WEBVTT 00:00:00.000 --> 00:00:03.680 Welcome back to CSE 316 — Data Communication and Networking. 00:00:03.730 --> 00:00:06.920 This is the detailed video version of Session 8. 00:00:06.970 --> 00:00:12.440 Last session your laptop shouted at the room, and you learned why it had to. 00:00:12.490 --> 00:00:19.610 Today we climb one more floor, to the layer that decides where the shout is aimed. 00:00:19.660 --> 00:00:23.950 It opens with a network layer whose specification promises nothing. 00:00:24.000 --> 00:00:32.217 There is a checkpoint at the end of each section, so you can tell whether to go on or go back. 00:00:32.267 --> 00:00:37.827 Two claims to start with, and both of them sound wrong. 00:00:37.877 --> 00:00:41.367 Claim one. The Internet's network layer will happily lose your packet. 00:00:41.417 --> 00:00:45.317 It will shuffle your packets, so they arrive in the wrong order. 00:00:45.367 --> 00:00:49.657 It will delay them without apology, and it will warn you of none of it. 00:00:49.707 --> 00:00:55.297 That is not a bug list. That is the specification, written down and agreed to. 00:00:55.347 --> 00:01:01.057 Claim two. Adding a faster router to a network can make everyone's delay worse. 00:01:01.107 --> 00:01:05.457 Not worse for that router. Worse for everybody on the path. 00:01:05.507 --> 00:01:09.027 So write two answers down before we begin. 00:01:09.077 --> 00:01:13.687 First: why would anyone deliberately build a core that promises nothing? 00:01:13.737 --> 00:01:18.487 And second: how can extra capacity possibly make a network slower? 00:01:18.537 --> 00:01:22.337 Slide forty-three marks both of them against yours. 00:01:22.387 --> 00:01:27.877 Both halves resolve on the same idea, and here it is in two sentences. 00:01:27.927 --> 00:01:33.397 A promise costs the core state, and state in the core dies with the router that holds it. 00:01:33.447 --> 00:01:39.977 A dropped packet comes back as a retransmission, so extra load past the knee delivers less. 00:01:40.027 --> 00:01:48.250 Section one earns the first sentence; section four earns the second. 00:01:48.300 --> 00:01:49.580 Section one. 00:01:49.630 --> 00:01:52.100 Packetize, route, forward. 00:01:52.150 --> 00:01:59.450 Three duties — one of which happens before your traffic ever arrives. 00:01:59.500 --> 00:02:06.190 Chapter eighteen opens by telling you everything the network layer does, and it fits in three verbs. 00:02:06.240 --> 00:02:09.800 Packetizing. Carry, don't touch. 00:02:09.850 --> 00:02:15.920 The payload is encapsulated in a datagram at the source and taken back out at the destination. 00:02:15.970 --> 00:02:20.690 In between, routers read the destination address to make a decision. 00:02:20.740 --> 00:02:25.430 They never change that address, and they have no business inside the payload. 00:02:25.480 --> 00:02:28.280 Routing. Map-making. 00:02:28.330 --> 00:02:32.320 Which of the many possible routes through this network is the best one? 00:02:32.370 --> 00:02:40.020 Routing protocols gossip in the background — before your traffic flows — until every table in the network agrees. 00:02:40.070 --> 00:02:45.980 Network-wide, and on a slow timescale. Seconds, not microseconds. 00:02:46.030 --> 00:02:48.930 Forwarding. The junction. 00:02:48.980 --> 00:02:53.670 One packet has arrived, right now. Which interface does it leave by? 00:02:53.720 --> 00:02:58.340 A table lookup, at one router, in microseconds. Per-packet, and fast. 00:02:58.390 --> 00:03:02.840 Routing writes the table. Forwarding reads it. 00:03:02.890 --> 00:03:10.368 If you can produce those five words in an exam you have already banked a mark. 00:03:10.418 --> 00:03:16.968 Packetizing in detail, because it is where last session's lesson becomes this session's rule. 00:03:17.018 --> 00:03:19.458 Encapsulate once. 00:03:19.508 --> 00:03:26.228 The transport-layer segment goes inside a datagram at the source, and it comes back out at the destination. 00:03:26.278 --> 00:03:30.978 Nowhere in between. Not at any router, not at any switch. 00:03:31.028 --> 00:03:34.528 Routers inspect, but they never edit. 00:03:34.578 --> 00:03:39.008 A router reads the destination address because it has a decision to make. 00:03:39.058 --> 00:03:45.128 Reading is not writing. The address it read is the address the next router will read. 00:03:45.178 --> 00:03:49.818 There is exactly one exception, and it is fragmentation. 00:03:49.868 --> 00:03:54.888 If a datagram is too big for the next link, it is cut into fragments. 00:03:54.938 --> 00:04:00.118 Even then the addresses are copied into every fragment. They are not edited. 00:04:00.168 --> 00:04:03.108 The next three slides do it properly. 00:04:03.158 --> 00:04:05.828 Now compare that with the frame. 00:04:05.878 --> 00:04:12.178 The frame's MAC pair was rewritten at every single hop. The datagram's IP pair is not. 00:04:12.228 --> 00:04:21.178 Two layers, two lifetimes. The datagram survives the journey; the frame does not survive the hop. 00:04:22.381 --> 00:04:31.331 Let us do that exception properly, because it is the only thing on the list that changes a packet at all — and because Session twenty spends ten minutes on what IPv6 took away, and you cannot miss something you were never shown. 00:04:36.411 --> 00:04:45.361 Every link has a maximum transmission unit: the largest payload one frame on that link can carry. For Ethernet it is one thousand five hundred bytes. 00:04:47.861 --> 00:04:56.811 The MTU belongs to the link. Not to your packet, not to your application, and not to anything the source can see from where it sits. 00:04:57.081 --> 00:05:05.371 When a datagram is larger than the next link's MTU, the router cuts it — and it cuts the data, never the header. 00:05:05.421 --> 00:05:14.241 Every piece has to carry its own twenty-byte header, so only one thousand four hundred and eighty bytes of data fit inside one frame. 00:05:14.291 --> 00:05:23.241 And the cut has to land on a multiple of eight. That is not tidiness: the fragment-offset field counts in eight-byte units, which the next slide's arithmetic shows. 00:05:25.521 --> 00:05:34.471 Now the part worth memorising. Three fields differ between the fragments: total length, fragment offset, and the more-fragments bit. The checksum is recomputed, because the header changed. 00:05:40.551 --> 00:05:46.451 Everything else is copied. Identification, source address, destination address — identical in all of them. 00:05:46.501 --> 00:05:52.401 That is what makes reassembly possible at all. The pieces are recognisably siblings. 00:05:52.451 --> 00:05:57.831 And now the rule about reassembly. No router reassembles anything. 00:05:57.881 --> 00:06:05.411 Each fragment is forwarded on its own, may take a different path, and may be cut again by a smaller link further along. 