Thread: Optimizing a simple program

Code:

#include <math.h>
#include <stdlib.h>
#include <stdio.h>
int main(void)
{
    float totalPaid=0.0, r=0.22, B=5000.0, P=0.0, firstMonth = 0.0, nthMonth = 0.0, n=1.0;
    printf("\nEnter your monthly payment: $");
    scanf("%f", &P);
    printf("\nAnnual Interest Rate (r) = %-5.2f%%\nAmount Borrowed (B) = $%-6.2f\nPayment Amount (P) = $%-6.2f\n", r*100, B, P);
    firstMonth = (r/12)*(B);
    totalPaid = totalPaid + firstMonth;
    printf("\n%-2.0f %-5.2f %-6.2f \n", n, firstMonth, B);
    nthMonth=firstMonth; 
    while(1){
        n=n+1;
        B = (B-(P-nthMonth));
        if (n>2) {
            totalPaid = totalPaid + nthMonth;
            if (B < 0) { break; } }
        else if(n<=2) {
            if (B < 0) { break; } }
        nthMonth = (r/12)*(B);
        printf("%-2.0f %-5.2f %-6.2f \n", n, nthMonth, B); }
    printf("\nTotal interest paid: $%-6.2f\n", totalPaid);
        exit(0);        
}

I already got the code to do exactly what it's needed to do, the only thing that is bothering me is my professor said that this should be a 10-15 line program and mine is double that. Are there any glaringly obvious ways that I can simplify this (while keeping the output the exact same)? I got it down to the point where I don't see anything more that I can optimize.

Can you use the I = P r t formula instead? The only variable missing from your current program are the terms, or the length of the loan. More useful than the monthly payment, here, would be the terms. You can derive the monthly payments from yearly interest.

Code:

#include <stdlib.h>
#include <stdio.h>
int main(void)
{
  float totalPaid=0.0, r=0.22, B=5000.0, P=0.0, nthMonth = 0.0, n=0.0;
  printf("\nEnter your monthly payment: $");
  scanf("%f", &P);
  printf("\nAnnual Interest Rate (r) = %-5.2f%%\nAmount Borrowed (B) = $%-6.2f\nPayment Amount (P) = $%-6.2f\n\n", r*100, B, P);
  while (B>0)
  {
    n++;
    nthMonth = r/12 * B;
    totalPaid = totalPaid + nthMonth;
    printf("%-2.0f %-5.2f %-6.2f \n", n, nthMonth, B);
    B = B - P + nthMonth;
  }
  printf("\nTotal interest paid: $%-6.2f\n", totalPaid);
  exit(0);       
}

Code:

#include <stdlib.h>
#include <stdio.h>
int main(void)
{
  float totalPaid=0.0, r=(0.22/12), B=5000.0, P=0.0;
  printf("\nEnter your monthly payment: $");
  scanf("%f", &P);
  printf("\nAnnual Interest Rate (r) = %-5.2f%%\nAmount Borrowed (B) = $%-6.2f\nPayment Amount (P) = $%-6.2f\n\n", r*12*100, B, P);
  for (float n=1.0, nthMonth=0.0;B>0;n++)
  {
    nthMonth = r/12 * B;
    totalPaid += nthMonth;
    printf("%-2.0f %-5.2f %-6.2f \n", n, nthMonth, B);
    B -= (P + nthMonth);
  }
  printf("\nTotal interest paid: $%-6.2f\n", totalPaid);
  exit(0);      
}

What about the above one?

Wow, thanks for all the replies! I'll mess around with my code and post a revised version here later today.

I based it mostly off of ApolCor's suggestion, although I switched some things around. I threw in a catch in case the user enters a monthly payment that happens to be less than the first month's interest payment (which creates an infinite loop inside of the while). I'm pretty happy with it at this point.

Code:

#include <stdlib.h>
#include <stdio.h>
int main(void)
{
  float totalPaid=0.0, r=0.22, B=5000.0, P=0.0, nthMonth = 0.0, n=1.0;
  printf("\nEnter your monthly payment: $");
  scanf("%f", &P);
  printf("\nAnnual Interest Rate (r) = %-5.2f%%\nAmount Borrowed (B) = $%-6.2f\nPayment Amount (P) = $%-6.2f\n\n", r*100, B, P);
  while (B>0)
  {
    nthMonth = r/12 * B;
    totalPaid = totalPaid + nthMonth;
    if (nthMonth > P)
     {
         printf("Monthly payment cannot be less than the owed monthy interest: $%-6.2f\n", nthMonth);
        exit(0);
     }
    printf("%-2.0f %-5.2f %-6.2f \n", n++, nthMonth, B);
     B = B - P + nthMonth;
  }
  printf("\nTotal interest paid: $%-6.2f over a course of %-6.2fmonths.\n", totalPaid, n);
  exit(0);      
}

Originally Posted by Malcolm McLean

I'd change that catch to "you have got yourself into an irrecoverable debt spiral. You are now Greece." Then run for a four year period just to show what happens.

But you only need to do the test once. The balance is either increasing or it is going down.

that's the kind of humor that would score me extra credit with my professor...but yes, I suppose it would make more sense to move the if statement above the while loop so it's not having to go past it every pass. But since nthMonth isn't given the correct value until inside the while loop, would it work to write the for loop like this:

Code:

    if (r/12 * B > P)

     {
         printf("Monthly payment cannot be less than the owed monthy interest: $%-6.2f\n", nthMonth);

        exit(0);
     }

Would that follow proper order of operations, or would I have to partition the if statement with more parentheses? I'm on my phone right now, otherwise I would check myself.

Originally Posted by iMalc

You actually introduced a bug with that line (and unneeded brackets).

Code:

B = B - P + nthMonth;

could be written as:

Code:

B -= P - nthMonth;

// or

B += nthMonth - P;

I'm interested, what bug would that be? I understand how it is faster to use the -= but I don't understand why

Code:

B -= (P + nthMonth);

introduces a bug.

Originally Posted by Paul Omans

I'm interested, what bug would that be? I understand how it is faster to use the -= but I don't understand why

Code:

B -= (P + nthMonth);

introduces a bug.

Basic math:

Code:

B -= (P + nthMonth);

is equivalent to

Code:

B = B - (P + nthMonth);

which after removing the parentheses becomes

Code:

B = B - P - nthMonth;

Bye, Andreas