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WK 13. COE 341: Data & Computer Communications Dr. Radwan E. Abdel-Aal. Chapter 7: Data Link Control Protocols. Where are we:. Chapter 7: Data Link: Flow and Error control. Data Link. Chapter 8: Improved utilization: Multiplexing. Physical Layer.
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WK 13 COE 341: Data & Computer Communications Dr. Radwan E. Abdel-Aal Chapter 7: Data Link Control Protocols
Where are we: Chapter 7: Data Link: Flow and Error control Data Link Chapter 8: Improved utilization: Multiplexing Physical Layer Chapter 6: Data Communication: Synchronization, Error detection and correction Chapter 4: Transmission Media Transmission Medium Chapter 5: Encoding: From data to signals Chapter 3: Signals and their transmission over media, Impairments
Contents • Flow Control • Stop-and-Wait flow control • Sliding-Window flow control • Error Control • Stop-and-Wait ARQ • Sliding-Window ARQ - Go-Back-N ARQ - Selective-Reject ARQ • High-Level Data Link (HDLC) Protocol • Basic Characteristics • Frame Structure • Operation
What is Data Link Control? • The logic or procedures used to convert the raw stream of bits handled by the physical layer into a “reliable” data link • Performed by the Data Link Control Protocol (Layer) • Requirements and Objectives: • Frame-level synchronization: Recognize frame start and end • Flow control: Regulate sending of frames to match the ability of RX to absorb them • Error control: Retransmission of damaged or unacknowledged frames • Addressing: Identify stations on a multipoint link • Allow control information to go with data on same link • Link management: To initiate, maintain, and terminate data exchange
Flow Control • Required to avoid the TX overwhelming the RX by the flow of data it sends • RX does not‘absorb’ the received data instantly! • It buffers (temporarily stores) the data it receives in a finite-size buffer to do some processing before sending it upward to higher layers • Without flow control, the RX buffer may overflow and data gets lost…
Flow Control over a link (assume no error) • For now, assume: • No frames lost (loss over a single link is a kind of error in recognizing the frame start) • No frames arrive in error • Frames arrive in the same order they were sent, after a propagation delay
Model for Frame Transmission over a link Frame lost: (error in start flag) Do you allow only one Or multiple frames to travel on the link at any given time Frame damaged: (error in data)
Stop and Wait • Frames sent and acknowledged one at a time: • Source transmits frame and waits for ACK • Destination receives frame and replies with an acknowledgement ACK • When source gets ACK, it sends next frame • Disadvantages: • Destination can stop the flow by not sending ACK (but we can use timeout to overcome this) • Not efficient - wastes time in waiting until short data frames arrive on long links (frame does not ‘fill’ the link) i.e. I would like to make frames long in time (large in size for a given data rate) Tf = LTb = L/R
Data Fragmentation (smaller frames!) • However, large blocks of data are often split into several smaller frames- Why? • Limited frame buffer size at RX • To reduce frame error rate: remember? FER = 1-(1-BER)F • Errors are detected sooner (when frame arrives) • On error, we need to retransmit a smaller amount of data • On a shared medium, e.g. a LAN, this ensures that a transmitting station does not occupy the medium for a long time • “Stop and wait” is inadequate in such situations where frames are “short” • Link utilization depends on the frame length in time relative to the link propagation time.
