Subnetting question: do I got the process down correctly?

Deathmage

Making sure I got this down myself.....

To me a Class A is (/9 to /15)
Class B is (/17 to /23)
Class C is (/25 to /30)

Now each one of these such as /9 to /15 go down the subnets list like /9 = 128, /10 = 192, /11 = 224 etc.

But were do the bits come into play exactly? - correct me if I'm wrong but say the ticks in the bit go to 32 and the subnet mask goes to say 128 that would give me say this if it was a class C: x.x.x.1 to x.x.x.30 | x.x.x.32 to x.x.x.62 | x.x.x.64 to x.x.x.95 | x.x.x.97 to x.x.x.126 | - Since .128 is the subnet to me it can't go past .128 for the bits. - (also as far as I've learned the last bit is the broadcast, ie .31, .63, .96, .127, also I do recall the 1st bit can't be used but I can't remember the term right now) did I do that correctly?




bits

128

64

32

16

8

4

2







subnets

X

X

X











128

X













192














224














240














248














252














254














255

Deathmage

bump.

Jon_Cisco

It sounds like you are trying to follow the subnets of a /27.

It's a bit confusing the way you have written it out. I'm not sure it's wrong I think your just making more work of it then is needed.

What I would say is that if you understand it then that is the most important part. I would suggest testing your idea with practice questions on Subnetting Practice Questions

mikeybinec

I'm not sure I understand your question. On your chart you left out the one column. I don't use charts, I use block sizes, and they all correspond no matter which octet you are working in

With that said, here;s a couple of examples;

192.168.1.0 /27

this is a classic Class C address.. The interesting octet is the fourth octet. Now three bits are carved out of the fourth to create some subnets (networks) How many? 2^3 = 8 networks... What;s left? 5 bits 2^5 =32 The first bit has to be for the network ID and the last is for the broadcast address, so you really have 30 hosts available

Range is 1.1 -1.30 1.0 is network ID 1.15 is broadcast
1.32 - 1.62 1.32 is network ID 1.63 is broadcast
(add 6 more subnets here)
etc etc.


172.16.16.0 /20

Classic Class B address and a member of the RFC 1918 crowd

The third octet is the interesting octet, that gets carved out to decide the boundary between networks and hosts. Four bits are added to the network portions: 2^4 = 16 neworks.. the rest are hosts. Well, 4 bits from whats left in the 3rd octet and the 8 in the fourth octet equals 12. 2^12 = 4096. subtract 2 for network ID and broadcast and you have 4094 hosts per network

I'll do the first two, you can do the remaining 14 hee hee hee

First network is 16.1 - 31.254 (16.0 is net ID, 31.255 is broadcast)
2nd network is 32.1 - 47.254 (32.0 is net ID, 47.255 is broadcast)

Interesting piece of information in the above. Because there are over 4000 hosts, you will see host addresses such as 20.0, and 21.255.. The 0's and 255's are not always delegated to Network IDs and broadcasts, respectively.

Hope this helps

Deathmage

mikeybinec wrote: »

I'm not sure I understand your question. On your chart you left out the one column. I don't use charts, I use block sizes, and they all correspond no matter which octet you are working in

With that said, here;s a couple of examples;

192.168.1.0 /27

this is a classic Class C address.. The interesting octet is the fourth octet. Now three bits are carved out of the fourth to create some subnets (networks) How many? 2^3 = 8 networks... What;s left? 5 bits 2^5 =32 The first bit has to be for the network ID and the last is for the broadcast address, so you really have 30 hosts available

Range is 1.1 -1.30 1.0 is network ID 1.15 is broadcast
1.32 - 1.62 1.32 is network ID 1.63 is broadcast
(add 6 more subnets here)
etc etc.


