Stationary Single-Server Queuing Processes with a Finite Number of Sources Author(s): Gerald Harrison Source: Operations Research, Vol. 7, No. 4 (Jul. - Aug., 1959), pp. 458-467 Published by: INFORMS Stable URL: http://www.jstor.org/stable/166944 . Accessed: 08/05/2014 17:54 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. . INFORMS is collaborating with JSTOR to digitize, preserve and extend access to Operations Research. http://www.jstor.org This content downloaded from 169.229.32.137 on Thu, 8 May 2014 17:54:17 PM All use subject to JSTOR Terms and Conditions
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Stationary Single-Server Queuing Processes with a Finite Number of SourcesAuthor(s): Gerald HarrisonSource: Operations Research, Vol. 7, No. 4 (Jul. - Aug., 1959), pp. 458-467Published by: INFORMSStable URL: http://www.jstor.org/stable/166944 .
Accessed: 08/05/2014 17:54
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp
.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].
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The Teleregister Corporation, Stamford, Connecticut
(Received October 6, 1958)
Some basic relations are obtained that apply to finite single-server sta- tionary queuing processes. These relations depend only on the mean service time and mean source idle time and not on the form of the distribu- tions. When the source idle time follows a negative exponential distribu- tion, the service time distribution being arbitrary, the length of the waiting line at instants of completion of service is a Markov chain. Its distribution as well as the waiting-time distribution is obtained. The results are spe- cialized to the cases of negative exponential and constant service times.
Q UEUING analyses of stationary single-server systems having a finite number of sources (objects or individuals that initiate re-
quests for service) have been made under the assumption that the service- time distribution is negative exponential[3' 5, 6, 9, 10] or chi-square. 2' ASHCROFT[1] considered the general case of an arbitrary service-time dis- tribution and showed how to calculate the load factor of the server. This can be used to determine the mean delay and the proportion of calls de- layed. However, it remained to obtain the distributions of the waiting- line length and the waiting time. We do this by applying the method of the imbedded Markov chain used by D. G. KENDALL[7] in the case of an infinite number of sources. This method also applies to the case of a finite number of sources if the idle periods of the individual sources have a negative exponential distribution.
FINITE STATIONARY QUEUING PROCESSES
CONSIDER a service system consisting of N sources and M servers. To begin with we assume only that the serving process is stationary (see ref. 4, Chap. 2, Sec. 8 and ref. 8, Chap. 9, Sec. 30.3), and make no assump- tions regarding the service time or source idle time distributions. A source may be in any one of three states, namely, an idle state, a waiting state which occurs after it has requested service but the server is occupied with another source, and a service state when it is being served. These states follow one another in the order mentioned, with the understanding that a source occupies a waiting state for a zero time interval when it finds a server free. No defections from the waiting line are permitted.
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If a is the mean source idle time, d the mean delay and h the mean service time, we have
f=a+d+h, (1)
where f is the mean intercall time of a source. Then the mean interar- rival time of all calls is f/N and if the calls are apportioned equally among the servers the mean interarrival time of calls at one of the servers is Mf/N. A server alternates between an idle state and a busy state, and moreover, each such alternation corresponds to the arrival of a call that experiences no delay. Hence
Mf/N=p(O) (b+c), (2)
where p(O) is the proportion of nondelayed calls, and b and c are the mean server busy and idle times, respectively. Now each server busy period consists of some number of services, the first of which is a nondelayed service. It follows that
h=p(O) b. (3)
The three basic relations (1), (2), and (3) are a consequence of the law of large numbers applied to the stationary process under consideration. To avoid trivialities we assume the mean source idle time a and the mean service time h are finite and nonzero, O<a< oo, O<h< oo. If b= oo then p(O) =0 and equation (2) must be written
Mf/N = h. (4)
Eliminating f between (1) and (4), we find
d/h = (N/M)-1-(a/h). (5)
Henceforth we assume 0< b < oo, from which it follows that 0 <p(O)_ 1, O<c< co.
