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.
Zewdu said:
10 years ago
Matching and work out equation place including.
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.
Fariha 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. Please solve it as early as possible.
You need to specify the network and broadcast address of each subnet and the valid host range in each of the subnet. Please solve it as early as possible.
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
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
S .HEMANATHAN said:
9 years ago
class c
192.168.2.0
Subnets = 5.
Hosts = 16.
128 64 32 16 8 4 2 1
1 1 1 0 0 0 0 0.
and
2^5 - 2=30 host
2^3 = 8 subnets.
So,
255.255.255.224.
192.168.2.0
Subnets = 5.
Hosts = 16.
128 64 32 16 8 4 2 1
1 1 1 0 0 0 0 0.
and
2^5 - 2=30 host
2^3 = 8 subnets.
So,
255.255.255.224.
(4)
Siri said:
9 years ago
Prove if a host on a network has the address 172.16.45.14/30, the broadcast address of the subnet is 172.16.45.15.
(1)
Shahnawaz said:
9 years ago
Thank you @Peeyush.
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.
Chaithanya katari said:
8 years ago
225.255.255.192=/26 so 24+2(2N we need to borrow) formula 2^2=4 subnets.
for hosts 256-192=64 hosts.
255.255.255.224=/27 so 24+3(3n we need to borrow)formula 2^n=2^3=8 sub.
for hosts 256-224=32 hosts.
255.255.255.240=/28 so 24+4(4N we need to borrow) formula 2^N=16 subnets.
for hosts 256-240=16 hosts.
for hosts 256-192=64 hosts.
255.255.255.224=/27 so 24+3(3n we need to borrow)formula 2^n=2^3=8 sub.
for hosts 256-224=32 hosts.
255.255.255.240=/28 so 24+4(4N we need to borrow) formula 2^N=16 subnets.
for hosts 256-240=16 hosts.
(1)
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