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A subnetted Class B network has the following broadcast address : 144.16.95.255. Its subnet mask

(a) is necessarily 255.255.224.0

(b) is necessarily 255.255.240.0

(c) is necessarily 255.255.248.0

(d) could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0


My attempt :

For class B network broadcast address first two octet or 16 bits should be all are 1 for preserving class B network address (255.255.0.0) , now given address 144.16.95.255 is same as 144.16.010 11111.1111 1111 , clearly last 13 bits all are 1 contiguously , it shows that last 13 bits for hosts addresses , in third octet first 3 bits should be for subnet address . It should be preserve in broadcast address . So that 3 bits should all are 1 in subnet mask . therefore , resultant subnet mask will be 1111 1111. 1111 1111. 111 00000. 0000 0000 = 255.255.224.0

Hence , option (a) is true as per my calculation.

Can you explain it in a formal way, please?

  • 1
    You need to actually use the mask against the addresses. Convert both the address and the mask to binary, perform a logical AND to get the subnet. You can perform a logical NOT on the mask to get the inverse mask. Add the inverse mask to the subnet to get the broadcast address. Convert back to decimal to see the broadcast address. – Ron Maupin Feb 10 '16 at 15:14
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The answer is (d).

The explanation:

When netmask is 255.255.224.0 we have 5 bits of third octect for host, then the networks are:

  • 144.16.0.0
  • 144.16.32.0
  • 144.16.64.0

this last one goes from 144.16.64.0 to 144.16.95.255


When netmask is 255.255.240.0 we have 4 bits of third octect for host, then the networks are:

  • 144.16.0.0
  • 144.16.16.0
  • 144.16.32.0
  • 144.16.48.0

And so on until we got to 144.16.80.0 that goes to 144.16.95.255


When netmask is 255.255.248.0 we have 3 bits of third octect for host, then the networks are:

  • 144.16.0.0
  • 144.16.8.0
  • 144.16.16.0
  • 144.16.24.0

And so on until we got to 144.16.88.0 that goes to 144.16.95.255

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