State N 2.6 The M/M/1/N Queueing System: The Finite Buffer Case

The blocking probability for various values of offered load, λ/μ, when N=5

The blocking probability for various values of offered load, λ/μ=0.9 and buffer size is varied

2.7 The M/M/∞ Queueing System: Infinite Number of Servers

The state dependent arrival and service rates are

The M/M/m Queueing Systems: m Parallel Servers with a Queue

The state dependent arrival and service rates m Parallel Servers

Summing up over all the states one has

In the above ρ=λ/mμ

In telephone engineering one quantity of interest is the probability that all the server in this system are busy

Substituting for p 0 this becomes

2.9 The M/M/m/m Queue: A Loss System

As Figure 2.22 shows, the queueing system simply consists of m parallel servers

Erlang’s B formula: Erlang formula of the first kind, in Europe Summing over all the m +1states leads to

2.10 Central Server CPU Model

The mean waiting to get through the queueing system

2.11 Transient Solution of the M/M/1/∞ Queueing System 1.Equilibrium case: A set of linear equations 2. Transient case: A set of linear differential equations

The moment-generating function The state probabilities form a sequence of numbers that can be transformed, like any numerical sequence, into a “frequency” like domain. It turn out to relatively easy to obtain the frequency domain solution.

The details of the solution techniques are:

The Transient Solution We first take the differential equations of the previous section, multiply both sides by z n, and sum from n=1 to∞:

Recall that to get proper moment generating functions the indexes must start at n=0 so we have

But three of the items here are just the boundary differential equation at n=0 and so drop out, leaving:

We can rewrite this further

Now identifying the moment-generating functions (which sometimes requires adding and subtracting terms) one has

The initial condition to this equation involves i customers at time 0:

To solve this partial differential equation, one can use Laplace transform which are defined as For the left-hand side of the boxed equation one can integrate by parts to find its transform:

Now take the Laplace transform of the boxed differential equation and multiply out the constants so that

We can now solve for P * (z,s): At this point one must determine P 0 * (s) and invert the transform. The interested reader can [GROS] (Page 99~103) for details. The well-known result for the time evolution of the state probabilities is

Where ρ=λ/μ, n(0)=i customers, And I l () is the modified Bessel function of the first kind of order l:

Speeding up the Computation An efficient procedure for the calculation of P n (t) This is done by first letting a=(2μt) 1/2 and b=(2λt) 1/2

Up to seven times faster

2.12 The M/G/1 Queueing System M/G/1: Arrivals  Poisson process, Service  General (arbitrary) probability distribution Mean number in the M/G/1 queueing system  mean waiting time, 

Mean number in the M/G/1 Queueing System Pollaczck-Khinchin mean value formula A key point is that that the state probabilities as these departure instant are in fact equal to the state probabilities at any time instant. GROS, Kendall’s approach

The recursion the determination of a succession of elements The number in the system immediately after the (i+1) st departure instant is n i+1. This number includes both those customers in the queue and the customer in service. This number is also equal to the number in the system immediately after the ith departure instant minus 1 (because of the departure of a customer at the (i+1) st departure instant. This number of arrival we will call a i+1. For this equation to work there must be at least one customer in the system all the departure instant. Thus

We still have to take care of the case n i =0. This is actually simpler than the case n i >0. The number in the queueing system at the (i+1) st departure instant is just the number of arrivals during the service period of the first customer to arrive to the empty queueing system. This first customer is the one that departs at the (i+1) st departures instant and is not counted in determining a i+1

Bringing this together we have We would like to express this as a single equation. To do this, we will introduce the unit step function:

Now we can rewrite the equation for n i+1 as If we square both side of this recursion and take expectations we have 6 items

In order to solve for the Pollaczek-Khinchin formula we need to solve for each of these terms, one by one, we will now do this Sundry (miscellaneous) Results E[(n i+1 ) 2 ], E[n i 2 ] These two terms should be equal in equilibrium so that they will cancel in (2.189). E[(u(n i )) 2 ]: Because of the nature of the step function u(n i ) 2 =u(n i ), we can replace E[(u(n i )) 2 ] with E[(u(n i ))]. But what is E[(u(n i ))]?