172.16.16.0 /20

Classic Class B address and a member of the RFC 1918 crowd

The third octet is the interesting octet, that gets carved out to decide the boundary between networks and hosts. Four bits are added to the network portions: 2^4 = 16 neworks.. the rest are hosts. Well, 4 bits from whats left in the 3rd octet and the 8 in the fourth octet equals 12. 2^12 = 4096. subtract 2 for network ID and broadcast and you have 4094 hosts per network

I'll do the first two, you can do the remaining 14 hee hee hee

First network is 16.1 - 31.254 (16.0 is net ID, 31.255 is broadcast)
2nd network is 32.1 - 47.254 (32.0 is net ID, 47.255 is broadcast)

Interesting piece of information in the above. Because there are over 4000 hosts, you will see host addresses such as 20.0, and 21.255.. The 0's and 255's are not always delegated to Network IDs and broadcasts, respectively.

Hope this helps

wouldn't it be 1.31 and not 1.15?

Maybe I also got it confused: to me I always thought like you said the 1st one in the Network ID and the last in the broadcast.... so if that's the case wouldn't it be:

Range is 1.1 - 1.30 | 1.0 is network ID | 1.31 is broadcast
Range is 1.33 - 1.62 | 1.32 is network ID | 1.63 is broadcast
Range is 1.65 - 1. 95 | 1.64 is network ID | 1.96 is broadcast
Range is 1.98 - 1.126 | 1.97 is network ID | 1.127 is broadcast
(add 4 more subnets here)

as for the /23 and so on (drawing a brain fart with the term on the tip of my brain but I can't recall the actual name!!!!)

But a.....

Class A is 8 bits or /8
Class B is 16 bits or /16
Class C is 24 bits or /24

192.168.104.0/30

so my understanding of this is if say its like a /30 you know it's a Class C since that's the closest and since it's a Class C you borrow 6 bits from that (since 30 - 6 = 24) so that 2^6=64 networks and what is left over is 2 bits so that's 2^2= 8 hosts.

192.168.104.1 - 192.168.104.7 | network ID = 192.168.104.0 | broadcast = 192.168.104.8
192.168.104.10 - 192.168.104.15 | netwokr ID = 192.168.104.9 | broadcast = 192.168.104.16
192.168.104.18 - 192.168.104.24 | Netwokr ID = 192.168.104.17 | broadcast = 192.168.104.25
so on and so forth.....

Is my understanding good?

xnx

Deathmage wrote: »

wouldn't it be 1.31 and not 1.15?

Maybe I also got it confused: to me I always thought like you said the 1st one in the Network ID and the last in the broadcast.... so if that's the case wouldn't it be:

Range is 1.1 - 1.30 | 1.0 is network ID | 1.31 is broadcast
Range is 1.33 - 1.62 | 1.32 is network ID | 1.63 is broadcast
Range is 1.65 - 1. 95 | 1.64 is network ID | 1.96 is broadcast
Range is 1.98 - 1.126 | 1.97 is network ID | 1.127 is broadcast
(add 4 more subnets here)

Correct

Deathmage

wow who would have thought a forum post would make more sense than a book explained it!

now just to understand subnet masks more better...

I kind of added the 2nd part of the post, the one post above this one, while the other post was posted by xnx, but is that a good understanding of it?

Thanks guys!

xnx

All there is to working out subnets is what octet changes and by what increment, like the /27 example above goes up in 32's on the 4th octet, a /19 would do the same but on the 3rd octet.

Deathmage

xnx wrote: »

All there is to working out subnets is what octet changes and by what increment, like the /27 example above goes up in 32's on the 4th octet, a /19 would do the same but on the 3rd octet.

so for shits and giggles.... it would be like this:

172.16.16.0/18

so that's 18 - 16 = 2 so that would equal 2^2 = 8 networks and since their is 6 bits left 2^6 = 64 hosts so then it would be something like this...

172.16.1.0 - 172.16.63.0 | network ID = 172.16.0.0 | broadcast = 172.16.64.0
172.16.66.0 - 172.16.127.0 | network ID = 172.16.65.0 | broadcast = 172.16.128.0
172.16.130.0 - 172.16.191.0 | network ID = 172.16.129.0 | broadcast = 172.16.192.0
add 5 more networks....

now as for the subnet mask would it be 255 - 64 = 255.255.255.192?

is that correct?

mikeybinec

Deathmage wrote: »

so for shits and giggles.... it would be like this:

172.16.16.0/18

so that's 18 - 16 = 2 so that would equal 2^2 = 8 networks and since their is 6 bits left 2^6 = 64 hosts so then it would be something like this...

