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Given an IP address 156.233.42.56 with a subnet mask of 7 bits. How many hosts and Subnets are Possible ?

1. 126 hosts and 510 subnet
2. 128 host and 512 subnet
3. 510 hosts and 126 subnet
4. 512 hosts and 128 Subnet
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@Na462 is there any info about the NID part? because if subnet mask is 7 bits that means NID +SID part=7bits so rest of the 32-7 bits are there for the hosts .. i.e 25 bits for the host ..

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That's exactly was my doubt i guess something is indeed wrong with the question :)
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Option C..???
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Is it correct?
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Yes it's correct actually they considered as 7 extra bits on netmask as its class B so now we have 16+7 = 23. I didn't understand that how number of subnets are 126 it should be 128 and host then should be 510.
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it is class B ip so so 16 NID bits and 16 HID bits and 7 bits are given as subnet bits so total no. of subnets=$2^7-2=126$ and now remaining are hid bits so total hosts are $2^9-2=510$ so option B is right.
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@Na462

Given ip address corresponds to class B so 16 bit are for net id and for rest 16 bit given 7 bit corresponds to subnet Id and 9 bit to host id part.

=126

Similarly for host=2^9-2=510 host for each subnet

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Why you guys are subtracting 2 from total subnet 2^7 ??

Its used in host becasue when we divide a network into subnets within a subnet we cant use the first and last ip address as it would be the netid and broadcast id. If u subtract 2^7 - 2 then u are actually wasting two subnets ( Every subnet has 510 usabke address)
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https://gateoverflow.in/1265/gate2007-67-isro2016-72

As per previous standard we used to subtract 2 addresses but as per new standard 2^n subnet are there for n bit subnet.

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@Soumya :)

Read the comment very above by Bikram sir, The question was ambiguous and this convention is not standard we will not subtract 2 from subnets according to RFC,IEEE :)
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We will follow the RFC,IEEE standard always so 2^n will be used.
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@Na462 yes we should follow 2^n for subnetting but Gate is following 2^n-1 you can reffer previous year question also .

+1 vote
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+1 vote