subnetting

[jaye1234]

Can anyone help.
I have a basic understanding of subnetting but an having difficulty working out the number of subnets required, e.g.
The question I am puzzeled over is:
Given the network address 10.0.0.0
create a mask for 1600 networks & identify the 127th network
vlsm this network address for a further 122 subnets

using my subnet chart I come up with a /21 prefix or mask 255.255.248.0
which results in the networks increasing by 8 in the third octet
0 = 10.0.0.0 - 10.0.7.255
1 = 10.0.8.0 - 10.0.15.255
2 = 10.0.16.0 - 10.0.23.255

& so on

my confusion is, from my understanding, is this provides 32 subnets from:
10.0.0.0 - 10.0.248.0 = 32 subnets
then
10.1.0.0 - 10.1.248.0 = 32 subnets
if the value of the second octet is 0 - 255 then surely that means 255 x 32 for the 32 subnets in each of the third octet but this equals 8192 subnets not 2048 which is the result from using /21 mask.

where am i going wrong?

[Prometheus_Fire]

Hi jaye1234,

Thanks for starting a new thread. It gives your question better visibility.

I think you've mixed up your bit boundary a little, so I'll start from the beginning. Please feel free to check my numbers as I rushed though this and didn't compare the common bits at the end. I think the basic idea here is sound though!

1. 10.0.0.0 is a class A IP network so the first 8 bits are out right away:
   > NNNNNNNN./HHHHHHHH.HHHHHHHH.HHHHHHHH
   > 11111111./00000000.00000000.00000000
   
   NB: N = Network bit, S = Subnet bit, H = Host bit & / = your bit boundary.
   
2. That leaves 24 bits to work with and the number of subnets available, just like the number of hosts goes up in powers of 2. E.g.
   • 1 borrowed bit = 2 subnets
   • 2 borrowed bits = 4 subnets
   • 3 borrowed bits = 8 subnets
   • 4 borrowed bits = 16 subnets
   • 5 borrowed bits = 32 subnets
   • 6 borrowed bits = 64 subnets
   • 7 borrowed bits = 128 subnets
   • 8 borrowed bits = 256 subnets
   • 9 borrowed bits = 512 subnets
   • 10 borrowed bits = 1024 subnets
   • 11 borrowed bits = 2048 subnets
3. If you are given a minimum number of subnets always go for the next highest available number, which in this case is 2048 as 10 borrowed bits doesn't provide 1600 subnets.
4. So what you're left with is this:
   > NNNNNNNN.SSSSSSSS.SSS/HHHHH.HHHHHHHH
   > 11111111.11111111.111/00000.00000000
   This makes your last subnet mask: 255.255.224.0 and leaves you with 13 host bits.
5. Working on the basis above your borrowed bits table works out like so:
   • 1 borrowed bit = 2 subnets
     > Subnet mask = 255.128.0.0
   • 2 borrowed bits = 4 subnets
     > Subnet mask = 255.192.0.0
   • 3 borrowed bits = 8 subnets
     > Subnet mask = 255.224.0.0
   • 4 borrowed bits = 16 subnets
     > Subnet mask = 255.240.0.0
   • 5 borrowed bits = 32 subnets
     > Subnet mask = 255.248.0.0
   • 6 borrowed bits = 64 subnets
     > Subnet mask = 255.252.0.0
   • 7 borrowed bits = 128 subnets
     > Subnet mask = 255.254.0.0
   • 8 borrowed bits = 256 subnets
     > Subnet mask = 255.255.0.0
   • 9 borrowed bits = 512 subnets
     > Subnet mask = 255.255.128.0
   • 10 borrowed bits = 1024 subnets
     > Subnet mask = 255.255.192.0
   • 11 borrowed bits = 2048 subnets
     > Subnet mask = 255.255.224.0
6. Looking at the list above you can see that 7 borrowed bits gives us 128 subnets so that's where the 127th network lives.
7. To find the exact network we subtract the rightmost non-255 octet of the subnet mask from 256, which in this case is 254. This gives us a remainder of 2.
8. We know that the subnets rise in increments of 2, so they should go as follows:
   • Subnet 1 = 10.0.0.0/15
   • Subnet 2 = 10.2.0.0/15
   • Subnet 3 = 10.4.0.0/15
   • Subnet 4 = 10.6.0.0/15
   • Subnet 5 = 10.8.0.0/15
   • Subnet 6 = 10.10.0.0/15
   • Subnet 7 = 10.12.0.0/15
   • Subnet 8 = 10.14.0.0/15
   • Subnet 9 = 10.16.0.0/15
   • Subnet 10 = 10.18.0.0/15
     And so on until you reach the last few
   • Subnet 125 = 10.250.0.0/15
   • Subnet 126 = 10.252.0.0/15
   • Subnet 127 = 10.254.0.0/15
9. This leaves you with a network of 10.254.0.0 and a subnet mask of 255.254.0.0, meaning that you have 16 bits left to play with.
10. As per the table above we want a further 7 borrowed bits for at least a further 122 subnets, which gives us 14 borrowed bits, or:
    > NNNNNNNN.SSSSSSSS.SSSSSS/00.00000000
    > 11111111.11111111.111111/00.00000000
    > Subnet mask = 255.255.252.0