172.16.1.0 - 172.16.63.0 | network ID = 172.16.0.0 | broadcast = 172.16.64.0
172.16.66.0 - 172.16.127.0 | network ID = 172.16.65.0 | broadcast = 172.16.128.0
172.16.130.0 - 172.16.191.0 | network ID = 172.16.129.0 | broadcast = 172.16.192.0
add 5 more networks....

now as for the subnet mask would it be 255 - 64 = 255.255.255.192?

is that correct?

8

---

8

---

8

---

8
11111111 11111111 11111111 11111111 (128 64 32 16 8 4 2 1)

Use 256 - subnet mask So lets' take your 172.16.16.0/18.. The interesting octet is the third octet. Carving out 2 bits from puts you on that 128+64-192.. 256-192 does equal 64.. That block size determines your network size in regards to host.. You got pretty much everything correct except for the 255 size..Should be 256... One more thing Broadcast addresses are always odd numbered and network IDs are always even numbered with respect to a network

Now there is another issue in the fact that you really cant start a network anywhere you want. They must be multiples of the blocks beginning at 0.. In other words, if you have a block size of 16, then you do 0-15 then 16- 31 32- 47 etc You cant start at 4. ( NET IDs are always even and broadcast are odd!!)

We're gonna change your 172.16.16.0 to 172.16.0.0 /18

So, with the above, here's the result: We take the first 2 bits on the left of the 3rd octet.. this is a size of 192 (first bit is 128, second is 64) 256 - 192 = 64-- You now have your network size Recall, the first bit is the net ID (even) and the last is the broadcast

172.16.0.1 -- 172.16.63.254 network ID is, as you know 172.16.0.0 broadcast is 172.16.63.255

Next network?

172.16.64.1 -- 172.16.127.254 Net ID is 172.16.64.0 Broadcast is 172.16.127.255

I promise you, just think block sizes in the interesting octet, and you can subnet in less than a minute

Deathmage

mikeybinec wrote: »

8

---

8

---

8

---

8
11111111 11111111 11111111 11111111 (128 64 32 16 8 4 2 1)

Use 256 - subnet mask So lets' take your 172.16.16.0/18.. The interesting octet is the third octet. Carving out 2 bits from puts you on that 128+64-192.. 256-192 does equal 64.. That block size determines your network size in regards to host.. You got pretty much everything correct except for the 255 size..Should be 256... One more thing Broadcast addresses are always odd numbered and network IDs are always even numbered with respect to a network

Now there is another issue in the fact that you really cant start a network anywhere you want. They must be multiples of the blocks beginning at 0.. In other words, if you have a block size of 16, then you do 0-15 then 16- 31 32- 47 etc You cant start at 4. ( NET IDs are always even and broadcast are odd!!)

We're gonna change your 172.16.16.0 to 172.16.0.0 /18

So, with the above, here's the result: We take the first 2 bits on the left of the 3rd octet.. this is a size of 192 (first bit is 128, second is 64) 256 - 192 = 64-- You now have your network size Recall, the first bit is the net ID (even) and the last is the broadcast

172.16.0.1 -- 172.16.63.254 network ID is, as you know 172.16.0.0 broadcast is 172.16.63.255

Next network?

172.16.64.1 -- 172.16.127.254 Net ID is 172.16.64.0 Broadcast is 172.16.127.255

I promise you, just think block sizes in the interesting octet, and you can subnet in less than a minute

so if broadcasts are always odd and network ID's are always even.... does that mean what I did down below is off by one digit on each? - cause it seems I got them flipped if this is the case.

