Dec 192017
Text file on : Financial Modeling With C, the Advantages and Pitfalls.
File FINAGOL.ZIP from The Programmer’s Corner in
Category C Source Code
Text file on : Financial Modeling With C, the Advantages and Pitfalls.
File Name File Size Zip Size Zip Type
FINANCE.TXT 16295 5898 deflated

Download File FINAGOL.ZIP Here

Contents of the FINANCE.TXT file

Eddy C. Vasile
73317,701 @ COMPUSERVE

Financial Modeling With C
Advantages And Pitfalls

While the world is awash in specialized financial software
ranging from spread sheets and data bases to expert systems,
there will always be a need for low level programming because:
1) Data provided by real events can be counted on to baffle
existing applications. Lotus Signal crashed along with the
market because it was not able to handle three digit drops.
2) For repetitive tasks, user interface, calculation speed and
accuracy can be better served by a specialized routine instead
of a 4GL environment that comes prepared to solve a plethora of
other tasks as well.

Although a discussion of computational difficulties is
likely to elicit a snicker from those involved in the
aero-space industry, the truth is that Cobol is not powerful
enough to deal with some of the complexities of financial
models. The language of choice should provide:
1) Very fast and accurate math functions such as exponentiation
and logarithms.
2) A rich selection of numeric data types such as double size
reals for more significant digits.
3) Recursion for iterative and goal seeking routines.
4) Dynamic data structures.
5) Portability to all popular operating systems and hardware.
6) Good control over the operating system internals for
customized I/O and user interface.

C is the best equipped language to meet the criteria
described above. However, like any power tool, C should be
approached with caution. Users of forgiving environments should
be even more alert because C is notorious for following
instructions blindly, even when the programmer didn't really
"mean" them.

The code of the following four programs will introduce
several financial modeling concepts ranged from very simple to
complex (recursion) and their solutions. The compiler I've
chosen is Turbo C because of its speed and detailed warnings
that come in handy for the novice C programmer.

Future Value
If one makes a series of regular deposits for a certain interest
rate for a certain number of periods, the amount of money
accrued at the end of the periods is the future value. The FV
formula is: FV = PMT * ((1+i)^N-1)/i where PMT is the periodic
deposit, i is the interest rate and N is the number of periods.
The program should prompt for the monthly deposit, the annual
interest rate (e.g. 8 for 8%) and then divide that number by 12
and by 100 for the true monthly interest rate. The number of
periods is the number of years multiplied by 12.
Since the power function takes doubles as arguments (along with
exp, log and most of the other math functions), all variables
are doubles (although type casting is possible). Since the scanf
input routine takes only floats as arguments, a temp
variable of type float is introduced. Also note that when
reading with the scanf routine, one must pass the address of the
variable to the routine (i.e. scanf("%f",&temp)) and not the
variable itself. Some compilers such as Turbo C will alert you
if you forget to pass by reference but most of them will simply
let you suffer. Also you must remember to include math.h or else
pow will return garbage cleverly disguised as a valid number.
This is the code for the future value program:

double interest, pmt, periods, fv;
float temp;
printf("Monthly deposit : ");
scanf("%f",&temp); pmt=temp;
printf("Annual Interest (e.g. 8 for 8%) : ");
scanf("%f",&temp); interest=temp/100/12;
printf("Number of years : ");
scanf("%f",&temp); periods=temp*12;
printf("Future value of investment : %1.2f\n",fv);

Present Value

The present value is the amount of the loan that can be obtained
for the i interest rate, PMT monthly payment and N number of
periods. The formula is PV=PMT * (1-(1+i)^-N)/i.

double interest, pmt, periods, pv;
float temp;
printf("Monthly payment : ");
scanf("%f",&temp); pmt=temp;
printf("Annual Interest (e.g. 8 for 8%) : ");
scanf("%f",&temp); interest=temp/100/12;
printf("Number of years : ");
scanf("%f",&temp); periods=temp*12;
printf("Amount to be borrowed : %1.2f\n",pv);

