Measures n n 3 i0 i results in the

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Unformatted text preview: 9, 6/19, 6/19, 4/19}. The probability of a caller arriving to see all three lines busy equals π3 , so ￿ ￿ 4 15 λeff = λ (1 − π3 ) = 1 1 − = per minute 19 19 Because there are three servers, there is no queue, and therefore the waiting time is zero. Wq = 0 minute 2. This queuing system is an example of a CTMC called a birth-death process with birth rates λi = λ/(i + 1) and death rates µi = µ. The Rate diagram can be written as follows By writing the equations for the cuts, and setting η0 = 1, the following is a valid invariant measure can be constructed. ￿ n− 1 =0 λi ηn = ￿in ∀n = 1, 2, . . . i=1 µi Applying this formula to our instance yields, with ρ = λ/µ, ￿ n− 1 i=0 ηn = ￿n λ i+1 i=1 Recalling the formula ex = ￿∞ n=0 µi = λn ρn = µ n n! n! xn /n!, yields ∞ ￿ ηn = n=0 ∞ ￿ ρn = eρ n! n=0 Using the relationship between the stationary distribution and the invariant measure, ηn ρn e − ρ πn = ￿ ∞ = n! ( i=0 ηi ) Notice that πn is equal to the probability mass function for a Poisson random variable with mean ρ, so the expected number customers in the long run is ρ. 3 3. (a) The generator is 1 − 12 1 15 0 0 0 1 12 0 0 3 − 20 1 12 0 2 15 − 13 60 1 12 0 2 15 − 13 60 0 0 2 15 0 0 0 1 12 2 − 15 (b) The rate diagram is given by Using the cut method generates the following equations 1 1 12 η0 = 15 η1 1 η = 2η 12 1 15 2 1 2 12 η2 = 15 η3 1 2 12 η3 = 15 η4 Substituting η0 = 1 gives η = {1, 1.25, 0.7813, 0.4883, 0.3052} with The stationary distribution is given by the relation, ηn πn = ￿ ￿ 4 i=0 ηi ￿4 n=0 ηn = 3.8247. ￿ This calculation gives the stationary distribution π = {0.2615, 0.3268, 0.2043, 0.07979}. (c) The steady state probability the call is accepted is the complement of the probability that the call will be transferred. A call is transferred if and only if four calls are currently in the system. P(Call Accepted) = 1 − P(Call Transferred) = 1 − π4 = 0.07979 (d) A call does not have to wait if the an ambulance is currently available, or equivalently, at most a single ambulance is responding to a call. P(No Waiting) = P(At Most One Ambulance Responding) = π0 + π1 = 0.5882 (e) Because there are two ambulances, the number of calls waiting is equal to the number in the system minus two. Nq = 1 π3 + 2 π4 = 0.2872 call 4 (f) The throughput of the system is the arrival rate times the probability that a call is accepted. 1 λeff = λ (1 − π4 ) = (1 − 0.07979) = 0.07668 per minute 12 Using Little’s Law the average waiting time for a call to be dispached is Wq = Nq = 3.7458 minutes λe ff (g) Caller X will have an ambulance dispatched as soon as two ambulances finish responding to current calls. Let N be a homogeneous Poisson process with rate 2/15 per minute, then this probability is equivalent to having at least two service completions in 20 minutes. P(Caller X has ambulance dispatched in less than 20 minutes) = P(At least two service completions in 20 minutes) = P(N (0, 20) ≥ 2) = 1 − P(N (0, 20) = 0) − P(N (0, 20) = 1) 2 2 2 = 1 − e−20· 15 − 20 · e−20· 15 15 = 0.7452 5...
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