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18 Frame 5 Equilibrium calculations for some typical cases The Poisson case The simplest assumption is that the number of sources (N) and the number of servers (n) are both unlimited, while each source j contributes with an infinitesimal call rate (λj), so that the overall call rate is state independent λi = λ = N ⋅λj . The departure rate in state i is proportional to the number in service, which is of course i: µ i = i ⋅ µ= i/s, where s is the mean holding time per call. The parameters λ and µ are constants, and equation (43) becomes p(i) ⋅λ= p(i + 1) ⋅ µ ⋅ (i + 1) (48) leading to p(1) = (λ/µ) ⋅ p(0) p(2) = (λ/2µ) ⋅ p(1) = (λ/µ) 2 /2! ⋅ p(0) : : : p(i) = (λ/µ) i /i! ⋅ p(0) : : Thus, using equation (44) we get ∞ ∑ p(i) = p(0)⋅ (λ / µ ) i=0 i ∞ ∑ / i! i=0 = p(0)⋅e λ / µ −λ / µ = 1 ⇒ p(0) = e and p(i) = e –λ/µ ⋅ (λ/µ) i /i! Using Little’s formula: A = λ ⋅ s = λ/µ, we obtain p(i) = e –A ⋅ Ai /i! (49) which we recognise as the Poisson distribution with parameter A. The mean value of the distribution is also A, as it ought to be. Recalling the interpretation of Little’s formula we can state that A = λ/µ = λ ⋅ s = average number of customers being served = average traffic intensity. The truncated Poisson case (the Erlang case) In practice all server groups are limited, so the pure Poisson case is not realistic. A truncation by limiting the number of servers to n leads to a recursive equation set of n equations (plus one superfluous) of the same form as that of the unlimited Poisson case above. The normalising condition is different, as n ∞ ∑ p(i) = 1and ∑ p(i) = 0 i=0 i=n+1 The truncated Poisson distribution thus becomes 0 ≤ i ≤ n p(i) = 0, i > n (50) Any call arriving when i < n will find a free server. Since λ is independent of i, the probability of an arbitrary call finding all servers busy is equal to the probability that i = n: (51) This is the Erlang congestion (loss) formula for an unlimited number of sources. The formula is not very practical for calculations, so there are numerous tabulations and diagrams available. In principle one must distinguish between time congestion and call congestion, time congestion being the probability of finding all servers busy at an arbitrary point in time, and call congestion being the probability that an arbitrary call finds all servers busy. In the Erlang case the two probabilities are identical, since all calls are Poisson arrivals, independent of state. The binomial (Bernoulli) case We assume a limited number of traffic sources (N) and enough servers to cover any need, i.e. n ≥ N. Furthermore, each free traffic source generates λ calls per time unit (λ is here the individual rate, rather than the total rate, and busy sources naturally do not generate any calls). Departure rate per call in progress is unchanged µ = 1/s. The equilibrium equations then take the form p(i) ⋅ (N – i) ⋅ λ = p(i + 1) ⋅ µ ⋅ (i + 1) 0 ≤ i ≤ N (52) By recursion from i = 0 upwards we get p(i) = p(0)⋅ b = λ/µ = offered traffic per free source N i ⎛ ⎞ ⎛ λ ⎞ ⎝ i ⋅ ⎠ ⎜ ⎟ = p(0)⋅ ⎝ µ ⎠ N ⎛ ⎞ ⎝ i ⎠ ⋅bi The normalising condition is (53) The p(i)-expression above is seen to consist of the common factor p(0) and the ith binomial term of (1 + b) N . Thus, by the normalising condition, we obtain p(0) = (1 + b) –N and p(i) = P(n, A) = E(n, A) = N ∞ ∑ p(i) = 1and ∑ p(i) = 0 i=0 i= N +1 p(i) = N ⎛ ⎝ i A i i! A j n ∑ j =0 j! ⎞ ⎠ ⋅bi ⋅ 1+ b A n n! A j n ∑ j = 0 j! ( ) − N = N ⎛ ⎞ ⎝ i ⎠ ⋅ai N −i ⋅( 1− a) Here, a = b/(1 + b) = λ /(λ + µ) = offered traffic per source. (54)
Offered traffic per source takes into account that a busy source does not initiate any calls. Since there are at least as many servers as there are sources, there will be no lost calls. This fact is a clear illustration of the difference between time congestion and call congestion when N = n: Time congestion = P{all servers busy} = p(n) = p(N) = aN Call congestion = P{all servers busy when call arrives} = 0 For N < n, time congestion = call congestion = 0 The truncated binomial case (the Engset case) The truncated binomial case is identical to the binomial case, with the only difference being that N > n. This condition implies that there are still free sources that can generate calls when all servers are occupied, and there will be call congestion as well as time congestion. The statistical equilibrium equations are the same, whereas the normalising condition is n ∞ ∑ p(i) = 1and ∑ p(i) = 0 i=0 i=n+1 This leads to an incomplete binomial sum and a not quite so simple distribution: ⎛ N⎞ ⎝ i ⎠ p(i) = ⋅bi n ⎛ N⎞ j ∑ ⎝ j ⋅b ⎠ j =0 Time congestion can then be expressed by holding times are very short (
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