DPC Computation: Part B

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Networks Illustrated: Principles without Calculus

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Princeton University

Networks Illustrated: Principles without Calculus

40 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. All the features of this course are available for free. It does not offer a certificate upon completion.

从本节课中

Power Control in Cellular Networks

How is it possible that we can all communicate effectively without disrupting each other's calls, messages, or Internet usage? In this lesson, we will take a look at some of the methods that have been developed for letting us "share" the air over which our phones communicate.

与讲师见面

Christopher Brinton
Lecturer
Electrical Engineering
Mung Chiang
Professor
Electrical Engineering

So returning now to our example, let's do the SIRs for the first iteration, using

the, using the power levels that we just changed to.

So, we know that the noise in the receivers or their values that we had

before, and here are the transit powers that we did before, 1.4, 1.5 and 2.68.

So now the transfer powers aren't same at the each of them anymore, so we're not

going to have 2 milliwatts everywhere. They're going to be different values now

and we are going to see how that plays out.

So with our equation to measure SIR at link A and if we start off remember again

we have direct times the transmit power so 0.9 times 1.4.

That's the signal and interference is 0.1 not times 1.4 but times 1.5 because

that's B's transfer power right now, 0.1 times 1.50 plus 0.2 times 2.68, and then

we add 0.1, which is the noise in the receiver.

And I'm not going to write the milliwatts out, because we saw before how they

cancel out. So then we do this, we're going to get

1.26 divided by 0.786, which is going to give us 1.60.

So now we can move on to B. We saw the direct is .8, time 1.5.

[SOUND] The indirect from A is .1. We multiply that by 1.4, plus the

indirect from C, which is .1. We multiply that by 2.68, [SOUND] and

then we add 0.2 onto that. And if we do that out, we get 1.20

divided by 0.608, which is 1.97. And now let's move on to C, the direct

transfer power is 0.9 times the gain which is 2.68 divided by 0.2 which is

indirect from A times A's transmit that is 1.40 plus the indirect from B which is

0.2 times 1.50 direct transmit plus the noise in the receiver which is Upon 3.

And we get 2.412 divided by 0.880 which is 2.75.

So now we'll move onto the second iteration and we'll update the power

levels used in the SIR stat we'll just compute it.

So A would feed back the value of 1.6 to transmitter A, which has a desired SIR of

1.8. And remember, our equation is that the

next power is equal to the ratio times the current power.

So first let's calculate the ratio for each of these guys in here.

So for A, the ratio is going to be the desired over the measured, and the

desired is 1.8 and the measured is 1.6. For B, the desired is 2.0 and the

measured is 1.97. So we have 2.0 divided by 1.97.

For C, the desired is 2.2, and the measured is 2.75.

So it's 2.2 divided by 2.75. So this is interesting because now, for

A, the ratio is greater than one, whereas before, we had it as less than one.

For B It's greater than one, whereas before it was less than one.

And for C, it's less than one, whereas before it was greater than one.

So they've switched, and now we expect A and B's transmit powers to go up, because

their SIRs are too low, and C's transmit power to go down because its SIR right

now is too high. So this is a process of Overshooting and

undershooting, which will likely occur many times till we hit the equilibrium

levels. So we've undershot in for A and B because

we're too low, and we've overshot for C because the SIR is too high, and so this

we expect it to look something like this, where each of the SIRs will kind of go

like this over time, until the eventually hit where they need to be.

So now let's do these power updates, so we can, so the transmit powers for A, B,

and C are 1.4, 1.5 and 2.68 respectively. So you take 1.8 divided by 1.6, multiply

that by 1.40, and we would get 1.58 millawatts, to be the new power level.

Multiply this by 1.50 and we get 1.52, so really close.

So you see B is very close to where it needs to be.

This power level didn't change much at all, because of that.

And, then for C, multiply this by 2.69, and we get 2.15 milliwatts.

So that's how the updates are computed. So now, iteration two will move on and we

will compute now the SIRs, just like we did before.

So we have our noise values, and we have our transit powers.

And so we'll start off with, like, A and we'll do that again.

So, for A we have direct gain 0.9 times 1.58 and at the interference we have 0.1

times 1.52 from B plus 0.2 times 2.15 from C plus 0.1 milli watts In the

receiver, and this we get 1.422 divided by 0.682, which is 2.08.

Now, for B, we have a direct of 0.8 times 1.52, indirect, or interference of .8

times 1.52 plus .1 for C times 2.15 plus .2 of noise.

And for that we get 1.216 divided by .573, which is 2.13, and then for C, we

have a direct of 0.9 times 2.15, and we divide that by the indirect interference,

0.2 times 1.58 plus 0.2 times 1.52 [lus the noise of 0.3, and we do that out, we

get 1.935 divided by 0.920, which is 2.10.

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