Multiplexer And Stop-and-Wait ARQ Analysis

Efficiency of the System

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

1. (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

1. DO a retransmission of data in case of lost framer
   1. Sender sends an information frame to receiver.
   2. Sender waits for an ACK before sending the next frame.
   3. Receiver sends an ACK if frame is correctly received.
   4. If no ACK arrives within time-out, sender will resend the frame.
2. 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.
3. 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.

1. 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.
2. Only selected damaged frames are transmitted.
3. The sender can only send frames received by NAK
4. When a frame is corrupted during transit, NAK is returned and the frame is resent out of sequence.
5. The sender keeps all data that have not been acknowledged.
6.  Receiver must sort frames in its possession. Inserts the re-transmitted frames to their appropriate                  place.

Go-Back-n ARQ

1. The Receiver window size is 1.
2. Whenever a frame is lost or corrupted, every frame sent since the last frame acknowledged is transmitted again.
3. 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

1. (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)

1. 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;

1. 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

1. a) Maximum number of stations = 23.
2. b) New number of stations supported after reducing R by half.

=45 Stations

1. c) Total nodes supported by the accountant’s proposal.

= 23 stations

1. (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.