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HELP (:

  1. Oct 13, 2011 #1
    HELP plz (:

    Ernie is always keen to go to the Family on Saturday nights and usually arrives around 11pm. On a particular night, Bert decides to start drinking early and he is already inside the Family while Ernie is on his way. Ernie has noticed over time that the expected number of people in the queue to get into the family is 10.5 around this time on a Saturday. He has also noticed that the security guard usually checks IDs at an average rate of 1 every 2 seconds. However the security guard is sometimes distracted by his friend Lady Moneypenny who seems to be around 20% of the time chatting to the security guard. During this time the security guard does not check the IDs as carefully, at an average rate of 2 per second. Ernie remembers that Bert has his wallet with his ID card. Just before Ernie arrives at the queue he calls Bert to bring down his ID. Assuming that the time between people arriving at the queue is Exponentially distributed with an expectation of 2 seconds, how long does Ernie expect to have to wait in the queue before having to show his ID?
     
  2. jcsd
  3. Oct 22, 2011 #2
    Re: HELP plz (:

    i solve the problem like this can you tell me is i am in the right trick:


    Since the queue is Exponentially distributed with an expectation of 2 seconds
    Then E(z)=1/μ
    so 2=1/μ therefor μ = 0.5
    and from the Little’s Formula :
    E(N)=μ E(D)
    we find that 10.5=μ E(D)
    so E(D)= 21
    and since the delay time in queue = Waiting in the queue time + Service times
    E(D) = E(Wq) + E(s)
    E(s)=1/(average time in the services)
    average time in the services = μ/P = 0.5/(20%) = 2.5
    so: E(s)1/2.5 =0.4

    sup in E(D) = E(Wq) + E(s)
    21 = E(Wq) + 0.4
    so :

    E(Wq) = 20.6
     
  4. Oct 23, 2011 #3

    Ray Vickson

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    Re: HELP plz (:

    Your expected service time is incorrect. The service time per customer is expl(1/2) [mu = 1/2] with probability 8/10 and is expl(2) [mu = 2] with probability 2/10 (times in seconds), so the expected service time is ET = (8/10)*2 + (2/10)*(1/2) = 17/10 (sec), and lambda = 1/2 (customers/sec), so the traffic intensity is rho = lambda*ET = 17/20 = 0.85.

    RGV
     
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