Re: [IPsec] DH keys calculation performance

["Dan Harkins" <[e-mail]>]

  Hello,

On Tue, July 26, 2011 6:03 am, Prashant Batra (prbatra) wrote:
> Thanks Yoav and Yaron  for the suggestions.
>
> Even I was thinking and tried generating and storing the key pair  well
> in the beginning,.  This helped to some extent.
>
>
>
> The secret calculation is also very expensive, but this has to be done
> in midst of the exchange only.

  You could pick one secret x and then for IKE exchanges do this:

0th exchange: y = g^x mod p
1st exchange: y = g^(x+1) mod p
2nd exchange: y = g^(x+2) mod p
.
.
.
nth exchange: y = g^(x+n) mod p

Getting from exchange i to exchange i+1, then, is just a single modular
multiply, which should be "cheaper" for you.

  Knowing n, y, g and p and that y = g^(x+n) mod p does not really give
an advantage (above the discrete logarithm problem) in finding x. That
said, I still would not suggest doing many more than a few of these (and
I am not qualified to quantify that statement) but for a few-- i.e. keep
n small and after n choose a new x and repeat-- it should be OK.

  Maybe this technique will allow you to "cheapen" your exchange a bit.
I think throwing hardware at this problem is your best bet though.

  regards,

  Dan.