I believe that the resultant network comes out to 10.254.0.0/22. This all seems rather non-intuitive until you remember that each time the bit boundary changes it's filtering traffic through to a different network or device.

Hope this makes sense to you. If I've made a mistake anywhere, then please feel free to correct me! It's left me feeling rather confused!

[chrishunt]

check out the TCP/IP guide subnetting topic

http://www.tcpipguide.com/free/t_IPS...ngConcepts.htm

I think where your going wrong is thinking in decimal, rather than base 2. An ip address is a 32 bit address that is REPRESENTED as 4 decimal octects conventionally. With subnetting it is in my opinion easier to work in base 2 so

255.255.248.0

is

11111111.11111111.11111000.00000000

now rember the subnet mask is AND'd with the IP address in order to determine which are the network bits and which are the host bits. So with that network mask for an IP address of say 10.0.16.100, represented in binary is

00001010.00000000.00010000.01100100

The red part indicates the network and the remainder is the host. So you have from that netmask 11 bits for the host and 21 bits for the network, so 2 to the 11 hosts so 2048 hosts (minus a couple of reservered values). Note 2048 hosts not subnets!

Of course this then means that

00001010.00000000.0001000 1.01100100

Is also a host on the same network but would have IP address in decimal

10.0.17.100

So in this representation even though the 3rd octed is 17 this is not a different network!! Given that most people use 255.255.255.0 the fact that the 3rd octet is a different decimal number can lead folks to the wrong conclusion that its a host on a different network.

Now given your saying the network address is 10.0.0.0. Checking IANA

http://www.iana.org/assignments/ipv4...ress-space.xml

this says that the 10 prefix (10/8) is for private use, therefore not allocated by IANA or a registry. So given that for 10.0.0.0. only the 1st octet is fixed then the 8 bits in the 2nd octect and the first 5 from the 3rd are available for the (variable) network part using the mask you stated, ie 13 bits which would give you 2 to the 13 networks, or 8192 with each allowed 2048 hosts (less the reservered network and multicast addresses)


Does that help?

[Prometheus_Fire]

Ok, I've just come back to this with a fresh pair of eyes and I have no idea what I was thinking earlier! I was fine up until step 5 then I apparently took complete leave of my senses and decided to move everything into the wrong octet!

1. 10.0.0.0 is a class A IP network so the first 8 bits are out right away:
   Bits................. > NNNNNNNN./HHHHHHHH.HHHHHHHH.HHHHHHHH
   Network Address > 00001010./00000000.00000000.00000000
   Subnet Mask..... > 11111111./00000000.00000000.00000000
   
   >NB: N = Network bit, S = Subnet bit, s = VLSM Subnet bit, H = Host bit & / = your bit boundary.
   
   
2. That leaves 24 bits to work with and the number of subnets available, just like the number of hosts goes up in powers of 2. E.g.
   
