Congestion Control. Topics
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1 Congestion Control Topics Congestion control what & why? Current congestion control algorithms TCP and UDP Ideal congestion control Resource allocation Distributed algorithms Relation current algorithms and resource allocation Broad implications 6.976/ESD.937 1
2 Congestion Control: What and Why? Internet is used by many independent users Resources (link capacity) are finite If every user sends data at very high rate, it will cause congestion packets will be dropped, i.e. unreliable transmission seemingly high utilization of resource may be actually very low! If every user sends data at very low rate, resource will not be well-utilized Users need to send data at the correct rate so that Resources are well-utilized Users get reliable data transfer Resources should be shared fairly This is the primary goal of congestion control 6.976/ESD.937 2
3 Congestion Control Main question: How to implement congestion control in the network With the help of only distributed protocols at the users is it even possible? if not, what is the best possible thing to do? An example: Sender link Receiver ( unknown capacity for sender/receiver ) 6.976/ESD.937 3
4 Congestion Control If senders get acknowledgment of receipt of packets They know if packets are dropped or not Based on this, senders can infer the following If packets are not dropped sender is sending at rate lower than the capacity If packets are dropped sender is sending at rate higher than capacity Use drops of packets as signal of congestion Change rate as a reaction to packet drop so as to achieve fair-share Congestion control is essentially feedback-control 6.976/ESD.937 4
5 Congestion Control Mechanism To perform congestion control, we need two basic protocols Algorithm I: Drop detection ask receiver to acknowledge receipt of packets if a sent packet is not acknowledged fast enough the packet is assumed to be dropped Algorithm II: Rate-control if packets are not dropped, i.e. no congestion and hence increase rate by certain amount else there is congestion and hence decrease rate by certain amount Algorithms I and II are key ideas behind current congestion control protocols 6.976/ESD.937 5
6 Congestion Control Recall, Internet has layered architecture Congestion control is essentially required for reliable transmission at fair-rate with high-resource utilization Implemented in Network Layer The congestion control protocol is also called transport protocol Transmission Control Protocol (TCP) is a popular protocol provides reliable transmission when all users exhibit good-citizen behavior but has higher delay (cost of reliability) User Data-gram Protocol (UDP) is another protocol unreliable and user selfish protocol but has lower delay useful for media communication 6.976/ESD.937 6
7 TCP Brief History TCP (some version) has been around since ARPANET Older versions(s) required users to send probes to network to detect level of congestion led network to always operate under congestion often led to congestion collapse Popular well-documented example Congestion collapse in October 1986 Rate between Lawrence-Berkeley Lab and UCB went down from 32 kbps 40 bytes! 6.976/ESD.937 7
8 TCP Post-mortem of congestion collapse Older TCP sent more and more packets without confirming their delivery violation of packet conservation principle was key reason for failure Did not account for inhomogeneity in network bandwidth Lack of rate-control Current TCP Packet conservation: inject new packet when old packet wave has reached destination Slow-start: search for capacity starting from zero Rate-control: control rate via packet drop feedback, and be good user 6.976/ESD.937 8
