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In a class B network subnet mask for $200$ subnets each with $220$ systems will be

1. $255.255.255.0$
2. $255.255.0.0$
3. $255.255.224.0$
4. $255.0.0.0$

Given it is a class B network, so the subnet mask will be something like 255. 255. _ . _

The question says that we need 220 systems in each subnet.

Using 8 bits, we can have $2^8 - 2 = 254$ systems(hosts). While with 7 bits we can have only $2^7 - 2 = 126$ systems.

So, we need to have atleast 8 bits for host address in each subnet. This means that the last octet of IP address should be 0. Our answer now will be something like: 255. 255. _ . 0

Given that we need 200 subnets. Using 7 bits we can only have $2^7 = 128$ subnets, so we need atleast 8 bits for subnet $(2^8 = 256 > 200)$.

This means that atleast 8 bits are required for subnet addressing.

Therefore subnet address should be 255. 255. 255. 0

(Last 8bits are used for host addresses).

Rishabh Gupta 2 The calculation should be made first in accordance with subnets and then for systems ! You didn`t mention the exact same calculation we need for subnets as well.

@saxena0612 I edited the answer. Is it correct now?

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