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Transmission & Error control

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Presentation on theme: "Transmission & Error control"— Presentation transcript:

1 Transmission & Error control

2 Two scenarios network End to End Hop by Hop (transport protocol issue)
(datalink protocol issue)

3 Coping with rx errors Forward Error Correction Retransmission
Extra overhead capable of CORRECTING errors Large overhead depending on protection capability Retransmission Extra overhead capable of DETECTING errors Small constant overhead (2-4 bytes) Called: Frame Check Sequence Cyclic Redundance Check

4 Retransmission scenarios referred to as ARQ schemes (Automatic Retransmission reQuest)
COMPONENTS: a) error checking at receiver; b) feedback to sender; c) retx SRC DST SRC DST DATA DATA Error Check: OK Error Check: corrupted ACK NACK Automatic retransmit DATA Basic ACK idea Basic NACK idea SRC DST SRC DST SRC DST DATA DATA DATA Retx Timeout (RTO) Error Check: corrupted ACK DATA DATA DATA Basic ACK/Timeout idea

5 Sequence numbers – a must
Sender side: Receiver side: DATA DATA RTO ACK DATALINK or NETWORK (ACK lost) rtx DATA DATA New data? Old data? Need to univocally “label” all packets circulating in the network between two end points. 1 bit (0-1) enough for Stop-and-wait

6 Link-level model C bits/sec
In e2e scenario, C approximated by bottleneck link rate Stop & Wait  one frame/packet at a time Pipelining (continuous ARQ)  more than one frame/packet

7 Stop & Wait

8 stop-and-wait sender receiver One way delay MSG/C RTT e ACK/C e
For simplicity all MSG of same size; extension to != size  use mean value RTT e ACK/C e REMARK: throughput always lower than Available link rate! time time

9 Upper bound No processing e = 0 ACK size negligible ACK = 0

10 stop-and-wait: upper bound /1 MSG = 1500 bytes
Under-utilization with: 1) high capacity links, 2) large RTT links

11 stop-and-wait: upper bound /2 MSG = 1500 bytes
Under-utilization with: 1) high capacity links, 2) large RTT links

12 Dealing with errors sender receiver
IMPORTANT ISSUE: setting the Time Out right! Too short  unnecessary rtx Too short  inconsistent protocol operation Too long  waste of time Ideally: TO = RTT+ACK/C (+2 e) Easy to say, but harder to do if RTT is not known (e.g. in e2e scenario) e TO e

13 Inconsistent protocol operation
sender receiver M=1 M=2  will be NEVER received! Consequence: ACK MUST be also numbered To always guarantee consistent operation TO M=1 M=2 M=3

14 Performance with loss (upper bound)
No processing time, negligible ACK Per packet loss probability P Assumed indipendent P(immediate success) = (1-P) P(success at second tx) = P(1-P) P(success at third tx) = P2(1-P)

15 Pipelining (Continuous ARQ)

16 Pipelining idea Up to W>1 frames can be “in fly”
In fly = frames transmitted but not yet ACKed Sliding Window Size W “Slides” forward at each received ACK

17 Pipelining cases W=4 RTT (+1tx) ? W=10 time time

18 Esercizio (fatelo) MSG=500 bytes W=4 C= 2 Mbps Propagation = 16 ms
How much time to transmit 6 messages? Which message size for continuous tx?

19 Continuous transmission
Condition in which link rate is fully utilized Time to transmit W frames Time to receive Ack of first frame We may elaborate: This means that full link utilization is possible when window size (in bits) is Greater than the bandwidth (C bit/s) delay (RTT s) product!

20 Bandwidth-delay product
Network: like a pipe C [bit/s] x D [s] number of bits “flying” in the network number of bits injected in the network by the tx, before that the first bit is rxed 64Kbps A (64000x0.240) bits “worm” in the air!! bandwidth-delay product = no of bytes that saturate network pipe

21 Long Fat Networks LFNs (el-ef-an(t)s): large bandwidth-delay product
RTT (ms) rate (kbps) BxD (bytes) Ethernet 3 10.000 3.750 T1, transUS 60 1.544 11.580 T1 satellite 480 1.544 92.640 T3 transUS 60 45.000 Gigabit transUS 60

22 Throughput for pipelining MSS = 1500 bytes

23 Maximum achievable throughput (assuming infinite speed line…)
W = bytes

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