Network Tomography CS 552 Richard Martin. What is Network Tomography? Derive internal state of the network from: –external measurements (probes) –Some.

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Presentation transcript:

Network Tomography CS 552 Richard Martin

What is Network Tomography? Derive internal state of the network from: –external measurements (probes) –Some knowledge about networks Captured in simple models.

Why Perform Network Tomography? Can’t always see what’s going in the network! –Vs. direct measurement. Performance –Find bottlenecks, link characteristics Diagnosis –Find when something is broken/slow. Security. –How to know someone added a hub/sniffer?

This week’s papers J. C. Bolot –Finds bottleneck link bandwidth, average packet sizes using simple probes and analysis. R. Castro, et al. –Overview of Tomography Techniques M. Coats et. al. –Tries to derive topological structure of the network from probe measurements. –Tries to find the “most likely” structure from sets of delay measurements. Heidemann et. Al. –Recent survey and techniques (as of summer 2008)

Measurement Strategy Send stream of UDP packets (probes) to a target at regular intervals (every  ms) Target host echos packets to source Size of the packet is constant (32 bytes) Vary  (8,20,50, 100,200, 500 ms) Measure Round Trip Time (RTT) of each packet.

Definitions s n sending time of probe n r n : receiving time of probe n rtt n = r n -s n : probe’s RTT  : interval between probe sends Lost packets: r n undefined, define rtt n =0.

Time Series Analysis n (packet #) RTT n (ms) Min RTT: 140 ms Mean RTT: ? Loss rate: 9%

Classic Time series analysis Stochastic analysis –View RTT as a function of time (I.e. RTT as F(t)) –Model fitting –Model prediction What do we really want from out data? –Tomography: learn critical aspects of the network

Phase Plot: Novel Interpretation RTT n RTT n+1 View difference between RTT’s, not the RTT itself Structure of phase plot tells us: bandwidth of bottleneck!

Simple Model Probe Traffic DFIFO queue  Other Internet traffic Fixed delayVariable delay rtt n = D + w n + p/   : bottleneck router’s service rate k: buffer size p: size of the probe packet (bits) w n : waiting time for probe packet n

Expectation for light traffic What do we expect to see in the phase plot –when traffic is light –  is large enough and p small enough not to cause load. w n+1 = w n rtt n+1 = rtt n For small p, approximate w n = 0

Light Traffic Example RTT n (ms) RTT n+1 (ms) “corner” (D,D) D = 140 ms n=800  =50 ms

Heavy load expectation Probe Traffic D FIFO queue  P n+k PnPn P n+1 P n+2 Burst rtt n+1 = rtt n + B/  Probe compression effect rtt n+2 - rtt n+1 = (r n+2 - s n+2 ) - (r n+1 - s n+1 ) = (r n+2 - r n+1 ) - (s n+2 - s n+1 ) = p/  -  Time between compressed probes Time between probe sends

Heavy load, cont/ What does the entire burst look like? rtt n+3 - rtt n+2 = rtt n+k - rtt n+k-1 = p/  -  Rewrite: rtt n+1 = rtt n + (p/  -  General form: y= x + (p/  -  Should observe such a line in the phase plot.

Finding the bottleneck  y= x + (p/  -  Find intercept. Know p, , can compute  !

Average packet size Can use phase data to find the average packet size on the internet. Idea: large packets disrupt phase data –Disruption from constant stream d, can infer size of the disruption. –Use distribution of rtt’s

Average packet size –Lindley’s Recurrence equation –Relationship between the waiting time of two successive customers in a queue: w n : waiting time for customer n y n : service time for customer n x n : interarrival time between customers n, n+1 arrivals departures time n n+1 n-1 n wnwn ynyn w n+1 w n+1 = w n + y n -x n, if w n + y n -x n > 0 xnxn w n+1 = prev. packet wait + service - overlap

Finding the burst size Model a slotted time of arrival where slots are defined by probe boundaries wb n = max(w n + p/ , 0) Apply recurrence: w n+1 = w n + (p +b n )/  -  Substitute and solve for b n: Note: assume w n + (p +b n )/  -  > 0, then b n =  w n+1 - w n +  - p

