Class B subnet question

I've been unable to resolve this question about a broadcast address for a class B subnet. If one has an address of 172.16.0.0 with a CIDR of /20, what would the broadcast address be on the 2nd available subnet? Also, if the CIDR was /26 what would the broadcast be on the 2nd available subnet? I'm working out the differences between using only the third octet and the fourth octet. Many thanks.

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EdTheLad

/20 = 255.255.240.0 256-240 = 16 subnets 0,16,32,48 etc second is 16
broadcast is 172.16.31.255

/26 = 255.255.255.192 256-192 = 64 => subnets 0,64,128,192 second is 64
broadcast is 172.16.0.63

Simple when you know how,if you dont understand use the search engine! This has been explained over and over again.

Tricon7

EdTheLad wrote:

/20 = 255.255.240.0 256-240 = 16 subnets 0,16,32,48 etc second is 16
broadcast is 172.16.31.255
/

I thought the second is 32, which would give a broadcast, if I'm right, of 172.16.47.255, wouldn't it?

james_

172.16.0.0 /20 (255.255.240.0)

172.16.0.0 - 172.16.15.255 is the first subnet
172.16.16.0 - 172.16.31.255 is the second subnet

Providing Subnet Zero is allowed, 31.255 is the second subnet's broadcast address.

Hope this helps,
James.

georgemc

EdTheLad wrote:

/20 = 255.255.240.0 256-240 = 16 subnets 0,16,32,48 etc second is 16
broadcast is 172.16.31.255

/26 = 255.255.255.192 256-192 = 64 => subnets 0,64,128,192 second is 64
broadcast is 172.16.0.63

Simple when you know how,if you dont understand use the search engine! This has been explained over and over again.

Tricon7,
I believe he meant broadcast is 172.16.0.127 for the secound subnet of /26.

Also, ip subnet zero is enabled by default, so the broadcast for the second subnet for /20 is x.x.31.255.

Tricon7

james_ wrote:

172.16.0.0 /20 (255.255.240.0)

172.16.0.0 - 172.16.15.255 is the first subnet
172.16.16.0 - 172.16.31.255 is the second subnet

Providing Subnet Zero is allowed, 31.255 is the second subnet's broadcast address.

Hope this helps,
James.

Thanks. I think I understand now. But I thought that - in this case - 31.255 would actually be the next subnet (32.0) and that the broadcast should be 31.254 instead?

Man, I hate class B.

remyforbes777

In order for it to be a broadcast the IP host address has to consist of all contigous (spelling) 1's. so 31.254 would not be all ones. It would end with a 0.

james_

Hi Tricon,

172.16.31.254 would be the last "usable" ip address by a host on that subnet, but 172.16.31.255 would be the broadcast. The next IP address would be 172.16.32.0, which falls into the next subnet.

We are working with a 255.255.240.0 subnet. I don't know how much you understand about subnetting, but the way I do it is to subtract the interesting Octet from 256 (in this case subtracting 240 from 256 gives us 16). This gives us the block number.

Now apply the block sizes to the IP address 172.16.0.0, we are looking at subnets of:

1: 172.16.0.0
2: 172.16.16.0
3: 172.16.32.0 etc

The broadcast in each case is 1 IP address before the next subnet begins, ie

1: 172.16.15.255
2: 172.16.31.255
3: 172.16.47.255 etc

Remember the interesting octet is the third octet, so we need to subnet the ip address on the third octet if this makes sense?