00:06:05.461 --> 00:06:14.411 Only the destination puts them back together — because a router that reassembles is a router holding state, and you have just spent ten minutes on why this layer refuses to hold any. 00:06:19.700 --> 00:06:23.650 Thirty-six seconds, and two things to watch. 00:06:23.700 --> 00:06:29.050 First, the header lifting off — the router reads it before anything is cut. 00:06:29.100 --> 00:06:37.350 The cut lands on a multiple of eight, and the piece size came from the MTU minus the header. 00:06:37.400 --> 00:06:46.350 And there is the second thing. Three fragments, three complete headers. Same identification, same addresses, different offsets. 00:06:50.200 --> 00:06:59.150 They become one datagram again only at the far end, and only when the piece with MF equal to zero arrives. 00:07:01.603 --> 00:07:05.913 Now the arithmetic, because this is the shape the question comes in. 00:07:05.963 --> 00:07:14.913 Four thousand bytes, twenty of them header, so three thousand nine hundred and eighty bytes of data. The next link carries fifteen hundred. 00:07:16.813 --> 00:07:25.763 Fragment one. Data bytes zero to fourteen seventy-nine. Total length fifteen hundred — twenty of header plus fourteen eighty of data. Offset zero. More fragments follow, so MF is one. 00:07:28.833 --> 00:07:37.783 Fragment two. Data bytes fourteen eighty to twenty-nine fifty-nine. Total length fifteen hundred again. 00:07:38.173 --> 00:07:47.123 And the offset is one hundred and eighty-five. Not fourteen eighty. One hundred and eighty-five, because the field counts eight-byte units and fourteen eighty divided by eight is one eighty-five. 00:07:56.573 --> 00:08:05.523 Fragment three. Bytes twenty-nine sixty to thirty-nine seventy-nine — whatever is left. Total length one thousand and forty. Offset three hundred and seventy. And MF is zero, which is the only thing telling the destination that this is the end. 00:08:11.583 --> 00:08:20.533 Write fourteen eighty where the field wants one hundred and eighty-five and the mark is gone. Divide by eight. Every time. 00:08:22.970 --> 00:08:28.040 These two words get swapped in exam scripts, so here they are side by side. 00:08:28.090 --> 00:08:33.720 Routing is network-wide, it happens before your traffic, and it is slow. 00:08:33.770 --> 00:08:40.100 Routers exchange what they know about their neighbourhood, over and over, until the tables converge. 00:08:40.150 --> 00:08:42.900 Nothing of yours is moving while that happens. 00:08:42.950 --> 00:08:46.920 And when a link dies, this is the thing that runs again. 00:08:46.970 --> 00:08:51.600 Forwarding is one router, one packet, and fast. 00:08:51.650 --> 00:08:59.300 A packet is here now. Look up its destination in the table that routing already built, and push it out of the right interface. 00:08:59.350 --> 00:09:06.100 No conversation with anybody. No waiting for anyone to agree. 00:09:06.150 --> 00:09:10.040 And here is the test to apply when you are not sure. 00:09:10.090 --> 00:09:13.210 Ask: was anything exchanged with another router? 00:09:13.260 --> 00:09:22.210 If yes, it is routing. If the router acted alone, on one packet, using a table it already had, it is forwarding. 00:09:24.637 --> 00:09:27.437 Ninety seconds. 00:09:27.487 --> 00:09:31.447 Four events. Label each one routing or forwarding. 00:09:31.497 --> 00:09:36.697 One. Routers exchange neighbourhood information every thirty seconds. 00:09:36.747 --> 00:09:42.677 Two. A packet for ten dot five dot six dot seven is sent out of interface three. 00:09:42.727 --> 00:09:46.867 Three. After a link dies, the tables are recomputed. 00:09:46.917 --> 00:09:53.687 Four. Label two-forty is swapped for one-seventeen, and the packet leaves. 00:09:53.737 --> 00:09:59.477 Pause the video and write four words. 00:09:59.527 --> 00:10:05.097 For each one, ask yourself whether the router had to talk to anybody. 00:10:05.147 --> 00:10:08.397 Routing, forwarding, routing, forwarding. 00:10:08.447 --> 00:10:11.297 Number four is the one that catches people. 00:10:11.347 --> 00:10:17.277 Swapping a label feels like a change of identity, so it feels like something big is happening. 00:10:17.327 --> 00:10:26.277 But it is one router, one packet, one table lookup, and no conversation. That is the definition of forwarding. 00:10:28.025 --> 00:10:32.465 Now the terms and conditions, in fine print. 00:10:32.515 --> 00:10:34.225 No delivery guarantee. 00:10:34.275 --> 00:10:41.075 A router under pressure may discard your datagram. Nobody is told. No apology is issued. 00:10:41.125 --> 00:10:47.745 That is not a fault in the implementation. It is what the specification says will happen. 00:10:47.795 --> 00:10:49.435 No ordering guarantee. 00:10:49.485 --> 00:10:53.845 Packets of one message may take different roads and arrive shuffled. 00:10:53.895 --> 00:10:59.105 The network never promised to keep them in line, so it does not. 00:10:59.155 --> 00:11:00.785 No timing guarantee. 00:11:00.835 --> 00:11:03.785 No promised delay. No promised jitter. 00:11:03.835 --> 00:11:10.075 Other people's traffic happens to you, and nothing at this layer protects you from it. 00:11:10.125 --> 00:11:11.475 So who fixes it? 00:11:11.525 --> 00:11:14.735 The ends do. The two machines that actually care. 00:11:14.785 --> 00:11:23.735 Chapters twenty-three and twenty-four are one long answer to this slide: the transport layer rebuilds every guarantee the network layer declined to make. 00:11:27.658 --> 00:11:32.298 Which leaves the obvious question. Why refuse? 00:11:32.348 --> 00:11:36.138 Because a promise is state, and state has to live somewhere. 00:11:36.188 --> 00:11:42.758 To guarantee order, or delivery, or timing, some machine must remember your conversation. 00:11:42.808 --> 00:11:50.858 Put that memory in the core, and every router is holding a piece of every conversation crossing it. 00:11:50.908 --> 00:11:53.728 And state in the core is state that can die. 00:11:53.778 --> 00:11:58.328 A router holding your connection is a router whose failure destroys it. 00:11:58.378 --> 00:12:03.578 A router holding nothing can fail, and the next packet simply goes another way. 00:12:03.628 --> 00:12:07.108 That is the survivability argument. 00:12:07.158 --> 00:12:09.848 So the promises moved to the ends. 00:12:09.898 --> 00:12:14.998 The two machines that care are the two ends, and they were always going to be there. 00:12:15.048 --> 00:12:23.998 Put the memory in them, and the core stays simple, and cheap, and — because it remembers nothing — almost impossible to break permanently. 00:12:24.478 --> 00:12:27.378 Promise nothing, and you never break your word. 00:12:27.428 --> 00:12:33.944 That is the design decision the Internet is built on. 00:12:33.994 --> 00:12:40.004 Checkpoint one. Four questions, and I would like them on paper. 