The load factor of a server, denoted L, is the proportion of time that the server is busy. Thus
L=b/(b+c). (6)
The following equations are obtained by combining (1), (2), (3), and (6) in various ways:
f=Nh/ML, (7)
d/h= (N/ML)-1-(a/h), (8)
L L=h/[h+cp(O)]. (9)
Notice that (5) is the special case of (8) when L =1.
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We will henceforth assume there is only one server, M = 1. We will also assume that the source idle time has a negative exponential distribu- tion dA (t), i.e., the probability that a source be idle for less than t is
A (t) = 0) ~(1?0) (0 (0_t)
Then the server idle time distribution dC(t) is given by
C(t) = 1-[1-A(t)]Nv, (I 1)
and c=a/N. (12)
Equations (8) and (9) can now be written in the form
d/h=N-1-(a/h)[l-p(O)], (13)
L=N/[N+ (a/h) p(0)]. (14)
This pair of equations express the mean delay d/h and the load factor L in terms of N, a/h, and p(0). Ordinarily, N and a/h are known. The proportion of calls not delayed, p(0), depends on the service time dis- tribution. In the next section we show how p(0) is determined.
By eliminating a/h between (13) and (14) we obtain the equation
d/h=N[l-(1/L-1 ) (1/p(O)- )]-1. (15)
THE DISTRIBUTION OF THE LENGTH OF THE WAITING LINE
THE LENGTH of the waiting line is the number of sources awaiting service as well as the source being served, if any. At instants of request for serv- ice or termination of service the source just arriving or departing is not included in the waiting line. The length of the waiting line at instants of departure, denoted X, is a finite Markov chain. This is a consequence of the important property of the negative exponential distribution that the conditional probability that a source, already idle for a time t', remain idle for less than an additional time t is A (t), and does not depend on t'. Thus all the sources that are idle at the commencement of a service period have the same probability A(t) of requesting service within a time t, re- gardless of how long they have been previously idle. Hence the length of the waiting line at an instant of departure depends only on the length of the waiting line at the previous departure. The device may be used of considering the idle time of a source made up of a number n of phases, each having a negative exponential distribution, which results in a source idle time distribution that is chi-square with 2 n degrees of freedom. Here again X is a Markov chain, but now account must be taken of the phases of the idle sources, so that X has nN states. We will consider here only the simplest case n= 1.
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The number of requests for service that arrive during a service period for which there are i sources on the waiting line at the commencement of service will be denoted U(i) and has the probability distribution
P{U(i)=j} = fb[j,N-i,A(t)] dH(t), i =
_02 'NN_ (16)
where dH(t) is the service time distribution and b[r,n,x] is the binomial probability that there be r successes in n trials with probability x for suc- cess on each trial. The notation Pt } denotes the probability of the event within the braces.
If a departing source leaves behind i >1 sources on the waiting line, the next service begins with N-i initially idle sources, while if a depart- ing source leaves no source waiting, the next service commences with N-1 initially idle sources. Thus if p(i,j) denotes the probability that the length of the waiting line at a departure be j, given that the length of the waiting line was i at the previous departure, then
p(i,j) =P{ U(i) =j-i+l ( i=I2; i i N-1) (17)
and p(O,j) =p(l,j). (j=O, 1, * * , N-l1) (18)
The transition probabilities (17) and (18) are all positive, and the re- maining transition probabilities are zero,
p(i,j)-O. =0 1 '
. N- (19)
Note that p(i,j) is the element in row i-1 and column j-1 of the transi- tion matrix. The finite Markov chain X is irreducible and aperiodic since each state of X is accessible from its neighboring states. Moreover, by the general theory of such finite Markov chains (ref. 5, p. 356) X has a unique stationary distribution
P{X=i} ==p(i), (i=0, 1, *Ia , N-1) (20)
which is the solution of the system of N homogeneous linear equations
The ith equation of (21) may be used to express p(i) in terms of p(i-1),
p(i-2), ***, p(O). Thus all the p(i) may be determined as multiples of p(O), and then p(O) determined from (22).