Most spread sheets provide PV and FV functions. There are
situations where the FV and PV are known but either i or N need
to be calculated. FV and PV are linked by the formula:
FV = PV * (1+i)^N or (1+i)^N=FV/PV. Applying natural logarithm
to both sides the equation becomes N*ln(1+i)=ln(FV/PV).
N is therefore (ln(FV)-ln(PV))/ln(1+i).
The interest rate calculation is given by:
ln(1+i)=(ln(FV)-ln(PV))/N. After applying exp to both sides the
equation becomes 1+i=exp((ln(FV)-ln(PV))/N) or
The program that calculates i through the FV PV link is:

double interest, periods, fv, pv;
float temp;
printf("Future Value : ");
scanf("%f",&temp); fv=temp;
printf("Present Value : ");
scanf("%f",&temp); pv=temp;
printf("Number of years : ");
scanf("%f",&temp); periods=temp*12;
printf("Interest Rate : %1.2f\n",interest);

Net Present Value

It's difficult to imagine a bank account with nothing but
regular and equal deposits (i.e positive cash flows) and no
withdrawals (i.e. negative cash flows). Net present value
addresses the worth of a series of unequal and different sign
cash flows. The NPV formula is:
CF[1]/(1+i)^1+CF[2]/(1+i)^2+..CF[N]/(1+i)^N where CF are the
cash flows in the respective periods, N is the number of periods
and i is the considered interest rate.
The NPV function would look like this:

double pv(a,intrate,numflows)
double a[],intrate; int numflows;
int i;
double temp=0;
for (i=1;i<=numflows;i++) {
if (intrate==0) temp +=a[i-1];
else temp +=(a[i-1]/pow(1+intrate/100,i));

If the interest rate is 0 then the denominator is 1. Therefore,
if i=0 the net present value is the sum of cash flows. As i
becomes larger NPV gets smaller. Consider the following example:

You buy a mountain cabin for $25,000. During the first four
years you are able to rent it out and you collect $3,500, $100,
$4000 and $100 respectively. The fifth year you have a negative
cash flow of $300 due to repairs. The cash flows for years 6
through 11 are : 4000 2300 -150 5600 321 -100. In the twelfth
year you sell the cabin for $24,900. You ask yourself, would you
have been better off depositing your $25,000 at 6% or is the
series of cash flows generated by your investment greater in
value? Solving the NPV equation for 6% gives a net present value
of $26,675.49 or $1,675.49 more than your initial investment.
Plotting the Net Present Value as a function of i between -5%
and 100% will illustrate the inverse proportionality between
the two values.

- Net Present Value
as a function of interest
- .
. NPV(5)=$28,858.16
- NPV(6)=$26,675.49
- . . . NPV(20)=$11,091.59

| | | | |

Internal Rate of Return

When dealing with multiple cash flows, often times the required
value is the interest rate that would generate the NPV equal to
the initial investment. In our example, the interest rate that
would generate a $25,000 NPV to match the initial investment is
6.847%. This interest rate is called the Internal Rate of Return
(IRR) and it cannot be computed analytically. There are instances
when several solutions are possible and sometimes there are no

The program should look only for solutions that are
financially acceptable (i.e. between 0% and 200%) and it should
return the lowest if more than one are available. The irr
function should look for a solution by moving on the graph from
0% to the right in assigned increments while trying to match the
initial investment with the NPV generated by the current
interest rate. If the difference between Investment and NPV
changes signs, then the function has moved too far and it should
step back to the prior tick and continue with smaller
increments. This process should invoke itself until the
difference is smaller then a ten thousandth of the NPV.

This time the program will accept input from a file. Note
that the program expects the name of the input file as its
argument. However, if no arguments are passed the user is
prompted for the file name. Note that the getvals function uses
pointers to the variables in order to be able to modify their
values globally. The irr function keeps track of the increment
and the current interest rate in the same manner:
irr(a,invest,&intrate,&inc,numflows). However note that when the
function is invoking itself recursively the "&" address symbol
is not used because you are already dealing with a pointer. The
program reads the annual interest rate, the initial investment
and the cash flows. If the interest rate is not 0 the NPV is
calculated. Else, the programs assumes that the user is
searching for IRR. To run the program with the example above
create a file containing the values on separate lines or on the
same line. E.g. file "data.dat" contains the line:
0 -25000 3500 100 4000 100 -300 4000 2300 -150 5600 321 -100 24900

IRR and NPV program listing

#define MAX 500
main (argc,argv)
int argc;
char *argv[];
FILE *fopen(), *f;
char fn[80];
double a[MAX],invest,intrate,inc,pv(),irr();
int numflows;
if (argc==2) strcpy(fn,argv[1]);
else {
printf("File > ");
if ((f = fopen(fn,"r"))==NULL) {
printf("Can not open %s\n",fn);
if (intrate!=0) printf("pv=%1.3f\n",pv(a,intrate,numflows));

