02-13-2008 07:25 AM - edited 03-05-2019 09:07 PM
Hi all, can anyone please tell me how many subnets are in the address
10.112.64.0/23
from the calculator its saying
10.112.64.1 - 10.112.65.254
is this 2 subnets?
how is it worked out so that it goes from 10.112.64.1 to 10.112.65.255 ?
02-13-2008 08:11 AM
With the subnet mask you supplied /23 your subnet starts out at 10.112.64.0 and ends at 10.112.64.255. You can use IP addresses for your hosts at 10.112.64.1 to 10.112.64.254.
This is one subnet....Hope this helps...
02-13-2008 08:14 AM
I think you typo'd ... the host range is 10.112.64.1 - 10.112.65.254.
10.112.64.0 is the network, 10.112.65.255 is the broadcast.
Good Luck
Scott
02-13-2008 08:12 AM
It's the /23 mask.
The base value for the third octet is 64; the last bit of the third octet is given to the host-portion of the address.
Once the first 255 hosts have been counted (255 in the fourth octet), the last bit (the "least significant bit" of the third octet is incremented (from a 0 to a 1) and the value of the third octet changes to 65.
Also remember that in this case an address of 10.112.65.0 is *** NOT *** a network address, since one of the "host bits" is "1" (the last bit of the third octet).
In the same light, the address 10.112.64.255 is ***NOT*** a broadcast address, since one of the "host bits" is still a zero (the last bit of the third octet).
Remember that a "network address' is ALL ZEROs in the host portion and a broadcast is ALL ONES in the host portion.
If any bit in the host portion doesn't meet that criteria, even if it's in another octet, then it's a valid host address.
When in doubt, write it out in binary. There are a lot of shortcuts to IP addressing; you should still practice with binary until you understand it, so you'll better understand why the shortcuts work.
Good Luck
Scott
02-13-2008 10:56 AM
Since the subnet mask you chose is 255.255.254.0 or /23 this is a single subnet with 510 host apart from 10.112.64.0 & 10.112.65.255 being the subnet & broadcast addresses respectively.
02-14-2008 05:42 AM
Im still a little unsure on this one why we have the 64 and 65, is this a subnetted class a address ? and how do we work out how many networks and subnets we have in this case, I know for netwworks and hosts you count the bits to the power 2 etc, but how does it work in this case ?
02-22-2008 05:34 PM
Hi Carl,
Well - here's how I look at it.
subnetting is taking network bits and giving them to the hosts - moving the line to the right, so to speak. a Class C address has 24 bits for the network, and 8 bits for the hosts. When you subnet that, you move the line to the right - for example 255.255.255.240 or /28. This divides your 254 host bits in 1 subnet into 16 subnets of 14 hosts each. 2(4th) subnets, 2(4th-2) hosts/subnet.
Your question is a little tricky. If the question had asked you:
take 10.0.0.0 /23 and give the number of subnets and hosts - you would do this:
10 = class A address
8 = default # of bits for class A address
/23 = 15 bits taken from the hosts for the network -->
2(15th) = number of subnets
9 = number of bits left for hosts =
510 = number of hosts/subnet 2(9th-2) = 510
But the qestion has already subnetted, and chosen one particular subnet. So that is one subnet, with 510 hosts.
subnet address: 10.112.64.0
first host address: 10.112.64.1
last host address: 10.112.65.254
broadcast address: 10.112.65.255
Hope this helps.
Jim Woodward
02-27-2008 09:57 AM
hi there
can you please tell me when they say a /23 mask can have 32728 subnets, what are these subnets? ie where do they start ? as i though that the network part does not change, or will it just be the first octet that does not change, ie 10.0.0.0/23, would the networks be 10.0.1.0 then 10.0.2.0 etc ?
please help
02-28-2008 05:14 PM
Hi Carl,
To answer your question very simply - your example of 10.0.1.0, 10.0.2.0, 10.0.3.0, 10.0.4.0 etc - that would be 10.0.0.0/24. A /23 mask, on the other hand, increments by 2, because the mask terminates on the 23rd bit, which represents the "2s" position.
1st subnet: 10.0.0.0 - 10.0.1.255
2nd subnet - 10.0.2.0 - 10.0.3.255
3rd subnet - 10.0.4.0 - 10.0.5.255
ad nauseam up to
10.255.254.0 - 10.255.255.255
If the subnet mask was /22 - the mask would terminate on the 22nd bit, which in binary represents the "4s" - and thus the subnets would increment by 4s and not 2s:
1st subnet - 10.0.0.0 - 10.0.3.255
2nd subnet - 10.0.4.0 - 10.0.7.255
3rd subnet - 10.0.8.0 - 10.0.11.255
you would have 16,384 subnets (exactly half the # of subnets from the /23 mask, because you have exactly one less bit for subnetting.
Hope this helps.
Jim Woodward
02-29-2008 02:33 AM
hi Jim, can you tell me where you get 16384 from? what happens when you reach 10.0.255.0 network, does it then start from 10.1.0.0, then then 10.1.2.0 etc ?
02-29-2008 03:53 AM
Hi Jim, so does the second octet still move up in 1's ie 10.0.255.254 then goes to 10.1.0.0 then 10.1.2.0 etc ?
03-05-2008 12:00 PM
Hi Carl,
Check out the following link as its easier to understand. Its sometimes hard to put your own words on something and I think this article hits the nail on the head so to speak.
http://www.cisco.com/warp/public/701/3.html
Craig
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