On the Fractal Behavior of TCP Anna Gilbert and Howard Karloff∗
ABSTRACT
complicated than this model [13], having slow start, round trip timer, fast retransmit, etc. Several recent studies on the impact of TCP on traffic characteristics [6, 9, 14] suggest that TCP’s congestion control algorithm plays a significant role in the multifractal nature of network traffic and that ignoring the implicit feedback loop in this control algorithm yields misleading queueing performance predictions. To gain a more fundamental understanding of the small time-scale features of network traffic on a per-link basis, as well as a global picture of many sources interacting over large networks, we require a realistic model, yet one simple enough to be mathematically tractable. Before we propose and adopt changes to such a critical protocol, we must have a rigorous understanding of its current behavior. Veres and Boda [14] were the first to propose that TCP itself, even while the sources are sending infinitely large files, could generate fractal behavior. They made an empirical observation: they observed that a plot of suitably averaged rate pairs (r1 , r2 ) of two sources transmitting infinitely large files seems fractal. They proposed that TCP congestion control is a chaotic system, capable of producing “complicated” behavior. Their study was empirical, obtained by running the network simulator ns, and, as such, neither proposed a mathematical model nor proved fractal behavior, since fractals are defined (see below) by limiting behavior. Our contribution is two-fold. First, we propose a natural, mathematically tractable model of TCP which captures (only) TCP’s additive-increase, multiplicative-decrease behavior. While our model does not encompass much of the complexity of real TCP, we feel it captures its most important aspect. Second, we prove, for certain choices of the parameters, that the model leads to fractal long-term behavior. Knowing the set of all possible rate pairs sets the stage for analyzing the resulting aggregate traffic and its properties. This knowledge is critical for traffic engineering and performance management. Here is our model. Two sources 1 and 2 maintain rates r1 and r2 , respectively, starting with some positive values. A recipient has a buffer of size B ≥ d into which both sources try to inject packets. The buffer’s occupancy is represented by b (0 ≤ b ≤ B) and the buffer is initially full. In each time step:
We propose a natural, mathematically tractable model of TCP which captures both its additive-increase, multiplicative-decrease behavior and its feedback mechanism. Neither a fluid nor a mean-field model, our model does not explicitly model the loss process; the losses are entirely determined by the rates of the sources at the time of buffer overflow. The system involves two sources competing to send packets into one recipient buffer of size B, from which bytes are drained at the rate of d per step. We prove that for many choices of the pairs (B, d), the long term behavior of the system is fractal. We conjecture that this fact continues to hold for all B > d and d > 2.
Categories and Subject Descriptors C.2 [Network protocols]: Routing protocols
General Terms Algorithms, theory.
Keywords TCP, fractal, network protocol, Internet, algorithm.
1.
INTRODUCTION
The Transmission Control Protocol (TCP) is part of the TCP-IP suite used to control 90% or more of all Internet traffic. Fundamental to TCP is its additive-increase, multiplicative-decrease behavior: as long as packets are successfully acknowledged, the source additively increases the rate at which it sends packets. Once packets are dropped, the source “backs off” by cutting its transmission rate, usually by a factor of two. Of course, real TCP is far more ∗ AT&T Labs–Research, 180 Park Ave., Florham Park, NJ 07932. E-mail:
[email protected] and
[email protected].
Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. To copy otherwise, to republish, to post on servers or to redistribute to lists, requires prior specific permission and/or a fee. STOC ’03, June 9–11, 2003, San Diego, California, USA. Copyright 2003 ACM 1-58113-674-9/03/0006 ...$5.00.
1. The recipient drains d units from its buffer, but never lets the buffer occupancy b go negative: b := max{b − d, 0}. 2. Each source increases its rate by 1: r1 := r1 + 1, r2 := r2 + 1 (additive increase).
297
3. Pick an i ∈ {1, 2} arbitrarily. 4. Source i “fires,” which means that the buffer occupancy b is increased by ri , except, of course, that it can never exceed B: b := min{b + ri , B}. Source i now backs off if the buffer is currently full: if b = B, set ri := ri /2 (multiplicative decrease). 5. Now source j = 3 − i fires: b := min{b + rj , B}. If b = B, set rj := rj /2 (multiplicative decrease). This process is repeated ad infinitum.1 Note that B, d, r1 , r2 are real numbers. The state of the system at any given time is given by the triple (r1 , r2 , b). Observe that if the first source to fire fills the buffer in a given round, necessarily both sources will back off. This model most closely resembles sources transmitting infinitely large files over a network with infinite line speeds, so that acknowledgements, or lack thereof, are received instantaneously and not lost or delayed in transit. In reality, even though basic TCP is deterministic, the environment in which it is run, due to line delays, delays caused by other traffic in the network, etc., is not: the same two sources could, in practice, experience different delays tomorrow from today. Also, the order of packet arrivals at a queue depends on background traffic, delays, round trip times, the reception of preceding ACKs, etc. Thus it seems reasonable to consider a nondeterministic model where the nondeterminism enters only in the order of packet arrival at the buffer. It is important to note that our model is neither a fluid model of TCP (see [1]) nor a mean-field approximation [8]. Nor is there an external loss process or a fixed probability of packet loss [4]; a source backs off precisely if the buffer is full after it has injected packets into it. Furthermore, the sources “interact” with each other; if one causes congestion by sending too many packets, the other is affected immediately (i.e., its loss is “caused by” the other source). We will prove that for some pairs (B, d), the long-term behavior of the model is fractal. By this, we mean that the attractor AB,d , the set of possible rate pairs obtainable after a large number of fills of the buffer (see Section 2 for a definition), defines a fractal. (This is definitely not the same as saying that the traffic is fractal, in the sense of [11]. In that case, the traffic generated by such sources exhibits statistical self-similarity.) See Figure 1 for pictures of the possible rate pairs when (B, d) is (4, 3) and (3000, 51), respectively. But what is a fractal? Although there is no standard definition (and Mandelbrot, one of the founders of the study of fractals, thinks there shouldn’t be [12, page 361]), one frequently-used definition is a set of nonintegral box dimension. For a large range of pairs (B, d), we prove that the attractor of our dynamical system has nonintegral box dimension. While we do not address in this paper such features as self-similarity (or lack thereof) and multifractal scaling (in the sense of [7]) of the aggregate traffic that our model generates, we believe that the connection between additive-increase, multiplicative-decrease feedback systems and chaotic dynamical systems, as evidenced by Veres and Boda [14], is significant enough to merit a rigorous fractal analysis of the attractor.
Figure 1: Numerical simulation of the attractors AB,d for B = 4, d = 3 (above) and B = 3000, d = 51 (below), computed pseudorandomly, with exact rational arithmetic for the iterations. The nondeterministic choices were made uniformly at random for a total of 100 buffer fills. The paper is arranged as follows. To guide the reader, we begin with several mathematical tools and definitions which we employ throughout the paper. Then we tackle our main result. We start by proving that for any B, d, AB,d ⊆ [1, d]2 (Section 3). Next we prove a general upper bound on the upper box dimension of the attractor: for any B, d, the upper box dimension of AB,d is less than 2 (Section 4). Then we give two cases in which the lower box dimension exceeds 1: • For any d ∈ (19/7, 3) and any B ≥ d, the lower box dimension of AB,d exceeds 1 (Section 5). • For any d which is a sufficiently large (integral) multiple of 6 and any B > d2 , the lower box dimension of AB,d exceeds 1 (Section 6). Hence, in the two cases above, AB,d is a fractal.
1
Code to simulate the dynamical system described in the text is available at http://www.research.att.com/ ~agilbert/tcp.fractal.html.
2.
298
MATHEMATICAL TOOLS
1/ lg 3 = log 2/ log 3 = lim inf i→∞ [(lg ai )/(− lg 3−i )]. Hence C’s box dimension is log 2/ log 3, which is strictly between 0 and 1. We use a more complicated variant of this same technique to prove our main theorems.
