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Summary: Performance Analysis of TCP

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• Simplifying assumptions
  • To make the analysis tractable, the author made some simplifying assumptions:

    Constant RTT "Integral" increase of CWND "One loss, all loss" packet loss model

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• Analysis 1: Loss Indications are exclusively "3 Duplicate ACKs"
  • Notations:

    Nt = number of packets transmitted in time interval [0, t]       = TCP throughput Bt = Nt/t = average number of packets transmitted in time interval [0, t]       = average TCP throughput in interval [0, t] B = lim(t &rarr &infin) Bt       = average TCP throughput over a very very long period of time

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  • The TD Period :

    NOTE:

    Due to the one lost, all lost packet loss assumption (see: click here ), the duplicate ACKs will be generated when the packets in the next transmission round are received

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• Notations related to thr throughput analysis
  • Notations:

    Yi = number of packets sent in the ith TD period ( Yi includes packets that were lost) Ai = duration of the ith TD period Wi = window size at the end of the ith TD period

  • Graphically:

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  • Throughput of TCP Reno:

    E[ Y ] B = -------- E[ A ]

    The throughput depends on the number of transmission rounds inside one TD period

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• Refined TCP throughput expression
  • Looking closely at how CWND increase inside a TD period:

    we find:

    E[Y] = E[ α ] + E[W] - 1 . . . . . . . . . . . . . . . . (2a)     E[A] = (E[X] + 1) × RTT . . . . . . . . . . . . . . . . (3a)

    where:

    αi = the index of the first packet that was lost X = number of complete transmission rounds in a TD period

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• Summary of results
  • E[X]:

    b E[X] = --- E[W] 2

    b = number of delayed ACKs

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  • E[α]:

    Prob[ α = k] = Prob[ the first packet lost is packet #k ] = Prob[ first k-1 packets not lost and the next packet is lost ] = (1 - p)k-1 × p 1 E[α] = --- (computed with Maple) p

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  • E[W]:

    E[W] is solved using 2 equations: (1 - p) E[Y] = E[W] + ------- (from above after replacing E[α]) p b 3 E[Y] = --- E[W] ( --- E[W] - 1 ) + E[β] (By counting the packets in a TD period) 4 2 E[β] = avg. number of packet in the last (partial) transmission round ~= E[W]/2 Solution is: -------------------------- / 2+b / 8(1-p) 2+b 2 E[W] = ----- + \ / -------- + ( ----- ) 3b \/ 3bp 3b

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• Final expression for TCP Throughput when there are only TD periods
  • Summary:

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• Analysis 2: Loss Indications are "3 Duplicate ACKs" and Time outs
  • Actual TCP Reno cycle: exponential increase in CWND in slow start

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  • Simplified Reno cycle: skip (short) exponential increase

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  • Average # packets transmitted in a Reno cycle:

    Notes:

    E[M] = E[Yi1] + E[Yi2] + ... + E[Yi,ni] + 1 + 1 + ... + 1 | | | | +------------------------------+ +---------------+ sent in the TD periods sent in TO periods = E[Yi1] + E[Yi2] + ... + E[Yi,ni] + E[R] Therefore: E[M] = E[n] * E[Y] + E[R]

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  • Average duration of a Reno cycle:

    Notes:

    E[S] = E[Ai1] + E[Ai2] + ... + E[Ai,ni] + E[ ZiTO] | | | | +------------------------------+ +--------+ lengths of the TD periods length of the TO period Therefore: E[S] = E[n] * E[A] + E[Z] . . . . . . . . . . . . . . . . (3)

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  • Throughput:

    E[n] * E[Y] + E[R] TCP Throughput = ----------------------- E[n] * E[A] + E[Z]

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  • E[Y] and E[A]

    Have been computed priviously in the TD recovery analysis above

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• Computing E[n], E[R] and E[Z]
  • E[n] can be computed as follows:

    E[n] = average number of TD periods in one Reno cycle = 1 * P[ 1st TD period ends in Timeout recovery ] + 2 * P[ 2nd TD period ends in Timeout recovery ] + 3 * P[ 3rd TD period ends in Timeout recovery ] + 4 * P[ 4th TD period ends in Timeout recovery ] + ... = 1 * Q + 2 * (1-Q) * Q + 3 * (1-Q)2 * Q + 4 * (1-Q)3 * Q + ... = Q * ( 1 + 2 * (1-Q) + 3 * (1-Q)2 + 4 * (1-Q)3 + ... ) 1 = --- Q

    where:

    Q = the probability that a TD period is ended by a Timeout recovery

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  • Q is computed by considering all possible ways to received less than 3 duplicate ACKs:

    Case 1: ( W <= 3 ) Case 2: ( W >= 4 ) and (packet 1, 2 or 3 in the transmission round was lost) Case 3: ( W >= 4 ) and (packet 1, 2 and 3 in the transmission round were not lost), but (packet 1, 2 or 3 in the RE-transmission round was lost)

  • Computing Q by conditioning on the window size W:

    Q = Q(1) * Prob[ W = 1 ] + Q(2) * Prob[ W = 2 ] + Q(3) * Prob[ W = 3 ] + Q(4) * Prob[ W = 4 ] + .... (upto infinity) . . . . . . . . . . . (6)

    Where:

    Q = the probability that a TD period ends in Timeout Q(W) = the probability that a TD Period with a congestion window size W will end in Timeout

  • Q(W):

    When W ≤ 3 Q(W) = 1 (can never receive 3 dup AKCs) When W ≥ 4 Q(W) = Prob[ (packet 1 in window lost) OR (packet 2 in window lost) OR (packet 3 in window lost) OR (packet 4 in window lost AND (packet 1, 2 or 3 in retrans lost) OR (packet 5 in window lost AND (packet 1, 2 or 3 in retrans lost) OR ... OR (packet W in window lost AND (packet 1, 2 or 3 in retrans lost) ] Result: Q(W) = A(W, 0) + A(W, 1) + A(W, 2) W-1 2 ---- +- ---- -+ \ | \ | + > | A(W, k) * > C(k, m) | . . . . . . . (4b) / | / | ---- +- ---- -+ k=3 m=0 With Maple: 2 W W 2 3 p (3 - 3 p + p ) ((1 - p) - (1 - p) p - 2 + 3 p - 3 p + p ) Q(W) = -------------------------------------------------------------- . . (5) W -1 + (1 - p)

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  • Q cannot be computed exactly because Prob[W = k] is unknown

    Approximation:

    Q = E[Q(W)] ≅ Q( E[W] ) . . . . . . . . . . . (7)

    In other words:

    2 W W 2 3 p (3 - 3 p + p ) ((1 - p) - (1 - p) p - 2 + 3 p - 3 p + p ) Q(W) = -------------------------------------------------------------- ....... (5) W -1 + (1 - p)

    Replace W in the above equation with E[W] below:

    -------------------------- / 2+b / 8(1-p) 2+b 2 E[W] = ----- + \ / -------- + ( ----- ) 3b \/ 3bp 3b

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  • E[R]:

    E[R] = 1 * Prob[R=1] + 2 * 1 * Prob[R=2] + 3 * 1 * Prob[R=3] + ... = 1 * Prob[R=1] + 2 * Prob[R=2] + 3 * Prob[R=3] + ...

  • P[R = k]:

    Prob[ R = k ] = probability that there are k retransmission rounds = Prob[ ((k - 1) rounds unsuccessful) AND last round successful ] = pk-1 * (1 - p)1

  • Therefore E[R] :

    E[R] = 1*Prob[R=1] + 2*Prob[R=2] + 3*Prob[R=3] + ... = 1 * (1-p) + 2 * p*(1-p) + 3 * p2*(1-p) + .... 1 = ------- 1 - p

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  • E[Z]:

    E[Z] = L1 * Prob[R=1] + L2 * Prob[R=2] + L3 * Prob[R=3] + ... Li = length of the timeout period if R = i E[Z] = Prob[ R=1 ] * T0 + Prob[ R=2 ] * (3 * T0) + Prob[ R=3 ] * (7 * T0) + Prob[ R=4 ] * (15 * T0) + Prob[ R=5 ] * (31 * T0) + Prob[ R=6 ] * (63 * T0) + Prob[ R=7 ] * (63+1*64) * T0 + Prob[ R=8 ] * (63+2*64) * T0 + Prob[ R=9 ] * (63+3*64) * T0 + .... = p0 (1-p) * T0 + p1 (1-p) * (3 * T0) + p2 (1-p) * (7 * T0) + p3 (1-p) * (15 * T0) + p4 (1-p) * (31 * T0) + p5 (1-p) * (63 * T0) + p6 (1-p) * (63+1*64) * T0 + p7 (1-p) * (63+2*64) * T0 + p8 (1-p) * (63+2*64) * T0 + .... Using Maple: 2 3 4 5 6 1 + p +2 p + 4 p + 8 p + 16 p - 31 p E[Z] = TO ------------------------------------------ 1 - p

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• From Padhye's paper...
  • Throughput of TCP: