WEBVTT 00:00:00.000 --> 00:00:03.970 Welcome back to CSE 316 — Data Communication and Networking. 00:00:04.020 --> 00:00:10.250 This is the detailed video version of Session thirteen, and it is not a lecture. It is a clinic. 00:00:10.300 --> 00:00:19.250 Six problems, worked in full, sixteen marks of Part A revised, and then a hunt for the five errors that cost the most marks every term. 00:00:20.000 --> 00:00:28.740 This session only works if you stop the video. Every problem is posed before it is worked, and the pause is the point. 00:00:28.790 --> 00:00:37.740 If you watch it straight through you will feel prepared and you will not be. 00:00:37.821 --> 00:00:41.611 Here is the claim, and it is a strong one. 00:00:41.661 --> 00:00:46.421 Every computational question on the midterm paper is one of six problems. 00:00:46.471 --> 00:00:52.451 Not similar to. Not based on. One of six, with the numbers changed. 00:00:52.501 --> 00:00:57.881 So your task is simple. When a prompt slide appears, pause and do it on paper. 00:00:57.931 --> 00:01:04.801 A revealed answer ends thinking. Committing to an answer first is what makes the walkthrough stick. 00:01:04.851 --> 00:01:09.791 And the measure of success: you leave with a list of exactly what to drill tonight. 00:01:09.841 --> 00:01:18.791 One rule for the hour: a wrong answer you actually committed to is worth more than a right answer you watched go past. 00:01:19.654 --> 00:01:22.454 Section one. The map and the paper. 00:01:22.504 --> 00:01:24.384 Six weeks, one slide. 00:01:24.434 --> 00:01:33.384 Before the problems, the shape of what is examinable — and, just as importantly, where the boundary falls. 00:01:34.421 --> 00:01:40.871 Six weeks, left to right, and the paper walks this map in the same order. 00:01:40.921 --> 00:01:46.981 Chapter one: what a network is. Topologies, switching, and the performance vocabulary. 00:01:47.031 --> 00:01:55.731 That is Part A on the paper. Cheap marks — and the ones students throw away by skipping Week one revision. 00:01:55.781 --> 00:02:02.161 Chapter two: who does what. The layers, and encapsulation adding a header at each one. 00:02:02.211 --> 00:02:10.181 Also Part A. And note that encapsulation overhead is an arithmetic question, not a definition question. 00:02:10.231 --> 00:02:19.141 Chapter three: the wire taken seriously. Signals, decibels, Nyquist, Shannon, and the four performance numbers. 00:02:19.191 --> 00:02:26.051 That is Part B, twenty marks — and problems one, two and three all live here. 00:02:26.101 --> 00:02:32.801 Chapter nine: frames get names, and ARP finds them. The I/G bit and the ARP cache. 00:02:32.851 --> 00:02:35.901 Part C, and problem four. 00:02:35.951 --> 00:02:44.901 And eighteen point one through eighteen point four: the layer's three duties, packet switching, delay, the address, the mask, the ritual, and equal cuts. 00:02:45.301 --> 00:02:53.321 Now look at the boundary, because it matters: the paper runs through Chapter nine and through eighteen point four. 00:02:53.371 --> 00:03:02.321 Equal cuts, and nothing harder. Unequal subnetting is after the midterm, and it is not on this paper. 00:03:03.129 --> 00:03:08.989 The structure, briskly, because you will want to plan your time. 00:03:09.039 --> 00:03:17.989 Part A: sixteen marks of Week one and two work. Surviving a failure in a topology, mesh links and ports, circuit versus packet switching, encapsulation overhead. 00:03:21.179 --> 00:03:30.129 Part B: twenty marks of physical layer. The dB chain — that is problem one. Then one pooled item from Nyquist rate, Nyquist levels or Shannon. 00:03:31.439 --> 00:03:40.389 Then the bridge problem — problem two, the big one. Then one item from delay, throughput or bandwidth-delay product — problem three. 00:03:43.209 --> 00:03:52.159 Part C: twenty-four marks of addressing. The ARP bill, mask to slash and back, the block ritual — problem five — and one equal-subnetting problem. 00:03:54.399 --> 00:04:01.239 So forty-four of the sixty marks are Parts B and C, and every one of those is a problem we work today. 00:04:01.289 --> 00:04:08.489 Part A is the cheap part. Ten minutes if you have revised it, thirty if you have not — so revise it. 00:04:08.539 --> 00:04:17.489 And "pooled" means pooled: each version draws its own item from the pool. Preparing four of the five and hoping is a coin-flip you do not need. 00:04:20.229 --> 00:04:23.579 Three instructions for the next hour. 00:04:23.629 --> 00:04:31.039 Every problem is posed before it is worked. When the prompt appears, stop the video and do it on paper. 00:04:31.089 --> 00:04:39.449 Three minutes is enough for any of them once the method is yours — and if it is not enough, that is the diagnosis. 00:04:39.499 --> 00:04:44.889 The working is the point, not the answer. Method marks are read off what you write. 00:04:44.939 --> 00:04:53.889 A wrong final number with correct working still scores. A bare number with nothing under it scores what it is worth and no more. 00:04:54.199 --> 00:04:59.519 And watch for the ghost on each problem — the classic wrong answer that feels right. 00:04:59.569 --> 00:05:08.110 Every problem here has one named, and you will meet all of them before the end. 00:05:08.160 --> 00:05:10.000 Sixteen marks of Part A, and two slides for it. Chapter one first, in five rows. 00:05:10.050 --> 00:05:11.610 Mesh, star, bus, ring. The paper rarely asks you to define them. It asks what one break costs, and that is row three. Every topology is a bet on which failure you can afford — mesh you cannot afford to build, bus you cannot afford to break, which is why the world went star. 00:05:11.660 --> 00:05:12.940 Links are n, times n minus one, over two. Ports are n minus one per device, and n times n minus one in total. At eight devices: twenty-eight links, seven ports each, fifty-six port-ends. The divide-by-two belongs to the links and never to the ports, because a port sits at each cable end and nobody halves the ends. 00:05:12.990 --> 00:05:15.290 Star: one cable down isolates one device; the hub down kills the whole system. Bus: a break stops all transmission, because the damaged point reflects signals back as noise in both directions. Ring: everyone, since traffic is one-way and every path uses every link. Mesh: one pair loses its direct link, and nobody else notices. 