9 TCP: Current Algorithm Algorithm has two components Sender algorithm and receiver algorithm Sender algorithm: Parameters: STHRESH: threshold for slow-start W max : maximum window-size CW: current window size Initially: number the data to be sent 1, 2,...,N /ESD.937 9
10 TCP: Current Algorithm Slow-start: Set STHRESH = W max /2 Start with CW= 1, packet sent P= 0 Each time, send packets from P to P+CW When a packet P+1 is acknowledged, set P P+1 For every acknowledgment, increase CW to CW+1 till CW < STHRESH or Packet loss is detected if packet loss, then set STHRESH STHRESH/2 and CW CW/2 Continue above until CW=STHRESH go to congestion-avoidance phase If P=N ever, transmission is over! 6.976/ESD
11 TCP: Current Algorithm Congestion-avoidance For acknowledgement of P+1, CW CW + 1 [CW] If packet-loss then CW is decreased different versions have different ways of dealing with the window size decrease one version: set STHRESH CW/2 CW 1 go to slow-start phase 6.976/ESD
12 TCP: Current Algorithm Receiver algorithm: For every received packet, send acknowledgment With request for the next packet # required if packets 1-10 and 13 are received, then send request for packet number 11 (and not 14) Packet-loss detection: Essentially, packet is dropped when acknowledgment is not received Specific protocol if ack. not received within TIME-OUT, or 3 consecutive requests sent for the same packet # by receiver 6.976/ESD
13 TCP: An Example Suppose, CW = 10 at some-time T= packets 1 CW:( ) 1 CW:( ) T=4 T=6 ACK 1 ACK 10 CW = 11 T= packets ACK T=14 T=20 TIME-OUT packets Sender Receiver 6.976/ESD
14 TCP: Performance The main performance metric Throughput of TCP That is, what is the net equilibrium rate of users as a fraction of the total capacity Evaluation of TCP throughput Model: TCP dynamics effect of slow-start vs. congestion-avoidance Characterization of equilibrium of TCP First, we ll consider a model for TCP 6.976/ESD
15 TCP: Performance We ll consider a very simple situation single link and single user ignore extra complications Given this model, we will identify Key system dynamics affecting its performance Specifically, we ll try to compare effect of Slow-start vs. Congestion control And, use this insight to evaluate performance precisely 6.976/ESD
16 TCP: Performance Model: Single user accessing link of capacity C T be round-trip time B be the butter size at the link Maximum window size W max = ct + B STHRESH = W max /2 = ct+b 2 User wants to transmit infinite data Goal: Evaluate average rate of transmission 6.976/ESD
17 TCP: Performance TCP dynamics Periodic between slow-start and congestion avoidance slow-start congestion-avoidance slow-start t = 0 Tss Tss + Tca Nss Nca Rate = N ss + Nca Tss + Tca Next, we evalute Nss, Nca, Tss and Tca To obtain rate of TCP 6.976/ESD
18 TCP: Performance We first study the Slow-start phase Specifically, evolution of window size and queue-size Slow-start phase has cyclic behavior Divide time into cycles where window size doubles in each cycle: t = 0: first packet sent; W(0)= 1 t = T: packet ACK; W(T)= 2 t = 2T: both packets ACK; W(2T)= 3. each mini-cycle has length =T Window-size evolution: W(nT) 2 n /ESD
19 TCP: Performance Next, we study queue-length evolution In slow-start phase using window-evoluation Again, queue-size has the same cyclic behavior Queue-size at the beginning of cycle 0 During nth cycle, 2 n 2 packets are sent total packets sent (until nth cycle) = 2 n n = 2 n+1 total packets acknowledged by end of nth cycle W(nT) = 2 n 2 Max-queue length in nth cycle, Q(nT) 2 n n 2 2 n 2 W(nT)/ /ESD
20 TCP: Performance In slow-start phase, buffer-overflow happens if Q is larger than B But window can be at most STHRESH Now, Q = W 2 which is at most STHRESH 2 = B+cT 4 ( ) Hence, for overflow B+cT 4 > B That is, ct > 3B B < ct/3 Equivalently, no overflow when B ct/ /ESD
