Chapter 2: Batch service queues
2.2. The M/G(0,s)/1 queueing system
The M/G(0’5)/l queue (i.e. Markovian arrivals, general batch service times with a minimum batch size of zero and a maximum batch size of s, with one server) will be discussed in this Chapter, with consideration given to different service time distributions.
In many queueing systems, s (the maximum batch size) is pre-defined and fixed and the minimum batch size is zero. The minimum batch size is called the
quorum. This Chapter will only consider systems with a quorum of zero and a fixed maximum batch size of s. An assumption of this queueing system is that the server will begin service regardless of the number of customers waiting, including the case when there are no customers waiting.
Assume that customers arrive singly and at random at a mean rate X , and that batches of customers are served at a mean rate ju with maximum capacity s, then
_ Demand on system over long period of time T Capacity of system over long period of time T
The total demand on the system during a long time period T will be equal to the average rate at which the customers arrive (X) multiplied by the time period T.
The total capacity o f the system over a long period of time T will be equal to the average service ra te (/i), multiplied by the maximum number that could be served (5), multiplied by the time period T.
XT X
P = — = —
s/iT s / j
Hence, — = sp
M
To analyse this queue, the imbedded Markov chain technique is used and since we have random arrivals, it is appropriate to use the departure instants as regeneration points. However, since a new service starts immediately after the previous one has finished, the instants at which we link the probabilities need to be clearly defined.
(as indicated in Figure 2.1). However, it should be remembered that all three event instants at A and B in fact correspond to one time instant.
Batch service finishes Batch service finishes
Next batch service starts
Next batch service starts
A B
Figure 2.1: Regeneration points for the M/G(0,sVl system
Let p n be the probability that there are n customers in the system just after a batch service finished (e.g. at A or B). The interval AB represents the service time of a batch.
Let kj be the probability that j customers arrive during the batch service time
Through multiplication of appropriate powers of z, and summation, the following expression for G (z) is found:
AB.
The probabilities, p n are expressed as follows:
Define the probability generating functions, G (z) and K (z)to be
00
G( z ) = p„ + zpi + z 2p 2 +... = X z"pn
n=0
00
K ( z )
Z(zS- z><
G(Z)
[ " - * ( , ) ]G(*) = - i=0
K ( z ) -1
We note that the numerator contains s unknown constants,
p i9 where/ = 0,1,... (*?-!)
Rouche’s theorem can be applied to the denominator of the above expression for G (z ), since G (z) is an analytic function of z within and on the unit circle. We show that the denominator of G ( z ) has exactly s zeros within and on the unit circle.
Denote the s zeros of the denominator within and on the unit circle as
1, z,, z2, ..., zs_j (1 will always be a root of G (z)). For G (z) to remain analytic the numerator must also equal zero for these values of z. When each of the z( 's are individually substituted, s simultaneous equations are produced that can be solved for the unknowns, p 0,Pi,—>Ps-i •
Rouche’s theorem
If / (z) and g ( z ) are analytic within and on a closed contour, C, and if
on C, then / (z) and / (z) + g (z) have the same number of zeros within C.
—
j x = rcos0 y = rsmO
C : z = re10 =r(c os0 + isin0) where r = l + £ , S> 0 but small.
Choose / (z) = z s and g (z) = - K ( z ) . It is clear that both / and g are analytic within and on the unit circle. Thus the first condition of Rouche’s theorem is satisfied.
We now need to show that |/ (z)| > |g(z)| on C.
On C, |/ (z)| = |z s = \z\s = (l + <?)* = 1 + sS to first order in 5. = \k0+zkl + z 2k2 + ... = k0 + k, |z| + z21 + ... = fc0 + /t1(l + £) + £2(l + <?)2 +. = £(!+<?) By Taylor’s theorem: A :(i+<?)=.*:(i)+<y/r(i)+— £ '( i ) + ... K (1 + S) = •£(!) + SK' (1) to first order in 8.
k
£ '( l ) = —, since A^'(l) is the mean number of arrivals in a service time.
= s p So |g(z)| < 1 + spS onC
| / ( z ) l > k ( z )l
o n C
if \ + s S > \ + spS i.e. if p < 1So Rouche’s theorem is satisfied if p < 1. Hence / (z) has the same number of zeros as f ( z ) + g ( z ) within and on C.
Clearly z 5 has s zeros within and on the unit circle (all at z = 0) therefore z s -
K(z) has s zeros within and on the unit circle.
