Cosc 1111 Data Communication And Assessment Answers
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...
- 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)
- b)What is the frame rate (frame per second) and the frame duration (in microsecond)
- 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.
- 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.
- 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.
- a)What is the bandwidth-delay product
- 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).
- 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
Round your numerical answers to 3 decimal places where possible.
(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.
Answer:
400 kbps channels
Time slot 5 bits.
Frame rate = Number of channels/time slot
Channels = 400/5 =80kbps
80*100000=8000000
Frame rate= 8000000bps/5b= 1600000fps
= 1600000fps
Frame Duration = 1/1600000 = 0.000000625 seconds
=0.625 ms
c). MUX output bit rate = N*Maximum input Rate
5* 400=2000kbps
2000*1000= 2000000bps
=2000000bps
- (i)
Given Frame size =600 bytes
Overhead =47 bytes
ACK frame size= 78 bytes
? = Efficiency
- R = 1.5Mbps
Processing time = 1.2ms
Load size = 600+47 = 647 Bytes
Tx = Load size / R = 647*8/1.5*10^6
= 3.45 ms
Tax = ACK frame size /R
=78*8 / 1.5*10^6
=0.416 ms
When RTT = 1.5ms
Total time = Tx + RTT + Processing time + Tax
=3.45 ms + 1.5 ms + 1.2 ms + 0.416 ms
= 6.566 ms.
? = Tx/Total time
3.45/6.566
= 52.54%
When RTT= 13ms
Total time = Tx + RTT + Processing time + Tax
=3.45 ms+13 ms+1.2 ms+0.416 ms
=18.066 ms.
? = Tx/Total time
=3.45/18.066
=19.09%
When RTT= 117ms
Total time = Tx + RTT + Processing time + Tax
=3.45 ms+117 ms+1.2 ms+0.416 ms
=122.066 ms.
? = Tx/Total time
3.45/122.066
=2.83%
When RTT = 1.25 seconds
Total time = Tx + RTT + Processing time + Tax
=3.45ms +1250ms+1.2ms+0.416ms
= 1255.066 ms.
? = Tx/Total time
=3.45/1255.066
=0.275 %
- R = 1Gbps
Processing time = 1.2ms
Load size = 600+47 = 647 Bytes
Tx = Load size / R = 647*8/1*10^9
=0.0052 ms
Tax = ACK frame size/R
=78*8/ 1*10^9
=0.00062 ms
When RTT = 1.5 ms
Total time = Tx + RTT + Processing time + Tax
=0.0052+1.5+1.2+0.00062
=2.71 ms.
? = Tx/Total time
=0.0052/2.71
=0.19%
When RTT = 13ms
Total time = Tx + RTT + Processing time + Tax
=0.0052+13+1.2+0.00062
=14.21 ms
? = Tx/Total time
=0.0052/14.21
=0.036%
When RTT= 117ms
Total time = Tx + RTT + Processing time + Tax
=0.0052+117+1.2+0.00062
=118.20582 ms
? = Tx/Total time
=0.0052/118.20582
` =0.0044%
When RRT=1.25seconds
Total time = Tx + RTT + Processing time + Tax
=0.0052+1250+1.2+0.00062
=1251.20582ms
? = Tx/Total time
=0.0052/1251.20582
=0.0004%
(ii).
The Stop and Wait ARQ
- DO a retransmission of data in case of lost framer
- Sender sends an information frame to receiver.
- Sender waits for an ACK before sending the next frame.
- Receiver sends an ACK if frame is correctly received.
- If no ACK arrives within time-out, sender will resend the frame.
- When an error is found in the data frame during transit, a NAK frame is returned. These NAK frames then informs the sender to transmit again the last frame.
- Whenever there is a bidirectional communication, all the parties transmit and accept data. Any outstanding ACKs are put in the header of information frames and then piggybacking can save bandwidth since the overhead from a data frame and an ACK frame (addresses, CRC, etc) can be combined into just one frame (Malhotra, 2016).
The Selective Reject ARQ.
- This ARQ solves the problems of Go-Back-N ARQ. It does this by accepting error free frames that are out of order and only allows for the transmission of individual frames.
- Only selected damaged frames are transmitted.
- The sender can only send frames received by NAK
- When a frame is corrupted during transit, NAK is returned and the frame is resent out of sequence.
- The sender keeps all data that have not been acknowledged.
- Receiver must sort frames in its possession. Inserts the re-transmitted frames to their appropriate place.
Go-Back-n ARQ
- The Receiver window size is 1.
- Whenever a frame is lost or corrupted, every frame sent since the last frame acknowledged is transmitted again.
- When 1, 2, 3, 4 frames are sent but the sender just gets a NAK value of 3, The NAK will then ask for frame 3 and the other earlier and subsequent frames to be sent again.
Bandwidth 512kbps
1 bit takes 39ms for a round trip.
Bandwidth product delay = 512××=51200 bits
(b)
Utilization percentage=?
