dennis@rlgvax.UUCP (01/31/87)
Are you
- earning interest daily or quarterly in your favorite bank
because of one deposit or regular deposits?
- planning to save up for your children's education?
- buying US Savings bonds from your employer?
If so, this programs for YOU. It will help you answer
how much money it will be worth after a period of time.
Just extract it, compile it, and run it. All questions are
self-explanatory. There is also a man/help file in case you get
a bit more inquisitive.
-enjoy,
-dennis
#--------------- CUT HERE ---------------
#! /bin/sh
# This is a shell archive, meaning:
# 1. Remove everything above the #! /bin/sh line.
# 2. Save the resulting text in a file.
# 3. Execute the file with /bin/sh (not csh) to create the files:
# Makefile
# compound.c
# compound.man
# This archive created: Fri Jan 30 23:06:42 EST 1987
#
if test -f Makefile
then
echo shar: will not over-write existing file 'Makefile'
else
echo x - Makefile
# ............ F I L E B E G .......... Makefile
cat << '\SHAR_EOF' > Makefile
compound: compound.o
cc compound.o -lm -o compound
\SHAR_EOF
# ............ F I L E E N D .......... Makefile
fi # end of overwriting check
if test -f compound.c
then
echo shar: will not over-write existing file 'compound.c'
else
echo x - compound.c
# ............ F I L E B E G .......... compound.c
cat << '\SHAR_EOF' > compound.c
#if 0 /* 1 = enable debugging, 0 = disable debugging */
# define DEBUG 1 /* enable debugging code */
#else
# undef DEBUG /* disable debugging code */
#endif
/*
* compound.c
* dennis 01 15 87 dennis@rlgvax.UUCP
*
* Written by:
* Dennis Bednar
* Computer Consoles Inc
* 11490 Commerce Park Dr.
* Reston VA 22090
*
* determine money because of compound interest.
*
* TODO: Handle multi-shot when deposit period != compound period.
*
* 01 30 87: Ask for inflation rate, and besides printing the total, print
* what the total would be worth in today's dollars, given the inflation
* rate chosen by the user.
*
* One-shot means one deposit into a bank which gains because of compounded
* interest.
*
* Multi-shot means that you keep making deposits at regular intervals into
* a bank, and the bank compounds your interest. For example, if your
* employer deposits into a savings account every pay period.
*
* Note: For people who don't understand compound interest, here is a simple
* example:
* $100 for one year at 4% would yield $104.00 after one year if compounded
* annually (once per year). The money *must* be deposited before the beginning
* of a period that the bank chooses.
* $100 for one year at 4%, compounded quarterly, would yield 1% each quarter.
* 101.00 at end of first quarter ($100 * 1.01)
* 102.01 at end of 2nd quarter ($101 * 1.01 = 101 + 1.01)
* 103.0301 at end of 3rd quarter (102.01 * 1.01 = 102.01 + 1.0201)
* 104.060401 at end of 4th quarter ($103.0301 * 1.01 = 103.0301 + 1.030301)
* Because of floating point round-off, the number produced on a cci632
* is really 104.060402 (when float is used), but is 104.060401 when double
* is used.
*
*
* Now I derive the formulas in general:
*
* FORMULA for one-shot:
*
* One-shot deposit:
* t = (p) * (1 + r/(n*100))**u
* t = total after compounded
* p = principal (amount deposited)
* R = r/(n * 100)
* R = interest rate per pay period (if 4% compounded 356 days a year,
* R would be .04/365.
* r = interest per year as a number (if 7% interest, r = 7)
* n = number of times compounded per year
* u = number of pay units ( if compounded daily for y years, would be
* y*365).
*
* However, if compounded continuously, the value after 1 years is:
* t = p * e^(r/100) [see page 298, 556, of "Calculus",
* by Michael Spivak]
* and after n years is,
* t = p * (e ^ (r/100))^^n = p * (e ^ ((r*n)/100) )
*
*
* FORMULA for multi-shot:
*
* Multiple-shot deposit: assumes deposit money at the same
* rate as the compounding. This formula below is *not* general
* purpose enough to handle the case when the deposit period
* is not the same as the compound period. For example, it cannot
* handle the case when deposits occur every two weeks, but interest
* is compounded daily. Nor can it handle the case when deposits
* occur every two weeks, but interest is compounded quarterly.
*
* t=total
* t0 = initial total
* t1 = total after one unit of time
* tu = total after u units of time
* p = principal (amount deposited each time unit)
* r = annual interest (7 means 7% interest per year)
* R = (1.0 + r/(n*100) ) multiplier for each pay period.
* n = number of compoundings per year
* 1 = annual
* 2 = bi-annual
* 4 = quarterly
* 365 = daily
* u = number of units
*
* At time t0: t = p
* t1: t = t0*R + p
* = (p*R) + p,
* R is (1+ r/(n*100))
* t2: t = t1*R + p
* = (p*R + p)*R + p
* = p*(R**2) + p*R + p
* t3: t = t2*R + p
* = p*(R**3) + p*(R**2) +p*R + p
* tu: t = p*R**(u-1) + ... + p*R + p
* = p * [1 + R + R**2 + R**3 + R**4 + ... + R**(u) ]
* = p * A (1)
* where A = [1 + R + R**2 + ... + R**(u) ]
*
* Solving for A:
* A = [1 + R + R**2 + R**3 + R**4 + ... + R**(u) ] (2)
* multiplying both sides of (2) by R we obtain equation (3)
* A*R = [ + R + R**2 + R**3 + R**4 + ... + R**(u) + R**(u+1)] (3)
*
* Now subtracting (2) - (3) we obtain
* A - A*R = 1 - R**(u+1) (4)
*
* factoring,
* A (1-R) = 1- R**(u+1)
*
* dividing both sides by (1-R), we obtain:
*
* (1- R**(u+1))
* A = ------------------ (5)
* (1-R)
*
* (1 and 5 are the only two formulas that we are interested in now).
*
*/
#include <stdio.h>
#define DOUBLE double /* Sys 5 R2 exp(3) *must* use a double */
#define FLOAT double /* if your machine doesn't have double you */
/* may redefine FLOAT to be float. */
/* but double gives more accurate answers */
/* fw non-int functions */
FLOAT raise();
FLOAT ask();
extern DOUBLE exp(); /* exp(3M) - exp(x) returns e to the power of x */
#ifdef DEBUG
/* EXP() is the debug routine */
DOUBLE EXP(); /* fw reference */
#else
# define EXP exp /* use exp() in libm.a */
extern DOUBLE EXP();
#endif
main()
{
setbuf(stdout, (char *)NULL); /* so automatic fflush(stdout) without newlines */
if (yesno( "Do you want to make one deposit"))
oneshot();
else if (yesno( "Do you want to make multiple deposits, where interest is compounded at same period"))
multshot();
else
error("sorry cannot handle multiple deposits when 'deposit period != interest period'");
exit(0);
}
/*
* t = (p) * (1 + r/(n*100))**u ( t = total with no inflation)
*
* Now if inflation were factored in, we ask what principle today,
* if invested one shot, would be worth t dollars with inflation rate
* "r_inf", and that is the value of p_inf when solved in (same formula
* as above, only replace "r" with "r_inf" and replace "p" with "p_inf):
*
* t = (p_inf) * (1 + r_inf/(n*100))**u
*
* or solving for p_inf, we have
*
* t
* p_inf = ---------------------
* (1 + r_inf/(n*100) )**u
*
*
*/
oneshot()
{
FLOAT p, /* principal */
r, /* annual interest rate */
t; /* total amount after compounded interest */
FLOAT r_inf; /* inflation interest rate */
FLOAT p_inf; /* principle - non-inflationary */
int u; /* number of compound periods total */
int n; /* number of compounds per year */
p = ask("principal (amount deposited)");
r = ask("annual interest rate (percent)");
n = aski("number times compounded annually (1=yearly, 4=quarterly, 12=monthly, 26=bi-weekly, 52=weekly, 365=daily, 0=continuously)");
switch (n) {
case 365:
u = aski("total number of days");
break;
case 52:
u = aski("total number of weeks");
break;
case 26:
u = aski("total number of bi-weeks");
break;
case 12:
u = aski("total number of months");
break;
case 4:
u = aski("total number of quarters");
break;
case 2:
u = aski("total number of half-years");
break;
case 1:
u = aski("total number of years");
break;
case 0: /* compounded continuously */
u = aski("total number of years");
t = p * EXP( (((DOUBLE)r*(DOUBLE)u) / ((DOUBLE)100.0)) );
printf("%f\n", t);
return;
default:
u = aski("total number of compounded periods?");
break;
}
t = p * raise( (double) 1.0 + (r/(double)(n*100.0)), u);
r_inf = ask("annual inflation rate (percent)");
printf("Total = %f\n", t);
/* inflationary computation */
p_inf = t / raise( ( (double) 1.0 + (r_inf/(double)(n*100.0)) ), u);
printf("Total at today's dollars = %f\n", p_inf);
}
/*
* raise a real number to an non-negative integer power
* raise(r, n) returns r to the n power, where n >= 0.
* return -1 (error) if n < 0.
*/
FLOAT
raise(m, n)
FLOAT m;
int n;
{
FLOAT t; /* total returned */
register int N;
if (n < 0)
return -1;
N=n;
for (t = 1; N > 0; --N)
t = t*m;
#ifdef DEBUG
printf("DEBUG: raise(%f, %d) = %f\n", m, n, t);
#endif
return t;
}
FLOAT
ask(q)
char *q; /* question */
{
FLOAT r;
int rtn;
do {
printf("%s? ", q);
rtn = scanf("%f", &r);
if (rtn == EOF)
exit(0);
} while (rtn != 1);
return r;
}
int
aski(q)
char *q; /* question */
{
int r;
int rtn;
do {
printf("%s? ", q);
rtn = scanf("%d", &r);
if (rtn == EOF)
exit(0);
} while (rtn != 1);
return r;
}
/*
* R is (1+ r/(n*100))
* tu: t = p*R**(u-1) + ... + p*R + p
* = p * A
* (1- R**(u+1))
* A = ------------------
* (1-R)
*/
multshot()
{
FLOAT p; /* principal */
FLOAT r; /* annual interest rate */
int n; /* number of periods per year */
int u; /* total number of periods */
FLOAT R; /* intermediate, makes computation easier */
FLOAT A; /* intermediate, makes computation easier */
FLOAT t; /* total */
FLOAT Num; /* numerator in A */
FLOAT Den; /* denominator in A */
FLOAT r_inf; /* inflation interest rate */
FLOAT p_inf; /* principle - non-inflationary */
p = ask("amount deposited each time period");
r = ask("annual interest rate (percent)");
n = aski( "number of deposits per year");
if (n <= 0)
error("must be positive integer (1,2,3,...)");
R = (1.0 + (r/ ((FLOAT)n*100.0) ) );
#ifdef DEBUG
printf("DEBUG: R = %f\n", R);
#endif
u = aski( "total number of deposits");
Num = (1.0 - raise(R, u+1));
Den = (1.0 - R);
A = Num / Den;
#ifdef DEBUG
printf("DEBUG: Num = %f, Den = %f, Num/Den=%f\n", Num, Den, A);
#endif
t = p * A - p; /* don't include last deposit! */
r_inf = ask("annual inflation rate (percent)");
printf("Total = %f\n", t);
/* inflationary computation */
p_inf = t / raise( ( (double) 1.0 + (r_inf/(double)(n*100.0)) ), u);
printf("Total at today's dollars = %f\n", p_inf);
}
#ifdef DEBUG
DOUBLE
EXP(n)
DOUBLE n;
{
DOUBLE r;
r = exp(n);
printf("DEBUG: exp(%f) = %f\n", n, r);
return r;
}
#endif
/*
* ask question until yes or no is answered,
* return 1 iff yes,
* return 0 iff no.
*/
yesno( msg )
char *msg;
{
char answer[100];
int rtn;
while (1)
{
printf("%s ? ", msg);
rtn = scanf("%s", answer);
if (rtn == 0)
continue;
else if (rtn == EOF)
exit(10);
if (answer[0] == 'y' || answer[0] == 'Y')
return 1;
else if (answer[0] == 'n' || answer[0] == 'N')
return 0;
}
}
error( msg )
char *msg;
{
fprintf(stderr, "compound: error: %s, exitting\n", msg);
exit (1);
}
/*
* I derived the formula, when multiple contributions and time period
* of interest is less than time period of deposit.
* But I didn't finish writing the code to handle it.
*
* multiple compounding in general
*
* p = principal
* tc = time period of contribution
* ti = time period of interest
*
* if ti <= tc (eg interest compounded daily, deposit bi-weekly)
* time t0: p
* time t1: p * X + p,
* where X = (1 + r/(ni*100) )^(ti/tc)
* where r = annual interest rate,
* where ni = number of time periods ti per year
* where ti/tc is the number of times interest compounded between contributions
* time t2: t2 = t1 * X + p
* = (p * X + p) * X + p
* = p*X^2 + p*X + p
* time t3: t3 = t2 * X + p
* = (p*X^2 + p*X + p) * X + p
* = p*X^3 + p*X^2 + p*X + p
* ...
* time tn: tn = p*X^n + p*X^(n-1) + ... + p*X + p
* = p * (X^n + X^(n-1) + ... + 1)
* = p * A, where A represents (X^n + ... + 1).
* Solving for A:
* A = X^n + X^(n-1) + ... + 1
* A*X = X^(n+1) + ....... + X
* A(X-1) = X^(n+1) - 1.
