WEBVTT 00:00:00.000 --> 00:00:03.690 Welcome back to CSE 316 — Data Communication and Networking. 00:00:03.740 --> 00:00:06.950 This is the detailed video version of Session eleven. 00:00:07.000 --> 00:00:11.970 Last session you learned what a mask is. This session you get paid for it. 00:00:12.020 --> 00:00:20.970 Fair warning about the shape of this hour: it is deliberately the least conceptual lecture of the whole arc. One recipe, three timed drills, and one trap hammered until it cannot surprise you. 00:00:26.080 --> 00:00:35.030 Calculator away before you start. The exam will not let you have one, so your head has to be the table. 00:00:35.700 --> 00:00:39.720 Here is the situation, and it is a real one. 00:00:39.770 --> 00:00:46.760 Your ISP sends you exactly one line. Two zero five dot sixteen dot thirty-seven dot thirty-nine, slash twenty-eight. 00:00:46.810 --> 00:00:52.510 That is the whole message. There is no attachment, no second email, no phone call. 00:00:52.560 --> 00:01:01.510 And inside that one line, four separate facts are hiding — facts you need before you can configure a single thing. 00:01:01.770 --> 00:01:08.000 An engineer pulls all four of them out in under a minute, on paper, without a calculator. 00:01:08.050 --> 00:01:17.000 They are N, the mask, the first address and the last address. Write down which of the four you could already compute from that one line. 00:01:17.630 --> 00:01:20.570 And by the end of this hour, you will do it timed. 00:01:20.620 --> 00:01:28.717 That is the contract for the session, and I will hold up my end of it in the last five minutes. 00:01:28.767 --> 00:01:31.927 Section one. The four facts. 00:01:31.977 --> 00:01:34.117 Everything is in there already. 00:01:34.167 --> 00:01:43.117 The ISP did not tell you where your block starts. They handed you an address from the middle of it, and a number. That turns out to be enough — and why it is enough is the first real idea of the hour. 00:01:51.725 --> 00:01:54.705 So here they are, named. 00:01:54.755 --> 00:01:59.585 Fact one: how many addresses. We call it N. The size of the block. 00:01:59.635 --> 00:02:04.975 Everything else gets easier once you know it, which is why it goes first. 00:02:05.025 --> 00:02:13.405 Fact two: the mask. Which you can already build in your sleep — divide by eight, look up the remainder on the ladder. 00:02:13.455 --> 00:02:17.705 Fact three, and this is the crown jewel: the first address. 00:02:17.755 --> 00:02:26.705 It has another name you must know: the network address. It is the block's official name, and it is the thing routers actually forward on. 00:02:27.055 --> 00:02:31.365 And fact four: the last address. The far wall of the block. 00:02:31.415 --> 00:02:36.185 This is the one people get wrong by exactly one, and slide twelve is why. 00:02:36.235 --> 00:02:45.185 Four facts, four steps, always in this order. The order is not a preference — each step needs the one before it. 00:02:47.125 --> 00:02:52.525 Before the arithmetic, look at what the ISP did not give you. 00:02:52.575 --> 00:02:59.735 The ISP did not give you the first address of your block. They gave you dot thirty-nine, which is somewhere in the middle. 00:02:59.785 --> 00:03:03.845 Nothing in that line says where the block begins. 00:03:03.895 --> 00:03:09.455 And you do not need them to. Any address in the block, plus n, is enough. 00:03:09.505 --> 00:03:16.905 The AND throws away exactly the part that varies inside the block, and leaves the part that does not. 00:03:16.955 --> 00:03:25.905 Which is what the AND is for. That is not a side effect of masking — it is the entire reason masks are stored as thirty-two bits. 00:03:26.165 --> 00:03:30.675 One instruction recovers the whole block from any member of it. 00:03:30.725 --> 00:03:39.675 So if someone hands you an address and a slash, you are never short of information. You are one AND away from all four facts. 00:03:42.762 --> 00:03:46.422 Fact one costs you one subtraction. 00:03:46.472 --> 00:03:52.242 Thirty-two minus n is how many host bits you own — the bits the mask does not claim. 00:03:52.292 --> 00:03:58.832 Those bits are free to vary, and every combination of them is one address in your block. 00:03:58.882 --> 00:04:03.072 Each host bit doubles the block. It does not add — it doubles. 00:04:03.122 --> 00:04:08.852 One bit is two addresses. Two bits is four. Ten bits is a thousand and twenty-four. 00:04:08.902 --> 00:04:14.412 Which is why a block size is never a round decimal number, and never will be. 00:04:14.462 --> 00:04:21.222 Slash twenty-eight. Thirty-two minus twenty-eight is four host bits. Two to the fourth is sixteen. 00:04:21.272 --> 00:04:25.662 Sixteen addresses. That is fact one, done. 00:04:25.712 --> 00:04:32.632 And say the inverse rule out loud once. Getting it backwards is a common error in this arc. 00:04:32.682 --> 00:04:38.872 Big n, small block. A slash thirty is four addresses. A slash eight is sixteen million. 00:04:38.922 --> 00:04:44.945 The n is a dial, and it turns backwards. 00:04:44.995 --> 00:04:52.275 Now the table. Your head has to be the table — no exam on this topic gives you one. 00:04:52.325 --> 00:05:01.275 Two, four, eight, sixteen. The small end. These are your slash thirty-one down to slash twenty-eight — point-to-point links and tiny subnets. 00:05:04.375 --> 00:05:09.805 Thirty-two, sixty-four, a hundred and twenty-eight, two hundred and fifty-six. 00:05:09.855 --> 00:05:18.805 The working range: a slash twenty-seven, a slash twenty-six, a slash twenty-five, a slash twenty-four. You will see these nearly every week for the rest of term. 00:05:22.175 --> 00:05:30.955 Five hundred and twelve. A thousand and twenty-four. Two thousand and forty-eight. Four thousand and ninety-six. 00:05:31.005 --> 00:05:39.955 This is the crossover — above two fifty-six, a block no longer fits inside one octet, and it spills into the octet to the left. That spill is the second half of this lecture. 