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Subnetting question

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Subnetting question

Thanks to PN i have a block of addresses for my training purposes, thanks Cheesy

My question is i now have a mask ending 252 which is 11111000 in binary.
This means the last three digits are for the host; fine.
What i don't understand is how i have 4 addresses with only 3 availible for the host. (router, gateway, spare and broadcast) :shock:

Thanks for looking Cheesy
5 REPLIES
N/A

Subnetting question

Actually, 252 is 11111100 in binary

(you had 11111000 - that would be a mask to allow 8 addresses)

Remember that you have 4 addresses available, so they "end" 00, 01, 10, 11

I've never had a block of addresses, so remainder is from theory, not experience:

AFAIK, your router will take the first of the block, you can assign the next two to
PCs on your LAN, and the last is the "broadcast" address.
The_10th
Grafter
Posts: 1,090
Registered: 08-04-2007

Subnetting question

An IPv4 (128 bit) single octet address range is 256 digits long (0 to 255) so 256 - 252 = 4 addresses left. 00 is reserved for the network address and 11 for the broadcast address for your subnetwork. This leaves 2 host addresses 01 and 10.

When counting up in binary from you network address is as follows: 00, 01, 10, 11.

Hope this sheds some light on your question?
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Subnetting question

yes, thanks for you help both Cheesy
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Subnetting question

Ah... you're back..

So they let you out at weekends for "good behaviour" :lol:

Hope your testing goes well...
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Subnetting question

yes, i am back. Thanks, i'll let you know Cheesy