Welcome to WRC-19 here on the last day of the World Radio Communication Conference in Shamashake, Egypt, where I'm very pleased to be joining the studio today by Dr Amar Badawi, who is the chairman of WRC-19 and has been here for the last four weeks, diligently ensuring that spirit of consensus is reached here by all the delegates and that many resolutions are passed. Dr Badawi, welcome to the studio. Thank you all, Max. Now, it's been a few months since we last met in Geneva and a lot of conversations have passed since then. A lot of resolutions have been made now here at the WRC-19. I wanted to ask you what have been your impressions? Has this led up to your expectations and what have been your major challenges throughout this time? Well, I would say I'm quite happy with the outcome of the WRC-19. Things have moved sometimes with some bumps, but I believe at the end that the major issues were resolved. We have some very good results that are making everybody leave Shamashake happy with what they got, and that was my main goal. And my job actually dictates that. And for me, at least, I did not have to spend any nights here, which was nice. I was going to say, you're looking very refreshed for a conference that has lasted for many weeks here. What are some of the major outcomes which you think will be key for the future of radio communications? I think we've got several outcomes for IMT, which is International Mobile Telecommunications, which serves what people usually know as 5G, which is the commercial or common name, which is extremely important for the next phase of communications and telecommunications. 5G will make a big move in this sector. We are going to have new applications. We have Internet of Things, we'll expand. We have artificial intelligence used with 5G will really bring amazing products, amazing applications. Like for example, autonomous cars, autonomous vehicles, I mean, without 5G and artificial intelligence you won't have that product. And then you have intelligent cities without the Internet of Things and 5G serving Internet of Things because the technology will help us use this available spectrum in a more efficient way. This is the key. And that's what WRC is all about. We need to make sure that this limited amount of resource that we have named the spectrum, which in my opinion is the fuel of this industry needs to be used in the most efficient way. So this conference has been working towards that. There were so many agenda items and I may have mentioned in the previous interview that I feel that we had an opportunity this time. New technologies have availed a lot of spectrum bands to both satellite and fixed earth telecommunications. Whether it's be fixed or mobile, satellite have new amazing technologies now that will allow the satellites to serve rural areas at very high speeds, which could bring broadband to everywhere where regular fiber networks cannot go there. And what I liked in this conference is that for the first time I see that the satellite operators or vendors and the mobile operators are starting to see that they complement each other rather than compete with each other. This spirit is very important because this will help both grow and at the end of the day we're looking to provide the consumer the best possible service at the least possible cost. Having this could give us that chance and everywhere. We're trying to reach to the last billion or two who cannot have any broadband or smart phones. We're trying to reach that.  We will have wonders with 5G, with the enormous capabilities there. Did you find this harmonious collaboration from the beginning of the conference? Has it been an evolution as people have been conversing here, as they've been negotiating here? I think they have recognized it before coming to the conference. I even heard it from one of the major players there. And to me this was music to my ears because this means that they will accommodate each other rather than try to kill each other. I know that everything is now water under the bridge, but were there any major challenges, any major sticking points that were difficult? We've had like seven issues that were not so resolved until Wednesday. I think six issues were really resolved. The other one was, was there no change? To me no change is that just pushing the ball to the next WRC. Okay. Thank you. Well, I was just about to come on to that. So looking forward, there are a number of issues that are going through for 2023 now. But I mean, I think I presume that's always the case that things are carried over for another four years. But the most important issues were resolved here and that's a very good result. What difference do you think it's made to the region itself to have this enormous conference here? Well, I think it showed that countries like Egypt can really organize and host such a very important and very complex conference. I did not realize the complexity of this conference until I was really the chairman and kind of looking at every place to every committee.   I mean, we've had like some documents just given to the documentation committee by midnight on Wednesday and they're here in the final act now. I mean, this is to me, this is amazing in six languages. You have been amazing. I think everybody will agree as chair of this extremely important conference. Thank you so much for joining us in the studio and I'm sure we will catch up with you again at some stage in the near future. OK, thank you very much. Thank you very much indeed. Thank you.


This is stock number 10055A. We are here at Summit Automotive in Fond du Lac, Wisconsin. You're new and used Ram 1500 headquarters. Today we're checking out this super clean 2017 Ram 1500 Crew Cab Short Box. Actually it's a six foot four inch short box or medium box. This is a Lear Me Limited and from this HD video you will be able to tell that this truck is extremely clean all the way around. No dents or dings on the front bumper. Bright silver metallic is the color. We shoot all of our videos in 1080p so if you have HD capabilities on your computer tablet or smartphone device turn them on right now because it's like you're right here looking at the truck with me. Factory chrome step bars. They are wheeled wheel steps so they go just like it sounds wheeled wheel. 20 inch painted alloy rims. Painted pockets and polished aluminum rims. No major scuffs or scrapes on them. It has Goodyear Wrangler SRA 27560R20 tires and these tires have right around half the tread left maybe just a hair under. It has really nice fender flares, the projector lamp headlamps, the LED running lights, the front bumper parking sensors. You get the RAM stamped grille. Didn't see any dents or dings on the hood. Passengers side fender is an excellent condition as well. And the passenger side rim. No scuffs or scrapes on that. As you go down this side of the truck you can see just how clean the body is. How reflective and mirror like that paint is. We take these HD videos so if you are far away or even if you're close by and just can't make the trip down you can still see the truck hear the truck and have confidence in the vehicle that you're looking at before you even get here. Cab is a nice shape as well. And on this side I didn't see one single dent or ding. Have the wheeled side steps there really nice feature. Lower rockers all look really good. Back rim. Pretty nice condition as well. Back tires have just as much tread as the front tires and it does have the plastic wheel well liners which keeps your frame and underbody looking great. You can see it's very nice and clean all the way underneath. Something