SUBNETTING

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Hi What is the broadcast address for the IP subnet 192.168.100.32 given a subnet mask of 255.255.255.224. I want to know step by step procedure on how to get the right ans. Brgds Mark
ASKED: May 2, 2007  4:20 AM
UPDATED: April 13, 2011  6:21 AM

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Start finding the subnet.
192.168.100.32 255.255.255.224, the first address is 192.168.100.32, the number of addr. in the subnet is
.255 subtrackted with .224 equals 31, 31 added to .32 equals .63
So the first addr is .32 and the last is .63 which means you have 32 ip addr. in the subnet, but you can not use the first .32, because it’s the network addr. of the subnet. And you can not use the last, because it’s the broadcast addr. of the subnet. This means that tou can only use 30 ip addr. in the subnet.
The binary method is.:
there must NOT be any 1 after 0, that makes en invalid subnet mask eg.
255.255.255.242 – 11111111.11111111.11111111.11110010
Your subnet mask eg.
255.255.255.224 – 11111111.11111111.11111111.11100000

Try the IP sub calc. from Wildcats
“http://www.wildpackets.com/products/free_utilities/ipsubnetcalc/overview”

best regards
Michael

which is the last addr. in the subnet
and is the net addr. of the subnet.
, the last addr. is .32 plus 31 = .63.

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  • Rlm2173
    Ok, Here is the poor mans method 192.168.100.32 given a subnet mask of 255.255.255.224 Your subnet mask is 255.255.255.224 Take the 224 and subtract it from 256 256 -224 = 32 This tell you your subnet is on a 32 boundary 1st subnet is 0 2nd sub net is 32 this is your subnet 3rd subnet is 64, thus your broadcast addr is 63 To take this a step further. Lets say you are given a host addr of 192.168.100.112 255.255.255.224 What is the subnet of this host? We know that the subnet is still on a 32 boundary (subnet mask is still 224) Now take the host ip and divide it by 32 112 / 32 = 3.5 The third subnet is 64 thus this host subnet is 192.168.100.64 255.255.255.224 Hope this helps
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