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A subnet has been assigned a subnet mask of $255.255.255.192$. What is the maximum number of hosts that can belong to this subnet?

1. $14$
2. $30$
3. $62$
4. $126$

edited | 1.9k views

(C) is answer since you have $6$ zeroes so you can make $64-2$ hosts

by Active (2.5k points)
edited by
0
Here class is not given , how one could find the hostid

nid + hostID = All 1's

Without nid how we could be able to find the hostID .

Plz explain .
0
there is slight mistake. no. of 1's is equal to netid+subnetid
0

@ashwina here no need of class,

255.255.255.192 = 11111111.11111111.11111111.11000000

we have to find number of zero i.e 6  here,

maximum no. of hosts = 2^ (no. of bits in hid) - 2

= 2^6 - 2

= 64 - 2

= 62

0
its a classless ip address(if not mention explicitly in the question)

to calculate the network address(NID) if one of the host IP address belonging to that subnet is given,the one have to perform the bitwise '&' operation between  ip address and subnet mask

for eg: host ip address ==>192.168.1.0.134

&

------------------------

192.168.0.128

out of the 64 hosts ip address in the subnet ,one is given to network address(router inteface) and one is directed broadcast address in that network.

so the available host address are 64 - 2 = 62

maximum no. of hosts = 2(no. of bits in hid) - 2

= 26 - 2

= 64 - 2

= 62

by Junior (777 points) by Junior (513 points)
0
why any host cannot take base Id and broadcast Id??
+2

@Vishnathan your answer is the only one which explains why $2$ is being subtracted from $2^6$.

If we only write the last byte into binary, then we get,

255.255.255.11000000

∴ Number of 0's in subnet mask = Number of bits in host ID = 6

∴ Maximum number of hosts = 2^ (Number of bits in host Id) - 2 [since the 1st and last IP address is used for Network ID and Directed Broadcast Adddress]

= 2^6 - 2

= 64 - 2

= 62

Option (c) is the right answer

by Junior (507 points)