Outline

1
11. Physical Layer

• Outline
• Building Blocks
• Shannons Theorem
• Encoding

2
Building Blocks

• Node
• Network Adaptor (memory is the speed bottleneck)
• Router (input ports, switch fabric, output ports)
• Links
• C f ?
• Cables
• Leased Lines T1(DS1), T3(DS3),
  OC-1(STS-1),OC-3(STS-3),OC-48(STS-48,2.48Gbps)
• Last mile links modem, xDSL, ISDN, cable modem

3
Shannons Theorem

• C B log (1S/N)
• C channel capacity, in bps
• B bandwidth, in hertz
• S average signal power
• N average noise power
• dB 10 log (S/N)
• dB signal-noise ratio expressed in decibels

2
10
4
Shannons Theorem

• Example
• B 300 Hz to 3300 Hz
• N/S 30 dB
• C 3000 log (1001) 30 Kbps

2
5
Encoding

• Signals propagate over a physical medium
• modulate electromagnetic waves
• e.g., vary voltage
• Schemes
• NRZ
• NRZI
• Manchester
• 4B/5B

6
12. Data Link Layer

• Outline
• Framing
• Error Detection
• Sliding Window Algorithm
• Token rings (FDDI)
• Wireless

7
Data Link Layer

• Data Link Layer
• Framing
• Error Detection
• Sliding Window (reliable transmission, flow
  control)
• Medium Access Control (MAC) Sub-Layer
• Token ring (FDDI)
• Wireless
• Logical Link Control (LLC) IEEE802.2

8
Framing

• Break sequence of bits into a frame
• Typically implemented by network adaptor

9
Approaches

• Sentinel-based
• delineate frame with special pattern 01111110
• e.g., HDLC, SDLC, PPP
• problem special pattern appears in the payload
• solution bit stuffing
• sender insert 0 after five consecutive 1s
• receiver delete 0 that follows five consecutive
  1s

10
Approaches (cont)

• Counter-based
• include payload length in header
• e.g., DDCMP
• problem count field corrupted
• solution catch when CRC fails

11
Approaches (cont)

• Clock-based
• each frame is 125us long
• e.g., SONET Synchronous Optical Network
• STS-n (STS-1 51.84 Mbps)

12
Error Detection

• Parity
• Two-dimensional parity
• Internet Checksum
• CRC

13
Cyclic Redundancy Check (CRC)

• Add k bits of redundant data to an n-bit message
• want k ltlt n
• e.g., k 32 and n 12,000 (1500 bytes)
• Represent n-bit message as n-1 degree polynomial
• e.g., MSG10011010 as M(x) x7 x4 x3 x1
• Let k be the degree of some divisor polynomial
• e.g., C(x) x3 x2 1
• Modulo 2 operation (XOR)

14
CRC (cont)

• Transmit polynomial P(x) that is evenly divisible
  by C(x)
• shift left k bits, i.e., M(x)xk
• subtract remainder of M(x)xk / C(x) from M(x)xk
• Receiver polynomial P(x) E(x)
• E(x) 0 implies no errors
• Divide (P(x) E(x)) by C(x) remainder zero if
• E(x) was zero (no error), or
• E(x) is exactly divisible by C(x)

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Selecting C(x)

• All single-bit errors, as long as the xk and x0
  terms have non-zero coefficients.
• All double-bit errors, as long as C(x) contains a
  factor with at least three terms
• Any odd number of errors, as long as C(x)
  contains the factor (x 1)
• Any burst error (i.e., sequence of consecutive
  error bits) for which the length of the burst is
  less than k bits.
• Most burst errors of larger than k bits can also
  be detected
• See Table 2.6 on page 102 for common C(x)

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Sliding Window

• Stop and wait
• Selective repeat
• Go back N

17
Acknowledgements Timeouts
18
Stop-and-Wait
Sender
Receiver

• Two sequence numbers (0 and 1)
• Problem keeping the pipe full
• Example
• 1.5Mbps link x 45ms RTT 67.5Kb (8KB)
• 1KB frames imples 1/8th link utilization

19
Sliding Window

• Allow multiple outstanding (un-ACKed) frames
• Upper bound on un-ACKed frames, called window

20
SW Sender

• Assign sequence number to each frame (SeqNum)
• Maintain three state variables
• send window size (SWS)
• last acknowledgment received (LAR)
• last frame sent (LFS)
• Maintain invariant LFS - LAR lt SWS
• Advance LAR when ACK arrives
• Buffer up to SWS frames

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SW Receiver

• Maintain three state variables
• receive window size (RWS)
• largest frame acceptable (LFA)
• last frame received (LFR)
• Maintain invariant LFA - LFR lt RWS
• Frame SeqNum arrives
• if LFR lt SeqNum lt LFA accept
• if SeqNum lt LFR or SeqNum gt LFA
  discarded
• Send cumulative ACKs

RWS


LFR
LFA
22
Sequence Number Space

• SeqNum field is finite sequence numbers wrap
  around
• Sequence number space must be larger then number
  of outstanding frames
• SWS lt MaxSeqNum-1 is not sufficient
• suppose 3-bit SeqNum field (0..7)
• SWSRWS7
• sender transmit frames 0..6
• arrive successfully, but ACKs lost
• sender retransmits 0..6
• receiver expecting 7, 0..5, but receives second
  incarnation of 0..5
• SWS lt (MaxSeqNum1)/2 is correct rule
• Intuitively, SeqNum slides between two halves
  of sequence number space

23
Selective Repeat

• At the receiver, it will send an ack for each
  frame received, even if the frame is out of
  order.
• The sender sets a timeout value for each frame
  sent out, and will resend the frame upon timeout.
  The timeout flag is cleared when an ack for the
  frame is received.
• Only those frames (or acks for them) are lost are
  retransmitted.
• The receiver may need a large buffer to hold out
  of order frames.

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Go Back N

• At the receiver, it will send an ack for each
  frame received, only if the frame is out of
  order. Out of order frames are discarded. It also
  sends a NAK when it finds a frame is lost.
• The timer mechanism is the same.
• The acks are cumulative in the sense that an ack
  indicates that the receiver has got all the
  frames up to the frame acked.
• The sender will resend all the frames starting
  from the frame for which a NAK is received.
• The receiver does not need a large buffer to hold
  out of order frames.