00:05:15.340 --> 00:05:16.560 A LAN connects hosts and is privately owned. A WAN interconnects connecting devices — switches, routers, modems — and you lease it from a communication company. Small-i internet is any two or more joined networks; capital-I Internet is the specific global one. Both distinctions are one-sentence marks, and both are asked most terms. 00:05:16.610 --> 00:05:25.560 Circuit switching reserves a path before the first word and holds it until teardown, so capacity reserved is capacity idle whenever nobody speaks. Packet switching shares the line and queues instead, which is where variable delay comes from. And the vocabulary: performance, reliability and security, with throughput and delay as the two working numbers. 00:07:15.189 --> 00:07:16.319 Chapter two, and it ends in the arithmetic the paper asks for. 00:07:16.369 --> 00:07:20.509 Application, transport, network, data link, physical. Message, segment, datagram, frame, bit. OSI has seven floors: session and presentation are the two TCP/IP never built, because transport covers part of that job and applications build the rest themselves. OSI survives as vocabulary and as exam questions. 00:07:20.559 --> 00:07:21.469 Four address pairs, one per layer, and one layer with none. Names at the application layer. Port numbers, which pick the program and not the machine. IP addresses, which are global and survive end to end. MAC addresses, which are local and new at every hop. The physical layer has no address, because a single bit cannot carry one. 00:07:21.519 --> 00:07:22.419 Now the arithmetic. A thousand bytes of application data, and the question hands you the header sizes: twenty bytes of TCP, twenty bytes of IP, and Ethernet's fourteen-byte header with a four-byte trailer. Note the trailer. The data-link layer is the only layer that adds one, and leaving it out costs four bytes in every line below. 00:07:22.469 --> 00:07:23.249 Add on the way down. A thousand plus twenty is a one-thousand-and-twenty-byte segment. Plus twenty is a one-thousand-and-forty-byte datagram. Plus eighteen — fourteen in front and four behind — is one thousand and fifty-eight bytes in the frame. Nothing inside is opened on the way: each layer wraps what it was handed as one opaque payload. 00:07:23.299 --> 00:07:32.249 So fifty-eight bytes of overhead. As a fraction of the frame that is five point five per cent, which leaves ninety-four and a half per cent of the wire carrying payload. Now shrink the payload to a hundred bytes: the same fifty-eight bytes is thirty-seven per cent of the frame. Small packets are expensive, and that comparison is the version of the question that carries the marks. 00:09:22.297 --> 00:09:26.497 Eighteen point one to eighteen point three are inside the assigned range and are not one of the six problems. Four cards, and they are definition and formula marks. 00:09:26.547 --> 00:09:28.817 Three duties. Packetizing: the source wraps the payload in a datagram and the destination unwraps it, and routers in between may not open it or change the addresses. Routing: finding the best path, done in advance by routing protocols. Forwarding: the action one router applies when a packet arrives at an interface. Routing builds the table; forwarding uses it. 00:09:28.867 --> 00:09:29.817 The datagram approach, which is connectionless. Each packet is an independent entity with no relationship to the other packets of the same message. It carries source and destination addresses, and the router forwards on the destination address alone. Packets of one message may travel different paths, so they can arrive out of order. 00:09:29.867 --> 00:09:33.607 The virtual-circuit approach, which is connection-oriented. Three phases: setup, data transfer, teardown. Setup exchanges a request packet and an acknowledgment packet so that every router has a table entry. After that each packet carries a flow label, the forwarding decision is made on the label rather than the address, all packets follow one path, and they arrive in order. 00:09:33.657 --> 00:09:34.827 And the delay equation, exactly as Forouzan writes it. Total delay equals n plus one, times transmission plus propagation plus processing, plus n times queuing. n routers give n plus one links. The n plus one processing delays are related to the n routers and the destination; queuing alone is counted n times, because a packet waits in the queues of routers and nowhere else. 00:09:34.877 --> 00:09:43.827 Throughput, in the same section, is the bottleneck: the minimum transmission rate along the path. Delay, throughput and packet loss are the three performance measures the section names. 00:11:28.517 --> 00:11:32.927 Section two. Problems one and two: the physical layer. 00:11:32.977 --> 00:11:35.547 Decibels, and the two speed limits. 00:11:35.597 --> 00:11:44.547 These are the two biggest-mark shapes on the paper. One is about sign discipline; the other is about reading the question before touching a formula. 00:11:47.347 --> 00:11:49.417 Problem one. 00:11:49.467 --> 00:11:57.837 A signal passes minus three decibels, then plus seven, then minus three. A cable, an amplifier, another cable. 00:11:57.887 --> 00:12:01.327 That is the whole physical situation. 00:12:01.377 --> 00:12:08.857 Part (a): what is the overall gain, in decibels? One line of arithmetic, and the signs are the mark. 00:12:08.907 --> 00:12:15.907 Part (b): if two milliwatts goes in, what power comes out? And this is where the ghost lives. 00:12:15.957 --> 00:12:19.917 Stop the video. Three minutes, both parts, on paper. 00:12:19.967 --> 00:12:27.547 And before you calculate part (b), commit to a verdict: more than two milliwatts, or less? 00:12:27.597 --> 00:12:36.547 Committing to "more than two milliwatts" before you compute is the habit that catches the ghost in part (b). 00:12:37.629 --> 00:12:41.909 Five lines. Write all five, every time. 00:12:41.959 --> 00:12:47.649 Line one, sum. Minus three plus seven minus three is plus one decibel. 00:12:47.699 --> 00:12:56.649 Cascaded stages add in decibels — because a decibel is a logarithm, and logarithms turn multiplication into addition. That is the whole reason the unit exists. 00:12:59.839 --> 00:13:07.009 Line two, the promise. Plus one decibel is a gain, so the output must be more than two milliwatts. 