that you just don't get with pictures. See the light reflecting off the vehicle as we come around to the back of the vehicle. See it does have a full towing package which includes receiver hitch, four pin and seven pin wiring, backup parking sensors, ram stamp tailgate, locking tailgate, backup camera, has a very nice spraying bed liner. The bed is in excellent shape it does have the LED bed lighting. And as you go down this side of the truck just as clean as that passenger side. Didn't see any dents or dings on this box. Very nice and clean and for full disclosure this back rim is in nice shape as well. No major scuffs or scrapes. As you go down the rest of the side of the truck I didn't see any dents. I didn't see any dings. We want to make sure that you get the most accurate representation so when you get here or if you get the truck shipped to you you'll know exactly what you're getting has the chrome trim tow mirrors on this one. I believe they power they are manual fold in but they are power mirrors, power windows, power locks. Inside the Laramie limited package gives you the black leather interior with the gray piping, memory driver seat, driver side lumbar. Both these seats are power heated and cooled, factory all-weather floor mats, auto headlamps, and power pedals. As we hop inside here you can see it has the seven inch display. You get the digital speedometer, outside temperature display, compass display, trailer gain, range to empty and you can modify each one of those corners to say pretty much anything you want. Instrument cluster is very nice and clean and this one comes with an almost full tank of gas. Heated, leather wrapped, wood grain trim steering wheel, very nice condition, cruise controls on the right, Bluetooth and information center controls on the left. Comes with the 8.4 touchscreen radio with the factory navigation has AM, FM, and Sirius XM radio capabilities. You can do your heated and cooled seat buttons right here as well as your heated steering wheel and you can check out the backup camera without actually putting it in reverse. Has the 8-speed automatic transmission, dual climate control, factory brake controller. It has the four corner air suspension. You can actually raise that up to off-road mode two and all the way down for the parking so you can easily get in and out of the truck. Has the heated cooled seat buttons right there, heated steering wheel and you can do your parking sensors. You can turn those off and on. Has the enter and go system with the keyless entry. You can also lower the truck for easy access. When you get in 115 volt, 150 watt plug-in, passenger seat is very clean as well. No rips or tears on that. This truck smells incredibly clean inside. I don't think it's ever been smoked in. Headliner is an excellent condition. Power sliding rear window, home link buttons for your garage door security systems and lighting systems. And finally you get the 911 and assist buttons which are features of the vehicle that you never have to pay a subscription for. We'll take a quick look at the back seats and then check out under the hood. Super clean inside this truck. I'd give it a 9.5 out of 10, 9.9 out of 10. Super clean. No rips or tears on the seats. Smells very clean. Dash is clean. Has the power sliding rear window. Factory latch, child safety system for your car seats. Has the load floor which goes down nice and flat. So if you got pets or muddy boots or tools you can put them on there. Not wreck your nice Laramie limited seats. Has the all-weather floor mats back here. I say they're all weather because these cloth inserts actually come out. So you can make them all rubber. And then has the in-floor storage on both sides with the removable bin so you can you know put ice and drinks and stuff in there. Comes with these saddle bags on the back of the seats and the outboard seats in the back are heated as well in there your heated seat buttons. But very clean. The headliner is an excellent shape. You got those speakers in the ceiling part of the Alpine sound system. There's a subwoofer underneath the passenger seat. And we'll take a quick look under the hood. I would personally like to thank you for checking out the video today. Hopefully from this HD video you'll have been able to tell just how clean this truck is all the way around inside and out. Under the hood we have the 5.7 liter V8 Hemi motor. Engine bay is very clean, runs very smooth. This truck has been fully safety and inspected by our service shop. Has a fresh oil and filter change. All the fluids have been checked and topped off. Truck has been gone through mechanically 100% and is 100% ready to go. To see more pictures of this truck or one of our other 450 new and used cars, trucks, SUVs, minivans, Wranglers, halftones, recorder tons, one tons. You name it we got to go to our website www.summitauto.com. Full pictures and descriptions of every single vehicle on our lot videos of every single used vehicle that we have. All at summitauto.com and definitely check www.summitauto.com out because we have all of our inventory from two locations there. If you'd like to check out more HD videos you can go to www.summitauto.com. Remember to like, subscribe, and share on this video and all the videos that you see there. In fact any second you'll see a link to subscribe to our YouTube channel on your left. A link to more half-toned truck videos like this one on your right. If you have not been to our website on the bottom link to this vehicle on our website click those check us out. We really look forward to help you with this super clean 2017 Ram 1500 crew cab 6 foot 4 inch short box. Thanks again.


Ever looked at a math problem and wondered why there are letters mixed in with the numbers? What's up with these X's and Y's playing around in math equations? Join me as we explore these questions in today's episode of Mr. Marks, Math Adventures. Hey Math Learners, it's Mr. Marks, your friendly neighborhood math teacher. Remember, it's not just about getting the right answers, it's about learning and growing along the way. Before we go into today's adventure, remember to hit that subscribe button and ring the bell to stay tuned for future episodes. And today, we're going to embark on an exciting journey into the world of algebraic expressions. But first, let's talk about what algebraic expressions are. And as we start, always feel free to use the companion worksheet. It's a great tool to follow along with the episode and to keep track of your notes. The companion worksheet's free to download, link in the description.


�만 වීිළකා වභිට තහ්ූතාමා න්ණිතා ඔබිගබ ඍ peas, වැදෂේ ඇතශ් තලේ්ත්ගල්යක් අලිල් මැණිරසත එක් සැඥඃ trustingම්දක රිඍෆදස, සැඥඃිධයා� folding style for a bigger gym හැකක්ත්න අලින්ද  bracken waam welcoming to the honourable minister of India Shri Narin Damodi. Aay on behalf of the detached government both studying in class 6 would like to express our heartful gratitude to Honourable Shri Narin Damodi for visiting our country. We as a BUDNESS is indeed very privileged to have India as our background supporter. As a Bhutani student, I am very happy to see that the Prime Minister of India is here to visit our country. I think that we should be all happy since the Prime Minister is here to visit our country and he is very happy to be here. I am Varun Amar and I work here. We saw the Prime Minister. We were very excited to see him and he came to meet us. He invited us and saw the dance. It was a really good feeling and we were really honoured to see him so closely. Thanks to the Embassy of India and everyone Indian community for giving this opportunity to us.