Stop and Wait Link Utilization: The ‘a’ ratio • Total number of bits in a frame = L bits, Tb = bit duration • R = Data rate, bps • Frame transmission time, tf sec: Time taken by the TX to emit all the frame bits into medium tf is large for large frames and low data rates • d = Link physical length, m • V = Velocity of propagation over link, m/s • Link Propagation time, tp sec: Time for a bit to traverse the link (link length in time) • Link length in bits, B bits/link: The number of bits that ‘fill’ the link if data is transmitted continuously • a = Propagation time / Transmission time = tp/tf Smaller ‘a’ means better link utilization (with large frames) If tf = 1, a = Propagation time tp
Stop and Wait Link Utilization (Efficiency) • Let us define link utilization, U, as: • Will demonstrate on next slide that U is given by: • Where ‘a’ is • Utilization is high for small a: U approaches 1, i.e. 100% efficiency • Shorter links, Higher propagation velocities • Larger frames sizes, Lower data rates • Efficiency is poor for large a: • Longer links, Lower propagation velocities • Smaller frames sizes, Higher data rates
Stop and Wait Link Utilization: 2 cases: Link is longer than frame: tp > tf: a > 1 Link is shorter than frame: tp < tf: a < 1 tf = 1 tp = a Let frame transmission time tf = 1 a = Propagation time = tp Link ‘full’ most of the time Better utilized Link ‘empty’ most of the time ! Underutilized a < 1 a > 1 + Overhead of an ACK frame in both cases ! tp > tf tp < tf Useful TX time Elapsed time
Satellite link between 2 ground stations • d = 2 x 36,000 km, Data rate, R = 1 Mbps • Typical wave velocity, V = 3 x 108 m/s • Frame TX time, tf = L/R = 8000/(1x106) = 8 ms • Propagation time, tp = d/V = 2x36x106/(3x108) = 240 ms • End of first frame reaches RX after 8+240 = 248 ms from start • ACK takes 240 ms more to reach TX, i.e. it starts sending 2nd frame after 488 ms • Utilization = 8/488 = 1.6% (=1/(2a+1)) Stop and Wait Efficiency: Example • Compare the efficiency of stop-and-wait flow control for two links using the parameter ‘a’: • Fame size, L = 1000 characters of 8 bits each, = 8000 bits Link is shorter than frame: a < 1 Link is longer than frame: a > 1 • 200-m optical fiber link • Data rate, R = 1 Gbps • Typical wave velocity, V = 2 x 108 m/s • Frame TX time, tf = L/R = 8000/(1x109) = 8ms • Propagation time, tp = d/V = 200/(2x108) = 1 ms • End of first frame reaches RX after 8+1 = 9 ms from start • ACK takes 1 ms more to reach TX, i.e. it starts sending 2nd frame after 10ms • Utilization = 8/10 = 80% (=1/(2a+1))
Sliding Windows Flow Control • Avoids the low efficiency of Stop-and-wait when a > 1 • Allows multiple frames to be “in transit” simultaneously on the link • RX keeps a buffer store (in memory) for W frames • So, TX can send up to W frames without waiting for ACK • Each frame carries a sequence number • ACK from RX shows the number of next expected frame • TX keeps a list of frames it can send • RX keeps a list of frames it expects to receive • These lists form sliding windows at TX and RX that shrink/expand as frames are sent, and ACKs are sent/received • Hence, Sliding Windows Flow Control
k = 3 bits Frame Sequence Numbering • Frame sequence number is limited by the size of a corresponding field in the frame, e.g. k bits • Frames are numbered modulo 2k • e.g. for k = 3, frame sequence # is modulo 23 = 8, i.e.= 0,1,..., 7, 0,1,… • Window size (W) is limited to a maximum of 2k-1 i.e. Wmax = 7 in the above example Frame Sequence Number FCS HDLC Frame
Sliding Window Send/Receive Cycle Window covers frames to be received Window covers frames to be sent RX TX Flexible- (Not rigid) Windows When you want to ACK some frames: 1. Delete ACKed frames from buffer 2. Expand Window to receive more 3. Send Acknowledgement 1. Delete ACKed frames from buffer 2. Expand Window to send more ACKs When you want to send more: Send frames Shrink window past received frames Frames Receive Frames Shrink window past sent frames Acknowledging frames is a separate issue from receiving them
Send/Receive Frames Receive/Send ACKs Sliding-Window Diagram FCS Deletion Marker k = 3 bits, W = 7 At TX Window covers frames to be sent Expand Delete Shrink W = 7 Max # of frames TXed without ACKed Remove ACKed frames from buffer At RX Window covers frames to be received
Example Sliding Window Max window size of 7 RX TX Shrink RR = Receiver Ready Expand Delete Received up to 2 Ready to receive 7 frames starting with 3 Received up to 3 Ready to receive 7 frames starting with 4
Sliding Window Enhancements • RR n: Positive receive ACK that asks for more (received up to frame n-1 and ready for n) • Receiver can also acknowledge receiving frames without permitting further transmission: (Receive Not Ready RNR) • Example RNR 5: “Received frames up to 4, but not ready for 5 and beyond yet” • When it becomes ready, RX must send a normal acknowledge (RR 5) later for TX to resume sending frames
Sliding Window in a Duplex System • In a Duplex System, destination also transmits data back to source • Piggybacking: Utilizing data frames from destination to carry ACK signals back to source to improve channel utilization • Additional field in the data frame for use only by +ive (RR) ACK • If you have no data to send now or your ACK is not RR, use a normal (dedicated) ACK frame (e.g. RR or RNR) • If data is to be sent but no acknowledgement needed, insert the last acknowledgement number to prevent RX from using the number existing in the ACK field of the data frame. (When RX station receives aduplicate ACK, it ignores it)
Sliding Window Protocol: Efficiency • Much more efficient than Stop and wait for a>1 • Treats link as a pipeline to be filled with several frames in transit simultaneously- not just one by one • With window size W and assuming no error, link utilization, U, is given by (Appendix 7A) where a = Propagation time/Frame transmission time = tp/tf • i.e. Sliding window protocol can achieve 100% utilization for W (2a + 1). • The smaller the W needed for this the better! (Why?). This requires a small a(so small a is still advantageous!)