00:13:07.059 --> 00:13:14.989 Written before the calculator. It is a free mark, and a free check on every line below it. 00:13:15.039 --> 00:13:21.809 Line three, convert. P out over P in equals ten to the dB over ten — ten to the nought point one. 00:13:21.859 --> 00:13:26.169 You leave decibel-land before you multiply anything. 00:13:26.219 --> 00:13:35.169 Line four, multiply. Ten to the nought point one is one point two six, and one point two six times two milliwatts is about two and a half. 00:13:35.569 --> 00:13:42.309 Two conversions worth memorising: plus one dB is "about a quarter more", and minus three dB is "about half". 00:13:42.359 --> 00:13:48.649 Line five, check. Two point five is more than two. The promise held. 00:13:48.699 --> 00:13:57.649 And there is the payoff: if an exponent slip had given you one point six milliwatts, line two would have caught it — without redoing any arithmetic. 00:14:03.259 --> 00:14:08.259 Forty-two seconds, and the whole problem including its ghost. 00:14:08.309 --> 00:14:14.349 Three stages on the signal path. Two cables and an amplifier, and two milliwatts going in. 00:14:14.399 --> 00:14:22.199 Notice the output box is a question mark — that is exactly what you have at the start of the item. 00:14:22.249 --> 00:14:27.139 Minus three, plus seven, minus three: plus one decibel. 00:14:27.189 --> 00:14:36.139 And underneath, the reason they add at all — log of a times b is log a plus log b. That single identity is what decibels are for. 00:14:36.419 --> 00:14:44.449 Now the promise, on its own line, in amber. Plus one is a gain, so the output exceeds two milliwatts. 00:14:44.499 --> 00:14:47.789 Written before the calculator comes out. 00:14:47.839 --> 00:14:52.889 And there is the ghost, in red. Two milliwatts plus one equals three. 00:14:52.939 --> 00:15:00.279 Decibels added to milliwatts — and look at the note underneath: the promise still passes. Three is more than two. 00:15:00.329 --> 00:15:05.409 A sanity check applied to a wrong method is not a sanity check. 00:15:05.459 --> 00:15:14.409 So you leave decibel-land first. P out over P in is ten to the dB over ten. Ten to the nought point one is one point two six. 00:15:14.589 --> 00:15:19.189 Now there is something to multiply by, and not before. 00:15:19.239 --> 00:15:26.679 One point two six times two milliwatts: about two and a half milliwatts. And the promise from line two holds. 00:15:26.729 --> 00:15:34.489 If an exponent had slipped and produced one point six, the sign in line one already told you it was wrong. 00:15:34.539 --> 00:15:39.729 And there is the box. Sum, promise, convert, multiply, check. 00:15:39.779 --> 00:15:48.729 Photograph that frame. It is the single most reusable object on the paper, and the method marks are read off exactly those five lines. 00:15:51.229 --> 00:15:55.149 This ghost is the template for every one of them. 00:15:55.199 --> 00:16:00.139 Decibels are not milliwatts. They are a ratio, written in logarithms. 00:16:00.189 --> 00:16:07.439 There is nothing to add them to, and no situation in which adding a decibel to a power means anything. 00:16:07.489 --> 00:16:14.479 And the cruel part is that the check passes. Three milliwatts is more than two, so the promise appears to hold. 00:16:14.529 --> 00:16:19.619 A sanity check applied to a wrong method is not a sanity check. 00:16:19.669 --> 00:16:27.729 Which is why line three gets written out in full. "P out over P in equals ten to the dB over ten" is not showing off. 00:16:27.779 --> 00:16:32.009 It is the line that makes it impossible to add a decibel to a milliwatt. 00:16:32.059 --> 00:16:41.009 Every ghost in this session has that shape: it feels right, and it survives a lazy check. The defence is the written method, not intuition. 00:16:44.429 --> 00:16:48.529 Checkpoint one. Stop the video. 00:16:48.579 --> 00:16:57.529 One. A chain is minus two dB, plus nine, minus three point five, with four milliwatts in. Give the total in dB, the promise, and the output. 00:17:00.139 --> 00:17:08.329 Two. Why can decibels from separate stages be added, when the powers themselves must be multiplied? 00:17:08.379 --> 00:17:17.329 Three. A student computes a total of minus six dB and an output larger than the input. What went wrong — and how do you know without redoing the arithmetic? 00:17:19.069 --> 00:17:19.999 Answers. 00:17:20.049 --> 00:17:28.999 One. Total plus three point five dB. The promise: output greater than four milliwatts. Ten to the nought point three five is about two point two four, so about eight point nine milliwatts. 00:17:33.819 --> 00:17:42.769 Two. Because a decibel is a logarithm of a ratio, and log of a times b is log a plus log b. Adding in decibels is multiplying in power. 00:17:43.339 --> 00:17:52.289 Three. The promise was broken. A negative total means a loss, so the output must be smaller. The error is in the conversion or the multiplication — and line two catches it before you look at either. 00:17:59.057 --> 00:18:06.717 Problem two. This is the biggest-mark item on the paper, and it is the one worth the most preparation. 00:18:06.767 --> 00:18:11.807 A channel has bandwidth one megahertz and a signal-to-noise ratio of sixty-three. 00:18:11.857 --> 00:18:17.577 A plain ratio — not decibels, and that matters in about four minutes. 00:18:17.627 --> 00:18:23.977 You want to run the link at four megabits per second. A chosen rate, not a forced one. 00:18:24.027 --> 00:18:30.517 How many signal levels do you need? That is the whole question — and it needs two formulas. 00:18:30.567 --> 00:18:32.757 Stop the video. Four minutes. 00:18:32.807 --> 00:18:41.757 And before you compute anything, write down which formula you are using first, and why. That sentence is worth marks on its own. 00:18:44.584 --> 00:18:53.534 Before any arithmetic, classify the question. Ten seconds, and it is most of the marks. 00:18:53.724 --> 00:19:02.674 An SNR is given. So a ceiling can be computed — that is Shannon, and Shannon is the only formula that uses SNR. 00:19:03.014 --> 00:19:10.344 Levels are asked for. So Nyquist has to appear, because Nyquist is the only formula with L in it. 00:19:10.394 --> 00:19:13.734 Therefore: both formulas, and Shannon first. 00:19:13.784 --> 00:19:20.454 Reaching for Nyquist first is how the ceiling gets skipped — and a rate above the ceiling cannot be reached at all. 00:19:20.504 --> 00:19:29.454 Learn to classify in ten seconds: what is given, what is asked. That classification is worth more than either formula. 