Episode 5. Scheduling and Traffic Management
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1 Episode 5. Scheduling and Traffic Management Part 2 Baochun Li Department of Electrical and Computer Engineering University of Toronto
2 Outline What is scheduling? Why do we need it? Requirements of a scheduling discipline Fundamental choices Scheduling best effort connections Scheduling guaranteed-service connections Packet drop strategies
3 Scheduling best effort connections Main requirement is fairness Achievable using Generalized Processor Sharing (GPS) Visit each non-empty queue in turn Serve infinitesimal from each queue in a finite time interval Why is this max-min fair? How can we give weights to connections?
4 More on Generalized Processor Sharing
5 More on Generalized Processor Sharing GPS is not implementable! we cannot serve infinitesimals, only packets
6 More on Generalized Processor Sharing GPS is not implementable! we cannot serve infinitesimals, only packets No packet discipline can be as fair as GPS: but how closely? while a packet is being served, we are unfair to others
7 More on Generalized Processor Sharing GPS is not implementable! we cannot serve infinitesimals, only packets No packet discipline can be as fair as GPS: but how closely? while a packet is being served, we are unfair to others Define: work(i, a, b) = # bits transmitted for connection i in time [a,b]
8 More on Generalized Processor Sharing GPS is not implementable! we cannot serve infinitesimals, only packets No packet discipline can be as fair as GPS: but how closely? while a packet is being served, we are unfair to others Define: work(i, a, b) = # bits transmitted for connection i in time [a,b] Absolute fairness bound for scheduling discipline S max (work GPS (i, a, b) - work S (i, a, b))
9 More on Generalized Processor Sharing GPS is not implementable! we cannot serve infinitesimals, only packets No packet discipline can be as fair as GPS: but how closely? while a packet is being served, we are unfair to others Define: work(i, a, b) = # bits transmitted for connection i in time [a,b] Absolute fairness bound for scheduling discipline S max (work GPS (i, a, b) - work S (i, a, b)) Relative fairness bound for scheduling discipline S max (work S (i, a, b) - work S (j, a, b))
10 What is next? We can t implement GPS So, lets see how to emulate it We want to be as fair as possible But also have an efficient implementation
11 Weighted Round Robin
12 Weighted Round Robin Round Robin: Serve a packet from each non-empty queue in turn
13 Weighted Round Robin Round Robin: Serve a packet from each non-empty queue in turn Unfair if packets are of different length or weights are not equal Weighted Round Robin
14 Weighted Round Robin Round Robin: Serve a packet from each non-empty queue in turn Unfair if packets are of different length or weights are not equal Weighted Round Robin Different weights, fixed size packets serve more than one packet per visit, after normalizing to obtain integer weights
15 Weighted Round Robin Round Robin: Serve a packet from each non-empty queue in turn Unfair if packets are of different length or weights are not equal Weighted Round Robin Different weights, fixed size packets serve more than one packet per visit, after normalizing to obtain integer weights Different weights, variable size packets normalize weights by mean packet size e.g. weights {0.5, 0.75, 1.0}, mean packet sizes {50, 500, 1500} normalize weights: {0.5/50, 0.75/500, 1.0/1500} = {0.01, , }, normalize again {60, 9, 4}
16 Problems with Weighted Round Robin
17 Problems with Weighted Round Robin With variable size packets and different weights, need to know mean packet size in advance
18 Problems with Weighted Round Robin With variable size packets and different weights, need to know mean packet size in advance Fair only over time scales longer than a round time At shorter time scales, some connections may get more service than others If a connection has a small weight, or the number of connections is large, this may lead to long periods of unfairness
19 Problems with Weighted Round Robin With variable size packets and different weights, need to know mean packet size in advance Fair only over time scales longer than a round time At shorter time scales, some connections may get more service than others If a connection has a small weight, or the number of connections is large, this may lead to long periods of unfairness For example T3 trunk with 500 connections, each connection has a mean packet size of 500 bytes, 250 with weight 1, 250 with weight 10 Each packet takes 500 * 8/45 Mbps = 88.8 microseconds Round time =2750 * 88.8 = ms
20 Deficit Round Robin (DRR) DRR modifies weighted round-robin to allow it to handle variable packet sizes without knowing the mean packet size of each connection in advance It associates each connection with a deficit counter, initialized to 0 It visits each connection in turn, tries to serve one quantum worth of bits from each visited connection A packet at the head of queue is served if it is no larger than the sum of the quantum and the deficit counter (which is then adjusted) If it is larger, the quantum is added to the connection s deficit counter Critique paper (due today!) 8
21 Deficit Round Robin: Example 9
22 Weighted Fair Queueing (WFQ)
23 Weighted Fair Queueing (WFQ) Deals better with variable size packets and weights
24 Weighted Fair Queueing (WFQ) Deals better with variable size packets and weights Do not need to know a connection s mean packet size in advance
25 Weighted Fair Queueing (WFQ) Deals better with variable size packets and weights Do not need to know a connection s mean packet size in advance Basic idea
26 Weighted Fair Queueing (WFQ) Deals better with variable size packets and weights Do not need to know a connection s mean packet size in advance Basic idea Simulates GPS on the side and uses the results to determine service order