P[busy] Consider an interval I. The server will be busy for a time period equal to I – I p 0 The number of customers served is (I – I p 0 )μ Where μ is the mean service rate of the queueing system. In equilibrium the quantity above should be equal to the number of arrivals

Note that no restrictive assumptions have been made either the arrival process or the service times. Thus this result is valid for a G/G/1 queueing system, that is, one with an arbitrary process and service process time distribution. 2E[(n i, u(n i ))]: Clearly n i u(n i )=n i, so we can say

2E[ u(n i )a i+1 ]: The number of customers that arrive into the system between the ith and (i+1)st departure instant, a i+1, is independent of the number in the queueing system immediately after the ith departure instant, n i. Thus We already know that E[u(n i )]=ρ. To determine E[a i+1 ] one can look at the original recursion for E[n i+1 ] And take the expectation of both sides:

In equilibrium E[n i+1 ]=E[n i ], so one is left with One can now say that: Using the same argument as before, the two quantities are statistically independent, so 2 E[n i+1 a i+1 ]:

Putting It All Together Now we can substitute these sundry results into the equation:

Solve for E[n i ], we have We will assume for now, and prove in section , that the expected number of customers in the queueing system at the departure instants is the same as the expected number of customers at any instant of time: We can also say, because the input is Poisson, that the expected number of arrivals between two successive departure instants is a time-invariant quantity:

We thus have We almost have a simple expression for the mean number in the M/G/1 queueing system in equilibrium or E[n]. What is lacking is a simple expression for E[a 2 ]

What is E[a 2 ] ? The quantity “a” is the number of arrivals during a typical servicing period in equilibrium. Processing as in [GROS] we can write

We need to solve for VAR [a]. Let us the service time. Then an interesting relationship [GROS] for two random variables X and Y that we can make use of is [PARZ]

One can solve for the two components of this sum: Here σ s 2 is the variance of the service time. So we have

The Pollaczek-Khinchin Mean Value Formula Substituting (2.216) for E[a 2 ] lead to

We can get another form by expressing E[n] as E[n] = n as a function of the mean arrival rate, mean service rate, and variance of the service time distribution.

Little’s Law n: the mean customers in a queueing system λ: the mean arrival rate to the queueing system τ : the mean waiting time to get through the queueing system

Page 82

Why We Use Departure Instants The Pollaczek-Khinchin mean value formula: the assumption of the state probabilities at departure instants are the same as the state probabilities at any instant. We will take a realization of the queueing process over a long interval. The number in the queueing system at time t will be n(t). The interval will stretch from time 0 to time T.

Why We Use Departure Instants The number of arrivals in the interval (0, T) that occur when the number of customers in the system is n will be called a n (t). The number of departures bringing the system from state n+1 to state n in the interval (0, T) will be called d n (t).

It should also be true from first principles that Where d(T) and a(T) are the total number of departures and arrivals, respectively, over (0,T). One can now define the state probabilities for the departure instants as This empirical definition of the state probabilities becomes exact in the limit. Now

The denominator (the part of a fraction that is below the line) can be recognized as the equation we just discussed, (2.223). The numerator was created by adding and subtracting a n (T) to d n (T). As T  ∞, d n (T)-a n (T) will never be greater than one, as has been mentioned. As T  ∞, and a(T) and a n (T) will  ∞, and

Why We Use Departure Instants What this proves directly is that state probabilities seen at the departure instants are the same as the state probabilities seen at the arrival instants. But going back to our Poisson process coin-flipping analogy, the arrival instants are random points in time and independent of the queueing system state. Thus it should seem as the state probability seen at any time. This completes the proof that looking at state probabilities at departure instants for the M/G/1 queueing system is the same as looking at state probabilities for this queueing system at any instant.

Probability Distribution of the Number in the Queueing System Define a matrix of transition probabilities as

The equilibrium equations π: The state probabilities, P : A matrix of transition probabilities Service time density, b(s), the probability of j arrivals between the ith and (i+1)st departure instance.

π: The state probabilities

Given K, or K(z), we can obtain the transfer function π(z) or the state probability π