00:19:32.629 --> 00:19:41.579 Line one, Shannon. C equals B log two of one plus SNR — ten to the six, times log two of sixty-four. 00:19:41.759 --> 00:19:50.709 One plus sixty-three is sixty-four, a power of two. Exam SNRs are chosen kindly; if yours is not, that is a signal. 00:19:51.839 --> 00:19:57.239 Line two, the ceiling. Ten to the six times six is six megabits per second. 00:19:57.289 --> 00:20:03.049 And no cleverness ever beats it. That is physics, not a design choice. 00:20:03.099 --> 00:20:09.729 Line three, the choice. You want four megabits, which is below the ceiling. The plan is legal. 00:20:09.779 --> 00:20:18.729 Running below the limit is engineering judgement — and it is precisely why the answer is four levels and not eight. 00:20:19.949 --> 00:20:28.429 Line four, Nyquist. Four times ten to the six equals two times ten to the six, times log two of L. So log two of L is two. 00:20:28.479 --> 00:20:36.679 Only now, with a rate already chosen. Nyquist answers "what gear", never "how fast". 00:20:36.729 --> 00:20:40.689 Line five. L equals four levels — two bits per signal element. 00:20:40.739 --> 00:20:46.489 And the sentence: Shannon tells you the speed limit; Nyquist tells you what gear to be in. 00:20:46.539 --> 00:20:55.269 Write that sentence down in the exam before you compute anything. It earns method marks on its own. 00:20:56.369 --> 00:21:03.219 Forty-two seconds on the biggest-mark item, and most of it is about reading rather than arithmetic. 00:21:03.269 --> 00:21:09.309 Four boxes: the bandwidth, the SNR, the plan, and the thing being asked for. 00:21:09.359 --> 00:21:14.139 Two of those four are formulas in disguise, and deciding which one comes first is most of the marks. 00:21:14.189 --> 00:21:21.699 An SNR is given, so a ceiling can be computed. Levels are asked, so Nyquist has to be used. 00:21:21.749 --> 00:21:27.989 Therefore both formulas, in that order. Recognise the shape and the marks follow. 00:21:28.039 --> 00:21:34.379 Shannon draws a wall at six megabits — ten to the six, times log two of sixty-four. 00:21:34.429 --> 00:21:40.089 Everything to the right of that red line is unreachable, whatever equipment you buy. 00:21:40.139 --> 00:21:48.319 And the green bar is your plan: four megabits, comfortably inside the wall. Legal, with a margin. 00:21:48.369 --> 00:21:56.439 This is the step people skip — and skipping it is how you end up answering eight levels instead of four. 00:21:56.489 --> 00:22:05.439 Only now does Nyquist speak. C equals two B log two of L; four million equals two million times log two of L. 00:22:05.579 --> 00:22:13.999 So log two of L is two — one division, and the logarithm is exact because the numbers were chosen to be. 00:22:14.049 --> 00:22:22.999 Four levels. And underneath, the two formulas separated: Shannon is the speed limit, physics. Nyquist is what gear to be in, engineering. 00:22:25.629 --> 00:22:34.579 And the last frame is the one to photograph. If the SNR arrives in decibels, the first written line is ten to the dB over ten. 00:22:34.819 --> 00:22:43.769 Thirty-six dB is not thirty-six. It is about three thousand nine hundred and eighty-one — and every line downstream inherits the error if you skip it. 00:22:48.461 --> 00:22:55.711 If the SNR arrives in decibels, convert it first. SNR equals ten to the dB over ten. 00:22:55.761 --> 00:23:01.061 That is the first written line, before Shannon is even mentioned. 00:23:01.111 --> 00:23:07.331 Thirty-six dB is ten to the three point six, about three thousand nine hundred and eighty-one. 00:23:07.381 --> 00:23:16.331 So one plus SNR is about three thousand nine hundred and eighty-two, and the ceiling is around twelve megabits — not the five point two one you would get from log two of thirty-seven. 00:23:19.711 --> 00:23:23.461 The ghost is feeding thirty-six straight into the formula. 00:23:23.511 --> 00:23:32.121 And every line downstream inherits the error, including a Nyquist answer that looks perfectly reasonable on its own. 00:23:32.171 --> 00:23:39.501 This reflex is worth more than any single answer, because it is the difference between a whole question and none of it. 00:23:39.551 --> 00:23:48.501 And the converse matters too: if the question does not say "dB", the SNR is a plain ratio and you use it directly. Read the units before you read the number. 00:23:53.257 --> 00:23:59.257 The two formulas get confused under time pressure, so separate them cleanly. 00:23:59.307 --> 00:24:02.237 Shannon: C equals B log two of one plus SNR. 00:24:02.287 --> 00:24:09.257 A ceiling imposed by noise. It uses the SNR, it contains no L, and nothing you build can exceed it. 00:24:09.307 --> 00:24:13.507 Nyquist: C equals two B log two of L. 00:24:13.557 --> 00:24:21.617 A relationship between rate and levels for a noiseless channel. It contains L, it contains no SNR, and it answers "what gear". 00:24:21.667 --> 00:24:27.877 So the test is: does the question mention SNR, or levels, or both? 00:24:27.927 --> 00:24:36.687 SNR only, Shannon. Levels only, Nyquist. Both — Shannon first, then Nyquist, and say so in writing. 00:24:36.737 --> 00:24:45.687 Write both formulas at the top of the page in the first minute of the exam. They cost nothing to write and they stop you reaching for the wrong one. 00:24:52.368 --> 00:24:56.488 Checkpoint two. Stop the video. 00:24:56.538 --> 00:25:04.938 One. A channel has bandwidth two megahertz and SNR two hundred and fifty-five. What is the Shannon ceiling? 00:25:04.988 --> 00:25:13.018 Two. The same channel is to run at twelve megabits per second. How many levels does Nyquist require? 00:25:13.068 --> 00:25:21.198 Three. A question gives SNR equals thirty dB. What is your first written line, and what is one plus SNR? 00:25:21.248 --> 00:25:22.168 Answers. 00:25:22.218 --> 00:25:31.168 One. One plus two fifty-five is two fifty-six, so log two is eight, and C is two times ten to the six times eight — sixteen megabits per second. 00:25:33.118 --> 00:25:42.068 Two. Twelve million equals two times two million times log two of L, so log two of L is three and L is eight levels. And twelve is below the sixteen-megabit ceiling, so the plan is legal. 00:25:47.118 --> 00:25:56.068 Three. SNR equals ten to the thirty over ten, which is ten cubed, a thousand. So one plus SNR is one thousand and one. Never thirty-one. 00:26:00.468 --> 00:26:03.918 Section three. Problem three: the pipe. 00:26:03.968 --> 00:26:06.538 Three sub-parts, three formulas. 