Computer Science - Networking - Discussion
Discussion Forum : Networking - Section 5 (Q.No. 35)
35.
On a class B network, how many subnets are available with a subnet mask of 248?
Discussion:
7 comments Page 1 of 1.
Sonia said:
1 decade ago
Here the subnet mask is 248.0 because of class B network.
It's binary 11111000.00000000
5 bits for network
2^5 = 32 - 2(network id & broadcast id) = 30
Thus, 30 subnets are available.
It's binary 11111000.00000000
5 bits for network
2^5 = 32 - 2(network id & broadcast id) = 30
Thus, 30 subnets are available.
Abcd said:
1 decade ago
Network id and broadcast id belongs to host id. So, 32 is the right answer.
Shewangizaw Bogale said:
9 years ago
Network id and broadcast id belongs to a given subnet , subnet mask of 248 = 11111000 , N= 5 , Total subnets = 2^N , 2^5 = 32.
So the final and correct answer is Total subnets = 2^N, 2^5 = 32.
So the final and correct answer is Total subnets = 2^N, 2^5 = 32.
Iglooroy said:
9 years ago
Yes, the answer is 32.
Rubysharma said:
7 years ago
Subnet mask 248.
Block size=256-248=8.
Number of subnet formed=subnet mask/blocksize - 1.
= (248/8)-1 = 30.
Block size=256-248=8.
Number of subnet formed=subnet mask/blocksize - 1.
= (248/8)-1 = 30.
(2)
Sanjay rasgon said:
5 years ago
The right answer is 32.
(1)
Wasim sajjad said:
4 years ago
3rd octet so given subnet Mask is 248 while 256 is full bits so final dedicated bits like 2^0+2^1 = 3 is subtracted from 248.
then 248-3 = 5;
So 5 subnets masks are available and we know;
subnet = 2^n here the value of n is 5,
So 2^5=32.
then 248-3 = 5;
So 5 subnets masks are available and we know;
subnet = 2^n here the value of n is 5,
So 2^5=32.
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