paraphrase

Answer

1. The network layer is doing the service of a carrier such as the postal office, which is responsible for delivery of packages from a sender to a receiver without changing or using the contents.

2. The network layer at the source host receives the payload from the upper layer protocols, adds its header which contains the source and destination IP addresses and delivers the packet to the data-link layer.

3. At the destination the network layer receives the packet, decapsulate it and removes its header, and delivers the payload to the upper layer protocol.

4. If the packet is fragmented at the source or at routers along the path, the network layer is responsible for waiting until all fragments arrive, reassembling them, and delivering them to the upper-layer protocol.

5. The routers in the path between the source and the designation are not allowed to decapsulate the packets they received unless the packets need to be fragmented. The routers are not allowed to change source and destination addresses either.

Question Two

2 Marks

Learning Outcome(s):

Analyze & design Local and Wide Area Networks.

a) What is a transparent switch?

b) Do you have a loop problem if there are redundant transparent switches to connect LANs with broadcast domains?

c) If your answer is yes, then explain loop problem with solution.

* Do not provide any drawings in your answers. Otherwise your work is awarded a zero mark.

a) A transparent switch is a switch in which the stations are completely unaware of the switch’s existence.

b)

Yes, there will be a loop if there are redundant transparent switches

c)

Loop problem: 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.

The process for finding the spanning tree involves three steps:

1. The switch with the smallest ID is selected as the root of the tree.

2. The algorithm tries to find a path with the shortest cost from the root switch to every other switch or LAN.

3. The combination of the shortest paths creates the shortest tree .

4. Based on the spanning tree, we mark the forwarding ports which forward a frame that the switch receives and the blocking ports which block the frames received by the switch,

Question Three

1 Marks

Learning Outcome(s):

Illustrate network protocols including Transport Control Protocol / Internet Protocol.

What is the basic difference between IPv4 and ICMPv4?

IPv4 has no error-reporting or error-correcting mechanism; it lacks a mechanism for host and management queries while ICMP has error reporting mechanism. ICMP (Internet Control Messaging Protocol) is an addition to Internet Protocol to carry error, routing and control messages and data.

Question Four

2.5 Marks

Learning Outcome(s):

Demonstrate protocol configuration, network-addressing schemes and analyze packet transmission.

Below are the details of an IPv4 packet:

1. 1st row = 1st 8 bits as (01000110) 2

2. 2nd row = M bit value of “0” and Offset value of “175”

3. 3rd row = 1st 16 bits as (0000011100010001)2

Based on the above information and knowledge of IP Datagram Header Format, answer the following questions:

1. What is the “Version Number”? Give reason for your answer.

Packet version is 4

Since the left first 4 bits of the 1st row is (0100)2 that equals to 4 in decimal

2. What is the header length and how is it calculated? Is there any “option” Bytes? If yes, how many?

The header length is “0110” , it represented in 4 bytes word

The HLEN value is (0110)2 or 6 in decimal, which means the total number of bytes in the header is 6× 4, or 24 bytes. The first 20 bytes are the base header, so the options bytes = 24-20 =4 bytes

3. Which fragment is this: 1st or Last and why? What is the starting position of this fragment and why?

Given that M bit value of “0” and Offset value of “175”

If the M bit is 0, it means that there are no more fragments; the fragment is the last one

The starting position = Offset x 8 = 175 x 8 = 1400

4. How many “hops” will the “received packet” travel? Give reason for your answer.

The first eight bits of the 3rd row (The ninth byte) defines the time to live which represents the maximum number of hops (routers) visited by the datagram

The time-to-live field value is (00000111)2 = (7)10 which mean the packet can travel 7 hops before being dropped.

5. Which upper layer protocol is used? Give reason for your answer.

The last eight bits of the 3rd row represents the protocol field

The protocol field is (00010001)2 which is (17)10 in decimal and so the upper layer protocol is UDP .