00:12:40.054 --> 00:12:48.414 One. A router recomputes its table after a link failure. Routing or forwarding — and what tells you? 00:12:48.464 --> 00:12:55.864 Two. Name the three things IP declines to guarantee, and say who provides them instead. 00:12:55.914 --> 00:13:03.974 Three. Why does putting connection state in the core make the network less survivable, rather than more? 00:13:04.024 --> 00:13:12.974 Four. A three-thousand-byte datagram meets a link with an MTU of one thousand. How many fragments, and what is the offset of the last one? 00:13:14.774 --> 00:13:15.744 Answers. 00:13:15.794 --> 00:13:24.674 One. Routing. Information was exchanged with other routers, and a table was written. Forwarding never talks to anybody. 00:13:24.724 --> 00:13:32.574 Two. Delivery, order and timing. The transport layer, at the two end machines, rebuilds all three. 00:13:32.624 --> 00:13:41.574 Three. Because a router holding your connection is a router whose failure destroys it. A router holding nothing can fail without taking your conversation with it. 00:13:44.684 --> 00:13:49.574 Four. Four fragments, and the last offset is three hundred and sixty-six. 00:13:49.624 --> 00:13:58.574 Here is the reasoning, and the trap is in the first step. One thousand minus twenty of header leaves nine hundred and eighty bytes for data — but nine hundred and eighty is not a multiple of eight. So each fragment carries nine hundred and seventy-six. 00:14:07.214 --> 00:14:16.164 Two thousand nine hundred and eighty bytes of data, in pieces of nine seven six, is three full pieces and fifty-two bytes left over. Four fragments. 00:14:17.134 --> 00:14:26.084 And the offsets are zero, one twenty-two, two forty-four, and three sixty-six — each one the running byte count divided by eight. 00:14:28.018 --> 00:14:29.238 Section two. 00:14:29.288 --> 00:14:31.828 Datagram, or virtual circuit. 00:14:31.878 --> 00:14:40.828 Two answers to the same question, one table with six rows, and a router that catches fire. 00:14:40.951 --> 00:14:47.061 There have only ever been two ways to build a packet network, and Forouzan settles the argument in six rows. 00:14:47.111 --> 00:14:51.201 Here are the first three. 00:14:51.251 --> 00:14:56.481 Before data flows. In a datagram network: nothing. You just send. 00:14:56.531 --> 00:15:04.921 In a virtual-circuit network there is a setup phase that builds the path first, and only then does your data move. 00:15:04.971 --> 00:15:12.851 What the packet carries. In a datagram network, the full source and destination address, in every single packet. 00:15:12.901 --> 00:15:20.911 In a virtual circuit, a short label — plus the addresses, but it is the label that does the work. 00:15:20.961 --> 00:15:25.761 And what the router decides by. The destination address, in one case. 00:15:25.811 --> 00:15:31.891 The label in the other — and the label is swapped at every hop, which slide eighteen takes apart. 00:15:31.941 --> 00:15:40.891 The pattern is already forming. Everything the virtual circuit gains, it pays for with state held inside the network. 00:15:42.703 --> 00:15:47.133 And now the three rows that decide the argument. 00:15:47.183 --> 00:15:56.133 The path taken. In a datagram network each packet chooses independently, so two packets of one message can take different roads. 00:15:56.503 --> 00:16:03.173 In a virtual circuit every packet follows the one path that was built at setup. 00:16:03.223 --> 00:16:08.413 If a router dies. In a datagram network the survivors route around it. 00:16:08.463 --> 00:16:14.833 In a virtual circuit the circuit dies with it, and you set it up again from scratch. 00:16:14.883 --> 00:16:21.533 And state inside the network. None, in the datagram case. Routers remember nothing at all. 00:16:21.583 --> 00:16:26.523 One entry per circuit, in every router along the path, in the other. 00:16:26.573 --> 00:16:35.523 That last row decides the argument, and it is the row the exam builds its scenario around. 00:16:35.803 --> 00:16:43.143 The three phases in detail, because "setup phase" hides a lot of work in that table. 00:16:43.193 --> 00:16:46.463 Setup. Before a single byte of yours moves. 00:16:46.513 --> 00:16:50.893 A request walks the path, and every router on it writes an entry. 00:16:50.943 --> 00:16:58.023 Something like: label two-forty arriving on port two leaves by port five as label one-seventeen. 00:16:58.073 --> 00:17:02.793 Only when the whole path has agreed does your data start. 00:17:02.843 --> 00:17:08.633 Data transfer. This is the fast part, and it is where the design earns its keep. 00:17:08.683 --> 00:17:14.493 The packets now carry a short label instead of being matched against a table of addresses. 00:17:14.543 --> 00:17:19.313 Every router does one small lookup and one swap. 00:17:19.363 --> 00:17:21.633 Teardown. Give the state back. 00:17:21.683 --> 00:17:25.503 The circuit is dismantled and every router forgets it. 00:17:25.553 --> 00:17:32.343 And if it is not torn down, those entries just sit there, occupying room that another circuit needs. 00:17:32.393 --> 00:17:41.313 Setup and teardown are pure overhead. For a one-packet message they dominate the cost; for an hour-long flow they are negligible. 00:17:41.363 --> 00:17:47.321 That is the whole trade, in one sentence. 00:17:47.371 --> 00:17:51.851 So what does the label actually buy? 00:17:51.901 --> 00:17:57.781 It is small. An MPLS label is twenty bits, against a thirty-two-bit destination address. 00:17:57.831 --> 00:18:01.861 Less to match against, and less to carry. 00:18:01.911 --> 00:18:07.261 It is local. Label two-forty means something only on the link it arrived on. 00:18:07.311 --> 00:18:14.081 The same number can be in use all over the network, meaning something different every time. 00:18:14.131 --> 00:18:20.941 It is swapped, not carried through. Every router replaces the incoming label with the outgoing one. 00:18:20.991 --> 00:18:28.751 The packet's identity changes at every junction — which is exactly why teardown matters so much. 00:18:28.801 --> 00:18:35.321 And it is still forwarding. One packet, one router, one table lookup, no conversation. 00:18:35.371 --> 00:18:38.801 Do not let the word "circuit" persuade you otherwise. 00:18:38.851 --> 00:18:47.801 This is the machinery of MPLS, which gets Session eighteen to itself. Today, just recognise the shape. 00:18:49.603 --> 00:18:53.863 Two minutes, and this one is worth doing properly. 00:18:53.913 --> 00:18:58.813 A large transfer is running. Halfway through it, router R2 dies. 00:18:58.863 --> 00:19:05.083 What happens to the packets still coming — in a datagram network, and in a virtual-circuit network? 