subnetting

Unanswered Question
Feb 13th, 2008

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 ?

I have this problem too.
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pciaccio Wed, 02/13/2008 - 08:11

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...

scottmac Wed, 02/13/2008 - 08:14

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

scottmac Wed, 02/13/2008 - 08:12

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

ritesh-mahajan Wed, 02/13/2008 - 10:56

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.

carl_townshend Thu, 02/14/2008 - 05:42

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 ?

jamesawoodward Fri, 02/22/2008 - 17:34

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

carl_townshend Wed, 02/27/2008 - 09:57

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

jamesawoodward Thu, 02/28/2008 - 17:14

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

carl_townshend Fri, 02/29/2008 - 02:33

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 ?

carl_townshend Fri, 02/29/2008 - 03:53

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 ?

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