20   Dynamics of TCP

In this chapter we introduce, first and foremost, the possibility that there are other TCP connections out there competing with us for throughput. In 8.3   Linear Bottlenecks (and in 19.7   TCP and Bottleneck Link Utilization) we looked at the performance of TCP through an uncontested bottleneck; now we allow for competition.

Although the focus of this chapter is on TCP Reno, many, though not all, of the ideas presented here apply to non-Reno TCPs as well, and even to non-TCP mechanisms such as QUIC. Several non-Reno TCP alternatives are presented later in 22   Newer TCP Implementations.

The following chapter continues this thread, with some more examples of TCP-Reno large-scale behavior.

20.1   A First Look At Queuing

In what order do we transmit the packets in a router’s outbound-interface queue? The conventional answer is in the order of arrival; technically, this is FIFO (First-In, First-Out) queuing. What happens to a packet that arrives at a router whose queue for the desired outbound interface is full? The conventional answer is that it is dropped; technically, this is known as tail-drop.

While FIFO tail-drop remains very important, there are alternatives. In an admittedly entirely different context (the IPv6 equivalent of ARP), RFC 4681 states, “When a queue overflows, the new arrival SHOULD replace the oldest entry.” This might be called “head drop”; it is not used for router queues.

An alternative drop-policy mechanism that has been considered for router queues is random drop. Under this policy, if a packet arrives but the destination queue is full, with N packets waiting, then one of the N+1 packets in all – the N waiting plus the new arrival – is chosen at random for dropping. The most recent arrival has thus a very good chance of gaining an initial place in the queue, but also a reasonable chance of being dropped later on. See [AM90].

While random drop is seldom if ever put to production use its original form, it does resolve a peculiar synchronization problem related to TCP’s natural periodicity that can lead to starvation for one connection. This situation – known as phase effects – will be revisited in 31.3.4   Phase Effects. Mathematically, random-drop queuing is sometimes more tractable than tail-drop because a packet’s loss probability has little dependence on arrival-time race conditions with other packets.

20.1.1   Priority Queuing

A quite different alternative to FIFO is priority queuing. We will consider this in more detail in 23.3   Priority Queuing, but the basic idea is straightforward: whenever the router is ready to send the next packet, it looks first to see if it has any higher-priority packets to send; lower-priority packets are sent only when there is no waiting higher-priority traffic. This can, of course, lead to complete starvation for the lower-priority traffic, but often there are bandwidth constraints on the higher-priority traffic (eg that it amounts to less than 10% of the total available bandwidth) such that starvation does not occur.

In an environment of mixed real-time and bulk traffic, it is natural to use priority queuing to give the real-time traffic priority service, by assignment of such traffic to the higher-priority queue. This works quite well as long as, say, the real-time traffic is less than some fixed fraction of the total; we will return to this in 25   Quality of Service.

20.3   TCP Fairness with Synchronized Losses

This brings us to the question of just what is a “fair” division of bandwidth. A starting place is to assume that “fair” means “equal”, though, as we shall see below, the question does not end there.

For the moment, consider again two competing TCP connections: Connection 1 (in blue) from A to C and Connection 2 (in green) from B to D, through the same bottleneck router R, and with the same RTT. The router R will use tail-drop queuing.

_images/dumbbell.svg

The layout illustrated here, with the shared link somewhere in the middle of each path, is sometimes known as the dumbbell topology.

For the time being, we will also continue to assume the synchronized-loss hypothesis: that in any one RTT either both connections experience a loss or neither does. (This assumption is suspect; we explore it further in 20.3.3   TCP RTT bias and in 31.3   Two TCP Senders Competing). This was the model reviewed previously in 19.1.1.1   A first look at fairness; we argued there that in any RTT without a loss, the expression (cwnd1 - cwnd2) remained the same (both cwnds incremented by 1), while in any RTT with a loss the expression (cwnd1 - cwnd2) decreased by a factor of 2 (both cwnds decreased by factors of 2).

