6.033 Lecture 13 3/18/09Sam MaddenCongestion ControlLast time, we saw how the end to end layer ensures exactly once delivery,using windows to efficiently make use of the network.Couple of loose ends:- Reordering -- packets may arrive at the receiver out of orderrecv(p)slot = p.sno - lastif (slot > last + size)dropelsenew = slot is empty if (new) put p in slotack pif (slot == last and new)deliver prefix to appslot += size(prefix)Cumulative acksSo far, have presented acks as being selective -- for a specific packetIn Internet, sometimes cumulative acks are usedin figure above, when p1 arrives, would send ack(2), indicating that all packets up to 2 have been received. + insensitive to lost acks (if using selective acks, a dropped ack requires a packet retransmit. Here, if the ack for P2 was lost, when P0 arrives, receiver knows P2 arrived.)- sender doesn't know what was losthere, sender has sent p3, p4, and p5; p4 and p5 have been receiver, butonly receives an ack for up to p2.can lead to excessive retransmissionIn the TCP, both cumulative and selective acks are used:common case, use cumulative, so that a lost ack isn't problematicwhen acking a packet for which later packets have arrived,use selective acks, so that sender knows what has arrivedChoosing the window sizeLast time, we send that the window size should be:Window = Rate x RTTWhere Rate is the maximum rate the receiver can accept packets atThis will keep the receiver busy all the timeE.g., suppose receiver can handle 1 packet / ms , and RTT is 10 ms; then window size needs to be 10 packets for receiver toalways be busy2 questions:1) How does the receiver know how fast it can process packets?2) What if the network can't deliver packets as fast as the receiver can process?1) It doesn't. In practice it doesn't matter that much -- just set it to be big enough. For example, on 100 Mbit ethernet, with 10 ms RTT,window = 100 Mbit/sec * .01 sec = 1 Mbit = 128 KBSo what if the network can't handle packets at this rate -- is that a problem?Suppose sender and receiver can both communicate at 3 pkts / sec, but that there is a bottleneck router somewhere between the sender and receiver that can onlysend 1 pkts / sec. Suppose RTT is 2 sec ( so receive window is 6 pkts)After 1 sec, there will be 2 packets in this router's queuesAfter 2 seconds, there will be 4 packetsCongestion collapse = delivered load constantly exceeds available bandwidth diagram:Notice that if we could adapt RTT on sender fast enough, that would help, since we'd sender fewer duplicates.But we could still get into trouble if we have a big window and there is a slow bottleneck.Why does congestion arise?Routers in the middle of the Internet are typically big and fast -- much faster than the end hosts. So the scenario above shouldn't arise, should it?Issue: sharing. Sources aren't aware of each other, and so may all attempt to use some link at the same time (e.g., flash crowds). This can overwhelm even the beefiest routers.Example: 9/11 -- everyone goes to CNN. Apple -- releases new products. What do do about it:Avoid congestion:Increase resources? Expensive and slow to react; not clear it will helpAdmission control? Used in telephone network. Some web servers do this, to reduce load. But its hard to do this inside of the network, since routers don't really have a good model of applications or the load they may be presenting.Congestion control -- ask the senders to slow down.How do senders learn about congestion?Options:1) Router sends feedback (either directly to the sender, or by flagging packets, which receiver then notices and propagatesin its acks.)Works, but can be problematic if network is congested because thesemessages may not get through (or may be very delayed)2) Assume that packet timeouts indicate congestion.What to do about congestion:- Increase timeouts.- Decrease window size.Need to do this very aggressively -- otherwise it is easy for congestion to collapse to arise. CC is hard to recover from.TCP does both of these:Timeouts: exponential backoffOn retransmitted packet, set:set timeout = timeout * c ; c = 2up to some maximumtimeout increases exponentiallyAfter ack's start arriving (losses stop), timeout is reset based on RTT EWMA (as in last class)2 windows: receive window + "Congestion Window":Window size = min(congestion window, receiver windoow)On retransmitted packet:CW = CW/2(Min size = 1)On ack'd packet:CW = CW + 1 (up to receive window size)To initialize window quickly, use TCP "slow start":At the start of the connection, double CW up to receive window, unless losses occur:Slow start -- while connection is starting onlyOn each ack:CW = CW * 2 (max receive window)Show diagram:Is TCP "fair"?no ; applications aren't required to use TCP in which case they won't obey congestion rules. Applications can also use multiple TCP connections.So why does this work? Apparently because most traffic is
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