00:05:42.685 --> 00:05:51.635 Eight thousand one hundred and ninety-two. Sixty-five thousand five hundred and thirty-six. Sixteen million seven hundred and seventy-seven thousand two hundred and sixteen. 00:05:53.485 --> 00:05:59.575 The last two should already be familiar — those are the old class B and class A. 00:05:59.625 --> 00:06:03.785 And two to the zero is one. A slash thirty-two is a single address. 00:06:03.835 --> 00:06:11.835 It looks like a joke, but it is a legal block: one address, mask two fifty-five dot two fifty-five dot two fifty-five dot two fifty-five. 00:06:11.885 --> 00:06:20.835 Drill all of these until nobody hesitates. Not "I can work it out" — recognise them, the way you recognise seven times eight. 00:06:23.375 --> 00:06:29.335 Fact two is last session's skill, arriving as step two of a recipe. 00:06:29.385 --> 00:06:36.395 n divided by eight gives the full octets. Twenty-eight divided by eight is three, remainder four. 00:06:36.445 --> 00:06:42.135 Three octets of two fifty-five, then the fourth octet from the ladder. 00:06:42.185 --> 00:06:45.855 The ladder gives the remainder. Four ones is two forty. 00:06:45.905 --> 00:06:54.485 So the mask is two five five dot two five five dot two five five dot two forty. Nothing was written in binary. 00:06:54.535 --> 00:07:02.835 And every octet entirely right of the boundary is zero. If you find yourself thinking about them, stop — there is nothing there. 00:07:02.885 --> 00:07:11.835 If that is not automatic yet, the drill demo at the end of this session is exactly what it is for. 00:07:11.978 --> 00:07:17.598 Fact three. This is the expensive step, and it is where the marks are. 00:07:17.648 --> 00:07:25.948 Copy every octet under a two fifty-five. Two zero five stays two zero five. Sixteen stays sixteen. Thirty-seven stays thirty-seven. 00:07:25.998 --> 00:07:30.088 Three quarters of the answer, written without thinking. 00:07:30.138 --> 00:07:39.088 Zero every octet under a zero. There are none in a slash twenty-eight — but when there are, they cost you nothing at all. 00:07:41.468 --> 00:07:44.748 And then AND the single octet the boundary cuts. 00:07:44.798 --> 00:07:52.328 Thirty-nine is zero zero one zero zero one one one. The mask octet is two forty — one one one one zero zero zero zero. 00:07:52.378 --> 00:08:00.188 Keep the top four bits: zero zero one zero zero zero zero zero. That is thirty-two. 00:08:00.238 --> 00:08:06.318 So the first address is two zero five dot sixteen dot thirty-seven dot thirty-two. 00:08:06.368 --> 00:08:15.318 And it is not a machine. It is the name of the block — the street, not the house. 00:08:16.470 --> 00:08:22.120 Checkpoint one. Paper, pen, and stop the video. 00:08:22.170 --> 00:08:31.120 One. How many addresses are in a slash twenty-seven, a slash twenty-five and a slash twenty? Give the powers of two you used. 00:08:31.300 --> 00:08:37.600 Two. Why is any address in the block enough to recover the first address? 00:08:37.650 --> 00:08:45.430 Three. Write the mask for slash twenty-one without writing any binary, and say which two moves you used. 00:08:45.480 --> 00:08:46.410 Answers. 00:08:46.460 --> 00:08:55.410 One. Slash twenty-seven is two to the fifth, thirty-two. Slash twenty-five is two to the seventh, a hundred and twenty-eight. Slash twenty is two to the twelfth, four thousand and ninety-six. 00:09:00.440 --> 00:09:09.390 Two. Because ANDing with the mask erases exactly the bits that vary inside the block. Every member has the same network bits, so every member gives the same answer. 00:09:10.790 --> 00:09:19.740 Three. Twenty-one divided by eight is two remainder five. Two full octets of ones, and five ones on the ladder is two forty-eight. Two five five dot two five five dot two forty-eight dot zero. 00:09:27.101 --> 00:09:29.331 Section two. The ritual. 00:09:29.381 --> 00:09:32.161 Four steps, in this order, every time. 00:09:32.211 --> 00:09:41.161 N, then the mask, then the AND, then plus N minus one. The order is the method — each step needs the answer above it. 00:09:44.201 --> 00:09:48.711 Fact four, and the minus one is the whole slide. 00:09:48.761 --> 00:09:55.321 Thirty-two plus fifteen is forty-seven. So the block runs from dot thirty-two to dot forty-seven, 00:09:55.371 --> 00:10:00.821 and every address between them shares the same first twenty-eight bits. 00:10:00.871 --> 00:10:05.511 Why minus one? Because the first address is itself a member of the block. 00:10:05.561 --> 00:10:14.511 Count on your fingers if it ever wobbles: thirty-two, thirty-three, thirty-four, all the way to forty-seven. That is sixteen numbers, because thirty-two is one of them. 00:10:17.831 --> 00:10:24.901 Add the full sixteen and you land on forty-eight, which is the neighbour's doormat — the first address of the next block. 00:10:24.951 --> 00:10:30.041 This is the single most common wrong answer on this entire topic. 00:10:30.091 --> 00:10:34.221 The error has a name: plus-N. And it has a free test. 00:10:34.271 --> 00:10:40.761 If your last address divides exactly by the block size, you have caught it. Subtract one and try again. 00:10:40.811 --> 00:10:49.761 A slash twenty-six starting at dot sixty-four ends at dot one twenty-seven, not one twenty-eight. 00:10:49.895 --> 00:10:53.885 This is the card. Copy it into your notes. 00:10:53.935 --> 00:11:02.885 Step one, N equals two to the thirty-two minus n. Step two, the mask: n divided by eight, then the ladder. 00:11:03.045 --> 00:11:10.785 About five seconds each once they are automatic — and neither of them looks at the address at all. 00:11:10.835 --> 00:11:15.185 Step three, first address equals address AND mask, one octet only. 00:11:15.235 --> 00:11:22.205 The expensive step. About thirty seconds. This is where the marks are won. 00:11:22.255 --> 00:11:27.995 Step four, last address equals first plus N minus one, counted in two-fifty-sixes. 