00:26:06.588 --> 00:26:15.538 And each sub-part plants one number you must ignore. The examiner is testing whether you know what does not belong. 00:26:16.425 --> 00:26:18.615 Problem three. 00:26:18.665 --> 00:26:27.615 A link twelve thousand kilometres long, propagation speed two point four times ten to the eight metres per second, bandwidth five megabits. 00:26:27.955 --> 00:26:31.785 You send a two and a half megabyte file. 00:26:31.835 --> 00:26:37.465 Three parts: propagation delay, transmission time, and the bandwidth-delay product. 00:26:37.515 --> 00:26:42.465 Three answers, three formulas, and no shared numbers between (a) and (b). 00:26:42.515 --> 00:26:46.685 Stop the video. Four minutes. 00:26:46.735 --> 00:26:54.965 And watch the two conversion sins as you work: kilometres left as kilometres, and megabytes left as bytes. 00:26:55.015 --> 00:27:02.075 Those two sins account for most of the lost marks on this item — not the formulas, which almost everybody knows. 00:27:02.125 --> 00:27:11.075 Write the units next to every number before you divide anything. Ten seconds, and it removes the commonest failure on this question. 00:27:14.318 --> 00:27:21.168 Line one, convert. Twelve thousand kilometres is one point two times ten to the seven metres. 00:27:21.218 --> 00:27:26.998 The sin is leaving it in kilometres, which is off by a factor of a thousand. 00:27:27.048 --> 00:27:35.998 Line two, propagation. One point two times ten to the seven, over two point four times ten to the eight: nought point nought five seconds. Fifty milliseconds. 00:27:37.698 --> 00:27:43.718 And the file size never entered. One bit or one terabyte — same journey. 00:27:43.768 --> 00:27:50.018 Line three, convert again. Two and a half megabytes is twenty times ten to the six bits. 00:27:50.068 --> 00:27:56.388 Times eight, always. A byte is eight bits, and the formula wants bits. 00:27:56.438 --> 00:28:02.108 Line four, transmission. Twenty million over five million: four seconds. 00:28:02.158 --> 00:28:08.808 And the distance never entered. Transmission is about the pipe's width, not its length. 00:28:08.858 --> 00:28:15.668 Line five, the pipe. Five million times nought point nought five: two hundred and fifty thousand bits. 00:28:15.718 --> 00:28:22.448 That is how many bits are airborne when the link is full — Session six compressed into one rectangle. 00:28:22.498 --> 00:28:31.448 Fifty milliseconds, four seconds, two hundred and fifty thousand bits. And the only number used twice is the propagation delay, in part (c). 00:28:35.091 --> 00:28:38.791 Now the thing this item is really testing. 00:28:38.841 --> 00:28:45.711 Part (a) hands you a file size and does not want it. Propagation delay is distance over speed. 00:28:45.761 --> 00:28:54.711 If the file size appears anywhere in your working for (a), the working is wrong — even if the answer happens to be right. 00:28:55.511 --> 00:29:01.421 Part (b) hands you a distance and does not want it. Transmission time is size over rate. 00:29:01.471 --> 00:29:10.421 A longer wire does not make the sender slower. It makes the last bit arrive later, which is part (a)'s business. 00:29:10.471 --> 00:29:16.491 And part (c) is the only one that combines them: bandwidth times propagation delay. 00:29:16.541 --> 00:29:21.231 It is the newest of the three formulas, so it is worth drilling twice tonight. 00:29:21.281 --> 00:29:29.032 Say out loud which number you are dropping, and why, as you write each part. 00:29:29.082 --> 00:29:33.302 Checkpoint three. Stop the video. 00:29:33.352 --> 00:29:42.302 One. A ten megabyte file crosses a two megabit link three thousand kilometres long at two times ten to the eight metres per second. Give all three answers. 00:29:45.392 --> 00:29:54.332 Two. Which of the three answers changes if the file doubles in size, and which if the wire doubles in length? 00:29:54.382 --> 00:30:03.332 Three. A student writes "two point five megabytes over five megabits equals nought point five seconds". Name the error in five words. 00:30:03.852 --> 00:30:04.572 Answers. 00:30:04.622 --> 00:30:13.572 One. Propagation: three times ten to the six over two times ten to the eight — fifteen milliseconds. Transmission: eighty million over two million — forty seconds. The pipe: two million times nought point nought one five — thirty thousand bits. 00:30:20.362 --> 00:30:29.312 Two. Doubling the file doubles only the transmission time. Doubling the wire doubles the propagation delay, and therefore also the bandwidth-delay product. 00:30:31.902 --> 00:30:37.399 Three. The times-eight never happened. 00:30:37.449 --> 00:30:41.419 Section four. Problems four and five: addressing. 00:30:41.469 --> 00:30:44.089 The bit, the bill, and the ritual. 00:30:44.139 --> 00:30:53.089 Twenty-four marks of Part C — and every one of them is a method you already own. What is being tested is whether you apply it under time pressure. 00:30:57.186 --> 00:31:00.846 The I slash G bit is the last bit of the first byte. 00:31:00.896 --> 00:31:09.746 Which means it lives in the second hex digit, not the first. That single fact is the whole of part one. 00:31:09.796 --> 00:31:15.526 Even second digit, unicast. Odd, multicast. All Fs, broadcast. 00:31:15.576 --> 00:31:24.526 Four-A: A is even, so unicast. Four-seven: seven is odd, so multicast — however unicast the four makes it look. 00:31:25.626 --> 00:31:32.586 Derive it from the bit rather than memorising a list. Write the first byte in binary and read the final bit. 00:31:32.636 --> 00:31:41.586 Zero one zero zero, zero one one one is forty-seven. The last bit is one, so the group bit is set. A misprinted table cannot hurt you if you can draw the byte. 00:31:45.256 --> 00:31:51.266 Stop the video. Classify both addresses, then do part two — the ARP bill. 00:31:51.316 --> 00:32:00.266 Twenty-five systems on a LAN, A sends twenty datagrams to B. How many bystander interruptions with broadcast, and how many with ARP? 00:32:05.676 --> 00:32:10.346 Twenty-five systems on a LAN, and A sends twenty datagrams to B. 00:32:10.396 --> 00:32:19.086 The bystanders are twenty-five, minus A, minus B: twenty-three machines with no stake in the conversation. 