1:39

And then don't forget that we have to add the noise in the receiver.

So, if you multiply this out,

you get 1.8 milliwatts on top,

and on the bottom, you get 0.7 milliwatts.

So now, notice that these are going to cancel and so,

the same interference ratio is a unilist quantity.

It just tells us how much higher the signal is

than the noise relative to the interference.

And so, if we do this out and do this division one 1.8 divided by 0.7 gives us 2.57.

So now, let's move on to B.

So B has a direct gain of 0.8, which is lower than,

again remember the direct unit of A and C are both 0.9,

but this direct unit of 0.8 times its initial transpower of two milliwatts.

So for here, we have 0.8 times two milliwatts as the signal,

then the interference, we have coming in from C and from A times their power.

So, we have 0.1 times two milliwatts plus 0.1 times two milliwatts,

and then we want to add also the noise and the receiver, 0.2 milliwatts.

Remember, the noise is higher than it was for A.

So, in the top here then,

we have 1.6 milliwatts,

0.8 times two, divided by, on the bottom,

we get 0.6 milliwatts.

And remember, again, these guys cancel.

And we do that out,

we get 2.67 as a signal-to-interference ratio.

And we can do the same thing for C. We have 0.9 times two as the signal,

then for the first interference,

we have from A,

0.2 times two plus, from B,

we have 0.2 times 2,

plus them in the ways

in C is 0.3 milliwatts.

Remember, that has the highest noise out of any of the receivers.

So, this gives us 1.8 milliwatts

divided by 1.1 milliwatt.

And after we do that,

we get 1.64 to be the SIR.

And so, just to tabulate this equation for the measure SIR,

it's equal to the signal over the interference plus the noise as we just saw.

So that's what we need to do to compute these initial SIRs.

So now, we move on to the first iteration and we'll do

the power update given those SIR values.

So we can start at A and we see that the A's

receiver's going to feedback that value of 2.57 and A is going to look at it,

and it's going to adjust its transmit power of

two milliwatts by looking at the current or the desired SIR, which is 1.8.

So, let's look at 2.57 versus 1.8.

2.57 is higher which means right now,

the signal-to-interference ratio is too high,

and we want to go down towards a lower SIR because we want to get

everybody in the end to their desired SIRs so that nobody is asking for too much.

So we can subtract 2.57 and 1.8 to get 0.77.

And what that's saying is that right now,

the SIR is too high by 0.77.

So, in order to update the transmit of power,

DPC employs something that's very similar to

that single step TPC algorithm that we saw before.

Except that that ratio quantity that we've

said before is not defined in terms of power levels,

it's defined in terms of signal-to-interference quantities.

So, if we look now intuitively here,

we see that we're desiring a value of 1.8.

And we have a value of 2.57.

So it's too high and we want the power level to go down.

So, best way to do that,

the most reasonable way of doing that is by saying, well,

let's multiply this current transit power by the ratio,

1.8 divided by 2.57.

So we take 1.8 and divide that by 2.57.

We're going to get a quantity that's less than one.

And if that specific quantity which is telling us

how much higher the current SIR is than the desired SIR.

And then we multiply that by the current transmit power of two milliwatts.

And so, this quantity, the ratio here,

comes out to be 0.7 and we multiply that then by two milliwatts and we get 1.4

milliwatts as being the power level for the next iteration of A.

Notice, it is lower than what it was before because the SIR was measured to be higher.

So we're trying to compensate for that.

Now, we move on to B. So B is going to feed back the SIR of 2.67 to the transmitter,

and then it's going to look at its current transmit power of two milliwatts,

and its desired SIR of 2.0 and determine how it should update.

And the updated formula is going to be the same,

but notice now that we have the same situation where this is too high.

So, measured SIR is still too high right now.

So, you have 2.67 minus 2.0 gives us 0.67.

So, that's how much higher it is than what it should be.

And so, we'll do the same thing again where we take that ratio 2.0,

which is where we want to be,

divided by 2.67 which is where we are right now,

and multiply that by the current transmit power of two.

And we get that ratio quantity to be 0.75 which means

that we need to lower it by 25% what it is right now to get down to 2.0.

And so, 0.75 then times two is going to give us 1.50 milliwatts.

Notice again that this is lower than this because this was higher than this.

Now, we'll move on to C and C's going to feed back this

1.64 and look at it relative to the 2.2.

So now, we're in the exact opposite situation to what we were before.

Now, the SIR is too low.

So the power level needs to go up,

and it's too low by 2.2 minus 1.64 which is 0.56.

And so, we do the same equation again where we want to be

divided by where we are right now which is 2.2 divided by 1.64.

But notice now, this quantity is greater than one rather than less than one,

and we multiply that by the current transmit power.

This transmit power is going to go up,

and it's 1.34 times two which gives us 2.68 milliwatts.

And notice now that this is higher than this because this was too low.

So we're boosting that power level now.

And so, this is the equation,

this is the actual DPC update equation right here,

that the next power is equal to this thing called the ratio which as we said,

is the desired SIR over the measured SIR,

it's where we want to be over where we are right now,

times the current power.

And we do that at each iteration as we keep measuring the SIRs.

So next question is,

why do we have to do such an update?

It seems pretty complicated.

So, why do we have to bother with it?

Well, it turns out that this is looked at from a different standpoint.

We're actually aligning each of the cell phones' incentives here.

And basically, what I mean by that is that we're forcing each of them to internalize

the negative externality that they

impose on the network by being there in the first place.

And the negative externality here is in the form of interference.

So, the fact that A is in this network is going to cause

interference for B and for C. So,

by making that follows this courtesy procedure,

we make them say that if their SIR level is too high right now,

that they will lower their transmit power and cut back in order to

internalize the negative effect that they're imposing on everyone else in the network.

Now, negative externality will be a recurring theme throughout

this course and we'll see it many times in many different context.

It's one of the main themes of this course.

And the way that that negative externality, as we said,

is internalized is through a process called negative feedback,

which we will take a brief detour to look at.