00:11:28.045 --> 00:11:36.995 Ten seconds. And if N is two fifty-six or more you do not add at all — you count runs of the last octet, which is the second half of this hour. 00:11:37.885 --> 00:11:46.835 Five, ten, thirty, ten. That is fifty-five seconds, which is why the promise in minute one was sixty and not ninety. 00:11:49.555 --> 00:11:54.505 And a word on why the order matters, because people try to skip. 00:11:54.555 --> 00:12:03.185 Step four needs N, so step one comes first. You cannot write the last address until you know how wide the block is. 00:12:03.235 --> 00:12:11.895 Step three needs the mask, so step two comes before it. The AND is meaningless without something to AND against. 00:12:11.945 --> 00:12:16.825 Steps one and two never look at the address at all. They depend only on n. 00:12:16.875 --> 00:12:22.885 In a drill you can do both before you have even read the address — and you should. 00:12:22.935 --> 00:12:30.705 Which means the address only matters at step three. That is exactly where the trap lives, and exactly where you should slow down. 00:12:30.755 --> 00:12:39.705 Jumping straight to "the first address is…" is where the answers go wrong under time pressure. The ritual exists to stop you guessing. 00:12:43.408 --> 00:12:49.568 Forty-two seconds. The ritual, run once, on the line from minute one. 00:12:49.618 --> 00:12:53.518 There is the line, and four empty tiles waiting to be filled. 00:12:53.568 --> 00:12:59.818 Nothing has been calculated. This is exactly what you have at the start of an exam question. 00:12:59.868 --> 00:13:06.088 Fact one. Thirty-two minus twenty-eight is four host bits. Two to the fourth is sixteen. 00:13:06.138 --> 00:13:11.248 Notice how little it needed: the address was not consulted at all. 00:13:11.298 --> 00:13:18.388 Fact two. Twenty-eight divided by eight is three remainder four, and four ones on the ladder is two forty. 00:13:18.438 --> 00:13:26.098 Two five five dot two five five dot two five five dot two forty. Still nothing from the address. 00:13:26.148 --> 00:13:32.578 Fact three, and now the address finally matters. One octet, eight bits, one AND. 00:13:32.628 --> 00:13:39.228 Thirty-nine meets two forty and the bottom four bits are erased. Thirty-two. 00:13:39.278 --> 00:13:43.188 Fact four. Thirty-two plus fifteen is forty-seven. 00:13:43.238 --> 00:13:52.188 And look at the red line: thirty-two plus sixteen is forty-eight, which is next door. That single line is the most valuable thing on the screen. 00:13:53.368 --> 00:14:02.318 Now the picture that changes how you think about all of this. Slash twenty-eight blocks tile the whole space in strides of sixteen — zero, sixteen, thirty-two, forty-eight. 00:14:04.238 --> 00:14:13.188 Our address, thirty-nine, fell inside the thirty-two block. It did not create a block. It landed in one that was already there. 00:14:14.658 --> 00:14:23.608 And the rule the tiling forces. Every first address divides by its block size. Thirty-two divides by sixteen. Thirty-nine does not. 00:14:24.168 --> 00:14:33.118 So a block "starting at thirty-nine" cannot exist — not by configuration, not by argument. File that as the alignment rule; in Session fifteen it decides an entire exam question. 00:14:39.648 --> 00:14:45.978 Now the same thing standing still, so you can copy it into your notes. 00:14:46.028 --> 00:14:53.688 Step one. Thirty-two minus twenty-eight is four host bits, so two to the fourth. Sixteen. 00:14:53.738 --> 00:14:59.678 Step two. Twenty-eight divided by eight is three remainder four; four ones on the ladder is two forty. 00:14:59.728 --> 00:15:05.368 Two five five dot two five five dot two five five dot two forty. 00:15:05.418 --> 00:15:14.368 Step three. Three octets copy straight through. Thirty-nine is zero zero one zero zero one one one, ANDed with one one one one zero zero zero zero, keeps zero zero one zero zero zero zero zero. 00:15:20.288 --> 00:15:25.688 Two zero five dot sixteen dot thirty-seven dot thirty-two. 00:15:25.738 --> 00:15:34.688 Step four. First plus N minus one: thirty-two plus fifteen is forty-seven. Not thirty-two plus sixteen, which is the next block's first house. 00:15:37.398 --> 00:15:45.908 And there is the answer row. Sixteen. Two five five dot two five five dot two five five dot two forty. Dot thirty-two. Dot forty-seven. 00:15:45.958 --> 00:15:54.908 Box it in your notes. Every drill for the next three sessions is this shape with different numbers. 00:15:56.282 --> 00:16:02.012 One word matters here, because it changes how the rest of the arc reads. 00:16:02.062 --> 00:16:11.012 Slash twenty-eight blocks tile the whole space in strides of sixteen. Zero, sixteen, thirty-two, forty-eight, sixty-four, and on up. 00:16:11.612 --> 00:16:16.302 They are already there. Nobody creates them. 00:16:16.352 --> 00:16:24.402 The address dot thirty-nine fell inside the dot thirty-two block. It did not create a block starting at thirty-nine. 00:16:24.452 --> 00:16:28.842 So "the block starting at dot thirty-nine" is not a thing that exists. 00:16:28.892 --> 00:16:35.312 And if an exam answer of yours ever says that, you have skipped step three. 00:16:35.362 --> 00:16:42.102 Here is the test that makes this concrete. Is two zero five dot sixteen dot thirty-seven dot forty-eight in our block? 00:16:42.152 --> 00:16:49.602 No. It is the first house of the next one. Forty-eight AND two forty is forty-eight, which is not thirty-two. 00:16:49.652 --> 00:16:58.602 And that AND-and-compare is literally what a router does to every packet it forwards. Practise it on its own — it is one AND and one comparison, and it is the whole of forwarding. 00:17:05.355 --> 00:17:11.695 Thirty-two divides by sixteen. Sixty-four divides by sixteen. Thirty-nine does not. 00:17:11.745 --> 00:17:20.695 So a slash twenty-eight starting at thirty-nine cannot exist. The tiling forces it, and no configuration can talk its way around it. 00:17:20.755 --> 00:17:27.715 Which gives you something valuable under time pressure: finish a question, then divide your first address by N. 00:17:27.765 --> 00:17:35.645 If it does not come out whole, step three went wrong. That check takes two seconds. 