Probability, Minimal Telephone Lines needed

Click For Summary

Homework Help Overview

The problem involves determining the minimal number of telephone connections needed to connect two cities, A and B, given that city A has 2000 telephones. Each user in city A requires a connection to city B for an average of two minutes during a specified time period, with the goal of ensuring that no more than 1% of calls are unable to connect due to insufficient lines. The problem requires the use of a Gaussian distribution to approximate the probability of occupied lines.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the average number of lines needed based on call duration and working hours. There is an exploration of the probability that a given phone occupies a line and how to sum these probabilities across multiple phones. Questions arise regarding the application of Gaussian distribution and the interpretation of its axes.

Discussion Status

Participants are actively engaging with the problem, sharing their reasoning and calculations. Some have provided insights into the mean and variance of the random variables involved, while others express uncertainty about the next steps in applying the Gaussian distribution. There is no explicit consensus on the final approach, but productive discussion is ongoing.

Contextual Notes

Participants note the constraints of the problem, including the assumption of random call distribution and the requirement to limit the probability of call failures. There is also mention of potential oversights in the problem's assumptions regarding telecommunications efficiency.

VVS
Messages
85
Reaction score
0
Hello,

I hope somebody can give me a hint for this problem.

Homework Statement



A set of telephone connections should be installed in order to connect the cities A and B. City A has 2000 telephones. If every user of city A gets a direct connection to city B, 2000 lines would be needed, which is extremely wasteful.
We assume that each user in city A needs a telephone connection to city B for two minutes on average during the 10 hours working hours (8.00-18.00) and that these calls are completely random. What minimal number M of telephone connections to city B is needed in order to ensure that not more than 1% of the calls from city A to city B do not get a direct line.

(Instruction: The sum of the probabilities for occupied lines, which is higher than the number of lines, should
be less than 1%. Approximate the probability function by a Gaussian distribution.)

Homework Equations



Table of Gaussian Distribution is given.

The Attempt at a Solution



I have absolutely no clue how to approach this problem. But here's what I got so far:

In the ideal case, i.e. each user is followed by another one:
If every user needs the telephone line for 2 minutes, then that means that we need total 4000 minutes of idle telephone lines. 10 hours are 600 minutes therefore 1 line can be used for 600 minutes. Therefore we need on average 4000 minutes / 600 minutes of lines = 6.6 lines.

I don't know how to proceed from here. He wants use to approximate the probability function by a Gaussian distribution. I am not sure what should be the y and x-axis of this probability function. Should the x-axis of Gaussian distribution be the number of occupied lines and y the probability that this number of lines are occupied.


Please help me
 
Physics news on Phys.org
VVS said:
Hello,

I hope somebody can give me a hint for this problem.

Homework Statement



A set of telephone connections should be installed in order to connect the cities A and B. City A has 2000 telephones. If every user of city A gets a direct connection to city B, 2000 lines would be needed, which is extremely wasteful.
We assume that each user in city A needs a telephone connection to city B for two minutes on average during the 10 hours working hours (8.00-18.00) and that these calls are completely random. What minimal number M of telephone connections to city B is needed in order to ensure that not more than 1% of the calls from city A to city B do not get a direct line.

(Instruction: The sum of the probabilities for occupied lines, which is higher than the number of lines, should
be less than 1%. Approximate the probability function by a Gaussian distribution.)

Homework Equations



Table of Gaussian Distribution is given.

The Attempt at a Solution



I have absolutely no clue how to approach this problem. But here's what I got so far:

In the ideal case, i.e. each user is followed by another one:
If every user needs the telephone line for 2 minutes, then that means that we need total 4000 minutes of idle telephone lines. 10 hours are 600 minutes therefore 1 line can be used for 600 minutes. Therefore we need on average 4000 minutes / 600 minutes of lines = 6.6 lines.
I'll attempt to add some insight into the problem by doing what you already did, but in a different way.

This problem is about summing random variables. Our end goal is find the mean of the sum, and the variance of the sum. If you know the mean and variance of the sum, you can plug those into your Gaussian normal distribution table, and get your final result.

Like you already did, let's start with the mean. But with my method, let's get a little more basic first. Let's ask ourselves, what is the probability that a given phone is taking up a connection line?

A given phone can be represented by a random variable. That variable can take on one of two possible values, '0' and '1'. A '1' means it's taking up 1 connection line, and '0' means it's taking up 0 connection lines. Since it takes up 2 minutes out of every 600 minutes, the probability of a '1' is 2/600 that it's taking up a line. Below is the probability distribution function of a single phone.

attachment.php?attachmentid=56479&stc=1&d=1362682313.jpg


Let's call this random variable x

What's the mean of x? Well, it's

μ = (0)(598/600) + (1)(2/600) = 1/300.

Now what we want to do is sum up 2000 of these random variables.

You should know already that when you sum random variables, the mean of the sum is the sum of all the random variables' means.

In other words, when summing 2000 of these random variables together,

Mean of the sum = μ1 + μ2 + μ3 + ... + μ2000.

And since in this case, all the phones' individual probabilities are equal,

Mean of the sum = (2000)(μ) = 2000/300 = 6.667 connection lines.

I don't know how to proceed from here. He wants use to approximate the probability function by a Gaussian distribution. I am not sure what should be the y and x-axis of this probability function. Should the x-axis of Gaussian distribution be the number of occupied lines and y the probability that this number of lines are occupied.Please help me

I've given you the probability distribution function of a single phone taking up a line. I've also given you the mean of a single phone. That was step 1.

The next steps are
(2) Find the variance of a single phone taking up a communication line.
(3) Properly combine the variances of 2000 phones. Now you should have the mean and variance of the necessary communication lines for 2000 phones.
(4) Along with that mean and variance of the sum, use your Gaussian normal table to ensure that the number of phone line connections satisfy 99% probability.

<Rant>
I might save it for a future post, but the author of the problem doesn't seem to know much about telecommunications. Only 1 user per line? No digitizing the signals? No time division multiplexing? Gwahh!
'Talk about being wasteful!
</Rant>
 

Attachments

  • PDF_for_a_phone.jpg
    PDF_for_a_phone.jpg
    10 KB · Views: 632
Last edited:
Thanks a lot for your help.
I really appreciate the mathematical precision of your way.

I have done the rest of the problem (see attached file). I hope it is correct.

View attachment E1P2.pdf
 
VVS said:
Thanks a lot for your help.
I really appreciate the mathematical precision of your way.

I have done the rest of the problem (see attached file). I hope it is correct.

View attachment 56480
'Looks correct to me. :approve:
 

Similar threads

Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
6
Views
5K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 29 ·
Replies
29
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K