21 TCP: Performance First, we consider no overflow situation That is, B ct/3 When there is no overflow, we have the following Tss: time for slow-start W(Tss) B+cT 2 2 T ss /T B+cT ( 2 ) B+cT Tss = T log 2 2 Nss: # of packets transmitted = # of packets acknowledge = window-size = B+cT /ESD
22 TCP: Performance Now, iff overflow happens in slow-start, i.e. B < c T/3 There are two slow-start phases: Phase 1: at the overflow Q= B = W/2; but detection happens after T time, i.e., window has doubled W = 2 B at Tss 1 (time to overflow) 2 T ss 1 /T 2 B T ss 1 = T log 2 2B Nss 1 = 2B Set STHRESH = 2B 2 = B 6.976/ESD
23 TCP: Performance Phase 2: after overflow Now, the overflow does not happen as STHRESH is low enough! W = B at the end of this phase 2 T ss 2 /T B T ss 2 T logb Nss 2 = B In summary, when overflow happens Tss = Tss 1 + Tss 2 = 2T log B + T Nss = 3B And when overflow does not happen ( ) B+cT Tss = T log 2 2 Nss = B+cT /ESD
24 TCP: Performance Now, we study Congestion-avoidance phase Start of window-size W 0 = W max /2 = ct+ B 2 if B ct 3 B if B < ct 3 Next, we study evolution of window-size with time t = W(t) Let a(t) = # of acknowledgments received till time t Then, change in window-size is dw dt = dw da da dt { C if W is large, server is busy Rate change of a = W/T if W is smaller da That is, dt = min{w/t,c} Also: dw da = 1 (by definition of algorithm) W 6.976/ESD
25 TCP: Performance Hence, dw dt = { 1/T if W < ct c/w if W ct Congestion avoidance will end when W = W max Two-phases Phase 1: W 0 W < ct Phase 2: ct W W max First, we consider Phase /ESD
26 TCP: Performance Phase 1: W 0 W < ct Tca 1 = T (ct W 0 ), since dw dt = 1/T N ca1 = a(t ca1 ) = T ca1 0 da(t), where Tca1 0 da(t) = = = Tca1 0 Tca1 0 Tca1 0 Tca1 da(t) dt dt W(t) T dt W 0 + t/t dt T W 0 T ca1 = 0 T [ = W 0 (ct W 0 ) + T2 ca 1 2T ct W 0 2 ] 6.976/ESD
27 TCP: Performance Phase 2: W ct; dw dt = c W(t) W(t)dW = cdt W 2 (t) W 2 (0) = 2ct (where W(0) = ct) W 2 (t) = 2ct + c 2 T 2 Phase ends when W(T ca2 ) = W max W 2 max = 2cT ca2 + c 2 T 2 2cT ca2 = W 2 max c 2 T 2 T ca2 = W2 max c 2 T 2 2c N ca2 = ct ca2 because, node is running at capacity c 6.976/ESD
28 TCP: Performance In summary, in congestion avoidance phase Tca 1 = T (ct W 0 ), [ N ca1 = W 0 (ct W 0 ) 1 + ct W ] 0 2 T ca2 = W2 max c2 T 2 2c, and N ca2 = ct ca2 Let, compare the contribution of slow-start and congestion avoidance phases When, B = ct, for large c We ll find that congestion avoidance dominates Only concentrate on TCP-dynamics for congestion-avoidance We ll use this insight to find throughput for simple model 6.976/ESD
29 TCP: Performance We consider only congestion-avoidance dynamics Single-link Many-sources Modeling T r : RTT of user r W r (t): window size of user r at time t q r (t): fraction of packet list at time t for user r x r (t) = W r(t) : rate of user r at time t T a r (t) = acknowledgment until time t for user r Dynamics Acknowledgment: increase window by Drop: decrease window by βw r (t) 1 W r (t) 6.976/ESD
30 TCP: Performance To study throughput of TCP We need to study evolution of W r (t) Let q r (t) fraction of packets are dropped at time t for user r Then, the rate at which packets are dropped at t for user r x r (t T r ) q r (t) That is, drop rate at time t for user r is x r (t T r )q r (t) Acknowledgment rate at t: x r (t T r )(1 q r (t)) Then, W r (t) changes as follows: dw r (t) dt = (1 q r (t)) [ ] 1 W r (t) x r(t T r ) [ ] q r (t) βw r (t) x r (t T r ) Since, x r (t) = W r (t)/t r dx r (t) dt = (1 q r(t))x r (t T r ) T r X r (t) q r (t)βt r x r (t)x r (t T r ) 6.976/ESD
