
1. Internetwork IP addressing: The different classes of IP addresses
is as below:
 Network address of all zeros means
"This network or segment".  Network address of all 1s means
" all networks", same as hexadecimal of all Fs.  Network number 127 is reserved for
loopback tests.  Host (Node) address of all zeros mean "This Host
(Node)".  Host (Node) address of all 1s mean "all Hosts (Nodes) " on
the specified network. 2. The range of numbers from 224.0.0.0
to 239.255.255.255 are used for multicast packets. This is
known as Class D address range. Let us take a network ID of 168.8.0.0,
and find the maximum number of possible subnets and the
corresponding subnet mask that can accommodate at least
500 hosts. The steps involved are outlined below: I. Find the Class of the IP address, in this case it is a class B
network. Class B network has the form N.N.H.H. Therefore,
we have a total of 16 bits (two octets) for assigning to
internal networks and hosts. The minimum number of host
addresses required is 500. The last octet corresponds to
2^8 = 256 hosts which is still less than 500 Hosts..
Therefore, you have to borrow one more bit from the third
octet to make it 256*2 = 512 Hosts. This leaves 7 bits in
the third octet for assigning subnet addresses. This is
equal to 2^7=128 subnets. II. Write the 7 bits available for subnetting in third octet in the
form 11111110 (last bit being the Host bit). The decimal
equivalent of the first seven bits is
2^7+2^6+2^5+2^4+2^3+2^2+2^1
= 128 + 64 +32
+ 16 + 8 + 4 + 2 =
254. III. Therefore, the subnet mask required is 255.255.254.0. 6. How to maximize the number of hosts
for a given number of subnets:
Determining the subnet mask that
allows maximum number of hosts:
Let us consider an IP address 196.202.56.0 with
four subnets and maximize the number of host for the given
subnets. The steps involved are as below: I. The number of
subnets required are four. We need to add subnets of all
ones and all zeros to this. This is because all zeros and
all ones subnets belong to "this subnet" and
"all subnets" broadcasts and can not be used.
Therefore, the total number of subnets to be reserved is
4+2 = 6. II. We want to implement maximum possible Hosts. Therefore, we need
to minimize the number of subnets. This minimum number is
6 here. If we reserve 2 bits, it results in only 2^2=4
subnets which is less
than 6. Therefore, we have to reserve 3 bits for implementing
subnets, resulting in 2^3=8 subnets. This is now optimized
for maximum number of Hosts (as we have optimized for
minimum number of subnets). III. Write the 3 bits available for subnetting in fourth octet in
the form 11100000 (Five 0s being Host bits). The decimal
equivalent is 2^7+2^6+2^5 = 128 + 64 +32
= 224. IV. Therefore, the subnet mask required is 255.255.255.224. 7. 127.0.0.1 is the local loop back
address.
b. 7 bits of subnet mask corresponds to
(2^72)=1282 = 126 subnets. c. 9 bits (167) of host addresses
corresponds to (2^92)=5122 = 510 hosts. Some times, the subnet mask is specified with
the bits available in the default subnet mask. In this
case the bits available in default subnet mask is 16.
Therefore, total number of bits available in the subnet
mask are 16+7=23. If you are given a subnet mask of 23
bits long for a class B address, it is understood that it
contains the bits from the default subnet mask as well. 10. The directed broadcast should reach
all Hosts on the intended network (or subnet, if sub
netted). For example, the directed broadcast address for
an IP network 196.233.24.15 with default subnet mask is
196.233.24.255. This is arrived by putting all 1s for the
host potion of the IP address.