00:32:19.136 --> 00:32:26.616 Broadcast everything, and it is twenty frames times twenty-three bystanders. Four hundred and sixty interruptions. 00:32:26.666 --> 00:32:32.286 That is every CPU in the room paying attention to somebody else's traffic. 00:32:32.336 --> 00:32:37.756 With ARP: one broadcast request, times twenty-three receptions. Twenty-three. 00:32:37.806 --> 00:32:45.446 And then B replies unicast — because the address is now known — and A caches it. 00:32:45.496 --> 00:32:51.496 The other nineteen datagrams travel as unicast, and every other card drops them in hardware. 00:32:51.546 --> 00:32:56.586 The bystanders' CPUs are never told those frames existed. 00:32:56.636 --> 00:33:05.586 Four hundred and sixty against twenty-three. That ratio is the whole argument for the ARP cache, and the exam asks you to defend it with exactly this arithmetic. 00:33:07.016 --> 00:33:15.966 And watch the trap: four eighty, or twenty-four. That is billing B as a bystander — and B is the destination, not an interruption. 00:33:20.356 --> 00:33:26.236 The request is broadcast because no address is known. That is the entire justification. 00:33:26.286 --> 00:33:32.996 You cannot address a frame to a machine whose MAC address is precisely the thing you are asking for. 00:33:33.046 --> 00:33:36.816 The reply is unicast because the address is now known. 00:33:36.866 --> 00:33:44.716 B learned A's MAC address from the request itself — it was sitting in the frame. So there is no reason to shout back. 00:33:44.766 --> 00:33:47.906 One sentence, and it explains the whole protocol. 00:33:47.956 --> 00:33:55.936 If a question asks why ARP is asymmetric, that is the answer — and it is worth full marks without any arithmetic at all. 00:33:55.986 --> 00:34:04.936 Most "explain why" marks in Part C are one sentence like that one. Learn three of them and you have covered the explanatory questions. 00:34:07.222 --> 00:34:12.222 Problem five, and by week seven this must be reflex. 00:34:12.272 --> 00:34:21.222 Two zero nine dot one hundred dot seventy-seven dot ninety, slash twenty-seven. N, the mask, the first address, the last address — in that order, always. 00:34:23.072 --> 00:34:28.122 Steps one and two never look at the address at all. 00:34:28.172 --> 00:34:32.212 And then a second question: how many machines can it actually seat? 00:34:32.262 --> 00:34:38.242 Which is a different question from how many addresses it contains, and the paper knows it. 00:34:38.292 --> 00:34:40.542 Stop the video. Three minutes. 00:34:40.592 --> 00:34:45.682 If it takes longer than three, that is tonight's drill identified. 00:34:45.732 --> 00:34:54.682 And watch for the two classics: a first address of dot ninety, where the AND never happened; and a mask of dot two forty, where the ones were miscounted. 00:34:59.472 --> 00:35:06.002 N: two to the thirty-two minus twenty-seven — two to the fifth, thirty-two addresses. 00:35:06.052 --> 00:35:10.762 Computed before you have finished reading the address. 00:35:10.812 --> 00:35:17.182 Mask: twenty-seven divided by eight is three remainder three; three ones on the ladder is two twenty-four. 00:35:17.232 --> 00:35:26.162 Two five five dot two five five dot two five five dot two twenty-four. One division, one lookup, no binary. 00:35:26.212 --> 00:35:35.162 The AND: ninety is zero one zero one, one zero one zero. ANDed with one one one zero zero zero zero zero, that keeps zero one zero zero zero zero zero zero — sixty-four. 00:35:39.382 --> 00:35:48.332 So the first address is two zero nine dot one hundred dot seventy-seven dot sixty-four. Not dot ninety: the address in the question is a resident, not the front door. 00:35:52.772 --> 00:35:58.892 Last: sixty-four plus thirty-two minus one is ninety-five. Dot ninety-five. 00:35:58.942 --> 00:36:07.152 And the check: ninety-five is one below ninety-six, and ninety-six divides by thirty-two. It holds. 00:36:07.202 --> 00:36:14.852 And usable: thirty machines. Thirty-two addresses, minus the network's name, minus the broadcast address. 00:36:14.902 --> 00:36:20.502 The signal word is "usable", or "hosts". Without it, the answer is thirty-two. 00:36:20.552 --> 00:36:29.502 Equal cuts are the other half of what the paper asks, and problem six is exactly that, two slides on. 00:36:31.368 --> 00:36:37.498 Five of the six, in one page, with the working hidden until you ask for it. 00:36:37.548 --> 00:36:43.088 It opens on problem one with the working greyed out — "hidden, work it yourself first". 00:36:43.138 --> 00:36:48.088 That is deliberate. The tool is a marker, not a textbook. 00:36:48.138 --> 00:36:53.308 Open the working and there are the five lines, each with its own reason underneath. 00:36:53.358 --> 00:37:02.308 And on the right: what it tests, the check that costs nothing, and the classic wrong answer. Read the right-hand column before you read the left. 00:37:04.558 --> 00:37:12.228 Problem two, and notice the first line is not a formula — it is "SNR given, levels asked, therefore both formulas". 00:37:12.278 --> 00:37:17.108 That reading step is a line of working, and it earns marks. 00:37:17.158 --> 00:37:25.628 Problem three, with both conversions on their own lines, and the note that the file size never enters part (a). 00:37:25.678 --> 00:37:34.628 Problem four. The bit, the two classifications, and then the two products side by side: four sixty against twenty-three. 00:37:35.158 --> 00:37:44.108 And problem five, the ritual, with the check built in: ninety-five is one below ninety-six, and ninety-six divides by thirty-two. 00:37:45.408 --> 00:37:54.358 And the summary: five answers, five methods, and the shape of the paper underneath. Problem six is not in this tool — it is on the two slides you have just worked. 00:37:55.908 --> 00:38:04.858 Open it tonight, work each problem cold, and only then press "show the working". If you press it first, you have read a textbook rather than sat an exam. 00:38:09.854 --> 00:38:14.614 The block contains thirty-two addresses. Forouzan counts all of them. 00:38:14.664 --> 00:38:19.564 The block is a range, and the range includes both ends. 00:38:19.614 --> 00:38:27.794 Used as a subnet it houses thirty machines. The first address is the network's name and the last is its broadcast address. 00:38:27.844 --> 00:38:31.034 Neither can be assigned to a computer. 00:38:31.084 --> 00:38:36.234 And the paper asks both ways, deliberately, sometimes in the same question. 