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Alecsey [184]
3 years ago
12

A group of N stations share a 56-kbps pure ALOHA channel. Each station outputs a 1000-bit frame on average once every 100 sec, e

ven if the previous one has not yet been sent (e.g., the stations can buffer outgoing frames). What is the maximum value of N?
Computers and Technology
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Explanation:

Here, in the given statement, maximum flow capacity of the ALOHA channel = 18.4% = 0.184

Then, the stations N share 56 kbps

And, the channel rate = 0.184*56 = 10.304 kbps.  

Then, the outputs of the wach stations is 1000-bit frames/100 sec

\therefore, the output of the bits/ sec by each stations = \frac{1000}{100} bits/sec = 10 bits/sec

So, The output of the N stations is 10 bits/sec on the channel that having 10.304kb/sec

Finally, N = \frac{(10.304 \times 10^{3})}{10} = 1030 stations

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Write a program that: Takes the list lotsOfNumbers and uses a loop to find the sum of all of the odd numbers in the list (hint:
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Answer:

Following are the code to this question:

#include <iostream>//defining header file

using namespace std;

int main()//defining main method

{

int x[]={2,3,4,6,7,8,9,1,11,12};//defining 1-D array and assign value

int i,sum=0;//defining integer variable

for(i=0;i<10;i++)//defining loop for count value

{

   if(x[i]%2==1)//defining if block to check odd value

   {

       sum=sum+x[i];//add value in sum variable

   }

}

cout<<sum;//print sum

return 0;

}

Output:

31

Explanation:

In the above-given program, an integer array "x" is declared that holds some integer values, and in the next line two integer variable "i and sum" is defined which is used to calculate the value.

In the next line, a for loop is declared, that counts all array value, and it uses the if block to check the odd value and add all the value into the sum variable.

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