how many hosts and subnets?

liven

I pretty much have a method down for the other kinds of subnetting questions, but not sure how to approach this kind...

Suggestions for a general method, for quickly figuring this kind out:

You are designing a subnet mask for the 172.23.0.0 network. You want 29 subnets with up to 1800 hosts on each subnet. What subnet mask should you use?

protocol167

I think this might be correct. I am new to subnetting also and wanted to see if I got it right. But what I got was the below answer. I can't really explain it, sorry. But just kind of did it my way.


You are borrowing 11 bits for the host portion. So that will give you a remaing of 5 bits that you did not use.

Subnet rules is that whatever you did not borrow times the power of 2 will give you the amount of subnets. So you have 5 remaining bits times the power of 2 will equal 32 subnets.

Whatever you borrowed, which is 11 bits, to the power of 2 will give you 2046 usable host minus two.

1111111.11111000.00000000.0000000

255.255.249.0


172.23.0.0/11

172.23.0.1 - 172.23.7.254



2 to the power of 11 will give you 2046 host and 2 to the power of remaining bits (5) gives you 32 subnet.

mikearama

Close, bro. You've got it right, just off slightly in your binary mask, and subnet.

You are given a class B subnet (172.23.0.0), so 255.255.0.0 is already spoken for. Now you need to break that down to give you the required subnets/hosts.

You want 29 subnets... that'll take 5 bits into the class C octet... so now your mask is 255.255.248.0. That's 11111111.11111111.11111000.00000000 That's a mask of /21.

Now, verify that this mask provides for the number of hosts required... 2**11th - 2 = 2046 Yep, plenty.

So, your available subnets using 255.255.248.0 are:

172.23.0.0 /21
172.23.8.0 /21
172.23.16.0 /21
...
172.23.248.0 /21

Mike

protocol167

oopsy...didn't add up my bits correctly. Should of been 248 like you said and in the wrong binary. Yeah off by one. Must be a faulty calculator im using.

Stotic

Hey, not calculator! Its not allowed on the exam!

But I'm guilty of that too...it's a hard habit to break.

NeonNoodle

It's a class B address, so you have 16 bits to work with. If you want at most 1800 hosts then you need 11 host bits because 2^10<1800<2^11=2048. That leaves you with 5 subnet bits which will give you 32 subnets, more than enough for your requirements. Thus you would have the following subnets:

172.23.0.0/21
172.23.8.0/21
172.23.16.0/21
.
.
.
172.23.8n.0/21
.
.
.
172.23.248.0/21

where 0<=n<=31

Cessation

Subnetting will be my downfall