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Calculating the Subnets.
Calculating the Subnets.
I am using the online Subnet Calculator. Some Data displayed on the calculator and indicated by the Red arrows on the screenshot below need some explanation:
example:
how did they come with subnet bits: 12
Maximum subnets 4096
host per subnets :14
subnet ID :172.16.1.0
Broadcast address : 172.16.1.15
Thank you
I am using the online Subnet Calculator. Some Data displayed on the calculator and indicated by the Red arrows on the screenshot below need some explanation:
example:
how did they come with subnet bits: 12
Maximum subnets 4096
host per subnets :14
subnet ID :172.16.1.0
Broadcast address : 172.16.1.15
Thank you
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Subnetting is based on classful network masks.
Your example uses a class B private scope or 172.16.0.0/16 scope as it's classful base.
In the classful base you have 16 network bits and 16 host bits based on the classful mask of /16 or 255.255.0.0. Expressed in binary is 11111111.11111111.00000000
Map the above binary to network (n) and host (h)
nnnnnnnn.nnnnnnnn.hhhhhhhh
In your example you are subnetting the above /16 into a /28 or 255.255.255.240. This mask represented in binary is 11111111.11111111.11111111
The positions that you have moved from the base classful mask of /16 to the subnet of /28 create the subnet bits (/28-/16 = 12)
To clarify this a bit, we will do the Network (n), Host(h) representation of this new mask but we will add the Subnet(s) callout to it.
nnnnnnnn.nnnnnnnn.ssssssss
To get the number of networks in this subnet scheme, take the number of subnet bits to exp2 so s^2 or 12^2 = 4096 individual subnets. (maximum subnets)
Each of those subnets will contain x unique addresses determined by hosts bits to exp2, so h^2 or 4^2=16 unique addresses in each subnet.
Your first usable subnet will be 172.16.0.0/18.
First address in the subnet is 172.16.0.0
Last address in the subnet is 172.16.0.15
First address in a subnet is always the network address and unusable (subnet ID)
Last address in a subnet is always the broadcast address and unusable (Broadcast Address)
This leaves 14 usable addresses in the scope (hosts per subnet)
First usable is 172.16.0.1
Last usable is 172.16.0.14
Your example uses a class B private scope or 172.16.0.0/16 scope as it's classful base.
In the classful base you have 16 network bits and 16 host bits based on the classful mask of /16 or 255.255.0.0. Expressed in binary is 11111111.11111111.00000000
Map the above binary to network (n) and host (h)
nnnnnnnn.nnnnnnnn.hhhhhhhh
In your example you are subnetting the above /16 into a /28 or 255.255.255.240. This mask represented in binary is 11111111.11111111.11111111
The positions that you have moved from the base classful mask of /16 to the subnet of /28 create the subnet bits (/28-/16 = 12)
To clarify this a bit, we will do the Network (n), Host(h) representation of this new mask but we will add the Subnet(s) callout to it.
nnnnnnnn.nnnnnnnn.ssssssss
To get the number of networks in this subnet scheme, take the number of subnet bits to exp2 so s^2 or 12^2 = 4096 individual subnets. (maximum subnets)
Each of those subnets will contain x unique addresses determined by hosts bits to exp2, so h^2 or 4^2=16 unique addresses in each subnet.
Your first usable subnet will be 172.16.0.0/18.
First address in the subnet is 172.16.0.0
Last address in the subnet is 172.16.0.15
First address in a subnet is always the network address and unusable (subnet ID)
Last address in a subnet is always the broadcast address and unusable (Broadcast Address)
This leaves 14 usable addresses in the scope (hosts per subnet)
First usable is 172.16.0.1
Last usable is 172.16.0.14
how did they come with subnet bits: 12
You start with an address that has a 16-bit mask. You extend the mask to accommodate the number of subnets/hosts and you end up with a 28-bit mask.
28-16=12.
Actual mask minus the original (or starting) mask will give the size of the subnet field. Which in this case is 12.
You chose class B network mask, so your results of a /28 is 12 bits in from the /16. Now if you go back and choose class C network, you will only be 4 bits from the /24.
jsk -
Sorry - should say
Your first usable subnet will be 172.16.0.0/28.
First address in the subnet is 172.16.0.0
Last address in the subnet is 172.16.0.15
First address in a subnet is always the network address and unusable (subnet ID)
Last address in a subnet is always the broadcast address and unusable (Broadcast Address)
Does that clarify?
Sorry - should say
Your first usable subnet will be 172.16.0.0/28.
First address in the subnet is 172.16.0.0
Last address in the subnet is 172.16.0.15
First address in a subnet is always the network address and unusable (subnet ID)
Last address in a subnet is always the broadcast address and unusable (Broadcast Address)
Does that clarify?
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