Data frame length- 128 bytes = 128×8= 1024 bit
Link utilization= 1024/51200=0.02 = 2% utilization.
c).
Utilization Percentage if using Go-Back-N ARQ with window size of 9.
The system sends up to 9 frames which is 9*1024 = 9216 bits
That is 9216/51200 =0.18
0.18*100 = 18%
Utilization percentage=18
- (i) The answer to this will have two assumptions, weather the diameter of the network is set as dynamic or static.
If It is set to Dynamic, then YES this flooding approach will ensure that the packet is guaranteed to reach its destination. This is because the hop count is large enough to reach the farthest node and all paths available shall be used.
If the diameter is set to static, then NO flooding will not guarantee that the packet will reach its destination since the destination is now further away from the current diameter.
(ii)
- Dijkstra’s Algorithm
The dijkstra’s algorithm computes the least cost path from a single node to all other nodes in a network. It is a link state algorithm.
The least cost paths are made known to k destinations after the k th iteration. These k paths will have the k smallest costs in least cost paths to all node destination (Murota, 2014).
|
Step |
N’ |
D(A),p(A) |
D(B),p(B) |
D(C),p(D) |
D(E),p(E) |
D(F),p(F) |
D(G),p(G) |
|
0 |
H |
∞ |
10,H |
∞ |
∞ |
∞ |
5,H |
|
1 |
HG |
14,G |
10,H |
11,G |
5,H | ||
|
2 |
HGB |
14,G |
10,H |
8,G |
10,G |
11,G |
5,H |
|
3 |
HGBC |
14,G |
10,H |
8,G |
10,G |
11,G |
5,H |
|
4 |
HGBCE |
14,G |
10,H |
8,G |
10,G |
11,G |
5,H |
|
5 |
HGBCEF |
14,G |
10,H |
8,G |
10,G |
11,G |
5,H |
Thus shortest paths are: HGA=14; HB=10; HGBC=8; HE=10; HF=11; HG=5;
- Bellman-Ford algorithm
|
Iteration |
H(Src) |
A |
B |
C |
D |
E |
F |
G |
H |
|
0 |
0 |
inf |
inf |
inf |
inf |
inf |
inf |
inf |
inf |
|
1 |
0 |
inf |
10 |
inf |
inf |
inf |
inf |
5 |
0 |
|
2 |
0 |
14 |
inf |
inf |
9 |
inf |
11 |
Inf |
0 |
|
3 |
0 |
inf |
Inf |
inf |
inf |
12 |
inf |
Inf |
0 |
|
4 |
0 |
inf |
inf |
14 |
inf |
inf |
inf |
inf |
0 |
- a) Maximum number of stations = 23.
- b) New number of stations supported after reducing R by half.
=45 Stations
- c) Total nodes supported by the accountant’s proposal.
= 23 stations
- (i).
(a). N bits of data is being transferred in k time that is set to constant. For this reason the application transfers data for a very long time. The circuit switched network is thus suitable to use for this application since it will ensure that the application runs without any interruption.
(b). When packet switched network is utilized, there won’t be need for congestion control mechanism since every link is composed of big capacities and the application can simultaneously transmit data a single or multiple links and there enough bandwidth plus low data rates. For this reason, a small queue may be formed without congestion (Walsh et al., 2014).
(ii). (a).
Constant rate or transmission rate = 32kbps
=32000bps.
The time for filling a cell is the packetilization delay
So total cell size = 8.P bits
And Packetilization delay= = 0.25P msec
Therefore Packetilization delay is given by 0.25×P.
(b). (i) For P=1500 bytes
Packetilization delay = 0.25×1500=375msec
=375msec
For P=48 bytes
Packetilization delay= 0.25×48
=12msec
(c). Transmission delay = (P.8 + 40)/R
Link rate R = 189mbps.
For P =1500
Trans delay = (1500×8+40)/189Mbps
= (12000+40)/189Mbps
=6.37 × sec
For P = 48
Transmission delay =(48×8+40)/189Mbps
=(384+40)/189Mbps
=2.243 µ sec.
(d). Using a small AT cell will ensures that packetilization delays are not greater than 20msec so that unpleasant echoes are not produced.
References
Malhotra, A., & Chitre, K. (2016, March). Performance analysis of data link layer protocols with a special emphasis on improving the performance of Stop-and-Wait-ARQ. In Computing for Sustainable Global Development (INDIACom), 2016 3rd International Conference on (pp. 593-597). IEEE.
Murota, K., & Shioura, A. (2014). Dijkstra’s algorithm and L-concave function maximization. Mathematical Programming, 145(1-2), 163-177.
Walsh, A. J., Mountjoy, J., Fagan, A., Browning, C., Ellis, A. D., & Barry, L. P. (2014, September). Reduced waiting times using a fast switching dual-polarization DDQPSK receiver in a packet switched network. In Optical Communication (ECOC), 2014 European Conference on (pp. 1-3). IEEE.
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