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Spring 2001 CS 585 1
OutlineBuilding BlocksShannon’s TheoremEncoding
11. Physical Layer
Spring 2001 CS 585 2
Building Blocks
• Node– Network Adaptor (memory is the speed bottleneck)
– Router (input ports, switch fabric, output ports)
• Links– C = f λ
– 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
Spring 2001 CS 585 3
Shannon’s Theorem
• C = B log (1+S/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
Spring 2001 CS 585 4
Shannon’s Theorem
• Example:– B : 300 Hz to 3300 Hz
– N/S = 30 dB
– C = 3000 log (1001) = 30 Kbps2
Spring 2001 CS 585 5
Encoding
• Signals propagate over a physical medium– modulate electromagnetic waves
– e.g., vary voltage
• Schemes– NRZ
– NRZI
– Manchester
– 4B/5B
Spring 2001 CS 585 6
OutlineFramingError DetectionSliding Window AlgorithmToken rings (FDDI)Wireless
12. Data Link Layer
Spring 2001 CS 585 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
Spring 2001 CS 585 8
Framing
• Break sequence of bits into a frame• Typically implemented by network adaptor
Frames
BitsAdaptor Adaptor Node BNode A
Spring 2001 CS 585 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
Header Body
8 16 16 8
CRCBeginningsequence
Endingsequence
Spring 2001 CS 585 10
Approaches (cont)
• Counter-based– include payload length in header– e.g., DDCMP
– problem: count field corrupted– solution: catch when CRC fails
SY
N
Header Body
8 8 4214 168
SY
N
Cla
ss CRCCount
Spring 2001 CS 585 11
Approaches (cont)
• Clock-based– each frame is 125us long– e.g., SONET: Synchronous Optical Network– STS-n (STS-1 = 51.84 Mbps)
Overhead Payload
90 columns
9 rows
STS-1Hdr STS-1Hdr STS-1Hdr
STS-3cHdr
Spring 2001 CS 585 12
Error Detection
• Parity• Two-dimensional parity• Internet Checksum• CRC
Spring 2001 CS 585 13
Cyclic Redundancy Check (CRC)
• Add k bits of redundant data to an n-bit message– want k << n
– e.g., k = 32 and n = 12,000 (1500 bytes)
• Represent n-bit message as n-1 degree polynomial– e.g., MSG=10011010 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)
Spring 2001 CS 585 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)
Spring 2001 CS 585 15
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)
Spring 2001 CS 585 16
Sliding Window
• Stop and wait• Selective repeat• Go back N
Spring 2001 CS 585 17
Acknowledgements & TimeoutsSender Receiver
Frame
ACK
Tim
eout
Tim
e
Sender Receiver
Frame
ACK
Tim
eout
Frame
ACKTim
eout
Sender Receiver
Frame
ACKTim
eout
Frame
ACKTim
eout
Sender Receiver
Frame
Tim
eout
Frame
ACKTim
eout
(a) (c)
(b) (d)
Spring 2001 CS 585 18
Stop-and-Wait
• 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
Sender Receiver
Spring 2001 CS 585 19
Sliding Window• Allow multiple outstanding (un-ACKed) frames• Upper bound on un-ACKed frames, called window
Sender Receiver
Tim
e
……
Spring 2001 CS 585 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 <= SWS
• Advance LAR when ACK arrives • Buffer up to SWS frames
SWS
LAR LFS
… …
Spring 2001 CS 585 21
SW: Receiver• Maintain three state variables
– receive window size (RWS)– largest frame acceptable (LFA)– last frame received (LFR)
• Maintain invariant: LFA - LFR <= RWS
• Frame SeqNum arrives:– if LFR < SeqNum < = LFA accept– if SeqNum < = LFR or SeqNum > LFA discarded
• Send cumulative ACKs
RWS
LFR LFA
… …
Spring 2001 CS 585 22
Sequence Number Space• SeqNum field is finite; sequence numbers wrap around• Sequence number space must be larger then number of
outstanding frames• SWS <= MaxSeqNum-1 is not sufficient
– suppose 3-bit SeqNum field (0..7)– SWS=RWS=7– 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 < (MaxSeqNum+1)/2 is correct rule• Intuitively, SeqNum “slides” between two halves of sequence
number space
Spring 2001 CS 585 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.
Spring 2001 CS 585 24
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.