Previously, we had
G (z) = 1=0
K ( z ) -1
We noted that if G (z) was to remain analytic on the unit circle then the s zeros of the denominator 1, zp z2, ..., z5_,, must also be zeros of the numerator. If we write the numerator out in full, we have:
(z5 -l)p<> + (z 5 - z ) P i + (z s - z 2)p 2 + - + (z s - z s_i) p s-,
(z - l)(z - Zj)(z - z2)...(z - zsA) . However the numerator could also have a constant multiplier apart from its factors, i.e. numerator =
^ ( z - l ) ( z - z 1) ( z - z 2) ...( z - z J,1) where A is constant. Thus we may write G (z) in the form:
«(*) = ---
W ) ~ l
The problem has now been simplified - only the constant A needs to be found.
To find A, straightforward substitution of z = 1 is not possible since both numerator and denominator become zero. By using differentiation, the solution can be found.
K ( z ) -1
5 -1
G {z) = A {z- \ ) Y \ (z- zi )
i=l
Differentiating both sides with respect to z gives:
* ( * ) - i G '(z) + K ( z ) s z s- ' - z sK '( z )
[*w 2]
G( z ) = A (z - 1 ) ~ P + f K z - z <)1=1 Setting z = 1 gives G(1) = A 5 -1no-*.)
i=1Using AT(1) = 1 and G (l) = land K' (l ) = - = s p
s - s p - A
no-*.)
1=1 :.A = ^5 -1 Ano-*.)
Z=1 Hence K ( z ) - l /=1 f ^ £Z 5 \ l ~ z i jTo complete the picture for G(z), the following observations relating to K(z) may be made.
K(z) is the PGF of the kj probabilities that j arrivals occur during a service time. K(z) is related to the Laplace transform of the PDF of the service time
distribution.
* ( * ) = V » I A ') }
Substitution of the appropriate form of K(z) into G(z) gives an expression for
G(z) that can be expanded in ascending powers of z to pick out the probabilities p o , p i , - - However, the summary measures are usually required and the mean
number of customers in the system at A, B (i.e. immediately after a service time has been completed) is given as L+d = G'(l) . Therefore differentiation of G(z) is required.
Manipulation of Equation 2.2 gives:
G(z)
K ( z ) -1
z — z.
Differentiation gives: G'(z) K { z ) + K (z) V
I
d: 1 1 { \ \ z - z ; 1 — z,i JJSetting z = 1 in the above equation causes G '(z) to disappear. Thus differentiation a second time is required:
G’ (z) K ( z ) -1 +G(z ) - G ( z ) ( k ( z ) 2 [ j ( ^ ’( z ) z SI + a : ( z ) ( 5 - i ) z ^ 2 ) - ^ z ' ' a : ' ( z ) - z 5a : ' ( z ) ] ) ^ f i \ K { z ) s z s• ' - z sK ,( z ) \ l K ( z ) K ,(z)\ 2 [ K ( z ) s z ° - ' - z ‘K ' { z ) \ c K ( z )2 1 } /i \ ( / i \ ^ 2n ~ d n) _ . 2 _
Now, setting z = l gives:
([.(r(i)+(5-i))-,r(i)-r(i)]) ([,-r(i)]
2^(i)) | 2[,-r(i)]
i i + i ^ '
= - s ( l- p )
The final terms that are required are ^ 5 - and K ” ( l ) .
dz Firstly, dz 5 -1
n
/ = ! f w z - z t j-i i = Y — t r i - z ,i=1 1 - Z ,
Rearrangement then finally gives an expression for L+i:
s2p + K" (l) - s (sp + ( 5 - 1))
2 s ( \ - p ) (2.3)
As previously noted, the values of z which are zeros of the denominator of G(z) need to be found. The only zeros of interest are the ones which lie within the unit circle. Within the unit circle there should be s - 1 roots. If 5 is an even number,
one root will be real (less than 1) and the remaining roots will be complex conjugate pairs. If s is an odd number then all roots will be complex conjugate pairs. When considering a service time distribution from the Exponential family, and when considering small batch sizes, finding the zeros of the denominator of
G(z) is a fairly simple task, but if large batch sizes are considered or if the service
time distribution differs from Exponential, an iterative process is used to find the roots.
Equation 2.2 showed that:
Therefore, we need to find the roots of the equation:
— 1 = 0
K(z)
The computer software package, MAPLE, was used to solve (2.4) for various values of p and s (batch size).
Once the zeros were found, ILd could be easily found. A simulation model was built in Visual Basic (in Excel) to simulate the batch service queue system as a check. The summary measures b, Lq, W and Wq were calculated from the output. For the sake of comparison, and for completeness, the values L, Lq, W and Wq needed to be derived theoretically and compared with the simulation output. This derivation may be found in (Holland 1991):
L = L+d + s p
Lq =L+d- 1 - 1 + c s p