Question

We will use linear interpolation in a C program to estimate the population of NJ between the years of the census, which takes place every 10 years. For our program the x's are the years and the y's are the population. So given a year, we will use linear interpolation to calculate the population for that year, using the surrounding census years.
1. read the data from the file njpopulation.dat into two arrays, one for the years and one for the populations
2. use int for the years and float for the populations
3. create a loop to do the following, until the user enters 0 1. prompt the user to enter a year between 1790 and 2010 2. give an error message and reprompt if the user enters a value outside the valid range 3. use linear interpolation to approximate the population for the given year
4. print the year and the approximate population; print the population with two decimal places

Answer 1

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!!