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Probability Problem 1.15 From Statistical and Thermal Physics, Reif

  1. Aug 27, 2012 #1
    1. The problem statement, all variables and given/known data
    A set of telephone lines is to be installed so as to connect town A to town B. The town A has 2000 telephones. If each of the telephone users of A were to be guaranteed instant access to make calls to B, 2000 telephone lines would be needed. This would be rather extravagant. Suppose that during the busiest hour of the day each subscriber in A requires, on the average, a telephone connection to B for two minutes, and that these telephone calls are made at random. Find the minimum M of telephone lines to B which must be installed so that at most only 1 percent of the callers to town A will fail to have immediate access to a telephone line to B.

    2. Relevant equations
    Book mentions approximating the distribution by a Gaussian distribution to facilitate the arithmetic.

    3. The attempt at a solution
    What does the author mean by "the random telephone calls"? Does that mean that every 2 minutes a random number of subscriber lines is turned on (assuming 0 to 2000)? By the way if calls are evenly distributed in the hour every two minutes- I think we would need 2000/30 lines. If calls are made at random, it wouldn't be evenly distributed in the hour. Help is appreciated.
     
  2. jcsd
  3. Aug 27, 2012 #2
    I can answer your last question. The statement should be interpreted as "If a random phone call is made during the busiest hour of the day, it will require on average 2 minutes." This means that any randomly made call during these hours will require a line for ~2 minutes. If we assume 1 call is made then for 2 minutes n-1 lines are available for use. If it wasn't 1:30am I would provide you more help but I gotta be up at 6:30! Good luck.
     
  4. Aug 27, 2012 #3

    Ray Vickson

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    I guess that what is wanted is: individual calls are of random length with mean duration = 2 minutes. The starting times of the calls are random, perhaps uniformly distributed within the hour.

    Without knowing more about the call-length distribution it is not possible to say very much about the "congestion" at times during the hour.

    We might try to treat this as a queueing problem with Poisson arrivals of calls and exponential call duration. We could try, for example, to treat this as an M/M/K/N queue, where K = number of lines and N = capacity (making calls plus waiting for a line), so N = 2000. Since, on average, 2000 calls will be made during 60 minutes, the arrival rate is λ = 2000/60 = 100/3 (calls per minute). The average call duration is 2 minutes, so the service rate per server is μ = 1/2 (calls/minute). Standard queueing formulas allow one to compute the probability of no queueing (which is basically what the question asks for) as a function of K, and to vary K until the desired performance is achieved. (For the record: for K = 87 lines I get that 98.9% of callers do not wait, while for K = 88 lines, 99.2% do not wait. Of course, these are *averages*, so in any particular hour there may be more, or fewer, who need to wait.)

    There are several reasonable objections to such a queueing model, and I suspect that such a model is not what the author had in mind. I also doubt that a book on "statistical physics" will have the necessary material on queueing theory, because it plays essentially no role in physics. I can imagine a few other possible approaches, but all of them seem much harder to deal with than a queueing model.

    RGV
     
    Last edited: Aug 27, 2012
  5. Aug 27, 2012 #4

    Ray Vickson

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    Now that I have had a night's sleep, I realize that the problem better fits in with the Erlang B-formula, giving the probability of blocking when no queue is allowed---that is, if users try to place calls when no line is available, they go away and try again later. The actual formula is quite simple, and furthermore, the formula is valid for *any* call-length distribution (not just for exponential duration that is assumed in the M/M/K/N model). The same lambda and mu values apply as in the previous post.

    For a derivation of the Erlang loss formula (and others) in the case of exponential call-length distribution, see
    http://onlinelibrary.wiley.com/doi/10.1002/9780470758946.app1/pdf .

    RGV
     
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