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astraxan [27]
3 years ago
5

Consider the latency model for static congestion windows. If the server receives an acknowledgement for the first data segment i

n the first window before the server completes the transmission of the first window, then the latency for an object of size O is:
Computers and Technology
1 answer:
Angelina_Jolie [31]3 years ago
7 0

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]

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Answer:

Following are the code to this question:

public class Main//defining a class

{

public static void main(String[] arg)//defining main method

{

  String[] words={"Key","day", "Know", "kind"};//defining array of String words

  int x=0;//defining integer variable for count value

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  {

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  x=x+1;//increment the value of x

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System.out.print("The number of letters which starts from k is: "+ x);//print value with message

}

}

Output:

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Explanation:

In this code, inside the main method an array of String "words" is defined that hold a value, and in the next step an integer variable "x" is defined, which is used to count the letters, which starts from k.

For this, a for loop is used that counts the value and in this an, if block is defined that uses the "startsWith" method to check the value and increment the value of x and at the last, it prints its value.

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