Looking for some help: /25 and /24

Hello all, scheduled to take the icnd1 on 9/12. Been studying for it using cbtnuggets, lammle, and odom. Have a question about /24 and /25 masks.

1) /25, how is the increment block "1", shouldn't it be 128?
first three octets are 255, the fourth octet is 128 (10000000), looking at that, i only have '1' subnet bit and '7' host bits.

2) /24, similar question. can anyone explain how this mask is giving me '1' subnet? no subnet bits in the fourth octet.

just really trying to understand the scheme behind it.

thanks for reading.

/regards

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bcall64

T.Abbasi wrote: »

Hello all, scheduled to take the icnd1 on 9/12. Been studying for it using cbtnuggets, lammle, and odom. Have a question about /24 and /25 masks.

1) /25, how is the increment block "1", shouldn't it be 128?
first three octets are 255, the fourth octet is 128 (10000000), looking at that, i only have '1' subnet bit and '7' host bits.

2) /24, similar question. can anyone explain how this mask is giving me '1' subnet? no subnet bits in the fourth octet.

just really trying to understand the scheme behind it.

thanks for reading.

/regards

1. You're right. /25 gives you a block of 128. If you had 192.168.1.0 /25 your range would be 192.168.1.0-192.168.1.126 and your broadcast address is 192.168.1.127. Just think of it as your lowest bit being your increment. If you had /26 which makes your last octet 11000000 your lowest bit is 64, so your increment is also 64.

2. Yes that mask gives you 1 subnet because it's a class C subnet mask. It's just the wording that has you confused. It's really just 1 network because it's a typically class c subnet mask.

Another trick is use is this: I take 256- my lowest bit. So if I had /27 which is 11100000 = 224. I simply take 256-224=32. 32 is my increment. Notice how the lowest bit is 32 and I can also subtract 224 from 256 and get the same number? Once you understand that it becomes much easier.

rickinyorkshire

bcall64 wrote: »

Another trick is use is this: I take 256- my lowest bit. So if I had /27 which is 11100000 = 224. I simply take 256-224=32. 32 is my increment. Notice how the lowest bit is 32 and I can also subtract 224 from 256 and get the same number? Once you understand that it becomes much easier.

I find it much easier just working out which bit out of 8 is the last one in my head rather then trying to do equations.

e.g:

/27 = 8.8.8.3 = 11100000 = 32.
/16 = 8.8.0.0 = 11111111 = 1

Of course this means knowing the binary table really well, which is something very basic that's required. But I guess it's each to there own and you'll find your own way of working out the fastest that suits you.

bcall64

rickinyorkshire wrote: »

I find it much easier just working out which bit out of 8 is the last one in my head rather then trying to do equations.

e.g:

/27 = 8.8.8.3 = 11100000 = 32.
/16 = 8.8.0.0 = 11111111 = 1

Of course this means knowing the binary table really well, which is something very basic that's required. But I guess it's each to there own and you'll find your own way of working out the fastest that suits you.

Yeah I like to understand it from every possible angle. On the test I will probably use the lowest bit. I like ratios as well. An example of that would be: Subnet mask is /28 240 so the ratio is 16:16 meaning I have 16 networks and 16 hosts per block. /26 would be 192 meaning my ratio would be 4:64.

SdotLow

T.Abbasi wrote: »

Hello all, scheduled to take the icnd1 on 9/12. Been studying for it using cbtnuggets, lammle, and odom. Have a question about /24 and /25 masks.

1) /25, how is the increment block "1", shouldn't it be 128?
first three octets are 255, the fourth octet is 128 (10000000), looking at that, i only have '1' subnet bit and '7' host bits.

2) /24, similar question. can anyone explain how this mask is giving me '1' subnet? no subnet bits in the fourth octet.

just really trying to understand the scheme behind it.

thanks for reading.

/regards

1) It's 1 block of 128. This block size is related to the bit. Like Ricky posted above, it's pretty easy when you know the binary bit numbers. So if it's 255.255.255.192, that's 2 bits. 11000000. That last bit will represent your block size, or increase rate so to speak, which in this case is 64.

So if you have the following, 192.168.1.0 /25 or 255.255.255.192 -

Your host ranges would be -

192.168.1.1 to 192.168.1.62 - The network for the hosts in this range would be 192.168.1.0 and the broadcast would be 192.168.1.63

192.168.1.65 to 192.168.1.126 - Network address 192.168.1.64 and broadcast 192.168.1.127

192.168.129 to 192.168.1.190 - Network address 192.168.1.128 and broadcast 192.168.1.191

192.168.193 to 192.168.1.254 - Network address 192.168.1.192 and broadcast 192.168.1.255

It makes it a lot easier when you've memorized binary and know what each bit turned off or on means. Knowing this will also help you find host and subnet ranges.

I would go so far as to say you HAVE to know this information inside and out in order to pass your ICND1. You need to subnet really fast on the fly, and to do it to multiple network ranges on a simulation. The practice tests I've taken have some pretty tough questions that have 4-5 different networks listed with various subnet masks used. So when people say you need to be able to subnet in under 10 seconds, I'd say that's pretty accurate.

2) An easier way (maybe, lol) to look at is like this, if you have a 255 this means everything in that octet is a network. Now, the only time you have "subnets" is when you break it apart by dipping into a pool outside of the class of the address.

A Class B address of 172.16.1.1 /16 is a basic class B address represented with a mask of 255.255.0.0. Those last two octets with "0" are all hosts. The 255's are pure network bits. Now this is a class B address, so when you go past /16 and move to /24 (dipping outside of the class specific mask), that third octet in 255.255.255.0 is now all subnet bits. Meaning every single number in that octet will be a network bit, and will be a subnet. No hosts will be present at all in that octet.

You need to know what ranges are Class A (/

, Class B (/16), and Class C (/24) so you can know when you're dealing with something that is a class specific network or a subnet.

T.Abbasi

tyvm for the replies everyone, i really appreciate the help.

so if i'm understanding this correctly,

/25, just as is, default for Class C, would give me '2' subnets since we are only dealing with the 4th octet.

but if i were to use /25 on a Class B, I would get '512' subnets, because the 3rd octet, .255, is considered as all network, so 11111111, and then in the 3rd octet, 10000000, is giving me a total of 9 subnet bits.

SdotLow

T.Abbasi wrote: »

tyvm for the replies everyone, i really appreciate the help.

so if i'm understanding this correctly,

/25, just as is, default for Class C, would give me '2' subnets since we are only dealing with the 4th octet.

but if i were to use /25 on a Class B, I would get '512' subnets, because the 3rd octet, .255, is considered as all network, so 11111111, and then in the 3rd octet, 10000000, is giving me a total of 9 subnet bits.

You nailed it, other than using "3rd octet" for the 4th octet :P

Ltat42a

Sounds like you're getting it. Here's a handy little tool I made when I went through the Network + course/exam. It's a IP addressing/subnetting Excel spreadsheet.

Hope it helps...

http://www.lastdayrevelation.org/albums/ipsubnetting.zip

LT

andy4tech

T.Abassi, its seems you are getting it,well as per my understanding of it, for you to get an increment you start counting from left to right on any of the class giving you like the /25 ,255.255.255.128 which is a class you start counting on the last octet ,converting the above to 11111111.11111111.11111111 .10000000,you start counting from left to right on the last octet to get your increment which is 128 when you consider the binary 128,64,32,16,8,4,2,1 if you are given /26-255.255.255.192,you apply the same method on the last octet 255.255.255.11000000,looking at this again you see the increment is 64 counting from left to right using the binary above.