00:19:05.133 --> 00:19:09.173 And which of the two recovers without anyone noticing? 00:19:09.223 --> 00:19:18.173 Pause here. Two sentences, and be specific about what has to happen before traffic flows again. 00:19:19.633 --> 00:19:27.273 In the datagram network the tables reconverge and the packets take another road. The application may not even notice. 00:19:27.323 --> 00:19:34.163 In the virtual-circuit network, the state R2 was holding is gone — and no other router has a copy of it. 00:19:34.213 --> 00:19:38.163 The circuit is dead, and it must be set up again from scratch. 00:19:38.213 --> 00:19:41.523 That is the survivability argument, made concrete. 00:19:41.573 --> 00:19:48.497 The network that remembered nothing was the network that survived. 00:19:48.547 --> 00:19:50.897 So which one won? 00:19:50.947 --> 00:19:54.067 The core is datagram, and it always has been. 00:19:54.117 --> 00:19:57.557 No setup, no per-flow state, no teardown. 00:19:57.607 --> 00:20:05.517 A router can be rebooted, replaced, or destroyed, and the traffic finds another way within seconds. 00:20:05.567 --> 00:20:07.957 But the virtual circuit did not die. 00:20:08.007 --> 00:20:15.647 It survives inside the core as MPLS, where the label machinery buys fast lookups and traffic engineering — 00:20:15.697 --> 00:20:22.107 carried by an operator who accepts the state, on a network they control end to end. 00:20:22.157 --> 00:20:24.237 And it survives at the edges too. 00:20:24.287 --> 00:20:30.807 Every connection-oriented thing you use is a promise made at the ends, over a core that made none. 00:20:30.857 --> 00:20:36.517 Which is the distinction to keep: connection-oriented is not a synonym for virtual circuit. 00:20:36.567 --> 00:20:45.257 TCP is connection-oriented, and it runs over a core that holds no state at all. 00:20:45.307 --> 00:20:47.507 Checkpoint two. 00:20:47.557 --> 00:20:54.477 One. In which of the two designs can two packets of one message arrive out of order, and why? 00:20:54.527 --> 00:21:02.267 Two. A router in the middle of the path is rebooted. Describe what the user sees, in each design. 00:21:02.317 --> 00:21:09.137 Three. Is label swapping routing or forwarding? Defend the answer in one sentence. 00:21:09.187 --> 00:21:10.117 Answers. 00:21:10.167 --> 00:21:19.117 One. Datagram. Each packet chooses its own path, so two packets can take roads of different lengths. A virtual circuit pins every packet to one path. 00:21:20.877 --> 00:21:29.827 Two. Datagram: a pause of seconds while the tables reconverge, and then it continues. Virtual circuit: the connection is gone, and must be set up again from scratch. 00:21:32.487 --> 00:21:41.437 Three. Forwarding. One packet, one router, one table lookup, and no information exchanged with anybody. 00:21:43.060 --> 00:21:46.280 Section three. The price, measured. 00:21:46.330 --> 00:21:50.150 Three of the four delays are physics, and you can compute them. 00:21:50.200 --> 00:21:57.870 The fourth has no formula at all, and it is the one that ruins your day. 00:21:57.920 --> 00:22:05.640 You met these four on a single link in Session six. Everything that changes today is the multiplier. 00:22:05.690 --> 00:22:09.810 Transmission. L over B. Pushing the bits onto the line. 00:22:09.860 --> 00:22:14.820 Ten thousand bits at a hundred megabits per second is a hundred microseconds. 00:22:14.870 --> 00:22:19.980 And it is paid at the source and at every router that forwards the packet. 00:22:20.030 --> 00:22:24.400 Propagation. d over v. The bits travelling along the link. 00:22:24.450 --> 00:22:30.130 Two thousand metres at two times ten to the eight metres per second is ten microseconds. 00:22:30.180 --> 00:22:35.550 Paid once per link, and no purchase you can make will ever make it smaller. 00:22:35.600 --> 00:22:39.030 Processing. Reading the header and looking up the table. 00:22:39.080 --> 00:22:46.280 Twenty microseconds in our example. Paid at every router — and once more at the destination. 00:22:46.330 --> 00:22:49.610 And queuing. Waiting behind other people's packets. 00:22:49.660 --> 00:22:54.840 Paid only in routers. And it is the only one of the four with no formula at all. 00:22:54.890 --> 00:23:01.631 Four terms. One path. Now let us count them properly. 00:23:01.681 --> 00:23:06.491 Before the formula, a line down the middle of those four. 00:23:06.541 --> 00:23:11.321 On this side: physics. Transmission, propagation, processing. 00:23:11.371 --> 00:23:19.281 Give me the packet size, the bandwidth, the distance and the router, and I can write all three down before the packet is sent. 00:23:19.331 --> 00:23:23.861 They do not care who else is on the network. They are yours. 00:23:23.911 --> 00:23:26.901 And on this side: other people. Queuing. 00:23:26.951 --> 00:23:31.041 It depends entirely on how much traffic arrived just before yours. 00:23:31.091 --> 00:23:36.871 Two identical packets on the same path, one minute apart, can differ by milliseconds. 00:23:36.921 --> 00:23:42.101 That is where jitter comes from — the thing you spent Session six measuring. 00:23:42.151 --> 00:23:44.651 And this is why the exam loves the split. 00:23:44.701 --> 00:23:53.641 A question that hands you every number except the queuing delay is testing whether you know that the other three are computable and this one is not. 00:23:53.691 --> 00:24:01.656 If you are handed t-queuing, you are being handed a measurement. Not a formula. 00:24:01.706 --> 00:24:08.206 Here is the formula, and the next slide is the reason it looks like that. 00:24:08.256 --> 00:24:13.756 Total equals n plus one, times the sum of transmission, propagation and processing. 00:24:13.806 --> 00:24:16.226 Plus n, times the queuing delay. 00:24:16.276 --> 00:24:23.066 n is the number of routers on the path. That is the only input the formula really needs. 00:24:23.116 --> 00:24:26.306 Because n routers means n plus one links. 00:24:26.356 --> 00:24:33.906 A packet leaving the source crosses a link to reach R1. Another to reach R2. And one more to reach the destination. 00:24:33.956 --> 00:24:39.096 Four routers, five links. Draw it once before you substitute. 00:24:39.146 --> 00:24:42.816 So three of the terms are paid n plus one times. 00:24:42.866 --> 00:24:48.686 And one — queuing — is paid n times, because queues live in routers. 00:24:48.736 --> 00:24:55.106 The source is not waiting behind anybody else's packet. The destination is not forwarding anything. 00:24:55.156 --> 00:25:03.667 Draw the path and count the links before you substitute. The off-by-one is where the marks go. 00:25:03.717 --> 00:25:07.117 Four lines, one for each term. 00:25:07.167 --> 00:25:09.557 Transmissions: n plus one. 00:25:09.607 --> 00:25:15.297 The source pushes the packet onto the wire once, and each of the n routers pushes it out again. 