Hello, my name is Simon Benjamin and this is lecture 7 in my series on Fourier series, Fourier transforms and partial differential equations. This is the penultimate lecture and we're going to wrap up our thinking about diffusion, which includes diffusion of heat and also of matter such as gas or an itinerant species of atom that can move through another material. We've been looking at a bunch of scenarios there and we're going to look at the last of those today. So I'm going to start by covering something I didn't quite manage to get to in the last lecture, which relates to the proper way to extend a Fourier series when we want to take a situation that's defined only between two points only over a region. I want to make it into a periodic function. And then I'm going to think about the scenario where you have a pulse of energy that instantaneously heats up part of the material and then that heat dissipates.  It's not really too bad, but it's it's it's kind of the toughest thing in this course, which is a scenario that will lead us to something called the error function, which is the path to get there and dealing with the error function is kind of tougher than the other elements of mathematics in the course. So yeah, I drew this little sketch to just warn you that you may need to have a strong cup of coffee before we embark on that part of the journey. Now, I've just had a look at all the material that I'd like to tell you about in this lecture and it's a lot. And I'm a bit worried that some people like getting into the details of how tricky maths works and other people don't and would just like to understand the main physical messages, the main sort of science message. So what I'm going to do is I'm going to take some of the most complex math aspects and put it into a short additional lecture that is just for the connoisseurs. And I'll make it very clear when there's some material in that additional lecture so that if you like, you can juggle backwards and forwards between them, but we'll just keep the essentials in this lecture. Otherwise, I think it would overrun or I would have to present things at super speed and I don't want to do either of those. And the first thing that I will put into the supplementary lecture is the stuff that's left over from last week about further practice at taking a finite initial condition and making it into a periodic function. So that will be the first thing in the supplementary lecture seven, right? So just before we get into some new material, let me say as I usually do that the notes for this course can be found at this location, SimonB.info. Okay, so moving on, now I'd like to think about the following situation. I want to imagine that we are going to deliver a pulse of energy to a material at time t equals zero. So at that instant t equals zero, the material, say let's say an infinitely narrow region of the material could be at one end or it could be in the middle, receives a finite amount of energy. So its energy density will be, you know, enormous, technically infinite in that instant, but then it will spread out. Now, in order to get at this problem, I want to go back to the first thing that we did when we were thinking about diffusion, which was a related problem. And that's the one that's showing on the screen here with a little bit of extra coloring that I've done since where we thought about what would happen if we blow torch the middle of a bar for a finite amount of time. So that there was a distribution of excess heat energy in a finite region of the bar or focused on, let's say, x equals zero. And then what would happen? And we guessed that maybe if it was a Gaussian distribution, so the temperature had this Gaussian shape to it, then it would keep that shape and just spread out over time. And following that guess in the best tradition of solving differential equations by guessing an answer and then just seeing if it will work, we showed that it did work. And I also mentioned back then that because of the symmetry of the problem, and in particular the fact that the hottest point at all times would be at x equals zero, heat would never flow across the x equals zero line, it would only ever flow outward. And that therefore it was kind of a still point in the problem and you could slice the bar at that point and mathematically just half of the bar on its own such as the positive x half would still conform to exactly the same mathematics. And that's all summarized here on this slide. Now what I want to think about is what happens if we deliver our energy not into a sort of Gaussian distribution or rather not a Gaussian distribution with a finite spread to it. But all in one instance, which is perhaps easier to think about in the lower one of these two examples. Maybe we've used some very high energy source like a laser to pump energy into the surface of a material. But again, mathematically, it won't matter whether we consider it to be at one end or in one plane of the material as in the upper figure. So we'll probably just keep on using that upper figure because it's just slightly more the integrals that will do things like minus infinity to infinity is just easier to remember that than always only doing half the bar. Now here was the solution that we found when we guessed a solution and then worked through it. It's quite a long expression, but the key part of it was well up here, of course, so that the form of the distribution of where the temperature was raised remained a Gaussian centered at x equals zero. And also here so that the amplitude of that heat pulse, if you want to think of it that way, was being reduced over time gradually. So as time increases, we divide by root T and the height goes down. And then you remember that the trick we used was that we said, well, actually to keep our expression reasonably neat, we'll say that time we define the start point of when we're watching to be some finite time, which then gives us a finite width to our distribution. And we just figured out what to call that first moment on the stopwatch, given that we had a characteristic width of the Gaussian that was we were writing as L init initial.  This is a solution that works generally for any x and t variables we put in. So we can certainly run time further backwards. And of course, what we would see then is that the Gaussian, as we ran time backwards from our previous initial moment t in it, our Gaussian would become more and more concentrated and would be a more and more extreme peak. And so that's it. Infectively, what what we want, right? So if we ran t all the way back to zero, our Gaussian would have zero width, it would have essentially an infinite height. So it's a bit of an artificial limit, but still that's what we want. And would it work? Well, it will because the total amount of energy in our Gaussian pulse is the same at all times. That was one of the things that we verified when we derived this solution. And I'll just check it again now because we assume that no, he could escape from the bar after the start point of our process. So it would only diffuse through the bar, but it wouldn't escape out of the bar. And therefore the total amount of energy is conserved, which means we can run time all the way back to time t equals zero. And even though in that limit, we'll get a spike, still the the integral over that which would give us a measure of how much energy would put in is still just some constant. Let's look at that now. So we could say that the energy, we won't worry about the energy and sort of the background energy in the bar. So the bar away from or if we had not injected this heat, then the bars temperature would be just theta subscript C everywhere. So that's just kind of the default energy. We're not interested in that. We're talking about the extra energy we put in. So what we can say is that, let's write it like this, the energy that we've put into the thing must be some integral along the whole scope of the bar, the whole width of the thing, times, let's say, the heat capacity. I'll use subscript L to mean the heat capacity per unit length because I don't want to start worrying about the cross section area of the bar. And then times this temperature, the transient part of the temperature distribution. So that whole thing that's above. So I'll move things around a little bit. Put the constant in front. Because it's an integral over X, time is regarded as a constant. So we can put our two root alpha T in front and then we just have an integral over our Gaussian part here. Now my claim is that that is in fact not a function of time. This is the amount of energy, excess energy that's in the bar. It's always the same. It just spreads out more and more thinly. It looks like it's a function of time because it's got time in two places. But as you can probably guess, if we just change variable, we'll get rid of that. So let's do that. So there I've taken the opportunity to take the heat capacity and bring it out in front and you can see there's no longer a time coordinate in the problem. We were previously going between a negative infinity and positive infinity for X and with our scale factor in front for any finite time, and we'll be able to see how the T