Sliding Window Efficiency: Example Shorter links are better (small a) • Compare the efficiency of Sliding Window flow control for two links using the parameter ‘a’: • Fame size, L = 1000 characters of 8 bits each, = 8000 bits • Satellite link between 2 ground stations • d = 2 x 36,000 km, Data rate, R = 1 Mbps • Typical wave velocity, V = 3 x 108 m/s • Frame TX time, tf = L/R = 8 ms • Propagation time, tp = d/V = 240 ms • a = tp / tf = 30 • 100 % link utilization is achieved with window size W: W (2 a+1) (2 x 30 +1) 61 W = 61, k = 6 bit (Large window, large buffers at TX, RX) • For k = 3 bits, W = 7: Utilization U = W/(2a+1) = 7/(61) = 11.5% > 1.6% for Stop and wait. • 200-m optical fiber link • Data rate, R = 1 Gbps • Typical wave velocity, V = 2 x 108 m/s • Frame TX time, tf = L/R = 8ms • Propagation time, tp = d/V = 1 ms • a = tp / tf = 0.125 • 100 % link utilization is achieved with window size W: W (2 a+1) (2 x 0.125 +1) 1.25 i.e. W = 2 (A window of just 2 frames!) - easily achieved in practice!
Error Control WK 14 • Use of retransmission to handle errors detected in frames (Backward Error Handling) • This process is called Automatic Repeat Request (ARQ) • Types of Problems: • Damaged frames (Frame arrives at RX but in error) • Lost frames (Noise burst damages frame header beyond recognition- so not recognized by RX) • For connections across a network, frames arriving too late e.g. due to network congestion- Will be ignored or dropped by time-out.
Error Control Techniques: • Apply error check mechanism (chapter 6) • Send Positive acknowledgment: (for one or more frames) From RX for Error-free frames, e.g. RR i • Send Negative acknowledgement requesting retransmission of a lost or damaged frame: RX sends negative ACK for damaged or lost frames, requesting retransmission, e.g. REJ i • How does RX detect a “lost” frame? Through receiving the next frame “out of sequence” – Unexpected (frames are numbered!) • Retransmission after timeout: TX automatically retransmits a frame that has not been acknowledged following a predetermined time-out interval
Categories of Error Control Mechanisms Main types of ARQ-based standard error control mechanisms
Stop and Wait ARQ: Possible ScenariosScenario for Damaged/Lost Frame • TX transmits a single frame (keeping a copy) • Then waits for ACK from RX • If frame reaches RX damaged (in error): • RX discovers this through error detection • It then discards the frame, and does not send ACK • TX “times out” on waiting for ACK • … and then retransmits the frame again automatically • RX thus receives only one correct copy of the frame • From the RX side, this is identical to a lost frame scenario
Stop and Wait ARQ: Scenario for Lost ACK • ACK from RX for a correct frame is lost (reaches TX damaged beyond recognition): • TX will timeout and retransmit the same frame again! • RX gets two good copies of that frame! • Without numbering the frames, RX will consider both copies as two different valid frames (but data duplication is not “reliable” data transport!) • To avoid this, TX labels frames alternately as 0, 1 (enough for Stop and Wait) duplication detected at RX • RX uses ACK0 & ACK1, Similar to sliding window RRn: • ACK0: Received 1 and ready for 0 (better named RR 0) • ACK1: Received 0 and ready for 1 (better named RR 1)
Stop and Wait ARQ = RR 1 ACK = RR 0 Lost Frame Scenario Same scenario if F0 was received damaged (in error) but RX kept quiet about it! Lost ACK Scenario RX gets two good copies of F1. Labeling frames allows RX to detect this and discard one of them.