27 Weighted Fair Queueing (WFQ) Deals better with variable size packets and weights Do not need to know a connection s mean packet size in advance Basic idea Simulates GPS on the side and uses the results to determine service order Compute the time a packet would complete service (finish time), had we been serving packets with GPS, bit by bit
28 Weighted Fair Queueing (WFQ) Deals better with variable size packets and weights Do not need to know a connection s mean packet size in advance Basic idea Simulates GPS on the side and uses the results to determine service order Compute the time a packet would complete service (finish time), had we been serving packets with GPS, bit by bit Then serve packets in order of their finish times
29 Weighted Fair Queueing (WFQ) Deals better with variable size packets and weights Do not need to know a connection s mean packet size in advance Basic idea Simulates GPS on the side and uses the results to determine service order Compute the time a packet would complete service (finish time), had we been serving packets with GPS, bit by bit Then serve packets in order of their finish times The finish time is just a service tag, and has nothing to do with the actual time at which the packet is served
30 Weighted Fair Queueing (WFQ) Deals better with variable size packets and weights Do not need to know a connection s mean packet size in advance Basic idea Simulates GPS on the side and uses the results to determine service order Compute the time a packet would complete service (finish time), had we been serving packets with GPS, bit by bit Then serve packets in order of their finish times The finish time is just a service tag, and has nothing to do with the actual time at which the packet is served We can more appropriately call it a finish number
31 WFQ: a First Cut
32 WFQ: a First Cut Suppose, in each round, the server served one bit from each active connection
33 WFQ: a First Cut Suppose, in each round, the server served one bit from each active connection Round number is the number of rounds already completed can be fractional
34 WFQ: a First Cut Suppose, in each round, the server served one bit from each active connection Round number is the number of rounds already completed can be fractional If a packet of length p bits arrives to an empty queue when the round number is R, it will complete service when the round number is R + p => finish number is R + p independent of the number of other connections!
35 WFQ: a First Cut Suppose, in each round, the server served one bit from each active connection Round number is the number of rounds already completed can be fractional If a packet of length p bits arrives to an empty queue when the round number is R, it will complete service when the round number is R + p => finish number is R + p independent of the number of other connections! If a packet arrives to a non-empty queue, and the previous packet has a finish number of f, then the packet s finish number is f + p
36 WFQ: a First Cut Suppose, in each round, the server served one bit from each active connection Round number is the number of rounds already completed can be fractional If a packet of length p bits arrives to an empty queue when the round number is R, it will complete service when the round number is R + p => finish number is R + p independent of the number of other connections! If a packet arrives to a non-empty queue, and the previous packet has a finish number of f, then the packet s finish number is f + p Serve packets in the order of finish numbers
37 WFQ: A catch
38 WFQ: A catch A queue may need to be considered non-empty even if it has no packets in it e.g. packets of length 1 from connections A and B, on a link of speed 1 bit/sec at time 1, packet from A served, round number = 0.5 A has no packets in its queue, yet should be considered nonempty, because a packet arriving to it at time 1 should have finish number 1+ p
39 WFQ: A catch A queue may need to be considered non-empty even if it has no packets in it e.g. packets of length 1 from connections A and B, on a link of speed 1 bit/sec at time 1, packet from A served, round number = 0.5 A has no packets in its queue, yet should be considered nonempty, because a packet arriving to it at time 1 should have finish number 1+ p Solution: A connection is active if the last packet served from it, or in its queue, has a finish number greater than the current round number
40 WFQ: Summary so far
41 WFQ: Summary so far To sum up, assuming we know the current round number R
42 WFQ: Summary so far To sum up, assuming we know the current round number R Finish number of packet of length p If arriving to active connection = previous finish number + p If arriving to an inactive connection = R + p
43 WFQ: Summary so far To sum up, assuming we know the current round number R Finish number of packet of length p If arriving to active connection = previous finish number + p If arriving to an inactive connection = R + p To implement WFQ, we need to know two things: Is the connection active? If not, what is the current round number?
44 WFQ: Summary so far To sum up, assuming we know the current round number R Finish number of packet of length p If arriving to active connection = previous finish number + p If arriving to an inactive connection = R + p To implement WFQ, we need to know two things: Is the connection active? If not, what is the current round number? Answer to both questions depends on computing the current round number
45 WFQ: Computing the round number
46 WFQ: Computing the round number Naively: round number = number of rounds of service completed so far what if a server has not served all connections in a round? what if new connections join halfway through a round?
