Stuck on CIDR

random ip : 191.253.169.201 /22

1011111.11111101.10101001.11001001

now to find subnet mask....

/22

22/8=2 with 6 remainder

255.255.???.0

I know what it is only because i looked . . (255.255.252.0)
but. How do i get to that 252?

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Comments

jahaziel

128 + 64 + 32 + 16 + 8 + 4 = 252 since you have 6 bits left you can do backwards math. Start at 128 then divide by 2 you get 64. Keep going till you have 6 bits taken. Then add all of it together to get your answer.

Best way, I can explain it.

{davros}

so

128 is 1 bit 64 is 2nd bit etc. . 4 is the 6th bit, add them up 252.

say it was a remainder of 3 that would be

128+64+32= 224 lol wrong.

Jon_Cisco

well the /22 is the shorthand. To understand it you need to follow the binary.

N for network
H for host

NNNNNNNN.NNNNNNNN.NNNNNNHH.HHHHHHHH

so you have 22 network bits hence the /22

Now convert that to binary mask. 1s for network 0s for host

11111111.11111111.11111100.00000000

binary numbers
128/64/32/16/8/4/2/1 add them up and you get 255 for your octet.

so the first two are 255.255.?.?

next octet does not have the 2/1 because they are 0s so 255-3=252 or (128+64+32+16+8+4=252)

last octet is all 0s so the final subnet mask is 255.255.252.0

Keep in mind this is a long explanation just to describe the parts. It's pretty simple with a little practice but you need to remember your binary math.
try this site for practice

Subnetting Practice Questions

{davros}

/22

22/8=2r6 8-6=2 2^2=4 256-4=252

johnpfusco

The guy in the video below in my opinion explains CIDR better than anyone else. Once you think you got it use the practice site below to test your knowledge. Hope this helps.

Joshua Butcher CIDR video

https://www.youtube.com/channel/UC26h2eo1F4VwLpN1pjrl9Ew

Practice site:
Subnetting Quiz #2 (CIDR)