Question

Question 23

How many subnets and hosts per subnet can you get from the network 192.168.91.0/30?

Question 24

Which subnet does host 172.30.122.242 255.255.255.192 belong to?

Question 25

What is the last valid host on the subnetwork 172.27.174.64/26?

Question 26

What is the last valid host on the subnetwork 192.168.210.56 255.255.255.248?

Question 27

What is the broadcast address of the network 172.24.57.224/28?

Question 28

What is the last valid host on the subnetwork 172.19.197.0/25?

Question 29

What is the first valid host on the subnetwork that the node 192.168.83.14 255.255.255.248 belongs to?

Question 30

What is the first valid host on the subnetwork that the node 172.20.32.1/20 belongs to?

Answer 1

23)

Given,

Network Address is 192.168.91.0/30

This is class C address.

Default subnet mask is /24

Therefore number of subnet bits borrowed = 30 - 24 = 6.

Number of subnets = 2^x = 2⁶ = 64 subnets, where x is the number of subnet bits borrowed.

Subnet Mask is /30.

Therefore host id bits = 32 - 30 = 2 bits.

This can give 2² = 4 addresses.

Therefore, valid host addresses for subnet = 4 - 2 = 2 , two addresses are subtracted because they are reserved for network id and broadcast address.

Therefore, there are 64 subnets possible with each of 2 hosts/ subnet.

24)

Given,

Host Address is  172.30.122.242

Subnet Id is 255.255.255.192 = /26

since it is /26, fourth octet is the interesting octet.

Subtraction of 256 - 192 = 64 which is the size of each subnet.

Therefore, subnets and their range looks as follows:

172.30.0.0 - 172.30.0.63

172.30.0.64 - 172.30.0.127

172.30.0.128 - 172.30.0.191

172.30.0.192 - 172.30.0.255

......................

.....................

....................

172.30.122.0 - 172.30.122.63

172.30.122.64 - 172.30.122.127

172.30.122.128 - 172.30.122.191

172.30.122.192 - 172.30.122.255 ;; Given host belongs to  belongs to subnet 172.30.122.192

......................

.....................

....................

172.30.255.0 - 172.30.255.63

172.30.255.64 - 172.30.255.127

172.30.255.128 - 172.30.255.191

172.30.255.192 - 172.30.255.255

Therefore, Given host belongs to subnet  172.30.122.192 / 26

25)

Given subnet Address: 172.27.174.64 /26

/26 indicates that number of host bits = 32 - 26 = 6

Converting it into binary gives:

10101100.00011011.10101110.01 000000 ==>172.27.174.64

Broadcast address contains all 1's in host id bits.

Therefore,

10101100.00011011.10101110.01 111111 ==> 172.27.174.127

Last valid host address is broadcast address - 1 i.e, 172.27.174.126

26)

Given subnet address : 192.168.210.56

Subnet mask is 255.255.255.248 = /29

/29 indicates that number of host bits = 32 - 29 = 3

Converting it into binary gives:

11000000.10101000.11010010.00111 000 ==> 192.168.210.56

Broadcast address contains all 1's in host id bits.

Therefore,

11000000.10101000.11010010.00111 111 ==> 192.168.210.63

Last valid host address is broadcast address - 1 i.e,192.168.210.62