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A network has subnet mask $255.255.255.254$. Determine the maximum or number of Host
in this network. Also determine the broadcast address of this network.
edited | 261 views
0
Technically,

$255.255.255.254\Rightarrow 11111111.11111111.11111111.11111110$
There are only $1$ host bits.
Hence, there are $2^1 -1= 0$ hosts are possible
and the DBA will be $255.255.255.255$
+1
This question seems ambiguous,

Can i have the IP address
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IP address is not provided in the Question.
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@Sukanya Das this is not an ambiguous question.

255.255.255.254⇒11111111.11111111.11111111.11111110

Above 31-Bit prefix address

Directed broadcast is disabled automatically on /31 links. -Ref:https://learningnetwork.cisco.com/message/239554#239470
Using 31-Bit prefixes, the host-id of 0 represents one host(Also Subnet), and a host-id of 1 represents the other host of a point-to-point link.
So we have 2 hosts here

You could refer to this RFC for more info on 31-bit prefixes (/31 links)
https://tools.ietf.org/html/rfc3021

answered by Active (1.1k points)
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Yes, this is an exception
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Really informative answer. Thank you.

255.255.255.254⇒11111111.11111111.11111111.11111110

Above 31-Bit prefix address
number of one's  in network mask represent the total no bit used in the network id

means 31 bit are used for network id and 1 bit are used for host id . so  maximum no of host is 21=2.

answered by (277 points)
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2^2-2=0 hosts are possible
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when they ask about maximum host do't subtract 2 .and when they ask  about total host subtract 2
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source of the above line????

As per given subnet mask = > 31 bits already used we left with 1 bit.

so  #no of IP Possible with 1 bit => 21  = 2  but practically we can configure 21 - 2 Hosts only .

Final Answer => 0 Host and   255.255.255.255 is broadcast address.

answered by (109 points)

+1 vote