172.16.1.0 - 172.16.63.0 | network ID = 172.16.0.0 | broadcast = 172.16.64.0
172.16.66.0 - 172.16.127.0 | network ID = 172.16.65.0 | broadcast = 172.16.128.0
172.16.130.0 - 172.16.191.0 | network ID = 172.16.129.0 | broadcast = 172.16.192.0
add 5 more networks....


so basically it needs to be this? :


172.16.1.0 - 172.16.62.0 | network ID = 172.16.0.0 | broadcast = 172.16.63.0
172.16.65.0 - 172.16.126.0 | network ID = 172.16.64.0 | broadcast = 172.16.127.0
172.16.129.0 - 172.16.191.0 | network ID = 172.16.128.0 | broadcast = 172.16.193.0
172.16.195.0 - 172.16.254.0 | network ID = 172.16.194.0 | broadcast = 172.16.255.0
add 4 more networks....

Now here is one thing, since there is 4 more networks needed and by this point it's now used .255 in the 3rd octet what do you do at this point to keep making more networks? - do you borrow another bit or something....

and just I got this down correctly:

lets do 192.168.102.0/26

so that's 26-24 = 2 so 2^2 = 8 networks and with 6 bits left over 2^6 = 64 hosts.

so that would be:

192.168.102.0 - 192.168.102.62 | Network ID = 192.168.102.0 | broadcast = 192.168.102.63
192.168.102.65 - 192.168.102.126 | Network ID = 192.168.102.64 | broadcast = 192.168.102.127
192.168.102.129 - 192.168.102.190 | Network ID = 192.168.102.128 | broadcast = 192.168.102.191
192.168.102.193 - 192.168.102.254 | Network ID = 192.168.102.192 | broadcast = 192.168.102.255

and the subnet mask would be in accordance with my chart at the top of this post with the subnets since it's 2 bit it would equal 192. so the mask for 192.168.102.0/26 = 255.255.255.192.

is this example correct?

I'm still pondering how I would do the math for subnets past 256, like here at work we do a /23 so we have 512 hosts but I'd just love to know how that was done mathematically...

mikeybinec

I'm at work right now so I can't study your response. As for the /23, i.e. 510 hosts, you're in the Class B or A zone. The fourth octet, if all the bits are ones (11111111) is 256 hosts. So if you enter the 3rd octet zone and snag one of it's bits you are at 510 hosts. One more bit, and double it 1024, then 2048, then 4096 etc etc

At this point, you have now entered the VLSM zone. When we talked about creating 8 networks that has 30 hosts for each network, this was an example of classful addressing.

Nobody uses classful addressing unless you are tutoring RIP version 1 (but I could be wrong, I'll let one of the Techexams.net Gurus smack me around now )

And vlsm is when you tune your network sizes to accomodate your network..Start from largest block size down to smallest, and dont forget about the proper starting points 0-4 4- 8, 0-16 etc etc..

Deathmage

mikeybinec wrote: »

I'm at work right now so I can't study your response. As for the /23, i.e. 510 hosts, you're in the Class B or A zone. The fourth octet, if all the bits are ones (11111111) is 256 hosts. So if you enter the 3rd octet zone and snag one of it's bits you are at 510 hosts. One more bit, and double it 1024, then 2048, then 4096 etc etc

At this point, you have now entered the VLSM zone. When we talked about creating 8 networks that has 30 hosts for each network, this was an example of classful addressing.

Nobody uses classful addressing unless you are tutoring RIP version 1 (but I could be wrong, I'll let one of the Techexams.net Gurus smack me around now )

And vlsm is when you tune your network sizes to accomodate your network..Start from largest block size down to smallest, and dont forget about the proper starting points 0-4 4- 8, 0-16 etc etc..

Well I got the even for network ID and odd for broadcast down now. Let me know later if I did the above part correctly.

I'm just now reading Odom's book so I'm up to the IPv4 addressing and Routing so I'm wanting to get subnetting down now before advancing further into the book since I heard like 40% of the exam is subnetting. I want ot get a good understanding of it now and then re-cap on it at the end after learning all the cli commands and Cisco stuff.

