APPENDIX A: ELEMENTS OF QUEUEING THEORYAPPENDIX A: ELEMENTS OF QUEUEING THEORY In a packet radio network, packets/messages are forwarded from node to node through the network by entering

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APPENDIX A: ELEMENTS OF QUEUEING THEORY In a packet radio network, packets/messages are forwarded from node to node through the network by entering a buffer (queue) of a certain length in each node and waiting for their turn to be transmitted to the next node. Also, call access to a network with a given capacity is modeled as a process with certain distribution of call arrival times and certain distribution of call service times. A number of different classes of queueing models are used in this field to analyze queueing delays or server (or network access) blocking probabilities. In order to generalize the presentation the elements in the queue (packets, messages, calls, etc.) will be referred to as customers. The problems nowadays become even more challenging in wireless cellular networks with handoffs, multimedia traffic and differentiated QoS. The key element in this analysis is a proper model for these distributions. The very first results in this field are based on the most common stochastic queueing models which assume that message (call) interarrival and service times obey the exponential distribution or, equivalently, that the arrival rate and service rate follow a Poisson distribution. In other words, if the arrival rate is λ , the probability of having n arrivals in an interval t is given as

npn ( ) t = [(λt) / n!]exp( −λt) . Process with such exponential distribution also has Markovian or memoryless properties. In other words, the probability that a customer currently in service has t units of remaining service is independent of how long it has already been in service. The state of the queue is characterized by the state probability pn ( )t = Pr( n packets in the queue at timet ) . In more general terms this

is also referred to as population of size n at time t. Owing to Markovian assumption, instantaneous changes in the system state can only amount to an increase (birth) or decrease (death) of one hence the name birth-death process. Therefore, the probability of a birth occurring in a small interval of length ∆t , given the system is in state n, is assumed to be λn∆ + ∆t o( t) (n>0), and similarly a probability of a death in the same state is µn∆t o+ ∆( t) , where µn / λn is the

Advanced Wireless Networks: 4G Technologies Savo G. Glisic © 2006 John Wiley & Sons, Ltd

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service/arrival rate of the process at state n. As a starting point let us see what would be the probability that the system is in state n at time t + ∆t . This probability can be written as

( + ∆t) = p t − λ ∆t][1 − µ t] p ( )[ λ ∆t][ µ t]p t ( )[1 ∆ + t ∆ +n n n n n n n

pn+1( )[1 − λn+1∆t][µn+1 t] pn−1( )[ λn−1∆t][1 − µn−1 t] o(∆t); t ∆ + t ∆ + n ≥ 1 (C.1)

For n=0 we have p0 (t + ∆t) = p0 t − λ0 t] p1 t − λ1∆t][ µ1 t] o(( )[1 ∆ + ( )[1 ∆ + ∆t) (C.2)

By moving pn ( ) t from the right- to the left-hand side, dividing through by ∆t and taking the limit as ∆t → 0 we have dp

dtn ( )t

= −(λn + µn ) pn t + µn+1 pn+1( ) + λn−1 pn−1 t n ≥ 1( ) t ( ); (C.3)

dp0 ( )t = −λ p t( ) + µ p ( )t

dt 0 0 1 1

The stationary solution is obtained for dpn ( ) / dt = 0 resulting in t 0 = −(λn + µn ) pn + µn+1 pn+1 + λn−1 pn−1; n ≥ 1

(C.4)0 = −λ p + µ p0 0 1 1

Equation (C.4) can be rewritten as

pn+1 =λn + µn pn −

λn−1 pn−1;n ≥1 µn+1 µn+1 (C.5)

λ0p1 = p0µ1

In a special case when λn = λ and µn = µ we have 0 = −(λ + µ) pn + µ pn+1 + λ pn−1;n ≥ 1

(C.6)0 = −λ p + µ p0 0 1 1

and

pn+1 =λ + µ pn −

λ pn−1;n ≥ 1 µ µ

(C.7)λp1 = p0µ

2

In order to discuss the elements of traffic modeling for advanced wireless communication systems, we will first review the basic results of conventional queueing theory beginning with the classical Poisson input (Markovian-M), exponential-service (M), single-server (1) M/M/1 queue. C1 The M/M/1 model The density functions for the interarrival times and service times for the M/M/1 queue are given, respectively, as a t( ) = λe−λt

(C.8)b t( ) = µe−µt

where 1/ λ is the mean interarrival time and 1/ µ the mean service time. Interarrival times, as well as service times, are assumed to be statistically independent. Since interarrival and service times are exponential, and arrival and conditional service rates Poisson, we have

Pr{an arrival occurs in an infinitesimal interval of length ∆t } = λ∆t +ο(∆t) ; Pr{more than one arrival occurs in ∆t } = ο(∆t)Pr{a service completion in ∆t , given system is not empty} = µ∆t +ο(∆t); and Pr{more than one service completion in ∆t , given more than one in system} = ο (∆t) .

resulting in a birth-death process. There are a number of ways to solve the birth-death equations like the iterative method, by using generating function or using delay operator. We will not deal with these methods explicitly but rather explain them through different examples below. By using Equation (C.7) iteratively we get

n

pn = p0 ∏⎛⎜⎜

λ ⎞⎟⎟ = p0 (λ / µ )n (n ≥ 1) (C.9)

j=1 ⎝ µ ⎠where p0 is obtained from the fact that probabilities must sum to 1, resulting in

1 = ∑x

⎜⎜⎛ λ

⎟⎟⎞

n

p0 (C.10) n=0 ⎝ µ ⎠

3

At this stage we define traffic intensity or utilization rate ρ as the ratio λ / µ , for single-server queues and get from Equation (C.10)

1 p0 = ∝∑n=0 ρ n

∑∝

n=0 ρ n is the geometric series 1+ ρ + ρ 2 + ρ 3 + ... that converges to

1/(1 − ρ) for ρ < 1 so that we have npn = (1− ρ )ρ for ( ρ = λ µ/ < 1) (C.11)

C2 Measures of effectiveness The steady-state probability distribution for the system size allows us to calculate the system's measures of effectiveness. Two of immediate interest are the expected number in the system and the expected number in the queue at steady state. To derive these, let N represent the random variable ‘number of customers in the system in steady state’ and L represent its expected value. We can then write

∞ ∞nL E N = [ ] = ∑np n = (1− ρ )∑ np (C.12)

n=0 n=0

Consider the summation ∞ ∞

∑ npn = ρ + 2ρ 2 + 3ρ 3 + ... = ρ(1+ 2ρ + 3ρ 2 +L) = ρ∑ nρ n−1 . n=0 n=1

Since ∑∞ nρ n−1 is the derivative of ∑∞ ρ n with respect to ρ and n=1 n=0

∞n∑ ρ =1/(1 − ρ)

n=0

we have ∞

n 1∑ nρ − =d [1/(1 − ρ)]

=1/(1 − ρ)2

n=1 d ρand

ρ (1− ρ ) ρ λL = = = (C.13) − ρ µ λ(1− ρ )2 1 −

4

If the random variable ‘number in queue in steady state’ is denoted by Nq and its expected value by Lq, then we have

Lq = ∑∞

(n −1)p = ∑∞

npn − ∑∞

pn = L − (1− p0 ) =ρ

− ρn=1

nn=1 n=1 1− ρ

Lq = L − (1− p0 ) holds for all single-channel, one-at-a-time service queues, since no assumptions were made in the derivation as to the input and service distributions. Thus the mean queue length is

ρ 2 λ 2

Lq =1− ρ

=µ µ λ( − )

(C.14)

We are also interested in the expected queue size of nonempty queues, which we denote by L'q; that is, we wish to ignore the cases where the queue is empty. We can write

∞ ∞' ' 'Lq = E ⎣⎡Nq | Nq ≠ 0⎦⎤ = ∑ (n −1) pn = ∑ (n −1) pn

n=1 n=2'where p n is the conditional probability distribution of n in the system

' given the queue is not empty, or pn = Pr{n in system | n ≥ 2}. From the laws of conditional probability,

Pr{n in system and n ≥ 2} pn

' =Pr{n ≥ 2}

= ∑∞

pn

p (n ≥ 2)

n=2 n

pn pn= = 21− ( − ρ) ( ρ ρ1 − 1− ) ρ' The probability distribution {p } is the distribution {p }normalized n n

by omitting the cases n = 0 and 1. Thus ∞

Lq ' = ∑ (n −1) pn

2 =L − p1 − (1−

2

p0 − p1 ). n=2 ρ ρ

Hence

Lq' =

1 =

µ . (C.15) 1− ρ µ − λ

5

The expected steady-state system waiting time W and line delay Wq

can be found easily from L and Lq by using Little's formulas, L = λW and Lq = λWq . In the case of the M/M/1 queue, we have

L ρ 1W = = = (C.16) λ λ(1− ρ) µ − λ

and

Wq =L λ

q =λ(1

ρ−

2

ρ) =µ

ρ− λ

. (C.17)

C3 Waiting-time distributions We now consider the random variable ‘time spent waiting in the queue’ Tq and its cumulative probability distribution Wq (t) . The first-come, first-served (FCFS) queuing discipline is assumed. Waiting time is, for the most part, a continuous random variable, except that there is a nonzero probability that the delay will be zero, that is, a customer entering service immediately upon arrival. Hence we have Wq ( ) = Pr{ q ≤ 0}= Pr{Tq = 0}= Pr{system empty at an arrival} = q0 .0 T Let the conditional probability of n in the system given that an arrival is about to occur be qn . These probabilities are not always the same as the pn with which we have been working, since the pn are unconditional probabilities of n in the system at an arbitrary point in time. To find the distribution of virtual waiting time (i.e. the time a fictitious customer would have to wait were it to arrive at an arbitrary point in time), we would use pn. However, for Poisson input, qn = pn , as we will show later when deriving qn for the truncated M/M/c/K case where qn ≠ pn . Thus Wq (0) = p0 = 1− ρ . It then remains to find Wq (t) for t > 0. Consider Wq (t) , the probability of a customer waiting a time less than or equal to t for service. If there are n units in the system upon arrival, then in order for the customer to go into service at a time between 0 and t, all n units must have been served by time t. Since the service distribution is memoryless, the distribution of the time required for n completions is independent of

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the time of the current arrival and is the convolution of n exponential random variables, which is an Erlang type n distribution. In addition, since the input is Poisson, the arrival points are uniformly spaced and hence the probability that an arrival finds n in the system is identical to the stationary distribution of system size. Therefore we may write W t( ) = Pr{T ≤ t} =q q

∞

W 0 + Pr n completions in ≤ t |arrival found n in system pq ( ) ∑ { } n n=1

∞ t n−1

= 1− ρ + (1− ρ)∑ ρ n ∫µ(µx) e−µxdx

n=1 0 (n −1)!

t ∞ n−1 t

∫0

−µx

n=1

(µ(n

xρ−1

))! ∫

0

−µx(1−ρ= 1− ρ + (1− ρ) µe ∑ dx = 1− ρ + ρ(1− ρ ) µe )dx

= 1− ρe−µ (1−ρ )t (t>0). (C.18).

Similarly, we can get the probability distribution of the total time (including service) that a customer has to spend in an M/M/1 system. Denote this random variable by T, its CDF by W(t), its density by w(t), and its expected value by W. We have already derived W, and it can further be shown that

W ( )t = 1− e(µ −λ )t (t > 0)w( ) (t = µ − λ)e−(µ −λ )t (t > 0) (C.19)

The derivation of Equation (C.19) very much follows that of the line delay distribution except that n + 1 service completions are required in time ≤ t.

C4 Queues With Parallel Channels (M/M/c) Next we consider the case in which there are c servers. In this multiserver M/M/c model each server has an independently and identically distributed exponential service-time distribution, with the arrival process again assumed to be Poisson. Since the input is Poisson

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and service exponential, we have a birth-death process. Hence λn = λ for all n, and it remains to determine µn prior to being able to use the previous results. If there are more than c customers in the system, all c servers must be busy with each putting out at a mean rate µ , and the mean system output rate is thus equal to cµ . When there are fewer than c customers in the system, say n < c, only n of the c servers are busy and the system is putting out at a mean rate of µn . Hence µn may be written as

⎧nµ (1 ≤ n < c),µ = ⎨cµ ( )c .

(C.20) n ⎩ n ≥

Utilizing Equation (C.20) in Equation (C.7) and the fact that λn = λfor all n, we obtain

⎧ λ n

1 n c pn =

⎪⎨⎪n!µ n

n

p0 ( ≤ < ) (C.21)

⎪−

λ p ( )≥⎪ n c n 0 n c ⎩c c!µ

In order to find p0, we again use the condition that probabilities must sum to 1, which gives

−1 c−1 n ∞ n

p0 = ⎜⎛⎜∑ n!

λµ n + ∑ cn−c

λc!µ n

⎟⎞⎟

⎝ n=0 n=c ⎠If we now use notation r = λ / µ , and ρ = r / c = λ / cµ. we have

c−1 n ∞ n

p = ⎜ + ⎟0 ⎛⎜∑

rn! ∑ cn

r −cc!

⎞⎟

−1

⎝ n=0 n=c ⎠Now consider the infinite series in the above equation

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∞ n c ∞ n c− c ∞ m r rr r ⎛ ⎞ r ⎛ ⎞∑ n c = ⎜ ⎟ = ∑− ∑ ⎜ ⎟n c c c! c! n c c c! m=0 c= = ⎝ ⎠ ⎝ ⎠

rc 1 ( / = <ρ 1 . )= r c c! 1 − r c /

which results in ⎛ c−1 r n r c ⎞

−1

p0 = ⎜⎜∑ + ⎟⎟ (r / c = ρ < 1) (C.22)⎝ n=0 ! c ( )− ρ ⎠n ! 1

To consider measure of effectiveness we first consider the expected queue size Lq , as it is computationally easier to determine than L, since we have only to deal with pn for n ≥ c . Thus

∞ ∞ n

q ∑ ( ) ( ) n cL = n c p n = ∑ −r − p0− n c

n c 1 n c= +1 c c!= +

c c

=r p0 ∑

∞

mρ m =r ρ p0 ∑

∞

mρ m−1 ( C.23)c! m=1 c! m=1

c ∞ c cρ 0 m r pρ 0 d ⎛ 1 ⎞ r pρ 0=r p d ρ = 1 .∑ − =

c! d ρ m=1 c! dρ ⎝⎜1− ρ ⎠

⎟c! 1( )− ρ 2

To find L now, we employ Little's formula to get Wq , then use Wq to find W = Wq +1/ µ, and finally employ Little's formula again to calculate L = λW . Thus we get

Wq =Lq ⎜

⎛ rc

2

⎞0 ( ).24= ⎟ p C

⎜ c c( )( )1 ρ ⎟λ ⎝ ! µ − ⎠

1 ⎛ rc ⎞W = + ⎜ 2 ⎟ p0 ( )C.25

⎜ c c( )( )1 ρ ⎟µ ⎝ ! µ − ⎠and

⎛ rcρ ⎞L r= + ⎜ ⎟ p (C.26)

⎝ ( )2

⎠0⎜ c! 1− ρ ⎟

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Letting Tq represent the random variable ‘time spent waiting in queue’ and Wq (t) its CDF, we have

c−1 c−1 n

Wq ( ) = Pr{Tq = 0} = Pr{ c 1 in system = ∑ pn = p0 ∑ r0 ≤ − }n=0 n=0 n!

nNow to evaluate ∑ r / n!, recall that it appears in the expression for p0 as given in Equation (C.22), so that

c−1 rn 1 rc

∑ n! =

p0

−c ( − ρ )n=0 ! 1

thus giving c c

Wq ( )0 = p0 ⎝⎜⎜⎛

p 1

0

−(r

) ⎠

⎞1

c (r p

− ρ ) (C.27)⎟⎟ = − 0

c! 1− ρ ! 1 ForTq > 0 and assuming FCFS,

∞

q q q 0 ∑ n − c +1W ( )t = Pr{T ≤ t} = W ( )+ Pr{ n=c

completions in ≤ t | arrival found n in system} pn .

