Computer Science - Networking - Discussion
Discussion Forum : Networking - Section 2 (Q.No. 33)
33.
You are working with a network that has the network ID 192.168.10.0. What subnet should you use that supports up to 25 hosts and a maximum number of subnets?
Discussion:
14 comments Page 1 of 2.
Kaa said:
5 years ago
The formula for finding the host is 2n (where n indicates the number of host bits)
We required 5 host bits (2n = 25) = 32
Now, let's find the subnet mask.
Hence, 5 bits are reserved for the host. There are 3 bits remaining for the network. Count the value of all the remaining three bits (27 + 26 + 25 = 224).
Therefore, the new subnet mask you will get is 255.255.255.224.
We required 5 host bits (2n = 25) = 32
Now, let's find the subnet mask.
Hence, 5 bits are reserved for the host. There are 3 bits remaining for the network. Count the value of all the remaining three bits (27 + 26 + 25 = 224).
Therefore, the new subnet mask you will get is 255.255.255.224.
Patty said:
3 years ago
B is correct.
2^n-2> = number of host.
2^n-2> = 25 then what number exponent to 2 and near reach of 25 or above
2^5-2> = 25 were 5 match is remaining host bits on host portion, then the borrowed bits from host portion are 3bits...then 3 bits match to 224 in binary ..11100000 last 0 take 0, second 0 from 1, p to n 0 then take 2^5+2^6+2^7 = 224.
2^n-2> = number of host.
2^n-2> = 25 then what number exponent to 2 and near reach of 25 or above
2^5-2> = 25 were 5 match is remaining host bits on host portion, then the borrowed bits from host portion are 3bits...then 3 bits match to 224 in binary ..11100000 last 0 take 0, second 0 from 1, p to n 0 then take 2^5+2^6+2^7 = 224.
Sibasish said:
9 years ago
We required 25 subnets.
2^4<100<2^5 OR 16<100<32.
no of subnets are 2^5 = 32 (no of 1's are 5).
Since default subnet mask of class c is 255.255.255.0.
i.e. 11111111.11111111.11111111.00000000 so 5 more 1 will get add so it will be;
11111111.11111111.11111111.11111000 i.e. 255.255.255.248.
2^4<100<2^5 OR 16<100<32.
no of subnets are 2^5 = 32 (no of 1's are 5).
Since default subnet mask of class c is 255.255.255.0.
i.e. 11111111.11111111.11111111.00000000 so 5 more 1 will get add so it will be;
11111111.11111111.11111111.11111000 i.e. 255.255.255.248.
Prabhash Singh said:
1 decade ago
For 25 hosts you need minimum 5 bit i.e 2^5 = 32.
Therefore, the 5 bit of the last octet in ip address would be reserved for host and the remaining would act as subnet.
So, in bit wise it would be 11111111.11111111.11111111.111*****
Which is 255.255.255.224
Therefore, the 5 bit of the last octet in ip address would be reserved for host and the remaining would act as subnet.
So, in bit wise it would be 11111111.11111111.11111111.111*****
Which is 255.255.255.224
Praveen said:
1 decade ago
How it is possible? What is the logic behind this? Can anyone explain this? This kind of questions are frequently asking in gate.
Manik said:
9 years ago
Same doubt as Piyush, please explain us @Prabhash Singh.
Tharindu said:
7 years ago
B is correct.
It is about subnet @Sibasish.
It is about subnet @Sibasish.
Gee said:
7 years ago
Please, anyone,explain the correct answer.
Piyush said:
1 decade ago
How is that three 1's coming in the end?
Abhishek said:
9 years ago
Nice explanation, Thank you all.
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