   • 1 borrowed bit = 2 subnets
   • 2 borrowed bits = 4 subnets
   • 3 borrowed bits = 8 subnets
   • 4 borrowed bits = 16 subnets
   • 5 borrowed bits = 32 subnets
   • 6 borrowed bits = 64 subnets
   • 7 borrowed bits = 128 subnets
   • 8 borrowed bits = 256 subnets
   • 9 borrowed bits = 512 subnets
   • 10 borrowed bits = 1024 subnets
   • 11 borrowed bits = 2048 subnets
   
   
3. If you are given a minimum number of subnets always go for the next highest available number, which in this case is 2048 as 10 borrowed bits doesn't provide 1600 subnets.
   
   
4. So what you're left with is this:
   
   Bits................. > NNNNNNNN.SSSSSSSS.SSS/HHHHH.HHHHHHHH
   Network Address > 00001010.00000000.000/00000.00000000
   Subnet mask..... > 11111111.11111111.111/00000.00000000
   >This makes your subnet mask: 255.255.224.0 and leaves you with 13 host bits.
   
   
5. Working on the basis above your borrowed bits table works out like so:
   
   • 1 borrowed bit = 2 subnets
     > Subnet mask = 255.128.0.0
   • 2 borrowed bits = 4 subnets
     > Subnet mask = 255.192.0.0
   • 3 borrowed bits = 8 subnets
     > Subnet mask = 255.224.0.0
   • 4 borrowed bits = 16 subnets
     > Subnet mask = 255.240.0.0
   • 5 borrowed bits = 32 subnets
     > Subnet mask = 255.248.0.0
   • 6 borrowed bits = 64 subnets
     > Subnet mask = 255.252.0.0
   • 7 borrowed bits = 128 subnets
     > Subnet mask = 255.254.0.0
   • 8 borrowed bits = 256 subnets
     > Subnet mask = 255.255.0.0
   • 9 borrowed bits = 512 subnets
     > Subnet mask = 255.255.128.0
   • 10 borrowed bits = 1024 subnets
     > Subnet mask = 255.255.192.0
   • 11 borrowed bits = 2048 subnets
     > Subnet mask = 255.255.224.0
   
   My apologies, lets ignore everything in my previous post after step 5 and lets do it properly now shall we?
   
   
6. At step 6 what we have is:
   
   Bits................. > NNNNNNNN.SSSSSSSS.SSS/HHHHH.HHHHHHHH
   Network Address > 00001010.00000000.000/00000.00000000
   Subnet Mask..... > 11111111.11111111.111/00000.00000000
   > This makes your last subnet mask: 255.255.224.0 and leaves you with 13 host bits.
   
   > That gives us a supernet of 10.0.0.0/19 that we have to find the 127th subnet in.
   
   
7. The subnets go up in increments of 32 (256 - 224 = 32) as follows.
   
   • Subnet 1 = 10.0.0.0/19 (10.0.0.0 to 10.0.31.255)
   • Subnet 2 = 10.0.32.0/19 (10.0.32.0 to 10.0.63.255)
   • Subnet 3 = 10.0.64.0/19 (10.0.64.0 to 10.0.95.255)
   • Subnet 4 = 10.0.96.0/19 (10.0.96.0 to 10.0.127.255)
   • Subnet 5 = 10.0.128.0/19 (10.0.128.0 to 10.0.159.255)
   • Subnet 6 = 10.0.160.0/19 (10.0.160.0 to 10.0.191.255)
   • Subnet 7 = 10.0.192.0/19 (10.0.192.0 to 10.0.223.255)
   • Subnet 8 = 10.0.224.0/19 (10.0.224.0 to 10.0.255.255)
   • Subnet 9 = 10.1.0.0/19 (10.1.0.0 to 10.0.31.255)
   • Subnet 10 = 10.1.32.0/19 (10.1.32.0 to 10.0.63.255)
   • Subnet 11 = 10.1.64.0/19 (10.1.64.0 to 10.0.95.255)
   • Subnet 12 = 10.1.96.0/19 (10.1.96.0 to 10.0.127.255)
   • Subnet 13 = 10.1.128.0/19 (10.1.128.0 to 10.0.159.255)
   • Subnet 14 = 10.1.160.0/19 (10.1.160.0 to 10.0.191.255)
   • Subnet 15 = 10.1.192.0/19 (10.1.192.0 to 10.0.223.255)
   • Subnet 16 = 10.1.224.0/19 (10.1.224.0 to 10.0.255.255)
   • Subnet 17 = 10.2.0.0/19 (10.2.0.0 to 10.0.31.255)
   • Subnet 18 = 10.2.32.0/19 (10.2.32.0 to 10.0.63.255)
   • Subnet 19 = 10.2.64.0/19 (10.2.64.0 to 10.0.95.255)
   • Subnet 20 = 10.2.96.0/19 (10.2.96.0 to 10.0.127.255)
     And so on until you reach the last few
   • Subnet 121 = 10.15.0.0/19 (10.15.0.0 to 10.0.31.255)
   • Subnet 121 = 10.15.32.0/19 (10.15.32.0 to 10.0.63.255)
   • Subnet 122 = 10.15.64.0/19 (10.15.64.0 to 10.0.95.255)
   • Subnet 123 = 10.15.96.0/19 (10.15.96.0 to 10.0.127.255)
   • Subnet 124 = 10.15.128.0/19 (10.15.128.0 to 10.0.159.255)
   • Subnet 125 = 10.15.160.0/19 (10.15.160.0 to 10.0.191.255)
   • Subnet 126 = 10.15.192.0/19 (10.15.192.0 to 10.0.223.255)
   • Subnet 127 = 10.15.224.0/19 (10.15.224.0 to 10.0.255.255)
   