00:17:35.695 --> 00:17:41.285 And it is not only a check. In two sessions an ISP hands you one block and six customers to fit inside it. 00:17:41.335 --> 00:17:48.865 Alignment is what decides who gets which addresses, and it is the constraint that makes the problem hard rather than tedious. 00:17:48.915 --> 00:17:57.798 File this under rules that look pedantic today and decide entire questions in Session fifteen. 00:17:58.858 --> 00:18:07.088 First drill. One nine two dot one six eight dot five dot one hundred, slash twenty-six. All four facts. 00:18:07.138 --> 00:18:16.088 N, then the mask, then the AND, then plus N minus one. On paper, alone, no calculator. 00:18:16.248 --> 00:18:25.198 Ninety seconds on the clock. Write all four down, even the ones you are sure of — the ritual only works if you run all of it. 00:18:26.458 --> 00:18:31.858 Answers. Thirty-two minus twenty-six is six host bits, so sixty-four addresses. 00:18:31.908 --> 00:18:37.818 Twenty-six is three octets plus two ones, and two ones is one ninety-two. 00:18:37.868 --> 00:18:42.548 One hundred AND one ninety-two keeps the top two bits: sixty-four. 00:18:42.598 --> 00:18:46.118 And sixty-four plus sixty-three is one twenty-seven. 00:18:46.168 --> 00:18:52.632 That was a real exam question at real exam difficulty. 00:18:52.682 --> 00:18:57.712 Two wrong answers turn up on that drill, and they are worth naming. 00:18:57.762 --> 00:19:00.972 Wreck one: "the first address is one hundred." 00:19:01.022 --> 00:19:09.972 No. One hundred is just where the ISP happened to point. It is an address in the block, not the block's name. 00:19:10.182 --> 00:19:14.362 Wreck two: "the last address is one twenty-eight." 00:19:14.412 --> 00:19:23.362 That is the plus-N error. Sixty-four plus sixty-four is one twenty-eight, which is the next block's first house. Sixty-four plus sixty-three is one twenty-seven. 00:19:25.652 --> 00:19:34.602 And the free check catches both of them. One twenty-eight divides by sixty-four, so it cannot be a last address — last addresses are always one below a multiple. 00:19:36.692 --> 00:19:44.082 If you got both right inside ninety seconds, do four more tonight and it stops being a skill and starts being a reflex. 00:19:44.132 --> 00:19:53.082 Neither wreck is a misunderstanding of masks. Both are skipped steps. The ritual is the fix. 00:19:53.748 --> 00:19:57.868 Checkpoint two. Stop the video. 00:19:57.918 --> 00:20:06.868 One. A slash twenty-seven block has first address two hundred dot ten dot five dot sixty-four. What is its last address, and what is the free check? 00:20:08.918 --> 00:20:17.868 Two. Is two zero five dot sixteen dot thirty-seven dot forty-eight in the block two zero five dot sixteen dot thirty-seven dot thirty-two slash twenty-eight? Show the computation a router would do. 00:20:22.768 --> 00:20:31.718 Three. Why can no slash twenty-eight block have two zero five dot sixteen dot thirty-seven dot thirty-nine as its first address? 00:20:32.278 --> 00:20:33.208 Answers. 00:20:33.258 --> 00:20:42.208 One. Sixty-four plus thirty-two minus one is ninety-five. And the check: ninety-five is one below ninety-six, and ninety-six divides by thirty-two. 00:20:44.258 --> 00:20:53.208 Two. No. Forty-eight ANDed with the mask gives dot forty-eight, which is not dot thirty-two. Different first address, different block — and that comparison is exactly what forwarding is. 00:20:57.128 --> 00:21:06.078 Three. Because thirty-nine does not divide by sixteen. Blocks of sixteen tile the space at zero, sixteen, thirty-two, forty-eight — a block cannot begin between two tiles. 00:21:12.843 --> 00:21:15.213 Section three. Mid-octet. 00:21:15.263 --> 00:21:16.623 Decimal lies to you. 00:21:16.673 --> 00:21:25.623 Everything so far kept the fight in the last octet, where you are comfortable. The exam knows you are comfortable there — so it moves the boundary into the third. 00:21:29.505 --> 00:21:34.165 One eighty dot eight dot seventeen dot nine, slash twenty-two. 00:21:34.215 --> 00:21:39.675 Before I show you anything: predict the first address, out loud. 00:21:39.725 --> 00:21:46.555 Almost everyone says one eighty dot eight dot seventeen dot zero. And it is wrong. 00:21:46.605 --> 00:21:55.555 Run the ritual instead. N: thirty-two minus twenty-two is ten host bits, so two to the tenth — a thousand and twenty-four. 00:21:56.345 --> 00:22:05.295 Mask: twenty-two is eight plus eight plus six, and six ones is two fifty-two. Two five five dot two five five dot two fifty-two dot zero. 00:22:06.285 --> 00:22:11.165 And that puts the boundary six bits into the THIRD octet. 00:22:11.215 --> 00:22:20.165 So the fight has moved. Seventeen is zero zero zero one zero zero zero one. The mask octet is one one one one one one zero zero. 00:22:22.005 --> 00:22:26.925 Keep six bits: zero zero zero one zero zero zero zero. That is sixteen. 00:22:26.975 --> 00:22:33.665 The first address is one eighty dot eight dot sixteen dot zero. Not seventeen. 00:22:33.715 --> 00:22:41.965 The seventeen was never all yours. Only its top six bits were; the bottom two belong to the host part. 00:22:42.015 --> 00:22:50.655 When the boundary is mid-octet, decimal reasoning is not risky. It is wrong. 00:22:51.745 --> 00:22:58.285 Forty-two seconds, and it does the last address in a way you may not have seen before. 00:22:58.335 --> 00:23:02.685 There is the gut answer, in pink, side by side with the real one. 00:23:02.735 --> 00:23:09.045 Look at how similar they are. One octet, one digit, and about four marks. 