I stumped myself with this subnetting problem: how to do the "math"

Deathmage · October 2014

Hey guys I stumped myself with my own subnetting.

How do I do the "math" for a subnet of:

172.16.22.5
255.255.254.0

I do my math like this:

for the mask as you guys know 256 - mask = hosts and then of course you minus 2 for the network/broadcast ID later on.

subnet mask

128
192
224
240
248
252
254
255

hosts per subnet
255
128
64
32
16
8
4
2
1

2 power

2^7
2^6
2^5
2^4
2^3
2^2
2^1
2^0

For hosts per subnet I use the relations trick; 255 (hosts) is to 1 (subnets), 128 is to 2, 64 is to 4, 32 is to 8, and 16 is by itself.

my formula's are:

subnets equal 2^S= subnets
hosts equals 2^H-2 = hosts

So I get this far:

Subnets = 2^S= (256-254=2) so since that's in the 7 bit I'd go 2^S= 2^7= 254 which is 2 hosts per subnet which has a relationship to 128 so its 128 subnets with two hosts per subnet...

So with 7 bits of 8 being used for subnets, I only have 1 bit left over for hosts but I've found I can't subtract 2 from well 2 like so....

Hosts = 2^H-2 so that would be with the above math 2^1-2 = 0.

So I know this won't work, I think I need to make this a /23 and make two /24 subnets but I don't know how to do the math....

Can someone please help me out; I've been subnetting for weeks and then I make this one I made and stump myself, lol!!!!

....I want to be prepared for anything...

aftereffector · October 2014

I'm not quite sure I'm following. What are you looking to pull out of the given IP and mask? Network address, broadcast address, and number of IPs/hosts?

For a 172.16.22.5 255.255.254.0, that's a /23. It's double the size of a /24, so it will give you 512 addresses and 510 hosts. The network ID would be 172.16.22.0 and the next usable network would be 172.16.24.0, so the broadcast address would be 172.16.23.255.

_____________________________________________________

Editing to show my work:

I base everything off of a /24 because I learned that one first. A /24 subnet mask looks like 255.255.255.0 and gives 256 IPs or 254 usable addresses (subtracting two for the subnet ID and the broadcast address). I just memorized those numbers verbatim, as well as the binary progression 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, and the octet "counting progression" 255, 254, 252, 248, 240, 224, 192, 128.

Now, to get where we need to go... the given subnet mask is 255.255.254.0, which is one "level" higher than a 255.255.255.0. If you write it out on a napkin or something you can count the host and network bits, but it's easier for me to just write out a table from memory. I did this on scratch paper for each of my exams until I could do it in my head, and then I wrote it out anyways because it made me feel better.

Mask

CIDR

Hosts

.0
/24

256

Let's walk the dog down the rows:

Mask

CIDR

IPs

.0
/24

256

128

/25

128

192

/26

64

224

/27

32

240

/28

16

248

/29

8

252

/30

4

254

/31

2

Now let's add a couple of rows above the /24 so we can see what happens when we work in the third octet.

Mask

CIDR

IPs

252.0

/22

1024

254.0

/23

512

255.0

/24

256

255.128

/25

128

255.192

/26

64

255.224

/27

32

255.240

/28

16

255.248

/29

8

255.252

/30

4

255.254

/31

2

So the example's 255.255.254.0 must be a /23. It has 512 addresses, which means that it will cover a range of 512 values starting before and ending after the given IP of 172.16.22.5. For instance, 10.0.0.0 to 10.0.1.0 covers 256 addresses, and 10.0.0.0 to 10.0.2.0 covers 512, so we know we will add two to whatever is in the third octet of the subnet ID to get the next valid subnet ID. For the example problem, 172.16.22.5 fits inside 172.16.22.0 (the first subnet address we could choose - it has to end in .0) and so the next one has to be 172.16.24.0, making our subnet broadcast address 172.16.23.255.

mikeybinec · October 2014

Always think blocks!!!! **** that grid and think blocks. A /23 is a block size of 2 and you work in the 3rd octet (called the interesting octet). You have a Class B address. Since you have carved out 9 bits for hosts, you have 7 bits left for networks, or subnets if you prefer.. Since the block size is 2, the 22 fits nicely as the network address for the first subnet