00:25:15.347 --> 00:25:18.427 That is n plus one shoves. 00:25:18.477 --> 00:25:24.017 Propagations: n plus one. One per link, and there are n plus one links. 00:25:24.067 --> 00:25:26.477 This is the easy one. 00:25:26.527 --> 00:25:32.617 Processings: n plus one. Every router reads the header — and so does the destination. 00:25:32.667 --> 00:25:36.297 n routers, plus one destination. 00:25:36.347 --> 00:25:38.237 And queues: n. Only n. 00:25:38.287 --> 00:25:45.177 The source is not waiting behind anyone else's packet, and the destination is not forwarding. 00:25:45.227 --> 00:25:48.987 Queues exist where traffic converges, and that is routers. 00:25:49.037 --> 00:25:57.987 If your answer has n queuing delays and n transmission delays, you have made the single most common mistake in this section. 00:25:59.798 --> 00:26:02.988 Now let us do it with real numbers. 00:26:03.038 --> 00:26:08.058 A ten-thousand-bit packet crosses a path with four routers, so five links. 00:26:08.108 --> 00:26:16.688 Every link runs at a hundred megabits per second, over two thousand metres of cable, at two times ten to the eight metres per second. 00:26:16.738 --> 00:26:23.798 Processing is twenty microseconds per node, and each router queues the packet for forty microseconds. 00:26:23.848 --> 00:26:27.508 Find the total source-to-destination delay. 00:26:27.558 --> 00:26:32.738 Transmission. Ten thousand bits divided by ten to the eight bits per second. 00:26:32.788 --> 00:26:38.498 A hundred microseconds. Paid once per link, and there are five links. 00:26:38.548 --> 00:26:43.498 Propagation. Two thousand metres divided by two times ten to the eight. 00:26:43.548 --> 00:26:47.488 Ten microseconds. Paid once per link. 00:26:47.538 --> 00:26:50.758 Processing. Twenty microseconds, given. 00:26:50.808 --> 00:26:55.938 Paid at each of the four routers, and once more at the destination. 00:26:55.988 --> 00:26:58.978 And queuing. Forty microseconds per router, given. 00:26:59.028 --> 00:27:01.788 Four routers, so four of them. 00:27:01.838 --> 00:27:05.328 No formula produced that number. Somebody measured it. 00:27:05.378 --> 00:27:14.328 Write the four ingredients down before you substitute anything. The arithmetic is trivial; the bookkeeping is where the marks are. 00:27:17.190 --> 00:27:23.310 Forty-eight seconds, and nothing to write down. Watch the bar at the bottom fill up. 00:27:23.360 --> 00:27:29.230 There is the path. Source, four routers, destination — and therefore five links. 00:27:29.280 --> 00:27:38.230 The budget bar underneath is empty, and there is a dashed line on it at eight hundred and ten microseconds. Remember that line. 00:27:38.400 --> 00:27:43.560 Transmission first. A hundred microseconds to shovel the packet onto the wire. 00:27:43.610 --> 00:27:49.130 That is the orange block, and it is the biggest single charge on this path. 00:27:49.180 --> 00:27:54.390 Then propagation. Ten microseconds for the bits to travel two kilometres. 00:27:54.440 --> 00:28:00.690 Barely visible next to the orange. On this path, distance is almost free. 00:28:00.740 --> 00:28:07.230 Then processing at R1. Twenty microseconds of reading the header and looking up the table. 00:28:07.280 --> 00:28:10.330 And then queuing. Forty microseconds of waiting. 00:28:10.380 --> 00:28:16.350 One hop has cost a hundred and seventy microseconds, and we have four more to go. 00:28:16.400 --> 00:28:19.930 So watch the same four charges happen four more times. 00:28:19.980 --> 00:28:25.720 Five transmissions, five propagations, five processings — and four queues. 00:28:25.770 --> 00:28:32.480 And there it is, landing exactly on the dashed line. Eight hundred and ten microseconds. 00:28:32.530 --> 00:28:37.870 Now run it again. Same path, same cable, same routers, same packet. 00:28:37.920 --> 00:28:45.600 The only difference is that R2 is congested, and it holds the packet for fifteen hundred microseconds instead of forty. 00:28:45.650 --> 00:28:50.900 Watch the red block: forty microseconds becomes fifteen hundred. 00:28:50.950 --> 00:28:54.100 Two thousand two hundred and seventy microseconds. 00:28:54.150 --> 00:29:03.100 And look at the four tiles. Transmission, propagation and processing are identical on both runs. Five hundred, fifty, and a hundred. 00:29:03.880 --> 00:29:07.210 Six hundred and fifty microseconds of honest physics, unchanged. 00:29:07.260 --> 00:29:12.620 Queuing went from twenty per cent of the journey to seventy-one per cent of it. 00:29:12.670 --> 00:29:20.596 Nothing physical changed at all. That is why the network layer cannot promise you a delay. 00:29:20.646 --> 00:29:25.786 Both runs on one slide, with the arithmetic written out. 00:29:25.836 --> 00:29:30.576 Run one. Five times a hundred and thirty, which is six hundred and fifty. 00:29:30.626 --> 00:29:34.056 Plus four times forty, which is a hundred and sixty. 00:29:34.106 --> 00:29:40.016 Eight hundred and ten microseconds. Queuing is twenty per cent of the journey. 00:29:40.066 --> 00:29:45.126 Run two. The same six hundred and fifty, because nothing physical changed. 00:29:45.176 --> 00:29:47.666 Plus sixteen hundred and twenty of queuing. 00:29:47.716 --> 00:29:53.936 Two thousand two hundred and seventy microseconds. Nearly three times the journey. 00:29:53.986 --> 00:30:00.196 And what changed? Not the cable. Not the bandwidth. Not the routers' processing. 00:30:00.246 --> 00:30:02.186 One number, contributed by strangers. 00:30:02.236 --> 00:30:11.186 Three of the four terms are honest. The fourth belongs to other people, and it is the one that decides your afternoon. 00:30:13.057 --> 00:30:17.817 Here is the same calculation as a thing you can open yourself. 00:30:17.867 --> 00:30:24.887 It opens on the worked example. Four routers, ten thousand bits, a hundred megabits, two kilometres. 00:30:24.937 --> 00:30:30.007 Eight hundred and ten microseconds, and one bar per hop, coloured by term. 00:30:30.057 --> 00:30:35.957 Look at the tile on the right: transmission owns sixty-two per cent of it. 00:30:36.007 --> 00:30:43.877 Now drop it to one router. Watch every number fall — not because the links got faster, but because the multiplier shrank. 00:30:43.927 --> 00:30:49.897 Two links instead of five. This is the n plus one, made visible. 00:30:49.947 --> 00:30:57.667 Back to four. Five bars, and five identical stacks — plus the queue in each of the four routers. 00:30:57.717 --> 00:31:02.347 Now ten times the bandwidth. A gigabit instead of a hundred megabits. 00:31:02.397 --> 00:31:04.987 The orange collapses to almost nothing. 00:31:05.037 --> 00:31:12.127 Propagation and processing did not move at all. You bought away one term, and only one. 00:31:12.177 --> 00:31:18.247 Now the opposite. Leave the bandwidth alone and stretch the link to five thousand kilometres. 