31 TCP: Performance To obtain long-term effective throughput We evaluate the equilibrium point: dx r(t) = 0 dt Ignoring the delay Tr in equation usually, can not ignored we ll study when is this justified This gives us the following: In equilibrium That is, 0 = (1 ˆq r) T 1 βˆx 2 rt rˆq r ˆx r = 1 T r 1 ˆq r βˆq r 6.976/ESD
32 TCP: Performance The analysis has following main message Throughput is mainly affected by congestion avoidance phase, and not by slow-start Qualitatively, throughput is inversely proportional to RTT (T r ) square root inversely proportional to drop-probability 6.976/ESD
33 TCP: Performance Questions: We assumed deterministic dynamics how valid is it? The Law-of-Large-Numbers or Fluid-models provide justification We ignored the effect of other users implicit in q r ( ) What happens when many links? naturally, hard to quantify exact relation however, its useful to ask the following basic question: what do we really what to achieve from TCP and has it achieved that? next, we address this basic question 6.976/ESD
34 Resource Allocation Suppose a piece of cake is to be shared between two people in a fair manner: How should we do it? Well, divide it into half each, assuming both care for size only. What if one cares only for cherry but other cares about actual cake? Division scheme should care about utility of cake to the people. Known literature of cake-cutting algorithm has inspired a lot of interesting work in Game Theory and Algorithms 6.976/ESD
35 Resource Allocation Consider a single link with capacity C R users want to use it and let f r, 1 r R be rate at which user r wants to send data If f f R C allocate demanded rate to each user But, if f f R > C, then we need an allocation mechanism to allocate rates x 1,...,x R to all users such that 0 x r f r ; 1 r R R x r C r=1 and, allocation should maximize the overall system utility 6.976/ESD
36 Resource Allocation Let u r (x r ) be utility of rate x r to user r Then, allocation (x r ) should be solution of R max u r (y r ) r=1 subject to 0 y r f r ; 1 r R R y r C Now, a bad user can have u r very high can t rely on users utility for fair allocation Question: what should be u r ( ) so that allocation is fair? r= /ESD
37 Fair Allocation Let s consider a simple example C = 10 ; f 1 = 4, f 2 = 20. x 1 = 4 ; x 2 = 6 makes sense because allocation is as much equal as possible Max-min fair: (x r ) is max-min fair iff For any (r, r ); x r > x r only if x r = f r In the above example, x 1 = 3 ; x 2 = 1 is not max-min fair because x 2 > x 2 but x 1 f 1. Philosophy of max-min fair: At the fair allocation, the only way to become richer is to make a poor, poorer /ESD
38 Max-Min Fair Allocation Network resource allocation Question: Consider network with L links with capacities C 1,...,C L Sources 1,...,R with demands f 1,...f r. Let S denote set of all sources. Source r sends data from node s r to d r using some links (according to routing algorithm) Rates (x r ) are feasible if data transmission, according to the rates, satisfies link capacities constraints that is, no link is over-subscribed What about fair allocation? Similar definition generalizes /ESD
39 Max-Min Fair Allocation Definition (max-min fair): Question: Let (x r ) be max-min fair for a given capacitated network only if (1) (x r ) is feasible (2) for any other feasible (y r ), if y s > x s for some source s, then there exists source p s.t. x p x s and y p < x p How to find such an allocation? Next, we will see a simple centralized algorithm. Later we will consider distributed algorithms and their relationship to current TCP! 6.976/ESD