Distribution plot distribution of w n+1 - w n +  ms 1st peak w n+1 -w n = p/  2nd: w n+1 =w n 3rd: b n =  w n+1 -w n +  -p know, , p solve for b n

Inter-arrival times A packet arrived in a slot if: w n+1 - w n > p /  -  Choose a small  Avoid false positives Count a packet arrival if: w n+1 - w n >0

Fraction of arrival slots slot Fitted to p(1-p) k-1, p=0.37

Packet loss What is unconditional likelihood of loss? –ulp = P(rtt n =0) Given a lost packet, what is conditional likelihood will lose the next one? –clp = P(rtt n+1 =0 | rtt n =0 ) Packet loss gap: –The number of packets lost in a burst –plg = 1/(1-clp)

Loss probabilities  (ms) ulp clp plg

Tomography Overview Basic idea Methods Formal analysis Future directions

Traffic Matrix Approaches Cast problem of the form: –Yt= Axt+et

Traffic Matrix example Send multicast packet Measure delay of packet at receivers Shared paths result in shared delay Find the “most likely” tree given the observations

Traffic Matrix Example Source node Destination nodes Intermediate routes

Problem Set-up YAX End observed delayRouting matrixLink delays

Introduction –Performance optimization of high-end applications –Spatially localized information about network performance Two gathering approaches: Internal: impractical(CPU load, scalability, administration…) External: network tomography –Cooperative conditions: increasingly uncommon –Assumption: the routers from the sender to the receiver are fixed during the measurement period

Contributions A novel measurement scheme based on special-purpose unicast “”sandwich” probes –Only delay differences are measured, clock synchronization is not required A new, penalized likelihood framework for topology identification –A special Markov Chain Monte Carlo (MCMC) procedure that efficiently searches the space of topologies

Sandwich Probe Measurements Sandwich: two small packets destined for one receiver separated by a larger packet destined for another receiver

Sandwich Probe Measurements Three steps –End-to-end measurements are made –A set of metrics are estimated based on the measurements –Network topology is estimated by an inference algorithm based on the metric

Step 1: Measuring (Pairwise delay measurements)

Step 1: Measuring (Continue) Each time a pair of receivers are selected Unicast is used to send packets to receivers Two small packets are sent to one of the two receivers A larger packet separates the two small ones and is sent to the other receiver The difference between the starting times of the two small packets should be large enough to make sure that the second one arrives the receiver after the first one Cross-traffic has a zero-mean effect on the measurements (d is large enough)

Step 1: Measuring (Continued)  35 is resulted from the queuing delay on the shared path

Step 1: Measuring (Continued) More shared queues  larger    

Step 2: Metric Estimation More measurements, more reliable the logical topology identification is. The choice of metric affects how fast the percentage of successful identification improves as the number of measurements increases Metrics should make every measurement as informative as possible Mean Delay Differences are used as metrics –Measured locally –No need for global clock synchronization

Step 2: Metric Estimation(Continued) The difference between the arrival times of the two small packets at the receiver is related to the bandwidth on the portion of the path shared with the other receiver A metric estimation is generated for each pair of receivers.

Step 2: Metric Estimation(Continued) Formalization of end-to-end metric construction –N receivers  N(N-1) different types of measurements –K measurements, independent and identically distributed –δ(k) – difference between arrival times of the 2 small packets in the k th measurement –Get the sample mean and sample variance of the measurement for each pair (i,j): x i,j and  i,j 2 (Sample mean of sample X = (X 1, X 2,...) is M n (X) = (X 1 + X 2 + ··· + X n ) / n (arithmetic mean) Sample variance is (1 / n)Σ i=1..n (X i − μ) 2 E(M n ) = μ )

Step 3: Topology Estimation Assumption: tree-structured graph Logical links Maximum likelihood criterion: –find the true topology tree T* out of the possible trees (forest) F based on x Note: other ways to find trees based on common delay differences (follow references) Probability model for delay difference –Central Limit Theorem  x i,j ~ N(γ i,j,σ i.j /n i,j ) –y i,j is the the theoretical value of x i,j –That is, sample mean be approximately normally distributed with mean y i,j and variance s i.j /n i,j – The larger n i,j is, the better the approximation is.

