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Feb 13th, 2008
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Hi all, can anyone please tell me how many subnets are in the address

from the calculator its saying -

is this 2 subnets?

how is it worked out so that it goes from to ?

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pciaccio Wed, 02/13/2008 - 08:11
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With the subnet mask you supplied /23 your subnet starts out at and ends at You can use IP addresses for your hosts at to

This is one subnet....Hope this helps...

scottmac Wed, 02/13/2008 - 08:14
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I think you typo'd ... the host range is - is the network, is the broadcast.

Good Luck


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


ritesh-mahajan Wed, 02/13/2008 - 10:56
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Since the subnet mask you chose is or /23 this is a single subnet with 510 host apart from & being the subnet & broadcast addresses respectively.

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

first host address:

last host address:

broadcast address:

Hope this helps.

Jim Woodward

carl_townshend Wed, 02/27/2008 - 09:57
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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, would the networks be then etc ?

please help

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

To answer your question very simply - your example of,,, etc - that would be 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: -

2nd subnet - -

3rd subnet - -

ad nauseam up to -

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 - -

2nd subnet - -

3rd subnet - -

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
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hi Jim, can you tell me where you get 16384 from? what happens when you reach network, does it then start from, then then etc ?

carl_townshend Fri, 02/29/2008 - 03:53
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Hi Jim, so does the second octet still move up in 1's ie then goes to then etc ?


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