172.16.22.0 range is 172.16.22.1 - 172.16.23.254 first subnet

172.16.24.0 range is 172.16.24.1 - 172.16.25.254 2nd subnet

172.16.26.0 range is 172.16.26.1 - 172.16.27.254 3rd subnet

..and so on 125 more times (2^7 = 12

) <=supposed to be an 8 at the end but the forum board software is acting goofy right now

srabiee · October 2014

I don't know if this will help you, but it certainly helped me when I was trying to learn how to subnet quickly for exam purposes:

https://www.youtube.com/watch?v=rs39FWDhzDs

Deathmage · October 2014

first off thanks for the response; this makes sense too me now. I see that I forgot about the CIDR and that kind of helps; just need to practice more with this knowledge of how CIDR's work!

so in an effort to see if I truly understand let me do this:

172.16.15.22
255.255.224.0

okie so by follow the above table, this would mean that the interesting octet is the 3rd octet since 172.16 is a class B.

So....

8+6=13 ... 8 being the complete 4th octet and counting reverse from 255 down to 224 is 6 bits, so the hosts would be 2 to the 13th power which equals 8192.

hosts equals 8192.

now something I found by chance the total hosts divided by 256 seems to give me the subnets not sure if this is legit or just dumb luck, dunno. So 8192 / 256 = 32.

subnets equals 32.

so if the IP above is 172.16.15.22 the first ip would be 172.16.15.0 (plus 32 subnets, not writing them out lol!) 172.16.47.255 as the last IP....

is this correct?

even though dividing works to get 32, I don't really know how to figure out the subnets any other way since the 2 power with the left over amount doesn't equal 32 or maybe i'm doing it wrong, ugh! I've only really mastered 8 bit subnetting like the 4th octet, adding another 8 bits is making my brain hurt, adding 16 more omg....

I get the IP addressing part but the subnets I'm really confused on...

I'm just thinking I'm making this way to hard, going to go watch that video. I'm not giving up but this is just difficult for me, dunno why...

is it normal to struggle this much?

TechGuru80 · October 2014

Deathmage wrote: »

my formula's are:

subnets equal 2^S= subnets
hosts equals 2^H-2 = hosts

Not sure where H-2 comes from. Think of it as:

Subnets = 2^S
Hosts per subnet = 2^(8-S) and then - 2

So:
4 subnets
2^2 = 4 ... 2 bits for network
hosts....2^(8-2)...(2^6) = 64 - 2 = 62 hosts per subnet.

Mask would be .192 because 2 bits for the network.

Subnetting can be very confusing but when it clicks...it clicks. I would also stick to class C addresses until you are positive you have them down. Once you start getting into B and A you will get high numbers that are so unpractical for an exam that it does no good.

aftereffector · October 2014

Forget about class A, B, and C when you are dealing with a custom subnet mask like 255.255.224.0. A Class C subnet always has 256 IPs, 254 hosts, and a 255.255.255.0 subnet. Always! A class B is always 255.255.0.0, and a class A is 255.0.0.0. What you're dealing with right now is CIDR and has nothing to do with classful subnet masks - you'll just get more confused if you think of 172.16.15.22 as a Class B address. I really wish introductory networking courses would do away with classful IP ranges...

If a 255.255.255.0 is /24, and a 255.255.254.0 is a /23, you can follow the progression from 255 > 254 > 252 > 248 > 240 > 224 and see that it's a /19 subnet mask in CIDR notation. If you write it out in binary, you will see 11111111.11111111.11100000.00000000. Counting the zeroes, that's 13 (eight in the fourth octet and five in the third), so your number of hosts is 2^13 or 8,192. Now, you are looking for the number of subnets, which is where I'm not following you. You would have to specify the address range to determine how many /19 subnets there are. Are you looking for the number of /19 subnets in a Class B (see, there it is again) address space?

fredrikjj · October 2014

Deathmage wrote: »

is it normal to struggle this much?