Here is a graphical version of the same argument, as originally introduced in [CJ89]. We plot cwnd1 on the x-axis and cwnd2 on the y-axis. An additive increase of both (in equal amounts) moves the point (x,y) = (cwnd1,cwnd2) along the line parallel to the 45° line y=x; equal multiplicative decreases of both moves the point (x,y) along a line straight back towards the origin. If the maximum network capacity is Max, then a loss occurs whenever x+y exceeds Max, that is, the point (x,y) crosses the line x+y=Max.

_images/cjgraph.svg

Beginning at the initial state, additive increase moves the state at a 45° angle up to the line x+y=Max, in small increments denoted by the small arrowheads. At this point a loss would occur, and the state jumps back halfway towards the origin. The state then moves at 45° incrementally back to the line x+y=Max, and continues to zigzag slowly towards the equal-shares line y=x.

Any attempt to increase cwnd faster than linear will mean that the increase phase is not parallel to the line y=x, but in fact veers away from it. This will slow down the process of convergence to equal shares.

Finally, here is a timeline version of the argument. We will assume that the A–C path capacity, the B–D path capacity and R’s queue size all add up to 24 packets, and that in any RTT in which cwnd1 + cwnd2 > 24, both connections experience a packet loss. We also assume that, initially, the first connection has cwnd=20, and the second has cwnd=1.

T A–C B–D  
0 20 1  
1 21 2  
2 22 3 total cwnd is 25; packet loss
3 11 1  
4 12 2  
5 13 3  
6 14 4  
7 15 5  
8 16 6  
9 17 7  
10 18 8 second packet loss
11 9 4  
12 10 5  
13 11 6  
14 12 7  
15 13 8  
16 14 9  
17 15 10 third packet loss
18 7 5  
19 8 6  
20 9 7  
21 10 8  
22 11 9  
23 12 10  
24 13 11  
25 14 12 fourth loss
26 7 6 cwnds are quite close
     
32 13 12 loss
33 6 6 cwnds are equal

So far, fairness seems to be winning.

20.3.1   Example 2: Faster additive increase

Here is the same kind of timeline – again with the synchronized-loss hypothesis – but with the additive-increase increment changed from 1 to 2 for the B–D connection (but not for A–C); both connections start with cwnd=1. Again, we assume a loss occurs when cwnd1 + cwnd2 > 24

T A–C B–D  
0 1 1  
1 2 3  
2 3 5  
3 4 7  
4 5 9  
5 6 11  
6 7 13  
7 8 15  
8 9 17 first packet loss
9 4 8  
10 5 10  
11 6 12  
12 7 14  
13 8 16  
14 9 18 second loss
15 4 9 essentially where we were at T=9

The effect here is that the second connection’s average cwnd, and thus its throughput, is double that of the first connection. Thus, changes to the additive-increase increment lead to very significant changes in fairness. In general, an additive-increase value of α increases throughput, relative to TCP Reno, by a factor of α.

20.3.2   Example 3: Longer RTT

For the next example, we will return to standard TCP Reno, with an increase increment of 1. But here we assume that the RTT of the A–C connection is double that of the B–D connection, perhaps because of additional delay in the A–R link. The longer RTT means that the first connection sends packet flights only when T is even. Here is the timeline, where we allow the first connection a hefty head-start. As before, we assume a loss occurs when cwnd1 + cwnd2 > 24.

T A–C B–D  
0 20 1  
1 2  
2 21 3  
3 4  
4 22 5 first loss
5 2  
6 11 3  
7 4  
8 12 5  
9 6  
10 13 7  
11 8  
12 14 9  
13 10  
14 15 11 second loss
15 5  
16 7 6  
17 7  
18 8 8 B–D has caught up
20 9 10 from here on only even values for T shown
22 10 12  
24 11 14 third loss
26 5 8 B–D is now ahead
28 6 10  
30 7 12  
32 8 14  
34 9 16 fourth loss
35 8  
36 4 9  
38 5 11  
40 6 13  
42 7 15  
44 8 17 fifth loss
45 8  
46 4 9 exactly where we were at T=36

The interval 36≤T<46 represents the steady state here; the first connection’s average cwnd is 6 while the second connection’s average is (8+9+…+16+17)/10 = 12.5. Worse, the first connection sends a windowful only half as often. In the interval 36≤T<46 the first connection sends 4+5+6+7+8 = 30 packets; the second connection sends 125. The cost of the first connection’s longer RTT is quadratic; in general, as we argue more formally below, if the first connection has RTT = λ > 1 relative to the second’s, then its bandwidth will be reduced by a factor of 1/λ2.

