How does the Host portion of the subnet effect the final IP

[SamWireT]

Please see attached image
Can someone please explain to me why the host portion of the network address is 112, how does the host part for the subnet effect the final network IP?

[Thomas Schmitt]

Take a look on your binary subnet mask. 1 means the corresponding bit of your address belongs to subnet and 0 means it is host part. In your case, you have /28 subnet, so first 28 bits is your network, and last 4 is host part. 4 bits allow 16 values (0-15), so each multiple of 16 is next subnet (0, 16,32,48,64,80,96,112 and so on). your host address ends with 115, so it is within 112-127 subnet and you are allowed to use 113-126 for your Hosts in this subnet 

[SamWireT]

Hi thank you for the reply

so how did you get the host part being 28? i know that binary is 0 1 2 4 8 16..... so i dont understand were the 28 comes from?

[Joseph W. Doherty]

"Can someone please explain to me why the host portion of the network address is 112 . . ."

It's not.

With a mask of 240 (decimal), the network number is the first four bits and the host number is the last four bits.  Given an address of 115 (decimal), the network number is 112 (decimal) and the host number is 3 (decimal).

". . . how does the host part for the subnet effect the final network IP?"

If provides the value of the host portion/part of the IP.  Remember, an IP has two portions/parts, the network number and the host number.  Also remember, the network number and host number are back-to-back, i.e. where one of them ends, its bits usage, the other begins, and each might use a different number of bits from IP address to IP address but the combination always being usage of 32 bits.

Two things that often cause confusion, I believe, are quad dotted decimal representation of a 32 bit binary numbers and "subnets" (from classful IP addressing - classless IP just has a network number and a host number).

[SamWireT]

Oh Right so the first 4 number in the octet are related to the network the last 4 in the octet are for the host, then i just pick the first useable one? 

[Joseph W. Doherty]

Unsure I fully understand "then i just pick the first useable one?"

That aside, yes, correct, in this case the first four bits in the octet are part of the network ID while the last 4 bits in the octet are the host ID.

Since four bits are available for a host ID, you can use 1..14 (as zero is set aside for network and 15 is for network broadcast).

Again, don't let the quad dotted decimal IP representation mislead you.  For example, if you had a /23, it's bit alignment, for the host value, would span two octets, i.e. NNNNNNNN.NNNNNNNN.NNNNNNNH.HHHHHHHH.

BTW, and further, subnets only really come into play with classful networks, where network numbers are always on octet boundaries.  The prior /23 if using a Class B address, would be: NNNNNNNN.NNNNNNNN.SSSSSSSH.HHHHHHHH.

[Spooster IT Services]

Hi SamWireT, 

Host Address :  172.71.135.115  

Subnet Mask : 255.255.255.240 (8+8+8+4) == /28

(128+64+32+16 = 240 , Which means in the last Octet Only four bits will be on ) 

The Block Size :  256 - 240 = 16 

Network Bits : 2^(borrowed Bits) = 2^4 = 16 

1) 172.71.135.0 ------ 1st Network Address 

Keep on adding 16 in the last octet  (the block size )

2) 172.71.135.16   3) 172.71.135.32. 4) 172.71.135.48  5) 172.71.135.64 6) 172.71.135.80  7) 172.71.135.96 172.71.135.112 9) 172.71.135.128 

Now, As you can see entry is 172.71.135.112 Network Address 

So now First Valid host for this network address will be : 172.71.135.113 

second valid host : 172.71.135.114

third valid host : 172.71.135.115

Therefore , Host Address 172.71.135.115 , will have 172.71.135.112 as the network address