UDP and TCP use 1s complement for their checksums. Suppose you have the following three 8-bit bytes: 01010011, 01100110, 0111010
0. What is the 1s complement of the sum of these 8-bit bytes? (Note that although UDP and TCP use 16-bit words in computing the checksum, for this problem you are being asked to consider 8-bit sums.) Show all work. Why is it that UDP takes the 1s complement of the sum; that is, why not just use the sum? With the 1s complement scheme, how does the receiver detect errors? Is it possible that a 1-bit error will go undetected? How about a 2-bit error?
1 answer:
Answer:
One's complement = 1 1 0 1 0 0 0 1.
To detect errors, the receiver adds the four words (the three original words and the checksum). If the sum contains a zero, the receiver knows there has been an error. All one-bit errors will be detected, but two-bit errors can be undetected (e.g., if the last digit of the first word is converted to a 0 and the last digit of the second word is converted to a 1).
Explanation:
See attached pictures.
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Answer:
Following are the code to this question:
#include<iostream>//declaring header file
using namespace std;
int main()//main method
{
int n= 6,j=0;//declaring integer variable
int X[n];//defining an array
for(j=0;j<=n;j++)//defining a loop for input value
cin>>X[j];//input value from the user
if(j==0) //defining if block that checks value at beginning
{
if(X[j]>=X[j+1])//defining if block to check to compare first and second value
{
cout<<"A peak is at array index "<<j<<" and the value is "<<X[j];//use print method to largest value with index number
}
}
else//defining else block
{
for(j=0;j<=n;j++)//defining for loop for compare other value
{
if(j==n-1) //use if block that checks next index
{
if(X[j]>=X[j-1])//use if block to compare value
cout<<"A peak is at array index "<<j<<" and the value is "<<X[j];//use print method to largest value with index number
}
else
{
if(X[j]>=X[j-1] && X[j]>=X[j+1])//comapre value
cout<<"A peak is at array index "<<j<<" and the value is "<<X[j];//use print method to largest value with index number
}
}
}
return 0;
}
Output:
please find the attached file.
Explanation:
In the given code, inside the main method two integer variable "n and j", is declared, in the next step, an array "x"is defined which input the value from the user end.
- In the next step, multiple if block is used, in the first if block it comapre the first and second value if it grater then it will print the value with its index number.
- In the next if block, it comapre is next value and if it grater then it will print the value with its index number.
