I usuallly get through any question from subnettingquestions.com but this Seems a little tricky for me
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 :
I put pipe symbol where the network and host portions meet. Then I can make all the host zeroes and recalculate.
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
The subnet ID is : 172.16.14.176 Number of host is: 232-28 -2 = 14
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.
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.
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.
/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.
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
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?
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?
No, thats 14.
Nevermind thought you were assigning ips for a second.
It is /28 255.255.255.240
256-240 = 16
14.176 (network id)
14.177 to 14.190 (useable host range)
14.191 (subnet broadcast address)
not sure why your so low in the replies but this is a good breakdown that i was too lazy to write out!