Explanation of the broadcast address of this network example

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• gojericho0 Member Posts: 1,059 ■■■□□□□□□□

  March 2009

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

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• amp2030 Member Posts: 253

  March 2009

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

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• dublinguy Member Posts: 22 ■□□□□□□□□□

  March 2009

  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

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• colink24 Member Posts: 43 ■■□□□□□□□□

  March 2009

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

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• kctxau Member Posts: 130

  March 2009

  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.

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