Data communication

1. 1
Assignment:
Subnetting - Part 1 - Solutions
Objective of the Lab:
- Carry out addressing of a Computer Network
- Understand how the messages are traversing network configuration
- Understand the influence of subnetting to the operation of a network
Consider the Network Configuration in Appendix 1.
1. Calculate the following Subnet-Addresses and Subnet-Masks with the following
assumptions:
Subnet_G contains a maximum of 8 Stations
1.1 Subnet_B = 192.168.0.248
1.2 Subnet_C = 192.168.0.128
1.3 Subnet_D = 192.168.0.64
1.4 Subnet_F = 192.168.0.192
1.5 Subnet_G = 192.168.0.224
1.6 Mask_G = 255.255.255.240
2. Please assign the following IP-Addresses:
2.1 IP0-R3 = 192.168.0.66
2.2 IP0-R0 = 192.168.0.1
2.3 IP0-R5 = 192.168.0.130
2.4 IP1-R0 = 192.168.0.250
2.5 IP0-R4 = 192.168.0.2
2.6 IPFTP-Server = 192.168.0.100
2.7 IPHTTP-Server = 192.168.0.202
Note: You can assign any IP-Address which is still available and inside the given IP-range.

2. 2
3. Please answer following questions:
3.1 What is the correct Netmask for a 32 IP address-network?
A: 255.255.255.0
B: 255.255.255.64
C: 255.255.255.128
D: 255.255.255.192
E: 255.255.255.224
F: 255.255.255.240
Answer: E
Explain your answer:
The Subnet mask 255.255.255.224 can also be written like follows:
255.255.255. 1 1 1 0 0 0 0 0
i.e. the netmask specifies an address area of 32 addresses
The other netmasks are specifying different subnet-sizes.
Note: Answer B is not a valid subnet
3.2 Which of the following IP addresses could be the broadcast address for a 32 IP
address subnetwork?
A: 193.10.2.0
B: 193.10.2.47
C: 193.10.2.79
D: 193.10.2.95
E: 193.10.2.159
F: 193.10.2.255
Answer: D, E, F
Explain your answer:
Last address of a subnet area is the broadcast address.
In that case following subnets can be considered as being created, all of them
having a /27 netmask.
D:
Subnet-Address: 193.10.2.64
Broadcast-Address:193.10.2.95
E:
Subnet-Address: 193.10.2.128
Broadcast-Address:193.10.2.159
F:
Subnet-Address: 193.10.2.224
Broadcast-Address:193.10.2.255

3. 3
3.3 Which of the following IP-addresses could be the subnet-address for a 64 IP-address
subnetwork?
A: 143.17.99.0
B: 143.17.99.32
C: 143.17.99.64
D: 143.17.99.96
E: 143.17.99.127
F: 143.17.99.192
Answer: A, C, F
Explain your answer:
A subnet can only start at a multiple of its own size.
3.4 Why can a subnet only start at subnet-address which is a multiple of its own length?
Here we have to look at the binary representation of a subnet-mask.
Let us look at a subnet-mask which creates a subnetwork with 64 IP-addresses:
255.255.255.192
is in binary
11111111 . 11111111 . 11111111 . 11000000
Here we have 26 bits which indicate the subnet-part of the mask and 6 bits which
indicate the host-part of the mask. The subnet-part tells us which subnet the IP-
address belongs to (changing a bit in the subnet-part also changes the subnet the
host is located in) and the host-part tells us the specific IP-address inside the
subnet (changing a bit in the host-part does NOT change the subnet but changes the
IP-address inside a subnet).
Since we can only use 6 bits to specify the IP-address inside the subnet we are
limited to 64 different IP-addresses (000000 = 0, 111111 = 63).

4. 4
SWITCH (Layer 2)
Sub._C = 192.168.0.???
SM_C = 255.255.255.192
PC 2
PC 1 PC n
192.168.0.190 192.168.0.129
Hugo 2
Sub_G = 192.168.0.???
SM_G = 255.255.255.???
Wireshark
Analyzer
B
Wireshark
Analyzer
C
IP1 = 192.168.0.65
IP0 = ???
IP0 = ???
IP0 = ???
Hugo 1
Wireshark
Analyzer
A
R6
Router
IP1 = ???
R2
Router
Sub_F = 192.168.0.???
SM_F = 255.255.255.224
R4
Sub_B = 192.168.0.???
SM_B = 255.255.255.252
R0
Backbone
Router
Sub_A = 192.168.0.0/26
IP0 = ???
R3
Router
HTTP
Server
FTP
Server
Sub_E = 192.168.0.96
SM_E = 255.255.255.224
Sub_D = 192.168.0.???
SM_D = 255.255.255.224
Router
R5
Router
Sub_H = 192.168.0.252
SM_H = 255.255.255.252
Appendix 1
Internet-SP
IP1 = 192.168.0.249