Let f(i,j) be the probability that at least j- i+ 1 requests for service are made during a service time for which the length of the waiting line is i at the commencement of service, that is
N-1i=I 2~1...ATN 1\
f(i)j) E p (ilk), (-,2*;N 1)(23)
N-i
or f(i,j)= E PtU(i)-k}. (24) k=j-i+l
Thus f(i,j) is the tail of the distribution of U(i), and
EZ'J f(ij) = E[U(i)], (25)
where E[ ] denotes the mean of the random variable in brackets. Now let us form a set of N-1 equations by adding the last N - i equations of (21 ) for i= 1, 2, ***, N-1. We obtain
The system of equations (26) may be used in place of (21) to solve for the distribution of X. Here the ith equation may be used to express p(i) in terms of p(i-1), * *, p(2), [p(O)+p(l)]. Thus all the p(i) may be determined as multiples of p(O) +p( 1), and then p(O) +p( 1) determined from (22).
The system of equations (26) has an additional significance. For the right hand side of the ith equation is the probability that an arriving source find a waiting line of length i, and its equality with p(i) states that the distribution of the length of the waiting line at instants of arrival is iden- tical with the distribution of the length of the waiting line at instants of departure. Interestingly, the former random variable is not in general a Markov chain, while the latter is. In fact, these two random variables, namely the length of the waiting line at instants of arrival and at instants of departure, have identical distributions under quite general circum-
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stances, namely whenever the length of the waiting line is a stationary stochastic process whose state values are 0, 1, 2, ***, having only one step transitions, and for which the zero state is persistent and nonnull. For then as time proceeds the number of increases of the waiting line (ar- rivals) will tend to be equal to the number of decreases of the waiting line (departures). Moreover, every change of state from i to i+1 is certain to be followed at some later time by a change of state i+ 1 to i. The iden- tity of the two distributions in question follows from the law of large num- bers.
If the equations of (26) are added, and note is taken of (25), we have
But, referring to (16) and the well-known properties of the binomial dis- tribution,
E[U(i)]= (N-i) f A dH. (i=1, 2, ** *, N-1) (28)
Substituting into (27) we have
I-p(0)={[p(0)+p(1)] (N-1)+p(2) (N-2)+. +p(N-1)} f A dH,
or 1-p(O) = {N-p(O) -E[X]} f A dH.
Hence E[X]-N-p(O)-[1-p(O)]/f A dH. (29)
The fraction in (29) is the ratio of the probability that a call be delayed to the probability that an initially idle source request service during a service period of another source. Both of these probabilities are small when the load factor is small, as is also E[XI, so that
1-p(O) (N-1) f A dH. (30)
On the other hand, when the load factor is high p(0) is small while f A dH is close to unity, and
E[X]zN-1, (31) as would be expected.
The method of generating functions may also be used to calculate E[X] as well as the higher moments of X. Let G(s) be the generating function of X and G(i,s) the generating function of U(i), that is
EACH CALL arrives during a service period whose initial state is a random variable Y, where Y O for nondelayed calls. Thus Y associates calls with the length of the waiting line at the beginning of the service period in which they arrive. The distribution of Y is given by
Pt Y=0} =P(O),
P{ Y= 1} = [p(O) +p(l)] E[U(1)], (35)
PI Y=i} =p(i) E[U(i)]. (i=2, 3, *, N-i)
For suppose that a total of k calls arrive during a time T for which there are mi service periods that have an initial state i, and suppose ul, u2, *.*,
umi are the number of calls that arrive in each of the mi service periods. Then allow T-* oo in the identity
Z=1j uJlk = (EZ1 uj/mi) (mi/k).
We have E11 uj/k--P{ Y =-i
57111 uj/mj->E[U(i)],
milk-1p(i). (i >2)
This is a consequence of the law of large numbers. The case i 1 requires the additional comment that when a departing source leaves a waiting line of length 0 or 1, the following service period commences in state 1. Thus the proportion of service periods that commence with a waiting line of length 1 is p(O)+p(1).
The proof given above for equations (35) applies to any stationary single-server process, and is not restricted to the case in which the source idle time follows a negative exponential distribution. We note that a verification that the sum of the probabilities (35) is unity when the source idle time follows a negative exponential distribution is contained in (27).