Step 3: Topology Estimation(Cont.) Probability density of x is p(x|T,  (T)), means  (T) is computed from the measurements x Maximum Likelihood Estimator (MLE) estimates the value of  (T) that maximizes p(x|T,  (T)), that is, Log likelihood of T is Maximum Likelihood Tree (MLT) T* T* = argmax TЄF

Step 3: Topology Estimation(Cont.) Over fitting problem: the more degrees of freedom in a model, the more closely the model can fit the data Penalized likelihood criteria: –Tradeoff between fitting the data and controlling the number of links in the tree Maximum Penalized Likelihood Tree(MPLT) is

When N is large, it is infeasible to exhaustively compute the penalized likelihood value of each tree in F. A better way is concentrating on a small set of likely trees Given: Posterior density = x can be used as a guide for searching F. Posterior density is peaked near highly likely trees, so stochastic search focuses the exploration Finding the Tallest Tree in the Forest

Stochastic Search Methodology Reversible Jump Markov Chain Monte Carlo –Target distribution: –Basic idea: simulate an ergodic markov chain whose samples are asymptotically distributed according to the target distribution –Transition kernel: transition probability from one state to another –Moves: birth step, death step and  -step

Birth Step A new node l* is added  extra parameter  l * The dimension of the model is increased Transformation (non-deterministic)  l * = r x min(  c (l,1),  c (l,2))  ’ c (l,1) =  c (l,1) –  l *  ’ c (l,2) =  c (l,2) -  l *

Death Step A node l* is deleted The dimension of the model is reduced by 1 Transformation (deterministic)  c (l,1) =  ’ c (l,1) +  l *  c (l,2) =  ’ c (l,2) +  l *

 –step Choose a link l and change the value of  l New value of  l is drawn from the conditional posterior distribution

The Algorithm Choose a starting state s 0 Propose a move to another state s 1 –Probability = Repeat these two steps and evaluate the log- likelihood of each encountered tree Why restart?

Penalty parameter Penalty = 1/2log 2 N N: number of receivers

Simulation Experiments Compare the performance of DBT(Deterministic Binary Tree) and MPLT Penalty = 0 (both will produce binary trees) 50 probes for each pair in one experiment, 1000 independent experiments When the variability of the delay difference measurements differ on different links, MPLT performs better than DBT Maximum Likelihood criteria can provide significantly better identification results than DBT

ns Experiment Topology used for the experiment

Experiment Results

Internet Experiment Source host: data collection and inference Receivers: a low overhead receiver task 8 minutes/experiment, 6 independent experiments 1 sandwitch probe / 50ms Penalty = 1.7 topology

Experiment Result Estimated topology

Conclusions and Future work Conclusions: –Delay-based measurement without the need for synchronization –MCMC algorithm to explore forest and identify maximum (penalized) likelihood tree –Foundation for multi-sender topology identification –Localization of layer-two elements Future work –Adaptive methods for selecting penalty parameter –Adaptivity in the probing scheme

Extra Credit Log into planetLab nodes –Use SSH with class-provided key Pick a set of hosts to perform the experiment –A set of 2 given hosts posted for the class You pick 3 more: –East Asia -> North America –North America -> Europe –Europe -> East Asia Generate & record a 1 minute ping sequence with different  (6 in all) 1, 5, 15, 50, 100, 200 ms

Extra Credit (cont) For each trace (30 in all): –Plot the phase plot –Find the equation of the line y= x + (p/  -  ) –Plot the distribution plot –Find the first three peaks; find b n For a set of traces between 2 hosts: –Provide the table of ulp, clp, plg

Extra Credit (cont) What to hand in: –Short paragraph describing the experiment, and problems you had –Phase plots + equations –Distribution plots + positions of peaks, Bn –Probability table –Label plots with source, destination host names, time of experiment, length of experiment