In this section, we define precisely what the attractor AB,d is and how to calculate its box dimension. We are interested in stopping the system after some large number, say, 100, of buffer fills, and examining the rate pair (r1 , r2 ) at that time. Which rate pairs could we possibly see after 100 buffer fills? A rate pair may occur after 100 buffer (0) (0) fills if, and only if, there are an initial pair (r1 , r2 ) and (0) (0) a sequence of choices so that, starting from (r1 , r2 , B) and choosing the given sources to fire first, the final state is (r1 , r2 , B). Since there’s nothing special about 100, it seems natural to study the set of rate pairs (r1 , r2 ) that can occur in the limit, i.e., can occur after N buffer fills, for all N . (If a pair can occur after N buffer fills, then it can occur after M , for all M ≤ N .) Specifically, say (0) (0) (r1 , r2 ) is an N th order preimage of (r1 , r2 ) if, starting (0) (0) in state (r1 , r2 , B), there is a sequence of choices which fills the buffer exactly N times and leaves the system in state (r1 , r2 , B). Insensitive to transient effects, the attractor A of the system is defined as follows: A = AB,d = {(r1 , r2 )|(r1 , r2 ) has an N th order preimage for all N }. We will show, for certain pairs (B, d), that AB,d is a fractal; that is, we will show that AB,d has nonintegral box dimension, where box dimension is defined as follows [3, 5]. Given a subset S of 2 and two positive reals a, b, let Na×b (S) be the minimum number of axis-aligned a × b (closed) rectangles (length a on the x-axis, b on the y) whose union is a superset of S.
3.
THE ATTRACTOR IS A SUBSET OF [1, d]2
In this section, we show that for any pair (B, d), the attractor AB,d is a subset of [1, d]2 . Definition 2. The Poincar´e section is the set of states (r1 , r2 , B) when the buffer is full. Lemma 3. The system always returns to the Poincar´e section in a finite number of steps. Proof. Otherwise, r1 and r2 would grow without bound until their sum exceeded d+1, and each later iteration would increase the buffer occupancy b by at least 1. This can clearly happen at most dBe times. Let us define the supersystem to be the system induced by the Poincar´e return map on the original system. That is, we observe the rates of the sources only when the buffer is full (when the map returns to the section). The following lemma shows that the attractors on the section for these two systems are the same and that analysis of the return map is equivalent to that of the original map on the Poincar´e section.
R
Lemma 4. The attractors are the same. (Proof omitted.)
Definition 1. The upper box dimension of S is lim supi→∞ [lg(N2−i ×2−i (S))/i] and its lower box dimension is lim inf i → ∞[lg(N2−i ×2−i (S))/i]. If (and only if ) the lim sup equals the lim inf, the common value is called the box dimension of S.
Definition 5. Given (r1 , r2 ), define i = i(r1 , r2 ) to be the number of iterations, starting from state (r1 , r2 , B), until the system is next in a state of the form (r10 , r20 , B), i.e., the number of iterations until the buffer is next full. (We will often omit the “(r1 , r2 )” from “i(r1 , r2 )”.)
(Later we will need Na×b (S) for a 6= b.) There are also other definitions in the literature. For example, one can use ²i × ²i boxes, where ²i goes to 0 geometrically in i, instead of 2−i × 2−i boxes, dividing by − lg ²i instead of i = − lg 2−i , or insist that the corners of the a × b rectangles have x-coordinates which are multiples of a and ycoordinates which are multiples of b, but, since they all give the same values for the upper and lower box dimension, one can use whichever definition one chooses. As an illustration, we calculate the box dimension of the Cantor set. All the box dimension estimates in this paper are derived via the technique used here. Recall that the Cantor set C is the set of all real numbers in [0, 1) whose ternary representation contains no 1’s. Equivalently, C is obtained from [0, 1) by removing its middle third [1/3, 2/3), then removing the middle thirds [1/9, 2/9), [7/9, 8/9) from the two pieces that remains, and then removing the middle third from each of the four remaining pieces, etc. We choose to use boxes with side length 3−i instead of 2−i . Let ai = N3−i ×3−i (C). Since C 6= ∅, a0 = 1. (We could use 1-dimensional boxes here, since C ⊆ 1 , if we wished.) Since the leftmost third and rightmost third of C are scale copies of C each of which we’ll call “(1/3)C,” we can write N3−i ×3−i (C) = 2N3−i ×3−i ((1/3)C) provided i ≥ 1. Now we make the key observation that N3−i ×3−i ((1/3)C) = N3−(i−1) ×3−(i−1) (C). We have ai = 2ai−1 if i ≥ 1, and a0 = 1, to which the solution is ai = 2i . It follows that lim supi→∞ [(lg ai )/(− lg 3−i )] = lim supi→∞ [i/(i lg 3)] =
Lemma 6. If B > d2 , then the buffer is never empty. This lemma, whose proof we omit, says that a buffer size exceeding d2 is effectively infinity. Even though B, d are arbitrary here, we need a lemma detailing i = i(r1 , r2 ) for the “B > d2 ” case. Lemma 7. If B > d2 , then i = max{1, dd−(r1 +r2 )−1e}. (Proof omitted.) Nonetheless, note that if d is integral and r1 + r2 is not, then i = d − dr1 + r2 e if r1 + r2 < d − 2, and i = 1 otherwise. Theorem 8. The attractor AB,d is a subset of [1, d]2 . We will prove three lemmas, which together imply that AB,d ⊆ [1, d]2 (Theorem 8). Lemma 9. If (r1 , r2 ) is in the attractor, then r1 , r2 ≥ 1. Proof. Whatever the current value of ri , after one more iteration its value is at least (ri + 1)/2, which is at least 1 if ri ≥ 1 and at least half the distance to 1 otherwise. Hence if (r1 , r2 ) is in the attractor, r1 , r2 ≥ 1. It is easy to prove that i = 1 if and only if r1 + r2 ≥ d − 2.
R
Lemma 10. If r1 , r2 ≤ d on the Poincar´e section, then at the next visit to the Poincar´e section, the same will be true. (Proof below.)
299
(r10 , r20 ) = f1 (r1 , r2 ), we first calculate i = i(r1 , r2 ), and set r20 = (r2 + i)/2 = r2 /2 + i/2; r10 = r1 + i usually, but might equal (r1 + i)/2. We observe first that the set of points on the Poincar´e section over which i(r1 , r2 ) is uniform and a given number (one or two) of sources is reset is a polygon. More formally, given any i, let I(i) = {(r1 , r2 )|i(r1 , r2 ) = i and only one source (i.e., r2 ) is reset under the map f1 }. Given reals α, β, and γ, let P (α, β, γ) denote the open polygonal region {(r1 , r2 )|α < r1 + r2 < β, r1 < γ}.
Lemma 11. Let (r1 , r2 , B) be a point on the Poincar´e section, let M = max{r1 , r2 }, and suppose that M > d. Suppose that r10 , r20 denote the values upon visiting the Poincar´e section the next time, and let M 0 = max{r10 , r20 }. Then M 0 ≤ max{d, (M + 1)/2}. (Proof below.) Corollary 12. If M = max{r1 , r2 } exceeds d, then after at most a finite number of iterations, r1 , r2 ∈ [1, d]. Proof.
Use M > 2 ⇒ (M + 1)/2 ≤ (3/4)M .
To prove Lemmas 10 and 11, we need one more lemma.
Lemma 14. Given any i, there exist α, β, γ such that P (α, β, γ) ⊆ I(i) ⊆ closure(P (α, β, γ)). There are similar statements for the map f2 and for the two cases in which both sources are reset. (Proof omitted.)
Lemma 13. Given (B, d) with B ≤ d2 and the current rates r1 , r2 , consider the identical system with “infinite” B, i.e., any B 0 > d2 , with the same drain rate d and current rates r1 and r2 . Let i, i0 be the number of iterations till the Poincar´e section is next reached in the initial and modified system, respectively. Then i ≤ i0 . (Proof omitted.)
(Since the boundary has box dimension 1, we can be a bit sloppy as to what part of the boundary A includes.) With this lemma, for a given map fj , we can partition [1, d]2 into a union of sets where each part is the intersection of [1, d]2 with the polygon over which i and k (k equaling the number of sources reset) are uniform for the map fj . To simplify notation, we let ~j = (j1 , . . . , jn ), ~i = (i1 , . . . , in ), and ~k = (k1 , . . . , kn ), kl ∈ {1, 2} for all l, signify a sequence of maps, values of i, and numbers of sources reset, respectively, at each application of the map. Define f~j = fjn ◦ fjn−1 ◦ · · · ◦ fj1 .
