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In-Reply-To: <HCEDJGLAJJLEAAGLKDIGCEHGEOAA.MikeSuesserott@xxxxxxxxxxx>
Oh... that looks complicated....
I think the formula, as I have it, is:
A=B+(C*D)
What I need to know is C. It looks simple algebra (not one of my fortes).
Could someone help with the formula, please, in a format like:
C= blah blah blah
Ta!
Cheers,
Ian
> Hi Ian,
>
> try this:
>
> Floor(0.1*(Sqrt(5)*Sqrt(8*Acct-75)+5))
>
> Floor might be called Int in EL, I don't remember now.
>
> Best regards,
>
> Michael Suesserott
>
>
> > -----Ursprüngliche Nachricht-----
> > Von: Ian Waugh [mailto:ianwaugh@xxxxxxxxxxxxxxxxxxx]
> > Gesendet: Thursday, October 25, 2001 17:24
> > An: omega-list@xxxxxxxxxx
> > Cc: ianwaugh@xxxxxxxxxxxxxxxxxxx
> > Betreff: Hard sums, easy formula
> >
> >
> > Would be most grateful if one of the math-types could help with a simple
> > formula. It's for a MM system which works like this:
> >
> > Contracts Account
> > 1 10
> > 2 15
> > 3 25
> > 4 40
> > 5 60
> > 6 85
> >
> > All that happens is that the account size must increase by 5, 10,
> > 15, 20 and
> > so on before adding another contract.
> >
> > In order to backtest it, I need to know how to work out how many
> > contracts
> > you can trade for a particular account size - do the sums backwards, in
> > other words, but it's got me foxed.
> >
> > Can someone give me a little formula that I can plug the account
> > size into
> > and that will give me the number of contracts? So if you plugged
> > any value
> > from 40-59 in, you'd get 4, a value from 60-84 would give you 5 and so
> > on.
> >
> > Ta!
> >
> > Cheers,
> > Ian
> >
>
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