Now when n ≥ c, the system output is Poisson with mean rate cµ , so that the time between successive completions is exponential with mean 1/( )cµ , and the distribution of the time for the n-c+1 completions is Erlang type n - c + 1. Thus we can write

−

W t( ) = W ( ) + p ∞

n c

n

∫t c c x( )n c

e−c xdxq q 0 0 ∑ r −

µ µ µ

n c c c! (n c− )!= 0

Wq ( ) +r p0 ∫

t

µ − µ ∑∞ ( rx)n c−

dx= 0 c

e c x µ

0 n c(c −1 !) = (n c− )! c t c

= Wq (r p0

) ∫0

−µx c r( − ) = q 0 (r p0

) ( −(cµ λ− )t )( )0 + µe dx W ( ) + 1− e c −1 ! c! 1− ρ

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Putting this result together with (C.27), we have

Wq ( )t = 1−c!

r(1

c

−p0

ρ)e−(cµ −λ )t . (C.28)

From Equation (C.28), c

Pr{ Tq > = −t} 1 W t ( ) =c! 1(

r p −

0

ρ )e−(cµ−λ )t

q

so that the conditional probability Pr{Tq > t | Tq > 0}= e−(cµ −λ )t . To find the formula for the CDF of the system waiting time, we first split the situation into two separate possibilities, namely, those customers having no line wait [probability Wq (0)] and those whose system wait is a line delay plus a service time [probability 1 - Wq (0)]. The first of these two classes of customers has a CDF which is identical to the exponential service-time distribution, with mean1/ µ ; the second has a CDF found as the convolution of the service-time distribution with a second exponential distribution having mean 1/(cµ − λ) , the latter representing the CDF of the line waiting time, given that Tq > 0 (see earlier in this section). This convolution can be also written as the difference of the two exponential functions

−Pr ≤ = ({T t}c (

c 1 (−1−

ρρ) −

)1(1− e−µt ) −

c (1−1 ρ ) −1

1− e−(cµ λ)t )

Thus the overall CDF of M/M/c system waits may be written as ( ) = W (0 1) ⎡ − e−µt ⎤ + −⎡ q 0 ⎤W t q ⎣ ⎦ ⎣1 W ( )⎦

c 1 ρ − ⎞×

⎛⎜

( −

)) (1− e−µt ) −

(1 (1− e−(cµ λ )t )⎟ ( C.29)⎜ c 1− ρ −1 c 1− ρ −1 ⎟

⎝ ( ) ⎠c ( − ρ ) −W 0 − 1−Wq ( )0 −(cµ λ )tq µt −=

1 c (1− ρ ) −1

( ) (1− e ) −c (1− ρ ) −1(1− e )

C5 M/M/c/K queue

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We now consider an M/M/c/K queue, in which there is a limit K placed on the number allowed in the system at any time. The approach here is identical to that of the infinite-capacity M/M/c except that the arrival rate λn , must now be 0 whenever n ≥ K . Equation (C.21) now becomes

⎧ λ n

1 n c ⎪n!µ n p0 ( ≤ < )pn = ⎨

⎪ (C.30)⎪ λ n

(c n K )⎪ n c n p0 ≤ ≤−⎩c c!µ

The usual boundary condition that the probabilities must sum to 1 will yield p0 . Again, the computation is nearly identical to that for the M/M/c, except that now both series in the computation are finite and thus there will to be no requirement that the traffic intensity ρ be less than 1. So

⎛ c−1 λ n K λ n ⎞−1

p0 = ⎜∑ n + ∑ n c − n ⎟⎝ n=0 n!µ n c c c !µ ⎠=

To simplify, consider the second summation above, with r = λ / µ and ρ = r / c :

− +⎧rc 1− ρ K c 1

(ρ ≠ 1)K rn

=rc K

ρ n c = ⎨⎪⎪ c! 1 − ρ∑ n c− ∑ −

n c= c c! c! n c= ⎪rc

( − + ) (ρ =1)⎪ K c 1 ⎩ c!

resulting in

⎧ c− − +⎛ 1 rn rc 1− ρ K c 1 ⎞−1

⎪⎜∑ + ⎟ (ρ ≠ 1)⎪⎝ n=0 n! c! 1− ρ ⎠

p0 = ⎨⎪ (C.31)⎪ −1 ⎪⎛ c−1 rn rc

( − + )⎞ρ = 1)⎪⎜∑ + K c 1 ⎟ (

⎩⎝ n=0 n! c! ⎠To find the expected queue length (ρ ≠ 1) we start with

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K c Kp r −L = (n − c) pn = −0 ∑ (n − c) rn c

q ∑ n c n c 1 c c! n c= +1= +

p rc ρ K p rc ρ K −c 0 ∑ ( − ) ρ n c− −1 = 0 ∑ iρ i−1= n cc! n c= +1 c! i=1

K cp rc ρ d ⎛1− ρ − −1 ⎞= 0

⎜ ⎟c! d ρ ⎝ 1− ρ ⎠or

qc! 1(

0

−

c

ρρ

)2 ⎣K c 1 − −1 − + ) ρ K − ⎤⎦L =

p r ⎡1− ρ − + ( ρ )(K c 1 c (C.32)

For ρ = 1, it is necessary to employ L'Hopital's rule (differentiation) twice. To obtain the expected system size, we use, like in the unrestricted M/M/c model, L = qL + r. However, for the finite-waiting-space case, we need to adjust this result (and Little's formula, as well), since a fraction pk of the arrivals do not join the system, because they have come when there is no waiting space left. Thus the actual rate of arrivals to join the system must be adjusted accordingly. Since Poisson arrivals see time averages, it follows that the effective arrival rate seen by the servers is λ(1− pK ). We henceforth denote any such adjusted input rate as λeff . The relationship between L and Lq must therefore be reframed for this model to be L = L +q

λeff / µ = Lq + λ(1− pK )/ µ = Lq + r(1− pK ).We know that the quantity r(1− pK ) must be less than c, since the average number of customers in service must be less than the total number of available servers. This suggests the definition of a parameter ρeff = λeff / cµ, which would thus have to be less than 1 for any M/M/c model even though no such restriction exists on the value of ρ = λ / cµ . Expected values for waiting times can now be obtained by Little's formula as

13

L LW = =λeff λ (1− pK )

1 Lq

(C.33) W = W − =q µ λeff

For M/ M/1/K, all of the above measures of effectiveness reduce to considerably simpler expressions, with key results of

⎧ 1− ρ (ρ ≠ 1)K +1⎪⎪1− ρp0 = ⎨ (C.34) ⎪ 1 (ρ =1)⎪⎩ K +1

n⎧(1− ρ ρ)pn =

⎪⎪⎨

1− ρ K +1 (ρ ≠ 1)( C.35)

⎪ 1 (ρ =1)⎪⎩ K +1 and

⎧ ρ ρ (K ρ K +1)⎪ − k +1 (ρ ≠ 1)⎪1− ρ 1− ρLq = ⎨ ( C.36)

⎪( − ) ( )⎪ K K 1

ρ =1 ⎩ 2(K +1)

with L = L + (1− p0 ) . The derivation of the waiting-time CDF is now q

complicated, since the series are finite, although they can be expressed in terms of cumulative Poisson sums, as we shall show. Also, it is now necessary to derive the arrival-point probabilities { qn }, since the input is no longer Poisson because of the size truncation at K, and qn ≠ pn.

We use Bayes' theorem to determine the qn , so that

qn = Pr{n in system | arrival about to occur }

14

Pr{arrivalabout tooccur n in system }⋅ pn =∑k Pr{arrivalabout tooccur n in system }⋅ pnn=0

⎪ ⎡λ t ο ( ) pn ⎪ ⎡λ ο+ ( ) / ∆t⎤ p ⎪= lim ⎧⎨ K

⎣−1

∆ + ∆t ⎦⎤ ⎫⎬ = lim

⎧⎪⎨ K

⎣−1

∆t ⎦ n ⎫⎬

∆ →0 ⎪⎩∑n=0 ⎣λ t ο ( )⎦ pn ⎪⎭∆ →t 0 ⎪⎩∑n=0 ⎣λ ο+ ( ) / ∆t⎤⎦ n ⎪⎭

t ⎡ ∆ + ∆t ⎤ ⎡ ∆t p

= ≤=λ∑

λ

n

K

p −n 1 p 1−

pn

pK

(n K −1)=0 n

To get the CDF Wq (t) we note that

( ) = Pr{ ≤ t = W ( )Wq t Tq } q 0 +K −1

∑ Pr{n-c+1 completions in ≤t| arrival found n in system }· qn n=c

since there cannot be arrivals joining the system whenever they encounter K customers. It follows that

−

W t( ) = W ( ) +K −1

qt µ µ( x)n c

e−c xµ dx q q 0 ∑ n ∫c c

n c=0 ( − )!n c

−K −1 ⎛ ∞ µ µ( x)n c −c x

⎞c c µ= Wq ( )0 + ∑=

qn

⎝⎜⎜1− ∫t ( − )!

e dx ⎠⎟

n c n c ⎟

by using

m

∫t

( )x m

e dx = ∑ ( )λ i −λt∞ λ λ −λx t e

m! i=0 i! and m = n - c and λ = cµ gives

∞ cµ(cµx)n−c n−c (cµx)i e−cµt

∫t (n − c)! e−cµxdx = ∑ i!i=0

and hence

15

t q 0 K −1

q −K −1

qn

n−c (cµt)i e−cµt

Wq ( ) = W ( )+ ∑ n ∑ ∑n=c n=c i=0 i!

= 1− ∑ qn ∑ ( )cµt e ( C.37)

K −1 n−c i −cµt

n=c i=0 i!

C6 Erlang's Formula (M/M/c/c) The special case of the truncated queue M/M/c/K for which K = c, that is, where no line is allowed to form, results in a stationary distribution which is known as Erlang's first formula and can be readily obtained from Equations (C.30) and (C.31) with K =c

/(λ µ )n

pn =

∑c (

n!/ ) (0 n c) (C.38 )≤ ≤

λ µ

i=0 i! The resultant formula for pc is itself called Erlang's loss formula and corresponds to the probability of a full system at any time in the steady state, namely,

r c pc = c

c / ! (r = λ / µ ) (C.39)i∑ i=0

r i/ ! Since the input to the M/M/c/c is Poisson, the probability that an arrival is lost is equal to the probability that all channels are busy.

C7 Queues with unlimited service (M/M/ ∞ ) If an infinite number of servers is available we have a model for self-service situation. Using the general birth-death results with λn = λ and µn = nµ , for all n, yields

−1

pn =rn

p0 , p0 = ⎜⎛∑

∞ rn

⎟⎞

n! ⎝ n=0 n! ⎠The infinite series in the expression for p0 is identical to the representation of er resulting in

16

n −r

pn =r e ( )≥ 0n (C.40)

n! so that the steady-state probability distribution of n in the system is Poisson with parameter r = λ µ/ .

C8 Finite source queues In this section we assume that the calling population is finite, say of size M, and future event occurrence probabilities are functions of system state. A typical application would be analysis of a network with relatively small number of terminals. We assume c servers (e.g. shared wireless channel capacity is c) are available, that the service times are identical exponential random variables with mean 1/ µ , and that the arrival process is described as follows. If a calling unit is not in the system at time t, the probability it will have entered by time t + ∆t is λ∆t o t ( )+ ∆ ; that is, the time a calling unit spends outside the system is exponential with mean 1/ λ . Because of these assumptions, we can use the birth-death theory developed previously, with the modified birth and death rates given by

λn =⎩⎨⎧0(M − n)λ

(n (0

≥≤Mn

)< M )

and

⎪nµ (0 ≤ < )⎧ n c µn = ⎨

⎪c ( )≥⎩ µ n c Using Equation (C.5) with r= λ / µ yields

⎧ !( − )! n⎪

M M n r p (1 ≤ < )n c

⎪ n! 0

pn = ⎨M M n ⎪ !( − )! n

⎪ − (c n M )r p ≤ ≤n c 0⎩ c c! or equivalently,

17

M n⎧( ) r p0 ( ≤ < c)1 n p = ⎪ n

(C.41)n ⎨M n! n⎪( ) r p (c ≤ ≤n M )n − 0n c⎩ c c!

The algebraic form of the { pn } does not allow the closed-form calculation of p0 , L , L, W and W . Instead, we must calculate each q q

of the coefficients multiplying p0 in Equation (C.41) (call them { an , n = 1.2,3,..., M}) and then complete the computation as

(1 a1 a a3 ... )−1 p = + + + + + a0 2 M

To find the average number of customers in the system (in our example the number of terminals with backlogged packet waiting for transmission), we get

M M

L = ∑np n = p0 ∑na n n=1 n=1

However, to obtain Lq and W and Wq, we must first find the effective mean rate of arrivals into the system. The mean arrival rate when the system is in state n is (M - n) λ and we get

M −1

λeff = ∑ (M − n)λ pn = λ (M − L)n=0

From our earlier work we have

Lq = −λeff L r (M − L)L = − . (C.42) µ

and from Little's formula

W =λ (M

L − L)

and Wq = λ (MLq

− L)(C.43)

For the single-server version of this problem, Equation (C.41) reduces to

M np = ( )n r 0 (0 ≤ ≤ Mn ! p n )n

and the rest of the analysis is identical.

18

C9 Network with dynamic spectra sharing As mentioned in Chapter 1, 4G networks will allow for spectra (resource) sharing between different systems (operators). Let us assume that up to Y additional active terminals from another system can be temporally permitted to contend for c channels in the system. The sharing model is defined by modifying λn as

⎧Mλ (0 ≤ n < Υ)λn =

⎪⎨(M − n + Υ)λ (Υ ≤ n < Υ + M )⎪⎩0 (n ≥ Υ + M )

For c available channels, we have

⎧nµ (0 ≤ n < c)µn = ⎨

⎩cµ (n ≥ c)

For c ≤ Y Equation (C.5) with r = λ / µ gives

⎧M n

⎪ rn p0

⎪ n! (0 ≤ n < c)⎪ M n

npn = ⎨ n−c r p0 (c ≤ n < Υ) (C.44) ⎪c c! (Υ ≤ n ≤ Υ + M )⎪ M ΥM! n⎪ n−c r p0⎩(M − n + Υ)!c c!

If Y is very large, we essentially have an infinite calling population with mean arrival rate M λ . Letting Y go to infinity in Equation (C.44) yields the M / M / c / ∞ results of Equation (C.21) with M λ for λ . When c > Y, we have

19

⎧M n⎪n

r p0 (0 ≤ n ≤ Υ)

⎪ n! npn =

⎪⎨⎪⎪

(M −M

n

Υ

+M

Υ!