Proof. (Lemma 10) Let r10 , r20 be the rates at the next return to the Poincar´e section. We deal first with the case in which i ≥ 2. By Lemmas 7 and 13, we know that i ≤ dd−(r1 +r2 )−1e. Hence r1 +i ≤ r1 +(1+(d−r1 −r2 −1)) = d−r2 ≤ d and, similarly, r2 +i ≤ r2 +(1+(d−r1 −r2 −1)) = d − r1 ≤ d. One or both of the rates are cut at the end of the phase, leaving us with r10 , r20 ∈ [1, d]. Now we deal with the case in which i = 1, i.e., r1 + r2 ≥ d − 2. At the beginning of the first (and only) iteration of the phase, we go from (r1 , r2 , B) to (r1 + 1, r2 + 1, B − d), since B − d ≥ 0, and then to (r10 , r20 , B) (because i = 1). If source 1 sent first, then r20 = (r2 + 1)/2 ≤ d. How about r10 ? If r1 + 1 ≥ d, source 1 fills the buffer when it sends and r10 = (r1 + 1)/2 ≤ d, because r1 + 1 ≤ d + 1 ≤ 2d. If, on the other hand, r1 + 1 < d, source 1 doesn’t fill the buffer, source 1’s rate doesn’t get halved, and r10 = r1 + 1 < d. In either case, we have (r10 , r20 ) ∈ [1, d]2 .
Lemma 15. Given ~j, ~i, and ~k, there are values an > 0, bn , cn > 0, and dn such that f~j (r1 , r2 ) = (an r1 +bn , cn r2 + dn ). Hence f~jn is an affine map. ~ Definition 16. For each ~j,~i, ~k, let W~i,j~k be the set of
points (r1 , r2 ) ∈ [1, d]2 such that when applying f~j to (r1 , r2 ), the sequence of i’s achieved is ~i and the sequence of the number of sources reset is ~k.
Proof. (Lemma 11) Suppose either r1 or r2 exceeds d. Clearly r1 + r2 > d − 2 and therefore i = 1. Now note that whichever source sends first, for both l = 1, 2, if rl > d − 1, then rl0 = (rl + 1)/2. Hence if M = max{r1 , r2 } > d − 1, the new maximum is at most max{d, (M + 1)/2}. Since d ≥ 2 and M > 2, we have (M + 1)/2 ≤ (3/4)M , and hence the number of iterations till the maximum is at most d is at most dlog4/3 (M/d)e.
4.
The two lemmas above, the simple observations that an affine map maps polygons to polygons and that the intersection of two polygons is another polygon, and induction on n give us Lemma 17. For each ~j,~i, ~k, there is an open polygon P ~ such that P ⊆ W~i,j~k ⊆ closure(P ).
THE UPPER BOX DIMENSION IN THE GENERAL CASE
If we begin with a set X in one of these polygonal re~ gions W = W~i,j~k and apply f1 to this set, we calculate the associated i(W ) and apply the map as described above to each point (r1 , r2 ) in X. Because we add a uniform value of i to each point in X, we say that f1 halves and translates the second coordinate, and hence f1 (X) shrinks X in the vertical direction (regardless of its behavior in the first coordinate). We will speak of this operation as “halving” the second coordinate. Likewise, f2 “halves” the first coordinate and f2 (X) shrinks X in the horizontal direction. As we will examine repeated applications of the maps below, we let V (X), H(X), R(X) denote shrinking X by a factor of two in the vertical, the horizontal, and both directions, respectively. Observe that V (H(X)) = R(X).
Again, in this section, we focus on the general case of all B, d and we show that the upper box dimension of the attractor A is less than 2. We observe that our model consists of two maps f1 and f2 on the Poincar´e section that determine where the rates of the sources go at each buffer fill. The maps are chosen nondeterministically, as which source “fires” first is chosen arbitrarily. Thus, one iteration of the supersystem comprises an arbitrary choice of a map (either f1 or f2 ) and the application of that map which yields the dynamics of the system in that step. (We are actually referring to the map used in the iteration that filled the buffer.) We let f1 be the map in which source 1 fires first and f2 be the map in which source 2 fires first. Since f1 is now a map on the Poincar´e section, for a given (r1 , r2 ), in order to calculate
Theorem 18. The upper box dimension of A = AB,d is less than 2.
300
0
0
f~j (x(0) , y (0) ) = (x, y). Hence, if we take all 2d functions of the form f~j and apply each to A, the union of the results is a 0
superset of A: A ⊆ ∪~j [f~j (A)]. For each of the 2d sequences of functions f~j we might apply to A, we partition [1, d]2 into ~
a finite union of sets W~i,j~k , as in Lemma 17, to ensure that
∗
~
partitioning A as A = ∪~i,~k A~ji,~k gives us
~
Lemma 19. A ⊆ ∪~j f~j ∪~i,~k A~ji,~k
∗
∗
Take an optimal covering C of g~j (A) by 2−j × 2−j boxes.
#
~
~
Note that g~j (W~i,j~k ) is polygonal and {g~j (W~i,j~k )} defines a partition. Take an optimal covering C0√of the set of points in the plane at distance at most 2−j 2 from the image, under g~j , of the union of the boundaries of all the polygons ~ W j over all ~i, ~k. Because there are only finitely many ~i and
.
Let Nj (X) = N2−j ×2−j (X). We have this simple consequence of the fact that A ⊆ [1, d]2 :2
~i,~ k
6= ∅ implies that at Lemma 20. For l = 1, 2, A~ i,~ k ~ least one component of k is 2. Proof.
∗
Nj (g~j (S)) = Nj (H s V r (S)) for any S (where s∗ , r∗ depend on ~j). Hence it is sufficient to prove that there is a C~j such P ∗ ∗ ~ that ~i,~k Nj (g~j (A~ji,~k )) ≤ Nj (H s V r (A)) + C~j 2j for all j.
each set is a set of uniform i and k values at each of the d0 ~ ~ steps. Define A~ji,~k = A ∩ W~i,j~k . Expanding the partition and "
0
(x, y) 7→ (x/2s , y/2r ). Define g~j (x, y) = (x/2s , y/2r ) for all (x, y) ∈ [1, d]2 . (Integers s0 , r0 , s∗ , r∗ all depend on ~j.) By the fact that r0 ≥ r∗ , s0 ≥ s∗ in Lemma 21 for ~ ~ nonempty sets, Nj (f~j (A~ji,~k )) ≤ Nj (g~j (A~ji,~k )), since if N δ × δ boxes cover S, N boxes of the same size cover the “horizontal halving” H(S) of S and the “vertical halving” V (S) of S also. (The difference between f~j and g~j is that f~j may halve a coordinate more times than g~j . “Halved” sets require no more boxes to cover than the original set.) P P ~ ~ Hence ~i,~k Nj (f~j (A~ji,~k )) ≤ ~i,~k Nj (g~j (A~ji,~k )). We also have
Proof. For ease of notation, let us define d0 = dde or 1 + dde, whichever is odd. If (x, y) ∈ A, then consider a d0 th-order preimage (x(0) , y (0) ) which is in the attractor (which must exist, though we omit the proof). Furthermore, there is a function f~j for ~j =< j1 , j2 , ..., jd0 > such that
~k and each polygon has a finite-length boundary, there is a c such that the number of boxes in C0 is at most c2j . ~ To cover g~j (A~ji,~k ), use the boxes of C which lie entirely
A~ 6= ∅ and k =< 1, 1, ..., 1 > imply that i,~ k
~
in the interior of g~j (W~i,j~k ), together with all the boxes of C0 . Over all ~i, ~k, no box of C is used twice. Since there are only finitely many pairs (~i, ~k), there is a C~j such P ~ that ~i,~k Nj (g~j (A~ji,~k )) ≤ (number of boxes in C) + C~j 2j =
there is an (x, y) ∈ A~ , yet f (x, y) = (x0 , y 0 ) i,~ k with x0 ≥ x + d0 > d or y 0 ≥ y + d0 > d, which contradicts the fact that (x0 , y 0 ) ∈ A ⊆ [1, d]2 .
Lemma 21. (1) Given ~j of dimension d0 with r 1’s and s 2’s, for all ~i, ~k there are r0 ≥ r and s0 ≥ s and constants γ, ∆ 0 0 ~ such that for all (x, y) ∈ W~i,j~k , f~j (x, y) = (x/2s + γ, y/2r + ∆). (2) A~i, 6= ∅ implies that s0 ≥ 1; symmetrically, ~ k
∗
∗
Nj (g~j (A)) + C~j 2j = Nj (H s V r (A)) + C~j 2j . From Lemma 19, we have 3 2 X X ~ j 4 Nj (f~j (A~i,~k ))5 . Nj (A) ≤ ~ j
A~ 6= ∅ implies that r0 ≥ 1. i,~ k
(1)
~i,~ k
P ∗ ∗ ~ By Lemma 22, ~i,~k Nj (f~j (A~ji,~k )) ≤ Nj (H s V r (A))+C~j 2j , where ~j has r 1’s and s 2’s. By the fact that r∗ , s∗ ≥ 1, we
Proof. Part (1) follows from the fact that f1 always halves r2 (and may halve r1 , also) and that f2 always halves r1 . Part (2) follows from Lemma 20, since if ~j =< l, l, ..., l >, then some iteration halves both r1 and r2 .
