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STA107 Lecture 31 Bernoulli Queue

STA107 Lecture 31 Bernoulli Queue. Queue – a stochastic process that has arrivals and services (departures) Bernoulli Queue - Arrivals happen according to Bernoulli counting process with probability of arrival in a frame p A.

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STA107 Lecture 31 Bernoulli Queue

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  1. STA107 Lecture 31Bernoulli Queue Queue – a stochastic process that has arrivals and services (departures) Bernoulli Queue - Arrivals happen according to Bernoulli counting process with probability of arrival in a frame pA. Services are completed (ie departures) happen according to Bernoulli counting process with probability of departure in a frame pS

  2. Bernoulli Queue In a frame, there can be: • 1 arrival, 0 departure  queue length increases by 1 • 0 arrival, 1 departure  queue length decreases by 1 • 0 arrival, 0 departures  queue length stays the same • 1 arrival, 1 departures  queue length stays the same

  3. n = # frames in one unit of (standard) time • D = time length of each frame • nD = length of one (standard) time unit Arrivals • pA is P(Arrival) in one frame • lAis success rate in one (standard) unit of time • lA = npA • lA = pA /D Services/Departures • pS is P(Service/Departure) in one frame • lSis success rate in one (standard) unit of time • lS = npS • lS = pS/D

  4.  Let i be number of individuals in the queue in any particular frame. Transitions to the next frame have probabilities: • P(ii+1) = pA(1- pS) • P(ii-1) = pS(1- pA) • P(ii) = pA pS + (1- pA) (1- pS)

  5. Example 1 • Arrival rate 24/hour; expected service time 1.5 minutes; set up as single server Bernoulli queue with 30second frames. Soln: lA = 24/hour  for 1 frame pA = 0.2; E(S) =1.5min = 3 frames  pS = 1/3 • If there are i in the queue in frame m , what is the probability there are still i in the queue in frame m+1? Soln: pA pS + (1- pA)(1- pS) = (0.2)(1/3) + (0.8)(2/3) = 0.6

  6. Example 2 • Bank machine gets an average of 260 customers per business day (9-5); average service time is 120 seconds. Set this up as single server Bernoulli queue with 15second frames. Soln: lA = 260 per business day  for one 15second frame pA = 0.1354; E(S) = 120sec = 8 frames  pS = 1/8 = 0.125 • If there is nobody in the queue between 2pm and 2:00:15 pm, what is the probability there is still nobody in the queue 30 seconds after 2pm? Soln: 1- pA= 1-0.1354 = 0.8646

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