)!n! r p0 (Υ +1 ≤ n < c) (C.45)

⎪ M ΥM! n⎪ n−c r p0 (c ≤ n ≤ Υ + M )⎪(M − n + Υ)!c c!⎩

The empty-system probability, p0, can be found as previously by once more using the fact that the probabilities must sum to 1, so that the computation of p0 is made up of finite sums. The same is true for L and Lq . To obtain results for W and Wq we must again obtain the effective mean arrival rate λeff . To obtain λeff , one can use Equation (C.42) directly or obtain it using logic similar to that used previously namely,

Υ−1 Υ+M ⎛ Υ+M ⎞λeff = ∑ M λ pn + ∑ (M − n + Υ)λ pn = λ ⎜ M − ∑ (n − Υ) pn ⎟n=0 n=Υ ⎝ n=Υ ⎠

(C.46)

C10 Waiting time distribution for finite source queues Before we can obtain the CDF of the line wait, we have to

relate the general-time probability pn to the probability qn that an arrival finds n in the system. For the general finite-source queue, the two probabilities are related as qn = (M − n p n /) k , where k is an appropriate normalizing constant determined from summing the {qn} to 1. To prove this, we again use Bayes’ theorem as in the M/M/c/K situation.

Pr{n in system | arrival is about to occur}

20

Pr{n in system}Pr{arrival is about to occur | n in system}=

∑n(Pr{n in system}Pr{arrival is about to occur | n in system})

pn[(M − n)λ∆t + o(∆t)]= lim

∆t →0 ∑ pn[(M − n)λ∆t + o(∆t)]n

(M − n) pn (M − n) pn= =∑ (M − n) pn M − L

n

For the spectra sharing system , qn(M) can be shown to be

⎧ Mpn +⎪ M − ∑ y M (n Y p − ) n= (0 ≤ ≤ −n Y 1) ⎪ n yqn = ⎨

⎪ M n Y p n (( − + ) Y n Y M ≤ ≤ + −1) +⎪ M −

y M (n Y p − )⎩ ∑n y n=

Now we can write W t ≤ =( ) = Pr {T t} W (0) +q q q

Y M 1+ −

∑ [Pr{ n c 1 completions in ≤ t | arrival found n in system} ⋅ q ]− + nn c=

+ −1 n c−Y M µ µ −c x= Wq (0) + ∑ qn ∫t c c x ( ) e µ dx

n c 0 (n c− )!=

+ −1 n c−Y M ⎡ ∞ c c x c x= Wq (0) + ∑ qn ⎢1− ∫t

µ µ( ) e− µ dx ⎤⎥

n c= ⎣ (n c)! ⎦−+ −1 n c iY M − µ −c t= −1 ∑ qn∑ (c t) e µ

n c i=0 i= ! (C.47)

C11 Reconfigurable systems and state-dependent service As a response to an increased congestion in the system an

advanced wireless communications system may adapt the required service time either by making the packets shorter at the MAC layer or by changing the source coding rate on application layer.

21

The first model we consider is one in which a single server has two mean rates, say slow and fast. Work is performed at the slow rate until there are k in the system, at which point there is a switch to the fast rate. We still assume the service times are Markovian, but the mean rate µn now explicitly depends on the system state n. Furthermore, no limit on the number in the system is imposed. Thus µn is given as

≤ < µn = ⎨

⎧µ1 (1 n k) (C.48)

⎩ (n k)µ ≥

Assuming the arrival process is Poisson with parameter λ and utilizing Equation (C.5) we have

≤ < pn = ⎨

⎧ k −1

ρ1 n

n

p −k

0 +1

(0 ≥ n k)

(C.49) ⎩ρ ρ1 p0 (n k)

where ρ1 = λ / µ1 and ρ = λ / µ < 1. Because the probabilities must sum to 1, we have

k −1 ∞ −1

0 ⎜⎛∑ ∑ 1

n 1 k −1 n−k +1

⎟⎞ p = ρ + ρ ρ

⎝ n=0 n=k ⎠

and

k k −1 −1⎧ ⎛1− ρ ρρ ⎞⎪ 1 + 1 (ρ ≠ 1, ρ <1) ⎪

⎜⎝ 1− ρ1 1− ρ ⎟

⎠ 1

p0 = ⎨ −1 (C.50)

⎪⎛ ρ ⎞ ⎪⎜ k + ⎟ (ρ1 =1, ρ <1) ⎩⎝ 1+ ρ ⎠

To find the expected system size, we assume ρ1 ≠ 1

22

∞ k −1 ∞

∑ n = 0 ⎜⎛∑ ∑ 1

n 1 k −1 n−k +1

⎟⎞L = np p nρ + nρ ρ

n=0 ⎝ n=0 n=k ⎠ ⎡ k −1 ρ

k −2 ∞ ⎤⎛ ⎞ = p ⎢ρ nρ n−1 + ρ 1 nρ n−1 ⎥0 1 ∑ 1 1 ⎜ ⎟ ∑ρ⎢ n=0 ⎝ ⎠ n=k ⎥⎣ ⎦

k −1 ⎛ ⎞ ∞

= p0

⎡⎢ρ1

d ∑ ρ ρ 1 n + 1

ρ1 k −2

d ∑ ρ n ⎤⎥⎜ ⎟ dρ ρ dρ⎢ 1 n=0 ⎝ ⎠ n=k ⎥⎣ ⎦

⎡ k k −2 k ⎤d ⎛1− ρ ⎞ ⎛ ⎞ d ⎛ 1 1 ρ ⎞ρ − = p0 ⎢ρ1 ⎜ ⎟ + ρ1 ⎜ ⎟ ⎜ − ⎥

dρ 1− ρ 1

ρ 1

dρ 1− ρ 1− ρ ⎟⎢⎣ 1 ⎝ 1 ⎠ ⎝ ⎠ ⎝ ⎠⎥⎦

This results in

ρ ⎡ + − 1 − kρ ρ (L p0 ⎜

⎜⎝ (1− ρ1 )

2 (1− ρ)2 ⎟⎠⎟ (C.51)

⎛1 ⎣1 ( ) k ρ1

k 1 k −1 ⎤⎦ ρ 1

k −1 [k k − − 1) ρ] ⎞ = +

We can find Lq as Lq = L − (1− p0 ), and W and Wq from Little’s formulas as W = L / λ and Wq = Lq / λ. Note that the relation W = Wq +1/ µ cannot be used here, since µ is not constant but depends on the system-state switch point k. However, by combining the above equations, we see that W W= q (1 p0 ) / λ which implies that the + −

expected service time is (1− p0 )/ λ .

C12 QoS provisioning and queues with impatience Customers are said to be impatient if they tend to join the

queue only when a short wait is expected and tend to remain in line if the wait has been sufficiently small. These situations are typical for the systems with constrained delays where QoS parameter is the maximum allowable delay. Impatience generally takes three forms. The first is balking, the reluctance of a customer to join a queue upon arrival; the second reneging, the reluctance to remain in line after

23

joining and waiting; and the third jockeying between lines when each of a number of parallel lines has its own queue.

C12.1 M/M/1 Balking

In real practice, it often happens that arrivals become discouraged when the queue is long and do not wish to wait. One such model is the M/M/c/K; that is, if the messages see K ahead of them in the system, they do not join. Rarely do all messages have exactly the same discouragement limit all the time. This might depend on the remaining number of hops to their final destination.

An alternative approach to balking is to employ a series of monotonically decreasing functions of the system size multiplying the average rate λ. Let bn be this function, so that λn = bnλ and

0 ≤ bn+1 ≤ bn ≤ 1 (n > 0, b0 ≡ 1).

Now Equation (C.5) when c = 1 gives

n ⎛ ⎞n nλ λ pn = p0 ∏ i−1 = p0 ⎜ ⎟ ∏bi−1 (C.52)

µ µi=1 i ⎝ ⎠ i=1

Possible examples that may be useful for the discouragement function bn, are 1/(n + 1), 1/(n2 + 1), and e−αn . Messages are not discouraged only because of queue size, but may rather attempt to estimate how long they would have to wait. If the queue is moving quickly, then the message may join a long one. On the other hand, if the queue is slow-moving, a message may become discouraged even if the line is short. Now if n messages are in the system, an estimate for the average waiting time might be n/µ, if the customer had an idea of µ. We usually do, so a plausible balking function might thus be bn = e−αn / µ . Also note that the M/M/1/K model is a special case of balking where bi = 1 for 0 ≤ i ≤ K −1 and 0 otherwise.

C12.2 M/M/1 Reneging

Messages that tend to be impatient may not always be discouraged by excessive queue size, but may instead join the queue to see how long their wait may become, all the time retaining the

24

prerogative to renege if their estimate of their total wait is intolerable. We now consider a single-channel birth-death model where both reneging and the simple balking of the previous section exist, which gives rise to a reneging function r(n) defined by

( ) = lim Pr{unit reneges during ∆t n customers present}

r n ∆ →t 0 ∆t

r (0) = r (1) ≡ 0

This new process is still birth-death, but the death rate must now be adjusted to µn = µ + r(n) . Thus it follows from Equation (C.45) that

n n

pn = p0 ∏ λi−1 = p0 λn ∏ bi−1 (n ≥1)

i=1 µi i=1 µ + r i( ) where

−1 ∞ n

p0 ⎜⎜⎛ 1 ∑ n ∏ bi−1 ⎟⎟

⎞ (C.53)= + λ

⎝ n=1 i=1 µ + r i( ) ⎠ α µA good possibility for the reneging function r(n) is e n / , n ≥ 2 . A

waiting customer would probably estimate the average system waiting time as n/µ if n – 1 customers were in front of him, assuming an estimate for µ were available. Again, the probability of a renege would be estimated by a function of the form eαn / µ .

C13 Transient behavior of M / M /1/1 {pn (t)} are probabilities that at arbitrary time t there are n

customers in a single-channel system with Poisson input, exponential service, and no waiting room. In this system, pn (t) = 0 for all n > 1 . We start with Equation (C.3), with λ0 = λ, λn = 0 , n > 0, and µ1 = µ :

d p ( )t = −µ p ( )t + λ p (t (C.54)

d t 1 0

25

1

d p0 ( )t = −λ p ( )t + µ p ( ) t

d t 0 1

Since p0 ( )t + p1( )t = 1 Equation (C.54) is equivalent to d p1( )t

≡ p1 ′( )t = −µ p1( )t + λ [1− p1( )t ]d t

resulting in

p1 ′( ) ( t + λ + µ) ( ) p1 t = λ

This is an ordinary first-order linear differential equation with constant coefficients, whose solution is

p1( )t = C e−(λ +µ )t +λ

λ + µ To determine C, we use the boundary value of p1(t) at t = 0 , which is p0( )t resulting in

λC p 0 −= 1 ( ) λ + µ

and

− +(λ µ )t − λ µ p t( ) = λ ⎡1− e + p ( )0 e ( + )t ⎤

1 λ µ ⎣ 1 ⎦+ (C.55)

( )λ µ ⎣

− +(λ µ )t 0 ( ) −(λ µ

⎦p t =µ ⎡1− e + p 0 e + )t ⎤

0 + since p0 ( )t = 1− p1( ) t for all t. The stationary solution can be found directly from Equation (C.54) in the usual way by letting the derivatives equal zero and then, with the use of the fact that p0 + p1 = 1, solving for p0 and p1 . Also, the limiting (steady-state,

equilibrium) solution can be found as the limit of the transient solution of Equation (C.55) as t goes to ∞, resulting in

26

ρ 1 p1 = and p0 = ρ +1 ρ +1

C14 Transient behavior of M/M/l/∞ Let us assumed that the initial system size N (0) = i where

N ( )t denotes the number in the system at time t . The differential-difference equations governing the system size are given in Equation (C.3) as

( ) = −( + µ p t + λ p ( ) µ p t (n > 0)p′ t λ ) ( ) t + ( ) (C.56)n n n−1 n+1

p0 ′ ( ) t = −λ p0 ( )t + µ p1( )t

We start with probability generating function defined as

∞

P( )z, t = ∑ pn ( ) t zn (z complex) n=0

such that the summation is convergent for z ≤ 1, with its Laplace transform defined as

∞

( ) = ∫ e P( )z, t d t ( s > 0)P z, t −s t Re 0

Starting from Equation (C.56), we get

zi+1 − µ (1− z p) (s)P z t ( , ) =(λ + +µ ) − −

0

z2 (C.57) s z µ λ

where p0 (s) is the Laplace transform of p0 (t) .

Since the Laplace transform P(z, t) converges in the region z ≤ 1, Re s > 0 , wherever the denominator of the right-hand side of Equation (C.57) has zeros in that region, so must the numerator. This

27

fact is used to evaluate p0 (s) . The denominator has two zeros, since it is a quadratic in z and they are (as functions of s)

λ + + − µ s (λ µ s)2 −+ + 4λµ z1 =

2λ

+ + + s (λ µ s)2 − 4λ µ + + λµ z2 =

2λ

One can check that < , z1 + z2 = (λ + µ + s)/ λ, and z1 z2 = µ / λ .z1 z2

At this point Rouche's theorem is used that can be expressed as: If f(z) and g(z) are functions analytic inside and on a closed contour C and if

( )g z < f ( )z on C, then f(z) and f(z) + g(z) have the same number of zeros inside C. For z = 1 and Re s > 0 we see that

z ≡ ( + + s) zλ µ = λ µ s λ µ+ + > + ≥ µ λ z2+ ≡ g ( z )f ( )

2Hence, from Rouche's theorem, (λ + µ + s)z − µ − λz has only one zero in the unit circle. This zero is obviously z1 , since < . Thus z1 z2

equating the numerator of the right-hand side of Equation (C.57) to zero for z = z1 gives

i +1z1s =p0 ( )µ( )z11−

When this transform for p0 (t) is inserted into Equation (C.57) and the result written in infinite series form, we find

⎛ ⎞ k

z zk

1 i j i− j

∞ z 1 i+1 ∞ ⎛ ⎞

P z s ( ) = ∑ z1 z ∑⎜ ⎟ + ∑⎜ ⎟ (, z / z <1 )2 0 k =0 2 ( ) k =0 2λ z z λ z 1− z z2 j= ⎝ ⎠ 2 1 ⎝ ⎠

28

)⎥tλµ

where = ( ) . Now, the transform of ( ) , p (s) is the coefficienti N 0 pn t n

of zn in the Laplace transform of the generating function P z, t( ) , P(z, s). So the next step in the process is to find pn (s) , and then this is, in turn, inverted to get pn (t) , utilizing key properties of the transforms of Bessel functions in this last step. The final result is, in terms of modified Bessel functions of the first kind, In (y) , and is

pn ( )t = e−(λ +µ )t ⎡⎢⎜⎜⎛ λ

⎟⎟⎞

(n−i ) / 2

In−i (2 λµ t)+⎛⎜⎜

λ ⎞⎟⎟

(n−l −1) / 2

In+i+1(2 λµ t)+ ⎢⎝ µ ⎠⎣ ⎝ µ ⎠

n ∞ − j / 2 ⎤ + ⎜⎜

⎛1−λ

⎟⎟⎞⎜⎜⎛ λ

⎟⎟⎞ ∑ ⎜⎜

⎛ λ⎟⎟⎞ Ii (2

⎝ µ ⎠⎝ µ ⎠ j =n+i+2 ⎝ µ ⎠ ⎦⎥ (C.58)

for all n ≥ 0 , where

∞ n+2k

( ) k =0

( ky !( / n 2 +)

k )! (n > −1) (C.59)In y = ∑

C15 Transient behavior of M/M/ ∞ Let us assume that the initial system size at time 0 is 0, so that