You are struggling because you aren't converting to binary. The quick methods are probably useful for cert exams, but they make it harder to understand what's actually going on. Additionally, if you know how things actually work, you can do IPv6 subnetting as well without much trouble.

so in an effort to see if I truly understand let me do this:

172.16.15.22
255.255.224.0

okie so by follow the above table, this would mean that the interesting octet is the 3rd octet since 172.16 is a class B.

The class terminology isn't relevant anymore since it's the subnet mask that determines how long the network prefix is, not the class. A mask of 255.255.224.0 is 11111111.11111111.11100000.00000000 when converted into binary. This is used to indicate that the network prefix is 19 bits long (there are 19 1s in a row there). This information is necessary because otherwise the computer is unable to know whether a particular IP address is on the local network or not. Originally this wasn't necessary because we had fixed prefix lengths (the classes).

So, based on the mask, you know that the first 19 bits of the IP address are the network prefix. From that follows that the last 13 must be the host portion of the address (19+13 = 32). You now convert the IP address into binary:
10101100.00010000.00001111.00010110. Because the subnet mask's first 2 octets are 255 we're not really interested in the first 2 octets of the binary representation since they will remain unchanged, but in these examples I'll keep them in to avoid any confusion.

We now have the mask and the address in binary:

11111111.11111111.11100000.00000000 Mask
10101100.00010000.00001111.00010110 IP address

I've colored the first 19 bits green. These are the prefix bits because the mask has 19 1s. To get the network-id to you simply add zeroes in the host portion of the address:

10101100.00010000.00000000.00000000 Network-ID

To get the broadcast address you instead add 1s which will give you the last address:

10101100.00010000.00011111.11111111 broadcast address

Converting back to decimal you get 172.16.0.0 and 172.16.31.255.255

(This isn't a completely generalized solution though since just adding 0s/1s to the host portion doesn't take into account that you could theoretically have discontiguous networks, but no one does that so we can safely disregard that possibility).

The number of hosts follows directly from the size of the host portion of the address. Since it's 13 bits, there are 2^13 possible combination. You could subtract 2 for the network id and broadcast address.

subnets equals 32.

My opinion is that questions that rely on using an IP address' class are completely obsolete and dumb. It's 2014, not 1991. You could ask someone something like:

"You have 172.16.15.22 with the subnet mask 225.255.224.0. How many subnets are there?"

The idea is then to know that 172.16.0.0 is a "class B" address. Since we have 19 bits for our prefix and the class B address uses a 16 bit prefix, we have 3 bits to play with. The answer would then be 2^3 = 8 subnets. However, we don't use classes anymore so the question is nonsensical really.

Jon_Cisco · October 2014

From the original post you were mixing up the classless as you figured out.
In your new example I think 224 would be 5 bits for hosts. It's early so please forgive any mistakes.

I always recommend this site subnetting.net - Subnet Questions and Answers
I spent maybe an hour doing questions and comparing what I did wrong to their answer. It just started to click after a while.
I don't do math shortcuts because in the exam you don't want to over think it. Just remember your blocks.

128 - 128
192 - 64
224 - 32
240 - 16
248 - 8
252 - 4
254 - 2
255 - 0

If you want something to practice binary cisco has a free app on the iphone that helps with speed a lot. I play it like 10 minutes a week and it keeps my fairly quick at converting to binary. Usually while waiting for a class to start.

Deathmage · October 2014

I will study these posts at lunch. Doing a VMware 5.5 build-out for a customer on a dual R720 dual xeon (16 core's per cpu) cluster with a Equalogic 6100xv SAN fabric atm and I'm having sooo much fun even though this is work...

Legacy User · October 2014

Reading the OP post confused the crap out of me. When I'm subnetting I always quickly chart it out and start from Class C and work backwards into Class B if need be.

This is how I chart it out:

CLASS C

Available ip addresses subtract 2 for network id and broadcast id
256 - 2
128 - 2
64 - 2
32 - 2
16 - 2
8 - 2
4 - 2
2 -2

Class C BITS
128
64
32
16
8
4
2
1

Subnet mask

128
192
224
240
248
252
254
255

CIDR Notation
/25
/26
/27
/28
/29
/30
/31
/32

Remember when you looking for a specific quantity of ip addresses do not select the bit it falls on *always select the bit that is right before it.

