This preview shows page 1. Sign up to view the full content.
Unformatted text preview: next departure will be exponential(10). However, if the output process
is Poisson, then the number of departures in disjoint intervals are independent.
Proof for s = 1. Our ﬁrst step in making the result in Theorem 4.8 seem
reasonable is to check by hand that if there is one server and the queue is
in equilibrium, then the time of the ﬁrst departure, D, has an exponential
distribution with rate . There are two cases to consider.
Case 1. If there are n 1 customers in the queue, then the time to the next
departure has an exponential distribution with rate µ, i.e.,
fD (t) = µe µt Case 2. If there are n = 0 customers in the queue, then we have to wait an
exponential( ) amount of time until the ﬁrst arrival, and then an independent
exponential(µ) for that customer to depart. If we let T1 and T2 be the waiting 142 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS times for the arrival and for the departure, then breaking things down according
to the value of T1 = s, the density of D = T1 + T2 in this case is
fD (t) =
e s · µe µ(t s) ds = µe µt
e ( µ)s ds
0 = µt µe µ ⇣ 1 ( e µ)t...
View Full Document
- Spring '10
- The Land