N (0) = 0. The differential-difference equations governing the system size are derived from Equation (C.3) with λn = λ and µn = nµ as p t′ ( ) = −(λ + nµ ) p t( ) + λ p (t ) + (n +1) µ p (t ) (n > 0)n n n−1 n+1

po ′ ( )t = −λp0 ( )t + µp1 ( )t (C.60) It can be shown that the generating function of the probabilitie{pn ( )t } is

∞

( , ) = p t z = exp (z −1)(1P z t ∑ n ( ) n ⎢⎡

− eµt ) λ ⎥⎤

n o= ⎣ µ ⎦ (C.61)

29

So if we expand Equation (C.61) in a power series, a0 + a1z + a2 z

2 + ... the coefficients an then are the transient probabilities pn ( )t that we desire. To do this we can use a Taylor series expansion about zero (Maclau-rin series), and we find that

n ( ) = n 1!⎝⎜⎛ (1− e−µt ) µ

λ ⎟⎠

⎞n

exp ⎝⎜⎛ (1 exp −µt ) µ

λ

⎠⎟⎞ n ≥ 0 (C.62)p t − −

C16 Composite queueing models for multimedia systems In multimedia wireless communications like WCDMA, a

customer may be using a high data rate requiring a multicode transmission. This can be seen as using a number of parallel channels simultaneously. An adaptive system may even change the rate on packet by packet basis. To model these applications, in this section we assume that the actual number of customers in any arriving module is a random variable X, which may take on any positive integral value less than ∞ with probability c . This queueing problem, referred to asx

xM [ ] / M /1 , is still Markovian in the sense that future behavior is a function only of the present and not the past. If λx is the arrival rate of a Poisson process of batches of size X, then cx = λx / λ where λ is the composite arrival rate of all batches and is equal to ∑i

∞=1 λ . This total process, which arises from the overlap of

the set of Poisson processes with rates {λx , x = 1,2,...}, is a multiple or compound Poisson process. By using the same logic as in derivation of Equations (C.1)-( C.6) equation (C.6) now becomes

n

(λ µ ) pn + µ pn +1 + λ ∑ pn − k ck ( (B.63)

0 = − + n ≥ 1) k =1

0 = −λ p0 + µ p1

The last term in Equation (C.63) states that the process can get to state n from any state n-k with probability ck. In the sequel we will need to define

30

∞ nP( )z = ∑ pn z ( z ≤ 1)

and n=0

∞

( ) = ∑cn z ( zC z n ≤ 1) n=1

as the generating functions of the steady-state probabilities { }pn and the batch-size distribution {cn } respectively. By multiplying Equation

n(C.63) by the appropriate z , and summing up we have

∞ ∞ ∞ ∞ n

0 = −λ∑ p zn − µ∑ p z n +µ ∑ p z n + λ∑∑ p c z n (C.64)n n n k k n −

n=0 n=1 z n=1 n=1 k =1

Because ∞ n ∞ ∞

n k n−k∑∑ pn−k ck z = ∑ck z ∑ pn−k z = ( ) ( ) C z P z n=1 k =1 k =1 n=k

Equation (C.64) may be rewritten as

= λ ( )− [ ( ) p0 ]+ µ [ ( ) po + λ ( ) ( ) 0 − P z µ P z − P z − ] C z P z and thus

( ) = µp0 (1− z)

z

( z (C.65)

P z ≤1)µ ( ) ](1− z)− λz[1− C z

To find p0 from P 1 and the average number in the system, L , from ( ) ′( ) generating function Equation (C.65), P 1 , let us rewrite the

with r = λ µ , as 0 C z .P( )z =

1− rzpC ( )z

, ( ) = 1−

1− C

z (z)

( ) = [1− ( ) / 1− z is the generating function of theC z C z ] [ ] complementary batch-size probabilities Pr{X f x}= 1− Cx = Cx , since 1/(1− z) is the generating function of 1 and C(z)/(1− z)is the generating function of the cumulative probabilities C , which can bex

seen from

31

∞ ∞ x ∞ ∞ ∞ x ⎛ ⎞ x ⎛ i x i − ⎞ ⎛ i ⎞ 1C z = ⎜ c z = ∑⎜c z = z = c z ∑ x ∑ ∑ i ⎟ i ∑ ⎟ ⎜ ∑ i

= = 1 = x i ⎠ ⎝ 1 ⎟1− zx 1 x 1 ⎝ i= ⎠ i 1 ⎝ = i= ⎠

It follows after one application of l’Hopital’s rule that C ( ) = [ ]and 1 E X after two applications that C ′(1) = E[X (X −1)]/ 2 , resulting in

r C 1 ′ ⎤ 1 = P( ) =

1− rpC0

( ) and P′ 1 ⎣⎡

1 (−

) rC + C

1 (1)⎦1

1 ( ) =

( ) Therefore

p0 [ ] = − ρ= −1 rE X 1 and

r E X + E X 2 ⎡{ [ ] ⎡ ⎤} ρ + rE X 2 ⎤ =

2 1( − ρ⎣)

⎦ = ( −

⎣ρ )

⎦ (ρ = λ [ ] /L E X µ ) (C.66)2 1

The remaining usual measures of effectiveness may be found by using Lq = L − (1− p0 ) = L − ρ and then Little's formulas. The individual state probabilities { } can often also be obtained by the direct pn

inversion of the generating function of Equation (C.65). Two interesting useful examples of these results would be to let the batch sizes be either constant (each of size K) or geometrically distributed. For the constant case,

L = ρ + Kρ

= K +1 ρ (ρ = λ K µ) (C.67)

2(1− ρ) 2 1− ρ and

2ρ 2 + (K −1) ρLq = −L ρ = (C.68)

2 1− ρ The inversion of ( ) to get the individual {pn }

( ) P z is a reasonable task

when K is small. When the batch sizes are instead distributed geometrically, c 1 α α x−1; 0 α 1 it follows that = − < <x ( ) ( )

∞

( ) ( ) n−1 n z)[ ]/ µ = r ( −α , C z 1 α ∑α z = z ( −α /(1 ρ = λE X / 1 ) = − 1 ) −α n=1

and from Equation (C.65)

32

(1− ρ )(1− z )P z( ) = 1− −z rz ⎣1− ( )⎤⎦⎡ C z

(1− ρ )(1− z ) (1− ρ )(1−α z )= =

1− −z rz ⎡⎣1− z (1−α ) ( −α z )⎦ 1− z ⎡⎣α (1 ) ⎦/ 1 ⎤ + − α ρ ⎤

= −(1 ρ )⎜⎛ 1

−α z

⎟⎞

⎜1− z ⎡α + − (1 α ρ ) ⎤ 1− z α (1 ) ⎤ ⎟⎝ ⎣ ⎦ ⎡⎣ + − α ρ ⎦ ⎠

Utilizing the formula for the sum of a geometric series, we have ⎛ ∞ n ∞ n ⎞ n+( ) ( = −1 ρ )⎜ α + − ) } − ⎡α (1 α ) p z ⎟⎟P z ⎜∑{⎡⎣ (1 α ρ ⎤⎦ z ∑{⎣ + − ⎤⎦

1}⎝ n=0 n=0 ⎠

from which we get

= − ⎤n

α α 1 n−1

p 1 ρ ⎡⎣α + − 1 α ρ ⎦ − ⎡⎣ + − α ρ ⎤⎦n ( ){ ( ) ( ) } (C.69)

n−1(1 ρ α ) ⎡ + − 1 ) ⎤ ⎡ 1−α ρ) (n > 0)= − ⎣ ( α ρ ⎦ ⎣ ( ⎤⎦

C16.1 Bulk service M / M [X ] /1 Let us assume that arrivals occur at a single-channel facility as

an ordinary Poisson process, that they are served FCFS, that there is no waiting-capacity constraint, and that these customers are served K at a time, except when less than K are in the system and ready for service, at which time all units are served. This would correspond to the case when a number of signals are multiplexed prior to transmission. Further, if less than K are in service, new arrivals immediately enter service up to the limit K, and finish with the others, regardless of the time into service after service begins. The amount of time required for the service of any batch is an exponentially distributed random variable, whether or not the batch is of full size K. For this model we use notation M / M [K ] /1 . A special case is a model in which the batch size for service must be exactly K and if the number present when the server becomes idle is less than K, it waits until K accumulates. The basic model is, of

33

course, a non-birth-death Markovian problem. The stochastic balance equations are now given as

0 = − λ µ (( + ) pn + µ pn+K + λ pn−1 n ≥ 1) (C.70)

0 = −λ p0 + µ p1 + µ p2 +L+ µ pK −1 + µ pK

By using the operator D defined as pn = Dpn−1 (B.70) becomes ⎡µDK +1 − (λ + µ ) D + λ⎤ p = 0 (n ≥ 0 ; ) (C.71)⎣ ⎦ n

If (r1,KrK +1 )are the roots of the operator equation, then we have

K +1

pn = ∑C r (n ≥ 0)i ii=0

Since ∑∞ p = 1, each ri must be less than 1 or Ci = 0 for all ri not n=0 n

less than 1. So let us now determine the number of roots less than one. By using Rouche's theorem it is found that there is, in fact, one and only one root (say r0 ) in (0, 1). So we have

npn = Cr0 (n ≥ 0, 0 < r0 < 1). Using the boundary condition that ∑ pn must total 1, we find that C = p0 = 1− r0 ; hence

npn = (1− r0 )r0 (n ≥ 0), (0 < r0 < 1) (C.72)

Measures of effectiveness for this model can be obtained in the usual manner. Since the stationary solution has the same geometric form as that of the M / M /1 (with r0 replacing ρ ), we can immediately write

= 0 0 q 0 ⎣ ( 0 )⎦ Wq = W −1/ µL r /(1 − r ), L = L − λ / µ , W r= / ⎡λ 1− r ⎤ , and

C16.2 Modified bulk service Let us now assume that the batch size must be exactly K, and

if not, the server waits until such time to start. Then Equation (C.70) becomes

34

0 = − + n n K n 1 n K )(λ µ ) p + µ p + + λ p − ( ≥

0 = −λ pn + µ pn K + λ pn 1 (1 ≤ n K ) (C.73)≤+ −

0 = −λ p0 + µ pK

The first line of Equation (C.73) is identical to Equation (C.70); nhence pn = Cr0 , (n K −1, 0 0≥ < r < 1) . The obtaining of C (and

p0 ) is a more complicated procedure here, since the foregoing formula for pn is valid only for n ≥ K −1.From the steady-state equations, p K ( / ) p0 = Cr0

K , and therefore C = (λ p0 ) /( µr0 K )= λ µ . To

get p0 now, we must use the K - 1 stationary equations given in Equation (C.73) as 0 = µ pn K − λ pn + λ pn 1 .+ −

Substituting the above geometric formula for pn , n ≥ K , into these K −1 equations, it can be seen that p rn = p − p , 1 ≤ n K ≤ .0 0 n n−1 ( ) These equations can be solved by iteration starting with n = 1 or we can note that these are nonhomogeneous linear difference equations whose solutions are pn = C1 + C2 r0

n , where C1 = p0 − C2 and C2 = − p0 r0 /(1− r0 ) , and

n+1⎧ p (1− r )⎪ 0 0 (1 ≤ ≤n K )

pn = ⎨⎪ 1− r0 (C.74) ⎪ p rn K−

0 0 n K⎪λ ( ≥ )

⎩ µ

To get p we use ∑∞

0 pn = 1. Hence from Equation (C.74), 0 n=

35

⎛ K −11− rn+1 λ ∞ ⎞−1

p0 ⎜1 ∑ 0 + ∑ 0 n K

⎟= + r −

⎝ n=1 1− r0 µ n=K ⎠2 K −1⎛ K −1 r0 (1− r0 ) λ ⎞

−1

⎜ ⎟ ⎜ 1− r 2 µ (1− r ) ⎟

= +1 − + ⎝ (1− r0 ) 0 ⎠

−1

=⎛⎜

µr0 K +1 − ( + ) r0 λ µ K 1− r0 ) ⎞

⎟λ µ + + (

⎜ 2 ⎟⎝ µ (1− r0 ) ⎠

From Equation (C.71) we know that µr0 K 1 (λ µ + λ = 0.+ − + ) So, the

previous equation becomes

p0 = µµ K (1

(− 1−

r0

r )2

)=

1− K

r0 (C.75) 0

The formulas for the { }could have also been obtained via thepn

probability generating function as in the previous section.

C16.3 Erlangian Models (M / Ek /1, Ek / M /1, E j / Ek /1) In this section, we allow for a more general probability

distribution for describing the input process or the service mechanism. In many practical situations, the exponential assumptions used so far may be rather limiting, especially the assumption concerning service times being distributed exponentially.

The Erlang distribution To start with, consider a random variable T which has the

gamma probability density

f ( )t =Γ( )

1 α tα −1e−t / β (α , β > 0 0 > t > ∞),

α β

where Γ( )α = ∫∞

tα −1e−t dt is the usual gamma function, and α and β 0

are the parameters of the distribution. The mean E[T ] and variance

36

Var [T ] are given as E T[ ] = αβ and Var [T ] = αβ 2. If we further consider a special class of these distributions where α and β are related by α = k and β =1/ kµ , where k is any arbitrary positive integer and µ any arbitrary positive constant, we obtain the Erlang family of probability distributions, namely,

( ) ( ) µk k k 1 − µf t =

(k −1 !)t − e k t (0 t )< < ∞

The parameters of the Erlang are thus k and µ , and the mean and variance are given by E T[ ] =1/ µ and Var [T ] = 1/ kµ 2.

Figure C.1 Erlang type k ( Ek ) distribution.

For a given k , the resulting Erlang is referred to as an Erlang type k or Ek distribution. Figure C.1 illustrates the effect of k on the Erlang family of distributions. Parameter k is often called the shape parameter. The Erlang family of probability distributions provides much more modeling flexibility than does the exponential. In fact, the exponential is a special Erlang, namely type 1. As k increases, the Erlang becomes more symmetrical, and as it approaches ∞ , the Erlang becomes deterministic with value 1/ µ . Thus in practical situations where observed data might not bear out the exponential­

37

distribution assumption, the Erlang can provide greater flexibility by being better able to represent the real world. Another reason why the Erlang is useful in queueing analyses is its relation to the exponential distribution and the Markovian property. The Erlang distribution itself is of course non-Markovian. However, as shown later, the sum of k iid exponential random variables with mean 1 / kµ yields an Erlang type-k distribution. It is this relation, as we shall see a little later, that allows the analysis of queueing models with Erlangian input or service to be performed.

Erlang service model (M/Ek/1) We now consider a model in which the service time has an Erlang

type-k distribution. Even though the service may not actually consist of k phases, it is convenient in analyzing this model to consider the Erlang as being made up of k exponential phases, each with mean 1/k µ . Let p n,i (t) represent the probability of n in the system and the customer in service being in phase i (i = 1,2,. . ., k), where we now number the phases backward; that is, k is the first phase of service and 1 is the last (a customer leaving phase 1 actually leaves the system). We can write the following steady-state balance equations

0 = -( λ + kµ )p n,i + k µ p n,i+1 + λ p n−1,i (n ≥ 2,1 ≤ i ≤ k −1), 0 = -( λ + µ n,k + k µ p n+1,1 + λ p −1,k (n ≥ 2)k )p n

0 =-( λ + µ 1,i + k µp1,i+1 (1 ≤ i < k −1)k )p (C.76) 0 = -( λ + kµ )p1,k + k µp2,1 + λ p 0

0=- λp0 + kµp1,1.

