Still confused on how to get block size

JockVSJock

Ok, I'm still working on subnetting here. Here is a question from subnetting.org that I'm lost on:

Question: You have the following subnetted network: 172.16.0.0/21. You need to assign your router the first usable host address on the third subnet. What address would you use?

Answer: 172.16.16.1

I'm still confused on how to do the blocking for this problem. I did the following:

Tried to get the block size by subtracting 32 - 21. The 21 is the prefix notation.

I get 11

Once I convert the ip address and subnet into binary, I get the following network address: 172.16.0.0

Then I tried to add the 11 into the third octect to create by subnets:

172.16.0.0
172.16.11.0
172.16.22.0

This can't be correct because the answer has 172.16.16.1 and that is falls under the 2nd subnet...

IEWANNABE

JockVSJock wrote: »

Ok, I'm still working on subnetting here. Here is a question from subnetting.org that I'm lost on:



I'm still confused on how to do the blocking for this problem. I did the following:

Tried to get the block size by subtracting 32 - 21. The 21 is the prefix notation.

I get 11

Once I convert the ip address and subnet into binary, I get the following network address: 172.16.0.0

Then I tried to add the 11 into the third octect to create by subnets:

172.16.0.0
172.16.11.0
172.16.22.0

This can't be correct because the answer has 172.16.16.1 and that is falls under the 2nd subnet...
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I found this info on this site....worked well for...

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

osJoe

First, you need to know that a /21 is 248. After you find that out you get your block size by subtracting 256 by 248, which gives you 8, so your subnets are...

subnet 0.0 | 8.0 | 16.0 ect....
first host 0.1 | 8.1 |16.1
last host 7.254 | 15.254 | 23.254
broadcast 7.255 | 15.255 | 23.255

JockVSJock

osJoe wrote: »

First, you need to know that a /21 is 248. After you find that out you get your block size by subtracting 256 by 248, which gives you 8, so your subnets are...

subnet 0.0 | 8.0 | 16.0 ect....
first host 0.1 | 8.1 |16.1
last host 7.254 | 15.254 | 31.254
broadcast 7.255 | 15.255 | 31.255

Ok, so I can't subtract the prefix number (/21). I have to convert to the subnet mask number and then subtract from 32, correct?

I'll have to practice some more and post my results back.

osJoe

JockVSJock wrote: »

Ok, so I can't subtract the prefix number (/21). I have to convert to the subnet mask number and then subtract from 32, correct?

I'll have to practice some more and post my results back.

I made i mistake in the table, subnet 24.0 comes after 16.0, not 32. \

There are many ways to subnet, I have never tried doing it that way so I'll let someone more knowledgeable to reply on that part.

JockVSJock

So if I am understanding you correct, then this is the logic behind it.

If the number is < 7 then subtract 8
If the number is < 15 and >= 8 then subtract 16
If the number is < 31 and >= 24 then subtract 32

osJoe

I think you are looking at it from a totally different perspective, or just thinking too hard..

You will always subtract from 256. To find the number that you need to subtract just look at the CIDR notation that is given.

For example... If we have a 172.16.10.66/27, we see that the subnet comes out to be 255.255.255.224

To find your interesting octet, which is where the subnetting is taking place, line up the IP address and the subnet mask..

172 .16 .10 .66
255.255.255.224

Now, subtract 224 from 256, you get 32. So you line up the subnets starting from 0, adding 32.

0,32,64,96,128,160... ect

From there you can fill out the host ranges, and broadcasts. The very first ip (network address) and last (broadcast) cannot be used as a valid host.

(this table skips networks 0 and 32)


Network
172.16.10.64
172.16.10.96
172.16.10.128
172.16.10.160


First Host
172.16.10.65
172.16.10.97
172.16.10.129
172.16.10.161


Last Host
172.16.10.94
172.16.10.126
172.16.10.158
172.16.10.190


Broadcast
172.16.10.95
172.16.10.127
172.16.10.159
172.16.10.191






Hope i explained well enough. I highly recommend Todd Lammle's way of subnetting!

onesaint

As there are 8 million ways to calculate a subnet, I looked at it a different way as well.

I see the /21 is less than 24, putting it in the 3rd octet. I go with a magic number from there:

16+5=21

I need to move 5 bits over in the 3rd octet: 128, 64, 32, 16, 8, 4, 2, 1
So, now I know my subnets will just be an increment of 8.xxx which leads to:
subnet 1: xxx.xxx.0.0 - xxx.xxx.7.255
subnet 2: xxx.xxx.8.0 - xxx.xxx.15.255
subnet 3: xxx.xxx.16.0 - xxx.xxx.23.255
Etc.