Is this fair?

Early thinking was that there was something to fix here; see [F91] and [FJ92], §3.3 where the Constant-Rate window-increase algorithm is discussed. A more recent attempt to address this problem is TCP Hybla, [CF04]; discussed later in 22.12   TCP Hybla.

Alternatively, we may simply define TCP Reno’s bandwidth allocation as “fair”, at least in some contexts. This approach is particularly common when the issue at hand is making sure other TCP implementations – and non-TCP flows – compete for bandwidth in roughly the same way that TCP Reno does. While TCP Reno’s strategy is now understood to be “greedy” in some respects, “fixing” it in the Internet at large is generally recognized as a very difficult option.

20.3.3   TCP RTT bias

Let us consider more carefully the way TCP allocates bandwidth between two connections sharing a bottleneck link with relative RTTs of 1 and λ>1. We claimed above that the slower connection’s bandwidth will be reduced by a factor of 1/λ2; we will now show this under some assumptions. First, uncontroversially, we will assume FIFO droptail queuing at the bottleneck router, and also that the network ceiling (and hence cwnd at the point of loss) is “sufficiently” large. We will also assume, for simplicity, that the network ceiling C is constant.

We need one more assumption: that most loss events are experienced by both connections. This is the synchronized losses hypothesis, and is the most debatable; we will explore it further in the next section. But first, here is the general argument with this assumption.

Let connection 1 be the faster connection, and assume a steady state has been reached. Both connections experience loss when cwnd1+cwnd2 ≥ C, because of the synchronized-loss hypothesis. Let c1 and c2 denote the respective window sizes at the point just before the loss. Both cwnd values are then halved. Let N be the number of RTTs for connection 1 before the network ceiling is reached again. During this time c1 increases by N; c2 increases by approximately N/λ if N is reasonably large. Each of these increases represents half the corresponding cwnd; we thus have c1/2 = N and c2/2 = N/λ. Taking ratios of respective sides, we get c1/c2 = N/(N/λ) = λ, and from that we can solve to get c1 = Cλ/(1+λ) and c2 = C/(1+λ).

To get the relative bandwidths, we have to count packets sent during the interval between losses. Both connections have cwnd averaging about 3/4 of the maximum value; that is, the average cwnds are 3/4 c1 and 3/4 c2 respectively. Connection 1 has N RTTs and so sends about 3/4 c1×N packets. Connection 2, with its slower RTT, has only about N/λ RTTs (again we use the assumption that N is reasonably large), and so sends about 3/4 c2×N/λ packets. The ratio of these is c1/(c2/λ) = λ2. Connection 1 sends fraction λ2/(1+λ2) of the packets; connection 2 sends fraction 1/(1+λ2).

20.3.4   Synchronized-Loss Hypothesis

The synchronized-loss hypothesis is based on the idea that, if the queue is full, late-arriving packets from each connection will find it so, and be dropped. Once the queue becomes full, in other words, it stays full for long enough for each connection to experience a packet loss.

That said, it is certainly possible to come up with hypothetical situations where losses are not synchronized. Recall that a TCP Reno connection’s cwnd is incremented by only 1 each RTT; losses generally occur when this single extra packet generated by the increment to cwnd arrives to find a full queue. Generally speaking, packets are leaving the queue about as fast as they are arriving; actual overfull-queue instants may be rare. It is certainly conceivable that, at least some of the time, one connection would overflow the queue by one packet, and halve its cwnd, in a short enough time interval that the other connection misses the queue-full moment entirely. Alternatively, if queue overflows lead to effectively random selection of lost packets (as would certainly be true for random-drop queuing, and might be true for tail-drop if there were sufficient randomness in packet arrival times), then there is a finite probability that all the lost packets at a given loss event come from the same connection.

The synchronized-loss hypothesis is still valid if either or both connection experiences more than one packet loss, within a single RTT; the hypothesis fails only when one connection experiences no losses.