00:31:18.297 --> 00:31:24.487 Propagation takes over completely, and the verdict changes: this path is distance-bound. 00:31:24.537 --> 00:31:29.617 No amount of bandwidth will help. Nobody sells a faster metre. 00:31:29.667 --> 00:31:33.617 Back to the textbook example. 00:31:33.667 --> 00:31:38.267 And now press the one button that is not physics at all. R2 gets busy. 00:31:38.317 --> 00:31:43.717 Two point two seven milliseconds, and queuing owns seventy-one per cent of a path whose physics never changed. 00:31:43.767 --> 00:31:52.717 Open it after the lecture and try to make queuing small by buying something. You cannot. That is the whole point of the tool. 00:31:55.873 --> 00:31:58.153 Checkpoint three. 00:31:58.203 --> 00:32:05.613 One. A path has two routers. How many transmission delays, and how many queuing delays? 00:32:05.663 --> 00:32:13.663 Two. Which term would you attack first, on a path where the packets are large and the links are short? 00:32:13.713 --> 00:32:22.303 Three. Two identical packets take the same path one minute apart, and differ by three milliseconds. Which term explains it? 00:32:22.353 --> 00:32:23.273 Answers. 00:32:23.323 --> 00:32:32.273 One. Three transmissions, three propagations and three processings — but only two queuing delays. n is two, so n plus one is three. 00:32:34.063 --> 00:32:43.013 Two. Transmission, L over B. It is the term that grows with packet size and shrinks with bandwidth, and propagation is already tiny on a short link. 00:32:44.863 --> 00:32:53.813 Three. Queuing, and only queuing. The other three are fixed by the packet, the link and the router — and none of those changed. 00:32:56.055 --> 00:33:01.615 Delay was the first yardstick. Here is the second, and it is one line. 00:33:01.665 --> 00:33:05.135 The throughput of a path is the minimum of the link rates. 00:33:05.185 --> 00:33:09.395 A chain of a hundred, forty and two hundred megabits carries forty. 00:33:09.445 --> 00:33:15.125 Not the average. Not the last one. Not the first one. The minimum. 00:33:15.175 --> 00:33:19.695 Think of water through three pipes. The narrowest pipe sets the flow. 00:33:19.745 --> 00:33:26.265 The two-hundred-megabit link cannot push more water than the forty-megabit link is handing it. 00:33:26.315 --> 00:33:29.685 Which means upgrading the wrong link buys you nothing at all. 00:33:29.735 --> 00:33:34.355 Turn the hundred into a thousand, and the path still carries forty. 00:33:34.405 --> 00:33:40.565 The bottleneck did not move. You just paid for capacity that will never be used. 00:33:40.615 --> 00:33:43.585 And in real life that narrow link is shared. 00:33:43.635 --> 00:33:49.585 If two other flows cross the forty-megabit link with you, your slice is nearer thirteen. 00:33:49.635 --> 00:33:56.572 The min rule gives you the ceiling. It does not give you the allocation. 00:33:56.622 --> 00:33:57.582 Sixty seconds. 00:33:57.632 --> 00:34:05.352 A path runs at a gigabit, then a hundred megabits, then a gigabit, then fifty megabits, then a gigabit. 00:34:05.402 --> 00:34:14.352 What is the throughput of the path? And if you could upgrade exactly one link, which one — and what would the new throughput be? 00:34:15.322 --> 00:34:22.372 Pause. Two numbers and one choice. 00:34:22.422 --> 00:34:27.362 Fifty megabits. Upgrade the fifty. And the new throughput is a hundred. 00:34:27.412 --> 00:34:31.142 Because fixing the fifty promotes the hundred to bottleneck. 00:34:31.192 --> 00:34:34.332 You bought a doubling, not a twenty-fold increase. 00:34:34.382 --> 00:34:43.332 Every capacity upgrade moves the bottleneck. None of them removes it. There is always a narrowest pipe. 00:34:44.256 --> 00:34:46.506 Section four. Congestion. 00:34:46.556 --> 00:34:53.346 A finite buffer, a retransmission, and the two curves every engineer ends up drawing from memory. 00:34:53.396 --> 00:35:00.073 This is where the second half of the opening question gets its answer. 00:35:00.123 --> 00:35:06.363 The third yardstick is packet loss, and it follows from one fact about buffers. 00:35:06.413 --> 00:35:09.033 A router's input buffer is finite. 00:35:09.083 --> 00:35:13.543 That is it. That one sentence is behind everything else in this section. 00:35:13.593 --> 00:35:18.563 A packet arriving while the router is busy waits in the buffer. 00:35:18.613 --> 00:35:22.353 And when the buffer is full, the next packet is discarded. 00:35:22.403 --> 00:35:30.353 No apology. No notification. Best-effort delivery, exactly as advertised back in section one. 00:35:30.403 --> 00:35:32.813 And drops become retransmissions. 00:35:32.863 --> 00:35:35.863 The Internet's answer to loss is "send it again". 00:35:35.913 --> 00:35:43.333 So the dropped packet does not disappear from the network. It comes back, later, as new traffic. 00:35:43.383 --> 00:35:48.243 More traffic, fuller buffers, more drops, more retransmissions, more traffic. 00:35:48.293 --> 00:35:51.633 Loss does not relieve the congestion. It feeds it. 00:35:51.683 --> 00:35:59.837 There is no brake in that list. Nothing in it gets smaller on its own. 00:36:00.927 --> 00:36:06.767 Forty-four seconds inside one router. Count the boxes arriving at the top. 00:36:06.817 --> 00:36:12.257 Light load. Six packets arriving, one slot of the buffer in use, nothing dropped. 00:36:12.307 --> 00:36:17.677 The buffer is a shock absorber, and at this load it is barely doing anything. 00:36:17.727 --> 00:36:20.677 Now the load rises. The queue is filling. 00:36:20.727 --> 00:36:28.937 Nothing is lost yet — but every packet is now waiting behind others, and the delay curve has started to bend. 00:36:28.987 --> 00:36:31.377 And there it is. All ten slots. 00:36:31.427 --> 00:36:36.937 The buffer is full, which means the next arrival has nowhere at all to go. 00:36:36.987 --> 00:36:41.107 So it is discarded. Watch the red crosses appear along the bottom. 00:36:41.157 --> 00:36:46.827 Nobody is told. The sender does not know, the receiver does not know. 00:36:46.877 --> 00:36:51.667 Which is why the next thing that happens takes so long. The source waits for a timeout. 00:36:51.717 --> 00:36:57.167 And then it sends the same packet again. Those are the brown boxes with the arrow. 00:36:57.217 --> 00:37:04.487 And now look at the arrivals row. Seventeen packets, and six of them are things the network already carried once. 00:37:04.537 --> 00:37:08.177 Every drop has come back as new traffic. 00:37:08.227 --> 00:37:11.537 Twenty packets arriving. Nine of them are echoes. 00:37:11.587 --> 00:37:19.517 The router is being offered four times what it can carry, and nine of the twenty are packets it has already delivered once and thrown away. 00:37:19.567 --> 00:37:24.687 Dropping packets did not relieve the congestion. The load came back with interest. 