SUBNETTING / VLSM

I understand the pre 12.0 ip subnet zero thing. What I am confused about is this ....

172.16.0.0 / 30 - What is the first pair of host addresses.

Sybex CCNP page 50 contradicts itself ...

the book says .1 and .2 are the first hosts,

I say .4 or .5 because 256-252 = 4 = increment and first net.

Either way, when taking the test if I am asked this question whats right ?

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

forbesl

vidalch wrote:

the book says .1 and .2 are the first hosts,

The book is correct

Network 172.16.0.0 with a 30 bit mask (/30 - 255.255.255.252) is hosts 172.16.0.1 and 172.16.0.2 with the broadcast IP being 172.16.0.3. The next network (if you continue subnetting with that mask) would be 172.16.0.4/30.

Humper

vidalch wrote:

I understand the pre 12.0 ip subnet zero thing. What I am confused about is this ....

172.16.0.0 / 30 - What is the first pair of host addresses.

Sybex CCNP page 50 contradicts itself ...

the book says .1 and .2 are the first hosts,

I say .4 or .5 because 256-252 = 4 = increment and first net.

Either way, when taking the test if I am asked this question whats right ?

You are kinda on the right track. The 256-252 will give you the block size of the subnet. The first subnet always starts at zero and works its way up. In this example above the subnet id or number starts at 0 then 4,8,12,16..etc. The first subnet is 0, so the usable host addresses is 1 and 2 and then 3 is the broadcast. The next subnet id is 4, then 5&6 are the usable host addresses and 7 is the broadcast. When trying to determinne how many hosts there are in a subnet take the size of the subnet (4,8,16,32..etc addresses) and subtract 2. One for the subnet id and one for broadcast.

Example:

A /26 is 255.255.255.192... that is a 64 address block (256-192=64). Now to determine the amount of hosts ---- it is address block - 2, so 64-2=62 ACTUAL host addresses in a /26. The other 2 addresses (subnet id and broadcast addresses) are NOT usable!

I hope this helped!

vidalch

Heres the whole story ... the question is actually related to VLSM

I need ip blocks as follows 300, 200, 4, 2

zero is only used if ios >= 12.0

I can either figure out the VLSM ranges using Hosts or the prefix.

But, depending on the IOS and ip subnet zero, assignments will vary.

The real question is how do I answer the VLSM question on the test.

The only possible answer is: if the question mentions subnet zero net then Ill use subnet zero if not, I wont.

Thats the only approach I can use because every printed doc Ive seen is inconclusive.

Thanks Regards
Chris

lwwarner

I think you are making this harder than it is.

vidalch wrote:

zero is only used if ios >= 12.0

Not quite true. ip subnet-zero is on by default in IOS >= 12.0, but it can also be turned on manually in IOS < 12.0. See:
IP Addressing Services - Subnet Zero and the All-Ones Subnet

The real question is how do I answer the VLSM question on the test.

IMO, this is the kind of thing that is not worth worring about. If you know what you are doing and read the questions carefully the correct answer(s) will be apparent. There are some sample Cisco VLSM questions in the Prep Center:
CCNA Prep Center - Practice Questions

EdTheLad

Hi Vidalch,
On the exam you will be informed per question whether subnet-zero is enabled or disabled, so since you understand the difference you'll have no problem.

yaniv_sharma

vidalch wrote: »

I understand the pre 12.0 ip subnet zero thing. What I am confused about is this ....

172.16.0.0 / 30 - What is the first pair of host addresses.

Sybex CCNP page 50 contradicts itself ...

the book says .1 and .2 are the first hosts,

I say .4 or .5 because 256-252 = 4 = increment and first net.

Either way, when taking the test if I am asked this question whats right ?

Answer: you are right that the block size will be 4, but if we see /30 it means it has 2 bits only for host. that means.
2*2 - 2 = 2
so their are only two valid host can create in each subnet.
172.16.0.0 (1st subnet)

172.16.0.1 (Ist valid I.p) 172.16.0.2 (last valid I.P) so the broadcast Ip will be 172.16.0.3

172.16.0.4 (2nd subnet)