Computer Science - Networking - Discussion

Given condition is for 8 subnets and a maximum number of hosts. so 2^3 = 8 subnets. This implies 3 bits in subnet ID part. so subnet mask is 224. Max number of hosts are 2^13-2= 1024*8-2=8192-2=8190.

So, the correct answer is 255.255.224.0.

Now if we consider 2^4=16 subnets because usable subnets are 2^4-2=14 subnets > 8 subnets then a number of hosts would be 2^12-2=4096-2=4094 hosts. Here the condition of subnets is satisfied but the condition of max hosts is not satisfied.

So, 255.255.240.0 is not the correct choice.

It is a class B address.

We need 8 subnets i.e., 8=2^3.
So we need to borrow 3 bits from the 3rd octet. That is 11100000=224.
Hence our mask is 255.255.224.0.
B is the correct answer.

@Ajay.

2^4=16 not 8, and the answer is B.

The Number of Subnet is = 8, 2^N = 8, N = 3, So We need 3 borrowed network Bits only. 11100000 ----> 224.

So, the correct & final answer is : [B]. 255.255.224.0.

Number of Subnet = 8.
So, We need 4Bit.

Then, the answer is 255.255.224.0.

8 = 1001 (4 digits are need).

So the answer is 11110000 = 240.

Answer is B, because, we want 8 subnets, 2^3=8.

Its a B class address.

8 subnet means 2 ^4 >=8.

So remaining 4 bytes are vacant for subnet.

The answer is 192+64+32+16+0+0+0+0 = 240.

255.255.240.0.