Ex: If you want 30 ip addreses you would choose /27 not a /28 because using a /27 means everything up to /27 is no longer available and you leave the ip addresses from /28 available.

If you accidentally choose /28 you are only leaving the ip address from /29 which is 14 hosts available.

Deathmage · October 2014

Well this bring subnetting into a new way of thinking; let me go about learning them in binary then and using the CIDR method for now. Once I understand it more thoroughly I guess I'll see if there is quicker equations to follow for exams but without a base of understanding those shortcuts won't mean much. I bounced the question also off of the other CCNA's in-house here and they have been helping me understanding it better too. Thank you all for your feedback.

I made this graph, I did this all up to /16 then I needed a cacuilator, I didn't do from /8 to /1 becasue I hardly doubt I'll ever touch a network that large or ever have a need for that so that seems unrealistic. on this topic how much should I memorize, lol! ...If I need to memorize a /8 I may need to find another line of work, lol!

not sure how class bits apply in the 2nd and 3rd octets. but I'm just using the 128,62,32,16,8,4,2,1 since each octet is 8 bits so that's what makes sense to me....

EDIT: oo look I broke the webpage

CIDR
X
X
X
X
X
X
X
/8
/9
/10
/11
/12
/13
/14
/15
/16
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
/31
/32

netmask
128.0.0
192.0.0
224.0.0
240.0.0
248.0.0
252.0.0
254.0.0.0
255.0.0.0
128.0.0
192.0.0
224.0.0
240.0.0
248.0.0
252.0.0
254.0.0
255.0.0
128.0
192.0
224.0
240.0
248.0
252.0
254.0
255.0
128
192
224
240
248
252
254
255

hosts possible minus 2 of course
X
X
X
X
X
X
X
16777214
8388608
4194304
2097152
1048576
524288
262144
131072
65536
32768
16384
8192
4096
2048
1048
512
256
128
64
32
16
8
4
2
1

class bits

128
64
32
16
8
4
2
1

Legacy User · October 2014

Looks like you have a better idea but the you have the available ip addresses for Class C incorrect:

You have

Available ip addresses
128
64
32
16
8
4
2
1

Class C BITS
128
64
32
16
8
4
2
1

When you should have:

Available ip addresses *subtract 2 for net id and broadcast id
256 - 2
128 - 2
64 - 2
32 - 2
16 - 2
8 - 2
4 - 2
2 -2

Class C BITS
128
64
32
16
8
4
2
1

To avoid any confusion I just write out the C Class Bits and double each bit which reflects ip addresses.

If you look in my Class C chart 128 bit is doubled to 256, 64 bit is doubled to 128, 32 bit is doubled to 64, hope you are seeing the pattern.

In binary class c is represented as:
11111111.1111111. 11111111. 00000000

Don't let the binary trip you up either all it is 4 sections of 8 bits each which represent
128, 64, 32, 16, 8, 4, 2, 1 in each section. Once you understand that each bit represents those numbers you can address the given chart much easier.

Once you go into class B you continue to double the previous ip addresses

Available ip addresses
131072 - 2
65536 - 2
32768 - 2
16384 - 2
8192 - 2
2048 - 2
1024 - 2
512 -2

Class B BITS
128
64
32
16
8
4
2
1

Subnet
128
192
224
240
248
252
254
255

CIDR Notation
/17

/18

/19

/20

/21

/22

/23

/24

In binary class b is represented as:
11111111.1111111.0000000.00000000

Deathmage · October 2014

dmarcisco wrote: »

Looks like you have a better idea but the you have the available ip addresses for Class C incorrect:

You have

Available ip addresses
128
64
32
16
8
4
2
1

Class C BITS
128
64
32
16
8
4
2
1

When you should have:

Available ip addresses *subtract 2 for net id and broadcast id
256 - 2
128 - 2
64 - 2
32 - 2
16 - 2
8 - 2
4 - 2
2 -2

Class C BITS
128
64
32
16
8
4
2
1

In binary class c is represented as:
11111111.1111111. 11111111. 00000000

To avoid any confusion I just write out the C Class Bits and double each bit which reflects ip addresses.

