Practicing subnetting

Hey all,

im practicing subnetting, mainly doing it in my head only and as quickly as possible. I am using the method @ http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html because it makes pretty good sense

now using that method, i get this question:

Question: How many subnets and hosts per subnet can you get from the network 172.18.0.0 255.255.255.0?
Answer: 256 subnets and 254 hosts

i get the hosts, theres 8 host bits, 2^8 = 256-2 = 254

how did they arrive at the 256 subnets though? using the method, you normally are left with network bits as leftovers. i can see why they got the answer, but cant formulate it in my head.

any help is appreciated

Find more posts tagged with

Save $250 on 2025 certification boot camps from Infosec!

Book now with code EOY2025

Button

Comments

j-man

Being this is a class B network, there are 16 Network bits which are never touched. The subnet mask of 255.255.255.0 = 16 Network bits, 8 subnet bits and 8 host bits. So really it will look like the following NNNNNNNN.NNNNNNNN.SSSSSSSS.HHHHHHHH

jtdk985

j-man wrote: »

Being this is a class B network, there are 16 Network bits which are never touched. The subnet mask of 255.255.255.0 = 16 Network bits, 8 subnet bits and 8 host bits.

and the simple things are always the hardest.. lol

thanks, i get it, i was looking at it as a class C network, which is why i was thinking i had no net bits left

j-man

jtdk985 wrote: »

and the simple things are always the hardest.. lol

Ain't that the truth!

Whenever I'm subnetting, I always determine the class straight away. I just find that a little easier if I can narrow down the number of octets I've got to work with.

jtdk985

j-man wrote: »

Ain't that the truth!

Whenever I'm subnetting, I always determine the class straight away. I just find that a little easier if I can narrow down the number of octets I've got to work with.

ill make sure to do that first thing

rajithshan

can someone help me with this subnet question.

what is the first valid host on the subnetwork that the node 172.30.243.54/23 belongs to.
I would be thankful if anyone could explain me the answer.

thanks ..

miller811

rajithshan wrote: »

can someone help me with this subnet question.

What is the first valid host on the subnetwork that the node 172.30.243.54/23 belongs to.
I would be thankful if anyone could explain me the answer.

Thanks ..

172.30.242.1

capitanuionut

miller811 wrote: »

172.30.242.1

Yes this is the correct answer....I will post an explanation for you:

172.30.243.54/23

First I see that is a class be network.
Then I look at the number of hosts : 2^9(32-23) = 512 addresses = 510 hosts.
That means the networks will be : 172.30.0.1 - 172.30.1.254
172.30.2.1 - 172.30.2.254
and so on
Based on this we get the answer 172.30.242.1

martell1000

i invented a method by counting with one hand down from 30 till i reach the subnet i need and then go with the other hand and from 4 and always double the number.

if you keep in mind of broadcast and network address its a very fast way for class C and B networks

jtdk985

have another one, feeling stupid about it

Question: What is the broadcast address of the network 172.20.161.128 255.255.255.224?
Answer: 172.20.161.159

class b network correct? thought i should be using the 3rd octet for my calculations? not seeing how they got this answer

jtdk985

and one more, i feel like im starting these out wrong somehow

Question: What is the first valid host on the subnetwork that the node 10.85.176.55 255.255.240.0 belongs to?
Answer: 10.85.176.1

capitanuionut

jtdk985 wrote: »

have another one, feeling stupid about it

Question: What is the broadcast address of the network 172.20.161.128 255.255.255.224?
Answer: 172.20.161.159

class b network correct? thought i should be using the 3rd octet for my calculations? not seeing how they got this answer

this is quite simple...just look at the subnet mask..../27...you have 5 bits for hosts..so 2^5 = 32
128+32 = 160 - next network
160- 1 = 159 - your networks broadcast address..

jtdk985

capitanuionut wrote: »

this is quite simple...just look at the subnet mask..../27...you have 5 bits for hosts..so 2^5 = 32
128+32 = 160 - next network
160- 1 = 159 - your networks broadcast address..

ty i think i made that one a lot harder than it needed to be. i started from the very bottom for no reason, and even added wrong. simple mistakes, but ty again for showing me the light

any ideas for the other that i posted?

actually looking at it, i think im doing the wrong thing again. for the other one i posted i mean, im looking that its a class A network, and starting there counting up with my host bits, when i need to be starting at the last interesting octet of my mask.

miller811

jtdk985 wrote: »

have another one, feeling stupid about it

Question: What is the broadcast address of the network 172.20.161.128 255.255.255.224?
Answer: 172.20.161.159

class b network correct? thought i should be using the 3rd octet for my calculations? not seeing how they got this answer

always search for the interesting octet. the first non 255...

256 - interesting octet tells you the block size
256-224 = 32

the broadcast address of a subnet will be one less than the next subnet.
your subnet is 172.20.161.128
next subnet is 172.20.161.160
broadcast is 172.20.161.159

CodeBlox

capitanuionut wrote: »

Yes this is the correct answer....I will post an explanation for you:

172.30.243.54/23

First I see that is a class be network.
Then I look at the number of hosts : 2^9(32-23) = 512 addresses = 510 hosts.
That means the networks will be : 172.30.0.1 - 172.30.1.254
172.30.2.1 - 172.30.2.254
and so on
Based on this we get the answer 172.30.242.1

For the sake of completeness, that part in red is wrong. Should be 172.30.2.1 - 172.30.3.254

Easy way to find subnet boundaries is to check "the place value of the least significant subnet bit" in the special octet.

If you have 172.18.33.0 /19

There are 3 bits set in the third octet with places values 128, 64, and 32. 32 is the value of the least significant subnet bit so that is your multiplier. Your subnets would be:

172.18.0.0
172.18.32.0
172.18.64.0
172.18.96.0
172.18.128.0
172.18.160.0
172.18.192.0
172.18.224.0

ccnaomkar

jtdk985 wrote: »

and one more, i feel like im starting these out wrong somehow

Question: What is the first valid host on the subnetwork that the node 10.85.176.55 255.255.240.0 belongs to?
Answer: 10.85.176.1

question:10.85.176.55

255.255.240.0

answer:240=128+64+32+16

block size of 16

10.85.0.0
10.85.16.0
10.85.32.0
|
|
|
|
10.85.160.
10.85.176.0-10.85.191.255 valid first host 10.85.176.1
10.85.192.0

ccnaomkar

jtdk985 wrote: »

have another one, feeling stupid about it

Question: What is the broadcast address of the network 172.20.161.128 255.255.255.224?
Answer: 172.20.161.159

class b network correct? thought i should be using the 3rd octet for my calculations? not seeing how they got this answer

Question: What is the broadcast address of the network 172.20.161.128 255.255.255.224?
Answer: 172.20.161.159

224=128+64+32 or 256-224=32
block size of 32

172.20.161.0

172.20.161.32

172.20.161.64

172.20.161.96

172.20.161.128 - 172.20.161.159

172.20.161.160

ccnaomkar

capitanuionut wrote: »

Yes this is the correct answer....I will post an explanation for you:

172.30.243.54/23

First I see that is a class be network.
Then I look at the number of hosts : 2^9(32-23) = 512 addresses = 510 hosts.
That means the networks will be : 172.30.0.1 - 172.30.1.254
172.30.2.1 - 172.30.2.254
and so on
Based on this we get the answer 172.30.242.1

172.30.243.54/23

11111111.11111111.111111110.00000000
block size of 2

172.30.0.0
172.30.2.0
172.30.4.0
172.30.240.0-
172.30.242.0-172.30.243.255
172.30.244.0

answer:172.30.242.1