bonl1@CFSMO.HONEYWELL.COM (Boniface Lee #1) (01/03/91)
Does anyone has a source program for mortgage payment calculations? I like to run it on my IBM PC...please e-mail the program...thanks! Boniface Lee (bonl1@cfsmo.honeywell.com) Honeywell Commercial Flight Systems - Minneapolis Operations (612) 785-4629
sushrut@ee.rochester.edu (Sushrut Mehta) (01/04/91)
In article <9101022144.AA13152@hp370b.CFSMO.Honeywell.COM> bonl1@CFSMO.HONEYWELL.COM (Boniface Lee #1) writes: > >Does anyone has a source program for mortgage payment calculations? I like >to run it on my IBM PC...please e-mail the program...thanks! > > >Boniface Lee (bonl1@cfsmo.honeywell.com) >Honeywell Commercial Flight Systems - Minneapolis Operations >(612) 785-4629 Me too, please. Thanx.
laird@chinet.chi.il.us (Laird J. Heal) (01/04/91)
>Does anyone has a source program for mortgage payment calculations? I like >to run it on my IBM PC...please e-mail the program...thanks! Well, I make a habit of deriving the formula every so often - not on the net before, so here goes: P=principal, n=number of payments, i=interest, x=payment: month Principal Interest payment new Principal 1 P iP x P+iP-x 2 P-iP+x i(P+iP+x) x P+iP-x+i(P+iP-x)-x iP+(i**2)P-ix P+2iP+(i**2)P-ix-2x 3 P+2iP+(i**2)P-ix-2x iP+2(i**2)P+(i**3)P-(i**2)x-2ix x P+3iP+3(i**2)P+(i**3)P-(i**2)x-3ix-3x P(1+i)**3-((i**3)x+3i(**2)x+3ix)/i P(1+i)**3-((i**3)x+3i(**2)x+3ix+x - x)/i P(1+i)**3-((i+1)**3) - 1)x/i and unless you misbelieve the patterns, to set "new Principal" to zero after n months, we have to set P(1+i)**n=((i+1)**n) - 1)x/i and x = Pi((1+i)**n)/(((1+i)**n) - 1) which can be further reduced, by means of such modern tools as pencil and paper, to x=Pi(1/(1-(1/(1+i)**n))) in order to only make one exponentiation. Well, I didn't write the program for you (I usually just use a pocket calculator) but the only thorny problem is really the one: given the monthly payment I can afford and the price/interest-rate I am facing, how long will it take to make to payoff? as x=Pi((1+i)**n)/(1 - (1+i)**n) x - x(1+i)**n = Pi(1+i)**n x=(Pi+x)(1+i)**n x/(Pi+x)=(1+i)**n n=(log(x/(Pi+x)))/log(1+i) Now maybe if I get really ambitious I will put together a little bit of scanf() and printf() and...nah. -- Laird J. Heal The Usenet is dead! Here: laird@chinet.chi.il.us Long Live the Usenet!
carl@p4tustin.UUCP (Carl W. Bergerson) (01/04/91)
In article <9101022144.AA13152@hp370b.CFSMO.Honeywell.COM> bonl1@CFSMO.HONEYWELL.COM (Boniface Lee #1) writes: > >Does anyone has a source program for mortgage payment calculations? I like >to run it on my IBM PC...please e-mail the program...thanks! At the risk of being the bizillionth responder, have you tried MORTGAGE.BAS? This program has been included with every MS/PC DOS release that I have used. -- Carl Bergerson uunet!p4tustin!carl Point 4 Data Corporation carl@point4.com 15442 Del Amo Avenue Voice: (714) 259 0777 Tustin, CA 92680-6445 Fax: (714) 259 0921
simon@castle.ed.ac.uk (Simon Brown) (01/07/91)
In article <1991Jan04.112213.5193@chinet.chi.il.us> laird@chinet.chi.il.us (Laird J. Heal) writes: >>Does anyone has a source program for mortgage payment calculations? I like >>to run it on my IBM PC...please e-mail the program...thanks! > >Well, I make a habit of deriving the formula every so often - not on the net >before, so here goes: > >P=principal, n=number of payments, i=interest, x=payment: > (etc) ... But remember not to try this one in Britain, where the interest rate increases every few months and throws all your calculations down the sewer... ------------------------------------------------------------------------------- Simon Brown simon@meiko.co.uk Meiko Scientific Ltd. simon@uk.ac.ed (Edinburgh Office, Edinburgh Parallel Computing Centre)
jog@hpcupt1.cup.hp.com (Rajeev Jog) (01/08/91)
> / hpcupt1:alt.sources.wanted / laird@chinet.chi.il.us (Laird J. Heal) / 3:22 am Jan 4, 1991 / > >Does anyone has a source program for mortgage payment calculations? I like > >to run it on my IBM PC...please e-mail the program...thanks! > > Well, I make a habit of deriving the formula every so often - not on the net > before, so here goes: > > P=principal, n=number of payments, i=interest, x=payment: > > really the one: given the monthly payment I can afford > and the price/interest-rate I am facing, how long will > it take to make to payoff? > > as x=Pi((1+i)**n)/(1 - (1+i)**n) > x - x(1+i)**n = Pi(1+i)**n > x=(Pi+x)(1+i)**n > x/(Pi+x)=(1+i)**n > n=(log(x/(Pi+x)))/log(1+i) > > Now maybe if I get really ambitious I will put together > a little bit of scanf() and printf() and...nah. > -- > Laird J. Heal The Usenet is dead! > Here: laird@chinet.chi.il.us Long Live the Usenet! Might help to try all the scanf(), and printf() and actually use it; For all positive x, P, and i, the formula n= (log (x / (Pi+x)) ) / log(1+i) above yields a negative numerator and a positive denominator, hence a negative number of payments. I wish the same were true of my mortgage. Rajeev Jog ******************************************************************************* Hardware Systems Performance * voice : +1 408 447 0220 Hewlett-Packard * e-mail: jog@hpda.cup.HP.COM 19447 Pruneridge Avenue, MS 42LX * jog%hpda@hplabs.HP.COM Cupertino, CA. 95014-9974 USA * fax : +1 408 447 4907 *******************************************************************************