Stop and Wait - Pros and Cons • Simple • Inefficient (As seen with flow control) • For improved efficiency, we use sliding-window based ARQ (Continuous ARQ)
Sliding Window ARQ • Improves line utilization by sending up to W frames before worrying about ACK • A form of Pipelining (several tasks started before 1st task is finished) • TX uses a window to mark frames to be transmitted until they are sent and acknowledged • The window size W should be ≤ 2k – 1, k is the size of the frame sequence field in the frame header, • Frame are given sequence numbers modulo 2k, i.e. for k = 3: 0,1,2,3,4,5,6,7,0,1,2,….. • W is fixed in this protocol, but may be variable in others • Each time a proper ACK is received for a number of frames the TX window slides past them, hence the name Sliding Window. This: • Releases those frames for deletion from TX buffer memory • Introduces new frames for transmission
TX Sliding Window - Window now has a fixed width (W) and slides rigidly as one piece upon receiving ACKs. - Within the window, frame sending is handled using a send (S) pointer Frames already sent but not yet ACKed Frames that can be sent S Pointer to next frame to be sent F0, F1 Positively ACKed Slide the window as a whole (Important: Pointer is not pushed with window)
RX Sliding Window Error Control • Size of the RX window for error control is always 1 • The RX window contains the sequence number for the frame expected to be received next • If a different frame arrives (i.e. out of sequence arrival), it is immediately discarded and the window does not slide • Once the expected frame arrives correctly, the window slides one step to point to the next expected frame
RX Sliding Window F0 now expected F0 Received Correctly F1 now expected
Sliding Window ARQ: Summary • Uses sliding windows (now rigid- moves as a whole) at TX and RX to track frame movement • TX uses timeout on waiting for ACK • If no error: RX acknowledges with RR i, where i is number of the next frame expected As you receive expected frame without error k As you receive ACKs Frames received correctly and have been or will soon be acknowledged Frames waiting to be received Frames received correctly. Have or will soon be acknowledged As you transmit Expected next frame to be received Next frame to be sent Size = 1 Size ≤ 2k-1
Sliding Window ARQ: • Two main standard approaches: • Go Back N • Selective Reject
Sliding Window ARQ: Go Back N • Error Scenarios: Will consider the following error scenarios • Damaged Frame • Lost Frame • Lost ACK • Lost Positive ACK (RR) • Lost Negative ACK (REJ)
Go Back N ARQ: Error Scenarios • Damaged Frame: RX received frame i damaged • RX discards frame i and all subsequent frames until frame i is received correctly • RX either: • Scenario 1.A: Sends a negative ACK (REJ i ) • TX must go back(hence the name go back N) and retransmit that frame and allsubsequent frames that were transmitted in the mean time • Scenario 1.B: Does not send REJ (relies on TX time out) Handled as a lost frame (next) (and also as a lost negative ACK)
Go Back N ARQ: Error Scenarios • Lost Frame: • RX expects frame i but does not get it, • So TX does not get any ACK for it • Scenario 2.A: TX can send more frames • TX carries on sending subsequent frames, i+1, … • RX gets frame (i+1)out of sequence (as it did not get frame i). This allows RX to detect the problem RX then either: • Sends REJ i (Note i, not i+1) to TX • Or: Takes no further action (relies on TX time out) • In both cases, TX goes back and retransmit frame i and allsubsequent frames transmitted in the mean time
Go Back N ARQ: Error Scenarios • Lost Frame, i: TX kept sending, RX does nothing, TX times out LHS of TX window: Frames sent but not yet ACKed RX frame pointer does not leave ‘2’ until F2 is correctly received RR 2 Slide W to Uncover F0, F1 F3 Arriving out of sequence (Not the next expected frame ), RX ignored it, but did nothing Go back 1 On F2 S points to next frame to be TXed R points to next frame to be RXed
Go Back N ARQ: Error Scenarios Lost Frame, Contd. • Scenario 2.B: • TX does not send further frames after i, and RX does not send any RR or REJ (RX could not detect the problem!. It does not know that a frame was sent and got lost) • TX times out waiting for RX response • Scenario B-1: • TX sends a polling command (RR with P bit = 1) to force RX to report its receiving status by sending RR i • When TX receives the RR i response, it retransmits frame I and all subsequently sent frames • Scenario B-2: • TX retransmits frame i after timeout
Go Back N ARQ: Error Scenarios • Lost Positive ACK (RR) from RX: • RX gets frame i OK and sends a positive ACK: RR (i+1) • This RR frame is lost on its way to TX • Two Scenarios: • Scenario 3.A: • A later Acknowledgement from RX, e.g. RR (i+n) manages to reach TX before TX times out. This solves the problem, since ACKs are cumulative • Scenario 3.B: • TX times out before receiving any subsequent RR acks: • TX sends a polling command (RR with P bit = 1) to force RX to report its receiving status by sending RR i. • TX Repeats step above a few times, if no success it initiates a Reset procedure
Go Back N ARQ: Error Scenarios • Lost Negative ACK (REJ) from RX: Similar to RX not sending REJ i.e. Scenario 1.B for Damaged Frame.