00:37:24.737 --> 00:37:32.124 That is congestion collapse, and there is no brake anywhere in the loop. 00:37:32.174 --> 00:37:39.574 Now the two curves. Draw these once tonight, from memory, and label them. 00:37:39.624 --> 00:37:40.904 Delay against load. 00:37:40.954 --> 00:37:47.524 Below capacity, the queue drains as fast as it fills, so delay is very nearly just propagation plus processing. 00:37:47.574 --> 00:37:56.074 As the load approaches capacity, the queue takes longer and longer to drain — and the curve heads for infinity. 00:37:56.124 --> 00:37:57.554 Throughput against load. 00:37:57.604 --> 00:38:03.024 It rises along the ideal line until the knee, where the routers begin to drop. 00:38:03.074 --> 00:38:12.024 And past the knee the sources resend what was discarded, the offered load climbs, and the throughput collapses. It bends back down. 00:38:12.744 --> 00:38:15.154 The knee is the whole point. 00:38:15.204 --> 00:38:18.544 Before it, extra load buys extra throughput. 00:38:18.594 --> 00:38:21.904 After it, extra load buys less throughput than you had before. 00:38:21.954 --> 00:38:30.216 A network running past its knee is a network doing more work and delivering less. 00:38:31.236 --> 00:38:38.306 One router, one capacity, and one slider that says how much the applications want to send. 00:38:38.356 --> 00:38:46.336 Half capacity. Both curves are in the comfortable region, nothing is dropped, and everything offered is delivered. 00:38:46.386 --> 00:38:50.976 Ninety per cent. Still nothing lost — but look at the left-hand curve. 00:38:51.026 --> 00:38:57.416 The delay has already started climbing steeply, and we have not lost a single packet yet. 00:38:57.466 --> 00:39:00.806 At capacity. The delay curve has gone vertical. 00:39:00.856 --> 00:39:05.736 This is the knee, and this is the best the network will ever do. 00:39:05.786 --> 00:39:11.136 Now push to a hundred and thirty per cent, with the sources resending what they lose. 00:39:11.186 --> 00:39:15.726 The applications asked for a hundred and thirty and are receiving thirty-two. 00:39:15.776 --> 00:39:20.936 The offered load is four times capacity, because three quarters of it is echoes. 00:39:20.986 --> 00:39:25.636 And look at the right-hand curve: it has turned back on itself. 00:39:25.686 --> 00:39:30.946 Now the control. Same load, but this time the sources give up on what is lost. 00:39:30.996 --> 00:39:36.576 The throughput simply saturates at the capacity line. Flat. No collapse at all. 00:39:36.626 --> 00:39:41.286 That is the curve everybody expects — and it is not what the Internet does. 00:39:41.336 --> 00:39:49.286 Retransmission is what turns saturation into collapse. That is the whole difference between the two pictures. 00:39:49.336 --> 00:39:55.046 So here is prevention. A window policy that never asks for more than the network can take. 00:39:55.096 --> 00:40:04.046 No queue builds, nothing is dropped, nothing is resent. And nobody had to detect anything, which is what makes it open loop. 00:40:04.176 --> 00:40:09.236 And here is removal. The applications still ask for a hundred and thirty per cent. 00:40:09.286 --> 00:40:14.926 But the router noticed, and fed back, and the offered load was pulled down to a hundred and four. 00:40:14.976 --> 00:40:23.897 Congestion happened, was detected, and was cured. That is closed loop, in one screen. 00:40:24.147 --> 00:40:29.887 Forouzan sorts the techniques into two buckets, and the sorting is what is examinable. 00:40:29.937 --> 00:40:33.227 Open loop first. Prevention. 00:40:33.277 --> 00:40:35.347 Good retransmission timers. 00:40:35.397 --> 00:40:43.377 A timer that fires too early resends a packet that was merely slow, and manufactures traffic out of nothing. 00:40:43.427 --> 00:40:45.337 A sensible window policy. 00:40:45.387 --> 00:40:52.337 Selective-Repeat resends the one packet that was lost. Go-Back-N resends it and everything after it. 00:40:52.387 --> 00:40:57.187 Chapter twenty-three makes that concrete, with numbers. 00:40:57.237 --> 00:40:58.797 Acknowledge less often. 00:40:58.847 --> 00:41:05.037 Every acknowledgement is itself a packet, on a network you are trying to relieve. 00:41:05.087 --> 00:41:10.667 And admission control. Refuse a new virtual circuit when the network is already full. 00:41:10.717 --> 00:41:16.127 The bouncer at the door — and it needs a network that has circuits to refuse. 00:41:16.177 --> 00:41:25.127 Now: the defining feature of this list is not the policies. It is that nothing anywhere is detecting congestion or sending a signal about it. 00:41:27.656 --> 00:41:34.616 Closed loop. Removal. Something detects congestion, and something else is told about it. 00:41:34.666 --> 00:41:39.426 Backpressure. A congested router tells its upstream neighbour to slow down. 00:41:39.476 --> 00:41:44.916 That neighbour tells its own upstream neighbour, hop by hop, back towards the source. 00:41:44.966 --> 00:41:53.206 It needs each router to know where a flow came from — which is per-flow state. So: virtual-circuit networks only. 00:41:53.256 --> 00:41:59.986 The choke packet. The congested router skips the queue of neighbours entirely and warns the source directly. 00:42:00.036 --> 00:42:04.616 One message, straight to the machine that can actually act. 00:42:04.666 --> 00:42:08.016 Implicit signalling. Nobody sends anything at all. 00:42:08.066 --> 00:42:12.876 The source notices its own timeouts and infers that the network is congested. 00:42:12.926 --> 00:42:18.676 This is what TCP actually does, and Session twenty-three is about it. 00:42:18.726 --> 00:42:24.166 And explicit signalling. A warning bit rides in packets that are passing through anyway — 00:42:24.216 --> 00:42:31.406 forward towards the destination, or backward towards the source. 00:42:31.456 --> 00:42:39.886 So here is the test, because the exam question is always the same shape: sort this list into two buckets. 00:42:39.936 --> 00:42:44.066 Ask one question. Did anything detect congestion and report it? 00:42:44.116 --> 00:42:51.906 If nothing did — if the technique is just a policy the ends follow whether the network is busy or not — it is open loop. Prevention. 00:42:51.956 --> 00:42:56.466 And if something did, it is closed loop. 00:42:56.516 --> 00:42:59.346 A router noticed, and a signal went somewhere. 00:42:59.396 --> 00:43:08.056 Backpressure, choke packets, warning bits — and even a source inferring from its own timeouts. All closed loop. 00:43:08.106 --> 00:43:12.756 Open loop prevents by policy. Closed loop removes by feedback. 00:43:12.806 --> 00:43:16.436 And admission control is the one most often misfiled. 