But as for classless addressing I remember that from my CCNA class in 2006 when I had my CCNA but I didn't use it for 6 years and let it go so now I'm re-learning it over again but stuff is coming back to me now...

so for right now subnetting from 0 to 256 is about the norm of what I should know for now, vlsm is getting a bit more advanced or should I know it for the CCENT/CCNA? - I'm sure I'll be shooting myself in the foot with this comment if it is needed since I'm only on chapter 4 in odoms book. I gave up on CCNA in 60 days books since it just didn't teach me much just CLI stuff....

mikeybinec

Let's do one for jokes.. Pick an IP address, any IP address.. Let's see, how about

100.200.210.5 /20 I just picked this for jokes.. So we want the Network ID this host belongs to. The broadcast and the host range.. READY!!! START THE CLOCK
A /20 means we're entering the 3rd octet.. we use 8 bits of the first octet, 8 of the second, and 4 of the third (8+8+4=20)

A /20 is a block size of 16..So how many 16s can we fit in 210 before going over it? (A /28 is also a block size of 16..A /12 is a block size of 16)

well 10 x 16 =160 NOT EN0UGH.. How about 13? Yeah that's enough!!! 13 * 16 =208..That's our network Id!!!

100.200.208.0 Network ID

100.200.208.1 first host (Now add 15 to the net id number to get to the end of your network range .Yes, I did say 15, not 16)

100.200 223.254 last host

100.200.208.255 broadcast id

Remember, the block size was 16, and we do that in the 3rd octet..

For extra credit, what class is this address in. And, if we were doing classful addressing, how many subnets can you have?

And yes, you do need to know vlsm. The reason mainly is because on WAN links, you only need 2 IP addresses, i.e. from the, let;s say ISP to you. That;s a block size of 4.. (because there is still a network ID and a broadcast address) Now after you connect to the outside world, you can then venture into NAT and RFC 1918 (private, non routable IP addresses to expand your network) Your LAN could have lots of addresses.. But you might want to take an address block, split it up and say give human resources 12 addresses, engineering, 30 addresses, and sales 10 addresses.. that's where your vlsm comes into play.. the HR block will be 16 (gives them room for a little expansion) engineering will stay put at the /27.. no room for expansion for them. and sales, we'll give them a block size of 16 also /28 (255.255.255.240)

Deathmage

mikeybinec wrote: »

Let's do one for jokes.. Pick an IP address, any IP address.. Let's see, how about

100.200.210.5 /20 I just picked this for jokes.. So we want the Network ID this host belongs to. The broadcast and the host range.. READY!!! START THE CLOCK
A /20 means we're entering the 3rd octet.. we use 8 bits of the first octet, 8 of the second, and 4 of the third (8+8+4=20)

A /20 is a block size of 16..So how many 16s can we fit in 210 before going over it? (A /28 is also a block size of 16..A /12 is a block size of 16)

well 10 x 16 =160 NOT EN0UGH.. How about 13? Yeah that's enough!!! 13 * 16 =208..That's our network Id!!!

100.200.208.0 Network ID

100.200.208.1 first host (Now add 15 to the net id number to get to the end of your network range .Yes, I did say 15, not 16)

100.200 223.254 last host

100.200.208.255 broadcast id

Remember, the block size was 16, and we do that in the 3rd octet..

For extra credit, what class is this address in. And, if we were doing classful addressing, how many subnets can you have?

And yes, you do need to know vlsm. The reason mainly is because on WAN links, you only need 2 IP addresses, i.e. from the, let;s say ISP to you. That;s a block size of 4.. (because there is still a network ID and a broadcast address) Now after you connect to the outside world, you can then venture into NAT and RFC 1918 (private, non routable IP addresses to expand your network) Your LAN could have lots of addresses.. But you might want to take an address block, split it up and say give human resources 12 addresses, engineering, 30 addresses, and sales 10 addresses.. that's where your vlsm comes into play.. the HR block will be 16 (gives them room for a little expansion) engineering will stay put at the /27.. no room for expansion for them. and sales, we'll give them a block size of 16 also /28 (255.255.255.240)

now my logic is all thrown out of wack. I'm completely lost how you think the /20 is a 3rd octet. I'm trying to figure out how you got 13 * 16, I get the 16 but not the 13. I'm so lost now...