equals zero is a limit of this. The limits are still minus infinity and plus infinity. And conveniently our one over root alpha T was exactly what we needed to bring into the integral to allow us to change variable. So now in this expression we see that it is a constant that doesn't depend on time and only depends on characteristics like if we take a snapshot at one moment when the peak temperature is theta H and the characteristic width is L in it, then that tells us along with the heat capacity how much energy we put in. In fact, this integral over here on the right, it now has none of our constants in it. It must be equal to just some mathematical constant. It is. It's actually equal to root pi. And I'll show you how to get that later on in the lecture, but for now I just want to rock on. So having got that expression for the energy we put into the problem, we could rewrite our original expression at the top here to focus on the fact that we're putting a certain amount of energy in. Let's do that. So there we are. We can just about fit everything we've been doing on the screen. There's the old way of thinking about things when we specified that at some time we were calling the first interesting moment there were certain characteristics in terms of the temperature difference and the characteristic width of the Gaussian. And the new way of thinking about things just focuses on the fact that we put a certain amount of energy, capital delta E in and there's a heat capacity. And then I've written the whole interesting part of the function here in brackets. And there's a particular reason why I've written it like this and even brought in the root pi. And it's the following reason. This particular thing inside brackets as we integrate from minus infinity to infinity through that, it always gives us one. We've derived it that way, but just to check. So if I'm calling that thing f of x and t just for the time being. And now I want to just instantly persuade you that the integral for over the complete range of x will always be one regardless of the value of t. Well, it's really very similar to what we just did, but let's do it quickly. Here we see all we need to do is the trick we did before, which is the same change of variables to x prime where we absorb in that x over to root alpha t. So give us then our integral prime squared that should be e to the minus x prime squared. So and I told you and we will be seeing later that that integral is equal to root pi. So the whole thing is equal to just one. Now, why is that interesting? It's interesting for the following reason. If I now would like to ask what does my temperature distribution look like as I wind t all the way back to zero. As I concentrate all the energy into one infinitely thin slice or region, then I can. Well, this is still the correct expression, but now we need to take the limit of it as time goes to zero. And it turns out that there's a well known function that is exactly the thing we want. We want a function with the following properties. So at time t equals zero. Of course, we've always got our boring background theta c temperature, which is throughout the entire bar. But now I want this, I want a function here that I'll give it its symbol and then we'll discuss it. So there's a delta symbol, lowercase delta and x. And what properties do I want from that? Well, I would like it to be equal to infinity at x is equal to zero because I put all the energy at x equals zero. But I'd like it to be equal to zero for x anywhere else positive or negative. And crucially, I would like that the integral of my function delta x dx over the complete range, but crucially over that zero point is actually equal to one. So even though my function goes to infinity, it's only doing that in an infinitely narrow region. And so when I integrate across it, I just get some constant. Now that thing is what's called the Dirac delta function. It's very, very interesting. Yes, and what I meant to say is it's very, very useful in mathematical physics in analysis of problems. It's a very handy thing to be able to use as a tool. Here, we found that the Dirac delta function is exactly how we can describe our heat distribution as time goes to zero. How does that help us? First, let's make sure we really understand what we're saying. So here, just one more time is the difference between what we're doing now and what we did last time we investigated this kind of thing. At the top is when we said we'll start the clock at a particular time when there's a Gaussian of finite width. And then we'll see what happens after that. And then we had, of course, our characteristics, which was what is the width at that particular time we're calling the start time. What is the peak temperature? And of course, it drops to the lower temperature elsewhere. Now we're winding the clock back to time t equals zero when we're saying that's actually what I now want to focus on. That initial moment when all the energy is concentrated. So I sort of drawn the diagram at some very, very small time, more than zero, because I've sort of indicated here as ever so, oops, and ever such a slight sort of curve to this thing, but really the spike is infinitely tall and infinitely narrow at exactly time t equals zero. And here is our function again. So we just have the amount of energy we've dumped in and divided by the specific heat capacity per unit length. So that makes sense to turn the energy into a temperature and then our direct delta function, which will actually boost that up to infinity at x equals zero, but zero everywhere else. So everywhere else, the bar is just at the default temperature, e to c. Now the reason to do this is we can actually solve that initial condition, even though it's sort of taken strictly, it's unphysical because we put infinite energy in there. But we can understand that this is an approximation to a scenario where we have put a lot of energy into a very, very narrow region. Mathematically, we can tackle this as it turns out. And the way we'll tackle it is with a Fourier series. Excuse me, a Fourier transform. We can't use a Fourier series. Why not? It's because our problem doesn't have either of the two conditions we need for it to be a Fourier series problem. It's not periodic. We just have a bar here and we've put some heat into one point and it's not defined over just a finite range. This is an infinite bar, or if we do the half problem, then a semi-infinite bar. So we can't use either of our Fourier series tricks and we have to go for the Fourier transform. But fortunately, this is a fairly easy use of the Fourier transform. So here's the plan. What we're going to do is try to figure out how to use the Fourier transform representation to describe our initial condition. That will be a way to show how an infinite continuous range of frequencies can build our initial condition. And then we'll hope that we can use the same tricks we've been using for the Fourier series to upgrade that initial condition that is only for time t equals 0 Instantly upgrade it to the full solution for all time. But first we need to describe the initial condition properly as a Fourier transform. So the only interesting part of the initial condition is certainly not this offset temperature. We can put that back in at the end, nor this constant, which isn't going to do anything. We can put that back in at the end. What we care about, of course, is this delta function. That's the interesting thing, this spike. How can we describe that? So what we're asking is really this. Our Fourier transform is defined as being the integral from minus infinity to infinity of some weighting function f of k e to the plus i kx dk. So the integral sweeps over all possible frequencies, and the weighting function tells us how much of each frequency we need. And then we also know what that weighting function is, or at least how to start out. We start out by writing that it is the integral of the function we're trying to build, and then minus i kx dx. So it looks over the whole range of the function and figures out how much of each frequency it needs in that way. Now it turns out that when you put the Dirac delta function into an integral, it just selects the value of the integral at 0. Very interesting. So what we're going to find here is that this is the easiest Fourier transform we've ever done. It's just going to be equal to e to the i minus i k times 0. That's it. In other words, it's just 1. So let me write that again just generally. So if you're working with the Dirac delta function, and you have any function at all, use g, because we're already talking about f of x. It just gives you back the value of the integrand at x is equal to 0. So that's what we're using there. And that's made it, as I say, extremely easy Fourier transform. So what we're saying then is that yes, we can build our initial condition, our initial spike, and put it in as an integral form, and it's this. It's just the integral of all possible frequencies. Excuse me, I was about to write dx. There we are. And now does that make sense? How can that be an integral representation of the thing we want? So we've written it down, but does it make sense? Can we just