Computer Science - Networking - Discussion
Discussion Forum : Networking - Section 2 (Q.No. 38)
38.
On a class B network, how many hosts are available at each site with a subnet mask of 248?
Discussion:
11 comments Page 1 of 2.
Ravi kumar said:
1 decade ago
As subnet mask given 248(class B)it means
255.255.248.0
It's binary form
11111111.11111111.11111000.00000000
It means we have 5bit(from 3rd octet) used for subnet (2^5 = 32 subnet)&
Remaining 3bit + 8bit = 11bit.
(from 3rd octet & 4th octet respectively)are used for HOST bit.
Hence Total No.of HOST = 2^11-2 = 2046.
255.255.248.0
It's binary form
11111111.11111111.11111000.00000000
It means we have 5bit(from 3rd octet) used for subnet (2^5 = 32 subnet)&
Remaining 3bit + 8bit = 11bit.
(from 3rd octet & 4th octet respectively)are used for HOST bit.
Hence Total No.of HOST = 2^11-2 = 2046.
(1)
Wasim sajjad said:
4 years ago
First, we have to convert all decimal to binary, we know the last two-octet in class B is reserved for Host but their is also 248 it 3rd octet so when it is convert to binary then only three bits are remaining for host , and the final octet is fully dedicated so their full 8 bit is free so
(2^3) * (2^8) - 2 = 246.
(2^3) * (2^8) - 2 = 246.
(1)
Arshad said:
4 years ago
The last 2 octets in class B for host address and here we have subnet mask 248 which mean that we have borrowed 5 bits from the second octet of the host side (8-3=3) and the final octet of the host address is fully dedicated.
Sso their full 8 bit is free now its easy just do (8+3=11, mean 2^11-2 = 2048 -2 = 2046.
Sso their full 8 bit is free now its easy just do (8+3=11, mean 2^11-2 = 2048 -2 = 2046.
(1)
Vasundhara said:
1 decade ago
Subnet mask is 255.255.248.0
then 2^8*2^3-2=2046
minus 2 is done since 2 spaces are reserved on every id's
then 2^8*2^3-2=2046
minus 2 is done since 2 spaces are reserved on every id's
Vishal said:
1 decade ago
How it is calculated ?
Subnet mask is 255.255.248.0
Then 2^8*2^3-2=2046
Minus 2 is done since 2 spaces are reserved on every id's.
Subnet mask is 255.255.248.0
Then 2^8*2^3-2=2046
Minus 2 is done since 2 spaces are reserved on every id's.
Praveen said:
1 decade ago
Why we should take 2^8*2^3-2 ? Can anyone explain the logic?
Raman said:
9 years ago
Thanks @Ravi Kumar. Good explanation.
Jonathan said:
8 years ago
I could not understand what is meant by "2 spaces are reserved on every id's". Remaining explanation is very good.
Varsha said:
8 years ago
@jonathan.
I think one is for network and another one is for the broadcast purpose.
It's right @ Ravi Kumar.
I think one is for network and another one is for the broadcast purpose.
It's right @ Ravi Kumar.
SKR said:
7 years ago
No, in 3rd and 4th octet, one bit is used for broadcasting.
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