Question 7.EX.35: Satellites are launched according to a Poisson process with ......

Satellites are launched according to a Poisson process with rate λ. Each satellite will, independently, orbit the earth for a random time having distribution F. Let X(t) denote the number of satellites orbiting at time t .
(a) Determine P{X(t) = k}.
Hint: Relate this to the M/G/∞ queue.

(b) If at least one satellite is orbiting, then messages can be transmitted and we say that the system is functional. If the first satellite is orbited at time t = 0, determine the expected time that the system remains functional.
Hint: Make use of part (a) when k = 0.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

(a) We can view this as an M / G / \infty system where a satellite launching corresponds to an arrival and F is the service distribution. Hence,
P\{X(t)=k\}=e^{-\lambda(t)}[\lambda(t)]^k / k !
where \lambda(t)=\lambda \int_0^t(1-F(s)) d s.
(b) By viewing the system as an alternating renewal process that is on when there is at least one satellite orbiting, we obtain
\lim P\{X(t)=0\}=\frac{1 / \lambda}{1 / \lambda+E[T]}
where T, the on time in a cycle, is the quantity of interest. From part (a)
\lim P\{X(t)=0\}=e^{-\lambda \mu}
where \mu=\int_0^{\infty}(1-F(s)) d s is the mean time that a satellite orbits. Hence,
e^{-\lambda \mu}=\frac{1 / \lambda}{1 / \lambda+E[T]}
so
E[T]=\frac{1-e^{-\lambda \mu}}{\lambda e^{-\lambda \mu}}

Related Answered Questions

Question: 7.EX.18

Verified Answer:

We can imagine that a renewal corresponds to a mac...
Question: 7.EX.3

Verified Answer:

By the one-to-one correspondence of m(t) and F, it...