It is easier to obtain the expected measures of effectiveness (L, L q , W, W q ) and the state probabilities from Equation (C.76) by working directly on the point process which instead counts phases of service in the system [with the bivariate state pair (n, i) transformed to (n-1)k + i] and then converts back to customers. This can be interpreted as the number of phases in the system requiring service, since n - 1 customers are waiting (n ≥ 1), each requiring k phases of service, and the customer in service requires i phases more. This is essentially equivalent to modeling the Erlang service problem as a constant bulk­

38

input model (the M [ K ] /M/1), where each input unit is considered to bring in K =k phases and the (phase) service rate µ is to be replaced by k µ . (The phase approach to the Erlang service problem and the analogous bulk-input model are not identical, since completed intermediate phases of service cannot leave the system Erlang queue, unlike what happens to an individual completed customer in the bulk model.) The first key result of this connection is that the average line delay for an M/E k /1 customer can be found from the average system size of the bulk-input model given in Equation (C.67) multiplied by 1/k µ [since the time to serve L phases is L(1/k µ )] with the service rate µ of the bulk model itself replaced by the phase service rate µk . This yields

k +1 ρWq = 2k µ(1− ρ)

, (ρ = λ / µ) (C.77)

It follows that k +1 ρ 2

Lq = λWq = (C.78) 2k 1− ρ

and L = Lq + ρ , W =W q +1/ µ . Regarding the steady-state probabilities themselves, we can imme­diately get the empty probability, since we know for all single-channel, one-at-a-time-service queues that p 0 = 1− ρ . Next, we shall convert Equation (C.76) to a (batch-input) system based on a single variable counting phase in the system using the aforementioned transformation ( n, )=i (n-1)k+1.

Making the above transformation in Equation (C.76) yields 0 = -( λ + kµ) p (n−1)k +i +k µp(n−1)k +i+1

+ λp (n−2)k +i (n ≥ 1, 1 ≤ i ≤ k ) (C.79) 0 =- λp0 + kµp1,

where any p turning out to have a negative subscript is assumed to be zero. By writing out the top equation in Equation (C.79) sequentially starting at n=1, i= 1, Equation (C.79) can be simplified to

39

0=-( λ + kµ) pn + kµpn+1 + λpn−k (n ≥1), (C.80)

0 =- λp0 + kµp1

Equation (C.63) would give the same result for a constant batch size k and service rate k µ . Now if we let pn

( P) represent the probability of n in the phase or bulk-input system defined by Equatin (C.80), then it follows that the probability of n in the Erlang service system pn is given by

nk

p = ∑ p (jP) ( n ≥ 1). (C.81)n

j =(n−1)k +1

The waiting-time CDF will be discussed later within the general theory of G/G/1 queues. While we have here utilized the relation between the M [k ] /M/1 and the M/E k /1 it is important to note that a similar partnership holds between the M/M [k ] /1 queue and the E k /M/1, so that the previous bulk results can be useful in deriving results about the Erlang arrival model to be treated in the following section.

Erlang arrival model (Ek/M/1) As mentioned in the previous section, we can use the results

of the bulk-service model to develop results for the Erlang input model. We assume that the interarrival times are Erlang type k distributed, with a mean of 1/ λ . We can look, therefore, at an arrival having passed through k phases, each with a mean time of 1/k λ , prior to actually entering the system. Here we number the phases frontwards from 0 to k - 1. Again, one should keep in mind that this is a convenient concept for analysis that does not have to correspond to the actual arrival mechanism; the only assumption on interarrival times is that they follow an Erlang type-k distribution with mean 1/ λ . We define the state variable as the number of arrival phases in the system, so that we want to find the probability of n arrival phases in the system in the steady state, which we denote by p ( n

P) . Once we have

40

this we use a relation similar to Equation (C.81), to obtain the probability of n customers in the system, that is,

nk k 1+ − ( )p n = ∑ p jP (C.82)

j nk =

We can get p (nP) from p (n

B) , the steady-state probability of n in the bulk-service model given by Equation (C.74) but with λ replaced by k λ . We know from Equations (C.74) and (C.75) that

p (P) = kλp0

( P)

r j −k (j ≥ k)j 0µ = ρ(1− r0 )r0

j−k ( = / )ρ λ µ So, for n > 0,

nk +k −1

p n = ∑ p ( jP) = ρ(1− r0 )(r0

nk −k + r0 nk −k +1 +K+ r0

nk −1)j=nk

= ρ(1− r0 )r0 nk −k (1+ r0 +K+ r0

k −1) = ρ(1− r0 k )(r0

k )n−1 (C.83)

This is a geometric distribution (as in M/M/1), but with r 0k as the

multiplier instead of ρ . It follows from Equation (C.83) that m

L = ρ(1− r0 k )∑n(r0

k )n−1 = ρ(1− r0 k ) 1

k 2 =ρ

kn=1 (1− r0 ) 1− r0

from which we can get L=L q +ρ ,W=L/ λ and W q = W-1/ µ . To derive the waiting-time distribution for this model denote the probability that an arrival into the system finds n customers already there by q , q = Pr{ n in system |arrival about to occur} = Pr { n inn n

system | arrival in phase k-1}

( )Pr{ in system and arrival in phase k-1} pnkP

+k −1qn = = Pr{arrival in phase k-1} 1/ k

since it is equally likely that an arrival is in any one of the k phases with probability 1/k. Thus we have

41

( P) kρ(1− r0 ) kn k ) kn q = kp = r = −(1 r r (C.84)n nk +k −1 0 0 0r0

where the final step follows from the characteristic Equation (C.71) with λ replaced by kλ . Now if there are n customers in the system upon arrival, the conditional waiting time is the time it takes to serve these n customers, which is the convolution of n exponentials, each with mean 1/ µ , This yields an Erlang type n distribution, and the unconditional line-delay distribution function can thus be written as

∞ n−1 ∞ n−1

W q (t) = q0 + ∑ µ (( n µ− x 1))!

e−µxdx = q0 + ∑ (1− r0 k )r0

kn ∫0

t µ (( n µ− x 1))!

e−µxdx n=1 n=1

k t k −µx ∞ (µxr0

k )n−1

= q0 + r0 ∫ (1− r0 )µe ∑ dx0

n=1 (n −1)!−µ −r0 ) x µ (1 r t= q0 + r0

k ∫0

t (1− r0

k )µe (1 k

dx = q0 + r0 k [1− e− − 0

k ) ]

The probability of no wait for service upon arrival is given by Equation (C.84) as q0 = 1− r0

k , , and thus

Wq (t) = 1− r0 k e−µ (1−r0

k )t , ( t ≥ 0 ) (C.85)

Queueing with differentiated QoS We will start with a discussion of a single exponential channel

with priorities. 4G systems will be also characterized by differentiated QoS (DQoS), so that queues with priorities will be a reality. Let us begin by assuming that customers arrive as a Poisson process to a single exponential channel and that upon arrival to the system each unit will be designated to be a member of one of two priority classes. A typical example is data and voice transmission where voice traffic is given higher priority due to the constraints on delay. The usual convention is to number the priority classes so that the smaller the number, the higher the priority. Let it further be assumed that the (Poisson) arrivals of the first or higher priority (voice) have mean rate λ1 and that those of the second (data) or lower priority have mean rate λ 2, such that λ = λ1 + λ2 . We also suppose that the first-priority

42

items have the right to be served ahead of the others, but that there is no preemption.

In view of the foregoing assumptions, a system of steady-state balance equations may be established for pmnr = Pr{in steady state, m units of priority 1 and n units of priority 2 are in the system, and a unit of priority r = 1 or 2 is in service}. These then lead to the following difference equations in the event that ρ = λ µ <1:

0 = −(λ + µ) pmn2 + λ1 pm−1,n,2 + λ2 pm,n−1,2 (m>0, n>1) 0 = −(λ + µ) pmn1 + λ1 pm−1,n,1 + λ2 pm,n−1,1 + µ( pm+1,n,1 + pm,n+1,2 )

(m>1,n>0) 0 = −(λ + µ) pm12 + λ1 pm−1,1,2

0 = −(λ + µ) p1n1 + λ2 p1,n−1,1 + µ( p2n1 + p1,n+1,2)

0 = −(λ + µ) p0 2 + λ2 p0, n 1,2 + µ( p n + p0, + ) (C.86)n − 1 1 n 1,2

0 = −(λ + µ) pm01 + λ1 pm−1,0,1 + µ( pm+1,0,1 + pm12 ) 0 = −(λ + µ) p012 + λ2 p0 + µ( p111 + p022 ) 0 = −(λ + µ) p101 + λ1 p0 + µ( p201 + p112 ) 0 = −λ p0 + µ( p101 + p012 ) p0 is still 1− ρ , since the ordering of service in no way affects the

probability of idleness, also n−1

npn = ∑ ( pn−m,m,1 + pm,n−m,2 ) = (1− ρ)ρ (n>0) m=0

Since the percentage of time the system is busy is ρ , the percentage of time it is busy with a type-r customer will be ρλr λ , so that

∞ ∞ ∞ ∞

∑∑ pmn1 =λ1 and ∑∑ pmn2 =

λ2

m=1 n=0 µ m=0 n=1 µ The most we can do in this case comfortably is obtain expected values via two-dimensional generating functions defined as

∞ ∞

(z) = z n p P z zn pPm1 ∑ mn1, m2 ( ) = ∑ mn 2 n=0 n=1

43

∞ mH1(y, z) = ∑ y Pm1(z) [with H1(1,1) = λ1 µ ]

m=1∞

mH 2 ( y.z) = ∑ y Pm2 (z) [with H 2 (1,1) = λ2 µ ] m=0

and H (y, z) = H1( y, z) + H 2 ( y, z) + p0

∞ ∞ ∞ ∞ m n m n= ∑∑ y z pmn1 + ∑∑ y z pmn2 + p0

m=1 n=0 m=0 n=1∞ ∞ ∞ ∞

m n m n= ∑∑ y z ( pmn1 + pmn 2 ) + ∑ y pm01 + ∑ z p0n2 + p0m=1 n=1 m=1 n=1

where H ( y, z) is the joint generating function for the two priorities, regardless of which type is in service. Note that H ( y, y) = p0 (1− ρy) [with H (1,1) =1], since H ( y, z) collapses to the generating function of M M 1 when z is set equal to y and thus no priority distinction is made . Hence if L1 and L2 are used to denote the mean number of customers present in system for each of the two priority classes, then ∂H ( y, z)

= L = L 1 +λ1 = λ W1 q 1 1∂y µy = z=1

and ∂H ( y, z) λ2= L2 = Lq2 + = λ2W2∂z y= z=1 µ

where Lq1 and Lq2 are the respective mean queue lengths. If we multiply Equation (C.86) by the appropriate powers of y and z and sum accordingly, we get

(1+ ρ −λ1 y

−λ2 z

− 1 )H1( y, z) =

H 2 ( y, z) +

λ1 yp0 − P11(z) − P02 (z)

µ µ y z µ z and

44

(1 ρ 1 − 2 ) 2 ( , ) = P11 ( ) + 02 (ρ 2 ) p0+ −λ y λ z H y z z P z( )

− −λ z (C.87)

µ µ z µ To find H1 and H 2 , we need to know the values of P11(z), P02 (z), and p0 . 11 02 ( ) may be found by One equation relating P z( ), P z and p0

summing zn (n = 2, 3, ...) times Equation (C.86), which involves p0n2, and then using the final three equations of Equation (B.86), resulting in

( ) (1 ρ λ2 z −

1)P ( ) (ρ λ2 z )P z = + − z + − p11 02 0µ z µ Using this equation in Equation (B.87) gives H1 and H 2 as functions of p0 and P , and thus also H ( y, z) as02

H ( y, z) = H1( y, z) + H 2 ( y, z) + p0 = (1− y) p0

1− y − ρy(1− z − λ1 y λ + λ1z λ)

+ (1+ ρ − ρy + λ1z µ)(z − y)P02 (z)

z[1+ ρ − λ1 y µ − λ2 z µ][1− y − ρy(1− z − λ1 y λ + λ1z λ)] By employing the condition that H (1,1) = 1, we have

µ (ρ = λ µ)P02 (1) = (1+ λ

λ

1

2 p µ

0

)(1− ρ) We next take the partial derivatives of H with respect to both y and z and then evaluate at (1,1) to find the means L1 and L2. In so doing, the exact functional relationship for P02 (z) turns out not to be needed, and P02 (1) alone suffices. Without presenting the details of the differentiation, the final results are

45

L =(λ1 µ)(1 ρ λ µ)+ − 1

1 1− λ µ1

ρλ µL = 1 q1 1− λ µ1

ρW = q1 −µ λ1 (C.88) L2 =

(λ2 µ)(1+ ρλ µ λ µ)1 − 1

(1− ρ)(1− λ µ)1

ρλ µL = 2 q2 (1− ρ)(1− λ µ)1

ρW = q2 (1− ρ µ λ)( − 1)

Using the more sophisticated theory of multidimensional birth-death processes, one can find the state probabilities for priority-1 customers as [1]

pn = (1− ρ)⎛ λ1 ⎞⎟⎟

n1

+λ2 ⎛ λ1 ⎞

n1

⎜⎛1−

(λ1 µ)n1

⎟⎞ ( n1 ≥ 0 ) (C.89)

1 ⎜⎜⎝ µ ⎠ λ1

⎜⎜⎝ µ ⎟⎟

⎠ ⎜⎝ (1+ λ1 µ)n1 +1

⎠⎟

Some systems with differentiated QoS or dynamic spectra sharing my be better modeled by the slight generalization of the previous problem by serving the priority-1 customers at the rate µ1 and the priority-2 customers at µ2 .These results are [2]

46

∧

+ ( 2 )( 12λ ρ λ λ λ λ µ µ 2

2 )L = 1 1 q1 µ 1− λ µ 11 1

∧

+ ( 2 )( 12(λ µ ρ) λ λ λ λ µ µ 2

2 )1 1L = 2 q2 1− λ µ 1− λ µ λ µ 2−1 1 1 1 2

Lq = Lq1 + Lq2

(C.90) ∧

2λ ρ µ λ λ µ 2 1 (− λ1 µ λ λ λ λ µ µ )]+ ( 2 )( 1+ 1)[ 1 2= 1 1 2

1− λ µ λ µ 1 2 1− 1− λ µ11 2 ∧

(ρ λ= µ1)

Using the theory of multidimensional birth-death processes, the state probabilities for priority-1 customers are [1]

1pn = (1− ρ * )

⎝⎜⎜⎛

µλ1

1 ⎠⎟⎟⎞

n1

+λ1 + µ

λ2

2 − µ1 ⎣⎢⎢⎡

⎝⎜⎜⎛

µλ1

1 ⎠⎟⎟⎞

n1

− (λ1

µ+

1

µλ1

n

2

1

)n+1

⎦⎥⎥⎤

( n1 ≥ 0 )