00:23:09.095 --> 00:23:18.045 N first: two to the tenth, a thousand and twenty-four. Then the mask: twenty-two over eight is two remainder six, and six ones is two fifty-two. 00:23:19.465 --> 00:23:24.095 Both of those were computed without looking at the address once. 00:23:24.145 --> 00:23:31.615 And there is the reason for the whole section. The ones stop inside octet three, so octet three is where the AND happens. 00:23:31.665 --> 00:23:37.415 Two bits of that octet — the last two — are host bits. Watch them. 00:23:37.465 --> 00:23:42.645 Seventeen meets two fifty-two, and the bottom two bits are erased. Sixteen. 00:23:42.695 --> 00:23:50.965 The seventeen was never all network. Nobody looked at those two bits, and that is where the mark went. 00:23:51.015 --> 00:23:56.015 Now the last address. Do not add a thousand and twenty-three to a dotted address. 00:23:56.065 --> 00:24:02.245 There is no honest arithmetic that does that — the dots are not decimal places. 00:24:02.295 --> 00:24:08.965 Instead, count two-fifty-sixes. A thousand and twenty-four divided by two fifty-six is four. 00:24:09.015 --> 00:24:17.965 Four clicks of octet three: sixteen, seventeen, eighteen, nineteen — each carrying a full zero-to-two-fifty-five run of the last octet. 00:24:20.425 --> 00:24:28.595 And the last address reads itself off the end of the last run: one eighty dot eight dot nineteen dot two fifty-five. 00:24:28.645 --> 00:24:37.595 One check before we move on: a slash twenty-three starting at ten dot one dot four dot zero. Last address? Two clicks of two fifty-six — ten dot one dot five dot two fifty-five. 00:24:44.464 --> 00:24:53.414 The rule: never add N minus one to a dotted address digit by digit. Count runs of two fifty-six instead. 00:24:53.554 --> 00:25:02.504 A dotted address is not a number you can add to. There is no honest way to "add a thousand and twenty-three" to one eighty dot eight dot sixteen dot zero by treating the dots as decimal places. 00:25:05.684 --> 00:25:12.134 Do not invent arithmetic. People do, and it is always wrong in the same direction. 00:25:12.184 --> 00:25:17.154 Every two hundred and fifty-six addresses is one click of the octet to the left. 00:25:17.204 --> 00:25:26.154 A thousand and twenty-four divided by two fifty-six is four. So four clicks of the third octet, each carrying a full run of the fourth. 00:25:27.324 --> 00:25:32.464 So the third octet walks sixteen, seventeen, eighteen, nineteen. 00:25:32.514 --> 00:25:41.464 And the last address reads itself off the end of the last run: one eighty dot eight dot nineteen dot two fifty-five. No carrying, no thirty-two-bit addition, no calculator. 00:25:47.859 --> 00:25:48.349 Step four of the ritual says last equals first plus N minus one. The book reaches the same wall a different way, using the mask, and an exam can ask for it in exactly those words. So here it is on a worked address. 00:25:48.399 --> 00:25:51.599 The rule as Forouzan writes it: the last address in the block equals any address in the block, OR the NOT of the mask. Two bit-wise operations and nothing else. The AND with the mask gives you the first address; the OR with the inverted mask gives you the last. 00:25:51.649 --> 00:25:52.889 The book's own example: one six seven dot one ninety-nine dot one seventy dot eighty-two, slash twenty-seven. Twenty-seven is three full octets plus three ones, so the mask is two fifty-five dot two fifty-five dot two fifty-five dot two hundred and twenty-four. The whole fight is in the last octet, where eighty-two is zero one zero one zero zero one zero and two twenty-four is one one one zero zero zero zero zero. 00:25:52.939 --> 00:25:55.959 NOT the mask flips every bit. The three octets of two fifty-five become zero, and one one one zero zero zero zero zero becomes zero zero zero one one one one one, which is thirty-one. So NOT mask is zero dot zero dot zero dot thirty-one. And fact one arrives free: NOT mask plus one is thirty-two addresses. 00:25:56.009 --> 00:25:57.209 Now OR the address with that. Zero one zero one zero zero one zero, OR zero zero zero one one one one one, is zero one zero one one one one one — ninety-five. The OR drives every host bit to one and leaves the prefix untouched, which is the definition of the last address. So the last address is one six seven dot one ninety-nine dot one seventy dot ninety-five. 00:25:57.259 --> 00:26:00.659 Check it against the ritual. Eighty-two AND two twenty-four is sixty-four, so the first address is one six seven dot one ninety-nine dot one seventy dot sixty-four. Sixty-four plus thirty-two minus one is ninety-five. Identical. Use plus N minus one for speed under the clock, and use the OR when a question names the mask operations — they are the same wall, reached two ways. 00:26:00.709 --> 00:26:09.659 One warning about the printed page. Example eighteen point two in your book prints the mask as two five six dot two five six dot two five six dot two twenty-four, which is impossible — no octet reaches two fifty-six. It then prints the first address as dot eighty-two and the last as dot two five five. Both are wrong, and Example eighteen point one on the facing page derives the correct dot sixty-four and dot ninety-five in binary. Trust the binary, and trust the mask operations. Do not trust that one boxed answer. 00:28:48.675 --> 00:28:55.045 Here they are written out, once, so the picture is in your head for good. 00:28:55.095 --> 00:29:04.045 Click one. Sixteen dot zero through sixteen dot two fifty-five. Two hundred and fifty-six addresses, and the first of them is the block's network address. 00:29:06.145 --> 00:29:10.215 Click two. Seventeen dot zero through seventeen dot two fifty-five. 00:29:10.265 --> 00:29:17.125 This is the octet the gut wanted to call the network. It is only the second run. 00:29:17.175 --> 00:29:25.925 Click three. Eighteen. Nothing special happens here — which is itself the point. The runs are uniform. 00:29:25.975 --> 00:29:32.135 Click four. Nineteen, and the very last address of that run is the block's last address. 