Telematics. 5th Tutorial - LLC vs. MAC, HDLC, Flow Control, E2E-Arguments

Transcription

1 Telematics 5th Tutorial - LLC vs. MAC, HDLC, Flow Control, E2E-Arguments Bastian Blywis Department of Mathematics and Computer Science Institute of Computer Science 18. November, 2010 Institute of Computer Science Telematics Tutorial 18. November,

2 Outline 1. Network Topologies 2. Reducing the Overhead 3. Efficiency of Stop-and-Wait 4. LLC vs. MAC 5. Flow Control 6. Sliding Window Mechanism 7. Sliding Window and Data Rate 8. Sliding Window 9. HDLC vs. PPP 10. The End to End Argument Institute of Computer Science Telematics Tutorial 18. November,

3 Network Topologies Consider the following four network topologies, each with n nodes: Star Fully Meshed Unidirectional Ring Bidirectional Ring Give a formula to calculate the minimum, maximum, and average number of hops between any two nodes for any number n.??? Institute of Computer Science Telematics Tutorial 18. November,

4 Network Topologies Minimum Average Maximum Star Institute of Computer Science Telematics Tutorial 18. November,

5 Network Topologies Minimum Average Maximum Fully Meshed Institute of Computer Science Telematics Tutorial 18. November,

6 Network Topologies Minimum Average Maximum 1 n-1 Unidirectional Ring n 1 1 i = n n 1 2 i=1 Institute of Computer Science Telematics Tutorial 18. November,

7 Network Topologies Bidirectional Ring Min. Average Maximum 1 if n even: 2 n 2 1 i=1 i + n 2 n 1 if n odd: 2 n 1 2 i=1 i n 1 = n 2 4n 4 = n if n even: n 2 if n odd: n 1 2 Institute of Computer Science Telematics Tutorial 18. November,

8 Reducing the Overhead As you have learned, frames (often) consist of data that is framed by special flag bytes (start and stop). Discuss if the stop flag can be omitted to save one byte as the start flag of the succeeding frame can be used to implicitly mark that the previous frame has ended. Start Data 1 Stop Start Data 2 Stop... =? Start Data 1 Start Data 2...?? Institute of Computer Science Telematics Tutorial 18. November,

9 Reducing the Overhead What happens when there is no next frame or a long gap? How can the receiver differentiate noise from data? Start Data 1 Start Data 2...?... Start Data 3 Some Alternative Approaches: Length field the header Three-state logic (pull-ups/downs, output enable) Out of band signaling, separate clock line Institute of Computer Science Telematics Tutorial 18. November,

10 Efficiency of Stop-and-Wait Assume a channel with a bit rate of 1 Mbps and a delay of 20 ms. A Stop-and-Wait protocol is used which unfortunately introduces waiting times and thus a low efficiency. The efficiency is dependent on the size of the frames. Determine the frame size for which the efficiency is 50%.??? Institute of Computer Science Telematics Tutorial 18. November,

11 Efficiency of Stop-and-Wait Using the Stop-and-Wait protocol, only one frame is on the line at a time Acknowledgement has to arrive before next frame is sent Assumptions Processing times are negligible Acknowledgements are always piggybacked Packets in both directions have the same size efficiency utilization in script (slide 4.70) efficiency = length/(length + bitrate RTT) (1) 0.5 = length/(length bps ) (2) 0.5 = length/(length bits) (3) length = bits (4) Institute of Computer Science Telematics Tutorial 18. November,

12 LLC vs. MAC Discuss the difference tasks of the LLC and MAC. Background Problem Statement: Several data link layer protocols exist Data link layer provides different services to the network layer Consistent primitives to the network layer are desired??? Institute of Computer Science Telematics Tutorial 18. November,

13 LLC vs. MAC Data link layer (often) consists of two sublayers: Logical Link Control (LLC) Media Access Control (MAC) LLC Specified in IEEE and adopted by the ISO/IEC and redesignated as ISO/IEC :1998 Hides differences between various kinds of networks, example: Token ring network can be connected (bridged) with Ethernet network Single format and interface to the network layer Format, interface, and protocol closely based on HDLC LLC header is prepended to network layer packet before inserting into payload field of, e.g., frame Services Unreliable datagram service (unacknowledged and connectionless) Acknowledged datagram service (connectionless) Acknowledged connection-oriented service Institute of Computer Science Telematics Tutorial 18. November,

14 LLC vs. MAC LLC Classes Class 1 Class 2 data-link-connectionless-mode service data-link-connectionless-mode service data-link-connection-mode service Class 3 data-link-connectionless-mode service acknowledged-connectionless-mode service Class 4 data-link-connectionless-mode data-link-connection-mode service acknowledged-connectionless-mode service Institute of Computer Science Telematics Tutorial 18. November,

15 LLC vs. MAC MAC Handles access to shared medium Contention for the medium access by all stations Different MAC approaches for different network technologies Network Allocation Vector (NAV) can be used for power management (if available) Institute of Computer Science Telematics Tutorial 18. November,

16 LLC vs. MAC Figure: IEEE Architecture Institute of Computer Science Telematics Tutorial 18. November,

17 Flow Control Repeat and discuss the task of flow control.??? Institute of Computer Science Telematics Tutorial 18. November,

18 Flow Control Control data flow between stations Protect slow receiver from fast sender Approaches Feedback-based flow control Rate-based flow control Different strategies when acknowledgments are not received until timeout Can enable reliable data communication Institute of Computer Science Telematics Tutorial 18. November,

19 Sliding Window Mechanism Consider some host-to-network technology where each frame contains a sequence number SEQ as well as an acknowledgement number ACK. The acknowledgement number acknowledges all frames up to ACK-1. Both numbers are represented by M bits and thus all calculations are modulo log 2 (M).??? Institute of Computer Science Telematics Tutorial 18. November,

20 Sliding Window Mechanism How many frames can be sent before at least one acknowledgement has to be received???? Institute of Computer Science Telematics Tutorial 18. November,

21 Sliding Window Mechanism The maximum window size is M 1. With a window size of M, the following two scenarios can not be distinguished by host A for W = M = 4: A 0 B A 0 B ACK ACK Timeout 0 ACK 1 ACK 1 0 Note: In real networks there is no guarantee that the delay equal in both directions! Institute of Computer Science Telematics Tutorial 18. November,

22 Sliding Window Mechanism Consider an example with M = 8 and window size W = 7. The frames with the sequence numbers 5, 6, 7, 0, 1 have been sent from A to B and no acknowledgement has been received yet. Which of the remaining sequence numbers may be used by A to sent further frames???? Institute of Computer Science Telematics Tutorial 18. November,

23 Sliding Window Mechanism Frames with the sequence numbers 5, 6, 7, 0, 1 have been sent The frames with sequence numbers 2 and 3 can be sent without getting an acknowledgement: M 1 = 7 5 frames have been sent last used sequence number + {1,2} Institute of Computer Science Telematics Tutorial 18. November,

24 Sliding Window Mechanism Which sequence numbers may be used for further frames, if acknowledgements are received with: 1. ACK = 2 2. ACK = 6 3. ACK = 5 List the acknowledged sequence numbers and the window of remaining sequence numbers.??? Institute of Computer Science Telematics Tutorial 18. November,

25 Sliding Window Mechanism Received ACK = 2 Acknowledged: 5, 6, 7, 0, 1 Window: 2, 3, 4, 5, 6, 7, (a) Previous State (b) Received Acknowledgement Institute of Computer Science Telematics Tutorial 18. November,

26 Sliding Window Mechanism Received ACK = 6 Acknowledged: 5 Window: 2, 3, (c) Previous State (d) Received Acknowledgement Institute of Computer Science Telematics Tutorial 18. November,

27 Received ACK = 5 Acknowledged: 4 Window: 2, 3 Sliding Window Mechanism (e) Previous State (f) Received Acknowledgement If a frame with sequence number 4 actually has been sent, the ACK=5 acknowledges this frame. The window doesn t change because sequence number 4 cannot be used. Alternatively, an error has happened Institute of Computer Science Telematics Tutorial 18. November,

28 Sliding Window and Data Rate Consider a host-to-network technology with a sliding window mechanism and a window size of W = 7. The frames can have a size of up to 1,500 bytes and the round trip time between two hosts is 50 ms. Calculate the maximum data rate that can be achieved.??? Institute of Computer Science Telematics Tutorial 18. November,

