IP Addressing/Subnetting : Chapter 3

After the tutorial and discussions in Chapter 1 and Chapter 2, I believe most of you should have an better idea on the IP addressing/subnetting. Now it's the time to apply the knowledge on some problems analyzing and designing.

Q : If you are given the situation below, and you should "separate" your network as accordingly. What should you do and how is the IP assignment?

Network ID : 203.158.30.0

Subnet Mask : 255.255.255.0

Number of networks needed : 4 different separated network

First of all, you need to know the total number of hosts available in the network given. Total number of IP address available

= ( 2 ^ unmasked bits ) - 2

= ( 2 ^ 8 ) - 2 = 256 - 2 = 254 hosts if default Class C is used

Divide the network into 4 different network, each network should have ( Including every network ID and broadcast IP )

= 256 / 4 = 64 hosts each

Thus, your subnet mask for each network

= 255.255.255. [ 254 hosts - 64 hosts each ]

= 255.255.255.192

Now, it is the time for you to "chop" your 203.158.30.x network. Your may get your Network ID for each network by adding 64 hosts each started from .0

The Broadcast IP for each network will be next Network ID minus 1 as shown in the "chop" diagram below. What remaining in between the Network ID and Broadcast IP are the available IP addresses can be used.

=== 1st Network ============================================

203.158.30.0 -------------------------------------   Network ID

203.158.30.1 ~ 203.158.30.62                    Available IP addresses

203.158.30.63 -----------------------------------    Broadcast IP

=== 2nd Network ============================================

203.158.30.64 ------------------------------------   Network ID

203.158.30.65 ~ 203.158.30.126                 Available IP addresses

203.158.30.127 ----------------------------------    Broadcast IP

=== 3rd Network ============================================

203.158.30.128 ----------------------------------   Network ID

203.158.30.129 ~ 203.158.30.190               Available IP addresses

203.158.30.191 ---------------------------------    Broadcast IP

=== 4th Network ============================================

203.158.30.192 -----------------------------------   Network ID

203.158.30.193 ~ 203.158.30.254                Available IP addresses

203.158.30.255 ----------------------------------    Broadcast IP

Now, you may try yourself on solving the question below.

Q : If you are given the situation below, and you should "separate" your network as accordingly. What should you do and how is the IP assignment?

Network ID : 192.168.4.0

Subnet Mask : 255.255.254.0

Number of networks needed : 16 different separated network

In the next post, I will continue in the C.I.D.R. Have a nice day~