Essentially, this is just shifting your focus from the 4th octet to the 3rd which made things a little easier for me. I x'ed out the first two octets as they don't matter because were only focusing on the 3rd octet (/21). You can drop in your 172.16.x.x as in your example and it works out just the same.

On a side note, in your logic you missed the 8 bits to an octet. So, 32-21 = 11 which is 3+8. 8 being the 4th octet and if you count 3 from right to left in the 3rd octet, your next number is the 8 bit (... 8,4,2,1), like the example above.

thedrama

JockVSJock wrote: »

Ok, I'm still working on subnetting here. Here is a question from subnetting.org that I'm lost on:



I'm still confused on how to do the blocking for this problem. I did the following:

Tried to get the block size by subtracting 32 - 21. The 21 is the prefix notation.

I get 11

Once I convert the ip address and subnet into binary, I get the following network address: 172.16.0.0

Then I tried to add the 11 into the third octect to create by subnets:

172.16.0.0
172.16.11.0
172.16.22.0

This can't be correct because the answer has 172.16.16.1 and that is falls under the 2nd subnet...
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My apologies but this is false from the beginning.

if IP address is 172.16.0.0/21 , you should substract the default value of class B (/16) from the given value(/21) in order to find how many
subnet bits borrowed.

So, 21-16 = 5 bits. Now, where to begin the subnetting?
Look, 172.16.x.x is the class B IP address by default. The default subnet mask for the class B is 255.255.0.0, however, it will change due to
subnetting.

the illustration for the class B is (network.network.host.host)

The result ? we had borrowed 5 bits. That does mean beginning from the left of the third octet because first and the second is for network thats why those should remain as they are, 11111000 is equal to 128+64+32+16+8 = 248

This leads us from 255.255.0.0 to 255.255.248.0

So,then we should substract this 248 from 256(the whole) in order to find the block size. And it is 8.

In this case, the first subnet will be 172.16.0.0, the second will be 172.16.8.0 the third will be 172.16.16.0 up to 172.16.240.0(included)

How many subnets do we have? 21-16 = 5 --> 2^5 = 32 lets verify this below

172.16.0.0
172.16.8.0
172.16.16.0
172.16.24.0
.
.
.
.
.
.
.
172.16.240.0


how many host bits per subnet? lets open the resulted subnet . 11111111.11111111.11111000.00000000

So, how many zeroes do you count? 3 bits on the third octet and 8 bits on the last octet resulted in 2^11 = 2048 host bits per subnet


Now your answer, lets get back to third subnet

if subnet zero rule is enabled, the first and the last subnet are counted as well.

172.16.0.0 --> the first subnet
( )
( ---- block size = 8 )
( )
172.16.8.0 --> the second subnet


172.16.16.0 --> the third subnet

172.16.24.0 --> the forth subnet

Thats why the valid hosts are 172.16.16.1 through 172.16.23.254 on the third subnet

Ltat42a

Here's a handy chart to help you with block sizes, this works for each of the octets -



Bit
1
1
1
1
1
1
1
1


Mask
128
192
224
240
248
252
254
255


Block size
128
64
32
16
8
4
2
1




hth

IEWANNABE

The info that I posted earlier in the link below is really good. For someone who has equally struggle with subnetting, the info in that link I posted earlier was the simplest and helped me greatly.

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

The main thing to get here is your barriers. The Octet is divided into four parts divided by a (.) or barrier...think of the "period" as a barrier.

at the end of the 1st peroid, or barrier is decimal # 8
at the end of the 2nd period, or barrier is decimal # 16
at the end of the 2nd period, or barrier is decimal # 24
at the end of the 2nd period, or barrier is decimal # 32

Now, whatever your /# or CIDR is, simply deduct that # from the next or upcoming barrier #, then use your power of 2's table. whatever you're left with after subtracting your CIDR # from your next barrier #, your block or increment will be that number x'S the power of 2.

Lets take for instance /26

You know that your next barrier number is 32....remember...8, 16, 24, 32

so in this instance, you simply subtract 26 from 32 which leaves you with 6

Now go to your power of 2's table... 2^6 = 64

64 is your block or increment

Very fast and easy way of determining your block if you have the CIDR.

