m n m 1 n i 2 m n c 1 1 b m

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Unformatted text preview: 0 µ = µ e µt t e The probability of 0 customers in equilibrium is 1 ( /µ) by (4.23). This implies the probability of 1 customer is /µ, so combining the two cases: fD (t) = µ µ · µ e µ µt t e + µ · µe µt At this point cancellations occur to produce the answer we claimed: e µt e t +e µt =e t We leave it to the adventurous reader to try to repeat the last calculation for the M/M/s queue with s > 1 where there is not a neat formula for the stationary distribution. Proof of Theorem 4.8. By repeating the proof of (1.13) one can show Lemma 4.9. Fix T and let Ys = XT chain with transition probability pt (i, j ) = ˆ s for 0 s T . Then Ys is a Markov ⇡ (j )pt (j, i) ⇡ (i) Proof. If s + t T then P (Ys+t = j |Ys = i) = = P (XT P (XT (s+t) = j, XT s = i) P (Ys+t = j, Ys = i) = P (Ys = i) P (XT s = i) = j )P (XT s = i|XT (s+t) = j ) ⇡ (j )pt (j, i) (s+t) = ⇡ (i) ⇡ (i) which is the desired result. If ⇡ satisfies the detailed balance condition ⇡ (i)q (i, j ) = ⇡ (j )q (j, i),...
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