****************************** Solution B ******************************

Tprime,lab: The raw time for a Linux lab computer to execute the
prime-testing program described below.

        real    0m0.003s

Tgauss,lab: The raw time for a Linux lab computer to execute the
Gaussian integral program described below.

        real    0m3.287s

Tprime,grendel: The raw time for grendel to execute the prime-testing
program described below.

        real    0m0.001s

Tgauss,grendel: The raw time for grendel to execute the Gaussian
integral program described below.
        real    0m12.239s

Tprime,ozark: The raw time for ozark to execute the prime-testing
program described below.

        real    0m0.003s

Tgauss,ozark: The raw time for ozark to execute the Gaussian integral
program described below.

        real    0m7.802s

Tprime,grendel / Tprime,lab

        0m0.001s/0m0.003s = .33

Tprime,ozark / Tprime,lab
            
        0m0.003s/0m0.003s = 1

Tgauss,grendel / Tgauss,lab

        0m12.239s/0m3.287s = 3.723
            
Tgauss,ozark / Tgauss,lab
        0m7.802s/0m3.287s =2.373

To obtain the time, real time as well as user time, both the prime
and gaussian integral programs take to run. I simply wrote the
two program in c, tested if they were working as anticipated using
the terminal. The command used to compile both programs was “gcc
prime.c” for the prime program and “gcc gaussian.c -lm” for
the gaussian program. To execute the two programs, i used the command
“./a.out”. I measured the time they take to use to run in the lab,
grendel, and ozark by using the command “time ./a.out”

#include<stdio.h>
#include <time.h>

main(){
    int n = 100;
    int d = 2;
    while((d*d) < n){
        if (n % d == 0){

            printf("false\n");
            return 0;
        }
        d=d+1;
    }
    printf("true\n");
    return 1;
}


#include<stdio.h>
#include<math.h>
#include<time.h>

double function(double x){
    return exp(-(x*x)/2);
}

int main(){
    double i ;
    double n = 100; 
    double a = 2;
    double delta = 4.0/n;
    double sum  = function(2.0) + function(-2.0);
    for(i; i<=n; i++){
        sum  = sum + 3 * function((-a + (i*delta)));
    }
    printf("%f",delta*sum/3 );
    return (int) delta*sum/3;
}