Finally, for a model having no priorities but unequal service rates for customers of two major types we have

∧

λ ρ1 − − µ µ λ µ 1 2 )1 (1 )( 2 2L = q1 µ 1− λ µ λ µ 2−1 1 1 2 ∧

2λ ρ µ µ 2 (1 µ µ λ µ 2 )+ − )( L = 2 1 2 1 2 1 q2 µ1 1− λ µ λ µ 1 2−1 2

∧ ∧

) + ( + ( 2 )( 12(λ µ ρ λ µ ρ) (µ µ )

=( )ρ* 2 λ λ λ λ µ µ 2

2 )1 2 2 1 2 1 qL = 1

1− λ µ λ µ 1− ρ* λ λ λ λ µ µ 2 )]2− + ( 2 )( 12 [ 11 1 2

∧ *ρ λ µ ρ λ µ λ µ 2 )( = 1, = 1 +1 2

(C.91)

47

Queueing with multiple differentiated QoS For the modeling of 4G networks multiple differentiated QoS

(MDQoS) models will be needed. Unfortunately when the number of priorities exceeds two, the determination of stationary probabilities in a nonpreemptive Markovian system is an exceedingly difficult problem. In this case, an alternative approach to obtaining the mean-value measures L and W can be used, namely, a direct expected-value procedure. Suppose that items of the kth priority (the smaller the number, the higher the priority) arrive before a single channel according to a Poisson distribution with parameter λk (k = 1,2, . . . ,r) and that these customers wait on a first-come, first-served basis within their respective priorities. Let the service distribution for the kth priority be exponential with mean 1 µk .Whatever the priority of a unit in service, it completes its service before another item is admitted. We begin by

k

defining k = k / k (1 ≤ k ≤ r), σ k = ∑ ρ i (σ 0 ≡ rρ λ µ 0,σ ≡ ρ). i=1

The system is stationary for σ r = ρ < 1. Then suppose that a customer of priority i arrives at the system at time t0 and enters service at time t1. Its line wait is thus Tq = t1 − t0. At t0 assume that there are n1 customers of priority 1 in the line ahead of this new arrival, n2 of priority 2, n3 of priority 3, and so on. Let S0 be the time required to finish the item already in service, and Sk be the total time required to

,serve nk . During the new customer's waiting time q , k items of T npriority k < i will arrive and go to service ahead of this current

, ,arrival. If Sk is the total service time of all the nk then i−1 i

,Tq = Sk + Sk S0.∑ ∑ + k =1 k =1

By taking the expected values of both sides of the above equation i−1 i

W E T = E ⎡ ⎤ + E S [ ] [ ] + E S ≡ ⎡ ⎤ Sq ( )i

⎣ ⎦q ∑ ⎣ ⎦k , ∑ k 0

k =1 k =1

Since σ i−1 < σ i for all i, ρ < 1 implies that σ i−1 < 1for all i.

48

To find E[S0 ], we observe that the combined service distribution is the mixed exponential, which is formed from the law of total probability as

r

B(t) = ∑ λk (1− e−µk t ),k =1 λ

where λ = ∑rk =1 λk . The random variable ''remaining time of

service," S0 , has the value 0 if the system is idle; hence [ ] = Pr{system is busy} ⋅ E[S | system is busy]. But the E S0 0

probability that the system is busy is

λ ⋅ (expected service time) = λ r λk 1

= ρk =1 k∑ λ µ

and E [ S0 system is busy]

r

= ∑ ( [ 0E S system is busy with priority- k customer] k =1

r

×Pr{ customer has priority k }) = ∑ 1 ρk

k =1 µk ρ resulting in

r r

[ ] = ρ∑ 1 ρk ρk (C.92) E S0 = ∑k =1 µk ρ k =1 µk

Since nk and the service times of individual customers, Sk (n) , are

independent, we have E S = E n S ( )n = E n E S ( ) = E n /[ k ] ⎣⎡ k k ⎦⎤ [ k ] ⎣⎡ kn

⎦⎤ [ k ] µk

kLittle's formula then gives E S[ k ] = λkWq ( )k / µk = ρkWq

( ) . Similarly, ` `⎡ ⎤ = ⎡ ⎤ µE S E n / , and then using the uniform property of thek ⎣ ⎦ k⎣ ⎦ k

` ( )i / .Poisson, we have E S⎡ ⎤ = λ W µ Therefore k k k⎣ ⎦ q

i−1 i

W W i + ρ W ( ) E S∑ ∑ q ( )i = q

( ) ρk k qk + [ ]0

k =1 k =1

49

or i

kW( ) + [ ]0∑ ρ qk E S

Wq ( )i = k =1 (C.93)

1−σ i−1

The solution to Equation (C.93) was found in [3] as E S

W ( )i =[ 0 ] (C.94)q (1−σ i−1)(1 −σ i )

Using Equation (C.92) now gives i

ρ µk∑ k

Wq ( )i = k =1 (C.95)

(1−σ i−1)(1 −σ i )r

and Equation (C.95) holds as long as σ r = ∑ ρk < 1. k =1

The analysis for the multiple-channel case is very similar to that of the preceding model except that it must now be assumed that service is governed by identical exponential distributions for each priority at each of c channels. In addition, in order to get a tractable equations, in this case we must assume no service-time distinction between

k

priorities. So, let us define ρk = (λk c µ for (1 ≤ k ≤ r), σ k ∑ i/ ) = ρ i=1

(σ r ≡ ρ = λ cµ). Again the system is completely stationary for ρ < 1, i−1 i

and W ( )i = E S k ' + E S k + E S , k⎡ ⎤ where, as before, S is theq ∑ ⎣ ⎦ ∑ [ ] [ ] 0

k =1 k =1

time required to serve nk items of the kth priority in the line ahead of ' 'the item, Sk is the service time of the nk items of priority k which

arrive during Wq(i) , and S0 is the amount of time remaining until the

next server becomes available. The first two terms of the right-hand side of the Wq

(i) equation are exactly the same as in the single-channel

50

case, except that the system service rate cµ is used in place of the single-service rate µk throughout the argument. To derive [ ] consider E[S0 ] = Pr{all channels busy} ⋅ E[S0 allE S0 , channels busy ]. From Equation (C.21) the probability that all channels are busy is

∞ ∞ n ccp∑ pn = p0 ∑ ( )cp − = p0

( ) n c

n c n= c c! c!(1 − ρ)= c

and E[S0 all channels busy] = 1 cµ from the memorylessness of the exponential. Thus from (B.22),

( )cp c ⎛ c−1 ( )cp n cp c −1

E S0 = + ⎟[ ]c ρ µ ⎜∑

=

( ))

⎞ !(1 − )( c ) ⎝ n 0 n! c!(1 − ρ ⎠

Therefore from Equation (C.95),

[ ] ⎢c!(1 − )( c )∑ (cp ) ( )i E S0 ⎣

⎡ ρ µ n

c

=

−1

0

(n c− ) n!+ cµ⎥⎦

⎤−1

W = = q (1−σ i−1)(1 −σ i ) (1−σ i−1)(1 −σ i ) and the expected line wait taken over all priorities is

r

Wq = ∑ λi Wq ( )i

i=1 λ Preemptive priorities

Some messages in 4G networks, like control commands for network reconfiguration, might require a high level of urgency. To model these situations let us now permit units of the higher priority to preempt. Since the customer served will always be of priority 1 when at least one unit of that priority is present, we may drop the use of the third subscript of pmnr as a service-customer indicator and instead use it to introduce a new priority, henceforth called 3 or lowest. So pmnr is now the steady-state probability that there are m units of priority 1 in the system with arrival rate λ1 and service rate µ1, n units of priority 2 in the system with arrival rate λ2 and service rate µ2, and r units of priority 3 with arrival rate λ3 and service rate µ3.

51

Under these assumptions, a system of difference equations may be derived for the stationary probabilities (λ = λ1 + λ2 + λ3 and ρ = λ µ + λ µ + λ µ 3 < 1) as1 1 2 2 3

0 = −λ p000 + µ1 p100 + µ2 p010 + µ3 p001

0 = −(λ + µ1) pm00 + λ1 pm−1,0,0 + µ1 pm+1,0,0

0 = −(λ + µ ) p n + µ p + λ p − + µ p2 0 0 1 1, n,0 2 0, n 1,0 2 0, n+1,0

0 = −(λ + µ3) p00 r + µ1 p10 r + µ2 p01 r + λ3 p0,0, r−1 + µ3 p0,0, r+1

0 = −(λ + µ1) pmn0 + λ1 pm−1, ,0 + λ2 p , − + µ1 pmn m n 1,0 +1, ,0 n

0 = −(λ + µ1) pm r0 + λ1 pm−1,0, r + λ3 pm,0, r−1 + µ1 pm+1,0, r (C.96) 0 = −(λ + µ2 ) p0nr + µ1 p1nr + λ2 p0, n−1, r + µ2 p0, n+1, r + λ3 p0, n r , −1

0 = − µ ) p n + λ n r λ pm n + λ m n r 1 + p n r (λ + 1 m r 1 pm−1, , + 2 , −1, r 3 p , , − µ1 m+1, ,

When the number of preemptive priorities is r there will be a total of 2r equations. In this case the steady-state generating function can be obtained and from it, the mean number of priority i waiting in the system. In the event that the service rate is the same for all items, it follows for ρ i = λi µ that

i

ρ i ∑ ρn 3 (i) n=1E[Nq ]= i−1 i (∑ ρi <1)

(1− ∑ ρn )(1− ∑ ρn ) i=1

n=1 n=1

with

E N⎡ ( )i ⎤ = E N⎡ ( )i ⎤ + ρi

⎣ ⎦ ⎣ q ⎦ i−1

1− ∑ ρi n=1

In addition, the variance can be found to be i−1 i

i−1 ρρi

2 + ρi (1− ∑ ρn )(1 − ∑ ρn ) Var ⎣⎡Nq

( )i ⎦⎤ = 2ρi

3 ∑ i−1 n

i + i−1 n=1

in=1

n=1 2(1− ∑ ρn )(1 − ∑ ρn ) (1− ∑ ρn ) (1 − ∑ ρn ) n=1 n=1 n=1 n=1

Both of these results generalize to r > 3 priorities.

52

Queueing with adaptive service In adaptive systems we may want to allow the number of

priorities to be continuous, so that the priority of a given unit is assigned according to some measure of the length of time needed to serve that unit. As an example, in active networks this might be the case in a situation such as the loading of programs onto a node. In particular, let the unit arrive as a Poisson process with mean rate

∞

λt (∫λt dt = λ) and be served according to an exponential distribution 0

with mean time 1 µt . Then the total service cumulative distribution is ∞ ∞λt − −µt t 1 λ −µttB(t) = ∫ (1 e )dt = −1 ∫ te dt 0 λ λ 0

The expected waiting time Wq(t ) of a customer with type t priority is

given by analogy with the discrete case Equation (C.94), as E S[ 0 ]W t( ) = q 2

⎛ t ⎞ ⎜1− ∫ ( ) λ ydB y ⎟ ⎝ 0 ⎠

It can be shown in general that

[ ] =λ E[S 2 ]E S0 2

for any single-channel queue with Poisson input. Thus 1 ∞∫ t

2 t

−µt tλ µt e dt 2

W ( )t = 0 q t

(1− λ µy e −µy ydt )2∫ y y 0

The expected number of customers in the line is thus ∞ ⎡λ ∞ ⎤ ⎡ t t ⎤W ( )t dt = t 2 e−µt tdt (1− y e −µy ydy)−2L = λ λ µ λ µ dtq ∫ q ⎢ ∫ t t ⎥ ⎢∫ ∫ y y ⎥

0 ⎣ 2 0 ⎦ ⎣ 0 0 ⎦ If µt is assumed to be the constant µ and λt to be λ, then Lq simplifies to

53

W (t ) = λ µ 2

= λ µ

−

2

µt 2q t µ)[1− e (1+ µt)]}(1− λ∫ yµe−µydy)2 {1− (λ

0

and λ

2 ∞ ⎞−2

Lq = ⎜ ⎟ ⎜ ∫ 1− ⎡⎣1− e ( ) ⎤⎦ ⎟ dt ⎛ ⎞ ⎛ λ −µt 1+ µt µ 0 µ ⎠⎝ ⎠ ⎝

C16.4 Queuing and routing: networks, series, and cyclic queues A routing process sees a data network as a group of nodes (say k

of them) where each node represents a service facility of some kind with, let us say, ci servers at node i, i = 1, 2,. . . , k. In this section we will limit our interest to queueing networks with the following characteristics:

(1) Arrivals from the "outside" to node i follow a Poisson process with mean rate γ i .

(2) Service (holding) times at each channel at node i are independent and exponentially distributed with parameter µi .

(3) To model adaptive systems with QoS provisioning, a node's service rate may be allowed to depend on its queue length.

(4) The probability that a customer who has completed service at node i will go next to node j (routing probability) is rij (independent of the state of the system), where i = 1, 2, . . . , k, j = 0, 1, 2, . . . , k, and ri0 indicates the probability that a customer will leave the system from node i. Networks that have these properties are called Jackson networks. As we shall see later, their steady-state probability distributions have a most interesting and useful product-form solution. Cases for which γ i = 0 for all i (no customer may enter the system from the outside) and ri0 = 0 for all i (no customer may leave the system) are referred to as closed Jackson networks. The general case described above will be referred to as open Jackson networks as illustrated in Figure C.2.

54

i j

ijr

Figure C.2 Open Jackson networks.

Series queues/source routing To start, in this section we consider models in which there are a series of service stations through which each calling unit must progress prior to leaving the system. An example is a source routing were a virtual path is established at the beginning of a session and all messages are supposed to follow the same route. This can be represented as

( j i +1,1 ≤ i k − )⎧1 = ≤ 1 , ⎪ ⎪

γ i = ⎧⎨

λ (i = 1 ,) and rij = ⎨1 (i = k j , = 0 , )

⎪ ⎪⎩0 (elsewhere )⎩0 (elsewhere ).

The first series model to be considered is a sequence of queues with no restriction on the waiting room's capacity between stations. Such a situation is pictured in Figure C.3. We further assume that the calling units arrive according to a Poisson process, mean λ , and the service time of each server at station i ( i = 1, 2, ..., n) is exponential with

55

mean 1/ µi . Since there is no restriction on waiting between stations, each station can be analyzed separately as a single stage (nonseries)queueing model. The first station is an M/M/ c1 / ∞ model. It is necessary to find the output distribution (distribution of times between successive departures) in order to find the input distribution (times between successive arrivals) to the next station. One can show that the interdeparture times are exponential with parameter λ .

Series queues with blocking We consider first a simple sequential two-station, single-server­

at-each-station model where no queue is allowed to form at either station. If a customer is in station 2 and service is completed at station 1, the station 1 customers must wait there until the station 2 customer is completed; that is, the system is blocked. Arrivals at station 1 when the system is blocked are turned away. Also, if a customer is in process at station 1, then even if station 2 is empty, arriving customers are turned away, since the system is a sequential one; that is, all customers require service at 1 and then service at 2. We wish to find the steady-state probability pn1,n2 of n1 in the first station and n2 in the second station. For this model, the possible states are

n1 , n2

0, 0 system empty; 1, 0 customer in process at 1 only; 0, 1 customer in process at 2 only; 1, 1 customer in process at 1 and 2; b, 1 customer in process at 2 and a customer finished at 1 but waiting for 2 to become available, i.e. system is blocked.

