Oct 12th, 2007

hi all,

not being the quickest of people to subnet etc could any advise on this theory i read in another post.

am trying to apply it but it i dont think it that clear.

Hi, I might have a solotion for you in finding more time during your exam.

Subnetting should cost you 10 sec a question there are some 10 questions about subnetting so that would not cost you more then 2 minutes. Subnetting in 10 seconds?

yes its simple.

Write down the bitnumbers of the last 2 octets and the sequence of 128 64 etc.

_17,18,19,20,21,22,23,24

_25,26,27,28,29,30,31,32

128,64,32,16,_8,_4,_2,_1

it's simple look at the table 22 is step 4

Divide 61 by step 4 and multiply the WHOLE NUMBER 15 by step 4 the subnet is 60 next subnet is 60 plus step 4 = 64 so broadcastadress is 63.255 (B-class)

61/4=15*4=60 next subnet 64 so subnet is 63.255

or 172.168.61.1/19 look at the table. 19 is step 32

61/32=1*32 subnet= 32 next subnet is 32+32=64 so broadcast is 63.255

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## Replies

sgregg Sat, 10/13/2007 - 10:05

Man everyone makes subnetting so much more difficult than it needs to be. I mean its good to understand how a router handles it by breaking it down to binary. But for practical use of subnetting how about something like this.

Understand that IP is a 32 bit address represented by 4 numbers of which can be represented by 8 bits each, 4 numbers times 8 bits each = 32 bits. The 128-1 that you see below is representative of the 8 bits which if you add them all together = 256 (0-255)

128/64/32/16/8/4/2/1

Did you know that in a class C address each of these bits define how many hosts there are too? Also did you know for every bit you borrow from the host bits you double the number of networks you can have and cut the number of clients you can have in half. Its so simple, take a look.

For instance

255.255.255.0 /24 (Class C 1 subnet of 256 ip's)

255.255.255.128 (/25 using 1host bit)= 2 networks of 128 IP's.

255.255.255.192 (/26 using 2host bit)= 4 networks of 64 IP's.

255.255.255.224 (/27 using 3host bit)= 8 networks of 32 IP's.

etc.

Also keep in mind to convert from bits to actual number for the mask just add the bits in question together. We know if you add all 8 bits it = 256 which is represented with a 255 because you start with 0. So a /24 would be 255.255.255.0 now for each bit you add from the host you add them together. For instance a /27 is using all 8 bits on the first 3 octets and then 3 bits from the last octet. The first 3 bits are 128, 64 and 32 which if you add together = 224. So a /27 = 255.255.255.224 Which we know uses 3 bits so thats 8 networks of 32 ip addresses each.

Did you know that if you subtract a bit from the network address you double your clients for each bit you subtract. For example a class c /24 = 256 IP's, a /23 = 512 IP's, a /22 = 1024 IP's etc. Also you calculate the mask the same but you pull bits from the other end starting with 1, 2, 4, 8, etc instead of starting with 128, 64, 32, etc. For example a /23 = 255.255.254.0

/22 = 255.255.252.0

/21 = 255.255.248.0

etc...

Also dont forget each network in a subnet starts on an even number and ends in an odd number and the starting IP will be a derivative of the number of hosts it can support.

For instance /28 (255.255.255.240) provides 16 networks of 16 IP's each. So the networks will look like this

x.x.x.0-15

x.x.x.16-31

x.x.x.32-47

x.x.x.48-63

etc

Also dont forget the first IP in every subnet is used as a network address and the last IP is used for a broadcast address so the number of usable IP addresses in a subnet is always 2 less then the actual number of IP's. WATCH FOR THIS in the test. Cisco is notorious about switching up whether they want the total number of IP's or the the total number of usable IP's/supported hosts.

Maybe this doesnt help you to understand but I hope it helps a little. To be honest I dont think there is an easier approach to come up with the right answers and understanding at least how to determine subnets with little effort.

-Shawn