Can anyone explain network and broadcast adress

NetAh

Can anyone explain what would be the network, hosts & broadcasts addresses for below:



Address:
10.149.18.171


Subnet Mask:
255.255.254.0

OctalDump

The network address is the lowest address in that range, and the broadcast is the highest. Host addresses are everything in between. The two things you need to know are that networks follow boundaries of the powers of two, and that subnet masks tell you where these boundaries are.

For 255.255.254.0, the boundary is every 512. So you take that address, find the closest 512 boundary below it (or above it), and then add (or subtract) 511 to get the broadcast (or network) address.

You can easily find the exact answers from an online calculator.

Priston

When you look at the host bits in binary: the network address is all zeros, broadcast address is all ones, and host address is everything in between.

E Double U

10.149.18.0 - network
10.149.19.255 - broacast
10.149.18.1 - 10.149.19.254 - hosts

mehutch355

This is simple subnetting. It looks weird because it is not a class C network which is what most people are used too. The network address will be 10.149.18.0 and the broadcast address will be 10.149.19.255. If you break this up into binary the ip will be:
address: 00001010.10010101.00010010.10101011
subnet : 11111111.11111111.11111110.00000000
^
This is whats called your CIDR line.


Now reset the IP address where at the end of the subnet. To do this make all of the bits in the host 0 where the subnet turns to 0 (At your CIDR Line). so your IP address for the host becomes:
Address: 00001010.10010101.00010010.00000000 <

---

This is your network address!
^
CIDR Line

After you find the network address, do the same thing but make all the 0's after the CIDR line, make all of the bits 1's.
Address: 00001010.10010101.00010011.11111111 <

---

This is your broadcast address
^
CIDR Line

Now the easy part. Every IP int between your Network Address and your Broadcast address is a usable host IP, unless it is reserved for something else.

mehutch355

Sorry the CIDR lines notation got messed up during the post for some reason. The CIDR line is where the subnet bits turn to 0.

james43026

mehutch355 wrote: »

This is simple subnetting. It looks weird because it is not a class C network which is what most people are used too. The network address will be 10.149.18.0 and the broadcast address will be 10.149.19.255. If you break this up into binary the ip will be:
address: 00001010.10010101.00010010.10101011
subnet : 11111111.11111111.11111110.00000000
^
This is whats called your CIDR line.


Now reset the IP address where at the end of the subnet. To do this make all of the bits in the host 0 where the subnet turns to 0 (At your CIDR Line). so your IP address for the host becomes:
Address: 00001010.10010101.00010010.00000000 <

---

This is your network address!
^
CIDR Line

After you find the network address, do the same thing but make all the 0's after the CIDR line, make all of the bits 1's.
Address: 00001010.10010101.00010011.11111111 <

---

This is your broadcast address
^
CIDR Line

Now the easy part. Every IP int between your Network Address and your Broadcast address is a usable host IP, unless it is reserved for something else.

This is a pretty good way to explain this in a simple way.

I would add that your other option, is to do a logical "and" between the given IP address and the subnet mask. Which is what a router actually does.

Rushlan

I am new to this, just learned subnetting, please correct me if I made any mistake.

So the ip address is 10.149.18.171 and the subnet mask is 255.255.254.0.

Set the ip address to 10.149.0.0 ( because the subnet mask is not 255 in the 3rd and 4th octet)

Now 256 - 254 = 2 , ( 254 as in the 3rd octet of the subnet mask )

2 is the increment, so you keep on adding 2 ....
10.149.0.0
10.149.2.0
10.149.4.0
. . .
. . .
10.149.18.0 So this is the network address

To find the broadcast address , use the formula: (network address) + (255 - subnet mask)

On the 3rd octet: 18 + (255 - 254) = 19
On the 4th octet 0 + ( 255 - 0) = 255

So the broadcast address is 10.149.19.255

Host address starts from the network address , in this case 10.149.18.1 and ends at the broadcast address, 10.149.19.254.

james43026

Not sure how many who have read or responded on this thread know and understand the concept of doing a logical "and" on the given IP address and the subnet mask. Below is an example. To perform a logical and, you simply line up the binary bits, you compare the bits of the subnet mask with the bits of the IP address, a logical and requires that both have a binary 1 in the same place for the operation to result in a binary 1, everything else would be a binary 0. For some this may be faster, for others it may not. For me it's faster than writing out a bunch of subnets, but what works for one person, may not work for another.



10.149.18.171= 00001010.10010101.00010010.10101011
255.255.254.0= 11111111.11111111.11111110.00000000

00001010.10010101.00010010.00000000= 10.149.18.0

alu408

This is the method i use as well, and it is very helpful!

Rushlan wrote: »

I am new to this, just learned subnetting, please correct me if I made any mistake.

So the ip address is 10.149.18.171 and the subnet mask is 255.255.254.0.

Set the ip address to 10.149.0.0 ( because the subnet mask is not 255 in the 3rd and 4th octet)

Now 256 - 254 = 2 , ( 254 as in the 3rd octet of the subnet mask )

2 is the increment, so you keep on adding 2 ....
10.149.0.0
10.149.2.0
10.149.4.0
. . .
. . .
10.149.18.0 So this is the network address

To find the broadcast address , use the formula: (network address) + (255 - subnet mask)

On the 3rd octet: 18 + (255 - 254) = 19
On the 4th octet 0 + ( 255 - 0) = 255

So the broadcast address is 10.149.19.255

Host address starts from the network address , in this case 10.149.18.1 and ends at the broadcast address, 10.149.19.254.