Now, if your not given the CIDR or slash # and you're given a subnet mask address instread, you simply convert the address into binary and count all of the 1's in the Octet that has 0's. For instance in the subnet mask 255.255.255.240, after you convert this address to binary, it will look like this: 11111111.11111111.11111111.11110000. You now have 28 Network bit, subtract that from your next barrier which is 32, and you're left with 4. You then go to your power of 2 table.... 2^4 =16
16 is your block or increment #

While we're here, you can use the binary address to figure out how many hosts and subnets you can get. Simply count the number of 1's in the last octet, then use your power of 2 tables. So you have four 1's... 2^4 = 16 for a total of 16 subnets. Likewise, to find out the number hosts per subnet, add up the # of zero's in that last octet and again use your power of 2 tables..in this case you have "four" zero's left... so 2^4 = 16 hosts per subnet.

This should work until you become familiar with the process and can memorize some of the helpful charts that guys have posted.

Anyone, please free free to critique what I've just posted, as I'm very new to subnetting and I certainly wouldn't want to mislead anyone.

billyr

Just look at the binary placement value of the last bit you turn on. It doesn't get much easier than that.

e.g

If it was a /27, the last bit you would have turned on. i.e the 27th bit along would be the 32 bit, thats your increment.

Alternatively:

If your mask is a /24 or above: Deduct from 32 and do the powers of 2 against the remainder.
If your mask is a /16 or above do the same but deduct from 24.
If your mask is a /8 or above again the same but deducting from 16.

e.g /29 : 32 -29= 3. 2^3 = increment of 8 in the last octet.
e.g. /19: 24 -19= 5. 2^5= increment of 32 in the 3rd octet.
e.g. /10: 16 -10= 6. 2^6= increment of 64 in the 2nd octet.

JockVSJock

billyr wrote: »

Just look at the binary placement value of the last bit you turn on. It doesn't get much easier than that.

e.g

If it was a /27, the last bit you would have turned on. i.e the 27th bit along would be the 32 bit, thats your increment.

Alternatively:

If your mask is a /24 or above: Deduct from 32 and do the powers of 2 against the remainder.
If your mask is a /16 or above do the same but deduct from 24.
If your mask is a /8 or above again the same but deducting from 16.

e.g /29 : 32 -29= 3. 2^3 = increment of 8 in the last octet.
e.g. /19: 24 -19= 5. 2^5= increment of 32 in the 3rd octet.
e.g. /10: 16 -10= 3. 2^6= increment of 64 in the 2nd octet.

If that is the case, then why for the following:

172.16.78.137/19

I take 24 - 19 which is 5, because that is the next bit to turn on is /24, right?

If I put this ip and subnet into the subnet calculator. It shows a block size of 32 with 8 subnets for the network.

EDIT: I keep screwing this up, I meant 256 - 224, which is a block size of 32.

billyr

JockVSJock wrote: »

If that is the case, then why for the following:

172.16.78.137/19

I take 24 - 19 which is 5, because that is the next bit to turn on is /24, right?

If I put this ip and subnet into the subnet calculator. It shows a block size of 32 with 8 subnets for the network.

EDIT: I keep screwing this up, I meant 256 - 224, which is a block size of 32.

Correct - a /19 would give you increments of 32.
The 19th bit turned on has a placement binary placement value of 32.
Or
24-19 = 5.....2^5 = an increment of 32
Or
256 - 224 = 32.

JockVSJock

Ok, I ran into another problem on subnetting.org, that involves block size.

Question: You need to assign a server the last valid host address of the fourth subnet on network 192.168.24.0/28. What address would you use?

Answer: 192.168.24.62

First I found the block size. First, need to subtrack the last octect which is 240 from 256.

This gives us 16, so this is the block size.

Then I break the following ip address down:

192.168.24.0
255.255.255.240

11000000.10101000.00011000.00000000
11111111.11111111.11111111.11110000

N: 192.168.24.0
FH: 192.168.24.1
LH: 192.168.24.14
B: 192.168.24.15

This is where I'm still getting confused. My first valid network address is 192.168.24.0.

From here I add 16 to the last octect, correct?

192.168.24.0
192.168.24.16
192.168.24.32
192.168.24.48
192.168.24.64
192.168.24.80

I think I count up to the fourth address, which is 192.168.24.48 and then assign the last valid ip address, correct?

thanks

onesaint

Yup. So, your host range is 49-62. 192.168.24.48 is the network number and 192.168.24.63 is the broadcast, making the last valid 192.168.24.62.