If you look in my chart 128 bit is doubled to 256, 64 bit is doubled to 128, 32 bit is doubled to 64, hope you are seeing the pattern.

Once you go into class B you continue to double the previous ip addresses

Available ip addresses
131072 - 2
65536 - 2
32768 - 2
16384 - 2
8192 - 2
2048 - 2
1024 - 2
512 -2

Class B BITS
128
64
32
16
8
4
2
1

Subnet
128
192
224
240
248
252
254
255

CIDR Notation
/25
/26
/27
/28
/29
/30
/31
/32

In binary class b is represented as:
11111111.1111111.0000000.00000000

now my original thinking has been depth charged into a million particles of water.......

O.o .... where does the doubling come from?

EDIT: after looking over your post like 3 time I understand it, just was never taught to me in that manner.....

I mean I do know there is 8 bits: 128,64,32,16,8,4,2,1 but that's the bit's and we all know a class C has 256 IP's. But I see were the double up is applied; I'm used to doing it with the 2nd power so it's just another way of seeing it so nothing has really changed....

now to me Class bits are the subnets and they go left to right ending it 128 (IE: 128, 64, 32, 16, 8, 4, 2, 1) and hosts range or subnet mask foes from right to left starting in 128 (IE: 128, 192, 224, 240, 248, 252, 254, 255) that's how I remember, now sure if that valid. But that's how I've learned....

Legacy User · October 2014

How about if I said its the rule of thumb would that be a good enough answer?

To be honest I don't remember the actual why the ip addresses double at each octet and I'm feeling lazy to google you up an answer but that is how it works.

Legacy User · October 2014

now to me Class bits are the subnets and they go left to right ending it 128 (IE: 128, 64, 32, 16, 8, 4, 2, 1) and hosts range or subnet mask foes from right to left starting in 128 (IE: 128, 192, 224, 240, 248, 252, 254, 255) that's how I remember, now sure if that valid. But that's how I've learned....

Lol sorry, I edited that post like 3 times. Recheck the CIDR notation on the CLass B I had a copy and paste error which I now corrected.

You are right all you need to know is where the ip addresses come into play. Everyone has there own method but when figuring out IP addresses if its under 254 hosts I like to start in the Class C using the chart I displayed in my earlier post.

If its anything greater then 512 I know /24 = 512 addresses which is the last bit in Class B and I double the 512 with every octet working from right to left to get to the desired ip addresses I need. I'm sure some people remember that /19 equals 131072 ip addresses but I sure don't so I use the method I mentioned.

Hope I helped you understand it a little better.

aftereffector · October 2014

They double because of the way binary counting works - it's a power of two.

With a subnet mask ending in 11111111.11111111.11111111.11111110, you can only change values in the last digit. So you could have an IP address written in binary with either a 0 or a 1, which is just two possibilities - or 2^1 = 2.

With a subnet mask ending in 11111111.11111111.11111111.11111100, you can change values in the last two digits, so you could have a binary IP address ending in 00, 01, 10, or 11. Four possibilities from two digits, or 2^2 = 4.

With a subnet mask ending in 11111111.11111111.11111111.11111000, you have three digits of freedom, so your binary IP address could end in 000, 001, 010, 011, 100, 101, 110, or 111. Three digits, eight possibilities, or 2^3 = 8.

And so on, and so forth. 2, 4, 8, 16, 32, 64, 128... you already know that a classful Class C address, subnet mask 255.255.255.0, has 256 addresses. That's because all the values in the fourth octet can be changed, so 2^8 = 256.

I think you already know all this so I'm not sure why I typed it all out

Legacy User · October 2014

aftereffector wrote: »

They double because of the way binary counting works - it's a power of two.

Well there you go

Deathmage · October 2014

aftereffector wrote: »

They double because of the way binary counting works - it's a power of two.

With a subnet mask ending in 11111111.11111111.11111111.11111110, you can only change values in the last digit. So you could have an IP address written in binary with either a 0 or a 1, which is just two possibilities - or 2^1 = 2.

With a subnet mask ending in 11111111.11111111.11111111.11111100, you can change values in the last two digits, so you could have a binary IP address ending in 00, 01, 10, or 11. Four possibilities from two digits, or 2^2 = 4.

With a subnet mask ending in 11111111.11111111.11111111.11111000, you have three digits of freedom, so your binary IP address could end in 000, 001, 010, 011, 100, 101, 110, or 111. Three digits, eight possibilities, or 2^3 = 8.

And so on, and so forth. 2, 4, 8, 16, 32, 64, 128... you already know that a classful Class C address, subnet mask 255.255.255.0, has 256 addresses. That's because all the values in the fourth octet can be changed, so 2^8 = 256.

I think you already know all this so I'm not sure why I typed it all out

it's more helpful than you know. Seeing other peoples logic helps me understand. The only thing I've learned with all of the CCNA study books is everyone has there own method of learning subnetting and I think that's what makes it so hard...

I think once I understand it, I'm just going to skip subnetting in the books so as to not confuse myself...

my current understanding is to understand class A and B; I know C pretty well. adding the 2nd and 3rd octets just makes the number so much larger and I suck at long math so that's why it's confusing....

I was the dork in math class that programmed his Ti-83 to do his trigonometry equation on exams, I played the clueless card and I was able to pass

....

I think once i get tons and tons of practise down like I have with class C it will be easier, but I do think the classless addressing is where I'm getting confused. So I do see the point of the above posts about using CIDR...

Deathmage · October 2014

So one of the CCNA Engineer here gave me a subnetting workbook that he'd like me to do this weekend.

Also he said for me to confirm to memory the Class A, B, and C Addressing guide [below] by writing it like 300 times so when it comes to the exam I just **** it. He looked at me figuring out subnetting and he was like your thinking too much; on the exam you have seconds to subnet not 40 seconds he said. So he said he just remember the table and understand the basics like the class C subnetting in binary. He said once you get to Class B and A memorizing binary and the digits is just going to make you waste time....

thoughts on this method?

ccnxjr · October 2014

It's similar to what I did .
However I wouldn't write out the # of usable hosts/subnets.
For the Class C section, i'd actually write out the subnets, and just the first host, last host and broadcast address eg for 192.168.1.0/26
192.168.1.0: First: 192.168.1.1 Last: 192.168.1.62 Broadcast: 192.168.1.63
192.168.1.64: First: 192.168.1.65 Last: 192.168.1.126 Broadcast: 192.168.1.127
192.168.1.128: First: 192.168.1.129 Last: 192.168.1.190 Broadcast: 192.168.1.191
192.168.1.192: First: 192.168.1.193: Last: 192.168.1.254 Broadcast: 192.168.1.255

If you try writing this out every day (take your time the first few days, and do the math, you'll get faster within a week)
Eventually you'll see a pattern, and see a relationship that makes sense to you.
For me, it's the lowest significant bit in the mask.
Eg, with 192.168.1.0/26, the lowest significant bit, the 26th bit corresponds to "64" th place, so your network number changes in multiples of 64.

That same number , 64 , can give you the total number of hosts if you subtract 2.
64-2 = 62 , so 62 hosts per block.
( I believe it's referred to as the "magic" number).

The Broadcast address is one less that the next netblock in the series.
ie, for 192.168.1.128/26, the next netblock is 192.168.1.192, so the broadcast address for 192.168.1.128 is 192.168.1.192-1 = 192.168.1.191

Deathmage · October 2014

Awesome, so this method isn't like the reinvention of the wheel!

Ya he gave me a subnetting workbook where is like 86 pages but its basically just repetition, repetition, repetition. So I think until I really master this I'll just keep labbing away at VMware 5.5 and knock out the VCP in like a month and then hopefully by then subnetting will be so engrained into my skull I'll get back on the band wagon for the CCENT.

all I'm doing right now is the work book while I wait for my vCenter appliance to install and for a VMware to Hyper-v Conversion to take place. I have a 750GB conversion VMware 5.0.0 (free) to Hyper-V 2012 (long-story customer is cheap and won't pay for VMware licensing) in two weeks sadly it will be over USB 3.0 so it will take forever!!!! ... so I want ot get the conversion down to a science beforehand.

I stumped myself with this subnetting problem: how to do the "math"

Comments