check? Well, it's always difficult to stare at a complex exponential, I think. Unfortunately, we don't have to here, although it's good to work with the complex exponential. It's very compact. But we can write this as just the exponential over the cos and sine parts of this e to the ikx. But then notice that the sine part will have to be 0, because it's an odd function integrated over symmetric limits. So we would just get cos of kx dx. Now, what does it mean when x is equal to 0? Then we are integrating over cos of 0, which is just 1, from minus infinity to infinity. So that will indeed give us the infinite value at x is equal to 0 that we want. And how about when x is not equal to 0? Well, that's difficult to sort of picture in one's mind, but we can at least see that if x is just equal to some value like 1.3, as we consider all different possible k's in this integral, we'll actually get all different possible values that cos can return, ranging from plus 1 to minus 1, and forever oscillating backwards and forwards between that, as we consider different k values. So it's perhaps, this hand-wavy word argument is far from approved, but it's perhaps intuitive to see that we get a kind of interference or cancellation everywhere except x is equal to 0. But that's by the by, we've shown that this expression is the correct one to use, and now we can ask, how does that help us with our heat flow problem? Well now, it may be helpful to remember what we would have done if we were just dealing with a Fourier series at this sort of moment. If we were dealing with a Fourier series, we would have had our temperature t equals 0 limit, or some interesting part of it, having ignored constants and so on. And it would be equal to some sum over our sines and causes. So let's say it would be an cos 2n pi over lx, something like that.  So this is where we have to decide is it a heat diffusion or a matter diffusion, because we use a symbol alpha for one and d for the other. This is a heat diffusion problem, so we should write minus alpha k squared, so that whole thing in brackets squared t, so that thing there. And as long as we do that, we've instantly upgraded our Fourier series. Here it was a cos-based one, but it could have cos n sign, we just follow the same rule. We instantly upgraded our series that describes the boundary condition to now be a series that describes the full time-dependent problem. So in other words, where we see a cos or a sign, or indeed an e to the i theta, so complex, exponential, any of these functions, all we need to do is throw in the appropriate damping term, the appropriate time term, and we've completed our process. Can we do that here? Well, instead of a sum, we've got ourselves an integral. But the answer is yes, we can do exactly the same thing. So if I want to now upgrade this object, if I want to now upgrade this from just the x to finite time, all I need to do is throw in that appropriate... Let's do it for cos first. Cos kx, it needs to go along with minus alpha k squared t, and I just do my integral. Excuse me, that looked a bit wrong. The integral, of course, is over k, because we're building it by sweeping over all possible frequencies. Just in the same way that this worked for a series, it also works for the integral, which we can think of as simply the limit of a series, as we had a whole lecture about this, where we would take our series to have a longer and longer cycle length, until eventually we ended up with a continuously varying frequency. So because of that, because the integral is just the limit of the series, we can still do this. We can put in our time dependence in this fashion, and we're done. So that is the solution. All we would need to do is remember what this f object was, where it was just the way we were writing our Dirac delta function. We need to go back and substitute in to our expression. Let's do that. So there we are. I've underlined that with a dashed line, because that is the solution. In a sense, we're done, and we could just walk away at this point. But it isn't the same kind of solution we've obtained before, where we're able to actually do the integral and leave it in a non-integral form, or a non-integral form. So can we do that in this case, or do we just have to stop? No, we can do it. So in fact, we saw that the causeway of writing things and the complex exponential are just the same. It's helpful to switch back to using the complex exponential in order to work on this a bit more and trying to get it into the final nicest form. So let's do that. Actually, I've realized the neat thing to do is first to change variables, and then to complete the square. So first we'll just change to, we'll call it k' just to absorb in some of the, just to rearrange things and make it look a bit neater. Okay, there's step one. I've just used a rescaling of our variable, and that's meant that I've needed to have a term out in front there in order to provide the constant I need to complete that change. Still I'm integrating from minus infinity to infinity, and now the messy part of the integral has appeared up here. So I would like to neaten it up a bit more, so let me write one more line out where I'll introduce a new symbol for that. Okay, that's got it looking about as neat as I can make it, I think. I've used a symbol gamma, which is x over alpha root t. That's the same ratio, by the way, of x to square root of t that we're seeing coming up again and again in these heat problems. And it's what I need to put in in order to make my integral as neat as possible. And now it is time to complete the square. I think I, because the k' is just a dummy variable and I can use anything at all, I'm now going to drop the prime because it's just messy to keep on writing primes. I hope that doesn't cause any confusion. Now let's do this complete the square business. What I've written down here is that I've, I note that because we've got two exponentials, we can just add the terms in the exponent of course. But I could write it even more neatly as the square of something up there in the exponent if I had one further term that I don't have, which would be, well, it would be the missing bit. There would be what I get if I squared the gamma, which would be minus, and then minus one from the i squared, gamma squared over four. So if I had that extra term, I would now be able to write the exponent in the form I have. But I can always give myself that extra term because it's just a constant by multiplying by some appropriate external factor. So here where I've written a question mark, we can see that what we need to put in. Is exactly the same thing as I've written here, but with a minus sign. So that would then, in order to cancel that, I would need to put that e to the plus gamma squared over four. And that's exactly the term that I need in order to legitimately rewrite the exponent with that squared form. So maybe just take a stare at that if that's a trick that you haven't seen before. But it's a good trick because now, of course, what I'm going to do is change variables once again, and hopefully now the integral will look very simple. Well, does it look simple? I've used a, I don't think we've used the symbol why yet. I've introduced that. It's just a shift to our variable K, but it's weird because it's a shift by adding an imaginary part. Now that makes, that's fine mathematically, I can still replace the dK is now just equal to dy because the constant goes away regardless that it's imaginary. It's still a constant and it will just go away. And that all seems very nice except when I write down the limits of the integral. Now it's going to look a little bit peculiar because it's going to be from minus infinity minus i gamma over two. Two plus infinity minus i gamma over two. Because when I put in K is equal to plus or minus infinity, I now have those shifts and that looks like an integral that maybe most people watching this won't have seen before. A strange integral. We normally only ever integrate over the real axis. What does it mean to integrate over the, with this sort of shift into the imaginary space? Well, what I'm going to do is I'm going to just tell you that in this case it doesn't matter and we can just drop those terms. But in a moment I'm going to justify why, but I know that some people maybe don't care and they'll be able to skip it. So let me just power on for now and say that I can just drop those little imaginary shifts to the limits. So there I've written it out again without the funky shifts in and we can see that that must just be equal to some quantity. In fact, we've met this before and I asserted that it was root pi. I'm going to assert that again, but again, momentarily I will justify that for those who are keen to feel that they've grasped at every step. But that will then give us one over root alpha t and then we're going to have a root pi, excuse me, a root pi on top, e to the minus m over 4. There we are. So we've done that integral. How has that helped us? It helps us because it allows us to go back and write our solution in a eta form. Let's remind ourselves why we were even doing this integral. We were doing it because it was this part, let's change colour, this part