0
can write the r = 0 and s = 0 terms as Nj (H d V (A)) as 0 Nj (V d H(A)), respectively. Simplifying inequality (1) by repeatedly collapsing a product of a V and an H into an R, we obtain Nj (A) ≤ P(d0 −1)/2 d0 0 0 2Nj (V d −1 R(A)) + 2 k=1 Nj (V d −2k Rk (A)) + C 0 2j k 0 for some C . Now repeatedly use Na×b (V (S)) = N2a×b (S) ≤ 0 2N2a×2b (S). We obtain Nj (A) ≤ 2(2d −1 )Nj−d0 (A) + P(d0 −1)/2 d0 d0 −2k 2 Nj−(d0 −k) (A) + C 0 2j . 2 k=1 k Let aj = N2−j (A). The above inequality is then aj ≤ P(d0 −1)/2 d0 −2k d0 0 2 2d aj−d0 + 2 k=1 aj−(d0 −k) + C 0 2j . It follows k from the theory of linear recurrences with constant coefficients [2] that aj is bounded above by a finite sum of terms, each of which is a polynomial in j times an exponential the base of which is a (complex) root of its characteristic polyP(d0 −1)/2 d0 −2k d0 k 0 0 nomial αd = 2d +2 k=1 2 α , plus a contribuk 00 j tion of the form C 2 from the “error term” C 0 2j . Assuming there is a root of the characteristic polynomial with norm at least 2, the upper box dimension of A is then bounded above by the limit, as j → ∞, of (lg aj )/j ≤ lg β, where β is the largest norm of a root of the characteristic polyno-
Lemma 22. Let ~j of dimension d0 be given with r 1’s and s 2’s, r + s = d0 . Let r∗ = max{r, 1}, s∗ = max{s, 1}. Then there is a constant C~j such that for all P ∗ ∗ ~ j, ~i,~k Nj (f~j (A~ji,~k )) ≤ Nj (H s V r (A)) + C~j 2j . The key point is that when ~j =< 1, 1, ..., 1 >, s0 > s = 0 (and when ~j =< 2, 2, ..., 2 >, r0 > r = 0). Hence r∗ = 1 when r = 0 and s∗ = 1 when s = 0. If this were not so, the only upper bound we could prove on the upper box dimension would be the trivial bound of 2. Proof. Whenever N boxes cover a set, translates of those N boxes will cover any fixed translate of the set. By Lemma 21, we may therefore think of f~j as the map 2 One can prove that the box dimension of A is less than 2 without using A ⊆ [1, d]2 by replacing d0 by some gross upper bound, like d + B + 1, on the value of any rate at any time.
301
mial. If we show that the characteristic polynomial has no complex root of norm at least 4, we will have shown that the upper box dimension is strictly less than lg 4 = 2. If there is no root with largest norm at least 2, then the error term determines the upper box dimension and trivially the dimension is less than 2. Let us divide both sides of the characteristic polynomial d0 0 by 2d and rearrange terms in the summation: α2 = P(d0 −1)/2 d0 α k 1 + 2 k=1 . Let x = α/4 and set p(x) = k 4 k P(d0 −1)/2 0 0 d0 x − 1. We will show that p has no 2d xd − 2 k=1 k complex root of norm ≥ 1. Suppose, for a contradiction, 0 0 that x is such a root. Set z = 1/x; we have 0 = 2d z −d − 0 0 P(d −1)/2 d0 −k P(d −1)/2 d0 d0 −k 0 0 2 k=1 z −1, 0 = 2d −2 k=1 z −z d , k k 0 P 0 0 0 0 (d −1)/2 d and 2d = 2 k=1 z d −k + z d . But |z| ≤ 1 and P(dk0 −1)/2 d0 d0 −k 0 hence the norm of 2 k=1 z + z d is at most P(d0 −1)/2 d0 P(d0 −1)/2 kd0 0 2 k=1 +1 = 2 k=0 −1 = 2d −1. But this k k P 0 (d0 −1)/2 d0 d0 −k d0 contradicts the fact that 2 = 2 k=1 z + zd , k completing the proof of Theorem 18.
5.
²/8) × (18/16, 19/16 − ²/16) →T (18/16, 19/16 − ²/16) × (21/8, 22/8 − ²/8) →1 (34/16, 35/16 − ²/16) × 1 th scale copy of W , trans(29/16, 30/16 − ²/16), a 16 posed. 7. Let L1 = (2, 3−2²)×(1, 2−²) and L2 = [3−2², 3−²)× (1, 2 − ²) so that W = L1 ∪ L2 . L1 ∪ L2 →1 ((3/2, 2 − ²)×(1, 3/2−²/2))∪([2−², 2−²/2)×(1, 3/2−²/2)) →1 ((5/2, 3 − ²) × (1, 5/4 − ²/4)) ∪ ([3/2 − ²/2, 3/2 − ²/4) × (1, 5/4−²/4)). The last rectangle is not a subset of W . However, its image D = [5/2−²/2, 5/2−²/4)×(1, 9/8− ²/8) under f1 is a subset of W . For C = (5/2, 3 − ²) × (1, 5/4 − ²/4) and D as defined, C = HV 2 (L1 ) and D = H 2 V 3 (L2 ). Lemma 24. The rectangular images given in the first six cases above and C, D, eight in total, have pairwise disjoint closures. (Proof omitted.) We have Nδ×δ (C) = Nδ×δ (HV 2 (L1 )) = N2δ×4δ (L1 ) ≥ N8δ×8δ (L1 ). Similarly, Nδ×δ (D) = Nδ×δ (H 2 V 3 (L2 )) = N4δ×8δ (L2 ) ≥ N8δ×8δ (L2 ). Therefore Nδ×δ (C)+Nδ×δ (D) ≥ N8δ×8δ (L1 ) + N8δ×8δ (L2 ) ≥ N8δ×8δ (W ). By Lemma 24 we have Nδ×δ (W ) ≥ 2Nδ×δ 14 W + 3Nδ×δ 18 W + 1 Nδ×δ 16 W + [Nδ×δ (C) + Nδ×δ (D)] ≥ 2N4δ×4δ (W ) + 3N8δ×8δ (W ) + N16δ×16δ (W ) + N8δ×8δ (W ). Writing ai = N2−i ×2−i (W ), we have ai ≥ 2ai−2 + 4ai−3 + ai−4 . Let p(α) = α4 −2α2 −4α1 −1. It is clear that p(2) = −1 and limα→∞ p(α) = ∞. It follows that there is a root α > 2 of p. We prove below in Lemma 25 that W ∩ A 6= ∅; hence a0 , a1 , a2 , a3 ≥ 1. Let ² = 1/α3 . It follows that ai ≥ ²αi for i = 0, 1, 2, 3. It follows by induction on i that ai ≥ ²αi for all i, for if ai ≥ ²αi for all i < k, then ak ≥ 2ak−2 + 4ak−3 + ak−4 ≥ ²αk−4 (2α2 + 4α + 1) = ²αk−4 (α4 ) = ²αk .
LOWER BOX DIMENSION FOR d = 3 − ²
Theorem 23. For all ² ∈ (0, 2/7), the lower box dimension of the attractor when d = 3 − ² (B ≥ d) exceeds 1. Proof. In this proof, we frequently use “(a, b) × (c, d)” to refer to the intersection of the open rectangle (a, b)×(c, d) ⊆ 2 and the attractor AB,d . Likewise, where S is a rectangle, “Nab (S)” really means “Nab (S ∩ A)”. Take d = 3 − ². Let W = (2, 3 − ²) × (1, 2 − ²). One key observation is that if (x, y) ∈ A, then f1 (x, y), f2 (x, y), and (y, x) are all in A. In a proof analogous to the proof for the Cantor set, we will show that W contains seven pairwise disjoint scale copies of itself or of its transpose (actually, one of them is shrunken differentially in the two dimensions). We need ² > 0 to make sure certain sets are nonempty and ² < 2/7 to make sure the mappings all go as stated. “→1 ” means “maps under f1 ,” “→2 ” means “maps under f2 ,” and “→T ” means “maps under the transpose.”
R
Lemma 25. W ∩ A 6= ∅ if 19/7 < d < 3. Proof. It is easy to see that (7/3, 5/3), which is in W , is in A, since (7/3, 5/3) is a fixed point of g = f1 ◦ f2 ◦ f1 . We conclude the proof of Theorem 23 by noting that N2−i ×2−i (A) ≥ N2−i ×2−i (W ∩ A) ≥ ²αi implies that lim inf(lg N2−i ×2−i (A))/i ≥ lim inf(lg ² + i lg α)/i > 1.
