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.