Assignment #

Deadline: Sunday 28/03/2021 @ 23:59
[Total Mark for this Assignment is 5]

Computer Networks

IT210

College of Computing and Informatics

Question One
1.25 Marks

Learning Outcome(s):
Demonstrate protocol configuration, network-addressing schemes and analyze packet transmission

Describe the complete process of how address subblocks are designed
support the description by showing the process of splitting 10.10.10.0/27 address space into two equal-sized subblocks?

Given address
10.10.10.0/27
Address: 10.10.10.0 00001010.00001010.00001010.000 00000
Netmask: 255.255.255.224 = 27 11111111.11111111.11111111.111 00000
Network: 10.10.10.0/27 00001010.00001010.00001010.000 00000 (Class A)
Broadcast: 10.10.10.31 00001010.00001010.00001010.000 11111
HostMin: 10.10.10.1 00001010.00001010.00001010.000 00001
HostMax: 10.10.10.30 00001010.00001010.00001010.000 11110
Hosts/Net: 30 (Private Internet)

Subnet address

Netmask

Range of addresses

Useable IPs

Hosts

Join

10.10.10.0/27

255.255.255.224

10.10.10.0 – 10.10.10.31

10.10.10.1 – 10.10.10.30

30

1. The number of addresses in the block N = NOT (mask) + 1.
2. The first address in the block = (Any address in the block) AND (mask).
3. The last address in the block = (Any address in the block) OR [(NOT (mask)].
4. The number of addresses in each subnetwork should be a power of 2.
5. The prefix length for each subnetwork should be found using the following formula:
nsub = 32 − log2Nsub
6. The starting address in each subnetwork should be divisible by the number of
addresses in that subnetwork.
Nsub = 16 Addresses (It’s a power of 2 so accepted)
Prefix length for each subnet:
nsub = 32 – 4 = 28
SUBNET 1
Address: 10.10.10.0 00001010.00001010.00001010.0000 0000
Netmask: 255.255.255.240 = 28 11111111.11111111.11111111.1111 0000
Wildcard: 0.0.0.15 00000000.00000000.00000000.0000 1111
=>
Network: 10.10.10.0/28 00001010.00001010.00001010.0000 0000 (Class A)
Broadcast: 10.10.10.15 00001010.00001010.00001010.0000 1111
HostMin: 10.10.10.1 00001010.00001010.00001010.0000 0001
HostMax: 10.10.10.14 00001010.00001010.00001010.0000 1110
Hosts/Net: 14 (Private Internet)
SUBNET 2
Address: 10.10.10.16 00001010.00001010.00001010.0001 0000
Netmask: 255.255.255.240 = 28 11111111.11111111.11111111.1111 0000
Wildcard: 0.0.0.15 00000000.00000000.00000000.0000 1111
=>
Network: 10.10.10.16/28 00001010.00001010.00001010.0001 0000 (Class A)
Broadcast: 10.10.10.31 00001010.00001010.00001010.0001 1111
HostMin: 10.10.10.17 00001010.00001010.00001010.0001 0001
HostMax: 10.10.10.30 00001010.00001010.00001010.0001 1110
Hosts/Net: 14 (Private Internet)

Subnet address

Netmask

Range of addresses

Useable IPs

Hosts

Join

10.10.10.0/28

255.255.255.240

10.10.10.0 – 10.10.10.15

10.10.10.1 – 10.10.10.14

14

10.10.10.16/28

255.255.255.240

10.10.10.16 – 10.10.10.31

10.10.10.17 – 10.10.10.30

14

https://www.davidc.net/sites/default/subnets/subnets.html
Question Two
1.25 Marks

Learning Outcome(s):
Explain networking principles, models and technologies.

The network’s performance can be measured in terms of latency (delay), throughput, and packet loss. Explain how each one of them can be used to measure the performance and elaborate with examples?
1. Latency
a packet, from its source to its destination, encounters delays. The delays in a network can be divided into four types: transmission delay, propagation delay, processing delay, and queuing delay.

Transmission Delay

A source host or a router cannot send a packet instantaneously. A sender needs to put the bits in a packet on the line one by one.
Delaytr = (Packet length) / (Transmission rate).

Propagation Delay

Propagation delay is the time it takes for a bit to travel from point A to point B in the transmission media. The propagation delay for a packet-switched network depends on the propagation delay of each network (LAN or WAN).
Delaypg = (Distance) / (Propagation speed).

Processing Delay

The processing delay is the time required for a router or a destination host to receive a packet from its input port, remove the header, perform an error detection procedure, and deliver the packet to the output port (in the case of a router) or deliver the packet to the upper-layer protocol (in the case of the destination host).

Queuing Delay

Queuing delay can normally happen in a router. As we discuss in the next section, a router has an input queue connected to each of its input ports to store packets waiting to be processed; the router also has an output queue connected to each of its output ports to store packets waiting to be transmitted.
Total delay = (n
+ 1) (Delaytr + Delaypg + Delaypr) + (n) (Delayqu)

Example:
A low transmission rate would lead into high transmission delay. A long distance for the message to travel would increate the propagation delay. A low resources router would drop lots of packets because of it’s small buffer and slow processing speed.
2. Throughput
Throughput at any point in a network is defined as the number of bits passing through the point in a second, which is actually the transmission rate of data at that point. In a path from source to destination, a packet may pass through several links (networks), each with a different transmission rate.
Example:
The throughput of three links is equal to their lowest as the link with the lowest throughput would act as a bottleneck for the entire transmission process.
3. Packetloss
Another issue that severely affects the performance of communication is the number of packets lost during transmission. When a router receives a packet while processing another packet, the received packet needs to be stored in the input buffer waiting for its turn. A router, however, has an input buffer with a limited size. A time may come when the buffer is full, and the next packet needs to be dropped. The effect of packet loss on the Internet network layer is that the packet needs to be resent, which in turn may create overflow and cause more packet loss.
Example:
Another scenario would be a faulty medium that corrupts messages or a low resources networking device that drops packets. The thing that will lead the same packets to show up on the network more than once due to the loss of the original copy of it.
Question Three
1.5 Marks

Learning Outcome(s):
Explain networking principles, models and technologies

Describe in your own words, the learning process of a switch table for five nodes connected to one switch.
Initalization

Mac address

Port

A sends a frame to C

Mac address

Port

Computer A mac address

1

Computer C mac address

3

B sends a frame to D

Mac address

Port

Computer A mac address

1

Computer C mac address

3

Computer B mac address

2

Computer D mac address

4

B sends a frame to E

Mac address

Port

Computer A mac address

1

Computer C mac address

3

Computer B mac address

2

Computer D mac address

4

Computer E mac address

5

Question Four
1 Mark

Learning Outcome(s):
Explain networking principles, models and technologies

Explain in your own words when the spanning-tree algorithm is used.
The reason why we use to spanning tree algorithm is to overcome the looping problem in case of the existence of more than one switch on the network. The existence of multiple switches enhances the reliability of the network yet forms the problem of looping.
Redundancy can create loops in the system, which is very undesirable. Loops can be created only when two or more broadcasting LANs (those using hubs, for example) are connected by more than one switch.
To solve the looping problem, the IEEE specification requires that switches use the spanning tree algorithm to create a loopless topology. In graph theory, a spanning tree is a graph in which there is no loop. In a switched LAN, this means creating a topology in which each LAN can be reached from any other LAN through one path only (no loop). We cannot change the physical topology of the system because of physical connections between cables and switches, but we can create a logical topology that overlays the physical one.