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What is the subnet number of host 172.16.14.188 /28

I usuallly get through any question from subnettingquestions.com but this Seems a little tricky for me

21 comments
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26 points · 5 days ago

The best way I found to learn this particular type of question is to use notepad and break it all out in binary. For instance :

172.16.14.188 172.16.14.1011|1100

I put pipe symbol where the network and host portions meet. Then I can make all the host zeroes and recalculate.

172.16.14.1011|0000 172.16.14.176

172.16.14.1011|1111 172.16.14.191

So you end up with a range of

172.16.14.176 - 191 . 176 is network,. 191 is broadcast Usable hosts of .177 through .190

Awesome breakdown!

[deleted]
6 points · 5 days ago

The subnet ID is : 172.16.14.176 Number of host is: 232-28 -2 = 14

Original Poster2 points · 5 days ago

Giving that block size (14) how can you find the subnet id?

The block size is 16 (256-240). 176 is the highest number that divides evenly by 16 without going over 188.

Original Poster1 point · 5 days ago

Aah okey thank you!! My fault was a bad block size calculation

in my opinion, for clarity, usable hosts calculation (block size) should come after magic number calculation (256-subnet mask) and then you use the magic number (in this case 16) and each subnet ID is a multiple of the magic number. determine the network, then determine the number of usable hosts.

[deleted]
1 point · 5 days ago

Block size is 14 Add 2 it’s 16

What’s the highest multiple of 16 that doesn’t surpass 188? (Because the magic number 16(256-240) tells us the network is divided by blocks of 16 this way 0—16—32—48—64—80—96.......—176–Host—

Learning your power of 2s makes subnetting a lot easier. Power of 2s = block sizes. Only so many. Learning to 20 - 216 is good enough.

CCNA
3 points · 4 days ago · edited 4 days ago

/25 - 128

/26 - 64

/27 - 32

/28 -16 is the magic number

so increments of 16, ultimately 16x11 = 176

176 + 16 = 192

so your range starts at 176 and ends 192 with 192 being the next network in the 172.16.14.x range with 188 in the middle.

PracticalNetworking.net
2 points · 5 days ago

This video shows you how to solve for all the Subnetting Attributes: https://www.youtube.com/watch?v=5-wlfAdcmFQ&t=44s

The one you are concerned with is the Network ID (Which you have referred to as a Subnet Number)

This video will explain what the Network ID (Subnet Number) is: https://www.youtube.com/watch?v=BWZ-MHIhqjM

Original Poster1 point · 4 days ago

Thank you!

by the way, what seemed tricky about this?

because 172.16.14.x lands in the Class B range but /28 lands in the 4th octect range?

Original Poster1 point · 4 days ago

I didn't calculate properly the block size, so everything was bad from the begining

I'm actually interested in this too, isn't the /28 the number of available ips at 16?

Avalable?

No, thats 14.

You're right

Nevermind thought you were assigning ips for a second.

It is /28 255.255.255.240

9 points · 5 days ago · edited 5 days ago

256-240 = 16

14.0

14.16

14.32

14.48

So on...

14.160

14.176

14.176 (network id)

14.177 to 14.190 (useable host range)

14.191 (subnet broadcast address)

14.192

CCNA
2 points · 4 days ago

not sure why your so low in the replies but this is a good breakdown that i was too lazy to write out!

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