40 Max-Min Fair Allocation (1) Let n l be number of sources in S that use link l. For each l s.t. c l 0, define fair share f l as: g l = c l n l (2) Define z r = min g l, which is the min rate over links that source r is using. l r (3) Define z min = min r z r. (4) Let R = {r : z r = z min } (5) For all r R; the max-min rate x r = min(z r,f r ) (6) Set S S\ R ; c l c l x r ; r R ;l r n l n l r R 1 {l r} (7) Repeat (1) (6) until S is empty /ESD
41 Max-Min Fair Allocation Algorithm Not distributed Not implementable Fairness There can be other fairness criteria Next, we will see A range of fairness criteria max-min fair as one of them Study distributed algorithm for allocation based on them /ESD
42 α-fair Allocation Let utility of user r be u r (x r ) = x 1 α r r w r 1 α r α r > 0 ; α r 1 w r log x r α r = 1. Also, f r =, i.e. everyone wants maximal rate. Fair allocation is solution to optimization problem subject to max (x r ) ur (x r ) x r c l ; l L l r x r 0 Next, we consider special examples of the above class of fair allocation /ESD
43 Examples (I) Minimum delay fair α r = 2 ; r x 1 α r r u r (x r ) = w r 1 α r = w r x r (II) Proportional fair α r = 1 ; r (III) Max-min fair u r (x r ) = +w r log x r α r = α ; r ;α An example 6.976/ESD
44 Resource Allocation Consider routing matrix M = [M re ] R L M rl = 1 if data of user r passes through link l y = Mx ; x = (x r ) rate vector Then, resource allocation problem (RAC) max ur (x r ) (x r ) subject to x r 0 ; y = Mx C Instead of solving RAC, we will first solve (RAC1) max u r (x r ) x 0 r l L 0 Here, constraint y = Mx C is absent s: l s x s Instead, f l ( ) is penalty (price) function s.t. y 0 f l(x)dx as y non-decreasing, continuum, non-negative f l (y)dy /ESD
45 RAC1 Note that, for all choice of (α r ) u r ( ) is strictly concave, increasing Hence, the new utility function of RAC1 V (x) = u r (x r ) r l is strictly concave s: l s x s Proof [pg. 24, [SRIKANT]] 0 f l (y)dy We also assume that u r (x r ) as x r /ESD
46 RAC1 For maximizing strictly concave function V ( ) V (x) as x 0 V (x) + as x There is a unique solution of RAC1 that lies in the interior of set x 0. Hence, optimal solution must satisfy dv dx r = 0 r u r(x r ) f l ( s ) = 0 r l:l r s:l sx Solution of these equations leads to optimal route But, difficult to solve in decentralized manner 6.976/ESD
47 Decentralize Solution for RAC1 Each node can compute traffic through outgoing link and corresponding price at time t; i.e. y l (t) = s:l s x s(t) ; p l (t) = f l (y)l(t)) price on a route = sum of prices on its links, i.e. q r (t) = l:l r p l (t) or q = Mp Then, optimality condition is u r(x r ) q r = 0 A natural gradient algorithm is ẋ r (t) = k r (x r )(u r(x r ) q r ), with k r ( ) non-negative, non-decreasing continuous function /ESD
48 RAC1 Then: The gradient algorithm converges to optimal solution starting from any initial condition Proof [pg. 26, [SRIKANT]] Implication: To reach modified resource allocation, simple radiant algorithm based on prices and utility function is sufficient Question: What prices/penalty lead to correct algorithm? well, penalty function are approximations for capacity constraints: f l (x) as x C l actual utilities are never exact Next, some exact penalty functions 6.976/ESD
49 Exact Penalty Function An explanation of penalty functions can be given in terms of dual variables [explained later]. Based on this idea, it is possible to introduce adaptive penalty function ( yl ) Bl Define, f l (y l, c l ) = ; c virtual capacity c l We will adapt c l during the course of algorithm as follows: d c l (A) dt = α l(c l y l ) + c l And, primal algorithm ẋ r = k r (x r )(u r(x r ) q r ) ; q r = l:l r p l = l:l r f l (y l1, c l ). The above combination provides a primal algorithm with adaptive price-function 6.976/ESD