FILE *f; double a[],*intrate,*invest; int *numflows;
float temp;
int i;
fscanf(f,"%f",&temp); *intrate=temp; /*fscanf reads only floats*/
fscanf(f,"%f",&temp); *invest=fabs((double) temp);
for (i=0;i i=0;
while (fscanf(f,"%f",&temp)!=EOF) a[i++]=temp;

double pv(a,intrate,numflows)
double a[],intrate; int numflows;
int i;
double temp=0;
for (i=1;i<=numflows;i++) {
if (intrate==0) temp +=a[i-1];
else temp +=(a[i-1]/pow(1+intrate/100,i));

double irr(a,invest,intrate,inc,numflows)
double a[],invest,*intrate,*inc; int numflows;
double tol,d1,d2;
tol=invest/10000; /*this will grant precision past 3rd decimal*/
d1=invest-pv(a,*intrate + *inc,numflows);
printf("Int Rate %1.3f generates delta=%1.3f\n",(*intrate + *inc),d1);
if (((*intrate+*inc)<0) || ((*intrate+*inc)>200)) {
printf("\nNo solution\n");
else {
if (fabs(d1)<=tol) return(*intrate+*inc);
else {
if (d2*d1<0) *inc /= 10;
else *intrate +=*inc;


Amortization comes from French and liberally translated means
taking something to death in tiny amounts. When you consider that
one must pay for a mortgage monthly for thirty years, the
description is fitting. When you borrow money from a bank you
must pay back both the principal and the interest. Each payment
you make contains a portion of the principal and a portion of the
interest. During the beginning periods one pays a very small
amount for the principal and a large amount for the interest.
Eventually, after building some equity, things change and a large
portion of the payment goes to the principal. The following
program calculates the payment for a certain loan, the principal
balance, the principal and the interest portion of each payment.
A table of these values is printed for each period. To make
things interesting, I'll request the program to bring in the
variables from the command line. Thus, the program must be
invoked like this: "AMORT 100000 10.5 30" in order to amortize a
$100,000 loan for a 10.50% annual interest rate lasting for 30
years. Modify the program to prompt you for the values of the
three variables. The payment formula is:
paymemt=loan amount*(monthly int/(1-1/(1+monint^periods)))


/*Simple amortization program Eddy C. Vasile
To be used with command line params for amount, annual
interest and years. Compile to exe.
Use it from inside dBase in the following manner:
set talk off
set echo off
input 'Enter loan amount (e.g. 100000) :' to amount
input 'Enter annual interest (e.g. 11.5):' to annint
input 'Enter number of years (e.g. 30) :' to years
paramline=' '+amount+' '+annint+' '+years
run amort ¶mline
main (argc,argv)
int argc;
char *argv[];
double amount,annint,years,periods,monint,pmt,intrate,
int curperiod;
if (argc!=4) {
printf("Format is \n");
else {
periods = years*12;
printf("Monthly Payment :%1.2f\n",pmt);
printf("Mon Balance Interest Principal\n\n");
while (curperiod <=periods) {
printf("%3d %9.2f %7.2f %7.2f\n",curperiod++,principal,

Now if after all this you're not convinced that C is a good
choice for programming in a business environment, then take the
equivalent dBase III code for the above program, run it and time


*Simple amortization table by Eddy C. Vasile
set talk off
set echo off
set decimal 2
set fixed on
input 'Enter loan amount (e.g. 100000) :' to amount
input 'Enter annual interest (e.g. 11.5):' to annint
input 'Enter number of years (e.g. 30) :' to years
periods = years*12
? 'Monthly Payment :',str(pmt,6,2)
? ' Period # Balance Interest Principal'
? '--------------------------------------------------------'
do while curperiod <=periods
? curperiod,principal,int,princpaid

Ok. This is all I have to say about C for now. To compile the
code for the various programs, pull this file into the editor and
save the code portions in separate files. Compile with your
compiler of choice since all the routines are simple and
portable. Send opinions, enhancements and general YAK YAK
electronically to EVASILE on MCI Mail or 73317,701 on CompuServe.
Hope you had fun reading this.

 December 19, 2017  Add comments

 Leave a Reply

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>