Exploring Network Topologies: Point to Point TCP — Transcript

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1. 0:05All right, welcome back to the board
2. 0:07everyone. Uh just wanted to remind you
3. 0:09what we did last time where we looked at
4. 0:12sort of two extremes of what we call
5. 0:14topologies as I said the structure of
6. 0:16networks. one which was the star with a
7. 0:19sender sending directly to different
8. 0:21receivers and the other extreme which
9. 0:23was what we're calling the daisy chain a
10. 0:25line with a sender sends to all the
11. 0:27receivers one after the other and then
12. 0:29just relay to the next receiver. Um so
13. 0:34what do actual networks look like? Well,
14. 0:36they don't tend to look like this. Uh
15. 0:39they don't tend to look like this. They
16. 0:41tend to look like something in the
17. 0:42middle which is what we mentioned last
18. 0:44time.
19. 0:46So maybe what you would have is a
20. 0:49topology. Say that we're still looking
21. 0:52at a sender and several receivers. You
22. 0:54would have a sender and you would have
23. 0:57several receivers and then you'd have
24. 0:59what we call a mesh. That's to say a
25. 1:01collection of different uh network nodes
26. 1:05in between. Not all of them needing to
27. 1:07be uh a receiver by themselves. So maybe
28. 1:11here's my sender
29. 1:15uh and here are the different receivers.
30. 1:16I'm going to put three of them are
31. 1:18receiver one, receiver two, receiver
32. 1:21three and then these middle nodes are
33. 1:23just nodes that are receiving data and
34. 1:26sending it around to different
35. 1:28receivers. Okay. Uh so how do we
36. 1:33actually manage uh in these kinds of
37. 1:35systems uh from the sender to the
38. 1:38receiver? There's sort of two different
39. 1:41uh aspects here. One is how do we manage
40. 1:44from a single sender to a single
41. 1:46receiver and then how do we manage from
42. 1:48a single sender to multiple receivers.
43. 1:50So let's just concentrate on this one
44. 1:53sender to receiver.
45. 1:56So I'm going to draw what's called a
46. 1:59route which is basically a path a single
47. 2:01path from this sender to this receiver.
48. 2:04What you can see is that it looks a bit
49. 2:06like this other topology we had except
50. 2:09that these intermediate nodes are not
51. 2:12actually receivers per se. They didn't
52. 2:14want necessarily the data. They're just
53. 2:16helping the data along and are agreeing
54. 2:19to actually participate in the
55. 2:21transmission of that data. So when we
56. 2:23look at this one route um how does
57. 2:27something like for instance
58. 2:30TCP IP manage these different losses
59. 2:34just like over here we're still going to
60. 2:36have say loss of epsilon which is excess
61. 2:40delay effectively over each of these
62. 2:42links
63. 2:44where the way it's going to manage it is
64. 2:47that it's going to manage it by having
65. 2:50what's called an acknowledgement that's
66. 2:52to Say that the sender is going to send
67. 2:54a packet which will get through this
68. 2:57first hop with probably 1 minus epsilon.
69. 3:00Then it's this node is going to send the
70. 3:02same packet over again 1 minus epsilon.
71. 3:06Then the next packet over 1 minus
72. 3:07epsilon. And then the receiver is going
73. 3:11to have another path which I'm not going
74. 3:13to draw explicitly here. Often the same
75. 3:16one running backwards is going to send
76. 3:19what's called an acknowledgement.
77. 3:21generally designated by just act.
78. 3:25So the receiver is going to tell the
79. 3:26sender and if the packet didn't make it
80. 3:29because it got lost here or here or
81. 3:31here, the sender is going to retransmit
82. 3:33the packet. What is the throughput in
83. 3:37this type of setting? Well, the
84. 3:39throughput is still going to be like
85. 3:41here is going to be 1 minus epsilon to
86. 3:44the n. In this case, n the number of
87. 3:46hops is three. Why? because a packet
88. 3:50still has to make it through all three
89. 3:52hops and if it doesn't make it through a
90. 3:55hop it has to be transmitted again and
91. 3:57those retransmissions lower the
92. 4:00throughput because I have to send
93. 4:02something yet again. Um so basically
94. 4:06what will have happen here is that that
95. 4:09transmission needs to then go again
96. 4:11through all of the three hops. As an
97. 4:14example, suppose that a packet got
98. 4:16through here, okay?
99. 4:18And it got through the second hop, okay?
100. 4:20And then it got lost in the third hop.
101. 4:23What's going to happen when you have a
102. 4:25retransmission from the sender is that
103. 4:27you're still going to have a system
104. 4:30where the packet is retransmitted over
105. 4:33the first hop even though it got there,
106. 4:35okay? And retransmitted over the second
107. 4:38hop, even though that had also gone
108. 4:40okay. So you're going to have these
109. 4:42wasteful extra transmissions.
110. 4:45Um if instead you code, even if you have
111. 4:49an acknowledgement from the receiver to
112. 4:51the sender, just like what we had before
113. 4:54here, you don't need to actually
114. 4:57transmit again things that had been
115. 5:00transmitted successfully before.
116. 5:03So you can code on a single sender,
117. 5:07single receiver system and still
118. 5:10do as well as what we talked about
119. 5:13before and get a throughput here which
120. 5:16was just the minimum of each of these
121. 5:18hops. In this case, all of the hops have
122. 5:21the same throughput each hop
123. 5:23individually. And so you would get 1
124. 5:26minus epsilon versus getting 1us epsilon
125. 5:31cubed. And this is larger than that
126. 5:34because 1 - epsilon is less than one.