CIDR question

If i had a block of two class C address, for example.

192.168.64.0/23 a block size of two class C address's

510 hosts?

Would the broadcast address of this block be 192.168.65.255/23 ?

TIA

Reply to
Jed
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Class C address starts at /24 and can not be a /23.

Here are some helpful links.

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Reply to
Freeride

Yes

Yes, the nine host bits are all 1s.

Aubrey

Reply to
Aubrey Adams

The question is about classless addressing (CIDR/VLSM) so the 2 contiguous "Class C" networks are summarised/supernetted with a /23 prefix.

IPv4 Classful addressing/routing is a legacy approach these days and will eventually become a historical footnote (except when configuring EIGRP and you have remember "no auto-summary").

Reply to
Aubrey Adams

yes

Reply to
BernieM

ok thanks all. :-)

Reply to
Jed

/23 = 255. 255. 254 (7+1).0 Bits left for hosts are 8+1 = 9

2^9=512 or 510 usable hosts

eager and beaver a.k.a master of subnets

Reply to
eager

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