00:29:32.185 --> 00:29:41.135 The answer row: a thousand and twenty-four, two five five dot two five five dot two fifty-two dot zero, one eighty dot eight dot sixteen dot zero, one eighty dot eight dot nineteen dot two fifty-five. 00:29:44.725 --> 00:29:52.235 And the check: one eighty dot eight dot twenty dot zero would be the next block, so nineteen dot two fifty-five is one below it. 00:29:52.285 --> 00:30:01.235 Write the runs out the first few times. After that you will see "one thousand and twenty-four" and think "four clicks" without drawing anything. 00:30:04.711 --> 00:30:10.121 Here is all of it as something you can type into, including the street picture. 00:30:10.171 --> 00:30:19.121 It opens on the flagship. Four tiles, numbered by step, and they match the board exactly: sixteen, two forty, dot thirty-two, dot forty-seven. 00:30:19.931 --> 00:30:28.881 Underneath, the street — and our block, highlighted, with the address that fell into it marked below. 00:30:28.991 --> 00:30:33.561 Now the membership test. Type dot forty-eight into the second box. 00:30:33.611 --> 00:30:41.091 Forty-eight ANDed with the mask is forty-eight, and forty-eight is not thirty-two. So: no. One door down. 00:30:41.141 --> 00:30:46.511 This AND-and-compare is literally what a router does to every packet. 00:30:46.561 --> 00:30:52.761 Change one digit. Dot forty. Forty AND two forty is thirty-two, which matches. Yes. 00:30:52.811 --> 00:30:58.481 Same test, different answer, and the difference is one bit. 00:30:58.531 --> 00:31:06.571 Here is drill one, so you can check yourself. Sixty-four, one ninety-two, dot sixty-four, dot one twenty-seven. 00:31:06.621 --> 00:31:13.481 And look at the binary table: three octets greyed out, one octet doing all the work. 00:31:13.531 --> 00:31:22.481 Now the mid-octet one. Watch the street change shape: the tiles are now steps of four in the THIRD octet, each one a thousand and twenty-four addresses wide. 00:31:24.691 --> 00:31:33.641 And the summit. Eight thousand one hundred and ninety-two, two five five dot two five five dot two twenty-four dot zero, dot one ninety-two dot zero, dot two twenty-three dot two fifty-five. 00:31:37.041 --> 00:31:43.001 That is the hardest shape the exam sets, and the explorer will not flinch at it. 00:31:43.051 --> 00:31:47.461 Last one, and this one goes in your actual notes, not your memory. 00:31:47.511 --> 00:31:53.991 Eleven dot ten dot zero dot zero, slash twenty-two. One thousand and twenty-four addresses. 00:31:54.041 --> 00:32:02.991 Next week an ISP hands you that block with six customers attached and twelve marks riding on it. 00:32:03.321 --> 00:32:07.621 Checkpoint three. Stop the video. 00:32:07.671 --> 00:32:16.011 One. Find all four facts for one thirty dot forty dot one ninety-eight dot sixteen, slash twenty-two. 00:32:16.061 --> 00:32:25.011 Two. A slash twenty-three block starts at ten dot one dot four dot zero. What is its last address, and how many clicks of two fifty-six did you count? 00:32:27.271 --> 00:32:36.221 Three. Why does decimal intuition fail on one eighty dot eight dot seventeen dot nine slash twenty-two, but not on one nine two dot one six eight dot five dot one hundred slash twenty-six? 00:32:41.051 --> 00:32:41.971 Answers. 00:32:42.021 --> 00:32:50.971 One. N is two to the tenth, a thousand and twenty-four. Mask two five five dot two five five dot two fifty-two dot zero. One ninety-eight AND two fifty-two is one ninety-six, so the first address is one thirty dot forty dot one ninety-six dot zero. Four runs, so the last is one thirty dot forty dot one ninety-nine dot two fifty-five. 00:33:06.201 --> 00:33:15.151 Two. Ten dot one dot five dot two fifty-five. Two clicks — four dot zero through four dot two fifty-five, and five dot zero through five dot two fifty-five. Five hundred and twelve addresses. 00:33:20.341 --> 00:33:29.291 Three. Because twenty-two is not a multiple of eight, so the boundary cuts a digit you cannot see in decimal. Twenty-six also cuts an octet — but the block still fits inside a single octet, so nothing spills. 00:33:37.018 --> 00:33:40.238 Section four. Drilled, and examinable. 00:33:40.288 --> 00:33:49.238 One more drill at full exam difficulty, a short table of addresses that are already spoken for, and one word that quietly changes the answer. 00:33:52.740 --> 00:34:01.480 Drill two. One thirty dot forty dot one ninety-eight dot sixteen, slash twenty-two. Two minutes, timed. 00:34:01.530 --> 00:34:10.480 It is deliberately the same shape as the worked slash twenty-two. That is not a kindness — it is how a reflex is built. 00:34:12.460 --> 00:34:21.410 N is a thousand and twenty-four. Twenty-two over eight is two remainder six, and six ones is two fifty-two. Two steps, ten seconds. 00:34:22.200 --> 00:34:31.150 One ninety-eight is one one zero zero zero one one zero. It loses its bottom two bits: one one zero zero zero one zero zero, which is one ninety-six. 00:34:33.370 --> 00:34:39.230 First address one thirty dot forty dot one ninety-six dot zero. 00:34:39.280 --> 00:34:48.230 And four runs: one ninety-six, one ninety-seven, one ninety-eight, one ninety-nine. Last address one thirty dot forty dot one ninety-nine dot two fifty-five. 00:34:51.280 --> 00:35:00.230 If you wrote one ninety-eight dot zero as the first address, you paid the decimal tax one last time. From here on, mid-octet means binary — the reflex, not the exception. 00:35:07.090 --> 00:35:14.930 Drill three, and this is the summit. One hundred dot seventy-two dot two zero three dot nine, slash nineteen. 00:35:14.980 --> 00:35:21.880 Full difficulty, nothing softened. Paper, no calculator, three minutes on the clock. 