6.
1. W = (2, 3 − ²) × (1, 2 − ²) →1 (3/2, 2 − ²/2) × (1, 3/2 − ²/2) →2 (5/4, 6/4−²/4)×(2, 5/2−²/2) →1 (9/4, 10/4− ²/4) × (6/4, 7/4 − ²/4), a 14 th scale copy of W .
LOWER BOX DIMENSION FOR 0(mod6), B > d2 , d LARGE
d ≡
Again we are working with the supersystem in which one step causes a return to the Poincar´e section. Throughout this section 6|d, d is sufficiently large, and B > d2 . B’s exact value is unimportant.
2. W →1 →2 (5/4, 6/4 − ²/4) × (2, 5/2 − ²/2) →2 (9/8, 10/8 − ²/8) × (6/4, 7/4 − ²/4) →1 (17/8, 18/8 − ²/8) × (10/8, 11/8 − ²/8), a 18 th scale copy of W .
Definition 26. Given r1 , r2 , both at least 1, let i = i(r1 , r2 ). For m ∈ {1, 2}, define fm (r1 , r2 ) = (r10 , r20 ) where 0 0 (1) r3−m = (r3−m + i)/2 and (2) rm = rm + i, except that 0 if i = 1 and rm ≥ d − 1, then rm = (rm + 1)/2. In other words, both sources’ rates are initially increased by i. The rate of the “other” source (not source m) is then halved. The new rate of source m is not halved, unless i = 1 and its rate before being increased was at least d − 1.
3. W →2 (3/2, 2 − ²/2) × (2, 3 − ²) →2 (5/4, 6/4 − ²/4) × (3/2, 2 − ²/2) →1 (9/4, 10/4 − ²/4) × (5/4, 6/4 − ²/4), a 14 th scale copy of W . 4. W →2 →2 →1 (9/4, 10/4 − ²/4) × (5/4, 6/4 − ²/4) →2 (13/8, 14/8 − ²/8) × (9/4, 10/4 − ²/4) →1 (21/8, 22/8 − ²/8) × (13/8, 14/8 − ²/8), a 18 th scale copy of W . 5. W →1 →2 →1 (9/4, 10/4 − ²/4) × (6/4, 7/4 − ²/4) →2 (13/8, 14/8−²/8)×(10/4, 11/4−²/4) →1 (21/8, 22/8− ²/8) × (14/8, 15/8 − ²/8), a 18 th scale copy of W .
Lemma 27. For m = 1, 2, suppose that the system is in state (r1 , r2 , B), that at the next return to the Poincar´e section the system is in state (r10 , r20 , B), and that in the iteration that returned to the Poincar´e section, source m fired first. Then (r10 , r20 ) = fm (r1 , r2 ).
6. W →1 →2 →1 (9/4, 10/4 − ²/4) × (6/4, 7/4 − ²/4) →1 (13/8, 14/8−²/8)×(10/8, 11/8−²/8) →1 (21/8, 22/8−
302
Proof. Let m0 = 3 − m. We always reset source m0 in the last, ith, iteration. We need to reset source m in the last iteration if and only if i = 1 and the rate of source m is at least d − 1, for the following reason. If i = 1, we halve the rate of source m in the last iteration if and only if its rate after being increased is at least d. If i > 1, consider the point in the ith iteration after the buffer has been drained by d packets and no source has sent packets. The occupancy at that point is b0 = B − i · d + (i − 1)(i + r1 + r2 ), r1 , r2 being the rates when last on the Poincar´e section, because B is huge. We halve m’s rate if and only if (rm + i) + [B − i · d + (i − 1) · (i + r1 + r2 )] ≥ B ⇐⇒ (rm + i) + (i − 1)(i + r1 + r2 ) ≥ i · d ⇐⇒ (r1 + r2 + i) + (i − 1)(i + r1 + r2 ) ≥ i · d + rm0 ⇐⇒ i(r1 + r2 + i) ≥ i · d + rm0 ⇐⇒ rm0 ≤ i(r1 + r2 + i − d). But i ≥ 2 means, by Lemma 7, that i+1 = dd−(r1 +r2 )e < d − (r1 + r2 ) + 1 and hence that r1 + r2 + i − d < 0, so “rm0 ≤ i(r1 + r2 + i − d)” never holds. So we halve exactly one source’s rate in the last iteration, unless that iteration is the first.
Proof. We prove only the first statement. Lemma 7 implies that i = d/3−1 > 1. Hence, by Lemma 27, (x, y) →1 (x + (d/3 − 1), (y + (d/3 − 1))/2). The new value of i is 1, since x + y/2 ≥ (d − 1)/2. Furthermore, if we apply f2 , we use (y + (d/3 − 1))/2 < d − 1 and Lemma 27 to conclude that (x + (d/3 − 1), (y + (d/3 − 1))/2) →2 ((x + d/3)/2, (y + (d/3 − 1))/2 + 1) = ((x + d/3)/2, (y + (d/3 + 1))/2). Definition 31. Relative to a positive integer l, define R0 to be the open rectangle {(x, y)| |x−y| < 1, 2d/3+(1−2−l ) < x + y < 2d/3 + 1} ⊆ R. Lemma 32. For any positive integer l, R0 contains 2l 2−l scale copies of R which are pairwise disjoint. Proof. (Sketch) For any sequence β ∈ {0, 1}l , construct a sequence β 0 ∈ {1, 2}2l by replacing each 1 by < 1, 2 > and each 0 by < 2, 1 >. Apply the 2l functions 0 0 , in that order. The result is fβ10 , fβ20 , fβ30 , fβ40 , ..., fβ2l−1 , fβ2l a 2−l -scale copy of R, and the 2l copies are pairwise disjoint. To prove this last point, it suffices to look at the images of the vertices of R.
Definition 28. Let R denote the region {(x, y)| |x−y| < 1, 2d/3 < x + y < 2d/3 + 1}.
Lemma 32, which is the reason for the “2l ei−l ” term in our recurrence, is enough to prove that the box dimension of A is at least 1. The next lemma, whose proof is deferred and which will not be used till it is proven, is the key to proving A’s box dimension exceeds 1.
R is an open rectangle with vertices (d/3, d/3 + 1), (d/3 + 1, d/3), (d/3 − 1/2, d/3 + 1/2), (d/3 + 1/2, d/3 − 1/2). Let A0 = A ∩ R. We will prove that A0 contains inside itself, for some positive integers l, m, 2l 2−l -scale copies of A0 , which are pairwise disjoint, and, disjoint from those 2l copies, one 2−m -scale copy of itself. Actually, we prove something slightly different from this, as we will see later. This fact will be almost enough to prove that the box dimension of A0 , and hence of A, exceeds 1, for the following reason. Instead of counting the minimum number of δ × δ boxes needed to cover A0 , for geometrically decreasing δ, we estimate the lower box dimension differently. R is an open 21/2 ×2−1/2 , non-axis-aligned rectangle. In the obvious way, divide R into 4i disjoint, 2−i+1/2 × 2−i−1/2 open rectangles. Replace each of the 4i rectangles by its closure. For any subset S of R, let Mi (S) be the number of these 4i closed rectangles which intersect S. Define ei = Mi (A0 ).