of our solution. And we wondered whether we could write it as a completed integral rather than just leaving it in integral form. Well, now we can. So I will now write out this line but with the solution we found in place. So there it is as just a direct substitution, but now let's tidy it up and put back in what that gamma-squared quantity is equal to. And there we are, just substituting back in and neatening up slightly. That is exactly the expression we found before by taking the limit of our former guest solution, the solution we guessed and filled in, winding it back in time so that all the energy concentrated to one location and then writing it down. That's what we found before and now we've derived it from that initial starting point of all the energy at one value of x but using the Fourier transform. So the point is this time we didn't have to guess. This time we just took that initial condition and systematically took the Fourier transform, put in the time factor, found that we could do that integral and we come to this solution. So we don't need to guess, we can just solve these things directly using the Fourier transform. Now in order to get to this nice solution I did take two very quick steps over some interesting stuff. Namely, there was the stuff about why is it okay to ignore these imaginary shifts to the axis of integration and also how do we work out what this integral here of the Gaussian is. Those things I think are good choices to put in the little supplementary lecture. So go and have a look there, that will be the second topic of that lecture if you're interested in knowing them and by the way they're really interesting, especially the one about coming off of the real axis. That hints at something called contour integration which is more or less my favorite thing that I learned in maths at university but it's not essential for understanding diffusion and getting the main physical message so also don't feel bad about skipping it. All right now the very last problem that we're going to meet while thinking about diffusion and we get to have some fun with some wave topics next in the next lecture is the following. The very last one seems like a very innocent one, a very simple one, but it isn't. So here is the story. We have a reservoir of some material. So again by the way this can be thought of either as a heat problem or as a matter diffusion problem as with all the things we've looked at but since we were just talking about heat of energy coming in now we'll talk about matter diffusion. What I'm about to describe comes up in material science for example in the carburization of steel where you want carbon to penetrate into steel a certain depth in order to alter the surface properties. So you can expose a piece of metal to a gas essentially and you need to know the conditions in which to do this so that you will get the right amount of impurity going into your metal to achieve the desired results. So it's very interesting what we're about to describe and as I say seems very simple. So we have a reservoir and it has a fixed concentration of some material. We could think of this as carbon so it has a fixed amount of carbon in the reservoir. That will never diminish its permanent and we call that in this problem C for concentration subscript s and then we have a block of material and it has its own concentration of the substance in question which here we've written as C in it throughout the entire block. Now the thing about this block of material is it goes on forever. So it starts at x is equal to zero and we can imagine it's a three-dimensional block so then we have an x-y plane but we only need to think of it in one dimension so the block is so big that the fact that it eventually has limits in the y and z direction doesn't matter to the physics of what goes on near the surface in near the origin of our coordinate system. So essentially it is a one-dimensional problem but the block has no end to it. It's essentially infinitely deep. In any real problem of course your piece of metal is not going to be infinitely deep but also you're only going to be penetrating into it a very shallow depth so in many realistic situations the block is effectively infinite in that no particles of carbon will permeate all the way through the material and get to the other side so it may as well be infinite and it is in the mathematically abstracted problem. Now we can assume that the density of the material in the reservoir is higher than the natural or initial density inside the material so what will happen is yes material will come in from the reservoir and it will enrich the concentration within the block of material but the interesting twist is it will never stop doing that. Material will keep coming in from the reservoir forever and the block is infinitely deep so it will never reach a steady state it will just keep on taking in more and more of this material. Now to tackle this problem we're going to be using I think some of the tricks we've already used in this lecture a couple of other ones and I would say that this problem is the most advanced one in the entire course. It's the first time I think that many viewers of this may end up meeting an integral that cannot be done. So maybe make yourself a strong cup of coffee or go to the window and take a last look out at the sunlit world and then come back here and we'll face this challenge of an integral that cannot be done. Maybe if there are any young people in the room ask them to leave so that they don't have to face the reality of this thing that's called the error function which we will get to quite quickly because of all the techniques that we've learned. Let's go. First I'm going to sketch the initial distribution of material and that's always how we start these things out by describing the time t equals 0 condition. There it is, pretty simple. The high concentration Cs in the reservoir region the low concentration C in it in the block region all at time t equals 0. So mathematically how would we write that down? Like this. As usual we want to remove the static and uninteresting parts of the problem and just put them back in later and focus our mathematical analysis on the simplest thing we can taking all constants and scaling out. So I've introduced some function S of x and t to capture what the interesting step-like behavior and the rest of the constants, well, we have the concentrations and the difference in concentration here. So what do I need this S function to be like? I want it to be equal to plus 1 where x is greater than 0 because if I put in 1 for the S value it will remove, the Cs will be taken away and I'll just get C in it, that works. I want it to be equal to 0 for x is equal to 0 because then I will get that at the surface of the block I have this higher concentration so I'm taking it that because the surface is in direct contact with the reservoir it even at time t equals 0 has the high concentration. Just because it will be mathematically convenient I'm going to choose minus 1 for x is less than equal to 0. Now that's actually, in a sense, wrong because that would give me the wrong concentration in the reservoir region but I'm doing this because I don't care I'm not trying to describe the reservoir region whatever the math says about the reservoir region I override that because I know that it's just fixed my challenge is to describe mathematically what's going on inside the block so it's only the first two conditions that matter from the point of view of getting it right getting it correct inside the block my choice for the third of these is just a mathematical convenience that will make my analysis a bit easier so there's a trick there this is my scaling function at time t equals 0 and I'll give it its own name I'll call it rx so rx is this boundary condition and it's a pretty simple one there we are so rx simply switches from minus 1 to plus 1 at x is equal to 0 a very nice simple function what I want to do as in the previous example we looked at is to describe this initial condition using well I can't use Fourier series so I'll need to use a Fourier transform but hopefully it won't be a difficult one to work out let's see well that's just the definition of a Fourier transform I'm hoping that I will be able to write rx as an integral over all the possible frequencies with a weighting function capital F of k that captures this particular case and I know that F of k oops I apologize I see that I put a dx up there but of course it's dk because when we write the function in terms of being built as a bunch of frequencies with a weighting function we must integrate over the frequency the spatial frequency but then the weighting function itself is as written down here as written down here our standard format but now we have rather a nice function here it's only either plus 1 or minus 1 but that may still give us a bit of a trick to integrate let's see what it looks like well it's simple enough to write down but it may still be a challenge for us I can tell you that these two integrals actually come to the same thing and you can convince yourself by change variables that they actually have to be the same thing so I'll choose the one that's