NO WORK SHOWN, NO MARKS
[1 mark for each answer box]
Note: Bits coloured in red are related to the subnets. Bits after the subnet ID are the host bits.
Problem
Number of needed usable subnets: 62
Network Address: 107.0.0.0
Answer
Show Your Work Here for Marks
Address class
A
107 in the first octet is between the range of 1-127 of class A network ID's
Default subnet mask
255.0.0.0
1111 1111.000 0000. 0000 0000.0000 0000
Custom subnet mask
255.252.0.0
1111 1111.1111 1100. 0000 0000. 0000 0000
Total number of subnets
64
26=64
Number of usable subnets
62
26= 64-2 =62
Total number of host addresses/subnet
262144
Since there are 18 bits left for the host,
218=262144
Number of usable host addresses/subnet
262142
218= 262144-2=262142
Number of bits borrowed
6
26 = 64
Problem with Solution
Number of needed usable subnets: 62
Number of needed usable hosts: 1022
Network Address: 190.168.0.0
Answer
Show Your Work Here for Marks
Address class
B
190 in the first octet is between the range of 128 and 191 of class B network ID's
Default subnet mask
255.255.0.0
1111 1111.1111 1111.0000 0000.0000 0000
Custom subnet mask
255.255.252.0
1111 1111.1111 1111.1111 1100.0000 0000
Total number of subnets
64
26 = 64
Number of usable subnets
62
26 = 64 - 2 = 62
Total number of host addresses/subnet
1k
Since there are 10 bits left for the hosts,
210 = 1k
Number of usable host addresses/subnet
1022
1k - 2 = 1022
Number of bits borrowed
6
26 yields a total of 64 subnets. Taking away the first and last unusable subnets yields a total of 64-2 = 62 required usable subnets
What is the 2nd usable subnet range, including the invalid host addresses
(state the starting and ending addresses)?
190.168.8.0 to
190.168.11.255
Usable subnet 3rd Octet 4th Octet
1st 0000 0100 0000 0000
2nd 0000 1000 0000 0000
3rd 0000 1100 0000 0000
4th 0001 0000 0000 0000
Thus, the 1st