   
8. OK, we're on the right track this time (I knew it didn't look right before!).
   
   
9. So now we have the 127th subnet 10.15.224.0/19, we need to get another 122 subnets out of it which means borrowing another 7 bits.
   
   
10. 7 borrowed bits = 128 subnets which makes our /19 a /26 and gives us:
    
    Bits................. > NNNNNNNN.SSSSSSSS.SSSsssss.ss/HHHHHH
    Network Address > 00001010.00001111.11100000.00/000000
    Subnet Mask..... > 11111111.11111111.11111111.11/000000
    > This makes your last subnet mask: 255.255.255.192 and leaves you with 6 host bits.
    
    
11. So the answer is 10.15.224.0/26, which gives you an extra 128 subnets from the 127th subnet in your 10.15.224.0/19 supernet and 62 host IP addresses in each subnet (6 host bits X ^2 - 2).

Clear as mud?

[chrishunt]

I'd be careful talking about a "Class A" address, didn't IANA move away from classful addressing almost 2 decades ago (although I accept many folks still talk about it) hence why IP addressing is normally written with the slash notation nowadays to indicate how many network bits there are?

ie

10.0.16.100/8

means that there are 8 bits for the network part of the address and as such the network mask written in dotted notation would be

255.0.0.0

Of course its up to the ip architect in an organisation to decide how they then use the host bits in the network they have been allocated. Even with a "/24" network, ie 8 bits for the host (24 bits for the fixed network part) if you only have a few host in each segment you could "subnet" that.

[Prometheus_Fire]

The question sounded like a certification question, is it Jaye1234?

Besides, even if it's not, there's no way you're gonna see a CIDR /8 in a production environment!

[chrishunt]

As the 10/8 network is private, ie not allocated by IANA or one of its registries and not routed in the internet you see it in many organisations :-) Along with 192.168.0.0 which is also private

[Prometheus_Fire]

Either way, it's always subnetted down and we're getting off topic.

Lets wait for Jaye1234 to clarify.

[jaye1234]

Hi guys,

Thanks for the pointers.
Prometheus fire, I was making the exact same mistake as you did in your first reply (after i realised i was useing the wrong mask) which is what was doing my head in.
The mistake I was making was useing the formula for calculating hosts to calculate subnets.
It took me looking at it the morning after to see it with fresh eyes too.
One query though, oh & yes this was an cisco accademy assesment question.
We were asked to use subnet 0 so, given that 10.0.0.0/19 is subnet 0 &
10.0.32.0/19 is subnet 1 etc this makes 10.15.192.0/19 subnet 126
But heres the thing... Its my understanding that subnet 0 is the 1st subnet so therefore subnet 126 is the 127th
When they are asking to identify the 127th subnet do they mean this or subnet 127 which would be 10.15.224.0/19
What do you recon?
I think I've covered both bases in the wording of my answer & have submitted already so time will tell, just curious what you guys think.

p.s. prometheus fire you used subnet number 121 twice above so the 127th you came up with was actually 128th. took me a while to spot, I was puzzled why the answer I got for the 127th was different to yours.