over the range 0 to positive infinity looks tricky I could do the integral as it stands but then the when I feed in the limits it won't make sense so what I need to do is again a trick that I don't know if the viewers of this on the whole world have met before it's another very powerful trick we're using a lot of powerful tricks a lot of heavy weaponry in this lecture we're going to use what's called an integrating factor so we're going to say that this is the limit of well I'll write it and then I'll explain it so I'm saying that the integral I want to do is the limit as delta which is now just some symbol I'm running out of symbols delta goes to 0 of this other integral so you can see that if I just substitute in 0 and I've made a sign slip, there we go now if I substitute in 0 I get my original integral and I can do this smoothly kind of sudden jump or discontinuity at 0 but now the point is the more complicated looking integral with the extra delta symbol I can do let's do it there we are the complex number delta plus ik is just some constant so I can just do that integral and now it will make sense when I put in the limits because so I put in the 0 limit where the e to the minus some constant times x is just 1 the infinite limit is now just going to give me a 0 because that delta as a finite I should have said positive so delta is a positive number but it comes down to 0 approaches 0 from the positive direction we would say and therefore when we put in x is infinite then that will give a 0 limit so now we can just substitute in delta is equal to 0 and find out what we've got there we are so it's simply 1 over ik or minus i over k so the answer to the integral is a simple expression but we needed that integrating factor in order to be able to work it out but we're nearly done I mentioned that both the integrals up here come to the same thing and so our weighting function f of k we can now write down done and having found the weighting function we can go back and write the integral integral form of our original function r of x there we are we've just substituted that straight in it doesn't look too bad and actually it can be made to look nicer we can cancel the tours of course but the interesting thing here is that anytime we want to we can expand our e to the ikx as cos and sin parts the cos part is an even function and the sin part is an odd function so integrating between symmetric limits the 1 over k is an odd function so in fact this means that only the sin part will survive that integral let me show you what I mean we are just eliminating the cos term and noticing that we have two i's i squared is minus 1 so that will get rid of the minus in front and give us a very very neat expression finally for our rx we've discovered that we haven't made a slip that it's just sin kx over k integrated over all k we could also write that from 0 to pi and double it if we want I'm not sure which is the nicer form but that's a very very elegant and compact expression and I remind you that what that's supposed to be is this switching function that just goes from minus 1 to plus 1 at x is equal to 0 difficult to tell just by looking at the thing so a good opportunity to just test because if I've got that wrong everything else is going to go wrong so let's just check is that a legitimate way to write down r of x well there we are I've typed it in I think that's what we we were expecting 1 over pi and then I've used mathematics as numerical integration so that it will do an approximate integral from k minus 100 to 100 I reckon that's going to be enough for us to see whether essentially we got the maths right or not and so we're now going to plot that thing from x is minus 5 to 5 let's execute that well it's taking a minute and there we are let me just make it a little smaller so it's perhaps a little easier to look at there we are so is that a function that switches from minus 1 at x is equal to 0 yes it is does it have some little ripples on it which are essentially this kind of Gibbs phenomenon thing that we've been seeing again and again wherever we take a Fourier series or now a Fourier transform and limit its range of course really it should be an infinite range well we see that but we're not surprised we know that those little ripples get infinitely thin as we increase the range and so they don't cause any practical difficulty to any calculation the point is this shows that I haven't made a slip or indicates that I haven't made a slip in terms of a pre-factor being wrong my 1 over pi factor was correct we built it so yes this is a way to build our switching function writing it as an integral how does that help let's go back and finish up good with that quick check we can say yes that builds the rx function that we wanted I remind you that that is the time t equals 0 limit of our setup and we know how to go from the time t equals 0 solution once it's in a cause or sign or complex exponential type form we know how to make it instantly the full time-dependent solution we just need to stick in that time factor and we can do it even into an integral doesn't have to be a sum so let's write that out we wrote that the sx of t the function that would catch all the interesting stuff of the problem as a full function of time was well it was rx was just the time t equals 0 limit but now if we put that back in the finite time we're going to have sine of kx over k and now e to the minus well it's a matter diffusion problem so I should use the letter d but then k squared t that's it that's the solution so we've managed to write down a mathematical expression for this function s which really captures all the interesting part of the spatial and temporal evolution of our problem we could then feed this into a computer and just get values out of it in a sense we've solved the problem but it's still written as an integral and we'd like to explore a bit more about whether we can get it into a more compact form that is actually a very non-trivial question and the sort of struggle to get it into a more compact form I think best belongs in the supplementary lecture as the I guess the final topic there what I'll write down now is the key sort of headlines of that struggle so what I've pasted in here is I think the only remarks I want to make in the main lecture here which is that we can change variables and always introduce symbols to collect up constants that kind of thing so what I have done with this function is I've changed variables here to this curvaceous n symbol is eta and so we can change to use that as our dummy variable but in that way absorb a root dt into the maths and we can also collect this suspect we've seen very often this x over root the constant times t just into a single symbol and that gets us a pretty compact form of our integral but it's still written as an integral let's see what the solution would look like at this point written fully there we are so if we wanted to give up that we would leave it there but what I will talk about in the supplementary lecture is that we can go a bit further let me write down the big result that is the sort of another 5 or 10 minutes of discussion there we are a classic it can be shown but I will show it in the supplementary lecture the bottom line is we cannot escape from writing the solution in a fashion that needs an integral to be present but we can quite fundamentally change the nature of the integral instead of the interesting stuff being so this beta being inside the integral hiding inside the sign there it can actually be in the limit of the integral that's the same thing you can see there that's beta over 2 so that's the quite sophisticated transformation we can do and that does lead to a nicer final expression but still not the kind of closed solution that we're used to so let's write that out now as the full solution in these terms right so this line was how we originally introduced our function s and now we've managed to figure out as best as we can what that s object is let's have a hard look at it and see what we're dealing with here quite interesting because for the first time we've ended up with a function whose x and t dependence is in the integral as a limit of the integral so that is an unusual state of affairs and the other thing is we can't do this integral we could have done the integral if it was from 0 to infinity that was one of the tricks we played earlier but this is an integral of the Gaussian function a very simple bell it's the bell shaped function but we're asked here to integrate to a certain point and there is no closed form for that so this is an integral that we can't go any further with it's a strange situation I suppose since pretty much everything you meet until this point you can always do an integral if you're clever enough and turn it into the standard functions that we use signs, causes, exponentials logarithms and so forth here we can't complete this integral the best we can do is give it a name and in fact that's what's been done and it is called as promised the error function so let me define the error function there it is that is the error function a much discussed and important function in mathematics in statistics probability and so on and it's