Assuming arrivals at the system (station 1) are Poisson with parameter λ and service is exponential with parameters µ1 and µ2 , respectively, the usual procedure leads to the steady-state equations for this multidimensional Markov chain:

56

L

L

Calling units

Station 1Servers

Output 1= input 2

Station 2

Output 2= input 3, etc.

Servers

1c

2c

L

L

Calling units

Station 1Servers

Output 1=input 2

Station 2

Output 2=input 3, etc.

Servers

1c

2c

MM

L Station nServersnc L Station nServersnc

Figure C.3 Series queue, infinite waiting room.

0 = −λ p0,0 + µ2 p0,1

0 = −µ p + µ p + λ p1 1,0 2 1,1 0,0

( + 2 ) p0,1 + µ1 p1,0 + µ2 pb,1 (C.97)0 = − λ µ

0 = −(µ1 + µ2 ) p1,1 + λ p0,1

0 = −µ2 pb,1 + µ1 p1,1

Using the boundary equation ∑∑ pn1,n2 = 1, we have six equations and five unknowns [there is some redundancy in Equation (C.97); hence we can solve for the five steady-state probabilities]. Equation (C.97) can be used to obtain all probabilities in terms of p0,0 , and the

57

boundary condition can be used to find p0,0 . If we let µ1 = µ2, the results are

λ λ + 2µ 2

p1,0 = 2µ 2 p0,0 , p0,1 = µ

p0,0 , p1,1 = 2µ 2 p0,0

( ) λ λ

(C.98)λ 2 2µ 2

pb,1 = 2µ 2 p0,0 , p0,0 =

3λ 2 + 4µλ + 2µ 2

It is easy to see how the problem expands if one allows limits other than zero on queue length or considers more stations. For example, if one customer is allowed to wait between stations, this results in seven state probabilities and we have to solve seven equations and a boundary condition.The complexity results from having to write a balance equation for each possible system state. For large numbers of equations, as long as we have a finite set, numerical techniques for solving these simultaneous equations can also be employed.

Open Jackson networks We now consider a network of k service facilities (nodes).

Customers can arrive from outside at any node according to a Poisson process. We will now represent the mean arrival rate to node i as γ i (instead of the familiar λi ) for reasons that will become clear shortly. All servers at node i work according to an exponential distribution with mean µi (so that all servers at a given node are identical). When customers complete service at node i, they go next to node j with probability rij (independent of the system state), i = 1, 2, . . . , k. There is a probability ri0 that customers will leave the network at node i upon completion of service. There is no limit on queue capacity at any node; that is, we never have a blocked system or node. Since we have a Markovian system, we can use our usual types of analyses to write the steady-state system equations. We first, however, must determine how to describe a system state. Since various numbers of customers can be at various nodes in the network, we desire the joint probability distribution for the number of customers at each node; that is, letting Ni be the random variable for the number of customers at node i in the steady state, we want to find

58

Pr{N1 = n1, N2 = n2 ,...., Nk = nk }≡ pn ,n ,...., . From this joint 1 2 nk

probability distribution, we can obtain the marginal distribution for numbers of customers at a particular node by appropriately summing over the other nodes. We shall use the method of stochastic balance to obtain the steady-state equations for this network. Rather than using the somewhat cumbersome k-component vector (n1, n2,..., nk) for describing a state, we employ the simplified notation given as

n n1, 2 ,....., n , nj ,..., n ⇒ ni k

n n, ,....., n +1, n ,....., n ⇒ n i; + 1 2 i j k

n1, n2 ,....., ni −1, nj ,....., nk ⇒ n i; −

n n n ⇒ ; + − 1, 2 ,....., ni +1, nj −1,......, k n i j

Using stochastic balance for equating flow into state n to flow out of state n , and assuming that ci = 1for all i (single-server nodes) and that ni ≥ 1 at each node (actually, the equation that results will also hold for ni = 0 if we set terms with negative subscripts and terms containing µi for which the subscript ni = 0 to zero), we obtain

k k k k k k

∑γ i pn;i − + ∑∑µi rij pn;i+ j + ∑µi ri,0 pn;i

= ∑µi (1− rii )pn + ∑γ i pn .− +

i=1 j =1 i=1 i=1 i=1 i=1

(C.99) Jackson [58,59] first showed that the solution to these steady-state balance equations is what has come to be generally called product form, where the joint probability distribution of system states can be written as

pn = Cρ1 n1 ρ2

n2 ⋅ ⋅ ⋅ ρk nk

To show that it satisfies Equation (C.99), let λi be the total mean flow rate into node i (from outside and from other nodes). Then, in order to satisfy flow balance at each node, we have the traffic equations

k

λi = γ i + ∑λ j rji (C.100a) j=1

59

or, in vector-matrix form, λ = γ + λR. (C.100b)

where λ = {λi}, γ = {γ i}, R = {rij} . We define ρi to be λi / forµi i = 1, 2, . . . ,k. Then, the steady-state solution to (6.99) is

p p = 1− ρ ρ n1 1− ρ ρ n2 ⋅ ⋅ ⋅ 1 n n ≡

1 , 2 ,....., nk 1 2 − ρ ρ kk (C.101)n n ( 1 ) ( 2 ) ( k )

What this result says is that the network acts as if each node could beviewed as an independent M/M/1 queue, with parameters λi and µi , so that the joint probability distribution can be written as a product of individual M/M/l's.To verify that Equation (C.101) does indeed satisfy Equation (C.99), we first show that pn = Cρ1

n1 ρ2 n2 ⋅ ⋅⋅ ρk

nk satisfies Equation (C.99) and

then that C turns out to be ∏k (1− ρi ) , in order to satisfy thei=1

summability-to-one criterion. We let ℜn = ρ1 n1 ρ2

n2 ⋅ ⋅⋅ ρk nk , and

plugging pn = Cℜn into Equation (C.99) gives

Cℜnk γ i C n

k k

µ r ρi + ℜ C nk

µ ρ∑ + ℜ ∑∑ i ij ∑ i i r 0 ii=1 ρi j=1 i=1 ρ f i=1

( ≠ )i j

? k kn n= Cℜ ∑µi ( − rii ) C ∑γ i1 + ℜ

i=1 i=1

Canceling out Cℜn , we have k k k λ µ k ? k

∑ γ µi i + ∑∑ µi irj i j + ∑ µi i r 0

λi ∑(µi − µi iri + γ li)= λ λ µ µi=1 i j=1 i=1 j i i=1 i i=1 i j( ≠ )

From Equation (C.100a) we have, k k

λ = γ + r λ + r λ ⇒ r λ = λ −γ − r λj j ∑ ij i jj j ∑ ij i j j jj ji=1 i=1(i j ) (i j ≠ )≠

so that i i i i∑

k γ µ + ∑

k µ (λ −γ − r λ ) + ∑ k

µ r λ ?

∑k

(µ − µ r + γ ) i=1 λi j=1 λ j

i i jj j i=1

i i 0 µi =

i=1 i i ii i

60

Changing the subscript on the second term on the left-hand side from j to i, we get

k ⎛ γ µ µ (λ γ r λ ) + λ r ⎞ ? k

µ µ r + γ )+ − − ( −∑⎜ i i ii i ii i i i0 ⎟=∑ i i ii i

i=1 ⎝ λi λi ⎠ i=1

After moving through with the summation and canceling we have

k ? k

∑λi ir 0 =∑γ i i=1 i=1

This is indeed true, since the left-hand side represents the total flow out of the network and the right-hand side represents the total flow in. For steady state, these must be equal.To obtained ρi , we need to obtain λi from the traffic Equation

(C.100b), which are solved as λ = γ (I R −1 . The invertibility of− )I − R is assured as long as there is at least one node releasing its output to the outside and no node is totally absorbing. Now to evaluate C, we use

∞ ∞ ∞

L n1 n2 knk∑ ∑ C ρ ρ1 2 Lρ = 1

nk =0 n2 =0 n1 =0

resulting in ⎛ ⎞

−1∞ ∞ ∞

n1 n2 nkC = ⎜⎜ L ρ ρ L ρ ⎟⎟ =∑ ∑ ∑ 1 2 k⎝ nk =0 n2 =0 n1 =0 ⎠

⎛ ∞ ∞ ∞ ⎞−1

⎛ 1 1 1 ⎞−1

⎜⎜ ∑ ρknk L ∑ ρ2

n2 ∑ ρ1 n1 ⎟⎟ = ⎜1− ρ

L1− ρ 1− ρ ⎟

⎝ nk =0 n2 =0 n1 =0 ⎠ ⎝ k 2 1 ⎠ k

= ∏ (1− ρi ) (ρi <1, i =1, 2, K, k ) i=1

We can obtain expected measures rather easily for individual nodes, since Li = ρi /(1− ρi ) and Wi = Li / λi . This is so because of the

61

product form of the solution for the joint probability distribution and again does not imply that the nodes are truly M / M /1 (which they may not be, although the system size processes are indeed independent M / M /1 s, since the joint probability distribution is the product of the marginals). The expected total wait within the network of any customer before its final departure would be ∑ i

Li /∑ γ ii

(Little's formula for the entire network). The above results for Jackson networks generalize easily to c-channel nodes. Let ci represent the number of servers at node i, each having exponential service time with parameter µi . Then Equation (C.101) becomes

p pn n = i p = λ µ (C.102)≡ 1, 2, K,n Π k rni

0i (ri i / i )n k i=1 ( )a ni i

where ⎧⎪n ! (n c )

a n( )i =⎩⎨

i

i i − i ci ( i

i

≥<

i

i

) (C.103)i n c

⎪c n c

and p0i is such that ∑∞ p0i rini / ai (ni ) = 1. One should note that ri is

ni

different parameter from the double-subscripted ri j , which are routing probabilities. Thus, again, what we have is a network that acts as if each node were an independent M / M / c .

It is a rather straightforward generalization to allow a customer of one type to have different routing probabilities than a customer of another type. The essential modification is to first solve the traffic equations separately for each customer type and then add the resulting λs. We will use a superscript to denote customer type, so that R(t ) is the routing probability matrix for a customer of type t (t = 1, 2,…, n). Solving Equation (C.100) yields a λ (t ) for each customer type t. We then obtain λ = ∑t

λ (t ) . We proceed as before to obtain the L fori

each of the nodes (i = 1,2,…, k) using M / M / c results. The average waiting time at each node (note that all customer types have the same

62

average waiting time, since they have identical service-time distributions and wait in the same first-come, first-served queue) can be obtained by Little's formula as before. The same is true for the average system sojourn time. We can also obtain the average system size for customer type t, by simply weighting the node average total size by customer type t’s relative flow rate, namely

( )t ( )t λ iLi =

λ (1) + λ (2) Lλ n Li( ) i i i

C16.5 Closed Jackson networks/mobile agent modeling If we set γ i = 0 and ri0 = 0 for all i , we have a closed Jackson

network, which is equivalent to a finite-source queue of, say, N items which continuously travel inside the network. This can be used for mobile agent modeling in advanced networks. If ci = 0 for all i , we can get the steady-state flow-balance equations from Equation (C.99), by settingγ i = ri0 = 0 . This yields

k k k

µ r pn i j

= µ 1− r p (C.104)∑ ∑ i ij &&&; + − ∑ i ( ii ) &&&n j=1 i=1 i=1 (i j )≠

Since this is a special case of a general Jackson network, once more we have a product-form solution,

p&&&n = Cρ1 n1 ρ2

n2 Lρknk ≡ Cϑ n (C.105)

where ρ i = λi / µi , must satisfy the balance equations for flow at each node i , so that the flows into and out of node i must be equal. This yields the closed network equivalent to the traffic equations of Equation (C.100a), namely,

k k

λ = µ ρ = λ r = µ r ρ (C.106)i i i ∑ j ji ∑ j ji jj=1 j=1

63

One of the equations of Equation (C.106) is redundant because the sum of the λi is known. Thus, we can arbitrarily set one ρi , equal to 1

when solving µi ρi = ∑k µ j rji ρ j . When verifying that Equationj

(C.105) is a solution by substituting it into Equation (C.104), C might not ‘break apart’ and must be evaluated by:

−1⎛ ⎞ n1 n2 nk n1 n2 nkCρ ρ Lρ 1 C = ρ ρ Lρ ⎟⎟∑ 1 2 k = ⇒ ⎜⎜ ∑ 1 2 k

n n1 + + +2 L k =N ⎝ n n2n 1 + +Lnk =N ⎠ where the sum is taken over all possible ways the N elements can be distributed among the k nodes. The constant C is often shown as C(N) to emphasize that it is a function of the total population size N.

−1Further, the solution is often written in terms C (N ) = G(N ), so that

pn n, 2L,n = 1 ρ1

n1 ρ2 n2 Lρk

nk (C.107)1 k ( )G N

where n1 n2 nkG N( ) = ∑ ρ1 ρ2 Lρk (C.108)

1 2 L n Nn n k+ + + =

Again, this closed network can easily be extended to ci servers at node i . The solution now becomes

p , , n = 1 k ρi

ni

(C.107a)n n1 2 L k G N ∏ 1 a n( ) i= ( )i i

where ai ( )ni is given by Equation (C.103) and

G N( ) = ∑ ∏ k ρi

ni

(C.108a)a nn n L k+ + + = 2 n N i=1 i ( )i1

For large N and k there are many possible ways to allocate the N

customers among the k nodes, in fact, ⎜⎛ N +k −1

⎟⎞ ways . An efficient

⎝ N ⎠ algorithm to calculate G(N )was developed in Buzen [28]. It is based

non the following consideration: if fi (ni ) = ρ ii / ai (ni ) , then

64

k

G ( ) (C.109)= f n∑ ∏ i i1 + + +2 L n N i=n n k = 1

An auxiliary function defined as m

gm ( )n = ∑ ∏ fi ( )ni (C.110) n1 +n2 +L+nm =n i=1

was used in [28]. Note that G(N ) = gk (N ) . We can now set up a recursive scheme for calculating G(N ) . Consider gm (n) and suppose we fix n = i first in the summation, so that we havem

⎛ m ⎞ n ⎛ m−1 ⎞f n = ∑ ig n( ) = ∑⎜⎜ ∏ i ( ) ∑ f i m ( ) ⎜⎜ ( )i ⎟⎟m ∑ i ⎟⎟ ∏ f n

⎝ n n1 2 L nm−1 + = 1 i=0 ⎝ 1 + + n n i =+ + + i n i= ⎠ n n2 L m− = − i 1 ⎠ n

∑ m ( ) m−1 ( ) (n i− n = 0,1,K, N= f i g ) i=0

(C.111) Note that from Equation (C.111) g1 (n) = f1 (n)and gm (0) = 1, so that we can use Equation (C.111) recursively to calculateG(N ) = gk N( ) . Further, these functions aid in calculating marginal distributions as well. Suppose we want the marginal distribution at node i , namely,

( ) = Pr{ = n . If we use notation = + + n + n +Lnp n N } S n n .. ,i i i 1 2 i−1 i+1 k

then we have

i ( ) K n = ∑ 1 k

f N( )p n = ∑ pn1 , , k G N ∏ i i S N n= − N nSi − ( ) i=1i

f n k

= i ( ) ∑ ∏ f ( )n (n = 0,1,KN )( ) i

i= −

( )G N S N n j=1

j i≠

This, however, is very cumbersome to compute, but for node k the expression simplifies to

k −1f n N n) k ( ) ( ) ∑ ∏

=1 i ( ) f n g( )

G N ( −

p n = k f n = k k −1 (n = 0,1,., N )G N( ) S N n= − i ( )k

(C.112)

65

To find the other pi ( ), in ≠ k , Buzen [28] suggests permuting the network to make the node i of interest = k . In the next section, we present a method for the calculation of expected-value performance measures for closed Jackson networks, called mean-value analysis, which also can yield marginal probability distributions.