00:35:21.930 --> 00:35:30.300 Run all four steps in order. Do not skip step one because the number looks big — step four needs it. 00:35:30.350 --> 00:35:37.100 Answers. Thirty-two minus nineteen is thirteen host bits, so eight thousand one hundred and ninety-two. 00:35:37.150 --> 00:35:42.110 Nineteen is eight plus eight plus three, and three ones is two twenty-four. 00:35:42.160 --> 00:35:50.830 Two zero three AND two twenty-four is one ninety-two, so the first address is one hundred dot seventy-two dot one ninety-two dot zero. 00:35:50.880 --> 00:35:59.830 And eight thousand one hundred and ninety-two is thirty-two runs of two fifty-six, so the third octet spans one ninety-two to two twenty-three. Last address dot two twenty-three dot two fifty-five. 00:36:03.730 --> 00:36:12.680 Four correct answers inside three minutes is exam pace for this item. If you are slower, the gap is drill, not understanding — and drill is homework. 00:36:16.719 --> 00:36:22.679 Three checks, five seconds, and they will tell you when you are wrong. 00:36:22.729 --> 00:36:31.679 One. Does the first address divide by N? One hundred dot seventy-two dot one ninety-two dot zero — and one ninety-two divides by thirty-two, which is the stride of a slash nineteen in the third octet. It checks out. 00:36:38.219 --> 00:36:47.169 Two. Is the last address one below a multiple? Two twenty-three is one below two twenty-four, and two twenty-four divides by thirty-two. It checks out. 00:36:49.059 --> 00:36:55.489 Three. Is N a power of two? Eight thousand one hundred and ninety-two is two to the thirteenth. 00:36:55.539 --> 00:37:04.079 If your N is not a power of two, you have made an arithmetic slip, not a conceptual one — and those are quick to fix. 00:37:04.129 --> 00:37:13.079 They will not tell you the right answer. They will tell you that yours is wrong, which under exam conditions is worth nearly as much. 00:37:13.949 --> 00:37:22.899 Build them into the ritual as a fifth beat. Five, ten, thirty, ten, five — still inside the minute. 00:37:23.153 --> 00:37:30.273 A quick preview, ten seconds a row. This is not today's lesson — Session sixteen owns it. 00:37:30.323 --> 00:37:39.273 Zero dot zero dot zero dot zero: this host, on this network. It is what a machine puts in the source field before DHCP has told it who it is. 00:37:41.393 --> 00:37:50.343 Two five five dot two five five dot two five five dot two five five: the limited broadcast. Everyone on this network — and routers slam the door on it rather than forward it. 00:37:54.183 --> 00:38:03.133 One twenty-seven slash eight: loopback. It never leaves your machine. Ping one twenty-seven dot zero dot zero dot one and you are talking to yourself, which is why it always works even when nothing else does. 00:38:08.743 --> 00:38:17.693 One sixty-nine dot two five four slash sixteen: link-local. This is what a host gives itself when DHCP never answers. It works on that one wire and nowhere else — so if you ever see one of these on your own machine, DHCP is what failed. 00:38:25.533 --> 00:38:34.483 Two twenty-four slash four: multicast. One packet, many listeners — the same idea you met one layer down in Session seven. 00:38:37.063 --> 00:38:46.013 And the private blocks: ten slash eight, one seventy-two dot sixteen slash twelve, and one ninety-two dot one sixty-eight slash sixteen. Reusable in every home and every campus, which is why NAT has to rewrite them on the way out. 00:38:54.273 --> 00:39:03.223 If block arithmetic ever lands you inside one of these on an exam, stop and re-read the question. 00:39:04.025 --> 00:39:10.915 Next, the difference between how many addresses a block has and how many hosts it can seat. 00:39:10.965 --> 00:39:17.625 A block has N addresses. Full stop. Forouzan counts all sixteen in a slash twenty-eight. 00:39:17.675 --> 00:39:22.555 The block is a range, and the range includes both ends. 00:39:22.605 --> 00:39:30.315 But used as a subnet, two of them are spoken for. The first is the network's name, and the last is the broadcast address. 00:39:30.365 --> 00:39:33.585 Neither can be handed to a machine. 00:39:33.635 --> 00:39:38.375 So a slash twenty-eight seats fourteen computers. Sixteen minus two. 00:39:38.425 --> 00:39:47.375 A slash twenty-six seats sixty-two. A slash twenty-four seats two hundred and fifty-four — and that number you have seen on every home router you have ever opened. 00:39:50.725 --> 00:39:59.375 The signal word is "usable". "How many addresses" means N. "How many hosts", or "how many usable", means N minus two. 00:39:59.425 --> 00:40:08.375 Exams ask both ways, deliberately, in the same question. Read it twice and underline the word. 00:40:10.339 --> 00:40:17.539 And here is where the ritual becomes a reflex. Four steps, one at a time, with a clock running. 00:40:17.589 --> 00:40:26.199 It opens with the flagship and nothing answered. Step one is highlighted, and the other three are greyed out — because you do not get to skip. 00:40:26.249 --> 00:40:35.199 The ladder and the powers of two sit across the top the whole time. That is not cheating; they are meant to be in front of you until they are inside you. 00:40:36.019 --> 00:40:43.779 Step one answered. Sixteen, marked green, with the reason underneath — two to the four, four host bits. 00:40:43.829 --> 00:40:48.129 Five seconds on the clock, and the tool moves you on. 00:40:48.179 --> 00:40:57.129 Step two. Two five five dot two five five dot two five five dot two forty, and the reason: twenty-eight over eight is three remainder four, and four ones is two forty. 00:40:59.899 --> 00:41:06.659 Notice the clock. Ten seconds for that step, and neither of the first two looked at the address. 00:41:06.709 --> 00:41:13.209 And here is the wrong turn everybody takes at least once: answering dot thirty-nine for the first address. 00:41:13.259 --> 00:41:21.179 Marked red, and labelled — that is the trap. The reason line names the mistake instead of just failing you. 00:41:21.229 --> 00:41:30.179 Step three, done properly. Dot thirty-two, with the AND written out. Thirty seconds — the expensive step, exactly as the recipe card said. 00:41:32.529 --> 00:41:40.609 Four facts, clean, in forty-one point four seconds. Under the minute, comfortably, and that is with the tool narrating. 