nav[aria-label="Primary Navigation"] { padding: 0; & ul { list-style: none; width: 100%; display: flex; flex-direction: row; justify-content: start; align-items: start; gap: 30px; padding: 0; & li { margin: 0; } & ul li { list-style: none; } } }

Explain this subnetting/host please

Sounds Good

I'm kind of stumped on this answer. How are the subnets/hosts concluded from this?

Thanks for any help

SG

Find more posts tagged with

SAVE $250 on Your CISCO Boot Camp — Use Code "EXAM26"

Exclusively for TechExams members for Infosec Boot Camps starting before April 30, 2026

View CISCO Boot Camps

Comments

Ltat42a

That looks like one of those matching questions.
In questions like these, look at the class of address you are subnetting from, then look at the mask you are given.
The 172.16.22.0 = Class B. The default mask for Class B is 255.255.0.0. You are given 255.255.255.192. 10 bits were taken from the host portion for subnets, which leaves 6 bits for hosts.

Using your subnets formula, 2(n) n= 10, you have 1024 subnets.

You have 6 bits remaining for hosts, using your hosts formula 2(n)-2, n = 6, 64-2 = 62 - you have 62 hosts per subnet.

Your answer is the 2nd one down from the top on the right.

hth

Carnby

This exact problem was stumping me for the past few weeks. Thanks for making this post, and that you Ltat42a for explaining.

goldenlight

Simple question once you understand the basics. 32 bits in one IPV4 classfull address. Network+ Subnet + Host= 32 bits

Given:

Class A = 8 Network Bits
Class B= 16 Network bits
Class C= 24 Network Bits

Subnet=2^x
Host= 2^x-2
x=bits

(H)Host = 32 - Prefix(P).
Subnet(S) = Prefix(P) - (N)Network.
Example 188.12.2.0 /22

Solution : This is a Class B Address so the Network = 16
By adding the Network and Host bits I get 26. Subtract 26-32 ANd we get 6 bits or 22-16=6

32-22= 10 bits

So we add these all up I should get 32. 16+6+10=32. NOw we are ready to calculate our answers.

Final answer
Subnet= 2^6= 64
Host=2^10= 1022

Sounds Good

Ltat42a wrote: »

That looks like one of those matching questions.
In questions like these, look at the class of address you are subnetting from, then look at the mask you are given.
The 172.16.22.0 = Class B. The default mask for Class B is 255.255.0.0. You are given 255.255.255.192. 10 bits were taken from the host portion for subnets, which leaves 6 bits for hosts.

Using your subnets formula, 2(n) n= 10, you have 1024 subnets.

You have 6 bits remaining for hosts, using your hosts formula 2(n)-2, n = 6, 64-2 = 62 - you have 62 hosts per subnet.

Your answer is the 2nd one down from the top on the right.

hth

Thanks for the explanation, but the answers are already paired with the questions. So the questions and answers are already in the right row.

I do get the idea behind it though. Thanks!