Lemma 33. There is a positive integer q such that there is an open rectangle I ⊆ R, of positive area, having (d/3, d/3 + 1) as a vertex, which maps linearly under f = (f2 ◦ f1 )q ◦ f12 to a quadrilateral whose closure is in R. To get the 2−m -scale copy of R, we’ll map R (via a sequence of the form (f1 , f2 )p for some p) to a very small scale copy I of R having (d/3, d/3 + 1), one of R’s vertices, as a vertex. Next, we will show that there is a sequence of maps (of the form f1 , f1 , (f1 , f2 )q , for some q, in fact) that maps I to a new quadrilateral J whose closure is a subset of R. The image J of I will be shrunken q + 2 times in the second dimension and only q in the first; counting from the beginning and letting s = p + q, there will be s shrinkings in the first dimension and s + 2 in the second. Because of the differential shrinkings in the two dimensions, J will not really be a scale copy of R (and a quadrilateral, not a rectangle), but we can use Nδ×δ (H s V s+2 (A0 )) = N2s δ×2s+2 δ (A0 ) 1 ≥ N2s+2 δ×2s+2 δ (A0 ) = Nδ×δ 2s+2 A0 . Now use the proof of Lemma 29 to infer that there is a γ > 0 such that 1 Mi (H s V s+2 (A0 )) ≥ γ · Mi 2s+2 A0 = γ · Mi−(s+2) (A0 ) = γ · ei−(s+2) ; this is the reason for the “γ · ei−m ” term in the recurrence. Of course we need J to be disjoint from R0 , which contains the 2l 2−l -scale copies of R. This we ensure simply by choosing a large enough l in Lemma 32, which we can do for the following reason. The fact that J’s closure is a subset of R means that there is some positive δ such that all points (x, y) of J satisfy x + y ≤ 2d/3 + (1 − δ). Since we can choose a large enough l that all points (x, y) of R0 satisfy x + y ≥ 2d/3 + (1 − 2−l ), we’re home free. Proving the existence of the sequence mapping I to a quadrilateral with closure in R is difficult. The hard part is the fact that the dynamical system is discontinuous, due, not only to its ceilings, but also to the fact that whether i = 1 or i = dd − (r1 + r2 ) − 1e depends on whether r1 + r2 ≥ d − 2
Lemma 29. The lower box dimension of A0 equals lim inf(lg ei )/i. Proof. There is a C > 0 such that for all i, 1 · N2−i ×2−i (A0 ) ≤ ei ≤ C · N2−i ×2−i (A0 ). C We will get a recurrence ei ≥ 2l ei−l + γ · ei−m for positive integers l ≥ m and a positive real γ, which, together with a later proof that ei ≥ 1 for all i, will be enough to prove that the box dimension of A0 exceeds 1. The crux of the proof will be proving that R has the required scale copies. Proving that R has the 2l 2−l -scale copies, for any l, is easy; it is proving that there’s a 2−m scale copy that is disjoint from the 2l 2−l -scale copies that is hard. Now we start the proof that R has the 2l copies. Lemma 30. Consider any point (x, y) ∈ R. Then (x, y) →1 (x + (d/3 − 1), (y + (d/3 − 1))/2) →2 ((x + d/3)/2, (y + (d/3 + 1))/2), the midpoint of (x, y) and the vertex (d/3, d/3 + 1) of R. Similarly, (x, y) →2 →1 ((x + (d/3 + 1))/2, (y + d/3)/2), the midpoint of (x, y) and the vertex (d/3 + 1, d/3) of R.
303
(k+2)
(k+1)
or not, respectively, and each of f1 and f2 varies according to whether i = 1 or not. The key technical goal here is to determine for which iterations i is 1. In fact, we will show that starting counting after the third iteration, and alternating f2 , f1 , f2 , f1 , ..., the lth iteration has i = 1 if and only if l is odd. Once we prove this fact, the rest will be easy. First, instead of showing that all points in an open rectangle I map as claimed, we’ll prove the statement for one point only. Afterward, we’ll show that all points in I map exactly as that one point maps. In fact, it will be convenient to skip the first three functions f1 , f1 , f1 ; later, we’ll deal with those f1 ’s.
Lemma 36. a1 +a2 d + ∆0j . (Proof omitted.)
6.1
(l−1) r2
bj = 2/3 + (1/6)4
(l−1)
(0)
(l)
= bj d − 1, a2
(4j+2)
= bj d − 1, a2
a1
(4j+1)
a1
(4j+3)
a1
(4j)
(4j+2)
6.2
(4j)
+ 1)/2, a2
(4j+2)
+ 1)/2, a2
= (a1
= (a1
(4j+1)
= a2
(4j+3)
= a2
1 −j 4 . 8
(4j)
+ 1,
(4j+2)
+ 1.
These last four definitions correspond to the fact that the (4j +1)st and (4j +3)rd iterations are applications of f2 , will have i = 1, and will have a2 < d − 1, so that r2 is increased by i = 1 and not halved, and that r1 is increased by i = 1 and then halved. (l) (l) (l) Define ²m = rm − am , the signed difference between the true rate and the approximate rate, for m = 1, 2 and all l. We need a simple lemma. (0)
Lemma 34. ²1 = 0. (0)
(0)
Starting With A Rectangle
Now, having dispensed with the case of the single ini(0) (0) tial point (r1 , r2 ), we proceed to study what happens for (0) (0) points in a neighborhood of (r1 , r2 ). We are still temporarily omitting the first three function applications, all f1 ’s. Our current goal is to show that starting from (x0 , y 0 ) leads to the same values of i and same numbers of sources reset (and hence that P (l) (x0 , y 0 ) is close to P (l) ) as starting (0) (0) (0) (0) from (r1 , r2 ), provided that (x0 , y 0 ) is close to (r1 , r2 ). (l) (l) (l) (l) (l) (l) To that end, define P = (r1 , r2 ) and p = r1 + r2 (l) (l) for all l. Let i be the value of i used in calculating P , i.e., i(l) = i(P (l−1) ). For any (x0 , y 0 ), define P (l) (x0 , y 0 ) = (l) (l) (r1 (x0 , y 0 ), r2 (x0 , y 0 )) to be the point reached when start0 0 ing at (x , y ) and applying the first l functions in the se(l) quence f2 , f1 , f2 , f1 , . . . ; define p(l) (x0 , y 0 ) = r1 (x0 , y 0 ) + (l) 0 0 (l) 0 0 r2 (x , y ). Let i (x , y ) be the value of i used in calculating P (l) (x0 , y 0 ), i.e., i(l) (x0 , y 0 ) = i(P (l−1) (x0 , y 0 )). Since the outcome of applying f1 or f2 depends crucially, in calculating i(r1 , r2 ), on the the ceiling of the sum of the arguments, it would be a disaster for us if any p(l) were (0) (0) integral, for then all neighborhoods of (r1 , r2 ) might have 0 0 (l) 0 0 (l) some points (x , y ) with dp (x , y )e = p and some with dp(l) (x0 , y 0 )e = p(l) + 1. Fortunately, we saw in Corollary 38 that when d ≡ 0(mod6), this doesn’t happen. By Lemma 39, there is a k > 0 such that P = P (4k+1) ∈ R. Since Corollary 38 says that p(0) , p(1) , ..., p(4k+1) are nonintegral, there is a δ2 > 0 such that dp(l) e−p(l) , p(l) −bp(l) c > δ2 for all l ≤ 4k + 1. Since part (1) of Lemma 37 says
1 −j 4 , 4
= dj d −
is nonintegral.
Lemma 39. There is a k > 0 such that (4k+1) (4k+1) (r1 , r2 ) ∈ R. (The proof, omitted, uses the proof of Lemma 37.)
,
= cj d −
(l)
Corollary 38. For all l ≥ 0, r1 + r2 (Proof omitted.)
dj = 1/3 − (1/24)4−j , (4j)
(l−1)
See the appendix for the difficult part of the inductive proof of the lemma. Lemma 37 is the only place where we need 6 to divide d. Elsewhere, a multiple of 3 would suffice.
cj = 1/3 − (1/6)4−j ,
a1
=
Lemma 37. (1) l odd implies that r1 + r2 > d − 2, (l) (l) < d − 1, and ²2 = ²1 . (l−1) is integral and that (2) l odd implies that 2(l−1)/2 r1 (l+3)/2 (l−1) is odd. 2 r2 (l−1) (l) (l−1) + r2 < d − 2 and ²2 = (3) l even implies that r1 (l) (1/2)²1 . (l−1) is integral and that (4) l even implies that 2l/2 r1 (l+2)/2 (l−1) is odd. 2 r2
We choose our initial point to be (r1 , r2 ) = (5d/6 − (l) (l) 1, d/6 − 1/4). For any l ≥ 1, define (r1 , r2 ) to be the state (0) (0) reached, when starting from (r1 , r2 ), after applying the first l functions in the sequence f2 , f1 , f2 , f1 , .... Our goal is (l) (l) to show that for some odd l, (r1 , r2 ) ∈ R. To prove this fact, we start by approximating the sequence of pairs (r1 , r2 ) by the pairs we would have gotten, had d been of the form 6 · 4t for an arbitrarily large integer t. For such a d (and B > d2 ), we can write a single linear function giving the value of i in any given iteration. For the true value of d, we cannot write one linear function that works for all d. Now we define the approximations. Define infinite se(0) (1) (2) (0) (1) (2) quences < a1 , a1 , a1 , ... > and < a2 , a2 , a2 , ... > as follows. For all nonnegative integers j, let −j
(k+3)
+a2
Define µ(x) = dxe − x for any real x and µ0 (l) = (l) (l) µ(r1 +r2 ) for any nonnegative integer l. Also define δ(l) = 1 0 µ (4l + 1) − µ0 (4l + 3) and δ 0 (l) = 12 µ0 (4l + 3) − µ0 (4l + 5) 2 for all l. (l−1) (l−1) Let i(l) = max{1, dd − (r1 + r2 ) − 1e} denote the number of iterations needed to go from the (l − 1)st to the lth fill of the buffer. Here is the main technical lemma of this section.
