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's510 hosts?
Would the broadcast address of this block be 192.168.65.255/23 ?
TIA
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's510 hosts?
Would the broadcast address of this block be 192.168.65.255/23 ?
TIA
Class C address starts at /24 and can not be a /23.
Here are some helpful links.
Yes
Yes, the nine host bits are all 1s.
Aubrey
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").
yes
ok thanks all. :-)
/23 = 255. 255. 254 (7+1).0 Bits left for hosts are 8+1 = 9
2^9=512 or 510 usable hostseager and beaver a.k.a master of subnets
Cabling-Design.com Forums website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.