1 00:00:00,000 --> 00:00:00,590 2 00:00:00,590 --> 00:00:03,420 Previously, we learned the concept of independent 3 00:00:03,420 --> 00:00:04,890 experiments. 4 00:00:04,890 --> 00:00:08,119 In this exercise, we'll see how the seemingly simple idea 5 00:00:08,119 --> 00:00:11,080 of independence can help us understand the behavior of 6 00:00:11,080 --> 00:00:13,240 quite complex systems. 7 00:00:13,240 --> 00:00:15,720 In particular, we'll combined the concept of independence 8 00:00:15,720 --> 00:00:19,430 with the idea of divide and conquer, where we break a 9 00:00:19,430 --> 00:00:22,630 larger system into smaller components, and then using 10 00:00:22,630 --> 00:00:26,310 independent properties to glue them back together. 11 00:00:26,310 --> 00:00:27,740 Now, let's take a look at the problem. 12 00:00:27,740 --> 00:00:30,950 We are given a network of connected components, and each 13 00:00:30,950 --> 00:00:32,299 component can be good with 14 00:00:32,299 --> 00:00:35,680 probability P or bad otherwise. 15 00:00:35,680 --> 00:00:38,370 All components are independent from each other. 16 00:00:38,370 --> 00:00:41,620 We say the system is operational if there exists a 17 00:00:41,620 --> 00:00:47,620 path connecting point A here to point B that go through 18 00:00:47,620 --> 00:00:49,690 only the good components. 19 00:00:49,690 --> 00:00:52,450 And we'd like to understand, what is the probability that 20 00:00:52,450 --> 00:00:54,490 system is operational? 21 00:00:54,490 --> 00:01:01,794 Which we'll denote by P of A to B. 22 00:01:01,794 --> 00:01:04,150 Although the problem might seem a little complicated at 23 00:01:04,150 --> 00:01:05,920 the beginning, it turns out only two 24 00:01:05,920 --> 00:01:07,420 structures really matter. 25 00:01:07,420 --> 00:01:10,800 So let's look at each of them. 26 00:01:10,800 --> 00:01:14,320 In the first structure, which we call the serial structure, 27 00:01:14,320 --> 00:01:18,640 we have a collection of k components, each one having 28 00:01:18,640 --> 00:01:22,290 probability P being good, connected one next to each 29 00:01:22,290 --> 00:01:24,230 other in a serial line. 30 00:01:24,230 --> 00:01:26,770 Now, in this structure, in order for there to be a good 31 00:01:26,770 --> 00:01:29,500 path from A to B, every single one of the 32 00:01:29,500 --> 00:01:31,380 components must be working. 33 00:01:31,380 --> 00:01:35,380 So the probability of having a good path from A to B is 34 00:01:35,380 --> 00:01:41,640 simply P times P, so on and so, repeated k times, which is 35 00:01:41,640 --> 00:01:43,940 P raised to the k power. 36 00:01:43,940 --> 00:01:46,320 Know that the reason we can write the probability this 37 00:01:46,320 --> 00:01:49,530 way, in terms of this product, is because of the 38 00:01:49,530 --> 00:01:51,210 independence property. 39 00:01:51,210 --> 00:01:58,720 40 00:01:58,720 --> 00:01:59,510 Now, the second useful 41 00:01:59,510 --> 00:02:01,750 structure is parallel structure. 42 00:02:01,750 --> 00:02:05,265 Here again, we have k components one, two, through 43 00:02:05,265 --> 00:02:08,419 k, but this time they're connected in parallel to each 44 00:02:08,419 --> 00:02:11,780 other, namely they start from one point here and ends at 45 00:02:11,780 --> 00:02:13,900 another point here, and this holds for 46 00:02:13,900 --> 00:02:15,420 every single component. 47 00:02:15,420 --> 00:02:18,250 Now, for the parallel structure to work, namely for 48 00:02:18,250 --> 00:02:22,370 there to exist a good path from A to B, it's easy to see 49 00:02:22,370 --> 00:02:25,330 that as long as one of these components works the whole 50 00:02:25,330 --> 00:02:26,760 thing will work. 51 00:02:26,760 --> 00:02:30,640 So the probability of A to B is the probability that at 52 00:02:30,640 --> 00:02:33,260 least one of these components works. 53 00:02:33,260 --> 00:02:36,260 Or in the other word, the probability of the complement 54 00:02:36,260 --> 00:02:39,125 of the event where all components fail. 55 00:02:39,125 --> 00:02:44,740 Now, if each component has probability P to be good, then 56 00:02:44,740 --> 00:02:49,920 the probability that all key components fail is 1 minus P 57 00:02:49,920 --> 00:02:51,740 raised to the kth power. 58 00:02:51,740 --> 00:02:55,000 Again, having this expression means that we have used the 59 00:02:55,000 --> 00:02:58,620 property of independence, and that is probability of having 60 00:02:58,620 --> 00:03:00,990 a good parallel structure. 61 00:03:00,990 --> 00:03:02,760 Now, there's one more observation that will be 62 00:03:02,760 --> 00:03:03,800 useful for us. 63 00:03:03,800 --> 00:03:07,360 Just like how we define two components to be independent, 64 00:03:07,360 --> 00:03:10,920 we can also find two collections of components to 65 00:03:10,920 --> 00:03:13,430 be independent from each other. 