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Calculate broadcast and host IP

October 22, 2005 at 11:53:30
Specs: WINDOWS XP, 512

Assume we have an IP address 101.35.15.10 with the subnet mask 255.248.0.0.
My question is:

a. Calculate Network Address.

b. Calculate Broadcast Address.

c. Calculate the 5th Host IP.

For the Network Address, it's very easy.

ANDING

101.35.15.10 35.15.10 (Class A)
255.248.0.0 248.0.0


35 = 00100011

248 = 11111000

ANS = 00100000 = 32

Network Address is 101.32.0.0

My problem is that I do not know how to calculate the broadcast address and the 5th Host IP.

Please explain me how to solve this issue.


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#1
October 25, 2005 at 08:34:37

You are on the right path there!

So let's see here:

101.35.15.10 = 1100101 00100011 00001111 00001010 (binary)
255.248.0.0 = 11111111 11111000 00000000 00000000 (binary)
This means you have 19 free bits.

So in effect, the first 13 bits stay "locked", and the last 19 bits change.

Like you did, to get the network address you changed all last bits to 0, which gave you 101.32.0.0.

The broadcast address is the same thing, except you put the free bits to 1. In this case, the last 19 bits will be 1, so you will get "1100101 00100111 11111111 11111111", which gives you 101.39.255.255

Now, to calculate the 5th host:

Since the network address is the first non-usable ip, you will simply need to increment the binary value 5 times, like this:

1100101 00100000 00000000 00000000 (Net addr)
1100101 00100000 00000000 00000001 (1st Host)
1100101 00100000 00000000 00000010 (2nd host)
1100101 00100000 00000000 00000011 (3rd host)
1100101 00100000 00000000 00000100 (4th host)
1100101 00100000 00000000 00000101 (5th host)
...
1100101 00100111 11111111 11111111 (bcast)

So when you convert the 5th host to decimal, you will get 101.32.0.5


I hope this clears things up for you ;) Don't worry, subnetting is hard at first, the more you to them the easier it gets.

- Alexandre P.


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#2
November 9, 2005 at 07:18:06

Hi,

Here is the output seen on my linux host for an interface

[root@tb03-02-host02 mint-1.2]# /sbin/ifconfig eth4
eth4 Link encap:Ethernet HWaddr 00:0D:88:53:08:8F
inet addr:20.10.10.3 Bcast:20.255.255.255 Mask:255.255.255.0
UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1
RX packets:13241 errors:0 dropped:0 overruns:0 frame:0
TX packets:15741 errors:0 dropped:0 overruns:0 carrier:119
collisions:0 txqueuelen:100
RX bytes:1021531 (997.5 Kb) TX bytes:1187686 (1.1 Mb)
Interrupt:10 Base address:0xcf00

ip address: 20.10.10.3
netmask: 255.255.255.0
So as per this, my network address is 20.10.10.0

Shouldn't the broadcast ip be obtained by making the last 8 bits as 1 which gives us 20.10.10.255

Why do I see 20.255.255.255 then?


Thanks.



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