Mean-value analysis The previously described methods of analyzing closed Jackson

queueing networks which require computing the normalizing constant G(N ) are often referred to as ‘convolution procedures’. Mean-value analysis is another approach which does not require evaluating G(N). It is built on two basic principles [23]

(1) The queue length observed by an arriving customer is the same as the general-time queue length in a closed network with one less customer, that is qn (N ) = pn (N −1) .

(2) Little's formula is applicable throughout the network. The first principle allows us to write the average waiting time at a node in terms of the mean service time and average number in the system found by an arriving customer. In addition, for the M / M /1 situation, using Equation (C.106) and (C.109), we have W = (1+ L) / µ . This implies that the average time an arriving customer must wait is the average time to serve the queue size as seen by an arriving customer plus itself. For M/M/c, qn = pn , so no adjustment need be made for the fact that L is based on p and not q . For our closedn n

network (we assume for the time being that all nodes have a single server), the equivalent equation becomes

L N −1) W N( ) = 1+ i ( (C.113)i µ

where Wi(N) is the mean waiting time at node i for a network containing N customers, µ i is the mean service rate for the single server at node i , and L (N - 1) is the mean number at node i in ai

network with N - 1 customers. Applying Little's formula throughout the network allows us to write

66

L N( ) = λ ( )N W N ( ) (B.114)i i i

where λi (N ) is the throughput (arrival rate) for node i in an N customer network. If we can find λi (N ) , Equations (C.113) and (C.114) give us a method for recursively calculating Li and Wi starting with an empty network [one with no customers, for which Li (0) = 0and Wi (1) = 1/ µ ], and building up to the network of interest having N customers. Now if we let Di (N ) represent the average delay per customer between successive visits to node i for a network with N customers, then by the laws of conservation we have λi (N ) = N / Di (N ) . This merely states that the number of arrivals at node i per unit time must equal the total number of customers in the system divided by the mean time it takes each customer between successive visits to node i, and is a form of Little's formula applied to the entire network, since the expected number of customers in the system is exactly N. To get Di (N ) , we use again Equation (C.106). Letting ϑi = µi ρi these equations become

k

ϑi = ∑ϑ j rji. (C.115) j=1

Since one of the above equations is redundant, we can arbitrarily set one ϑi ( say ϑl ) equal to 1 and solve for the others. The ϑi , then are relative throughputs through node i, that is, ϑi = λi / λl , assuming ϑl

is the one on which we normalize. Now we can write D N( ) = ∑k ϑW ( ); that is, DlN (N ) is a weighted average of thel i=1 i i

average delays at each node, weighted by the relative throughputs (arrival rates) of each node to node l , or equivalently weighted by the expected number of visits to each node prior to returning to the ‘normalized’ node, node l (note that ϑi can also be interpreted as the expected number of visits to node i after leaving node i prior to returning to node l ). For example, if we have a two-node network with ϑ1 = 1 and ϑ2 = 2 , then since the arrival rate at node 2 is twice

67

that at node 1, the expected number of visits to node 2 after leaving node 1 prior to returning to node 1 must be two.

We can now write the mean-value analysis (MVA) algorithm for finding Li (N ) and Wi (N ) in a k -node, single-server-per-node network with routing probability matrix R = {rij } as follows:

k

(1) Solve the traffic Equations (C.115), ϑi = ∑ϑ j rji (i = 1, 2,. . . , k), j=1

setting one of the ϑ j (say ϑl ) equal to 1. (2) Initialize Li (0) = 0 (i = 1, 2, . . . , k). (3) For n = 1 to N, calculate

(a) Wi (n) = 1+ Li (n −1) (i =1, 2,..., k)µi

(b) λi (n) = k

n ( assume ϑi = 1) ∑ϑiWI (n) i=1

(c) λi (n) = λl (n)ϑi (i =1, 2,..., k i , ≠ l)

(d) Li (n) = λi (n)Wi (n) (i =1, 2,... k)

It is also possible to obtain marginal steady-state probabilities at each node by recursion, and in fact, we would add to the MVA algorithm a recursive relationship similar in spirit to that of pn = ρpn−1 for M / M /1, the relation achieved from detailed (as opposed to global) stochastic balance,

pi (n, N ) = λi (N ) pi (n −1, N −1) ( ,n N ≥ 1) (C.116)µi

where pi (n, N ) is the marginal probability of n in an N-customer system at node i, and pi (0,0) = 1.

Multiple-server cases The MVA algorithm must be modified in step (3)(a). Since

there are multiple servers who work simultaneously on reducing the customer queue we have

68

1 1 n−1

i ( ) = + ∑ ( j ci 1) pi ( j n −1)W n − + ,µi ci µi j ci=

because if there are j > ci , customers at node i when a customer arrives, it must wait until j − ci +1are served at rate ci µi to get into service. This may be simplified to

1 ⎛ n−1 n−1 ⎞Wi (n) = ⎜⎜ci + ∑ jpi ( j, n −1) − (ci −1)∑ pi ( j,n −1)⎟⎟ci µi ⎝ j=ci j =ci ⎠

1 ⎡ ci −1 ⎛ ci −1 ⎞⎤ + L (n 1) ∑ jp ( j, n 1) ( ) ⎜1 p ( j, n −1)⎟⎥= ⎢ci i − − i − − ci −1 × − ∑ ici µi ⎣⎢ j=0 ⎝

⎜j =0 ⎠

⎟⎥⎦1 ⎛ ci −2 ⎞

= 1+ L n( − + (c − − ) ( ,⎜ i 1) ∑ i 1 j p i j n −1) ⎟c µ ⎝ j=0 ⎠i i

Thus in this case, even if we are interested in only the Wi and Li we still need to calculate the marginal probabilities pi ( j, n −1) for j = 0 , 1,. . . , ci − 2 . To calculate these recursively in the spirit of the M/M/c we now have

pi ( j, n) = λi (n) pi ( j −1, n −1) (i j n≤ ≤ −1) α i ( j)µi

where

αi ( )j = ( )

=⎧⎨

j j( ≤ c ) (C.117)a ji i

i 1) ⎩ i ( ≥ ci )a j( − c j Now the modified MVA algorithm can be represented as: (1) Solve the traffic Equations (C.115) as done previously (note these are the same regardless of the number of servers at a node). (2) Initialize for i = 1, 2, …k, Li (0) = 0; pi (0,0) =1; pi ( ,0) = 0, ( j ≠ 0).j (3) For n = 1 to N, calculate

ci

(a)Wi (n) = 1 ⎛

⎜⎜1+ Li (n −1) + ∑−2

(ci −1− j) pi ( j, n −1)⎞⎟⎟ , (i =1,2,...., k)

ci µi ⎝ j=0 ⎠

69

k

(b) λi (n) = n / ∑ϑiWi (n) (assume ϑl = 1) i=1

(c) λi (n) = λl (n)ϑi (i =1, 2,... k i ; ≠ l), (d) Li (n) = λi (n)Wi (n) (i =1, 2,..., k),

(e) pi ( j, n) = α

λ

i ( i ( jn )µ )

i

pi ( j −1, n −1)

( j = 1, 2,..., n i; = 1, 2,...., k)

Finally we prove Equation (C.116) and note that the proof for (C.117) is similar when the multiserver factor α i (ni ) is included. We let Ni represent the random variable ‘number of customers at node i (in the steady state)’, so that the marginal probability distribution pi (ni ; N ) is

pi (ni ; N ) ≡ Pr = {Ni = ni N customers in network }

= ∑ 1 ρ1 n1 ρ2

n2 ⋅ ⋅ ⋅ ρini ⋅ ⋅ ⋅ ρk

nk .n1 +n2 +⋅⋅⋅+ni−1 +ni+1 +⋅⋅⋅+nk = N −ni

G(N )The complementary marginal cumulative probability distribution is

Pi (ni ; N ) ≡ Pr{Ni ≥ ni N customers in network} ∞

= ∑ ∑ 1 ρ1 n1 ρ2

n2 ⋅ ⋅ ⋅ ρij ⋅ ⋅ ⋅ ρk

nk

j ni n1 +n2 +⋅⋅⋅+ ni−1 +ni+1 +⋅⋅⋅+ nk = N − j G N( )=

j n1 n2 ni−1 ni+1 nk= ∑∞

ρ i ∑ 1 ρ1 ρ2 ⋅⋅⋅ ρi−1 ρi+1 ⋅⋅⋅ ρkj=ni n1 +n2 +⋅⋅⋅+ni−1 +ni+1 +⋅⋅⋅+nk =N − j G(N )

= ∑∞

ρ ij 1 gk −1(N −1) [from Equation (C.110)]

j=ni G(N )

=ρ i

ni

∑∞

ρij −ni gk −1(N − j) = ρi

ni

∑∞

ρ il gk −1(N − ni − l)

G(N ) j=1 G(N ) l =0

=ρi

ni

gk (N − ni ) [from Equation (C.111)] =ρi

ni

G N − ni )(( )G(N ) G N

70

Now pi (ni ; N ) = Pi (ni ; N ) − Pi (ni +1; N )

ρ ni ρ ni +1

= i G(N − ni ) − i G(N − ni −1)G(N ) G(N )

=ρi

ni

[ ( − ni ) − ρiG N − ni −1)] G N (( )G N

and pi (ni ; N )

pi (ni −1; N −1)

ρ ni G(N −1) G(N − n ) − ρ G(N − n −1)= i i i i

G(N ) ρ ini −1 G(N −1− ni +1) − ρ iG(N −1− ni +1−1)

( −1) G N − n ) − ρ G N − n −1) ρ G N −1) ρiG N ( i i ( i i (= =

( ) G N( − ni ) − ρi ( − ni −1) (G N G N G N ) So,

ρ G N( ; ) = i (( )

−1) p n ( −1; p n N N −1) i i i iG N The throughput at node i is

λi (N ) =

Pr{ server busy at node i }⋅ µi = P (1; N )µi = G

ρ (N

i

) G(N −1)µi ,

so that ( −1)

=λ (N )ρiG N i

( ) µiG N and finally

p n N i ( ;i ) = λi ( )Ni ( i −1; N −p n 1)

µi

71

C16.6 Network management traffic/ cyclic queues If we consider a closed network of k nodes such that

i k −1) ⎧1 ( j i= +1,1 ≤ ≤ rij = ⎪⎨1 (i k j = , =1) (C.118)

⎪⎩0 (elsewhere)

then we have a cyclic queue. A cyclic queue is a series queue in a ‘circle’, where the output of the last node feeds back to the first node. An example of such traffic is a collection of mobile agents used for network management. This is a special case of a closed queueing network. So, for single servers at each node, Equations (C.105) and (C.106) give pn n ,..., n = Cρ1

n1 ρ2 n2 ⋅ ⋅ ⋅ ρk

nk (C.119)1, 2 k

with µ ρ = ∑k µ r ρ . Using Equation (C.118) in the traffic equationi i j=1 j ij j

results in ⎧µi−1ρ i−1 (i = 2,3,..., k)

µi ρ i = ⎨⎩µk ρk ( 1) =i

and µi )ρ i−1 (i = 2,3,..., k)

(C.120)⎧(µi−1ρ = ⎨ ( 1) i ⎩(µk µ1 )ρk i =

From Equation (C.120) we have

ρ2 =µ1 ρ1, ρ3 =

µ2 ρ2 =µ1 ρ1,..., ρk −1 =

µ1 ρ1, ρk =µ1 ρ1µ2 µ3 µ3 µk −1 µk

Owing to redundancy, one ρ can be set equal to 1. We select ρ1 = 1, and substitute into Equation (C.119) to obtain

−1 µ N n1

pn1 ,..., nk =

( ) µ2 n2 µ3

n 13 ⋅ ⋅ ⋅ k

nk (C.121)G N µ

Again, G(N ) can be found by summing over all cases n1 + n2 + ⋅⋅⋅+ nk = N or by Buzen's algorithm.

72

The multiple-server case can also be treated similarly. Of course, it is not necessary to develop special versions of Equations (C.105) and (C.107); these results can be used as they are, with the appropriate {rij } given by Equation (C.118).

C16.7 Active networks/dynamic spectra sharing/ multiclass Jackson networks

In this section we discuss multiclass Jackson networks, where, in addition to each class of customers having its own routing structure (active networks) , each class also has its own mean arrival rate, and the mean service times at a node may depend on the particular customer type (class to which the customer belongs) as well. The applications will be in the area of active networks and dynamic spectra sharing. In Basket at al. [10] such multiclass Jackson networks were treated, and product-form solutions for the following three queueing disciplines were obtained: (1) processor sharing (each customer gets a share of and is served simultaneously by a single server) which can be used for dynamic spectra sharing system modeling; (2) ample service; and (3) LCFS (last come first served) with preemptive-resume servicing. They allowed the network to be open for some classes of customers and closed for others. Customers may switch classes after finishing at a node according to a probability distribution; that is, there is a probability ris; jt , that a customer of class s completing service at node i next goes to node j as a class t customer. Exogenous ‘Poisson’ input can be state-dependent (a general birth process), and service distributions can be of the phase type. Basket at al. [10] also considered c-server FCFS nodes, but for these, service times for all classes must be IID exponential; that is, for these nodes all customer types look alike, and service times are exponentially distributed. Kelly [63-65] probably represent the most comprehensive generalization of Jackson networks. Kelly [63,64] also considers multiple-customer classes, and set up a notational structure which allows for unique class service times at multiserver FCFS nodes. In fact, his work is so general that it includes ‘most’ queueing disciplines

73

(for example, all those considered by Basket at al. [10]). Priority dependent on customer class is not considered in these references. However, a price must be paid for this generality in that the description of the state space becomes much more complex. Up until now, the state space could be described by a vector consisting of the number of each class of customer at each node, but now the state space must be described by a complete customer ordering, by type, at each node. Kelly [63-65] considers, as well as exponential service time, Erlang service, which further expands the state-space descriptor to include service phase. Nevertheless, it is proved that the solution is still of product form. Many of these results can be extended to include general service-time distributions. This conjecture is based on the fact that nonnegative probability distributions can always be well approximated by finite mixtures of gamma distributions. This conjecture is proved in Barbour [7]. In Gross and Ince [115] the above multiclass results were applied to a closed network and numerical solutions were obtained.

A great deal of effort has been invested in obtaining computational results for closed multiclass Jackson networks due to their use in modeling computer systems. The basic model generally considers a computer system with N terminals, one for each user logged on. While it is not strictly a closed system, since users log on and off, during busy periods one can assume all terminals are in use, so that there are always N customers (jobs) in the system. These can be at various stages in the system, such as ‘thinking’ at the terminal, waiting in the queue to enter the central processing unit (CPU), being served by the CPU, waiting or in service at input/output stations, and so on. Multiple job classes are an important part of any such model. In Bruell and Balbo [23] a number of computing algorithms are developed to treat such models.

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