Srinidhi Varadarajan 02/14/2000 Error Detection Codes Error Detection ■ Two types ■ Naïve scheme ◆ Error Detection Codes (e.g. CRC, ◆ Send a duplicate copy of the Parity, Checksums) message ◆ Error Correction Codes (e.g. ■ Problems Hamming, Reed Solomon) ◆ Takes up too much space ■ Basic Idea ◆ Poor performance. ◆ Add redundant information to ✦ Can’t even detect 2 bit errors determine if errors have been ■ Cyclic Redundancy Codes are introduced common ✦ Why redundant? Two Dimensional Parity Internet Checksum Algorithm Parity bits ■ Used by IP and TCP 0101001 1 ■ Algorithm treats data as 16 bit 1101001 0 unsigned quantities 1011110 1 ◆ Add the data in the frame using 1’s Data 0001110 1 complement arithmetic 0110100 1 ◆ Take the one’s complement of the result 1011111 0 ◆ IP and TCP store the sum as a 32 Parity 1111011 0 bit unsigned integer byte ✦ Note: The example in the book ■ Each byte is protected by a parity bit uses a 16 bit checksum ■ The entire frame is protected by a parity byte Cyclic Redundancy Check CRC Computation ■ Treat the (n+1) bit message as a polynomial ■ Given: of degree n. The bits are the coefficients of ◆ Message: M(x) the polynomial. ◆ 1101 = 1*x 3 + 1*x 2 + 0*x 1 + 1*x 0 ◆ Divisor: C(x) ■ Multiply M(x) by x k , i.e. add k zeroes to ■ Calculating CRC the end of the message. Call this T(x) ◆ Sender and transmitter choose a divisor polynomial of degree k. e.g x 3 + x 2 + 1 ■ Divide T(x) by C(x). ◆ Add k bits to the (n+1) bit message such that ■ Subtract the remainder from T(x) the n+1+k bit message is exactly divisible by the divisor ■ The result is the message including the ■ Choice of divisor is very important. CRC ◆ It determines the kind of errors that the CRC can guard against. Title goes here 1
Srinidhi Varadarajan 02/14/2000 CRC Computation Reliable Transmission 11111001 Generator 1101 10011010000 Message 1101 1001 ■ Why? 1101 ◆ Frame corruption can be severe 1000 1101 ◆ CRCs are not enough. Recall 1011 CRCs don’t correct errors 1101 ■ Two fundamental mechanisms 1100 1101 ◆ Acknowledgment 1000 1101 ◆ Timeout 101 Remainder ■ General idea is called ARQ ■ C(x) = x 3 + x 2 + 1 (Automatic Repeat Request) ■ M(x) = x 7 + x 4 + x 3 + x ■ Subtraction: logical XOR operation Stop and Wait: Possible Scenarios Stop and Wait Protocol Sender Receiver Sender Receiver Frame Frame ■ Simple operation Timeout Timeout Time ACK ACK ◆ Transmit a packet Frame ◆ Wait for an acknowledgement Timeout (ACK) ACK ◆ If no ACK arrives within a preset (a) (c) time interval, assume that the packet is lost and retransmit Sender Receiver Sender Receiver Frame Frame Timeout Timeout ■ Repeat the procedure until all ACK packets have been successfully Frame Timeout Frame transmitted Timeout ACK ACK (b) (d) Performance Problems Performance problems? ■ No more than one packet in flight. ■ Using the actual 10Mbps Ethernet RTT of 50us ◆ That’s usually bad, here’s why (roughly) ■ Take a 10Mbps network with a 50ms round trip time ■ Delay bandwidth = 10 7 * 50us = 500 bits ■ Delay bandwidth = 10 7 * 0.050 = 500 Kbits ■ In Stop and Wait, only frame can be in flight. The max Ethernet frame size is 1500 bytes ◆ Hence sending rate = ■ In Stop and Wait, only frame can be in flight. The ✦ 1500 * 8 ÷ 50us = 240 Mbps max frame size is 1500 bytes ◆ Hence sending rate = ◆ This is much greater than the link capacity of 10 ✦ 1500 * 8 ÷ 0.050 = 240 Kbps Mbps ✦ What happened?? ◆ This is much less than the link capacity of 10 Mbps Title goes here 2
