What is the subnet address of 172.16.159.159/22?
A.
172.16.0.0
B.
172.16.128.0
C.
172.16.156.0
D.
172.16.159.0
E.
172.16.159.128
F.
172.16.192.0
Explanation:
Converting to binary format it comes to 11111111.11111111.11111100.00000000 or
255.255.252.0
Starting with 172.16.0.0 and having increment of 4 we get.
4×39=156
0
0
/32 1/256
and so on
/26 1/4
/25 1/2
/24 1
/23 2
/22 4
/21 8
/20 16
and so on.
0
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can you please give more details i am little confused of getting the subnet address?
thanks
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wow guys its simple the subnet address is 255.255.252.0
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Know your power of 2’s
128, 64, 32, 16, 8, 4, 2, 1
128,192,224,240,248,252,254,255 << subnet purposes only and subnet can only be these numbers. Each one of these numbers is a 1 bit. Counting left to right till you get to the CIDR is on way of doing it. Remember every 8 bits is written down as a 255
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Simple metod:
172.16. 159. 159
255.255.252. 0
——————
172.16. 4(256-252).0
where is 255, repeat number above; where is 0, repeat the 0. Other number (in this case, 252, do: 256-252 = 4 ..multiple of 4 next 159) is 156!
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