****************************** Solution G ******************************

Tprime,lab: 0.000s
Tgauss,lab: 0.004s
Tprime,grendel: 0.001s
Tgauss,grendel: 0.004s
Tprime,ozark: 0.002s
Tgauss,ozark: 0.007s
Tprime,grendel / Tprime,lab: 0
Tprime,ozark / Tprime,lab: 0
Tgauss,grendel / Tgauss,lab: 1
Tgauss,ozark / Tgauss,lab: 1.75

To obtain my meausrements I used the Unix command time. This command
returns three unique time stamps, but for my results I solely looked at
the user time. The user time measures the amount of CPU time it takes
for the program to run through all of it's included instructions. Below
are the terminal instructions I used for each individual test case:

Tprime,lab terminal instructions:
gcc primetesting.c
time ./a.out
33880

Tprime,grendel terminal instructions:
ssh grendel
(entered password)
gcc primetesting.c
time ./a.out
33880

Tprime,ozark terminal instructions:
ssh ozark
(entered password)
cd www
gcc primetesting.c
time ./a.out
33880

Tgauss,lab terminal instructions:
gcc gaussian.c -lm
time ./a.out
33880

Tgauss,grendel terminal instructions:
ssh grendel
(entered password)
gcc gaussian.c -lm
time ./a.out
33880

Tgauss,ozark terminal instructions:
ssh ozark
(entered password)
cd www
gcc gaussian.c -lm
time ./a.out
33880

Appendix:

primetesting.c:

    #include <stdio.h>

    int main() {
        int n;
        int d = 2;
        printf("Please enter a number: \n");
        scanf("%d",&n);
        while((d*d) < n) {
            if((n % d) == 0) {
                printf("even\n");
                return 0;
            }
            d += 1;
        }
        printf("prime\n");
        return 1;
    }

gaussian.c:

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

    int main() {
        double n;
        int i;
        double answer;

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

        printf("Please enter a number: \n");
        scanf("%lf", &n);
        double sum = f(-2) + f(2);
        int d= 4;
        double delta = d/n;

        printf("n is equal to: %f\n", n);
        printf("f(-2) = %f and f(2) = %f \n", f(-2), f(2));
        printf("sum is: %f\n", sum); 
        printf("d is equal to: %d\n", d);
        printf("delta is: %f\n", d/n);

        for(i = 1; i <=(n-1); i++) { //for i from 1 to n-1, inclusive
            sum = sum + 3 * f(-2+i*delta);
        }
        printf("sum divided by 3 is: %f\n", sum/3);
        answer = delta *(sum/3);
        printf("%f \n", answer);
        return;
    }