00:38:36.284 --> 00:38:43.464 The word "usable", or "hosts", is your signal. Underline it as you read, before you start computing. 00:38:43.514 --> 00:38:49.356 This is the cheapest mark on the paper to protect. 00:38:49.406 --> 00:38:58.356 Problem six, and it is the item the paper names as its boundary. 00:38:58.596 --> 00:39:07.546 A building is granted one seven two dot twenty dot eight dot zero, slash twenty-two, and it needs eight networks. One block, eight equal subnets. Every piece has to be a legal block in its own right, which means it starts on a multiple of its own size. 00:39:13.466 --> 00:39:22.416 Part (a): the new prefix, the size of each piece, and the mask. All three fall out of the borrowed bits, and none of them requires you to look at an address — the same discipline as problem five, where steps one and two never touch the address either. 00:39:30.176 --> 00:39:39.126 Part (b): name subnet five, give its last address, and say how many machines it seats. The name is that piece's own first address carrying the new prefix. Writing slash twenty-two there names the whole block and is marked wrong. And seats is the signal word for N minus two. 00:39:49.376 --> 00:39:58.326 Stop the video. Three minutes, on paper. And watch the off-by-one: subnet k starts at k minus one, times the piece size. Subnet five begins four pieces along, not five. 00:40:05.596 --> 00:40:06.746 Five lines again, and four of them are last session's ritual. 00:40:06.796 --> 00:40:10.456 Borrow. Eight subnets is two cubed, so you borrow three bits, and the new prefix is twenty-two plus three — slash twenty-five. That is the whole formula: the new prefix is n plus log base two of the number of subnets. It works only for powers of two, because you borrow whole bits and there is no half a bit. 00:40:10.506 --> 00:40:11.246 Size. Two to the thirty-two minus twenty-five is two to the seventh — a hundred and twenty-eight addresses per subnet, mask two five five dot two five five dot two five five dot one twenty-eight. And the check: eight times a hundred and twenty-eight is one thousand and twenty-four, which is the whole slash twenty-two. Borrowing redistributes; it never mints. 00:40:11.296 --> 00:40:13.526 Count. Subnet k starts at k minus one, times the piece size. So subnet five starts at four times a hundred and twenty-eight — five hundred and twelve addresses into the block. Write five times a hundred and twenty-eight and you get six hundred and forty, which is subnet six. Engineers count from zero. 00:40:13.576 --> 00:40:15.486 Name it. Five hundred and twelve addresses is two whole octets, so subnet five begins at one seven two dot twenty dot ten dot zero, slash twenty-five. Its last address is that plus a hundred and twenty-seven — dot ten dot one twenty-seven. The name carries the new prefix; slash twenty-two there names the entire block, and it is marked wrong. 00:40:15.536 --> 00:40:24.486 Audit. A hundred and twenty-eight addresses seats a hundred and twenty-six machines: minus the subnet's own name, minus its broadcast address. Every start divides by a hundred and twenty-eight, and subnet eight ends flush at one seven two dot twenty dot eleven dot two five five, so the block is exactly spent. Outside the building nothing changed: still one route, slash twenty-two. 00:42:10.424 --> 00:42:17.064 Checkpoint four, and this one is the Part C rehearsal. Stop the video. 00:42:17.114 --> 00:42:26.064 One. Classify five-C colon zero zero, and thirty-three colon thirty-three — and say which bit decided each. 00:42:26.464 --> 00:42:35.124 Two. Thirty systems, fifty datagrams from A to B. Give the broadcast bill and the ARP bill. 00:42:35.174 --> 00:42:44.124 Three. Find all four facts for one nine two dot one six eight dot twenty dot two hundred, slash twenty-six — then say how many hosts it seats. 00:42:50.364 --> 00:42:51.294 Answers. 00:42:51.344 --> 00:43:00.294 One. Five-C: the second digit C is even, so unicast. Thirty-three: the second digit three is odd, so multicast. In both cases it is the I slash G bit, the last bit of the first byte. 00:43:05.274 --> 00:43:14.224 Two. Bystanders are twenty-eight. Broadcast: fifty times twenty-eight — one thousand four hundred interruptions. ARP: one broadcast times twenty-eight — twenty-eight receptions, then forty-nine unicast datagrams the bystanders never see. 00:43:20.004 --> 00:43:28.954 Three. N is sixty-four. Mask two five five dot two five five dot two five five dot one ninety-two. Two hundred AND one ninety-two is one ninety-two, so the first address is dot one ninety-two and the last is dot two fifty-five. And it seats sixty-two hosts. 00:43:41.500 --> 00:43:50.450 Five wrong answers, one per problem from P1 to P5. Read each one in a flat voice and recognise it. 00:43:51.840 --> 00:43:56.660 Problem one: "two milliwatts plus one equals three milliwatts." 00:43:56.710 --> 00:44:05.660 The fix: decibels are a ratio in log form. Convert with ten to the dB over ten, then multiply. Never add dB to a power. 00:44:07.930 --> 00:44:15.510 Problem two: "C equals ten to the six times log two of one plus thirty-six", with an SNR given in decibels. 00:44:15.560 --> 00:44:24.510 The fix: the first written line is SNR equals ten to the dB over ten. Thirty-six dB is about three thousand nine hundred and eighty-one, not thirty-six. 00:44:28.140 --> 00:44:34.780 Problem three: "two point five megabytes over five megabits equals nought point five seconds." 00:44:34.830 --> 00:44:43.660 The fix: times eight, always. Two and a half megabytes is twenty million bits, so the answer is four seconds. 00:44:43.710 --> 00:44:50.430 Problem four: "forty-seven starts with four, which is even, so unicast." 00:44:50.480 --> 00:44:58.580 The fix: the I slash G bit is in the second hex digit. Seven is odd, so forty-seven is multicast. 00:44:58.630 --> 00:45:05.750 And problem five: "the first address is two zero nine dot one hundred dot seventy-seven dot ninety." 00:45:05.800 --> 00:45:14.320 The fix: the AND never happened. Ninety AND two twenty-four is sixty-four — and ninety does not divide by thirty-two anyway. 00:45:14.370 --> 00:45:21.034 Name the fix for each one, and write it in your notes tonight. 00:45:22.034 --> 00:45:29.394 And here is the drill that makes those five stick: five worked solutions, each with one line wrong. 