Subnet Question related to Class B Network

[COLOR=rgba(0, 0, 0, 0.87058

][FONT=&quot]Question: What valid host range is the IP address 172.26.220.95 255.255.255.128 a part of?

The answer it show's as "[/FONT][/COLOR][COLOR=rgba(0, 0, 0, 0.87058

][FONT=&quot]172.26.220.1 through to 172.26.220.126".

Where as my calculation is getting different address as 172.26.220.0 through to 172.26.220.127.

The way of calculation is as below.

1. The address is of class B , So first two octets are of network bits + third octet 8 bits + 1'st bit of fourth octet.

2. So address range will start like this first n/w 172.26.0.0 - 172.26.0.127 , second n/w 172.26.0.128 172.26.0.255 , third n/w 172.26.1.0 - 172.26.1.127 .. so an..

3. In this manner the valid host range for 172.26.220.95 should be 172.26.220.0 - 172.26.220.127.

Can any one please let me know , where my understanding is wrong in this case.

thanks
Vlrk[/FONT][/COLOR]

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Comments

rob42

Your forgetting about the Network Address [172.26.220.0] and the Broadcast Address [172.26.220.127], so the 'Valid Host Range' is .1 to .126.

vlrk

thanks rob.. it skipped out of my mind.

dontstop

By "host" range they're referring to addresses that can be assigned to a host machine. The network (first address) and last address (subnet broadcast address) are reserved.

For instance: 192.168.0.0/24 has an address range of

192.168.0.0 <- reserved (Network Address)
192.168.0.1 <- First host address
.
.
.
192.168.0.254 <- Last host address
192.168.0.255 <- (Network Broadcast Address)

Uriah7

The host address is actually classless due to the subnet mask being provided in the question. Your step 2 is correct for a classful network, but in this case, you can look at the block size provided by the subnet mask and then figure out what host range the given host ip is a part of.

Step 1: Determine the block size given the subnet mask: 255.255.255.128 = block size of 128 - 256 = 128.
Step 2: Determine that the given host ip address exists within the first block size of the subnet given the "95" which yields a subnet range of 172.20.220.0 - 172.20.220.128
Step 3: As Rob42 mentions above, the actual host range of the subnet range determined in step 3 is 172.20.220.1 - 172.20.220.127 after removing the first ip address and last ip address because they are reserved for the network and broadcast addresses respectively.

rob42

vlrk wrote: »

thanks rob.. it skipped out of my mind.

You're welcome. Easy thing to forget, but it'll stick now, I'll bet...

The 'block size' method that Uriah7 explains is an excellent way to learn Subnetting, b.t.w.