[Prometheus_Fire]

Hi Jaye1234,

Ah, so you were removing 2 subnets? Thanks for pointing out my little errors, copy and paste is the lazy mans folly.

• I listed subnet 1 as 10.0.0.0/19, which is what I understand it to be (Going from 1 rather than 0).
• The way I actually worked out the 127th subnet was that knowing there are 8 increments of 32 before the next octet changes I divided 128 by 8 and got 16 then I just looked one subnet down.
• That gave me 10.15.224.0/19 as the 127th subnet.
• I meant to put 10.15.0.0/19 down as subnet 120 (Copy/Paste laziness again!)

I'm really glad you asked this question as I have my BSCI coming up and its always a good idea to keep going back over Cisco subnetting questions so I have it down cold for the exam!

[johnwill]

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[Prometheus_Fire]

Oops, sorry Johnwill.

[jaye1234]

Originally Posted by Prometheus_Fire

Hi Jaye1234,

Ah, so you were removing 2 subnets? Thanks for pointing out my little errors, copy and paste is the lazy mans folly.

• I listed subnet 1 as 10.0.0.0/19, which is what I understand it to be (Going from 1 rather than 0).
• The way I actually worked out the 127th subnet was that knowing there are 8 increments of 32 before the next octet changes I divided 128 by 8 and got 16 then I just looked one subnet down.
• That gave me 10.15.224.0/19 as the 127th subnet.
• I meant to put 10.15.0.0/19 down as subnet 120 (Copy/Paste laziness again!)

I'm really glad you asked this question as I have my BSCI coming up and its always a good idea to keep going back over Cisco subnetting questions so I have it down cold for the exam!

Prometheus fire,
it is the 9th subnet that changes the next octet & then the 17th & every 8th after that. it is the 129th subnet that changes the second octet to 16. Even though if you divide 128 by 8 you get 16 thats asuming that every 8th subnet the second octet changes i.e 8, 16, 24 but it is actuall 9, 17, 25 as there are 8 subnets in the third octet before the second octet changes.
So the 127th is 10.15.192.0
That make sense?

number____Subnet____ Network address______First____________Last
1st_________0__________10.0.0.0_________10.0.0.1_______10.0.31.254
2nd_________1_________ 10.0.32.0________10.0.32.1______10.0.63.254
3rd_________2__________10.0.64.0________10.0.64.1______10.0.95.254
4th_________3__________10.0.96.0________10.0.96.1______10.0.127.254
5th_________4__________10.0.128.0_______10.0.128.1_____10.0.159.254
6th_________5__________10.0.160.0_______10.0.160.1_____10.0.191.254
7th_________6__________10.0.192.0_______10.0.192.1_____10.0.223.254
8th_________7__________10.0.224.0_______10.0.224.1_____10.0.255.254
9th_________8__________10.1.0.0_________10.1.0.1_______10.1.31.254
* * * * * * * * * *
127th ______126_________10.15.192.0______10.15.192.1____10.15.223.254
128th ______127_________10.15.224.0______10.15.224.1____10.15.224.254
129th ______128_________10.16.0.0________10.16.0.1______10.16.31.254


My appolgies Johnwill for posting an assignment question on here, however I was not looking for a direct answer, I knew what to do but I knew I was initialy making an error in the calculation & was just looking for advise on this. In the end I did work it out but it has been a very useful excersize having this discussion on here.

[Prometheus_Fire]

Hi Jaye1234, while you are correct that the first subnet (10.0.0.0) is indeed known as subnet 0 it is still the first subnet. You always refer to subnet 0 as the first subnet. Not a the zeroth subnet.

Which makes the next subnet (10.0.32.0) the second subnet, etc. That still makes 10.15.224.0/19 the 127th subnet.

I'm pretty much sure about this one.