=1 1 r 1 r i=1,2, ••• k. Premultiplicating (7.2) by u = gives an equation where both sides become zero when z = 0 • ~ l'Hospitals rule, (7. 5) A A A A A k~1 (a*vK1 )' (o) (P-P(O)) = uc1 '(o )P(o )-i I: ua, • r=1 r A .... Consider (a*UK1 ) 1 (0)P(O). The process !Jnl is recurrent and therefore Jn converges in distribution to J , say. Since and uK1 = (bv K,v (K-F)) m m where - 23 -= (k (t),k1(t), ••• ,k 1(t)) , o m-k.(t) = L: k .. (t) , J . lJ l it follows that A "' A (a*bvmK)'(O) = i(E(S'(j))jJ=j)-iE(A)+iE(B) • Thus = i '"E(S'(j))jJ=j)Pr(J=j)-iE(A)+iE(B) • j By (3.11) we find E(S'(J)-A+B) = (p-1)E(A)+E(B) = kA- 1 (p-1)+E(B) • Equation (6.12) therefore reduces to A Ah k-1 (7.6) E(B)-k(1-p) = -iA(a*bK)'(O)P(O)+v 2 1Q(O)-A L: u~ • m- r=1 r Together with (7.4) we have a set of k+1 matrix equations to determine the k+1 un1mowns 'P(o), a. 1 , • •. ,a.k_ 1 • It is known that a probability distribution can be approximated by a linear combination of Erlang distributions. From Lemma 6.1 it is clear that the results in the last section can be general-ized to the case when a(o) is a linear combL~ation of Erlang distributions. Accordingly, it is possible to obtain approxi-mate solutions of the waiting time equation for general a(a) • - 24 -References [1] Cohen, J.W. (1969). The single se~rer gueue. North-Holland Publishing Company. [2] Keilson, J. (1962). The general bulk queue as a Hilbert problem. J. R. Statist. Soc .. B, .E.i 1 344-359. [3] Kingman, J.F.C. (1966). On the algebra of queues. J. Appl. Probab. ~' 285-326. [4] Lambotte, J.P. and Teghem, J.L. (1969). Modeles d'attente :r-1/G/1 et GI/M/1 a arrivees et services en groupes. Springer-Verlag, Berlin. [5] Le Gall, P. (1962). Les systemes avec ou sans attente et les processus stochastigues. Dunod, Paris. [6] Lindley, O.V. (1952). The theory of queues with a single server. ~Pr~o~c~._c_am~b~·~Ph~i~l~·~· ~S~o~c~. ~' 277-283. [7] Miller, R.G. (1959). A contribution to the theory of bulk queues. J. R. Statist. Soc. B, Z1' 320-337. [8] Takacs, L. (1962). Introduction to the theory of gueues. Oxford University press. London. 

ON THE WAITING TIME DISTRIBUTION OF BULK QUEUES

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