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Novay_Z [31]
2 years ago
15

⦁ Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes. ⦁ What is the maximum val

ue of L such that TCP sequence numbers are not exhausted? Recall that the TCP sequence number field has 4 bytes. ⦁ For the L you obtain in (a), find how long it takes to transmit the file. Assume that a total of 66 bytes of transport, network, and data-link header are added to each segment before the resulting packet is sent out over a 155 Mbps link. Ignore flow control and congestion control so A can pump out the segments back to back and continuously.
Computers and Technology
1 answer:
timurjin [86]2 years ago
8 0

Answer:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

Explanation:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

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Answer:

See explaination

Explanation:

PhoneLookup.java

import java.io.FileReader;

import java.io.IOException;

import java.util.Scanner;

public class PhoneLookup

{

public static void main(String[] args) throws IOException

{

Scanner in = new Scanner(System.in);

System.out.println("Enter the name of the phonebook file: ");

String fileName = in.nextLine();

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FileReader reader = new FileReader(fileName);

table.read(new Scanner(reader));

boolean more = true;

while (more)

{

System.out.println("Lookup N)ame, P)hone number, Q)uit?");

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if (cmd.equalsIgnoreCase("Q"))

more = false;

else if (cmd.equalsIgnoreCase("N"))

{

System.out.println("Enter name:");

String n = in.nextLine();

System.out.println("Phone number: " + table.lookup(n));

}

else if (cmd.equalsIgnoreCase("P"))

{

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String n = in.nextLine();

System.out.println("Name: " + table.reverseLookup(n));

}

}

}

}

LookupTable.java

import java.util.ArrayList;

import java.util.Collections;

import java.util.Scanner;

/**

A table for lookups and reverse lookups

*/

public class LookupTable

{

private ArrayList<Item> people;

/**

Constructs a LookupTable object.

*/

public LookupTable()

{

people = new ArrayList<Item>();

}

/**

Reads key/value pairs.

atparam in the scanner for reading the input

*/

public void read(Scanner in)

{

while(in.hasNext()){

String name = in.nextLine();

String number = in.nextLine();

people.add(new Item(name, number));

}

}

/**

Looks up an item in the table.

atparam k the key to find

atreturn the value with the given key, or null if no

such item was found.

*/

public String lookup(String k)

{

String output = null;

for(Item item: people){

if(k.equals(item.getName())){

output = item.getNumber();

}

}

return output;

}

/**

Looks up an item in the table.

atparam v the value to find

atreturn the key with the given value, or null if no

such item was found.

*/

public String reverseLookup(String v)

{

String output = null;

for(Item item: people){

if(v.equals(item.getNumber())){

output = item.getName();

}

}

return output;

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}

Item.java

public class Item {

private String name, number;

public Item(String aName, String aNumber){

name = aName;

number = aNumber;

}

public String getName(){

return name;

}

public String getNumber(){

return number;

}

}

input.txt

Abbott, Amy

408-924-1669

Abeyta, Ric

408-924-2185

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408-924-3159

Accardo, Dan

408-924-2236

Acevedo, Elvira

408-924-5200

Acevedo, Gloria

408-924-6556

Achtenhagen, Stephen

408-924-3522

Note: Replace all the "at" with at symbol

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