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DedPeter [7]
3 years ago
15

Suppose an IP packet is fragmented into 10 fragments, each with a 1% (independent) probability of loss. To a reasonable approxim

ation, this means there is a 10% chance of losing the whole packet due to loss of a fragment. What is the probability of net loss of the whole packet if the packet is transmitted twice?
a. assuming all fragments received must have been part of the same transmission?
b. assuming any given fragment may have been part of either transmission?
c. explain how use of the IDENT field might be applicable here.
Computers and Technology
1 answer:
Marysya12 [62]3 years ago
6 0

Answer:

a. 0.01

b. 0.001

c. The identification field of the packet fragment can be used to uniquely identify and collate the fragments lost in transmission.

Explanation:

The probability of losing a packet is 10% or 0.1, so the probability of losing the packet twice during transmission;

= 0.1 x 0.1 = 0.01

When any fragments have been part of the transmission, the probability of the packet is dependent on the fragments;

= 0.01 x 0.1 = 0.001

The identification field is a unique 16-bit value assigned to an IPv4 packet, when a packet is fragmented for transmission, its field is used to collate the unique fragments in the packet.

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Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) Something
yKpoI14uk [10]

Answer:

Let P(x) = x is in the correct place

Let Q(x) =  x is in the excellent place

R(x) denotes the tool

Explanation:

a) Something is not in the correct place.

P(x) is that x is in the correct place so negation of ¬P(x) will represent x is not in the correct place. ∃x is an existential quantifier used to represent "for some" and depicts something in the given statement. This statement can be translated into logical expression as follows:

                                                    ∃x¬P(x)

b) All tools are in the correct place and are in excellent condition.

R(x) represents the tool, P(x) represents x is in correct place and Q(x) shows x is in excellent place. ∀ is used to show that "all" tools and ∧ is used here because tools are in correct place AND are in excellent condition so it depicts both P(x) and Q(x). This statement can be translated into logical expression as follows:

                                       ∀ x ( R(x) → (P(x) ∧ Q(x))

c) Everything is in the correct place and in excellent condition.

Here P(x) represents correct place and Q(x) represents excellent condition ∀ represent all and here everything. ∧  means that both the P(x) and Q(x) exist. This statement can be translated into logical expression as follows:

                                              ∀ x (P(x) ∧ Q(x)

7 0
3 years ago
Who tryna play among us
harkovskaia [24]

Answer:

I'm doing a digital media test right now

8 0
3 years ago
Read 2 more answers
1. Write a while loop that lets the user enter a number. The number should be multiplied by 10, and the result assigned to a var
shutvik [7]

Answer:

import java.util.Scanner;

public class num6 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       int product=0;

       do{

           System.out.println("Enter a number");

           int num = in.nextInt();

           product = num*10;

           System.out.println("The product is " + product);

       }while (product<100);

  }

}

Explanation:

In the Java program above, The user will be prompted to enter a number.

The number is multiplied by 10 and assigned to a variable called product.

A do....while loop has been implemented to check the variable product

As long as this variable is less than 10, the user will be prompted to enter another number.

3 0
3 years ago
15. How many PC’s can be connected to a UPS?
Alex_Xolod [135]

Answer: Depends on the power of the computer /computers and the UPS

Explanation: For someone who had a big connected network of computers and printers we would plug the computer or server into the UPS as well as the monitor and sometimes a printer sometimes the monitor was plugged into the wall

When you get a UPS you must plug it in for a certain time could be hours or even a day.  A USP has a big battery in it and you must charge it (sometimes it comes semi-charged but you must plug it in to strengthen it. )

there are two connector wires that you affix to the battery and you must put a lot of strength into it to make it secure.

So in closing, you can get away with two like a desktop and a laptop.

Read your documentation that comes with the UPS

6 0
2 years ago
You are given an array x of string elements along with an int variable n that contains the number of elements in the array. You
jarptica [38.1K]

Answer and Explanation:

Make an algorithm of that which include array x of string where Mode of the value of the array means that you have the values such that 1.2,4,2,6,7,8,2 so 2 is the value of set as it appears more than other number similarly when a array of x has been given.First of all initialize a variable then declare it . so for initialization create a variable .Let suppose i create a variable of add and initialize it with 0 and given a it data type integer then

int add= 0;  // declaration

int add[]=0; // declaration and initialization in array

class add {  

void main(String array[]){  

int add[]=0;//declaration and instantiation  

int add[] = new int [3];

add[0]=30;//initialization  

add[1]=80;  

add[2]=20;  

output the given array of string

for ( int i = 0 ; i <add.length ;i++)

cout <<add[i];

 }

Taking an other example let initialization and declare array of string and print the greater value.For this create a variable with integer data type \

class Test{  

void main(String[] array){  

       //declaration of array

int test[][] = new int [2];

test[0]; new int [2];

test[1]; new int [4];

for initialization of array lets take variable a with data type integer

      int a=0;  

//taking loop for condition

       for (int i=0; i<test.length; i++)  

           for(int j=0; j<test[i].length; j++)  

//if the loops working properly then take increment in the variable a

               test[i][j] = a++;  

this process repeat until loops got false

   for (int i=0; i<test.length; i++)  

           for(int j=0; j<test[i].length; j++)  

         cout<<test[i][j]<<"+"<<"";  

           }  

           System.out.println();

       }  

   }  

6 0
4 years ago
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