00:45:29.444 --> 00:45:34.654 Case one, the dB chain. Four lines of working, and one of them is a ghost. 00:45:34.704 --> 00:45:42.704 Read them as an examiner would — not "does this look familiar", but "is this step legal". 00:45:42.754 --> 00:45:46.844 Found. Line three: two milliwatts plus one equals three. 00:45:46.894 --> 00:45:54.704 And look at line four, marked green: the check passes. That is the whole reason this ghost survives. 00:45:54.754 --> 00:46:00.574 Here is a miss, on the Shannon case. The line picked was correct; the ghost was elsewhere. 00:46:00.624 --> 00:46:09.574 Notice the tool explains why the picked line is fine, as well as where the real error is. Both halves are worth reading. 00:46:09.654 --> 00:46:14.884 Case three. The transmission line: two point five over five equals nought point five. 00:46:14.934 --> 00:46:22.424 Megabytes divided by megabits per second, and the times-eight never happened. That is a whole sub-part. 00:46:22.474 --> 00:46:31.424 Case four. "Forty-seven starts with four, which is even." The first hex digit was read, and the I slash G bit lives in the second. 00:46:32.844 --> 00:46:39.374 And a miss on the last case — the arithmetic line looked wrong, but it was the first-address line above it. 00:46:39.424 --> 00:46:46.414 The arithmetic was right; it was applied to the wrong starting point. That is how this one hides. 00:46:46.464 --> 00:46:52.074 And found: the first address was never ANDed. Ninety, straight from the question. 00:46:52.124 --> 00:47:01.074 Five cases. Do all five tonight, and name the ghost out loud before you click. Naming it is what stops you writing it. 00:47:02.670 --> 00:47:07.320 Four checks. None of them takes more than five seconds. 00:47:07.370 --> 00:47:12.240 Problem one: does the sign of the dB total match the direction of the answer? 00:47:12.290 --> 00:47:18.360 A positive total promises more power out. A negative total promises less. 00:47:18.410 --> 00:47:22.390 Problem two: is one plus SNR a power of two? Exam SNRs are chosen kindly. 00:47:22.440 --> 00:47:26.530 If yours is not clean, check whether it was given in decibels. 00:47:26.580 --> 00:47:35.530 Problem three: do the units cancel? Metres over metres-per-second gives seconds. Bits over bits-per-second gives seconds. 00:47:35.690 --> 00:47:38.510 If they do not cancel, a conversion is missing. 00:47:38.560 --> 00:47:46.670 And problem five: does the first address divide by N, and is the last address one below a multiple? 00:47:46.720 --> 00:47:51.770 Two divisions, five seconds, and they catch almost every slip in the AND. 00:47:51.820 --> 00:48:00.770 None of these tells you the right answer. All of them tell you that yours is wrong — which under exam conditions is worth nearly as much. 00:48:06.930 --> 00:48:15.880 One timed pass of all six, with fresh numbers. Change the chain to minus two, plus nine, minus three point five. Set the SNR to two fifty-five. 00:48:18.190 --> 00:48:24.480 Make the file ten megabytes on two megabits. Use thirty systems and fifty datagrams. Use a slash twenty-six. 00:48:24.530 --> 00:48:26.510 Same skeletons, new numbers. 00:48:26.560 --> 00:48:35.510 Then three passes of your weakest. You already know which one it is — it is the one where you had to restart the video. 00:48:36.750 --> 00:48:39.590 Do that one until it is boring. 00:48:39.640 --> 00:48:47.990 And ten minutes on Part A. Topologies, mesh links and ports, circuit versus packet switching, encapsulation overhead. 00:48:48.040 --> 00:48:51.210 Sixteen marks, and they are the cheapest on the paper. 00:48:51.260 --> 00:49:00.210 New numbers on the same skeleton is exactly what the exam versions do to you. Practising that way is practising the actual test. 00:49:02.436 --> 00:49:04.886 The practical part. 00:49:04.936 --> 00:49:13.146 Several versions circulate. The letter at the top of your paper decides your numbers, so your neighbour's answer is mechanically a wrong answer. 00:49:13.196 --> 00:49:20.316 The only thing worth copying is the method — and you have just been given all six of those for free. 00:49:20.366 --> 00:49:29.316 Everything is on paper, exactly like Quiz one. Your answers in the answer boxes, your working in the dashed box under each question. 00:49:29.826 --> 00:49:36.476 Show the working, because the method marks are read off it. The dB sum. The AND. The times-eight. 00:49:36.526 --> 00:49:43.656 A wrong final number with correct method still scores. A bare number with nothing under it does not. 00:49:43.706 --> 00:49:50.156 Bring a calculator that can do logarithms, a pen, and your student ID. There is no formula sheet. 00:49:50.206 --> 00:49:53.096 Every formula you need has to be memorised. 00:49:53.146 --> 00:50:02.096 Arrive ten minutes early. Room and time are announced by the department, and the exam is co-ordinated across all sections. 00:50:04.744 --> 00:50:13.694 Run all six methods cold, with the working written out. Not recognise them — run them, on fresh numbers, against a clock. 00:50:14.654 --> 00:50:23.604 Name the ghost for each one before you start. Saying "this is the one where people add decibels to milliwatts" takes two seconds and prevents the error. 00:50:25.834 --> 00:50:33.444 And check your own answer in five seconds. The promise, the power of two, the units, the divisibility. 00:50:33.494 --> 00:50:42.444 Forty-four of the sixty marks are those six problems. The remaining sixteen are Weeks one and two, revised earlier in this session, and they are cheaper than any of them. 00:50:46.918 --> 00:50:48.818 That is Session thirteen. 00:50:48.868 --> 00:50:53.128 Six problems, six methods, and a named ghost for each. 00:50:53.178 --> 00:51:00.788 Every computational question on the paper is one of the six, with the numbers changed — and you have now worked all six. 00:51:00.838 --> 00:51:07.148 Method marks are read off your working: the dB sum, the AND, the times-eight. Show it, even when you are sure. 00:51:07.198 --> 00:51:14.028 Tonight: one timed pass of each with fresh numbers, three passes of your weakest, and ten minutes on Part A. 00:51:14.078 --> 00:51:18.438 Next class is the midterm. Calculator, pen, ID, ten minutes early. 00:51:18.488 --> 00:51:26.468 And remember what your neighbour's paper is worth: nothing. Only the method transfers, and you have all of it. 00:51:26.518 --> 00:51:33.728 Good luck. I will see you on the other side of it.