Chapter 2: Direct Link Networks (continued)

1. Chapter 2: Direct Link Networks (continued) Slide Set 5

2. Reminder + Announcements • Quiz January 27th. • Chapters 1 and 2.1 to 2.6 (Upto this set of slides) • Bring Scantron sheets, calculators

3. In this set.. • Reliable Transmissions -- retransmissions. • Ethernet

4. Why retransmissions ? • Error correction although feasible, is not enough to handle all kinds of errors -- especially burst errors. • Corrupt frames cannot be deciphered and are therefore dropped. • Retransmissions needed to provide reliability.

5. ACKs and Time-outs • When frames are sent piggyback an acknowledgement (ACK) for received packets onto sent packets. • If no ACK received up to a preset time-out, resend frame. • Called ARQ -- Automatic Repeat request. 1 Ack 1 100

6. Stop and Wait • Allow only one outstanding packet at any given time. • If ACK not received within time-out, send again.

7. How efficient is Stop and Wait ? • Consider a 1.5 Mbps link with 45 ms RTT. • BW - Delay product = 67.5 Kb = 8KB. • You can fill 8 Kilo bytes of data prior to receiving an ACK. • However, if your frame size is 1 KB, you are using only 1/8 of the capacity. • Inefficient.

8. Sliding Window • What we really like is that the 9th frame be transmitted when ACK for the first frame arrives :). • Say we do this -- A window of packets sent -- as ACKs are received window slides i.e., more packets sent. • Now what do we need in addition ? • Need to know which packets have been received and which have not. • Packets labeled using sequence numbers.

9. Some definitions • We have a window of packets sent -- Send Window Size or SWS. • Last Acknowledgement received is denoted LAR. • LFS represents the last frame sent. • NOTE: LFS - LAR <= SWS.

10. < SWS ─ ■ ■ ■ ■ ■ ■ LAR LFS Sender Functions • When an ACK is received, the LAR moves to the right. • This allows for the transmission of an additional frame.

11. < ─ RWS ■ ■ ■ ■ ■ ■ LFR LAF Receiver functions Notation: RWS -- Receive Window Size LAF -- Largest Acceptable Frame. LFR -- Last Frame Received. • When frame with Seq_Num arrives: • If Seq_Num <= LFR or Seq Num > LAF, discard. • Frame is outside window. • If LFR < Seq_Num < LAF, accept frame. • Frame within window. • Let Seq_Num_to_Ack be the largest sequence number yet to be acked. This implies, all frames <= Seq_Num_to_Ack have been received. • LFR = Seq_Num_to_Ack • Adjust LAF = LFR + RWS.

12. An Example • Let LFR =5 and RWS = 4. • This implies LAF = 9 • If packets 7 and 8 arrive (not 6), they are buffered. • Note that they are out of order. • Typically, Receiver will resend an ACk for packet 5. • When 6 arrives, it can cumulatively ACK all buffered packets i.e., it ACKs 8 and moves LFR to 8 and LAF to 12.

13. Other possibilities • Send NAK (negative acknowledgement) for lost packets -- example for 6, when 7 is received. • Duplicate ACKs -- send an ACK for 5 again when 7 is received to trigger retransmission of 6. • Selective ACKs : Explicitly ACK frames that are received -- more complex.

14. Setting the Window Size • SWS -- Set considering the BW product. • RWS -- Receiver can set it to something appropriate -- may depend on buffering resources. • If RWS = 1 what happens to out of order frames ?

15. Sequence number wrapping • Sequence numbers are finite -- thus there is a need to reuse -- called wrap around. • What is the relationship between the SWS and MaxSeqNum ?

16. Maximum Sequence Number and SWS • Should it not be SWS <= MaxSeqNum + 1 ? • Let us consider an example: • Sender has eight Seq Nums from 0 to 7. • SWS = RWS = 7. • Sender transmits frames 0-6 • Receiver gets them, ACKs but ACKs get lost. • Receiver expects 7 and next 0...5. But sender sends the previous 0..5. • When the receiver gets these, he cannot distinguish. • Thus, when RWS = SWS, SWS < (MaxSeqNum+1)/2. (Verify that system works).

17. Sequence Numbers and SWS • For other cases i.e., when RWS is not equal to SWS, other rules may apply. • Depends on the specific case. • A different way with TCP -- we will see later. • Easy solution -- large Sequence number space.

18. Flow Control + • Feedback control that allows the receiver to throttle the sender. • Informs sender not only about what frames it has received but also how many more frames it can receive. • Need this in order to ensure that receiver buffer does not overflow. • Read rest of the parts on Sliding Window -- implementation etc.

19. Multiple Access • Nodes send and receive frames over a shared “direct link” network. • One way -- contention based i.e., nodes contend to send -- example Ethernet. • A second way is scheduled access -- everyone knows when and how to send -- i.e., they take turns -- example Token Ring. • Decentralized operations needed.

20. CSMA - CD • Stands for Carrier Sense Multiple Access with Collision Detection. • Carrier sensing -- nodes can distinguish between an idle and a busy line. • Collision detection -- nodes can detect when a transmission is “interfered” with. • CSMA-CD can work on a bus architecture as an example. • Forms the basis for Ethernet.