Generally, I just see that the 28 is 4 bits more than /24 and start counting up blocks by 16.

drkat

you can always look at the mask /21 next classful boundry is /24 24-21 = 3 2^3 = 8 block size is eight

your 11 is your hosts 2^11 = 2048

dazl1212

How does this work?
Its these i'm really struggling with 254 class B subnets
Question: Which subnet does host 172.16.89.2 255.255.254.0 belong to?
Answer: 172.16.88.0

Ltat42a

1. What Class network is this? B
2. Default mask for Class B is 255.255.0.0

The intersting octet here is x.x.254.0

A 255.255.254.0 mask gives you a block size of 2 in the third octet
Starting at zero, count up the subnets in increments of 2(in the third octet)
172.16.0.0
172.16.2.0
171.16.4.0
172.16.6.0
etc
etc
etc till you get to the subnet in which 172.16.89.2 belongs to (172.16.88.0)

This particular address is in this range -
Network - 172.16.88.0
1st Host - 172.16.88.1
Last host - 172.16.89.254
Broadcast - 172.16.89.255

Next network - 172.16.90.0

Does this help???

dazl1212

Ltat42a wrote: »

1. What Class network is this? B
2. Default mask for Class B is 255.255.0.0

The intersting octet here is x.x.254.0

A 255.255.254.0 mask gives you a block size of 2 in the third octet
Starting at zero, count up the subnets in increments of 2(in the third octet)
172.16.0.0
172.16.2.0
171.16.4.0
172.16.6.0
etc
etc
etc till you get to the subnet in which 172.16.89.2 belongs to (172.16.88.0)

This particular address is in this range -
Network - 172.16.88.0
1st Host - 172.16.88.1
Last host - 172.16.89.254
Broadcast - 172.16.89.255

Next network - 172.16.90.0

Does this help???

I think so, I will try a few more using that formula. I have also been reading Lammles book since I finished work 2 hours ago my head is hurting.
Think I need to take a break.
Will post how I get on.
Thank you

drkat

The problem is there are so many ways of doing this.

With an example:

172.16.89.6 255.255.254.0

We know two things

1) this is a class b address
2) we'll be working in the 3rd octet - this is because all the 255's arent usable the highest value in 1 octet is 255

since it is 254 we HAVE 1+8 available bits for hosts 2^9 = 512-2 so 510 hosts per subnet

now on to the network part

256 - 254 = 2 which is our block size (how big our sub-nets are - just reminder that our next total increment is the next subnet)


172.16.88.0 - 172.16.89.254
172.16.90.0 - 172.16.91.254 etc

so how many subnets does this give us?

lets look at what we've borrowed.

original class is /16 + 7 = 23 so this is a /23
7 ones added onto the original class
2^7 = 128 total networks with 510 hosts per network


I hope this clears it up

I personally dont like Lammle's method of subnetting.

IEWANNABE

Get it yet??

pham0329

I think you need to pick a subnetting method that works for you, and stick with it. If you looked at your posts about subnetting over the last few months, you'd see that you're flipping back and forth between different methods. Find one that you like (preferably one that doesn't involve converting everything to binaries) and practice it

dazl1212

I think so mate.
One thing thats confusing me now (is and this may seem silly so bare with me)
I think you only count the subnet bits as the subnets, so say you borrow to subnet bits you have 4 subnets and not 2 bits plus the 16bits already in the mask by default?

IEWANNABE

I agree with Pham0329, with so many ways to do subnetting, you'll have to find one that your comfortable with. I used the method that I posted earlier in this thread, because there was no binary conversion which made it EASY for me...and trust me when I say that I struggled before doing this method. Try re-reading it...also let it marinate... don't rush! Often I've had to just sit there and stare it for sometime before things started clicking.

dazl1212

IEWANNABE wrote: »

I agree with Pham0329, with so many ways to do subnetting, you'll have to find one that your comfortable with. I used the method that I posted earlier in this thread, because there was no binary conversion which made it EASY for me...and trust me when I say that I struggled before doing this method. Try re-reading it...also let it marinate... don't rush! Often I've had to just sit there and stare it for sometime before things started clicking.

