# Subnetting class b, network address 172.23.0.0/27

jmrseadogg
Posts:

**4**Member ■□□□□□□□□□
Subnetting is not my greatest strength

I need to subnet 172.23.0.0/27 the subnet mask is 255.255.255.224, I can not figure out what is the first and what is the last usable subnet.

The questions I have are: what is the last usable first subnet? What is the last usable second subnet? What is the last usable third, forth and fifth subnets? What are first usable subnets for the first, second, third, fourth and fifth?

can i get some help on this?

I need to subnet 172.23.0.0/27 the subnet mask is 255.255.255.224, I can not figure out what is the first and what is the last usable subnet.

The questions I have are: what is the last usable first subnet? What is the last usable second subnet? What is the last usable third, forth and fifth subnets? What are first usable subnets for the first, second, third, fourth and fifth?

can i get some help on this?

0

#### Welcome!

#### Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

## Quick Links

#### Categories

- 108.4K All Categories
- 34 Welcome Center
- 7 Announcements
- 2 Forum Rules of Engagement
- 10 Introduce Yourself
- 15 TechExams Support
- 86.5K Certification Preparation
- 128 Check Point: CCSA & CCSE
- 32.6K Cisco
- 21.5K CCNA & CCENT
- 364 CCDA & CCDP
- 8.9K CCNP
- 1.7K CCIE
- 361 Cloud Certifications
- 133 Amazon Web Services (AWS)
- 6 Azure
- 85 CCSP
- 73 Cloud+ & Cloud Essentials
- 15.2K CompTIA
- 5K A+
- 208 CASP
- 74 CySA+
- 958 Linux+
- 4.4K Network+
- 14 PenTest+
- 257 Project+
- 3.8K Security+
- 327 Server+
- 247 Other CompTIA Certifications
- 1.2K EC-Council
- 580 CHFI
- 581 CEH
- 1.2K GIAC
- 19 IAPP
- 895 ISACA
- 241 CISA
- 652 CISM
- 4.1K (ISC)²
- 496 CISSP
- 3.6K SSCP
- 914 Juniper
- 18.4K Microsoft
- 104 MCSA 2016 / MCSE 2016
- 148 Windows 10 exams
- 139 Windows 8 exams
- 1.4K Windows 7 exams
- 1.6K MCSA / MCSE on Windows 2012 General
- 2.3K MCTS / MCITP on Windows 2008 General
- 828 Exchange Server & Office Communications Server Exams
- 528 Other Microsoft Electives
- 349 MCSA/MCSE: Security
- 299 Microsoft Developers Certifications
- 482 SQL Server exams
- 129 Offensive Security: OSCP & OSCE
- 2K Other Security Certifications
- 2.8K Other Certifications
- 384 CWNP Certifications
- 698 LPI, RHCE, and SAIR
- 845 ITIL Certifications
- 207 Project Management Certifications
- 84 Apple Mac OS X Certifications
- 62 Novell Certification
- 79 Oracle Certifications
- 52 Sun Microsystems Java Certification
- 230 Citrix Certifications
- 111 Storage Certifications
- 6.6K General Certification
- 17.2K Education & Development
- 5 Colleges & Schools
- 7 Educational Resources
- 17.2K IT Jobs / Degrees
- 5 Professional Development
- 7 Cybersecurity
- Cloud Security & IoT
- 5 Security Awareness & Training
- 2 Security News & Breaches
- 19.8K General
- 7 Computer Gaming
- 1 Data Center
- 3 Classifieds
- 2 For Sale
- Wanted (ISO)
- 1 Help Wanted
- 2 Just for Fun
- 17.3K Off-Topic
- Scripting
- Show Us Your Tech!
- Troubleshooting
- 2.5K Virtualization

## Comments

3,353MemberThis means that, we'll use the 1st 3 bits when figuring out our IP range we can use.

172.23.0.xxx000000

Our starting subnet range would be:

172.23.0.00100000

Our ending subnet range would be:

172.23.0.11100000

So 172.23.0.32 would be the starting range while 172.23.0.224 would be the ending range.

The reason is because since we're working with only the 1st 3 octets, 001 would be a 1 which is the smallest # in binary, and all 1s would be the largest # we can have. To find out the 2nd subnet, we'd just do 010 since binary goes from 128 64 32 16 8

4 2 1. 3rd subnet would be 011, 4th would be 100, and so on.Does that help?

4Member ■□□□□□□□□□first usable subnet would be 001 32

second usable subnet would be 010 64

third usable subnet would be 011 96

fourth usable subnet would be 100 128

fifth usable subnet would be 110 192

would this be correct?

what woudl the last first subneet be then? Do you work backwords

4Member ■□□□□□□□□□So what is the last first usable Subnet, how do I reach that answer if the first usable first subnet is 32?

I am understanding this more, thanks for you time. I think I am understanding how your first the first usable, but not the last usable.

3,353MemberIt's when doing client IPs you do 2n = 2.

So all your subnets would be:

000 = 1st subnet

001 = 2nd subnet

010 = 3rd subnet

011 = 4th subnet

100 = 5th subnet

101 = 6th subnet

110 = 7th subnet

111 = 8th subnet

Man it's been a long time since I've subnetted, you're making me think!

4Member ■□□□□□□□□□So 8 subnets

So when I am asked,

last usable subnet of the first subnet, what does that mean?

3,353MemberThat doesn't make sense. All those questions ask something such as, last Client IP Address of the 1st subnet, etc...

230Membernetwork 0

1st useable = .1

broadcast address = .31

last useable = .30

network 32

1st useable = .33

broadcast = .63

last useable .62

2Member ■□□□□□□□□□So I'd agree with Royal that in practice you get the 8 subnets total - but in exam format the "usable subnets" is actually 8-2 = 6. This can be very mis-leading and I have been trying to find verification on what the official line in the CCNA is, but one of the practice exams I did definitely had the concept of "usable" subnets.

2Member ■□□□□□□□□□http://www.mcmcse.com/articles/subnetting.shtml

995MemberDetermining Subnets:Us the equation 2^x = s where x is the number of borrowed bits (the 1's) and s is the number of subnets you will have.

Let's look at it like this: 172.23.0.11100000

Since there are three 1's in the last octect; x = 3.

So, 2^3 = 8 total subnets

Determining Hosts per Subnet:To determine the number of hosts per subnet use the equation 2^n - 2 = y. N in this case is the number of unmasked bits (the 0's that follow the 1's in the last octect). So n in this case is 5.

So (2^5) - 2 = 30 usable hosts per subnet.

Example:

IP Address: 172.23.0.0 /24

Step 1: 2^3 = 8

Step 2: (2^5) - 2 = 30

So we will have 8 subnets which will have 30 usable IP addresses each.

The subnets are:

172.23.0.0

172.23.0.32

172.23.0.64

172.23.0.96

172.23.0.128

172.23.0.160

172.23.0.192

172.23.0.224

The IP's associated with your first network are:

Network Address: 172.23.0.0 /27

Usbale IP Range: 172.23.0.1 /27 to 172.23.0.30 /27

Broadcast IP: 172.23.0.31 /27

1Member ■□□□□□□□□□I am studying for the CCNA esxam 640-802 and 640-816: I attended Cisco lessons but nobody told me how to subnet a class b network.

I have no problem to subnet a class C but I really am not able to subnet a class C.

I have this network

172.16.0.0

255.255.128.0/17

I know there are 2 subnets, and 32,766 host

I do not know how to calculate the valid host range for each subnet

Is there anybody who can help me?

thank you

Francesco Ruosi

Naples (Italy)