# What area of study is this question related to?

• posted

This is CIDR or supernetting. They want you to combine three class C network, so just write them out in binary

00001100 = 12 00001101 = 13 00001110 = 14

------

11111100 is the common mask, so 255.255.252.0 is the best answer, but you have also combined 196.56.15.0 by doing this. Not a good question, IMHO.

Doan

• posted

So I got this question on a prep test, and IO have no idea? Reading material anyone?

> Nationalact has been assigned three class C network addresses: > > 196.56.12.0 > 196.56.13.0 > 196.56.14.0 > > You want to combine these addresses into one logical networkto

increase the number of hostIDs you can have on the network. Which network mask prvides for this?

> 255.255.252.0 > 255.255.255.254 > 255.255.254.0 > 255.255.255.252 >

I thought it was subnetting but couldnt come up with justification and ended up guessing at the answer...thx

• posted

I would chose 255.255.254.0, because it maximizes the the number of host ID's together in one subnet yet does not go outside the allocated range. Probably worded poorly, but it doesn't say you have to use all three class C's.

Kevin

So I got this question on a prep test, and IO have no idea? Reading material anyone?

increase the number of hostIDs you can have on the network. Which network mask prvides for this?

I thought it was subnetting but couldnt come up with justification and ended up guessing at the answer...thx

• posted

Instead of "subnetting" a Class C network into multiple subnets, the question asks you for "supernetting" multiple Class C networks into a supernet.

The netmask required can be found using the following steps:

1. Find the longest match of the leftmost bits of the network addresses.
2. Replace the matched bits by 1s and the remaining bits by 0s.

196.56.12.0 = 11000100 00111000 00001100 00000000

196.56.13.0 = 11000100 00111000 00001101 00000000 196.56.14.0 = 11000100 00111000 00001110 00000000

Netmask = 11111111 11111111 11111100 00000000 = 255.255.252.0

KPLAB

- Free CCNA Study Guide

• posted

This is CIDR / supernetting question.

..12. = 0000 1100 ..13. = 0000 1101 ..14. = 0000 1110

=> common address part for third octet is ".0000 11xx." => network mask for the third octet is ".1111 1100."

So the three networks summarizes or aggregates into

196.56.12.0/20