Subnetting: Accommodate more than 225 hosts

Hi,

I have a subnetting question. First I’ll give an example:

An organization has the IP 200.1.1.0 and needs 4 subnets with the following characteristics:

Subnet A =>72 hosts

Subnet B => 35 hosts

Subnet C => 20 hosts

Subnet D => 18 hosts

Provide the IPs per subnet.

Solution

Subnet A =72 host

For 72 hosts we need 2^7 - 2 = 128 - 2 = 126 host. We need 7 bits for the hosts.

Network => 200.1.1.0/25

Mask => 255.255.255.128

First Host => 200.1.1.1/25

Last Host => 200.1.1.126/25

Broadcast => 200.1.1.127/25

Subnet B =35 host

For 72 hosts we need 2^6 - 2 = 64 - 2 = 62 host. We need 6 bits for the hosts.

Network => 200.1.1.128/25

Mask => 255.255.255.192

First Host => 200.1.1.129/26

Last Host => 200.1.1.190/26

Broadcast => 200.1.1.191/26

Subnet C =72 host

For 72 hosts we need 2^5 - 2 = 32 - 2 = 30 host. We need 5 bits for the hosts.

Network => 200.1.1.192/27

Mask => 255.255.255.224

First Host => 200.1.1.193/27

Last Host => 200.1.1.222/27

Broadcast => 200.1.1.223/27

Subnet D =72 host

For 72 hosts we need 2^5 - 2 = 32 - 2 = 30 host. We need 5 bits for the hosts.

Network => 200.1.1.224/27

Mask => 255.255.255.224

First Host => 200.1.1.225/27

Last Host => 200.1.1.254/27

Broadcast => 200.1.1.255/27

So, according the above solution, how do we create subnets for the IP 155.52.48.0/23 with the following characteristics (provide Network, Mask First Host, Last Host and Broadcast IPs foe each subnet)?

Subnet A =>200 hosts

Subnet B => 100 hosts

Subnet C => 40 hosts

Subnet D => 40 hosts

I’m really stuck with that.

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

bornwith

[TH]Subnet Name[/TH]
[TH]Needed Size[/TH]
[TH]Allocated Size[/TH]
[TH]Address[/TH]
[TH]Mask[/TH]
[TH]Dec Mask[/TH]
[TH]Assignable Range[/TH]
[TH]Broadcast[/TH]

A
200
254
155.52.48.0
/24
255.255.255.0
155.52.48.1 - 155.52.48.254
155.52.48.255

B
100
126
155.52.49.0
/25
255.255.255.128
155.52.49.1 - 155.52.49.126
155.52.49.127

C
40
62
155.52.49.128
/26
255.255.255.192
155.52.49.129 - 155.52.49.190
155.52.49.191

D
40
62
155.52.49.192
/26
255.255.255.192
155.52.49.193 - 155.52.49.254
155.52.49.255

anarchos78

Thank you very much about your reply.
What I don't understand is the "Network Address" part.
Is it valid from address 155.52.48.xxx to go to 155.52.49.xxx?
Could you please explain me the "mechanics" of that approach?

rob42

I don't know if it'll help you, but I'm working on tying to explain the "mechanics" of Subnets and have just put together this URL="http://rob42.net/IPv4_Subnets1.htm"]Rob42.net - IPv4 Subnets[/URL. I'd appreciate any feedback on my work.

It's a work in progress...

volfkhat

anarchos78 wrote: »

Thank you very much about your reply.
What I don't understand is the "Network Address" part.
Is it valid from address 155.52.48.xxx to go to 155.52.49.xxx?
Could you please explain me the "mechanics" of that approach?

To answer your question: YES.

It's all about understanding how to identify the "octet of interest".
Once you know the octet, everything else falls into place.

155.52.48.0/23 = 155.52.48.0 - 155.52.49.255
its not much to explain; you just have to practice subnetting /16 - /23.

I learned subnetting from Danscourses:
https://www.youtube.com/playlist?list=PL33E07ECCA73C0755
Watch video# 47.
(after you "get" it; give him a Thumbs-UP)

As for the rest of your Q:
200 hosts = a blocksize 256.
aka: 2^8
aka: /24
This means it falls under the 4th octet.
so your 200 hosts would fit on the subnet: 155.52.48.[0-255]

Your next subnet starts at 155.52.49.0
(because there is no more room inside 155.52.48.x ).

B = 155.52.49.[0-127]
C = 155.52.49.[128-191]
D = 155.52.49.[192-255]

just takes practice..

anarchos78

Thank you very much!