21. The Ethernet Standard • Specified by IEEE. • Referred to as the 802.3 standard. • Initially started out as a standard for a bus topology -- 10 Mbps Ethernet. • Today, we have switched Ethernet -- Fast Ethernet with 100 Mbps and Gigabit Ethernet with 1000 Mbps.

22. Ethernet on a Coax • We consider the 10 Mbps Ethernet. • Ethernet implemented on a coax cable of up to 500m. • Hosts tap into this cable and must be at least 2.5 meters apart. • Transceiver which is attached detects idle line and drives signal.

23. É É É É Repeater Host Repeaters • Multiple Ethernet segments can be joined using “repeaters”. • Repeaters amplify and forward signals. • No more than four repeaters between any two hosts. • This ensures that maximum reach of Ethernet = 2500 meters. • We will see why this is needed later.

24. Ethernet -- Signal broadcasts • Signal broadcast over the multiple-access link • Broadcast means, sent to everyone. • Each node propagates signal in both directions. • Repeaters forward on all outgoing segments. • Note: Ethernet uses Manchester encoding.

25. Cables • Use of a 10base2 cable -- 10 Mbps, base --> it is baseband -- frequency shifting -- No more than 200 m (actually only 185 m). • 10base T -- T stands for twisted pair.

26. Ethernet Frame • Nuggets: (Other details refer book) • 64 bit preamble for synchronization. • 48 bit hardware or Ethernet address. • Ethernet frame size: up to 1500 bytes of data and “at least” 46 bytes --> padding may be necessary. • Note -- Body carries IP datagram -- length of IP header specified packet size -- can determine padding.

27. Ethernet Address • Ethernet address unique for every host. • Technically address belongs to adaptor and not host -- burned into ROM. • 6 bytes -- represented as six numbers separated by colons: 8:0:2b:e4:b:1:2. • Each number corresponds to a pair of hexadecimal units -- one for each of the nibbles in the byte.

28. Addressing (cont) • Uniqueness: Each manufacturer (e.g. AMD) is allocated a different prefix -- 8:0:2b --> 24 bits. • Every frame transmitted on the Ethernet is received by every adaptor on the network -- the address is recognized and only the particular adaptor recovers frame. • All 1’s implies broadcast -- every adaptor picks it up. • An address that has the 1st bit set to 1 but is not the broadcast address is used for multicast.

29. Transmitter Algorithm • Intelligence mainly on transmitter side -- receiver dumb. • Algorithm: • If line idle, transmit frame immediately; no negotiation with other adaptors. • If line busy, wait for the line to be idle and either • Transmit immediately (1-persistent version) • Transmit with a probability p (p-persistent version). With a probability q=1-p, defer the transmission for a later time.

30. Why p-persistency? • Deference is needed since multiple transmitters may be waiting to transmit. • If for example, p = 0.25, one could have up to 4 transmitters on average -- only one would transmit when the line is free. • If multiple transmitters send simultaneously, a collision will result.

31. Waiting time • Each node looks at a “slot” time. • Each slot is of a duration equal to that of a frame. • When node has something to send, in each idle slot, it sends with a probability p, and defers to the next slot with a probability q and so on. • Just a note: 1-persistency is good if nodes are lightly loaded -- has been found effective.

32. Upon Collision Detection • When an adaptor detects a collision, it transmits a 32 bit jamming sequence and stops transmission. • Minimal amount of data sent 64 bit preamble + 32 jamming sequence = 96 bits. • How does a node detect a collision ? As nodes send, they monitor the line for other transmissions.

33. Runt frame • If two hosts that simultaneously transmit are very close to each other, they can detect a collision quickly. • In this case, enough only the 96 bits are sent -- such a frame is called a runt frame.

34. Effects of distance on CD • What if nodes are far apart ? Let us consider the case where propagation time is d. • Let A transmit at time t. • A’s transmission jams that of node B which just begins transmission at “t+d”. • It then sends out a jamming sequence -- this takes another d seconds to reach A. • Thus, A knows of the collision at “t+2d”. • A has to be transmitting at this time in order to realize the collision i.e., it has to be transmitting for 2d seconds. • Note that 2d = RTT. • So transmission time should be at least equal to RTT.

35. Transmission time and RTT • To repeat : Transmission time at least equal to RTT. • Now since RTT is at most 51.2 ms given that maximum reach is 2500 meters. • Thus, Transmission time at least 51.2 ms. On a 10 Mbps link, transmitter needs to send at least 512 bits. • Thus, we have a minimum size of 46 bytes for the body (header makes frame 512 bits).

36. Ethernet back-off • Once adaptor detects collision, it waits before trying again. • First delay -- selected randomly from the pair 0, (21-1)51.2 ms = {0, 51.2} ms. • If a second collision is experienced, choose back-off time from 0, (22-1)51.2 ms i.e, from {0, 51.2, 102.4 and 153.6} ms. • Thus, after N collisions choose a back-off time from {0.. (2N-1)51.2} ms. • In reality, Ethernet gives up after a maximum number of attempts which is 16. However, maximum value of N is fixed; typically N = 10.

37. Key properties • Ethernet works best when loads are light -- small number of active nodes. • Typically utilization is less than 30 %. • Easy to administer and maintain; inexpensive. • Currently, switched Ethernet.

38. Next time • Token Ring, • FDDI, • Wireless LANs.