Good luck onthe CCNA!

paige1

The Mask 192 produces 2 Subnets 64 ( 01000000) and 128 (10000000)

Working with Class C would look like this:

Subnet 64
01 00000000 = 64 (Subnet Address) all host bits off
01 00000001 = 65 (First Host) first host bit on
01 11111110 = 126 (Last Host) first host bit off
01 11111111 = 127 (Broadcast Address) all host bits on

Subnet 128
1 0000000 = 128 (Subnet Address) all host bits off
1 0000001 = 129 (First Host) First host bit on
1 1111110 = 254 (Last Host) First host bit off
1 1111111 = 255 (Broadcast Address) All host bits on

Working with a Class B would look like this:

Subnet 64.0
01 000000.00000000 = 64.0
01 000000.00000001 = 64.1
01 111111.11111110 = 127.254
01 111111.11111111 = 127.255

Subnet 128.0
1 0000000.00000000 = 128.0 (Subnet Address) all host bits off
1 0000000.00000001 = 128.1 (First Host) First host bit on
1 1111111.11111110 = 255.254 (Last Host) First host bit off
1 1111111.11111111 = 255.255 (Broadcast Address) All host bits on

If you study this you will see the bit pattern (Off, On, Off, On,...) When working with Class B subnets you work on both sides of the period (decimal place) Once the bit pattern is done, just do the math.

georgemc

paige1 wrote:

The Mask 192 produces 2 Subnets 64 ( 01000000) and 128 (10000000)

Working with Class C would look like this:

Subnet 64
01 00000000 = 64 (Subnet Address) all host bits off
01 00000001 = 65 (First Host) first host bit on
01 11111110 = 126 (Last Host) first host bit off
01 11111111 = 127 (Broadcast Address) all host bits on

Subnet 128
1 0000000 = 128 (Subnet Address) all host bits off
1 0000001 = 129 (First Host) First host bit on
1 1111110 = 254 (Last Host) First host bit off
1 1111111 = 255 (Broadcast Address) All host bits on

Working with a Class B would look like this:

Subnet 64.0
01 000000.00000000 = 64.0
01 000000.00000001 = 64.1
01 111111.11111110 = 127.254
01 111111.11111111 = 127.255

Subnet 128.0
1 0000000.00000000 = 128.0 (Subnet Address) all host bits off
1 0000000.00000001 = 128.1 (First Host) First host bit on
1 1111111.11111110 = 255.254 (Last Host) First host bit off
1 1111111.11111111 = 255.255 (Broadcast Address) All host bits on

If you study this you will see the bit pattern (Off, On, Off, On,...) When working with Class B subnets you work on both sides of the period (decimal place) Once the bit pattern is done, just do the math.

Paige1,
You might want to look at your math again. You have a single mask (255.255.255.192 or /26) but are showing 2 subnets of different sizes. You would actually have 4 subnets of equal size (.0, .64, .128, and .192). This is because you have 2 bits available for subnets and we're assuming that subnet zero is enabled.

With subnet zero enabled 2^2=4
Without subnet zero enable 2^2 - 2 =2 (you lose the first and last subnet)

Subnet 0 (Boldface = subnet bits, Normal = host bits) binary 00000000=0
00000000 = 0 (Subnet Address) all host bits off
00000001 = 1 (First Host) first host bit on
00111110 = 62 (Last Host) first host bit off
00111111 = 63 (Broadcast Address) all host bits on

Subnet 64 (Boldface = subnet bits, Normal = host bits) binary 01000000=64
01000000 = 64 (Subnet Address) all host bits off
01000001 = 65 (First Host) first host bit on
01111110 = 126 (Last Host) first host bit off
01111111 = 127 (Broadcast Address) all host bits on

Subnet 128 (Boldface = subnet bits, Normal = host bits) binary 10000000=128
10000000 = 128 (Subnet Address) all host bits off
10000001 = 129 (First Host) first host bit on
10111110 = 190 (Last Host) first host bit off
10111111 = 191 (Broadcast Address) all host bits on

Subnet 192 (Boldface = subnet bits, Normal = host bits) binary 11000000=192
11000000 = 192 (Subnet Address) all host bits off
11000001 = 193 (First Host) first host bit on
11111110 = 254 (Last Host) first host bit off
11111111 = 255 (Broadcast Address) all host bits on

I hope this helps clarify things for someone.

Georgemc

paige1

It took me a minute, but yes, I have some bits in the wrong place (Subnet 12

and no, I did not enable subnet zero. If anyone is confused, I apologize. A lesson to me to double check my work.