# subnetting practice

Member Posts: 918
Does anyone know a site that offers random subbnetting practice? Something that has consistently different questions?
encrypt the encryption, never mind my brain hurts.

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Here you go...

http://www.subnettingquestions.com

Got it from someone here on the site and bookmarked it...

Enjoy...
CCNA (Expired...), MCSE, CWNA, BSc Computer Science
Working on renewing CCNA!
• Member Posts: 749
There are only 10 kinds of people... those who understand binary, and those that don't.

CCIE Studies: Written passed: Jan 21/12 Lab Prep: Hours reading: 385. Hours labbing: 110

Taking a time-out to add the CCVP. Capitalizing on a current IPT pilot project.
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does anyone feel that subnetting can be easier? if you know the /30-/25 are in Class C.... hmmmm trying to explain but hard to lol.

if you have the subnet mask of 252..../30 /22... theres exactly 8 between 30 and 22.... if you know 252 is 30 all you do is take 8 for /22 and take 8 for /14... you get the / prefix number of class B and A if you know your class C subnet masks.....

This then leads to knowing block size for each mask and number of subnets.... for /30 is block size 4 - number of subnets 64 for each Class....

Now if you think Class B has plus 8 bits and class A has plus 16 bits (bits at 0 = hosts -2) the /22 must have 10 bits at 0 if the /30 has 2 bits at 0 and 2x2x2x2x2x2x2x2x2x2=1024 - 2 =1022..... If you learn /22 host number of 1022 it's easier to work out other hosts /20 is 1022 x2 +2 =2042 x2 +2 = 4094 hosts for /20

All you need is Class C knowledge, Magic number 8... 8 for same mask number, 8 for bit numbers at 0 and the host number for the /22 as class C hosts can be done in your head.

Its seams clear to me, would like to hear your opinions.
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What I was taught was using block size is the easiest. But you need to understand the prefix notation and the dotted decimal notation for your subnet mask.

For example if you got a question that asks what network is the host 192.168.1.67 /28 in?

First you know this is a class C address therefore the default subnet mask is 255.255.255.0 or /24.

Second, you know there were 4 bits borrow on the 4th octet, that would mean it is 255.255.255.240.

Third, you get the block size by taking 256 -240 of the fourth octet = 16 block size. Therefore by counting as 0 to 16 to 32 to 48 to 64 to 80 to 96 etc. You can conclude that the host belongs in the network 192.168.1.64 and its broadcast address is always one number before the next network which is 192.168.1.79. There fore any host between those are the valid host.
On to MCSA 2003.