Priscilla Oppenheimer wrote:
> 
> Jason Steig wrote:
> > 
> > Hello all. I'am stumped on an access-list that i need to
> > create. What i did was i set up two routers using rip and put
> > loopbacks on one of them and advertised them in rip. I then
> > attempted to build an access-list allowing just these networks
> > to pass into the other router. The router with the loopbacks
> is
> > A the destination is B. so I know this will be a standard
> > access list (direction in) on router B's interface to router
> A.
> > 
> > The requirements are 
> > 
> > allow any packet originating from 192.17.77.0 /24 
> > allow any packet originating from 192.17.73.0 /24 
> > allow any packet originating from 192.81.77.0 /24 
> > allow any packet originating from 192.81.73.0 /24 
> > allow any packet originating from 176.17.77.0 /24 
> > allow any packet originating from 176.17.73.0 /24 
> > allow any packet originating from 176.81.77.0 /24 
> > allow any packet originating from 176.81.73.0 /24 
> > 
> > Hers what i think i can do 
> > 
> > with the 182 address i can do 
> > permit ip 192.17.73.0 0.64.4.0 
> 
> 17 is 0001 0001 in binary
> 81 is 0011 0001 in binary
> 
> The one place they DON'T agree is the bit in the 2^6 place, or
> 64.
> 
> So you don't want 64, you want the opposite. Reverse all the
> bits from the answer you came up with. Remember 0 means must
> match. 1 means don't care. Then put that result in decimal.

Oops. It looks like I didn't take my own advice, that 0 means must match and
1 means don't care. So you do want 64. Sorry.

You had it right.

Priscilla

> 
> It looks like you need to reverse the bits from the answer you
> got for the next octet too.
> 
> Priscilla
> 
> > 
> > because the 64 will increase the second octet to 81 then the 4
> > in the third bit will increase the network to 77. Is this how
> i
> > would impliment this filtering policy in just two statements?
> > The same way with the 176 networks?
> 
> 




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