Transmitter Receiver Go back 3 (4, 5, 6, 7) and Retransmit them Go Back N Example TX does not wait Got 0, 1 Ready for 2 Frame 4 lost on the way Got 3 Ready for 4 Got 5 not 4 REJ 4 and beyond 7 0 1 2 3 4 5 6 7 0 1 2 Go back in window until you meet the rejected Frame #, and resume transmission from there RR 7 (+ ive ACK) lost on the way. TX times out Before getting any subsequent RRs after lost Ack TX polls RX To send its Receiving status This is a command F bit = 1 Any corrective action needed here at TX?
Example: Two neighboring nodes (A and B) use a sliding-window go-back-N for error control with a 3-bit sequence number. The window size is 4. Assuming A is TX and B is RX, show the window positions at A for the following situations: • Before A sends any frames • After A sends frames 0, 1, 2 and receives RR 2 from B • Later, after A sends frames 3, 4, and 5 and receives RR 5 from B At TX Send Next
By mistake RX gets 2 copies of F0 Window Size limit for Go-back-N ARQ W 2k – 1 • Size of the window must be W 2k – 1 i.e. W < 2k where “k” is the number of bits reserved (in the control field) for the sequence number • Let k = 2, i.e. W should be < 4 • By comparing the figures opposite for W = 3 and W = 4, justify the need for this limit on window size in Go Back N ACK ACK on F0 TX goes back to send F0 F0 on F0 F0 F0 x TX goes back to send F0 F0 All happened before timeout F0 F0 sent twice in both cases. Mistake is detected only on the left (for the correct W size)
Selective Reject (Selective Repeat ARQ) • Also called selective retransmission • RX requests retransmission of only the rejected frame using SREJ i • Subsequent frames received after the rejected frame are buffered at RX (not thrown away as in Go Back N ARQ) • TX retransmits only the frame that was specifically rejected and those that timed out • Less retransmission traffic than Go Back N
Selective Reject: Example - Waiting for 4 - 4 is lost 5 received out of sequence, Problem detected, SREJ 4 sent 5, 6 • Only rejected frame (4) retransmitted • (not 4,5,6) • Normal transmission resumes where left (7) TX sends: 5,6,4,7,0 (complex sequencing) Acknowledge up to 6 (5 & 6 were kept!) + ive ACK lost on its way! TX times out before getting subsequent RRs to the lost ones So it polls RX for its receiving status F bit = 1 Things turned out OK
Selective Reject: Pros and Cons • Minimizes retransmission Better link utilization (Useful where link utilization is poor- sending a frame is an inefficient process) e.g. short frames on long (e.g. satellite) links • But more complex: (so, less common than Go back N) • Receiver: • Must maintain large enough buffer to save post-SREJ frames until missing frame arrives • Needs logic for inserting requested frame in place when it arrives later • Transmitter: • Needs logic to allow sending the requested frame out of normal sending sequence • Also, more restricted window size, W: • With k bits, max window size is 2(k-1) (vs 2k-1 for Go Back N), e.g. for k = 3: Wmax = 4 for SREJ, Wmax = 7 for GO Back N
How are such flow and error mechanisms implemented:High-Level Data Link Control Protocol (HDLC) • HDLC is the most common data link control protocol and forms the basis for many others • Runs in the Data Link Layer (Layer 2 in OSI) • Main Functions: • Flow Control: Data is transmitted by TX- only as fast as RX can absorb it. • Error Control: Objective: Pass data up to higher layer exactly as transmitted, i.e.: • Without error, • Without loss, • Without duplication, • and in the correct order