Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 5)
5.
You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Answer: Option
Explanation:
You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts-this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts. This is the best answer.
Discussion:
26 comments Page 2 of 3.
JaganPJ said:
8 years ago
@Roopa
Why the best value for h is 5?
Isn't it would be 9 because h=18/9.
Why the best value for h is 5?
Isn't it would be 9 because h=18/9.
Shahnawaz said:
9 years ago
Thank you @Peeyush.
Gopi Kanaparthi said:
9 years ago
Class c default subnet mask \24 like 255.255.255.0
Example:
Converting bit 1, n=2^1=2.
Converting bit 2, n=2^2=4.
Converting bit 3, n=2^3=8.
So first converting bit 1 is two subnets, second one is 4 subnets and third one is 8
Subnets support. We want 5 subnets so by using 3 converting bits.
Default network bits+converting bits, N+n = 24+3 = 27.
By using subnetmask 255.255.255.224
Example:
Converting bit 1, n=2^1=2.
Converting bit 2, n=2^2=4.
Converting bit 3, n=2^3=8.
So first converting bit 1 is two subnets, second one is 4 subnets and third one is 8
Subnets support. We want 5 subnets so by using 3 converting bits.
Default network bits+converting bits, N+n = 24+3 = 27.
By using subnetmask 255.255.255.224
Jaffar shaikh said:
1 decade ago
How 255.255.255.224 provides 8 subnets ,each with 30 hosts?
please tell me answer.
please tell me answer.
Angel said:
10 years ago
Suppose you are given the following class C IP address: 192.100.1.1. You need to perform Subnetting in such a way that you get 3(three) subnets such that each subnet contains 62 (sixty two) valid hosts. You need to specify the network and broadcast address of each subnet and the valid host range in each of the subnet.
solve it and give answer
solve it and give answer
Peeyush said:
10 years ago
As we require 16 Host in each subnet. We will first check Binary Table.
128 64 32 16 8 4 2 1.
If we take 16 we will not get 16 usable IP we will only get 14 usable IP. 2 ID will be taken for Subnet id and Broadcast ID.
So will take 32.
Now if we check mask it will be created like below:
11111111.11111111.11111111.11100000
By converting this into decimal it will then become: 255.255.255.224
This is the right answer.
128 64 32 16 8 4 2 1.
If we take 16 we will not get 16 usable IP we will only get 14 usable IP. 2 ID will be taken for Subnet id and Broadcast ID.
So will take 32.
Now if we check mask it will be created like below:
11111111.11111111.11111111.11100000
By converting this into decimal it will then become: 255.255.255.224
This is the right answer.
Zewdu said:
10 years ago
Matching and work out equation place including.
Mohamed said:
10 years ago
@Vinay and @Hagos explained well Thanks for your valuable inputs, I keep expect from your side.
Mack said:
10 years ago
Please tell me if I'm wrong, but another way to think of it is:
128 64 32 16 8 4 2 1.
1 1 1 0 0 0 0 0.
32 is the smallest bit used so.
32 - 1 (domain) - 1 (server) = 30.
Honestly, I think I remember seeing it this way in a tutorial, so if I'm wrong please correct and/or email me.
128 64 32 16 8 4 2 1.
1 1 1 0 0 0 0 0.
32 is the smallest bit used so.
32 - 1 (domain) - 1 (server) = 30.
Honestly, I think I remember seeing it this way in a tutorial, so if I'm wrong please correct and/or email me.
Pratik said:
1 decade ago
224 = 11100000 in binary.
So (2^5)-2 = 32-2 = 30.
So 30 hosts.
So (2^5)-2 = 32-2 = 30.
So 30 hosts.
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