Starting With One Point (0)
(k+4)
= d+∆j and a1
(0)
Proof. We have ²1 = r1 − a1 = (5d/6 − 1) − (b0 d − 1) = (5d/6 − 1) − (5d/6 − 1) = 0. Now let j be arbitrary and let k = 4j. We state a definition and then state a simple fact about the approximating sequences. Definition 35. Define ∆j = − 14 4−j and ∆0j = − 18 4−j .
304
(0)
(2)
(4)
(4k)
Lemma 44. A0 6= ∅.
that r2 , r2 , r2 , ..., r2 < d − 1, there is a δ3 > 0 (l) such that (d − 1) − r2 > δ3 for l = 0, 2, 4, ..., 4k. Let δ4 = (1/2) min{δ2 , δ3 }.
The proof, which is omitted, uses the contraction fixed point theorem [10, pp. 66-67] and Lemma 33.
Lemma 40. For all l, for all (x0 , y 0 ) such that |x0 − < δ4 , (1) i(l) (x0 , y 0 ) = i(l) and (2) each of the l applications of f2 or f1 , starting from (x0 , y 0 ), resets exactly (0) (0) one source (as is the case when starting from (r1 , r2 )).
Theorem 45. For any sufficiently large multiple d of 6 and B > d2 , the lower box dimension of AB,d exceeds 1.
(0) (0) r1 |, |y 0 − r2 |
Proof. From Lemma 44, A0 6= ∅; hence ei ≥ 1 for all i. Letting p(α) = αl − γαl−m − 2l , we see that p(2) < 0 and limα→∞ p(α) = ∞. Hence p has a root α > 2. Let ² = 1/αl−1 . From ei ≥ 1 for all i, it follows that ei ≥ ²αi for i = 0, 1, 2, ..., l − 1. It follows by induction on i that ei ≥ ²αi for all i, for if ei ≥ ²αi for all i < k, then ek ≥ 2l ek−l + γek−m ≥ 2l (²αk−l ) + γ(²αk−m ) = ²αk−l (2l + γαl−m ) = ²αk−l (αl ) = ²αk . Hence the lower box dimension lim inf(lg ei )/i ≥ lim inf(lg α) > 1, because α > 2.
Proof. By induction on l. Assume that (1) i(j) (x0 , y 0 ) = i(j) for j = 1, 2, ..., l−1 and that (2) applying each of the first l − 1 functions in the sequence f2 , f1 , f2 , f1 , ... starting with (x0 , y 0 ) resets exactly one source. There are γ, ∆ such that (0) (l−1) (0) (l−1) = r1 /2d(l−1)/2e + γ and r2 = r2 /2b(l−1)/2c + ∆. r1 (l−1) (l−1) (x0 , y 0 ) = x0 /2d(l−1)/2e + γ, r2 (x0 , y 0 ) = Then r1 y 0 /2b(l−1)/2c + ∆. Therefore p(l−1) (x0 , y 0 ) = x0 /2d(l−1)/2e + (0) y 0 /2b(l−1)/2c + γ + ∆. We have p(l−1) = r1 /2d(l−1)/2e + (0) r2 /2b(l−1)/2c + γ + ∆. Therefore |p(l−1) (x0 , y 0 ) − p(l−1) | ≤ δ4 /2dl/2e + δ4 /2bl/2c ≤ 2δ4 . We have dp(l−1) e − p(l−1) , p(l−1) − bp(l−1) c > δ2 . Since 2δ4 ≤ δ2 , p(l−1) (x0 , y 0 ) 6∈ , dp(l−1) (x0 , y 0 )e = dp(l−1) e, and therefore i(l) (x0 , y 0 ) = i(l) . If l is even, then i(l) (x0 , y 0 ) = i(l) > 1 and only one source (l−1) is reset. If l is odd, then i(l) = 1 and (d − 1) − r2 > δ3 . (l−1) (l−1) 0 0 For some m we have |r2 (x , y ) − r2 | = |(y 0 /2m + (0) (0) ∆) − (r2 /2m + ∆)| = |y 0 − r2 |/2m < δ4 < δ3 . Therefore (l−1) r2 (x0 , y 0 ) < d − 1 and only one source is reset.
7.
Z
(0)
Conjecture 1. For all B, d such that d > 2 and B > d, AB,d is a fractal.
8.
ACKNOWLEDGMENTS
The authors gratefully acknowledge the help of Konstantin Mischaikow, Jonathan Mattingly, Flip Korn, Andras Veres, Jennifer Rexford, Robin Pemantle, Michael Frame, and an anonymous STOC referee.
(0)
Corollary 41. If |x0 − r1 |, |y 0 − r2 | < δ4 , then for (l) (l) (0) (0) all l, |r1 (x0 , y 0 ) − r1 | = |r1 (x0 , y 0 ) − r1 |/2dl/2e and (l) (l) (0) (0) 0 0 0 0 bl/2c |r2 (x , y ) − r2 | = |r2 (x , y ) − r2 |/2 . Proof.
CONJECTURE
9. REFERENCES [1] F. Baccelli and D. Hong, “AIMD, Fairness and Fractal Scaling of TCP Traffic,” Proc. INFOCOM 2002. Also available from http://www.di.ens.fr/~hong /publication.html. [2] V. K. Balakrishnan, Introductory Discrete Mathematics, Dover Publications, New York, 1991, pp. 102–103. [3] M. Barnsley, Fractals Everywhere, Academic Press, New York, 1988, chapter 5. [4] V. Dumas, F. Guillemin, and P. Robert, “Limit Results for Markovian Models of TCP,” Internet Performance Symposium 2001. For both short and long versions, see http://www.icir.org/padhye/tcp-model.html. [5] K. Falconer, Fractal Geometry, John Wiley and Sons, New York, 1990, chapter 3. [6] A. Feldmann, A. C. Gilbert, P. Huang, and W. Willinger, “Dynamics of IP traffic: A study of the role of variability and the impact of control,” Proc. ACM SIGCOMM, Boston, MA, 1999, pp. 301–313. [7] A. Feldmann, A. C. Gilbert, and W. Willinger, “Data networks as cascades: Investigating the multifractal nature of Internet WAN traffic,” in Proc. ACM SIGCOMM, Vancouver, BC, Canada, 1998, pp. 25–38. [8] A. C. Gilbert, Y. Joo, and N. McKeown, “Congestion Control and Periodic Behavior,” in Proc. IEEE LANMAN Workshop, Boulder, CO, 2001. [9] Y. Joo, V. Ribeiro, A. Feldmann, A. C. Gilbert, and W. Willinger, “On the impact of variability on the buffer dynamics in IP networks,” in Proc. Allerton Conf., 1999. [10] A. Kolmogorov and S. Fomin, Introductory Real Analysis, Dover, New York, 1970.
Follows easily from Lemma 40.
Finally, we deal with the three omitted f1 ’s. Lemma 42. Choose 0 < δ < 1/2. Suppose (x, y) is such that (2d/3 + 1) − δ < x + y < (2d/3 + 1) and (x + 1) − δ < y < (x + 1). Then f1 (f1 (f1 (x, y))) = (x + (d/2 − 1), y/8 + (d/8 − 3/8)). (Tedious proof omitted.) Now we prove that (x + d/2 − 1, y/8 + d/8 − 3/8) is close to (0) (0) (5d/6 − 1, d/6 − 1/4) = (r1 , r2 ). (This is why we defined (0) (0) (r1 , r2 ) as we did.) Lemma 43. Under the hypotheses of Lemma 42, |(x + d/2 − 1) − (5d/6 − 1)| ≤ δ/2 and |(y/8 + d/8 − 3/8) − (d/6 − 1/4)| ≤ 3δ/16. (Tedious proof omitted.) Now we complete the proof of Lemma 33. Proof. (Lemma 33) Use Lemma 39 to choose k such (4k+1) (4k+1) that P = (r1 , r2 ) ∈ R. Suppose (x, y) satisfies (2d/3 + 1) − δ < x + y < (2d/3 + 1) and (x + 1) − δ < y < (x + 1) for a 0 < δ < 1/2 to be chosen later. Let (x0 , y 0 ) = f1 (f1 (f1 (x, y))) = (x + d/2 − 1, y/8 + d/8 − 3/8). By Lemma (0) (0) 43, |x0 − r1 |, |y 0 − r2 | ≤ δ/2. For δ/2 < δ4 , Corollary 41 (4k+1) (4k+1) (0) now implies that |r1 (x0 , y 0 )−r1 | ≤ |x0 −r1 | ≤ δ/2; (4k+1) (4k+1) (0) similarly, |r2 (x0 , y 0 ) − r2 | ≤ |y 0 − r2 | ≤ δ/2. Since P ∈ R and R is open, there is an ² > 0 such that the closed, axis-aligned square centered at P and of side length 2² is in R. Choose δ > 0 with δ/2 ≤ ², δ/2 < δ4 , and δ < 1/2.