Srinidhi Varadarajan 02/14/2000 Performance Analysis Performance Analysis ■ Putting in numbers for 10 Mbps ethernet ◆ Packet size: 1518 bytes ◆ ACK size: 64 bytes ◆ PROP: 51.2us ■ Efficiency = 92.22% ◆ More believable! ■ Moral: If frame size exceeds delay bandwidth product, efficiency computation should be used Significance of Delay Bandwidth Sliding Window Protocols ■ Delay bandwidth represents the amount of data that has left the transmitter and is still on the cable. ■ Think of the cable as a pipe. This keeps the pipe full ■ Delay bandwidth also represents the upper bound on stability. ■ More sophisticated ARQ algorithms try to match their sending rate to the dynamic delay • Keep the pipe full bandwidth product •Send N packets before expecting the first ACK ◆ Why is delay bandwidth dynamic? Sliding Window Protocol Sliding Window Protocol Send out N frames, each with a linearly ■ Receiver maintains 3 variables ■ increasing sequence number ◆ Receive window size: Upper bound on Sender uses 3 variables ■ out of order frames received ◆ Send window size: Upper bound on unACKed frames ◆ LAF: Largest acceptable frame ◆ LAR: Last ACK received ◆ LFR: Last frame (in sequence) received ◆ LFS: Last frame sent ◆ Invariant: LAF – LFR ≤ RWS Invariant: LFS – LAR ≤ SWS ■ ≤ RWS ≤ SWS … … … … LFR LAF LAR LFS Title goes here 3
Srinidhi Varadarajan 02/14/2000 Performance Improvements Operation ■ Sender sends “send window size” ■ Negative ACK (number of) frames ◆ Receiver NACK’s frames that were not ■ Receiver ACKs last in sequence frame received received. ✦ Additional complexity. ■ Error conditions: • Loss of NACKs, receiver timeout ◆ Timeout mechanism needed ■ Selective ACK: ◆ Receiver receives out of receive window frame ◆ Receiver ACKs specific frames. ACKs ✦ ACK it. Throw away the frame are not inclusive ◆ Sender receives out of “send window” ◆ Sender can use this to detect out of order ack arrival and retransmit ✦ Old delayed ACK. Throw it away Efficiency Efficiency ■ E.g.: ◆ Packet size = 2000 bits, ACK = 80 N TRANS = bits Efficiency min , 1 + + TRANS ACK 2 * PROP ◆ Bandwidth = 155 Mbps ATM over fibre ◆ Cable length: 30 km N 2000 = min , 1 Efficiency + + ■ Increases linearly with N until it reaches 100 % 2000 80 31000 N = 1; Eff = 5.95%; N = 16; Eff = 96.73% Window Size Settings Finite Sequence Numbers ■ Sequence numbers have a finite length. ■ Common modes They increment by modulo arithmetic ◆ RWS = 1, SWS = N ✦ Receiver does not buffer any out of ■ Cases: order frames ◆ RWS = 1 ◆ RWS = SWS = N ◆ MaxSeqNum >= SWS + 1 ✦ Receiver can buffer as many ◆ Why + 1 frames as the sender sends. ✦ Receiver receives SWS, sends ACK ◆ RWS > SWS? ✦ ACK is lost, receiver expects wrong frame Title goes here 4
Srinidhi Varadarajan 02/14/2000 Finite Sequence Numbers ■ RWS = SWS ◆ MaxSeq Number >= 2 * SWS ◆ Why? ✦ Hint: receiver ACK is lost ■ RWS < SWS ◆ MaxSeq Number >= max(SWS + 1, 2 * RWS) ◆ How? Title goes here 5
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