Subnetting Question

How many subnets and hosts per subnet can you get from the network 174.20.0.0/23 Remember for host part at least 2 bits are required?
Solve this question please and tell me the complete method ...
Thanks in advance

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fadhil

there are 128 subnets and 510 hosts from that network.

to get number of hosts
from /23 it means that 23 bits are on while others are off. so take number of bits which are off.
for there are 9 bits which are off
then 2^9 -2 =510 number of hosts.
for number of subnets
since it class B its default is /16.but according to the question is /23.
take 23-16=7
hence to get subnets =2^7 =128

union122

Fadhil, I like you method.. How do you get the bits off?

swild

its a 32 bit address with 23 (/23) bits for the network leaving 9 bits for the host.
The mask would be written in binary like this:
11111111.11111111.11111110.00000000
Ones being "on" and the zeros being the "off" bits.

gadav478

yeah bro that method is smoking fast... bookmarking this now.

fadhil

thanks so much Union122 and geoffrey.a.davenport

rphann

The answer is: 510 hosts per subnet and 128 subnets.

255.255.254.0 = subnet mask
11111111.11111111.11111110.00000000

Determine the hosts:
Formula: (2^h)-2=
(2^9)-2=510
Note: count the zeros for the hosts (h)

Determine the subnet:
Formula: 2^s=
(2^7)=128
Note: count the ones in the third octet.

Irfanshafiq

Good Day to All...!!!!

What will we say about the class of this IP

174.20.0.0/23

Either it is a Class B IP or example of Class less IP...

Looking for expert comments.

Steadfast Girl

thanks to all for this help .

fadhil

Irfanshafiq wrote: »

Good Day to All...!!!!

What will we say about the class of this IP

174.20.0.0/23

Either it is a Class B IP or example of Class less IP...

Looking for expert comments.

that is class B ip address since class B range from 128 to 191 and class C range from 192 to 223

FloOz

great help in this thread! thanks!

KenC

fadhil wrote: »

that is class B ip address since class B range from 128 to 191 and class C range from 192 to 223

I think this is considered a CIDR address (classless) because the default subnet mask of 255.255.0.0 for class B is not being used.

NetworkVeteran

Steadfast Girl wrote: »

How many subnets and hosts per subnet can you get from the network 174.20.0.0/23 Remember for host part at least 2 bits are required?
Solve this question please and tell me the complete method ...
Thanks in advance

A /24 has 256 addresses. A /23 must has one bit more (twice as many addresses), and 256x2 = 512.
You generally subtract the all-zeroes and all-ones host addresses, leaving 510 host addresses.

174.20.x.x is Class B (16 network bits), so there are 23-16=7 subnet bits.
2^7 = 128 subnets total.

NetworkVeteran

KenC wrote: »

I think this is considered a CIDR address (classless) because the default subnet mask of 255.255.0.0 for class B is not being used.

Subnetting predates and does not require CIDR. In this case, there are 16 network bits, 7 subnet bits, and 9 host bits. In fact, answering "How many subnets?" with a concrete number--as is typical--obligated us to make the classful assumption that there are 16 network bits!

KenC

I'm a little bit confused about this - will have to revisit the books.

So is CIDR got to do with dealing with public IP addresses or something?