We will return to possible failure of the synchronized-loss hypothesis in 21.2.2   Unsynchronized TCP Losses. In 31.3   Two TCP Senders Competing we will consider some TCP Reno simulations in which actual measurement does not entirely agree with the synchronized-loss model. Two problems will emerge. The first is that when two connections compete in isolation, a form of synchronization known as phase effects (31.3.4   Phase Effects) can introduce a persistent perhaps-unexpected bias. The second is that the longer-RTT connection often does manage to miss out on the full-queue moment entirely, as discussed above in the second paragraph of this section. This results in a larger cwnd than the synchronized-loss hypothesis would predict.

20.3.5   Loss Synchronization

The synchronized-loss hypothesis assumes all losses are synchronized. There is another side to this phenomenon that is an issue even if only some reasonable fraction of loss events are synchronized: synchronized losses may represent a collective inefficiency in the use of bandwidth. In the immediate aftermath of a synchronized loss, it is very likely that the bottleneck link will go underutilized, as (at least) two connections using it have just cut their sending rate in half. Better utilization would be achieved if the loss events could be staggered, so that at the point when connection 1 experiences a loss, connection 2 is only halfway to its next loss. For an example, see exercise 18.0 in the following chapter.

This loss synchronization is a very real effect on the Internet, even if losses are not necessarily all synchronized. A major contributing factor to synchronization is the relatively slow response of all parties involved to packet loss. In the diagram above at 20.3   TCP Fairness with Synchronized Losses, if A increments its cwnd leading to an overflow at R, the A–R link is likely still full of packets, and R’s queue remains full, and so there is a reasonable likelihood that sender B will also experience a loss, even if its cwnd was not particularly high, simply because its packets arrived at the wrong instant. Congestion, unfortunately, takes time to clear.

20.3.6   Extreme RTT Ratios

What happens to TCP fairness if one TCP connection has a 100-fold-larger RTT than another? The short answer is that the shorter connection may get 10,000 times the throughput. The longer answer is that this isn’t quite as easy to set up as one might imagine. For the arguments above, it is necessary for the two connections to have a common bottleneck link:

_images/extreme_rtt.svg

In the diagram above, the A–C connection wants its cwnd to be about 200 ms × 10 packets/ms = 2,000 packets; it is competing for the R–C link with the B–D connection which is happy with a cwnd of 22. If R’s queue capacity is also about 20, then with most of the bandwidth the B–C connection will experience a loss about every 20 RTTs, which is to say every 22 ms. If the A–C link shares even a modest fraction of those losses, it is indeed in trouble.

However, the A–C cwnd cannot fall below 1.0; to test the 10,000-fold hypothesis taking this constraint into account we would have to scale up the numbers on the B–C link so the transit capacity there was at least 10,000. This would mean a 400 Gbps R–C bandwidth, or else an unrealistically large A–R delay.

As a second issue, realistically the A–C link is much more likely to have its bottleneck somewhere in the middle of its long path. In a typical real scenario along the lines of that diagrammed above, B, C and R are all local to a site, and bandwidth of long-haul paths is almost always less than the local LAN bandwidth within a site. If the A–R path has a 1 packet/ms bottleneck somewhere, then it may be less likely to be as dramatically affected by B–C traffic.

A few actual simulations using the methods of 31.3   Two TCP Senders Competing resulted in an average cwnd for the A–C connection of between 1 and 2, versus a B–C cwnd of 20-25, regardless of whether the two links shared a bottleneck or if the A–C link had its bottleneck somewhere along the A–R path. This may suggest that the A–C connection was indeed saved by the 1.0 cwnd minimum.

20.4   Epilog

TCP Reno’s core congestion algorithm is based on algorithms in Jacobson and Karel’s 1988 paper [JK88], now twenty-five years old. There are concerns both that TCP Reno uses too much bandwidth (the greediness issue) and that it does not use enough (the high-bandwidth-TCP problem).

In the next chapter we consider alternative versions of TCP that attempt to solve some of the above problems associated with TCP Reno.

20.5   Exercises

Exercises are given fractional (floating point) numbers, to allow for interpolation of new exercises. Exercises marked with a ♢ have solutions or hints at 34.15   Solutions for Dynamics of TCP.

