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?
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?
Comments
Best way, I can explain it.
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.
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
22/8=2r6 8-6=2 2^2=4 256-4=252
Joshua Butcher CIDR video
https://www.youtube.com/channel/UC26h2eo1F4VwLpN1pjrl9Ew
Practice site:
Subnetting Quiz #2 (CIDR)