Subnetting question

in CCNA & CCENT
I was given a subnetting question which was 49.210.83.249/29(ip of host) and worked out that the subnet this is on is 49.210.83.248 with a broadcast address of 49.210.83.255.
I then thought i'd extrapolate on the question and try to calculate the next subnet after this, along with its associated broadcast and range.
I worked it out as
49.211.0.0 subnet
49.211.0.7 broadcast
49.211.0.1 to 49.211.0.6
can anyone confirm it or give me an idea where i went wrong please?
i tried checking it with a calculator but couldn't find any that would generate a whole list of subnets one after the other to find the next one after the 49.210.83.248 subnet. The best i found was solar winds program but i tried generating the list but as there's over 2000000 subnets it was taking a long time.I waited about 10 minutes then gave up as i was no where near the end of the list.
I then thought i'd extrapolate on the question and try to calculate the next subnet after this, along with its associated broadcast and range.
I worked it out as
49.211.0.0 subnet
49.211.0.7 broadcast
49.211.0.1 to 49.211.0.6
can anyone confirm it or give me an idea where i went wrong please?
i tried checking it with a calculator but couldn't find any that would generate a whole list of subnets one after the other to find the next one after the 49.210.83.248 subnet. The best i found was solar winds program but i tried generating the list but as there's over 2000000 subnets it was taking a long time.I waited about 10 minutes then gave up as i was no where near the end of the list.
I'm an Xpert at nothing apart from remembering useless information that nobody else cares about.
Comments
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MrBrian Member Posts: 520
Well since it's a Class A it has a TON of possible subnets..
I don't think the next subnet after 49.210.83.248 /29 would be 49.211.0.0 /29... going to 211 in the second octet would assume you exhausted all the possibilities in the 3rd and 4th octets.
I believe the next subnet would simply be 49.210.84.0 /29
with range: .1 - .6 in the last octet, and broadcast of .7...
and then the next subnet would be 49.210.84.8/29.. and after exhausting all the subnets in the 4th octet you'd once again just increment the 3rd octet to 85 and repeat filling up the 4th octetCurrently reading: Internet Routing Architectures by Halabi -
MrXpert Member Posts: 586 ■■■□□□□□□□
thanks for your help with this.I'm an Xpert at nothing apart from remembering useless information that nobody else cares about.