Cmit Subnetting Assignment
Subnet 90.0.0.0/8 into 260 subnets
In order to get 260 subnets the subnet mask will need to have a range of 255.0.0.0
The IP Range will start at 1-254 since 0 and 255 will already be utilized.
Netmask: 255.0.0.0 = 8 11111111 .00000000.00000000.00000000
Broadcast: 90.255.255.255 01011010 .11111111.11111111.11111111
Subnet 1
Network address: 90.0.0.0
Broadcast: 90.0.255.255
Available IPs: 90.0.0.1-90.0.255.254
Subnet 2
Network address: 90.1.0.0
Broadcast: 90.1.255.255
Available IPs: 90.1.0.1-90.1.255.254
Subnet 3
Network address: 90.2.0.0
Broadcast: 90.2.255.255
IP Range: 90.2.0.1-90.2.255.254
Subnet 4
Network address: 90.3.0.0
Broadcast: 90.3.255.255
IP Range: 90.3.0.1-90.3.255.254
Subnet assignment #32 80.0.0.0/8 250 subnets
New subnet mask after subnet 255.255.0.0
In order to get more subnets out of the current network, host bits will need to be used as network bits. To get 250 subnets the subnet mask will need to fit 250 subnets into a range of 0-255. 11111111=255 256-255=1 256/1=256 possible subnets therefore each network will increase in the second octet by 1 (1,2,3,4,5,6,7…) It would work similar if there was a need for say 124 subnets 11111110=254 256-254=2 256/2=128 possible subnets. So to get the adequate subnets from 255.0.0.0 the second octet or host bits from the second octet need to be used for network bits to increase the number of subnets thus giving the new subnet mask of 255.255.0.0
The first four subnets
(1)
Network address 80.0.0.0/16
01010000=80 and 00000000 x3 = 80.0.0.0 starting with the given network. Broadcast address 80.0.255.255
The last address in the subnet with subnet mask of 255.255.0.0 which for all networks in this case will be 80.xxx.255.255 where xxx=the current subnet Assignable range 80.0.0.1-80.0.255.254
The
In order to get 260 subnets the subnet mask will need to have a range of 255.0.0.0
The IP Range will start at 1-254 since 0 and 255 will already be utilized.
Netmask: 255.0.0.0 = 8 11111111 .00000000.00000000.00000000
Broadcast: 90.255.255.255 01011010 .11111111.11111111.11111111
Subnet 1
Network address: 90.0.0.0
Broadcast: 90.0.255.255
Available IPs: 90.0.0.1-90.0.255.254
Subnet 2
Network address: 90.1.0.0
Broadcast: 90.1.255.255
Available IPs: 90.1.0.1-90.1.255.254
Subnet 3
Network address: 90.2.0.0
Broadcast: 90.2.255.255
IP Range: 90.2.0.1-90.2.255.254
Subnet 4
Network address: 90.3.0.0
Broadcast: 90.3.255.255
IP Range: 90.3.0.1-90.3.255.254
Subnet assignment #32 80.0.0.0/8 250 subnets
New subnet mask after subnet 255.255.0.0
In order to get more subnets out of the current network, host bits will need to be used as network bits. To get 250 subnets the subnet mask will need to fit 250 subnets into a range of 0-255. 11111111=255 256-255=1 256/1=256 possible subnets therefore each network will increase in the second octet by 1 (1,2,3,4,5,6,7…) It would work similar if there was a need for say 124 subnets 11111110=254 256-254=2 256/2=128 possible subnets. So to get the adequate subnets from 255.0.0.0 the second octet or host bits from the second octet need to be used for network bits to increase the number of subnets thus giving the new subnet mask of 255.255.0.0
The first four subnets
(1)
Network address 80.0.0.0/16
01010000=80 and 00000000 x3 = 80.0.0.0 starting with the given network. Broadcast address 80.0.255.255
The last address in the subnet with subnet mask of 255.255.0.0 which for all networks in this case will be 80.xxx.255.255 where xxx=the current subnet Assignable range 80.0.0.1-80.0.255.254
The