Let W(t) denote the probability that a call be delayed more than t>O, and let W(tli) denote the conditional probability that a call that arrives in a service period whose initial state is i be delayed more than t > O. Then
Let W(tli,j) denote the probability that a call that arrives in a service period whose initial state is i and finds the length of the waiting line to be j be delayed more than t>0. Then
Now consider one of the initially idle sources s together with the remaining N - i-1 idle sources at the commencement of a .service period which begins with a waiting line of length i. The probability that s request service a time u within du after j-i of the other N-i- 1 initially idle sources re- quest service is b[j-i,N-i-1,A(u)]dA(u). Hence,
where the integral in the numerator is extended over all nonnegative values of u, vi, * * *, vj for which vl>u and vl+v2+ - - +vj>u+t.
NEGATIVE-EXPONENTIAL AND CONSTANT SERVICE TIMES
ALTHOUGH the case of negative exponential service time may be more expeditiously treated by conventional methods, we will give an outline of the present method to verify that identical results are obtained. The service time distribution is
{o, (t ? 0) H(t) = { ( ? (40)
01 -e- t1h (o < t)
The transition probabilities are given by
p(i,j)=afJ b[j-i+1,N-i,A] (1-A ) - dA, (41 )
where a = a/h. This may be evaluated to obtain
p(i,j) =a(N-j,j-i+l)/(N-j-l+a,j-i+2), (42)
where we adopt the notation (x,0)=1 and (x,n)=x (x+1) * (x+n-1) for x any number and n a positive integer. Then it is easily verified that
On substituting into the jth equation of (26) it reduces to
p(j)-=[(N-j) /(N-j+at)][p(j- 1) +p(j)] or a p(j) = (N-j) p(j- 1). (j=1, 2, *,N-1) (45)
This recurrence relation immediately leads to
p(j) = (N-j,j) c-j p(O), (46)
where l/p(0)-=j= (N-jj) a&j, (47)
as was to be shown. It should be noted that the relation between p(j) and Pj, where Pj denotes the proportion of time that the system is in state j, is given. by p (j)=Pj+1/ (1-PO ) where j-=0, 12, *.*, N-1.
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As for the determination of the waiting time distribution, the integral in the numerator of (39) splits into the product of two factors. The first factor is the probability that the (j-i+l)th request for service come before the termination of service of the source being served, and is given by f b[j-i,N-i-1,A](1-A)adA which may be calculated to be f(i,j)/(N-i). The second factor is the probability that the sum of j negative exponentially distributed service periods of mean h be greater than t, and this is the complement of the chi-square distribution function, denoted K(t/h,2 j), whose value of chi-square is t/h and having 2 j degrees of freedom. Thus
W(tfi,j) =f(i,j) K(t/h,2j)/[(N-i) f A dH]. (48)
On substituting into (38) and (37), we find N-1 N-1 N-1
W (t) = [p (O) + p l )] , f(i,j) K (t/h,2 j) + 5, E p (i) f (i,j) K (t/h,2 j) . j=l i=2 j=i
If the order of summation is reversed and (26) applied, we finally obtain
W(t) = EN 1(J) K(t/h,2j). (49)
In the case of constant service time we have
0), (t<h)
H(t) = 1 1. (h t) (50)
Then p(i,j)-b[j-i+1,N-i,A] (51)
and f(i,j) =B[j-i+l,N-i,A], (52)
where B[r,n,x] = El-n b[k,n,x] (53)
and A =A (h). The simple iterative method for solving (21) or (26) for the distribution of X has already been indicated. As for the waiting time distribution, we have
'((1/A )fA b[j-i,N-i-1,A (u)] dA (u), [O?t<(j-1) h]
+ E p(i) {B[k-i+l,N-i,A(kh-t)]+6s(k-i+2,N-i,A) } (57)
+ Ei'=kN+1 p(i) (3(1,N-i,A),
while for k=N-1,
W(t)=[p(O)+p(1)] B[k,N-1,A(kh-t)]
+ t=2 p(i) B[k-i+l1,N-i,A(ht]
Thus W[(N-1) h]=O and W(t)=O for (N-i) h?t. It can also be veri- fied that W(O) =1-p(O) by comparing the W(O) as given by (57) with the result of summing the equations of (26).
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