To prove that there is one we return to the general

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: wait for an exponential amount of service with rate µ2 . - µ1 - µ2 - Our main problem is to find conditions that guarantee that the queue stabilizes, i.e., has a stationary distribution. This is simple in the tandem queue. The first queue is not a↵ected by the second, so if < µ1 , then (4.23) tells us that the equilibrium probability of the number of customers in the first queue, 1 Xt , is given by the shifted geometric distribution ✓ ◆m ✓ ◆ 1 P (Xt = m) = 1 µ1 µ1 In the previous section we learned that the output process of an M/M/1 queue in equilibrium is a rate Poisson process. This means that if the first 2 queue is in equilibrium, then the number of customers in the queue, Xt , is itself an M/M/1 queue with arrivals at rate (the output rate for 1) and service rate µ2 . Using the results in (4.23) again, the number of individuals in the second queue has stationary distribution ✓ ◆n ✓ ◆ 2 P (Xt = n) = 1 µ2 µ2 To specify the stationary distribution of the system, we need to know the 1 2 joint distribution of Xt and Xt . The answer is somewhat remarkable: in 144 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS equilibrium the two queue lengths are independent. ✓ ◆m ✓ ◆ ✓ ◆n ✓ 1 2 P (Xt = m, Xt = n) = 1 · 1 µ1 µ1 µ2 µ2 ◆ (4.27) Why is this true? Theorem 4.11 implies that the queue length and the depart...
View Full Document

Ask a homework question - tutors are online