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DedPeter [7]
2 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]2 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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In C++:

void sortlist(char nums[],int charlent){    

   int Myarr[charlent];

const char s[4] = " ";  

char* tok;  

tok = strtok(nums, s);  

int i = 0;  

while (tok != 0) {  

 int y = atoi(tok);

 Myarr[i] = y;

 tok = strtok(0, s);  

 i++;

}  

int a;  

for (int i = 0; i < charlent; ++i) {

for (int j = i + 1; j < charlent; ++j) {

 if (Myarr[i] > Myarr[j]) {

  a =  Myarr[i];

  Myarr[i] = Myarr[j];

  Myarr[j] = a;

       }   }  }

 

for(int j = 0;j<charlent;j++){ printf(" %d",Myarr[j]); }  

}

Explanation:

This line defines the sortlist function. It receives chararray and its length as arguments

void sortlist(char nums[],int charlent){    

This declares an array

   int Myarr[len];

This declares a constant char s and also initializes it to space

const char s[4] = " ";  

This declares a token as a char pointer

char* tok;

This line gets the first token from the char

tok = strtok(nums, s);

This initializes variable i to 0

int i = 0;

The following while loop passes converts each token to integer and then passes the tokens to the array  

<em> while (tok != 0) {  </em>

<em>  int y = atoi(tok); </em><em>-> Convert token to integer</em><em> </em>

<em>  Myarr[i] = y; </em><em>-> Pass token to array</em><em> </em>

<em>  tok = strtok(0, s); </em><em>-> Read token</em><em> </em>

<em>  i++; </em>

<em> }  </em>

Next, is to sort the list.

int a;

This iterates through the list  

for (int i = 0; i < charlent; ++i) {

This iterates through every other elements of the list

for (int j = i + 1; j < charlent; ++j) {

This condition checks if the current element is greater than next element

 if (Myarr[i] > Myarr[j]) {

If true, swap both elements

<em>   a =  Myarr[i]; </em>

<em>   Myarr[i] = Myarr[j]; </em>

<em>   Myarr[j] = a; </em>

       }   }  }

The following iteration prints the sorted array

<em>for(int j = 0;j<charlent;j++){ printf(" %d",Myarr[j]); }  </em>

}

<em>See attachment for illustration of how to call the function from main</em>

Download cpp
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