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A problem set from a university course, ece 567, on communication network analysis, focusing on discrete-state markov processes and poisson processes. The problems cover topics such as poisson merger, poisson splitting, poisson method for coupon collector's problem, sum of a random number of random variables, mean hitting time for a simple markov process, a two-station pipeline in continuous time, and simple population growth models. Students are expected to solve these problems using concepts from markov processes, poisson processes, and probability theory.
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PROBLEM SET 1 Wednesday, September 6
Topic: Discrete-state Markov processes including Poisson processes
Assigned reading: Sections 1.1-1.5 of the notes. (Kleinrock, Queueing Systems Vol. I , Sections 2-2-2.5, has related material.)
Problems to be handed in:
limk→∞ P [A(k ln k + kc) ≥ k ln k + kc′^ ] =
0 if c < c′ 1 if c > c′ (d) Let d(k, n) denote the probability that the collection is complete after n coupon arrivals. (The letter “d” is used here because the number of coupons, n, is deterministic.) Show that for any k, t, and n fixed, d(k, n)P [A(t) ≥ n] ≤ p(k, t) ≤ P [A(t) ≥ n] + P [A(t) ≤ n]d(k, n). (e) Combine parts (c) and (d) to identify limk→∞ d(k, k ln k + kc).
1 − a 0 a 0 0 0. 5 0 0. 5 0 0 1 0
for some constant a with 0 ≤ a ≤ 1. (a) Sketch the transition probability diagram for X and give the equilibrium probability vector. If the equilibrium vector is not unique, describe all the equilibrium probability vectors. (b) Compute E[min{n ≥ 1 : X(n) = 3}|X(0) = 0].
P (z, t) = ze−λt 1 − z + ze−λt
(c) Find E[N(t)]. (d) Solve for π(t). (e) Another population evolves deterministically. The first cell splits at time λ −^1 , and thereafter each cell splits into two exactly λ −^1 time units after its creation. Find the size of the population of this population as a function of time, and compare to your answer in part (c).
i −q^ ii^ π^ i^ , where the sum is over all^ i^ in the state space, is finite.^ Show that the equilibrium distribution for the jump chain (X J^ (k) : k ≥ 0) is given by π (^) iJ = −Bq (^) ii π (^) i. (So π and π J^ are identical if and only if q (^) ii is the same for all i.)