00:43:16.486 --> 00:43:25.436 It sounds active, so it feels like feedback. But it is a policy applied at the door, before anything is congested. Open loop. 00:43:27.641 --> 00:43:32.131 Checkpoint four, and this is the last one. 00:43:32.181 --> 00:43:39.251 One. A source infers congestion from its own timeouts and slows down. Open loop or closed? 00:43:39.301 --> 00:43:44.451 Two. Why is backpressure unavailable on the Internet's core? 00:43:44.501 --> 00:43:52.341 Three. A network is running past its knee. What happens to throughput if the offered load rises further? 00:43:52.391 --> 00:43:53.321 Answers. 00:43:53.371 --> 00:44:02.291 One. Closed loop. Congestion was detected and acted upon after it happened. The signal was implicit, but there was one. 00:44:02.341 --> 00:44:11.291 Two. Because it needs each router to know which upstream neighbour a flow came from, and that is per-flow state. The datagram core holds none. 00:44:12.661 --> 00:44:21.611 Three. It falls. Past the knee the extra load is mostly retransmissions, so the network carries more packets and delivers fewer of them. 00:44:23.513 --> 00:44:28.803 So. The two claims from minute one, and the answers to both. 00:44:28.853 --> 00:44:30.093 Why promise nothing? 00:44:30.143 --> 00:44:36.223 Because every promise costs state, and state in the core is state that dies with the router holding it. 00:44:36.273 --> 00:44:44.193 Promise nothing, and the core never breaks its word — and never loses your conversation when a machine fails. 00:44:44.243 --> 00:44:46.923 And why can a faster router make things worse? 00:44:46.973 --> 00:44:51.193 The first reason is the min rule. The bottleneck did not move. 00:44:51.243 --> 00:44:58.883 Path throughput is the minimum of the links, so upgrading one router just feeds the narrow link faster. 00:44:58.933 --> 00:45:00.573 And then the drops come back. 00:45:00.623 --> 00:45:05.763 The full buffer discards, and every discard returns as a retransmission. 00:45:05.813 --> 00:45:11.743 Offered load rises, the buffer fills faster, and more is discarded. 00:45:11.793 --> 00:45:14.233 Past the knee, the curves take over. 00:45:14.283 --> 00:45:18.013 Delay heads for infinity, and throughput collapses. 00:45:18.063 --> 00:45:23.343 The upgrade shoved a network that was sitting near its knee straight past it. 00:45:23.393 --> 00:45:26.783 Promise nothing, and the core never breaks its word. 00:45:26.833 --> 00:45:33.728 Push it past the knee, and extra speed is just extra fuel. 00:45:33.968 --> 00:45:37.528 Five ways to lose marks on this material. 00:45:37.578 --> 00:45:41.568 Wrong: routing and forwarding are the same thing. 00:45:41.618 --> 00:45:50.568 Right: routing writes the table — network-wide, before traffic, by exchanging information. Forwarding reads it: one packet, one router, no conversation. 00:45:53.168 --> 00:45:57.178 Wrong: path throughput is the average of the link rates. 00:45:57.228 --> 00:46:06.178 Right: the minimum. One forty-megabit link caps a hundred-forty-two-hundred chain at forty — and sharing that link can cut your slice further still. 00:46:08.298 --> 00:46:13.938 Wrong: counting n queuing delays as n plus one, or n plus one transmissions as n. 00:46:13.988 --> 00:46:22.938 Right: n routers, n plus one links, n plus one transmissions, propagations and processings. But only n times t-queuing, because queues live in routers. 00:46:26.878 --> 00:46:30.848 Wrong: dropping packets relieves the congestion. 00:46:30.898 --> 00:46:39.848 Right: sources retransmit what routers discard, so the load comes back with interest. That is precisely why delay goes to infinity and throughput collapses. 00:46:42.188 --> 00:46:47.978 And wrong: filing admission control as closed loop, because it sounds active. 00:46:48.028 --> 00:46:56.978 Right: open loop. It is a policy applied at the door, before anything is congested, with no detection and no feedback signal anywhere. 00:46:59.874 --> 00:47:03.024 Four things you should be able to do now. 00:47:03.074 --> 00:47:11.394 Separate routing from forwarding on sight — and justify it by asking whether the router exchanged anything with anybody. 00:47:11.444 --> 00:47:15.744 State what best-effort refuses, and say who supplies it instead. 00:47:15.794 --> 00:47:22.624 Delivery, order, timing — all rebuilt by the transport layer at the two ends. 00:47:22.674 --> 00:47:26.254 Compute a total path delay without an off-by-one. 00:47:26.304 --> 00:47:34.744 n plus one times the three, plus n times the queuing — having drawn the path and counted the links first. 00:47:34.794 --> 00:47:40.024 And draw both congestion curves, and sort a list of techniques into two buckets. 00:47:40.074 --> 00:47:49.024 Delay to infinity, throughput collapsing past the knee. Open loop prevents by policy; closed loop removes by feedback. 00:47:51.861 --> 00:48:00.091 Five final checks. Pause after each one if you need to. 00:48:00.141 --> 00:48:09.091 One. A path has three routers. Write the total-delay formula with the numbers filled in, given L over B is fifty microseconds, d over v is five, processing is ten, and queuing is a hundred. 00:48:15.161 --> 00:48:21.541 Two. Name the three guarantees IP declines, and the layer that supplies them. 00:48:21.591 --> 00:48:29.631 Three. A virtual circuit's router is rebooted mid-transfer. What must happen before data flows again? 00:48:29.681 --> 00:48:38.631 Four. A chain runs two hundred, eighty, one fifty, eighty megabits. What is the throughput, and what does upgrading one of the eighties buy you? 00:48:41.161 --> 00:48:50.111 Five. Sort these into open and closed loop: choke packet, admission control, good timers, backpressure, warning bit. 00:48:50.701 --> 00:48:59.651 The answers are along the bottom of the slide. Number four is the one to look at twice — upgrading one eighty leaves the other one as the bottleneck, so the throughput is still eighty. 00:49:04.471 --> 00:49:06.031 That is Session eight. 00:49:06.081 --> 00:49:12.601 Three duties: packetize, route, forward. Routing writes the table; forwarding reads it. 00:49:12.651 --> 00:49:17.611 Three delays are physics. The fourth is other people, and it has no formula. 00:49:17.661 --> 00:49:22.551 Path throughput is the minimum of the links, and sharing cuts your slice further. 00:49:22.601 --> 00:49:28.401 And past the knee, every extra packet is fuel: more drops, more resends, less delivered. 00:49:28.451 --> 00:49:33.521 Before the next session, read Forouzan eighteen point four — IPv4 addresses. 00:49:33.571 --> 00:49:38.101 The addressing arc starts next class, and it runs for six straight sessions. 00:49:38.151 --> 00:49:47.101 Bring a calculator, and know your powers of two up to two to the tenth by heart. They will pay rent for the rest of the term. 00:49:47.191 --> 00:49:52.054 I will see you there.