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Optimizing a simple program

This is a discussion on Optimizing a simple program within the C Programming forums, part of the General Programming Boards category; 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 = ...

  1. #1
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    Question 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.

  2. #2
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Paul Omans
    my professor said that this should be a 10-15 line program and mine is double that.
    Don't worry too much about the number of lines. It should be just a guide. For example, you can combine two of your printf calls by a little re-ordering, but as-is they are fine.

    That said, I would write your while loop like this:
    Code:
    while (1) {
        ++n;
        B -= P - nthMonth;
        if (n > 2) {
            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);
    }
    Notice that I place the closing braces on their own lines, got rid of the really superfluous parentheses, and made use of the assignment operator forms where applicable.

    Next, this part:
    Code:
        if (n > 2) {
            totalPaid += nthMonth;
            if (B < 0) {
                break;
            }
        }
        else if(n <= 2) {
            if (B < 0) {
                break;
            }
        }
    can be simplified to:
    Code:
        if (n > 2) {
            totalPaid += nthMonth;
        }
        if (B < 0) {
            break;
        }
    Besides this, I notice that you don't use anything from <math.h>, so you should remove that header inclusion. Also, there is no need for the call to exit as the program will exit right after that line. If you do want to be explicit, return 0; instead. Of course, once you get rid of the exit call, you can also get rid of the inclusion of <stdlib.h>.
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  3. #3
    Maz
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    And using for instead while would combine 3 lines in a natural way.

  4. #4
    Registered User whiteflags's Avatar
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    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.

  5. #5
    Registered User ApolCor's Avatar
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    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);       
    }

  6. #6
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    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?

  7. #7
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    The logic isn't very neat.

    You don't need to handle the first month specially, as I see. Your condition is while the balance is greater than zero, so it's better to have that explicitly in the loop control statement. Only use break when you really need it.

    Then the loop is just B = (B - payment) * (1 + interest), a print, and maybe a loop counter for visual display.
    C allows you to tag on an increment to another statement like this:

    Code:
    printf("Month %d balance %f\n", i++, B);
    So you only need two lines in the loop.
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  8. #8
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    Wow, thanks for all the replies! I'll mess around with my code and post a revised version here later today.

  9. #9
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    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);      
    }

  10. #10
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    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.
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  11. #11
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by ArunS View Post
    Code:
        B -= (P + nthMonth);
    What about the above one?
    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;
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  12. #12
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    Quote Originally Posted by Malcolm McLean View Post
    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.


    Quote Originally Posted by iMalc View Post
    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.

  13. #13
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    Quote Originally Posted by Paul Omans View Post
    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
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