I think my understanding of the subnets is completely off or I'm making this too hard.

To me this is how a subnet class is:

Class A: /1 to /7
Class B: /9 to /14
Class C: /24 to /21

OK, so can someone help me here and correct the Class A, B, C listings for the hosts and for the networks.

I'm reading the subnetting.net tutorial and I'm trying to follow were you get (2^7, 2^14,2^21 for the networks):



Class
Purpose
First Octet Range
Maximum Hosts
Total Networks


Class A
Very Large Networks
1-126*
16,777,216 (2^24)
128 (2^7)


Class B
Large Enterprise
128-191
65,536 (2^16)
16,384 (2^14)


Class C
Small Business
192-223
256 (2^
2,097,152 (2^21)


Class D
Multicast
224-239
N/A
N/A


Class E
Experimental
240-255
N/A
N/A





Before I learn this incorrectly I just want to get a understanding of the math. I know this is easier said than done since subnetting can be really hard to explain. But I'm determined to figure it out.

I'm really just shocked I figured this out 3 weeks ago and I read the books and do no subnetting and now I've forgotten this is upsetting...

BerkshireHerd

all I can say is I'm glad I watched CBT Nuggets, now everytime I try to see someone else explain I have no idea what they are saying...

Deathmage

OK... so going over my notes from the beginning part of the book in subnetting I'm going to do this, please correct me where I'm wrong. I'm now coming back to re-do subnetting so I can make sure I know it...

Class A: 8
Class B: 16
Class C: 24

lets do 172.16.100.11/19

So since this is a /19 that would be a Class B so 16 + 3 = 19 so that dictates the below results:



bits
128
64
32
16
8
4
2






subnets
X
X
X










128
X












192
X












224
X












240













248













252













254













255












Now this is what I got so far:



Subnet
First Host
Last Host
Broadcast




172.16.0.0
172.16.0.1
172.16.31.254
172.16.31.255




172.16.32.0
172.16.32.1
172.16.63.254
172.16.63.255




172.16.64.0
172.16.64.1
172.16.95.254
172.16.95.255




172.16.96.0
172.16.96.1
172.16.127.254
172.16.127.255
XXX



172.16.128.0
172.16.128.1
172.16.159.254
172.16.159.255




172.16.160.0
172.16.160.1
172.16.191.254
172.16.191.255




172.16.192.0
172.16.192.1
172.16.223.254
172.16.223.255




172.16.224.0
172.16.224.1
172.16.256.254
172.16.256.255





























ooo yes I presume the subnet mask would be....255.255.224.0? and the one with the "XXX" is the range with the 172.16.100.11 I think...

it took me sigh 10 minutes to do one of these though so I'm not sure if that's normal, how long are exams, lol


Did I do this correctly? If i did do this correctly I guess I really do need to do subnetting daily leading up to the exam or else I'll forget it...

and gosh I love this chart! - this finally makes sense to me the shorthand!

Subnet Mask Chart


[TH] Decimal [/TH]
[TH] Shorthand [/TH]
[TH] Binary [/TH]


