# 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:

25 comments Page 1 of 3.
Abhishek Jha said:
2 years ago

16 hosts < 32, 32 = 2^5, so it takes 5 bits, and each subnet has 30 hosts.

32 " 5 = 27, it means the subnet-mask has /27 bits.

5 subnets < 8, 8 = 2^3,

therefore, 27 " 3 = 24 bits,

= 255.255.255.0.

8 subnets x 32 IPs " 2 = 254, the answer should be 255.255.255.0.

For example class c IP, 192.168.1.1/24 with subnet-mask 255.255.255.224.

32 " 5 = 27, it means the subnet-mask has /27 bits.

5 subnets < 8, 8 = 2^3,

therefore, 27 " 3 = 24 bits,

= 255.255.255.0.

8 subnets x 32 IPs " 2 = 254, the answer should be 255.255.255.0.

For example class c IP, 192.168.1.1/24 with subnet-mask 255.255.255.224.

Jayden said:
3 years ago

16 hosts < 32, 32 = 2^5, so it takes 5 bits, and each subnet has 30 hosts.

32 " 5 = 27, it means the subnet-mask has /27 bits.

5 subnets < 8, 8 = 2^3,

therefore, 27 " 3 = 24 bits,

= 255.255.255.0.

8 subnets x 32 IPs " 2 = 254, the answer should be 255.255.255.0.

For example class c IP, 192.168.1.1/24 with subnet-mask 255.255.255.224.

32 " 5 = 27, it means the subnet-mask has /27 bits.

5 subnets < 8, 8 = 2^3,

therefore, 27 " 3 = 24 bits,

= 255.255.255.0.

8 subnets x 32 IPs " 2 = 254, the answer should be 255.255.255.0.

For example class c IP, 192.168.1.1/24 with subnet-mask 255.255.255.224.

Pollob basak said:
3 years ago

4 subnets, each with at least 20 hosts. Which subnet mask? Please explain about it.

Czar said:
5 years ago

5 subnets 16 hosts

First, you have to meet the number of the host which is 16. By borrowing 4 bits we will meet 16 hosts but there will be no room for the network and broadcast address so we will use 5 bits that has 32 hosts. Using the 2^5 -2 will give us 30 which satisfy our required number of hosts.

Secondly, we will use the formula an actual number of network = 32 (by default) minus no. of borrowed bits so 32-5 equals 27.

Our new subnet mask is /27 which is equivalent to 255.255.255.224.

First, you have to meet the number of the host which is 16. By borrowing 4 bits we will meet 16 hosts but there will be no room for the network and broadcast address so we will use 5 bits that has 32 hosts. Using the 2^5 -2 will give us 30 which satisfy our required number of hosts.

Secondly, we will use the formula an actual number of network = 32 (by default) minus no. of borrowed bits so 32-5 equals 27.

Our new subnet mask is /27 which is equivalent to 255.255.255.224.

(3)

Manikandan said:
7 years ago

I need formula 2^16=32 value of network.

So I taken 128.64.32. value mention after another 0.0.0.0.0 calculate 128+64+32=224.

Ans 255.255.255.224.

So I taken 128.64.32. value mention after another 0.0.0.0.0 calculate 128+64+32=224.

Ans 255.255.255.224.

Chaithanya katari said:
7 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)

JaganPJ said:
7 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:
8 years ago

Thank you @Peeyush.

Siri said:
8 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)

S .HEMANATHAN said:
8 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.

(3)

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