47 WFQ: Computing the round number Naively: round number = number of rounds of service completed so far what if a server has not served all connections in a round? what if new connections join halfway through a round? Redefine round number as a real-valued variable that increases at a rate inversely proportional to the number of currently active connections this takes care of both problems
48 WFQ: Computing the round number Naively: round number = number of rounds of service completed so far what if a server has not served all connections in a round? what if new connections join halfway through a round? Redefine round number as a real-valued variable that increases at a rate inversely proportional to the number of currently active connections this takes care of both problems With this change, WFQ emulates GPS instead of bit-bybit Round Robin
49 WFQ: An Example 15
50 WFQ: An Example Packets of size 1, 2, and 2 units arrive at a WFQ scheduler at time 0, on equally weighted connections A, B, and C 15
51 WFQ: An Example Packets of size 1, 2, and 2 units arrive at a WFQ scheduler at time 0, on equally weighted connections A, B, and C A packet of size 2 arrives at connection A at time 4 15
52 WFQ: An Example Packets of size 1, 2, and 2 units arrive at a WFQ scheduler at time 0, on equally weighted connections A, B, and C A packet of size 2 arrives at connection A at time 4 The link service rate is 1 unit/second 15
53 WFQ: An Example Packets of size 1, 2, and 2 units arrive at a WFQ scheduler at time 0, on equally weighted connections A, B, and C A packet of size 2 arrives at connection A at time 4 The link service rate is 1 unit/second Questions What are the finish numbers of all packets? What is the round number when the system becomes idle? When does this happen? 15
54
55 WFQ: An Example, ')\ o 3 E- -n 'E o t l- A _--I- B.C ---*- Round number Time 16
56 Problem: Iterated Deletion 17
57 Problem: Iterated Deletion Suppose the round number at time 0 is 0, and has a slope of 1/5 We may incorrectly suppose that, at time 5, it would have a value of 0 + 1/5 * 5 = 1 17
58 Problem: Iterated Deletion Suppose the round number at time 0 is 0, and has a slope of 1/5 We may incorrectly suppose that, at time 5, it would have a value of 0 + 1/5 * 5 = 1 However, our computation may be wrong, because one of the five connections active at time 0 may become inactive before time 5 Say, connection 1 becomes inactive at time 4, so at time 4, the slope of the round number line increases from 1/5 to 1/4 At time 5, the round number would be 4 * 1/5 + 1 * 1/4 =
59 Problem: Iterated Deletion Suppose the round number at time 0 is 0, and has a slope of 1/5 We may incorrectly suppose that, at time 5, it would have a value of 0 + 1/5 * 5 = 1 However, our computation may be wrong, because one of the five connections active at time 0 may become inactive before time 5 Say, connection 1 becomes inactive at time 4, so at time 4, the slope of the round number line increases from 1/5 to 1/4 At time 5, the round number would be 4 * 1/5 + 1 * 1/4 = 1.05 Suppose connection 2 has a finish number smaller than 1.05 It would also become inactive before time 5, further affecting the slope of the round number which may delete more connections! 17
60 Problem: Iterated Deletion Round number # active connections
61 Problem: Iterated Deletion Round number # active connections A server recomputes the round number on each packet arrival
62 Problem: Iterated Deletion Round number # active connections A server recomputes the round number on each packet arrival At any re-computation, the number of connections can go up at most by one, but can go down to zero
63 Problem: Iterated Deletion Round number # active connections A server recomputes the round number on each packet arrival At any re-computation, the number of connections can go up at most by one, but can go down to zero To solve this problem: use previous count to compute round number if this makes some connection inactive, recompute repeat until no connections become inactive
64 WFQ implementation
65 WFQ implementation On packet arrival use the source and destination address to classify it to a connection, and look up the finish number of the last packet served (or waiting to be served) recompute round number compute finish number insert in a priority queue sorted by finish numbers
66 WFQ implementation On packet arrival use the source and destination address to classify it to a connection, and look up the finish number of the last packet served (or waiting to be served) recompute round number compute finish number insert in a priority queue sorted by finish numbers On service completion select the packet with the lowest finish number in the priority queue
67 WFQ: Analysis
68 WFQ: Analysis If GPS has served x bits from connection A by time t
69 WFQ: Analysis If GPS has served x bits from connection A by time t WFQ would have served at least x - P bits, where P is the largest possible packet in the network
70 WFQ: Evaluation
71 WFQ: Evaluation Pros like GPS, it provides protection can obtain worst-case end-to-end delay bound (more details later) gives users incentive to use intelligent flow control
72 WFQ: Evaluation Pros like GPS, it provides protection can obtain worst-case end-to-end delay bound (more details later) gives users incentive to use intelligent flow control Cons needs per-connection state iterated deletion is complicated requires explicit sorting of the output queue
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