66 00:03:13,430 --> 00:03:16,820 For example, in this diagram, if we call the components 67 00:03:16,820 --> 00:03:21,580 between points C and E as collection two, and the 68 00:03:21,580 --> 00:03:25,580 components between E and B as collection three. 69 00:03:25,580 --> 00:03:28,760 Now, if we assume that each component in both 70 00:03:28,760 --> 00:03:29,570 collections-- 71 00:03:29,570 --> 00:03:32,210 they're completely independent from each other, then it's not 72 00:03:32,210 --> 00:03:35,600 hard to see that collection two and three behave 73 00:03:35,600 --> 00:03:36,780 independently. 74 00:03:36,780 --> 00:03:41,250 And this will be very helpful in getting us the breakdown 75 00:03:41,250 --> 00:03:45,710 from complex networks to simpler elements. 76 00:03:45,710 --> 00:03:47,630 Now, let's go back to the original problem of 77 00:03:47,630 --> 00:03:50,990 calculating the probability of having a good path from point 78 00:03:50,990 --> 00:03:54,560 big A to point big B in this diagram. 79 00:03:54,560 --> 00:03:57,920 Based on that argument of independent collections, we 80 00:03:57,920 --> 00:04:00,070 can first divide the whole network into three 81 00:04:00,070 --> 00:04:04,870 collections, as you see here, from A to C, C to E and E to 82 00:04:04,870 --> 00:04:09,330 B. Now, because they're independent and in a serial 83 00:04:09,330 --> 00:04:12,180 structure, as seen by the definition of a serial 84 00:04:12,180 --> 00:04:16,140 structure here, we see that the probability of A to B can 85 00:04:16,140 --> 00:04:24,580 be written as a probability of A to C multiplied by C to E, 86 00:04:24,580 --> 00:04:29,230 and finally, E to B. 87 00:04:29,230 --> 00:04:33,880 Now, the probability of A to C is simply P because the 88 00:04:33,880 --> 00:04:36,510 collection contains only one element. 89 00:04:36,510 --> 00:04:41,090 And similarly, the probability of E to B is not that hard 90 00:04:41,090 --> 00:04:43,070 knowing the parallel structure here. 91 00:04:43,070 --> 00:04:47,010 We see that collection three has two components in 92 00:04:47,010 --> 00:04:53,500 parallel, so this probability will be given by 1 minus 1 93 00:04:53,500 --> 00:04:55,945 minus P squared. 94 00:04:55,945 --> 00:04:59,480 95 00:04:59,480 --> 00:05:02,930 And it remains to calculate just the probability of having 96 00:05:02,930 --> 00:05:08,650 a good path from point C to point E. To get a value for P 97 00:05:08,650 --> 00:05:16,710 C to E, we notice again, that this area can be treated as 98 00:05:16,710 --> 00:05:24,260 two components, C1 to E and C2 to E connected in parallel. 99 00:05:24,260 --> 00:05:27,570 And using the parallel law we get this probability is 1 100 00:05:27,570 --> 00:05:36,790 minus 1 minus P C1 to E multiplied by the 1 minus P C2 101 00:05:36,790 --> 00:05:43,010 to E. Know that I'm using two different characters, C1 and 102 00:05:43,010 --> 00:05:47,760 C2, to denote the same node, which is C. This is simply for 103 00:05:47,760 --> 00:05:51,610 making it easier to analyze two branches where they 104 00:05:51,610 --> 00:05:53,930 actually do note the same node. 105 00:05:53,930 --> 00:06:02,600 Now P C1 to E is another serial connection of these 106 00:06:02,600 --> 00:06:08,660 three elements here with another component. 107 00:06:08,660 --> 00:06:13,580 So the first three elements are connected in parallel, and 108 00:06:13,580 --> 00:06:16,580 we know the probability of that being successful is 1 109 00:06:16,580 --> 00:06:22,430 minus P3, and the last one is P. 110 00:06:22,430 --> 00:06:29,570 And finally, P C2 to E. It's just a single element 111 00:06:29,570 --> 00:06:34,400 component with probability of successful being P. At this 112 00:06:34,400 --> 00:06:37,720 point, there is no longer any unknown variables, and we have 113 00:06:37,720 --> 00:06:41,180 indeed obtained exact values for all the quantities that 114 00:06:41,180 --> 00:06:42,740 we're interested in. 115 00:06:42,740 --> 00:06:44,960 So starting from this equation, we can plug in the 116 00:06:44,960 --> 00:06:51,290 values for P C2 to E, P C1 to E back here, and then further 117 00:06:51,290 --> 00:06:54,520 plug in P C to E back here. 118 00:06:54,520 --> 00:06:57,880 That will give us the final solution, which is given by 119 00:06:57,880 --> 00:07:01,370 the following somewhat complicated formula. 120 00:07:01,370 --> 00:07:04,240 So in summary, in this problem, we learned how to use 121 00:07:04,240 --> 00:07:07,820 the independence property among different components to 122 00:07:07,820 --> 00:07:11,850 break down the entire fairly complex network into simple 123 00:07:11,850 --> 00:07:15,570 modular components, and use the law of serial and parallel 124 00:07:15,570 --> 00:07:19,580 connections to put the probabilities back together in 125 00:07:19,580 --> 00:07:22,640 common with the overall success probability of finding 126 00:07:22,640 --> 00:07:24,700 a path from A to B.