29 Sliding Window and Data Rate A host can send 7 1, 500 bytes = 10, 500 bytes without getting an acknowledgement. Assumption: Processing times are negligible Thus, a frame that is received by host B at time 25 ms will be acknowledged 25 ms later Host A receives the acknowledgement at time 50 ms The window is shifted and the next frame can be sent 10,500 byte can be sent in 50 ms. Host A can receive 20 acknowledgements in 1 sec Thus, 20 10, 500 byte can be sent 210, 000 byte/s 1.6 Mbps. Institute of Computer Science Telematics Tutorial 18. November,

30 Sliding Window and Data Rate In reality you have to consider the processing times of both hosts 1,500 byte = 12,000 bit take about 120 µs to get the frame on the wire with Fast Ethernet (100 MBit/s) The first frame is received by host B after 25.12ms. Assumption: The ACK is sent immediately without any additional delay Assumption: ACK frames without data are less than 100 bit long and it takes about 1µs sent the frame Thus, after ms host A can move the window by one position and sent the next frame... Institute of Computer Science Telematics Tutorial 18. November,

31 Assume a geostationary satellite sends frames of 1000 bits over a channel with a bit rate of 1 Mbps. The frame take 270 ms to arrive at the station on earth. Calculate the maximum achievable efficiency using a Stop-and-Wait protocol Sliding Window protocol with window size 13 The acknowledgements are always piggybacked on data frames.??? Institute of Computer Science Telematics Tutorial 18. November,

32 Transmission starts at t=0 Station on earth receives 1st frame at t=271 ms Station replies at t=272 ms Reply from station arrives at t=542 ms at the satellite k Frames can be sent in 542 ms until an acknowledgement is required to proceed Efficiency is k/542 k=1 (Stop-and-Wait): efficiency = 1/542 = 0.18% k=13 (Sliding Window): efficiency = 13/542 = 2.40% Institute of Computer Science Telematics Tutorial 18. November,

33 Sliding Window Why do we need preferably a full-duplex connection for sliding window protocols???? Institute of Computer Science Telematics Tutorial 18. November,

34 Sliding Window Without a full-duplex connection, there is only limited parallelization Data and acknowledgements should be on the medium at the same time The rate and delay at which the acknowledgements arrive, significantly determines the performance (remember this when we will discuss TCP!) In many scenarios both stations want to send data at the same time Parallelization is an important motivation for sliding window protocols Institute of Computer Science Telematics Tutorial 18. November,

35 HDLC vs. PPP Compare the HDLC and PPP protocols. Specify appropriate metrics for the comparison.??? Institute of Computer Science Telematics Tutorial 18. November,

36 HDLC vs. PPP HDLC PPP Origin based on SDLC Inspired by HDLC Standardization ISO IETF Implementation hardware software Framing Synchronous or asynchronous Stuffing bit byte Services Modes connection-oriented, connectionless point-to-point or point-tomultipoint depends on framing connection-oriented point-to-point Checksum CCITT CRC-16 depends on framing Discuss further differences! Institute of Computer Science Telematics Tutorial 18. November,

37 HDLC vs. PPP Examples for PPP encapsulation Simpson PPP in HDLC-like Framing RFC 1662, 1994 Mamakos et al. A Method for Transmitting PPP Over Ethernet (PPPoE) RFC 2516, 1999 Simpson PPP over SONET/SDH RFC 1619, 1999 Gross et al. PPP Over AAL5 RFC 2364, 1998 Simpson PPP over ISDN RFC 1618, 1994 PPP Connection Establishment Simpson PPP Challenge Handshake Authentication Protocol (CHAP) RFC 1994, 1996 Simpson PPP LCP Extensions RFC 1570, 1994 Institute of Computer Science Telematics Tutorial 18. November,

38 The End to End Argument Read the article End-to-end arguments in system design by Saltzer et al. Discuss the ideas of the end to end argument and which functions/services should be provided on particular layers of the reference model.??? Institute of Computer Science Telematics Tutorial 18. November,

39 Motivation: The End to End Argument Context: Protocol design Distributed system with different layers Functions that might be implemented on multiple levels of the protocol stack Key question: Where to place which function? End to End (E2) Argument: Implementing these functions at low level may be redundant or provides little usefulness Functions in question should be implemented at the endpoints of the communication system Example: Congestion control at endpoints Discussion: E2E-Argument does not tell at which layer to place the functions Consider a trade-off between costs and performance Institute of Computer Science Telematics Tutorial 18. November,

40 The Last Slide TM Thank you for your attention. Questions? Institute of Computer Science Telematics Tutorial 18. November,