*
* A = [X^(n+1) - 1] / (X - 1)
*/
#if 0
ilessc()
{
FLOAT p; /* principal */
FLOAT r; /* annual interest rate */
int ni; /* number of interest periods per year */
int cperi; /* number of compound periods per interest period */
int X; /* easier */
int A; /* easier */
p = ask(" amount deposited periodically" );
r = ask( "annual interest rate (percent)" );
ni = aski( "number of interest units per year" );
cperi = aski( "number of compound periods per interest period" );
/* unfinished !!! */
#endif 0
\SHAR_EOF
# ............ F I L E E N D .......... compound.c
fi # end of overwriting check
if test -f compound.man
then
echo shar: will not over-write existing file 'compound.man'
else
echo x - compound.man
# ............ F I L E B E G .......... compound.man
cat << '\SHAR_EOF' > compound.man
compound
Compound is an interactive program that asks you questions in order
to compute total money after compound interest has been accumulated.
There are two different parts (part 1 vs. part 2 computation occurs
based on how you answer the first question).
Part 1 assumes interest is compounded on a single deposit made.
Part 2 assumes interest is compounded on multiple deposits that are made
regularly at the same frequency. The total amount is then printed.
The amount printed does not include the final deposit at the end of
the final period.
In addition, compound asks you for the inflationary rate, so that you
may see how much that grand total would really be worth, but in buying
power measured in today's dollars. For example, that $17.6 million
($2000 deposited regularly for 40 years at 20% interest in Fidelity Mutuals)
dwindles to $2.2 million when there is 5% inflation.
Example - part 1
An example would be a deposit of a sum into a bank that compounds
money quarterly at 4% annually. Suppose you put $100 in on
Dec 31, 1969, and want to know how much there will be at the end of
Dec 31, 1970, one year later. Each interest period yields 1%
interest four times that year:
Deposit $100
3 months $100 * 1.01 = 101.00
6 months ($101.00 * 1.01) = 101.00 + 1.01 = $102.01
9 months $102.01 * 1.01 = 102.01 + 1.0201 = 103.0301
12 months $103.0301 * 1.01 = 103.0301 + 1.030301 = 104.060401
Because of floating point arithmetic round-off, the actual answer printed
(on a cci632 with FLOAT defined as float) is 104.060402, or
(on a cci632 with FLOAT defined as double) is 104.060401.
Example - part 2
For example, depositing $10 each quarter at 4%
compounded quarterly. At the end of 3 quarters you have:
for two quarters:
Start with $100
End of 1st quarter ($100 * 1.01 + 100) = 101.00 + 100 = 201.00
End of 2nd quarter 201 * 1.01 + 100 = 201 + 2.01 + 100 = 303.01
End of 3rd quarter 303.01 * 1.01 + 100 = 303.01 + 3.0301 + 100
= 406.0401
End of 4th quarter 406.0401 * 1.01 + 0 = 406.0401 + 4.060401
= 410.100501
(Note that each quarter is the previous quarter plus 1% (interest),
plus $100 (deposit), and note that $100 is not deposited at the end
of the 4th quarter.)
The actual number printed is slightly different because of floating point
round-offs (410.100616 on cci632).
Bugs:
Round-off errors in floating point arithmetic.
In real life, if you deposit into a bank that compounds quarterly,
your money earns no interest from the time of deposit until the
beginning of the next quarter. So if you deposit Jan 1, but
Dec 31st was the beginning of the quarter, you may have entered
the wrong number of periods. Check your bank for details.
(TODO) In part 2, it is assumed that the time period of the contributions
(deposits) is equal to the time period at which the compounding
occurs. In other words it is presently not possible to do either
of the two following:
(TODO) - interest period less than deposit period, for example,
interest compounded daily, but contributions bi-weekly
(TODO) - interest period greater than deposit period, for example,
every two weeks your employer directly deposits a fixed
amount of money into a bank that compounds quarterly.
(note that the bank is taking advantage of your money
every quarter, that is, the bank is borrowing your
money free-of-charge for a maximum time of 3 months
as long as you continue depositing there).
I assume that if interest is compounded monthly, that each
month yields the same r/12 % interest, where r = annual interest,
regardless of the number of days in a month. If a bank
weights monthly interest based on the length of the month,
this program will not agree with the bank. (I have never
heard of such a thing, but you never can tell).
Notes:
On a leap year with 366 days, and you are compounding interest
daily for one year, enter 366 as the number of periods per
year, and enter 366 as the total number of periods.
\SHAR_EOF
# ............ F I L E E N D .......... compound.man
fi # end of overwriting check
# end of shell archive
exit 0
--
FullName: Dennis Bednar
UUCP: {seismo|sundc}!rlgvax!dennis
USMail: CCI; 11490 Commerce Park Dr.; Reston VA 22091
Telephone: +1 703 648 3300