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U of I CS 438 - Performance Analysis

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Performance AnalysisPerformance Analysis TCP delay modeling With applications to HTTP Introduction to queueing theory With a probability refresher TCP throughput analysis (revisited)Delay modelingQ: How long does it taketo receive an objectfrom a Web serverafter sending arequest?Ignoring congestion,delay is influenced by: TCP connectionestablishment data transmission delay slow startNotation, assumptions: Assume one link between clientand server of rate R S: MSS (bits) O: object size (bits) no retransmissions (no loss, nocorruption)Window size: First assume: fixed congestionwindow, W segments Then dynamic window,modeling slow startFixed congestion window (1)First case:WS/R > RTT + S/R: ACK for firstsegment in window returnsbefore window’s worth of datasentdelay = 2RTT + O/RFixed congestion window (2)Second case: WS/R < RTT + S/R:wait for ACK aftersending window’sworth of data sentdelay = 2RTT + O/R+ (K-1)[S/R + RTT - WS/R]TCP Delay Modeling: Slow Start (1)Now suppose window grows according to slowstartWill show that the delay for one object is:RSRSRTTPRORTTLatencyP)12(2 !!"#$%&'+++=where P is the number of times TCP idles at server:}1,{min !=KQP- where Q is the number of times the server idles if the object were of infinite size.- and K is the number of windows that cover the object.TCP Delay Modeling: Slow Start (2)RTTinitiate TCPconnectionrequestobjectfirst window= S/Rsecond window= 2S/Rthird window= 4S/Rfourth window= 8S/Rcompletetransmissionobjectdeliveredtime atclienttime atserverDelay components:• 2 RTT for connectionestab and request• O/R to transmit object• time server idles due toslow startServer idles: P = min{K-1,Q} timesExample:• O/S = 15 segments• K = 4 windows• Q = 2• P = min{K-1,Q} = 2Server idles P=2 timesTCP Delay Modeling (3)RSRSRTTPRTTRORSRTTRSRTTROidleTimeRTTROPkPkPpp)12(][2]2[22delay111!!+++=!+++=++=!==""RTTinitiate TCPconnectionrequestobjectfirst window= S/Rsecond window= 2S/Rthird window= 4S/Rfourth window= 8S/Rcompletetransmissionobjectdeliveredtime atclienttime atserverTCP Delay Modeling (4)!!"##$+=+%=%&=%+++=%+++=&&)1(log)}1(log:{min}12:{min}/222:{min}222:{min22110110SOSOkkSOkSOkOSSSkKkkkLLCalculation of Q, number of idles for infinite-size object,is similar.Recall K = number of windows that cover objectHow do we calculate K ?HTTP Modeling Assume Web page consists of: 1 base HTML page (of size O bits) M images (each of size O bits) Non-persistent HTTP: M+1 TCP connections in series Response time = (M+1)O/R + (M+1)2RTT + sum of idle times Persistent HTTP: 2 RTT to request and receive base HTML file 1 RTT to request and receive M images Response time = (M+1)O/R + 3RTT + sum of idle times Non-persistent HTTP with X parallel connections Suppose M/X integer. 1 TCP connection for base file M/X sets of parallel connections for images. Response time = (M+1)O/R + (M/X + 1)2RTT + sum of idle timesHTTP Response time (in seconds)RTT = 100 msec, O = 5 Kbytes, M=10 and X=5For low bandwidth, connection & response time dominated bytransmission time.Persistent connections only give minor improvement over parallel connections.HTTP Response time (in seconds)RTT =1 sec, O = 5 Kbytes, M=10 and X=5For larger RTT, response time dominated by TCP establishment & slow start delays. Persistent connections now give important improvement: particularly in high delay·bandwidth networks.Queueing Theory Intro Recall router model Want to model Q(t) under real-world,probabilistic A(t) and D(t) Start with a probability refresherµA(t), λD(t)Model of FIFO router queueQ(t)A Quick Probability Refresher A random variable, X, can take on a number ofdifferent possible values Example: the number of students who came to officehours is a random variable with values 1,2,3,… Each time we observe (or sample) the randomvariable, it may take on a different value A random variable takes on each of these valueswith a specified probability Example: X = {0, 1, 2, 3, 4} P[X=0] = .1, P[X=1] = .2, P[X=2] = .4, P[X=3] = .1, P[X=4]= .2 The sum of the probabilities of all values equals 1 Σall values P[X=value] = 1A Quick Probability Refresher Example Suppose we throw two dice and the random variable, X, isthe sum of the two dice Possible values of X are {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} P[X=2] = P[X=12] = 1/36 P[X=3] = P[X=11] = 2/36 P[X=4] = P[X=10] = 3/36 P[X=5] = P[X=9] = 4/36 P[X=6] = P[X=8] = 5/36 P[X=7] = 6/36Note:12Σi=2P[X=i] = 1A Quick Probability Refresher A probability distribution function matcheseach possible value of a random variablewith its associated probabilityA Quick Probability Refresher The cumulative distribution function ofa random variable, X, is defined by CDF: P[X ≤ x] = Σall y=x P[x=y]A Quick Probability Refresher Expected Value Can be thought of a “long term average” ofobserving the random variable a large number oftimes Example: dice E[X] = 2*1/36 + 3*2/36 + 4*3/36 + 5*4/36 +6*5/36 + 7*6/36 + 8*5/36 + 9*4/36 + 10*3/36 +11*2/36 + 12*1/36E[X] = x =ΣAll possible values of xValue * P[X = value]A Quick Probability Refresher Average vs. Expected Value Short term average Suppose a random variable X is sampled N times Let ni = # of X = i was observed Average of samples = (n0*0 + n1*1 + n2*2 + n3*3 + …)/N = n0/N*0 + n1/N*1 + n2/N*2 + n3/N*3 + … As N → ∞, the ratio ni/N becomes pi Thus, E[X] = limN→∞ [n0/N*0 + n1/N*1 + n2/N*2 + n3/N*3 + …] = p0*0 + p1*1 + p2*2 + p3*3 + … = Σ∞i=0 i * piA Quick Probability Refresher Continuous Random Variables In many cases, a random variable takes a value drawnfrom a continuous interval Ex: processing time for a packet may be any real value [0, ∞) The distribution of possible values a continuous randomvariable can take is given by a probability density function,F(x) P(a ≤ x ≤ b) = ⌠ab F(x)dx = Σbi=a P(x = i) E[x] = ⌠-∞∞ xF(x)dx = Σ i * P(x = i)Basic Queueing Theory Elementary notions Things arrive at a queue according to some probabilitydistribution Things leave a queue according to a second probabilitydistribution Averaged over time Things arriving and things leaving must be equal Or the queue length will grow without bound Convenient to express probability distributions as averageratesLittle’s Law Goal Estimate relevant


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U of I CS 438 - Performance Analysis

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