ALOHA Throughput

Loading...

来自 Princeton University 的课程

Networks Illustrated: Principles without Calculus

40 个评分

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. Please note that per Princeton University policy, no certificates, credentials or reports are awarded in connection with this course.

从本节课中

Random Access in Wifi Networks

In this lesson, we will investigate WiFi, another type of wireless network. Rather than having stringent power control algorithms as we saw for cellular, WiFi relies on "random access" methods to manage interference among users in the same location.

Unlicensed Spectrum5:01

Traffic Analogy4:43

WiFi Standards5:34

WiFi Deployment6:24

Accessing WiFi10:55

Interference7:49

Controlled vs. Random Access6:49

Random Access Protocols & ALOHA7:00

ALOHA Successful Transmission5:15

ALOHA Throughput7:12

ALOHA Inscalability8:19

CSMA Carrier Sensing5:54

CSMA Backoff7:49

CSMA vs. ALOHA5:41

与讲师见面

Christopher Brinton
Lecturer
Electrical Engineering
Mung Chiang
Professor
Electrical Engineering

So now, let's look at the throughput. And we just came up with expressions for

the probability that a station's going to transmit and the probability that all the

other stations aren't going to transmit. So first, let's look at the per-station

throughput. So now, again, we said that when we have

to have things happen at the same time, we multiply them.

So, B is transmitting and A is not transmitting and D is not transmitting

and C is not transmitting and E is not transmitting.

So it's multiplication. So we basically take probability of

transmission, which is probability that A transmits And we multiply that by the

probability that none of the other stations are transmitting, and we just

saw that before. That was 1 minus PropTrans and then we

repeated that multiplication four times, and here we can just raise that to the

fourth power. So, now we know in that in general, then

if we had, suppose we had ten stations, well then we'd have one transmitting and

nine not transmitting, so this would be raised to the ninth power, and if we had

one transmitting but, and we had 20 stations it would be one and then 19, so

in general let's say that we have, we cal the number of stations in this numstat.

And that's we'll send that's the number of stations.

So this would be that probtrans times one minus probtrans.

[NOISE] And do that multiplication a bunch of times.

Times one minus Probtrans and this multiplication happens Numstat minus one

because one of those stations is this initial trans station over here.

So, that happens numstat minus 1 times, or you can raise it to the numstat minus

1 power. Now, this is the per station throughput.

So now, what is the total throughput? So, this is for instance if we looked at

B or A or C, D or E independently. But now if we want to look at the total

throughput we have to say that we have to add the throughputs of each of the

individual stations. So, we just simply add the throughput of

B plus throughput of A plus throughput of D plus throughput of C and so on.

So we get through of A plus the through-put of B and we keep adding up.

And since it's the same equation for each station, we then just get that the total

through-put is equal to the number of stations we have times the per-station

through-put, very simple. [SOUND] So, if we go back to our example

again. Suppose the probability of transmission,

[SOUND] was 0.8. So we have Was 20%.

So then we convert that into a decimal, we get 0.2.

And then the per-station throughput would be equal to the probability of

transmission, which is 0.2. Times 1 minus the probability of

transmission. And here we have five stations, we'll

just assume that [INAUDIBLE] five stations.

So we, then we'd have 1 minus 0.2. And we have that four times 1 minus 0.2,

times 1 minus 0.2, times 1 minus 0.2. And if you do that out, we get 0.2 times

0.8 times 0.8 times 0.8 times 0.8 again, [INAUDIBLE] broken record here and then

that is equal to 0.08192, which is 8.192%, so what that means is that every

station is going to get 8% through. Then the total throughput, we would get

by multiplying the per-station throughput.

By the total number of stations. So the total then, figure this out, would

be equal to the per-station times the total number of stations, which is 0.4096

or 40.96%. Lets also point out as well that here we

are dealing with throughput in percentages right so we are dealing with

them as percent's and in reality we said before that throughput is measured in

bytes per second, right so typically we'll see like mega bytes per second and

the reason that we are doing in a percent's here it's just comparatively

the same. To understand this you have to recall how

we get this percentage in the first place.

So we're basing this relative to the idea that the maximum throughput we could have

is one frame in one time slot. So in each time slot that we have, we

could just have one frame. Now, each frame is going to have a

certain amount of bits in it and each time slot is going to have a certain

amount of seconds. Or it's going to.

Actually be fractions of a second. So this is a time slot and this is a

frame up here rather than it being a bit and a second that we're talking about.

But it's just a multiplicative factor for us to get there.

We'll just have to multiply by the number of frames that we have.

Per bit, or the other way around. And then we have to multiply by the

number of time slots that we have per second in order to get bits per second.

And so, in this percentage rate here, when we say 40.96%, we mean that 40.9% of

the time slots we are going to have a successful message delivery.

And in the other 60%, approximately 60%, we're not going to have a successful

message delivery. And in this case right here, we're saying

that we have, this is, again, remember this is per station through put.

So we're saying Relative to the maximum that this station could get if it was the

only station in the network, which is not the case we are getting 8% of what we

could in maximum get if we were transmitting successful in each time

slot, so usually these percentages are relative to the maximum, or the

percentages of the maximum, which would be 100% through-put or meaning that we

get one frame in each time slot.

探索我们的目录

免费加入并获得个性化推荐、更新和优惠。

Coursera 致力于普及全世界最好的教育，它与全球一流大学和机构合作提供在线课程。

© 2018 Coursera Inc. 保留所有权利。

通过 App Store 下载

通过 Google Play 获取