50 Exact Penalty Function Then: If routing matrix M is full rank, then algorithm solves RAC, exactly. Proof [pg. 30, [SRIKANT]] Implications: It is possible to change rates and price function adaptively to obtain solution of Resource allocation problem The price has natural optimization interpretation in terms of dual variable 6.976/ESD
51 RAC: Dual RAC: subject to max x 0 ur (x r ) Mx c. Lagrangian of RAC: x 0 ; λ 0 L(x;λ) = r u r (x r ) λ T (Mx c). Dual D(λ) = where, y l = s:l s x s. max x 0;Mx c = max x 0;Mx c [ ] u r (x r ) λ T (Mx c) r [ u r (x r ) r l λ l y l ] + l λ l c l, 6.976/ESD
52 Dual and It s Properties Dual optimal By strong-duality: inf D(λ) = RAC λ 0 inf D(λ) λ 0 For given λ 0: maximizing x is s.t. for all r L = 0 u r(x r ) x r x r l:l r λ l = 0 Further, by strong duality, for the optimizing (λ, x) pair λ l (y l c l ) = 0. Based on above optimality conditions, a simple gradient algorithm: [pg. 28, [SRIKANT]] Set x r = u 1 r ( l:l r λ l), λ l = h l (λ l )(y l c l ) + λ l /ESD
53 RAC: Dual Then: The gradient descent algorithm described for dual of RAC converges to optimal solution if the routing matrix M has full rank Proof [pg. 29, [SRIKANT]] Implication: By adjusting prices (λ l = p l ) and rational use behavior, desired rate-allocation can be achieved What if both prices and rate are changed simultaneously? 6.976/ESD
54 RAC: Primal Dual Primal-dual algorithm Primal algorithm at sources ẋ r = k r (x r )(u r(x r ) q r ), Dual algorithm at links ṗ l = k l (p l )(y l c l ) + p l. Then: The algorithm converges to the solution of RAC, starting from initial condition Proof [pg. 28, [SRIKANT]] 6.976/ESD
55 Relation to Current Algorithms Primal algorithm ẋ r = k r (x r )(u r(x r ) q r ) Changing rate based on feedback Recall, TCP changes window-size w r (t) be windowsize T r be RTT Then, rate x r (t) = w r(t) T r Let q r (t) be drop probability Then, TCP-Reno increases window if no drop (prb. 1 q r (t)) decreases window if drop (prb. q r (t)) 6.976/ESD
56 TCP versus Primal Algorithm Specifically, TCP dynamics ẋ r = 1 q r(t)x r (t T r ) T 2 r x r (t) Let x r (t T r ) x r (t); then βq r (t)x r (t)x r (t T r ). ẋ r = 1 q r(t) βx 2 T r(t)q r 2 r (t). If q r (t) small, 1 q r (t) 1; then ẋ r = 1 βx 2 T r(t)q r 2 r (t) [ = βx 2 r(t) 1 βt 2 r x 2 r(t) q r(t) = k r (x r (t)) [u r(x r (t)) q r (t)] 1 Then, u r (t) = βtr 2 x r (t) That is: TCP weighted delay fair allocation! ] 6.976/ESD
57 What About Dual? TCP corresponds to primal algorithm What about dual algorithm? q r ( ): corresponds to drop probability Queue management algorithm corresponds to dual algorithm Popular queue-management algorithm Random Early Detection (RED) Mark (or drop) packet with probability proportional to queue-size If butter-size is B, then probability of marking packet when queue-size is Q is Q B. Let y be arrival rate 6.976/ESD
58 Dual = Active Queue Management Now, queue-dynamics Then, marking (drop) probability Q = (y c) + Q p = Q B ṗ = α(y c)+ p This has exactly the same dynamics as prices or Lagrange multiplier in dual or primal-dual algorithm Note that q r l:l r p l ; because if drops at links are independent then (1 q r ) = (1 p l ) 1 p l, l:l r l:l r when p l is small /ESD
59 Next We assumed that RCA algorithm or TCP has immediate feedback. In practice, feedback is always delayed Questions: (1) How does delayed feedback affect performance? (2) If delay is very large: feedback is useless. Then how large can delay be so as the network can still operate? Next, we will see how ideas from control theory can be useful. Professor Mitter s lectures 6.976/ESD
60 Other Issues If everything works well, then TCP and RED seem good algorithms Question: What if uses do not cooperate? Malicious users can lead to undesirable network performance. Next, we study network security /ESD
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