ENG 2003: INFORMATION TECHNOLOGY
Tutorial 4: IP Subnetting
1) Calculate the maximum number of subnets supported by the following
networks.
a. 192.168.1.0/26
b. 172.17.0.0/20
a) Number of subnet bits = 2, max. number of subnets = 22 -2= 2.
b) Number of subnet bits = 4, max. number of subnets = 24-2=14.
2) Continue from question 1, calculate the maximum number of interfaces
supported by each subnet.
a) Number of host bits = 32 – 26 = 6, max. number of supported
interfaces = 26 -2 = 62
b) Number of host bits = 32 – 20 = 12, max. number of supported
interfaces = 212 -2 = 4094
3) (a) Calculate the maximum number of interfaces that can be
accommodated in the IP network 68.0.0.0 (excluding all reserved IP
addresses).
This is a class A network, which can accommodate 224-2=16777214
interfaces
(b) Given the following IP networks.
For each of the networks, design a subnet mask to divide the network into
exactly 14 subnets of equal sizes. From your answer, calculate the number
of interfaces that can be accommodated in EACH of these subnets.
Hints: You should NOT use the all-zeros and all-ones subnets. All reserved
IP addresses should be excluded.
(i) 160.18.0.0
This is a class B network. It has a network ID of 16 bits long.
To divide the network into 14 equal-size subnets, a subnet ID
of 4 bits long is needed as 16-2=14.
Therefore, the subnet mask should be designed as follow
11111111 11111111 11110000 00000000 (255.255.240.0)
It has a host ID of 12 bits long. Therefore, each subnet can
accommodate 212-2=4094 interfaces.
(ii) 204.17.5.0
This is a class C network. It has a network ID of 24 bits long.
To divide the network into 14 equal-size subnets, a subnet ID
of 4 bits long is needed as 16-2=14.
Therefore, the subnet mask should be designed as follow
11111111 11111111 11111111 11110000 (255.255.255.240)
It has a host ID of 4 bits long. Therefore, each subnet can
accommodate 16-2=14 interfaces.
(c) Consider a router with the following routing table.
Destination IP Address Range Interface
From 10100000 00000000 00000000 00000000
To 10100000 00111111 11111111 11111111 0
From 10100000 01000000 00000000 00000000
To 10100000 01000000 11111111 11111111 1
From 10100000 01000001 00000000 00000000
To 10100001 01111111 11111111 11111111 2
Otherwise 3
Suppose the following packets have arrived at the router, which
interfaces will the packets be forwarded? Justify your answers.
(i) Packet A with a destination IP address 160.32.65.1
Packet A will be forwarded to interface 0, since
160.32.65.1 is in between 160.0.0.0 and
160.63.255.255.
(ii) Packet B with a destination IP address 161.130.0.5
Packet B will be forwarded to interface 3, since
161.130.0.5 is not covered by the other ranges.
(d) Consider the following scenario.
A client from eng2003.com is looking for the IP address of www.pear.hk.
Describe the corresponding DNS query process.
1) The client consults the DNS server of eng2003.com for the IP
address of www.pear.hk.
2) DNS server of eng2003.com does not have the information. It then
consults the DNS server of .hk for the IP of the DNS server of pear.hk
on behalf of the client.
3) DNS server of .hk returns the IP address of the DNS server of
pear.hk. to the DNS server of eng2003.com.
4) DNS server of eng2003.com consults the DNS server of pear.hk for
the IP of www.pear.hk on behalf of the client.
5) DNS server of pear.hk returns the IP of www.pear.hk to DNS
server of eng2003.com.
6) DNS server of eng2003.com returns the IP address of
www.pear.hk to the client.
4) Give the subnet mask for dividing Network 158.132.0.0 into 30 equal-size
subnets (excluding the all-one and all-zero subnets).
158.132.0.0 is a class B network.
Let n be the number of bits required to represent 30 subnets, we can obtain
2n-2 = 30
n=5
Therefore, the subnet mask should be 255.255.248.0/21
