Probability, Minimal Telephone Lines needed

In summary: Var(the sum of 2000 random variables) = (μ1 + μ2 + μ3 + ... + μ2000)^2= (9.8 + 12.7 + 15.4 + ... + 105.2)^2= 178.8^2= 17,880.So, the mean of the sum of 2000 random variables is 17,880.
  • #1
VVS
91
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
 
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  • #2
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>
 

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  • #3
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
 
  • #4
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:
 
  • #5
.


Hello,

Thank you for reaching out for help with this problem. I can offer some guidance on approaching this problem.

First, let's break down the problem into smaller parts and identify the key information given. We know that there are 2000 telephones in city A and we need to connect them to city B. However, connecting each user directly would require 2000 lines, which is wasteful. We also know that during the 10-hour working hours, each user needs a telephone connection to city B for an average of 2 minutes. Our goal is to find the minimal number of telephone connections (M) needed to ensure that no more than 1% of calls from city A to city B do not get a direct line.

To solve this problem, we can use probability theory. We can assume that the calls from city A to city B are completely random and that the probability of a call needing a direct line is equal for each user. This means that the probability of any given call needing a direct line is 1/2000 or 0.0005. We can also assume that the calls follow a normal distribution, which is why the Gaussian distribution is mentioned in the problem.

Now, let's look at the equation for a normal distribution:

f(x) = (1/σ√(2π))e^(-(x-μ)^2/2σ^2)

In this equation, σ is the standard deviation and μ is the mean. We can use this equation to approximate the probability function for this problem. The x-axis would represent the number of occupied lines and the y-axis would represent the probability of that number of lines being occupied. We can use the given table of the Gaussian distribution to find the corresponding probabilities for different values of x.

To find the minimal number of telephone connections needed (M), we need to find the value of x that corresponds to a probability of 0.01 or 1%. This value of x would represent the number of occupied lines that would ensure that no more than 1% of calls do not get a direct line. Once we have this value, we can add it to the 2000 telephones in city A to find the total number of telephone connections needed.

I hope this helps guide you in solving the problem. Please let me know if you have any further questions or need clarification on any of the concepts mentioned. Good luck!
 

1. What is probability and why is it important?

Probability is the measure of the likelihood of an event occurring. It is important because it helps us make predictions and informed decisions based on the available information.

2. How is probability calculated?

Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This can be represented as a fraction, decimal, or percentage.

3. What is the relationship between probability and telephone lines needed?

The relationship between probability and telephone lines needed is that the more probable an event is, the fewer telephone lines are needed to connect a certain number of people. This is because a higher probability means there is a greater chance of individuals being connected with fewer lines.

4. How does minimal telephone lines impact efficiency in communication?

Having minimal telephone lines can improve efficiency in communication by reducing the chances of busy signals or dropped calls. It also saves resources and costs associated with maintaining and installing additional telephone lines.

5. Can probability be used to determine the optimal number of telephone lines needed?

Yes, probability can be used to determine the optimal number of telephone lines needed. By calculating the probability of individuals being connected with a certain number of lines, we can determine the most efficient and cost-effective number of telephone lines needed for a given situation.

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