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adelina 88 [10]
3 years ago
10

g What field in the IPv4 datagram header can be used to ensure that a packet is forwarded through no more than N routers? When a

large datagram is fragmented into multiple smaller datagrams, where are these smaller datagrams reassembled into a single larger datagram? Do routers have IP addresses? If so, how many? How many bits do we have in an IPv6 address? We use hexadecimal digits (each with 4 bits) to represent an IPv6 address. How many hexadecimal digits do we need to represent an IPv6 address?
Computers and Technology
1 answer:
ad-work [718]3 years ago
4 0

Answer:

a) Time to live field

b) Destination

c) Yes, they have two ip addresses.

d) 128 bits

e) 32 hexadecimal digits

Explanation:

a) the time to live field (TTL) indicates how long a packet can survive in a network and whether the packet should be discarded. The TTL is filled to limit the number of packets passing through N routers.

b) When a large datagram is fragmented into multiple smaller datagrams, they are reassembled at the destination into a single large datagram before beung passed to the next layer.

c) Yes, each router has a unique IP address that can be used to identify it. Each router has two IP addresses, each assigned to the wide area network interface and the local area network interface.

d) IPv6 addresses are represented by eight our characters hexadecimal numbers. Each hexadecimal number have 16 bits making a total of 128 bits (8 × 16)  

e) IPv6 address has 32 hexadecimal digits with 4 bits/hex digit

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2 years ago
We will use linear interpolation in a C program to estimate the population of NJ between the years of the census, which takes pl
nikklg [1K]

Answer:

See explaination for program code

Explanation:

program code below:

#include<stdio.h>

int main()

{

//file pointer to read from the file

FILE *fptr;

//array to store the years

int years[50];

//array to store the population

float population[50];

int n = 0, j;

//variables for the linear interpolation formula

float x0,y0,x1,y1,xp,yp;

//opening the file

fptr = fopen("njpopulation.dat", "r");

if (fptr != NULL)

{

//reading data for the file

while(fscanf(fptr, "%d %f", &years[n], &population[n]) == 2)

n++;

//prompting the user

int year = -1;

printf("Enter the years for which the population is to be estimated. Enter 0 to stop.\n\n");

while(year != 0)

{

printf("Enter the year (1790 - 2010) : ");

scanf("%d", &year);

if (year >= 1790 && year <= 2010)

{

//calculating the estimation

xp = year;

for (j = 0; j < n; j++)

{

if (year == years[j])

{

//if the year is already in the table no nedd for calculation

yp = population[j];

}

else if (j != n - 1)

{

//finding the surrounding census years

if (year > years[j] && year < years[j + 1])

{

//point one

x0 = years[j];

y0 = population[j];

//point two

x1 = years[j + 1];

y1 = population[j + 1];

//interpolation formula

yp = y0 + ((y1-y0)/(x1-x0)) * (xp - x0);

}

}

}

printf("\nEstimated population for year %d : %.2f\n\n", year, yp);

}

else if (year != 0)

{

printf("\nInvalid chioce!!\n\n");

}

}

printf("\nAborting!!");

}

else

{

printf("File cannot be opened!!");

}

close(fptr);

return 0;

}

OUTPUT :

Enter the years for which the population is to be estimated. Enter 0 to stop.

Enter the year (1790 - 2010) : 1790

Estimated population for year 1790 : 184139.00

Enter the year (1790 - 2010) : 1855

Estimated population for year 1855 : 580795.00

Enter the year (1790 - 2010) : 2010

Estimated population for year 2010 : 8791894.00

Enter the year (1790 - 2010) : 4545

Invalid chioce!!

Enter the year (1790 - 2010) : 1992

Estimated population for year 1992 : 7867020.50

Enter the year (1790 - 2010) : 0

Aborting!!

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Answer:

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Answer:

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