Answer:
Latency of an object O is shown below.
Explanation:
W segment and stalled state transmits nothing and waits for acknowledgement. The latency is 2 R TT + the time required for server that are using to transmit the object + the amount of time when server is in stalled state. Let K be the number of window that is K= O/WS .The serveries stalled state where K-1 is period of ime with period lasting RTT-(W-1)S/R
latency = 2RTT +O/R +(K-1)[S/R +RTT - WS/R]
After combining the two case
latency = 2 RTT + O/R + (K-1)[S/R +RTT - WS/R]
where [x] means maximum of (x.0). This is the complete ananlysis of the static windows.
server time for transmit the object is (K-1)[S/R +RTT - WS/R]
Answer:
A wave that has been digitized can be played back as a wave over and over, and it will be the same every time. For that reason, digital signals are a very reliable way to record information—as long as the numbers in the digital signal don’t change, the information can be reproduced exactly over and over again.
Explanation:
Answer:
#include <iostream>
using namespace std;
int main() {
cout<<"My name is Rajat Sharma"<<endl<<"My address is Flat no=23 GH=5 Paschim Vihar New Delhi 110087 India"<<endl;
return 0;
}
Explanation:
The program is written in C++ language.In the program I have used cout to print my name and the address.First the name will be printed then the address in the new line endl is used for new line.To print any sentence just put them in double quotes.The same sentence in the program will be printed on the screen.
Answer:
the indexOf() method
Explanation:
The indexOf() method in java returns the first occurrence of the of the string or a character specified in it.It can be used to search both a character or a string.
for example:-
import java.util.*;
import java.lang.*;
import java.io.*;
class indexOf
{
public static void main (String[] args)
{
String s="IamNumber4";
int l=s.indexOf("Num");//using indexOf() method on string object..
System.out.println(l);
}
}
Output:-
3