305
[11] W. Leland, M. Taqqu, W. Willinger, and D. Wilson, “On the Self-Similar Nature of Ethernet Traffic,” Proc. ACM SIGCOMM, San Francisco, CA, USA, 1993, pp. 183–193. [12] B. Mandelbrot, “The Fractal Geometry of Nature,” W. H. Freeman and Company, New York, 1983. [13] R. Stevens and G. Wright, TCP Illustrated, vols. 1-3, Addison-Wesley, Reading, MA, 1994. [14] A. Veres and M. Boda, “The Chaotic Nature of TCP Congestion Control,” Proc. INFOCOM 2000. Also available at http://citeseer.nj.nec.com/veres00chaotic.html.
10.
From Lemma 36, we have (k+2)
a1
2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.
(m)
+ r2
(k)
> d − 2.
1 (0) 1 1 ² + [ j δ(0) + j−1 δ(1) 4j+1 1 4 4 1 1 1 + j−2 δ(2) + j−3 δ(3) + · · · + 0 δ(j)]. 4 4 4 (k+4)
²1
(k+1)
(k+4)
(k+2)
− µ0 (k + 1) − a1
.
(k+4)
(k+3)
= d − r2
(k+4)
− µ0 (k + 3) − a1
.
(k+4)
We have a1 + a2 = d − 1 − 14 4−(j+1) . From (0) (k+4) (k+4) ²1 = 0 (Lemma 34) and ²2 = (1/2)²1 (k+4) (by part 18 for ≥ h m = i k), we hinfer that i²1
1−4−(j+1) = − 23 1 − 4−(j+1) 3/4 (k+4) (k+4) (k+4) (k+4) (k+4) fore r1 + r2 + ²i 1 = a1 h + a2 ≥ d − 1 − 14 4−(j+1) + 32 32 −1 + 4−(j+1) = 1 + 4−(j+1) − 14 + 1 > d − 2. (0)
4−(j+1) ²1
(2)
The equation for iteration k + 4 mimics that for iteration k + 2: ²1
(4)
Proof. We want to prove that (1/2)µ0 (r + 1) − µ0 (r + (r+3) 3) ∈ {0, −1/2} for r = 0, 4, 8, 12, ..., k. We have r1 = (r+1) (r+3) (r+1) ((r1 + i(r+2) ) + 1)/2 and r2 = (r2 + i(r+2) )/2 + 1, because iteration r + 2 uses f1 , i(r+2) > 1 (by part 6 for m = r), iteration r + 3 uses f2 , i(r+3) = 1 (by part 10 for (r+2) m = r), and r2 < d − 1 (by part 11 for m = r). Where (r+1) (r+1) (r+3) (r+3) s denotes 1 + r1 + r2 , we have r1 + r2 = (r+2) 0 1 s+1+i . Hence µ (r + 3) = µ(s/2). Now there are 2 two cases. Case 1: dse is even. Then ds/2e = (1/2)dse and therefore µ0 (r + 3) = (1/2)dse − s/2 = (1/2)µ(s) = (1/2)µ0 (r + 1). It follows that (1/2)µ0 (r + 1) − µ0 (r + 3) = 0. Case 2: dse is odd. Then ds/2e = (1/2)dse + 1/2 and therefore µ0 (r + 3) = 1/2 + (1/2)dse = 1/2 + (1/2)µ0 (r + 1). It follows that (1/2)µ0 (r +1)−µ0 (r +3) = (1/2)µ0 (r +1)−1/2−(1/2)µ0 (r + 1) = −1/2.
Z
= d − r2
=
Lemma 46. δ(l) ∈ {−1/2, 0} for l = 0, 1, 2, ..., j.
We include here only the proof that part 1 for m = k + 4 follows from parts 1-18 for m = 0, 4, 8, ..., k, and omit the rest of the proof. Define j by k = 4j. We have i(k+2) > 1 (by part 6 for m = (k+2) (k+2) (k+2) (k+2) (k+1) k). We have ²1 = r1 − a1 . Now r1 = r1 + (k+1) (k+1) (k+2) (k+2) i where i = d−dr1 +r2 e, since parts 3 and 4 (k+1) (k+1) imply that r1 + r2 = integer/21+k/2 + odd/22+k/2 6∈ (k+1) (k+1) (k+1) (k+1) . So i(k+2) = d − r1 − r2 − µ(r1 + r2 ). So (k+2) (k+1) 0 r1 = d − r2 − µ (k + 1). Hence (k+2)
(k)
− 12 − 21 ²1 − µ0 (k + 1) − ∆j − µ0 (k + 3) − ∆0j = 14 ²1 + 1 0 µ (k + 1) − µ0 (k + 3) + 21 ∆j − ∆0j . However, 12 ∆j − 2 ∆0j = 0. Recalling that δ(l) = 12 µ0 (4l + 1) − µ0 (4l + 3), we (k+4) (k) (m+4) have ²1 = 14 ²1 + δ(k/4). We therefore have ²1 = 1 (m) ² + δ(m/4) for m = 0, 4, 8, 12, ..., k. A simple inductive 4 1 argument implies that
(m) r2 < d − 1. (m+1) 21+m/2 r1 is integral. 2+m/2 (m+1) 2 r2 is odd. (m+1) (m+1) ²2 = ²1 . (m+1) (m+1) r1 + r2 < d − 2. 1+m/2 (m+2) 2 r1 is integral. 3+m/2 (m+2) 2 r2 is odd. (m+2) (m+2) ²2 = (1/2)²1 . (m+2) (m+2) r1 + r2 > d − 2. (m+2) r2 < d − 1. 2+m/2 (m+3) 2 r1 is integral. (m+3) 23+m/2 r2 is odd. (m+3) (m+3) ²2 = ²1 . (m+3) (m+3) r1 + r2 < d − 2. 2+m/2 (m+4) 2 r1 is integral. (m+4) 24+m/2 r2 is odd. (m+4) (m+4) ²2 = (1/2)²1 .
²1
(k+1)
From equation (2), we see that = d − r2 − µ0 (k + (k+1) (k+1) 0 1) − d + a2 − ∆j = −µ (k + 1) − ²2 − ∆j . By part 5 (k+1) (k+1) (k+1) for m = k, ²2 = ²1 . By part 1 for m = k, r1 = (k) (k+1) (k) (1/2)(r1 + 1). Since a1 = (1/2)(a1 + 1), we infer that (k+1) (k) (k+1) (k) ²1 = (1/2)²1 and ²2 = (1/2)²1 . It follows that (k+2) 0 1 (k) ²1 = − 2 ²1 − µ (k + 1) − ∆j . (k+4) (k+3) From Lemma 36, we infer that a1 = d − a2 + ∆0j . (k+4) (k+3) 0 Now equation (3) says ²1 = d − r2 − µ (k + 3) − d + (k+3) (k+3) a2 − ∆0j = d − ²2 − µ0 (k + 3) − d − ∆0j . (k+3) (k+3) = ²1 . By part By part 14 for m = k, ²2 (k+2) (k+3) = (1/2)(r1 + 1). Since 10 for m = k, r1 (k+2) (k+3) (k+3) = (1/2)(a1 + 1), we infer that ²1 = a1 (k+3) (k+2) (k+2) and ²2 = (1/2)²1 . It follows that (1/2)²1 (k+4) (k+2) (k+4) ²1 h = − 12 ²1 − µ0 (k +i3) − ∆0j . Therefore ²1 =
APPENDIX
(m)
+ ∆j .
(k+2) ²1
Lemma 37 is proven by proving the following 18 statements for m = 0, 4, 8, 12, ... by induction, in the given order. Only the ones numbered 1 and 10 are difficult, but the proofs of those two are delicate. Parts 10-18 almost match parts 1-9, respectively. We omit the base case here. Let us pick a nonnegative k, 4|k, and assume the following 18 statements are true for m = 0, 4, 8, ..., k; our goal is to prove that they hold for m = k + 4. 1. r1
(k+1)
= d − a2
(3)
306
−
1 2
. There(k+4)
+ ²2
d−1−