255.0.0.0
/8
11111111.00000000.00000000.00000000


255.128.0.0
/9
11111111.10000000.00000000.00000000


255.192.0.0
/10
11111111.11000000.00000000.00000000


255.224.0.0
/11
11111111.11100000.00000000.00000000


255.240.0.0
/12
11111111.11110000.00000000.00000000


255.248.0.0
/13
11111111.11111000.00000000.00000000


255.252.0.0
/14
11111111.11111100.00000000.00000000


255.254.0.0
/15
11111111.11111110.00000000.00000000


255.255.0.0
/16
11111111.11111111.00000000.00000000


255.255.128.0
/17
11111111.11111111.10000000.00000000


255.255.192.0
/18
11111111.11111111.11000000.00000000


255.255.224.0
/19
11111111.11111111.11100000.00000000


255.255.240.0
/20
11111111.11111111.11110000.00000000


255.255.248.0
/21
11111111.11111111.11111000.00000000


255.255.252.0
/22
11111111.11111111.11111100.00000000


255.255.254.0
/23
11111111.11111111.11111110.00000000


255.255.255.0
/24
11111111.11111111.11111111.00000000


255.255.255.128
/25
11111111.11111111.11111111.10000000


255.255.255.192
/26
11111111.11111111.11111111.11000000


255.255.255.224
/27
11111111.11111111.11111111.11100000


255.255.255.240
/28
11111111.11111111.11111111.11110000


255.255.255.248
/29
11111111.11111111.11111111.11111000


255.255.255.252
/30
11111111.11111111.11111111.11111100

mikeybinec

I'll be back later tonight, but one of your questions was "How can a /20 be in the 3rd octet?" I think you have that figured out by now. And one thing: Yeah, building charts and filling in the boxes are gonna be time killers, Man..

Do the block method.. I promise, you can subnet in less than a minute, maybe even in less than 30 seconds

A /11 is a block size of 32 /19 is also 32 /27 is 32

rocdamike

mikeybinec wrote: »

I'll be back later tonight, but one of your questions was "How can a /20 be in the 3rd octet?" I think you have that figured out by now. And one thing: Yeah, building charts and filling in the boxes are gonna be time killers, Man..

Do the block method.. I promise, you can subnet in less than a minute, maybe even in less than 30 seconds

A /11 is a block size of 32 /19 is also 32 /27 is 32

Totally agree. Get confident with converting subnet masks from CIDR '/' notation to dotted decimal, finding the interesting octet and working out block sizes. You will find it so much easier in the long run. Plus, it beats having to carry a subnet chart in your pocket when you're out on the field

Deathmage

will I learn the block system if I do subnetting enough? ... or is their a method to the madness?

rocdamike

Here's how I do it. Essentially what I do is convert the CIDR prefix to a dotted decimal subnet mask. I then use the interesting octet of this mask to find my block size. Once you know the block size, everything else is a piece of cake. Let's try an example (I will use one of your example questions):

What network is 172.16.100.11/19 on?

step i) Convert CIDR to dotted decimal: /19 = 255.255.224.0. (I now know that the 3rd octet is the interesting octet as it's the first octet that isn't 255)
step ii) Convert 224 to the corresponding block size. 224 = 32 block size
step iii) I now know my subnets (start from 0 and add 32 each time to the 3rd octet). 172.16.32.0, 172.16.64.0, 172.16.96.0, 172.16.128.0
step iv) It's plain to see that IP address 172.16.100.11 falls within the 172.16.96.0 subnet. 172.16.96.0 is the subnet address, 172.16.127.255 is the broadcast address and everything in between is my Host IP range i.e. 172.16.96.1 - 172.16.127.254.



Mask (interesting octet)
128
192
224
240
248
252
254
255


Corresponding Block Size
128
64
32
16
8
4
2
1

Deathmage

rocdamike wrote: »

Here's how I do it. Essentially what I do is convert the CIDR prefix to a dotted decimal subnet mask. I then use the interesting octet of this mask to find my block size. Once you know the block size, everything else is a piece of cake. Let's try an example (I will use one of your example questions):

What network is 172.16.100.11/19 on?

step i) Convert CIDR to dotted decimal: /19 = 255.255.224.0. (I now know that the 3rd octet is the interesting octet as it's the first octet that isn't 255)
step ii) Convert 224 to the corresponding block size. 224 = 32 block size
step iii) I now know my subnets (start from 0 and add 32 each time to the 3rd octet). 172.16.32.0, 172.16.64.0, 172.16.96.0, 172.16.128.0
step iv) It's plain to see that IP address 172.16.100.11 falls within the 172.16.96.0 subnet. 172.16.96.0 is the subnet address, 172.16.127.255 is the broadcast address and everything in between is my Host IP range i.e. 172.16.96.1 - 172.16.127.254.



Mask (interesting octet)
128
192
224
240
248
252
254
255


Corresponding Block Size
128
64
32
16
8
4
2
1

I'll give it a go the next time I do a subnetting quiz on subnetting.net - Thanks!