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?
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 = 2x = 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
Question 23 How many subnets and hosts per subnet can you get from the network 192.168.91.0/30?...
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