# COSC 1111 Data Communication And Net Centric Computing

## Question:

A multiplexer combines five 400 kbps channels A, B, C, D, E (see Figure 1 below). Each frame has 1 synchronisation bit added to the beginning of the frame and has slot size of 5 bits. Synchronous time division multiplexing will start from Channel A, then B, C, D, E and then back to channel A, then B, C, D, E and so on...

1. a)Draw a diagram to show the content of the first two frames of the output for the inputs as shown in Figure 1. In your diagram, for each frame, show all the slots containing bits of each channel and the synchronisation bit. (2 marks)(It is up to you to choose what sync bit for each frame, as long as there is 1 sync bit at the beginning of each frame. The data generated by each channel is shown in Figure 1)
2. b)What is the frame rate (frame per second) and the frame duration (in microsecond)
3. c)What is the bit rate (bps) of the MUX output link?

Use 100 kbps = 100000 bps.

(i) Suppose that frames are 600 bytes long which includes 47 bytes of overhead. Also assume that ACK frames are 78 bytes long.

1. a)The transmission uses Stop-and-Wait ARQ. Let the transmission rate of the system be R where R = 1.5 Mbps. For convenience calculation, the Processing Times at each end is 1.2 ms. Calculate the efficiency of the system if RTT takes the following values: 1.5 ms, 13 ms, 117 ms, and 1.25 seconds.
1. b)Repeat if R = 1.5 Gbps.

[Note: this question can give some very small numbers. Round your results up to 5 decimal places if possible]

1 Mbps = 1,000,000 bps

1 Gbps = 1,000,000,000 bps

Efficiency of the system is calculated using the following formula

Where TFRAME is the transmission time of a frame, TOVERHEAD is the transmission time of the frame overhead and TO is the overall time. You need to fully understand how Stop-and-Wait ARQ works to be able to determine TFRAME and TOVERHEAD and calculate the overall time TO.

(ii) Three ARQ protocols are covered in this course. Discuss how each ARQ protocol will respond when it detects a frame with errors? Explain how recovery is achieved.

(iii) In a Stop-and-Wait ARQ system, the bandwidth of the line is 512 kbps, and 1 bit takes 37 ms to make a round trip.

1. a)What is the bandwidth-delay product
1. b)If the system data frames are 128 bytes in length, what is the utilization percentage of the link as estimated from the BDP in (a).
2. c)What is the utilization percentage of the link if the link uses Go-Back- N ARQ with window size of 9.

Note:

1 kbps = 1000 bps

(i) This is an analysis question.

You are a network engineer who is setting up a highly reliable data communication network for your company. To guarantee packet delivery, you have decided that your network will use Flooding as a routing method. One technical challenge you face when using Flooding is that packets can stay circulating around in circles on your network "forever", causing congestion in your network.

To address this challenge, you include a time to live (TTL) field in each packet. This value is an integer which takes into account the number of nodes that a packet may have to pass through on the way to its destination. You are thinking about what value you should set the TTL field for each packet.

Your colleague suggests an approach to set the TTL value based on Network Diameter. It means that when calculating a routing table, you will have a list of paths from one node to every other node in the network. From the routing table of every nodes in your network, you can determine the paths from every node to every other node. From this list of paths, the path with the highest hop count will be the network diameter.

If you choose to set the TTL field of each packet to the "diameter" of the network as suggested by your colleague, then as a packet traverses the network, the TTL field is decremented after each hop until the TTL value reaches zero, then the router processing that packet will discard that packet, unless that router is the destination of the packet.

Please answer the following question: Does this approach always ensure that a packet will reach its destination if there is at least one functioning path (to the destination) exists? Why or why not.

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