Computing the Performance Measures of a Multichannel Loss System

• Multichannel system with losses

  Let us consider the classical Erlang problem.
  There are n channels receiving a flow of requests with rate λ. The service flow has rate μ. Find the limiting probabilities of the system's states and its performance measures.
  System S (the QS) has the following states (numbered by the number of requests in the system): S0, S1, S2, …, Sk, …, Sn, where Sk — is the state of the system when it contains k requests, i.e., k channels are busy.
  The state transition diagram of the QS corresponds to a birth-and-death process and is shown in Fig. 7.
  
  Fig. 7
  The flow of requests successively moves the system from any left state to the neighboring right state with the same rate λ. The rate of the service flow, however, which moves the system from any right state to the neighboring left state, constantly changes depending on the state. Indeed, if the QS is in state S2 (two channels busy), it can move to state S1 (one channel busy) when either the first or the second channel finishes service, i.e., the combined rate of their service flows will be 2μ. Similarly, the combined service flow that moves the QS from state S3 (three channels busy) to S2 will have rate 3μ, i.e., any of the three channels can become free, and so on.
  In formula (16) for the birth-and-death scheme, we obtain for the limiting probability of the state
  (23)
  where the expansion terms represent the coefficients of p0 in the expressions for the limiting probabilities p1, p2, …, pk, …, pn. The quantity
  (24)
  is called the reduced arrival rate or channel load intensity.It expresses the mean number of requests arriving during the mean service time of one request. Now
  p1=ρ·p0 (25)
  (26)
  Formulas (25) and (26) for the limiting probabilities are called the Erlang formulas in honor of the founder of queueing theory.
  The loss probability of the QS is the limiting probability that all n channels of the system will be busy, i.e.
  (27)
  The relative throughput — the probability that a request will be served:
  (28)
  The absolute throughput:
  (29)
  The mean number of busy channels k is the expected value of the number of busy channels:
  
  where pk — are the limiting probabilities of the states, determined by formulas (25), (26).
  However, the mean number of busy channels can be found more simply, if we take into account that the absolute throughput of the system A is nothing other than the rate of the flow of requests served by the system (per unit time). Since each busy channel serves on average μ requests (per unit time), the mean number of busy channels
  (30)
  or, taking into account (29), (24):
  (31)

See also

• [[b1987]]

See also

• Computing the Performance Measures of a Single-Channel Loss System