Yes I think I'm finally getting there.
Gonna knock out about 10 practice sessions a day to try and keep it fresh

drkat

dazl1212 wrote: »

I think so mate.
One thing thats confusing me now (is and this may seem silly so bare with me)
I think you only count the subnet bits as the subnets, so say you borrow to subnet bits you have 4 subnets and not 2 bits plus the 16bits already in the mask by default?

/16 + 2 = /18 which would be 255.255.192.0 see you've "borrowed" two bits from the third octet? 255.255.0.0 defines the octet you're working in since 255.255.0.0 says we cant "work" in the first two octets we MUST work in the third.

Just remember your default classes. /8 /16 /24

/8 = we're subnetting in the second octet
/16 third
/24 fourth

etc and pay attention to the type of address and the subnet mask.

I WILL agree with everyone that you need to pick ONE method and stick with it. I went through a lot of methods trying to learn subnetting and if you just sit back and relax you will see that it is all a big pattern. (SEE BELOW)

You may also want to try drawing a chart?

so in this example say:

172.16.88.9
255.255.254.0

7 bits turned on so we count down the list till we see 254 so we know this is the right mask and then we count across to 7 which is 128 which is our # of subnets so since we're in the third octet and we have 7 on that must mean we have 9 0's for hosts so count across 9 spaces which is 512-2 = 510 and there is your answer:


128 1
192 2
224 3
240 4
248 5
252 6
254 7

2 4 8 16 32 64 128 256 512 1024 2048 4096

===============
if you're wondering why we have to go 256-subnet mask to find our block size it's simply because a single octet holds 0 - 255 or 256 possible values. So say for instance you didnt know the block size but knew we had 128 subnets.... 256/128 = 2 it's all relative.

Keep practicing you'll get it.

IEWANNABE

drkat wrote: »

/16 + 2 = /18 which would be 255.255.192.0 see you've "borrowed" two bits from the third octet? 255.255.0.0 defines the octet you're working in since 255.255.0.0 says we cant "work" in the first two octets we MUST work in the third.

Just remember your default classes. /8 /16 /24

/8 = we're subnetting in the second octet
/16 third
/24 fourth

etc and pay attention to the type of address and the subnet mask.

I WILL agree with everyone that you need to pick ONE method and stick with it. I went through a lot of methods trying to learn subnetting and if you just sit back and relax you will see that it is all a big pattern. (SEE BELOW)

You may also want to try drawing a chart?

so in this example say:

172.16.88.9
255.255.254.0

7 bits turned on so we count down the list till we see 254 so we know this is the right mask and then we count across to 7 which is 128 which is our # of subnets so since we're in the third octet and we have 7 on that must mean we have 9 0's for hosts so count across 9 spaces which is 512-2 = 510 and there is your answer:


128 1
192 2
224 3
240 4
248 5
252 6
254 7

2 4 8 16 32 64 128 256 512 1024 2048 4096

===============
if you're wondering why we have to go 256-subnet mask to find our block size it's simply because a single octet holds 0 - 255 or 256 possible values. So say for instance you didnt know the block size but knew we had 128 subnets.... 256/128 = 2 it's all relative.

Keep practicing you'll get it.

This^ good info! Here's also a full chart that will help illustrate what he's saying. Also for those seasoned networkers, the ruler in Figure B, is excellent for planning out networks:

IP subnetting made easy | TechRepublic

daz, check out the Figure C chart...capitlizes on what drkat was saying.

drkat

I would agree on the link. You just really need time to digest it. Take a break from it and when you come back you'll be able to see it in a different light. Unfortunately it's just like our times tables.. have to keep practicing - the math will make sense along the way

JockVSJock

Ok, had another problem (I wish there was a way to select only these problems and do these so I can get better at them).

I didn't get this correct

Question: Your router needs to be assigned the first valid host address of the 2nd subnet on network 192.168.4.0/28. What address would you assign?
Answer: 192.168.4.17

I though the block size was 16 (256 - 240) and got the following for the network address:

Network - 192.168.4.240
1st host - 192.168.26.240

drkat

You are correct the size is 16 but you're in the wrong octet.

Your router needs to be assigned the first valid host address of the 2nd subnet on network 192.168.4.0/28. What address would you assign?

192.168.4.0
255.255.255.240

256 - 240 = 16

192.168.4.0 network valid hosts= 1 - 14 .15 broadcast
192.168.4.16 network valid hosts 17 - 30 .31 broadcast

second subnet first host is .17


192 = class C
255.255.255.240 = 4th octet we're working in