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来自 Princeton University 的课程

Networks Illustrated: Principles without Calculus

39 个评分

Princeton University

Networks Illustrated: Principles without Calculus

39 个评分

What makes WiFi faster at home than at a coffee shop? How does Google order its search results from the trillions of webpages on the Internet? Why does Verizon charge $15 for every GB of data we use? Is it really true that we are connected in six social steps or less? These are just a few of the many intriguing questions we can ask about the social and technical networks that form integral parts of our daily lives. This course is about exploring the answers, using a language that anyone can understand. We will focus on fundamental principles like “sharing is hard”, “crowds are wise”, and “network of networks” that have guided the design and sustainability of today’s networks, and summarize the theories behind everything from the social connections we make on platforms like Facebook to the technology upon which these websites run. Unlike other networking courses, the mathematics included here are no more complicated than adding and multiplying numbers. While mathematical details are necessary to fully specify the algorithms and systems we investigate, they are not required to understand the main ideas. We use illustrations, analogies, and anecdotes about networks as pedagogical tools in lieu of detailed equations. Please note that per Princeton University policy, no certificates, credentials or reports are awarded in connection with this course.

从本节课中

Routing Traffic through the Internet

It is hard to overstate the impact that the Internet has had on society. In this lesson, we will overview the fundamental concepts behind the way the Internet is designed. We will also take a look at routing, which is the process of determining how packets of information are transported.

与讲师见面

Christopher Brinton
Lecturer
Electrical Engineering
Mung Chiang
Professor
Electrical Engineering

And so, we're actually done with that example.

we don't have to continue anymore because it

has already found all the minimum cost paths.

And here's just a description, after three hops, what we see here.

so again, this is if we're going to destination F the cost that we would see.

so A could get there in a cost of 12, and he would forward to C.

B could get there at a cost of 13, he

would forward to E, C forward to D at a cost

of 10, and so on.

D would forward directly to F and E would forward to D to get there.

So these are the paths that the packets would take getting to F.

So, in general, suppose we have Num, we'll say NumNodes,

it could be eight, nine, in this case it's six.

but not counting destination, so sorry in this case it's five.

So this is, it's very important to say we're not counting the destination here.

so suppose we have in this case five nodes,

so if we had 20 nodes counting the

destination it would be 19 without the destination.

then we can guarantee actually to have found a solution after num hops, right?

So if we take away the destination, count the number of

nodes, and we've already gone that many hops or just the

number of total hops counting the destination minus one, then we

know that we're done and that we don't have to go anymore.

because otherwise we'd have to make a cycle back, right so in this case,

right, if I wanted to go five nodes, right, I'd

need to cover all the other nodes in the graph.

And because there is only five of them, so obviously you're not going to want to

loop back you're not going to want to go to A to D to C.

Which that's not even possible. But, a loop back here.

It's hard to actually see any loops, or A to B, to E to D, to A.

you would never do that because that would only add cost, right?

because the loop is just going to have a higher cost than without

looping, so, if you've already gone to all of the different nodes then obviously you

must've, or if you've already had enough hops

that you could have covered all the nodes

of the graph, then you much have found

the minimum cost, because the maximum number of

hops you'd use would be the number of

nodes on the graph without the destination present.

So, in general with six nodes we would need to go five hops, like here.

But we're actually done already, we don't need

to keep going, because the cost won't keep updating.

You can see, right, that nothing is going to change now.

B's cost is nine and nine plus four is 13.

And you can verify all the costs are all equalized now.

so, in a real life imitation we never know the real amount of routers, Right?

so, we have to rely on the lack of really new costs coming in.

Alright, so the routers themselves would say, okay well

the cost aren't updating, I'm not seeing anything new, right?

I'm not seeing you can get there quicker, I'm

not seeing that If I'm B, for instance, I'm not

seeing E or C can get there any quicker, until

that happens I'm not going to update anymore at all.

so because we don't know the number of nodes, we just have to rely

on the lack of new costs or updated cost to determine that we're first.

But again these costs are going to change

all the time depending on upon what happens.

You know, if links go out then these costs go to infinity.

If they have to re-converge the algorithm in each re-converge and so on.

so here's a, here's the routing table or basically what

would appear to be a routing table for each router.

A through E coming down here.

I'm going to destination F and again these are, these are names they would be in

the routing table with the IP addresses, right,

so each of these routers has an IP address.

That's how they're addressed.

so router A, B, C, D, E has one IP address, then but the

destination F, which is really just one

entry in each of these router's routing table.

So each

router has a separate routing table. A would have one for B, C, D, E, and F.

we only did just the case of going totally to F.

But then if A receives a packet destination to F, he knows

that he sends out the port this port right here, which would

go next hop of C and really this would be written in

C's IP address too and The cost is 12 to get there, right?

So here we're showing symbolically,

destination next top, really in a routing table this would be IP address, and

then port number, which port am I going out for that router to know that.

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