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A Statistical Mechanics Problem

  • Thread starter basma
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  • #1
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Hi all,
I have this problem which I don't even know how to start it! Any hint will be appreciated.

"A set of telephone lines is to be installed so as to connect between town A and 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 supscriber 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 minium number M of telephone lines to B which must be installed so that at mostonly 1% of the callers of town A will fail to have immediate access to a telephone line to B. (Suggestion!: approximate the distribution by a Gaussian distribution to facilitate the arithmetic)"

This is it, I hope I will have some hints on this problem.

Basma
 

Answers and Replies

  • #2
Andrew Mason
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basma said:
Hi all,
I have this problem which I don't even know how to start it! Any hint will be appreciated.

"A set of telephone lines is to be installed so as to connect between town A and 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 supscriber 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 minium number M of telephone lines to B which must be installed so that at mostonly 1% of the callers of town A will fail to have immediate access to a telephone line to B. (Suggestion!: approximate the distribution by a Gaussian distribution to facilitate the arithmetic)"

This is it, I hope I will have some hints on this problem.

Basma
First work out the number of lines you would need if the calls were all evenly distributed throughout the busy hour.

AM
 
  • #3
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You mean I should divide 1980 by 30 min (average for one call is 2 min)?

basma

I would appreciate any OTHER hints for this problem really!
 
Last edited:
  • #4
Andrew Mason
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basma said:
You mean I should divide 1980 by 30 min (average for one call is 2 min)?
The average number of concurrent calls would be 2000/30 or 67. Assume you have 2000 lines. If you were to sample each second during the hour (to get a large enough number of samples) and plot the distribution (ie the number of occurrences of x concurrent calls vs. x) you would end up with a distribution about a peak of 67 calls. It is not gaussian (it is not centred on the 67 because it can go up to 2000 but not below 0) but it is approximately gaussian.

That is to say, of the 3600 samples, you would have more samples of 67 and fewer samples of all others getting progressively smaller sample sizes as the number of calls increases or decreases from 67 (min 0 and max 2000). You then have to find the number of concurrent calls (ie the point on the x axis) that corresponds to the area under the graph to its left (number of samples) being 99% of the total area (in this case the total area is the total number of samples = 3600). If you had that many lines, this would mean that only 1 percent of the time would the number of calls exceed the number of lines. So a person trying to get a line would find that they could get a line 99% of the time.

AM
 
Last edited:
  • #5
6
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Thank you!
I could do it. I even used the erf to find teh area.

Basma
 
  • #6
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I'm sorry to bring a post from the dead but I have problem with that question
(or that book for that matter)
I've figured out the 67 thingy but I didn't understand the area calculation.

can anyone help me here?
 
  • #7
1
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could anyone offer a little more help on the area of calculation for this problem?
 

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