Explanation of the broadcast address of this network example

kctxau

Can someone explain how this is 31? I couldn't arrive at that number.
What is the broadcast address of the network 192.168.193.0/27?
Answer: 192.168.193.31
Chances are, once its explained, I'll say to myself "oh yeh, noooow I remember"
Thank-you, for your help.

gojericho0

Do you know how to come up with the block size for that subnet mask?

amp2030

192.168.193.0/27
24 < 27 < 32
32 - 27 = 5
2^5 = 32
0 32 64 ...
32 - 1 = 31
192.168.193.31

dublinguy

kctxau wrote: »

Can someone explain how this is 31? I couldn't arrive at that number.
What is the broadcast address of the network 192.168.193.0/27?
Answer: 192.168.193.31
Chances are, once its explained, I'll say to myself "oh yeh, noooow I remember"
Thank-you, for your help.

I will calculate with the following method, As mask is 27, 32-27=5 , 2 ^5 = 32, So the block size is 32, now come to the ip

192.168.193.0 , 32 is the number to add so next one should be
192.168.193.32
192.168.193.64

192.168.193.0 - 192.168.193.31 (this is the broadbast address)
192.168.193.1 - 192.168.193.30 (Is the valid host range)

hope this helps, more details in the below link

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

colink24

3 bits of the host portion so
128+64+32 = 224
256 - x = 224
x = 32
Network addresses 0 32 64 96 etc

kctxau

Thanks for the replys. I arrived the 224 and had forgotten from there.
So now I'll say to myself "oh yeh, noooow I remember"
Again, thanks.