Quick Subnetting Question

I'm going through the questions on www.subnettingquestions.com

They have :

Question: How many subnets and hosts can you get from the network 172.16.0.0/24?

Answer: 256 subnets and 254 hosts

I understand that they are using classfull subnetting, so I guess my question is how am I supposed to know when to use classfull subnetting and when to use classless? Will the question mention CIDR? If the question reads:

How many subnets and hosts can you get from the CIDR block 172.16.0.0/24?

Would the answer be the same?

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Comments

kafifi13

I ran into the same problem. ON the exam will it tell you to subnet using classless?

ignign0kt

The network is classful if it uses a default network mask. /8 for class A, /16 for class B, and /24 for class C. Anything else is going to be classless and therefore using CIDR

malcybood

Following on from what ignidn0kt said, if the specified address combined with subnet mask, is anything other than the below default values then you are using a classless address. Being able to determine that the address is classless, knowing host ranges of subnets, network and broadcast addresses is part of the exam objectives.

Class A
Address Range = 0.0.0.0 - 126.255.255.255
CIDR = /8
Default Subnet Mask - 255.0.0.0

Class B
Address Range = 128.0.0.0 - 191.255.255.255
CIDR = /16
Default Subnet Mask - 255.255.0.0

Class C
192.0.0.0 - 223.255.255.255
CIDR = /24
Default Subnet Mask - 255.255.255.0

Once you learn subnetting properly you will be able to tell if an address is classful by combining network address & subnet mask.

For example a network address of 199.167.3.1 with subnet mask of 255.255.255.0 is a classful class C address as the first octet is 199 (falls inbetween 192 - 223). The network address here would be 199.167.3.0 and the broadcast 199.167.3.255 leaving .1 - .254 as usable hosts.

The same network address of 199.167.3.1 with a subnet mask of 255.255.255.252 is classless as you have used a /30 or .252 CIDR block. The network address would be 199.167.3.0 and the broadcast would be 199.167.3.3 leaving .1 and .2 as usable hosts