> > How many class C addresses can be summarized with 172.92.172.20/20?
> >
> > Is 16 the correct answers? Because there's 4 bit's left for host
addresses?
> > 2 powered by 4 is 16.
> >
> > Can one of you confirm this.
> 172.92.172.20 is a Class B address. How can you summarise a Class C
> addresses with a Class B?
Ugh, "classes" went out some time ago.
Let's work through this - I need the practice. :-)
172.92.172.20 /20 translates to: 172.92.172.20 255.255.240.0
Performing a logical "and" on the third octet of the address and the mask by
converting to binary:
172: 10101100
240: 11110000 =
= 160: 10100000
We get the following network number: 172.92.160.0 for 172.92.172.20/20
If we put all "1s" in the host address, we get a broadcast address of:
172.92.175.255
So, our "supernet" range is:
172.92.160.0 to 172.92.175.255
So, using a 20 bit mask does indeed give us a supernet of 16 class "c"
address blocks, but the network number is 172.92.160.0
Hope that helps,
Irwin
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