Subnetting question finding # of hosts and networks

I'm having problem figuring this.

How many networks and host can you get from 192.168.234.0 /26

I have no idea how to do this.

help is appreciated thank you!

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Comments

mikej412

Do you remember ever coming across any formulas that looked like 2^N or (2^N)-2 in your studies?

How many subnet bits do you have in that problem?

How many host bits do you have in that problem?

bubble2005

192.168.234.0 /26

In each octet there are 8 bits (self-explanatory). The /26 is a scheming method used for ip addressing called Classless Inter-domain Routing. If you look at the range of that ip address, you are working with a Class C which ranges from 192.x.x.x - 223.x.x.x. So the first three octets are /24 i.e. 192.168.234 therefore the /26 carries over into the last octet (00000000) turns into (11000000) turning on the first two bits to make 26. Now to determine the number of hosts and subnets are easy. There is an equation; for subnets its 2^n, n being the number of bits turned on. To determine the number of hosts for each subnet its, 2^n-2, that being the number of bits turned off ((the zeros *2)-2).

So for this ip address, you will have 4 subnets due to the /26 (two bits turned on) 2^2 = 4

The number of hosts for each subnet will be (2^6 = 64 -2 = 62) a total of 62 hosts per subnet.

The block size for each subnet would be the number of bits turned on in that last octet subtracted from 256. ie. 256-192 = 64

so your subnets would be

0-63
64-127
128-191
192-255

I hope it helps.

tim100

bubble2005 wrote: »

192.168.234.0 /26

In each octet there are 8 bits (self-explanatory). The /26 is a scheming method used for ip addressing called Classless Inter-domain Routing. If you look at the range of that ip address, you are working with a Class C which ranges from 192.x.x.x - 223.x.x.x. So the first three octets are /24 i.e. 192.168.234 therefore the /26 carries over into the last octet (00000000) turns into (11000000) turning on the first two bits to make 26. Now to determine the number of hosts and subnets are easy. There is an equation; for subnets its 2^n, n being the number of bits turned on. To determine the number of hosts for each subnet its, 2^n-2, that being the number of bits turned off ((the zeros *2)-2).

So for this ip address, you will have 4 subnets due to the /26 (two bits turned on) 2^2 = 4

The number of hosts for each subnet will be (2^6 = 64 -2 = 62) a total of 62 hosts per subnet.

The block size for each subnet would be the number of bits turned on in that last octet subtracted from 256. ie. 256-192 = 64

so your subnets would be

0-63
64-127
128-191
192-255

I hope it helps.

Building on top of this, a faster way to find out how many hosts are in a subnetted class C or any network address given to you with a
/25 - /30 is to take the rightmost bit - in this case 64 and subtract 2 instead of doing (2^N)-2.

bubble2005

Oh cool, thanks Tim, I'll use that method now.

SurferdudeHB

Is the method the same for class B network?

Question: How many subnets and hosts per subnet can you get from the network 172.16.0.0 255.255.248.0?

tim100

SurferdudeHB wrote: »

Is the method the same for class B network?

No. This would only work when you want to find out how many host addresses are available in a /25 - /30 subnet because your working with the last octet.

SurferdudeHB wrote: »

Question: How many subnets and hosts per subnet can you get from the network 172.16.0.0 255.255.248.0?

32 Subnets with 2046 hosts per subnet.

SurferdudeHB

tim100 wrote: »

No. This would only work when you want to find out how many host addresses are available in a /25 - /31 subnet because your working with the last octet.

32 Subnets with 2046 hosts per subnet.

So how would you find the solution for a class A and B IP address?