00:41:40.659 --> 00:41:45.559 The best time and the clean-run counter are the numbers to chase. 00:41:45.609 --> 00:41:51.259 And a second run, on the mid-octet slash twenty-two, in fifty-two point nine. 00:41:51.309 --> 00:42:00.259 Do five of these tonight, timed, and report your times to yourself honestly. That is the homework, and it is the whole homework. 00:42:02.407 --> 00:42:09.007 Checkpoint four, and this one is the exam rehearsal. Stop the video. 00:42:09.057 --> 00:42:16.817 One. How many usable host addresses are in a slash twenty-seven, and how many addresses does the block contain? 00:42:16.867 --> 00:42:23.327 Two. Find all four facts for one seventy-two dot sixteen dot thirty-five dot two hundred, slash twenty. 00:42:23.377 --> 00:42:32.327 Three. Your answer for a slash twenty-six gives a first address of two hundred dot one dot one dot one hundred. What is wrong, and how do you know before checking the arithmetic? 00:42:37.657 --> 00:42:38.587 Answers. 00:42:38.637 --> 00:42:47.587 One. The block contains thirty-two addresses. Thirty are usable, because the first is the network address and the last is the broadcast address. 00:42:49.637 --> 00:42:58.587 Two. N is two to the twelfth, four thousand and ninety-six. Mask two five five dot two five five dot two forty dot zero. Thirty-five AND two forty is thirty-two, so the first address is one seventy-two dot sixteen dot thirty-two dot zero. Four thousand and ninety-six is sixteen runs of two fifty-six, so the last is one seventy-two dot sixteen dot forty-seven dot two fifty-five. 00:43:15.007 --> 00:43:23.957 Three. One hundred does not divide by sixty-four, and every first address is a multiple of its block size. The alignment check catches it in two seconds, without redoing anything. 00:43:32.297 --> 00:43:35.667 Five ways to lose marks on this topic. 00:43:35.717 --> 00:43:44.457 Wrong: one eighty dot eight dot seventeen dot nine slash twenty-two is on network one eighty dot eight dot seventeen dot zero. 00:43:44.507 --> 00:43:53.457 Right: one eighty dot eight dot sixteen dot zero. Twenty-two cuts the third octet, and seventeen AND two fifty-two is sixteen. Mid-octet means binary, always. 00:43:57.597 --> 00:44:04.537 Wrong: the last address of a slash twenty-six starting at dot sixty-four is dot one twenty-eight. 00:44:04.587 --> 00:44:13.537 Right: dot one twenty-seven. Last equals first plus N minus one, because the first address is itself a member of the block. 00:44:14.087 --> 00:44:19.747 Wrong: the ISP gave me dot thirty-nine, so the block starts at dot thirty-nine. 00:44:19.797 --> 00:44:28.657 Right: the block starts at dot thirty-two. They gave you an address from the middle, and step three finds the block that contains it. 00:44:28.707 --> 00:44:33.107 Wrong: a slash twenty-eight gives me sixteen machines. 00:44:33.157 --> 00:44:42.107 Right: fourteen. The block holds sixteen addresses, but the first is the network name and the last is broadcast. Watch for the word "usable". 00:44:43.357 --> 00:44:46.667 And wrong: big slash number, big block. 00:44:46.717 --> 00:44:55.667 Right: backwards. A slash thirty is four addresses and a slash eight is sixteen million. Every extra network bit halves what is left. 00:44:59.247 --> 00:45:04.817 Back to minute one. And this time, with a clock. 00:45:04.867 --> 00:45:11.647 N sixteen. Mask two forty. Thirty-nine AND two forty is thirty-two. Plus fifteen, forty-seven. 00:45:11.697 --> 00:45:19.167 Five seconds, ten, thirty, ten. Fifty-five in total, and that is without hurrying. 00:45:19.217 --> 00:45:24.697 So the answer to minute one is yes. Four facts, one line, under a minute, no calculator. 00:45:24.747 --> 00:45:32.297 And now you have the ritual that does it — which is a different thing from having watched me do it. 00:45:32.347 --> 00:45:35.137 And here is the promise for the rest of the arc. 00:45:35.187 --> 00:45:42.307 Subnetting will feel new for about four minutes, until you notice it is this exact ritual, run once per subnet. 00:45:42.357 --> 00:45:46.527 The ISP allocation problem is this ritual six times, with a seating plan. 00:45:46.577 --> 00:45:55.527 You have already learned the hard part of the next three lectures. What remains is deciding where to cut. 00:45:56.660 --> 00:45:59.790 Four things you should be able to do now. 00:45:59.840 --> 00:46:08.790 Recover all four facts from an address and a slash, in under a minute. N, the mask, the AND, plus N minus one — in that order, without a calculator. 00:46:12.660 --> 00:46:21.610 Survive a mid-octet boundary. Find the octet the fence cuts, AND that one in binary, and treat the rest as copying. 00:46:21.940 --> 00:46:30.780 Find a last address by counting two-fifty-sixes — never by adding N minus one to a dotted address digit by digit. 00:46:30.830 --> 00:46:39.290 And check your own answer in five seconds. First divides by N, last is one below a multiple, N is a power of two. 00:46:39.340 --> 00:46:48.290 Homework: five blocks tonight, timed. And keep eleven dot ten dot zero dot zero slash twenty-two in your notes. 00:46:50.093 --> 00:46:51.733 That is Session eleven. 00:46:51.783 --> 00:46:54.933 Every address slash n carries four facts. 00:46:54.983 --> 00:47:01.473 N, the mask, the AND, plus N minus one. Sixty seconds, no calculator, every time. 00:47:01.523 --> 00:47:08.153 Blocks are already there — you do not build one, you find the one that contains the address. 00:47:08.203 --> 00:47:16.343 And when the boundary is mid-octet, decimal stops telling the truth, so you drop to binary and count in two-fifty-sixes. 00:47:16.393 --> 00:47:24.263 Before the next session: five timed runs, and read the subnetting part of eighteen point four, Example eighteen point five. 00:47:24.313 --> 00:47:29.033 Next class: one block, four labs, and the only real question — where do you cut? 00:47:29.083 --> 00:47:34.264 I will see you there.