1.0. In the section 20.2.3   Example 3: competition and queue utilization, we derived the formula

Q = wA + wB − 2d − 2(αdA+βdB)

under the assumption that the bottleneck bandwidth was 1 packet per unit time. Give the formula when the bottleneck bandwidth is r packets per unit time. Hint: the formula above will apply if we measure time in units of 1/r; only the delays d, dA and dB need to be re-scaled to refer to “normal” time. A delay d measured in “normal” time corresponds to a delay dʹ = r×d measured in 1/r units.

2.0. Consider the following network, where the bandwidths marked are all in packets/ms. C is sending to D using sliding windows and A and B are idle.

          C
          │
         100
          │
A───100───R1───5───R2───100───B
                   │
                  100
                   │
                   D

Suppose the one-way propagation delay on the 100 packet/ms links is 1 ms, and the one-way propagation delay on the R1–R2 link is 2 ms. The RTTnoLoad for the C–D path is thus about 8 ms, for a bandwidth×delay product of 40 packets. If C uses winsize = 50, then the queue at R1 will have size 10.

Now suppose A starts sending to B using sliding windows, also with winsize = 50. What will be the size of the queue at R1?

Hint: by symmetry, the queue will be equally divided between A’s packets and C’s, and A and C will each see a throughput of 2.5 packets/ms. RTTnoLoad, however, does not change. The number of packets in transit for each connection will be 2.5 packets/ms × RTTnoLoad.

3.0. In the previous exercise, give the average number of data packets (not ACKs) in transit on each individual link:

(a).♢ for the original case in which C is the only sender, with winsize = 50 (the only active links here are C–R1, R1–R2 and R2–D).
(b). for the new case in which B is also sending, also with winsize = 50. In this case all links are active.

Each link will also have an equal number of ACK packets in transit in the reverse direction. Hint: since winsize ≥ bandwidth×delay, packets are sent at the bottleneck rate.

4.0.♢ Consider the following network, with links labeled with one-way propagation delays in milliseconds (so, ignoring bandwidth delay, A’s RTTnoLoad is 40 ms and B’s is 20 ms). The bottleneck link is R–D, with a bandwidth of 6 packets/ms.

A────── 15 ──────┐
                 │
                 R──5──D
                 │
           B──5──┘

Initially B sends to D using a winsize of 120, the bandwidth×round-trip-delay product for the B–D path. A then begins sending as well, increasing its winsize until its share of the bandwidth is 2 packets/ms.

What is A’s winsize at this point? How many packets do A and B each have in the queue at R?

It is perhaps easiest to solve this by repeated use of the observation that the number of packets in transit on a connection is always equal to RTTnoLoad times the actual bandwidth received by that connection. The algebraic methods of 20.2.3   Example 3: competition and queue utilization can also be used, but bandwidth there was normalized to 1; all propagation delays given here would therefore need to be multiplied by 6.

5.0. Consider the C–D path from the diagram of 20.2.4   Example 4: cross traffic and RTT variation:

C───100───R1───5───R2───100───D

Link numbers are bandwidths in packets/ms. Assume C is the only sender.

(a).♢ Give propagation delays for the links C–R1 and R2–D so that there will be an average of 5 packets in transit on the C–R1 and R2–D links, in each direction, if C uses a winsize sufficient to saturate the bottleneck R1–R2 link.
(b). Give propagation delays for all three links so that, when C uses a winsize equal to the round-trip transit capacity, there are 5 packets each way on the C–R1 link, 10 on the R1–R2 link, and 20 on the R2–D link.

6.0. Suppose we have the network layout below of 20.2.4   Example 4: cross traffic and RTT variation, except that the R1–R2 bandwidth is 6 packets/ms and the R2–R3 bandwidth is 3 pkts/ms. The delays are as shown, making the C–D RTTnoLoad 10 ms and the A–B RTTnoLoad 16 ms. A connects to B and C connects to D.

