Network Routing and Switching

Network Routing and Switching

Question Number
Question 1: (3 marks)
a)
b)
c)
Classless address block
Address: 154.78.177.3 10011010.01001110.10110001.000 00011
Netmask: 255.255.255.224 = 27 11111111.11111111.11111111.111 00000
Wildcard: 0.0.0.31 00000000.00000000.00000000.000 11111
=>
Network: 154.78.177.0/27 10011010.01001110.10110001.000 00000 (Class B)
Broadcast: 154.78.177.31 10011010.01001110.10110001.000 11111
HostMin: 154.78.177.1 10011010.01001110.10110001.000 00001
HostMax: 154.78.177.30 10011010.01001110.10110001.000 11110
Hosts/Net: 30
Question 2: (8 marks)
a) Network Address
Address: 138.77.216.5 10001010.01001101.11011000 .00000101
Netmask: 255.255.255.0 = 24 11111111.11111111.11111111 .00000000
Wildcard: 0.0.0.255 00000000.00000000.00000000 .11111111
=>
Network: 138.77.216.0/24 10001010.01001101.11011000 .00000000 (Class B)
Broadcast: 138.77.216.255 10001010.01001101.11011000 .11111111
HostMin: 138.77.216.1 10001010.01001101.11011000 .00000001
HostMax: 138.77.216.254 10001010.01001101.11011000 .11111110
Hosts/Net: 254

b) Subnet No. addresses Subnet address Mask /n
1 32 138.77.216.6 255.255.255.255
2 32 138.77.216.7 255.255.255.255
3 16 138.77.216.8 255.255.0.0
4 16 138.77.216.9 255.255.0.0
5 16 138.77.216.10 255.255.0.0
6 16 138.77.216.11 255.255.0.0
c)
The company is granting the host address that can be used in the network. However, in the
address block, the first address will be used as the network address and special address is
assigned as the last address. The 1024 addresses are allocated from the existing block because
the company is granting host address in a block and also provide special addresses for each
subnet.Lifewire. (2018).
Question 3: (4 marks)
a)
i) Three horizontal display shows the following information's,
 Packet Lists
 Packet Details
 Packet Bytes
The Frame 2 is capturing the packet of 85 bytes. It captures the packets by using the User
datagram protocol. It has payload length of 31 and hop limit is 128. It scans the source port