Any software that will calculate mortgage repayments?

Gnuplot.

Not software, as such, but I've used the following Java applets before now:

- Note that the fonts get a bit messed up here, so you might need to select the entire contents of a numeric field and delete using the keyboard, before typing blind. Also note that the interest rate appears to default to 0% (which would be nice!).

Best Regards, Alex.

Have a look at a similar thread started by me 'Mortgage calculate (sic) spreadsheet' posted on this group back in August. I got some good responses.

Mike

There is an Excel Loans template which can do most/all of what you want - not sure if it has made it through to the later Excel versions.

Try the loan calculator with extra payments:

Thanks, Glyn, for the link for the updated versions.

To explain further, you don't really need any software to work out what you want. Just a pocket calculator. But graph drawing software, like gnuplot, which is free and available not just for unix but also for windows - see

- is a neat way of visualising the functions.

For instance, if you want to see how the debt reduces over the course of a standard 25 year repayment mortgage at 4.5%pa using monthly rests, here is what you need to type in to gnuplot:

f=4.5/1200+1 n00 d(x)=(1-f**(x-n))/(1-f**-n) plot [0:n] d(x)

This defines f to be the monthly interest factor, n the term length in months, and d(x) the size of the debt at month x, scaled to 1.0 corresponding to the amount borrowed. The graph is then plotted for the whole range of the term. It looks like a curve resembling a straight diagonal line from 1 at month 0 to 0 at month 300, but with a bit of a bulge on top. The bulge is more pronounced the higher the interest rate is. The curve is actually an upside down, origin-shifted, exponential.

Now, the effect of overpaying k each month is to amass a balance just as if putting k per month into a savings account which compounds monthly and pays interest tax free at the loan rate. The formula, in gnuplot form, for the balance so amassed after x months, is:

b(x,k)=k*(f**x-1)/(f-1)

And all we have to do to see how much would be amassed by overpaying one, two, or three thousandths per month of the initial amount borrowed, is to plot the three functions, together with the above debt function, as follows:

plot [0:n] d(x),b(x,.001),b(x,.002),b(x,.003)

Using k=0.001 corresponds to overpaying £100 a month on a £100k loan.

Where the b graphs (which are right-side-up origin-shifted exponentials) intersect the d graph, the balances amassed match the loan debt, and the mortgage is paid off. You can see approximately where in time these points lie, and you can get a more accurate idea by zooming in by modifying the range:

plot [220:240] d(x),b(x,.001)

Tells you that the k=.001 pay-off point lies at 227 months, i.e.

73 months (just over 6 years) early.

m=(f-1)/(1-f**-n) print m

Defines m to be the monthly repayment required, and prints it (it comes out as about 0.00556). If the loan is paid off 73 months early as a result of 227 overpayments of size .001, it means we save 73*m at the cost of .227.

print 73*m-.227

Tells us how much is saved in terms of raw cashflow. In the example, the answer given is about 0.17876, so you'd save nearly £18k on a £100k loan by paying £100 more each month than the normal £556 payment.

Incidentally, it is worth pointing out that a normal repayment plan

**is equivalent** to overpaying, relative to an interest-only mortgage, that amount by which the monthly repayment m exceeds the monthly interest f-1.

You can see this by doing

plot [0:n] b(x,m-f+1)

which shows that the balance amassed by saving £181 a month (the amount by which £556 exceeds the interest-only payments of £375) will in fact reach £100k after 300 months.

I usually use the BBC website

Andy