(a).♢ Find window sizes wA and wC so that the A–B and C–D connections share the bottleneck R1–R2 bandwidth equally, and there is no queue.
(b). Show that increasing each of wA and wC by 30 packets leaves each connection with 30 packets in R1’s queue – so the bandwidth is still shared equally – and none in R2’s. Hint: As in (a), the A–B bandwidth cannot exceed 3 packets/ms, and C’s packets can only accumulate at R1. To show A cannot have less than 50% of the bandwidth, observe that, if this happened, then A can have no queue at R2 (because packets now leave faster than they arrive), and so all of A’s extra packets must also queue at R1.
_images/exercise4.svg

7.0. Suppose we have the network layout of the previous exercise, 6.0. Suppose also that the A–B and C–D connections have settled upon window sizes as in 6.0(b), so that each contributes 30 packets to R1’s queue. Each connection thus has 50% of the R1–R2 bandwidth and there is no queue at R2.

Now A’s winsize is incremented by 10, initially, at least, leading to A contributing more than 50% of R1’s queue. When the steady state is reached, how will these extra 10 packets be distributed between R1 and R2? Hint: As A’s winsize increases, A’s overall throughput cannot rise due to the bandwidth restriction of the R2–R3 link.

8.0. Suppose we have the network layout of exercise 6.0, but modified so that the round-trip C–D RTTnoLoad is 5 ms. The round-trip A–B RTTnoLoad may be different.

The R1–R2 bandwidth is 6 packets/ms, so with A idle the C–D throughput is 6 packets/ms.

(a). Suppose that A and C have window sizes such that, with both transmitting, each has 30 packets in the queue at R1. What is C’s winsize? Hint: C’s throughput is now 3 packets/ms.
(b). Now suppose C’s winsize, with A idle, is 60. In this case the C–D transit capacity would be 5 ms × 6 packets/ms = 30 packets, and so C would have 60−30 = 30 packets in R1’s queue. A then begins sending, with a winsize chosen so that A and C’s contributions to R1’s queue are equal; C’s winsize remains at 60. What will be C’s (and thus A’s) queue usage at R1? Hint: find the transit capacity for a throughput of 3 packets/ms.
(c). Suppose the A–B RTTnoLoad is 10 ms. If C’s winsize is 60, find the winsize for A that makes A and C’s contributions to R1’s queue equal.

9.0. One way to address the reduced bandwidth TCP Reno gives to long-RTT connections is for all connections to use an increase increment of RTT2 instead of 1; that is, everyone uses AIMD(RTT2,1/2) instead of AIMD(1,1/2) (or AIMD(k×RTT2,1/2), where k is an arbitrary scaling factor that applies to everyone).

(a). Construct a table in the style of of 20.3.2   Example 3: Longer RTT above, showing the result of two connections using this strategy, where one connection has RTT = 1 and the other has RTT = 2. Start the connections with cwnd=RTT2, and assume a loss occurs when cwnd1 + cwnd2 > 24.
(b). Explain why this strategy might not be desirable if one connection is over a direct LAN with an RTT of 1 ms, while the second connection has a very long path and an RTT of 1.0 sec. (Hint: the cwnd-increment value for the short-RTT connection would have to apply whether or not the long-RTT connection was present.)

10.0. Suppose two 1 kB packets are sent as part of a packet-pair probe, and the minimum time measured between arrivals is 5 ms. What is the estimated bottleneck bandwidth?

11.0. Consider the following three causes of a 1-second network delay between A and B. In all cases, assume ACKs travel instantly from B back to A.

(i) An intermediate router with a 1-second-per-packet bandwidth delay; all other bandwidth delays negligible
(ii) An intermediate link with a 1-second propagation delay; all bandwidth delays negligible
(iii) An intermediate router with a 100-ms-per-packet bandwidth delay, and a steadily replenished queue of 10 packets, from another source (as in the diagram in 20.2.4   Example 4: cross traffic and RTT variation).

(a). Suppose that, in each of these cases, the packet-pair technique (20.2.6   Packet Pairs) is used to measure the bandwidth. Assuming no packet reordering, what is the minimum time interval we could expect in each case?

(b). What would be the corresponding values of the measured bandwidths? (For purposes of bandwidth measurement, you may assume that the “negligible” bandwidth delay in case (ii) is 0.01 sec.)

12.0. Suppose A sends packets to B using TCP Reno. The round-trip propagation delay is 1.0 seconds, and the bandwidth is 100 packets/sec (1 packet every 10 ms).

(a). Give RTTactual when the window size has reached 100 packets.

(b). Give RTTactual when the window size has reached 200 packets.