simply written like that so the error function is the error function of some argument z here is the integral from 0 to z of a Gaussian so if we were to draw it we would see this so it's the area clearly as z here increases we sweep out more and more of the area under this Gaussian as z goes to infinity what we would find is that the area goes to 2 root pi over 2 and so the error function itself because of its pre-factor just goes to 1 so the error function goes between 0 and 1 and that's it really that is the first time I imagine that you may have met the error function and it's come up for us in this solution of how a material will permeate into a block let's have a quick look at what this is saying about how that process of for example carburization of steel happens we can write out our solution to our diffusion problem now using the error function which will make it very straight forward to put it into a maths program and see what happens so first off let's have a look what does the error function actually look like well let's scale it down but as we said it must being that integral underneath the Gaussian it's equal to 0 when x is equal to 0 and it tends to 1 as x increases so pretty simple looking function now what does it mean for our diffusion problem right I've brought our solution into Mathematica to have a look at it I've made up some numbers I've said that the high concentration in the reservoir is 0.8 the initial value is 0.3 I've put in a value for the diffusion coefficient using double D because Mathematica won't let me it reserves D for another thing and there we are let's I've already typed in what happens for an extremely small time instance so essentially time t equals 0 and there we see exactly what we wanted which is that the very surface of our block the concentration is at the high value I should do it this way for you at the very surface of the block our concentration is at the high value but then it's at the lower value everywhere else so now let's run time forward a bit right not much happening yet but let's keep running time okay now we see a little bit I don't know if it will even show up for you let's go five times further so now we see a little bit of material has diffused into the block of course away from that interface significantly away from the interface we're still at the initial condition but let's boost up to a larger amount of time and we see a slow process as I increase by a factor of five so now we can clearly see that we've permeated into the material a bit but it's still quite a abruptly shutting down let's go for a much larger amount of time and we see that it's a slow process so that's what we expect because we have a square root of time entering into our solution so we should expect that in order to double how far we've penetrated into the material so that the shape of the curve looks roughly the same but it's gone twice as far along the x axis we should expect to have to put in four times the time so let's see if that works at the moment I would say that it's dropped down significantly by x is equal to a half and almost back to the initial at x is equal to one let's see if we can make it go twice as far by doubling excuse me by multiplying by four the time so that the square root will double and there we are I think that's exactly what we want that way what we expected to see right so it's a slow process and the rate at which the material propagates in as in if you were looking for a particular concentration and you tracked how fast that concentration point moved into the material it would be slower and slower but never completely halting because it's square root of time well there we are so that's the correct solution and it's a very compact expression because the error function comes up so often that it has its own it has its own symbol so that's it we finished at last we finished looking at diffusion problems and we've looked at essentially all of them we've looked at periodic problems which we related either to a series of pipes that would open or to a stack of material that would melt we've looked at problems where the initial distribution is a Gaussian and we've looked at problems where there's a pulse of energy or material that puts all the concentration at one point and now we've looked at a block an infinite block of material in contact with a reservoir and by the way that last one the one we've just looked at is isomorphic to the problem of two blocks two infinite blocks being brought together at the x equals zero and melting into one another you can have a think about how you would need to adjust the master maths to reflect that and you'll find that it's just a subtle change to the initial condition that when scaled means our solution immediately works so we've done it all it feels like it too so that we've also met in this lecture some pretty powerful techniques that we needed to use in order to solve some hairy integrals that came up now the next lecture is on waves a topic I really enjoy looking at and we'll see whether the long-promised analysis of a guitar string yields something that is physically correct or not but thanks for listening


All right, the chair becomes very tough, so I have to be. All right, this is the Ethiopian experience. Well, I'm from United Nations University, Merit, from Maastricht. Well, there is revival of interest in industrial policy, where we are all gathering here. But controversies still remain on the functional versus selective comparative advantage following the nature of state and business relationship. There is also renewed interest in industrial policy, making in Africa particularly, but there is little evidence so far, particularly after the reintroduction of industrial policies in recent years. So I am here to examine the choice, implementation process and outcome of the Ethiopian industrial policy. Well, it is one of the few African countries that have formulated and implemented full-fledged industrial policy even early in the 2000. Government has shown extraordinary commitment and ownership implemented through subsequent development plans and various sub-sector strategies. Well, the outcomes so far appear to be a bit mixed. I will see in detail what does that mean. I give you some numbers here. Well, the average growth of Ethiopia in comparison to the sub-Saharan Africa average, it's more or less more than double in terms of GDP growth in the last decade. Industry also grew by 10% and the manufacturing sector by 9.4%, particularly after the implementation of the industrial policy between 2004 and 2011. If you compare to the sub-Saharan Africa, this is more than double. But we lag behind in terms of level. So this is a country in catching up context even with the sub-Saharan average. So it is half of the industry value added into GDP in comparison to sub-Saharan Africa. Also the manufacturing value added contribution to GDP and as well as manufacturing exports as percentage of merchandise exports, it's almost like one-third of the African average. So we have to do a lot even if there is promising developments. In Ethiopia, manufacturing sector emerges in the 1920s. For example, in 27, there were 25 manufacturing factories, mostly established by foreigners. The sector starts to get momentum in the 50s. This was also a period that marks the start of a conscious plan to promote industrial development and the economy at large. Ethiopia has seen three regimes over the last eight decades, the imperial regime up to 74, the derg regime, 74, 91, and the aperitif regime since 91. So these successive governments had their own industrial policies. I wouldn't go into detail, but let me show you this table. The imperial regime, the industrial policy was market-friendly, market-oriented, private-led, but dominance of foreign-owned enterprises. Target industries were import substituting and labor-intensive industries where envisaged key players were foreign investment. Well, the policy instruments, as you see, protection of domestic market and at the same time provision of economic incentives. The derg regime period was a common economy. Every medium and large enterprise were nationalized. This were also reorganized under state-owned enterprises. So there was dominance of public-owned enterprises. But the whole strategy of industrial development of Ethiopia at that time was import substituting and also labor-intensive, at the same time basic industries were favored. The key player was the public sector at that time. There were also protection and financial subsidy to the public sector. The aperitif regime post 1992, market-oriented, private-led, but also the involvement of the state, well, strong state role, dominance of domestic-private. There is a change from import substituting to export-oriented. Well, the leading engine of the private, I mean, the manufacturing sector would be the domestic-private sector. There are also some policy instruments, direct support and provision of economic incentives. What you can see from this graph that particularly we have data from 1981 until 2011, you see that the regime period were declining period until 1991. After the government change, there was some improvements. But particularly the last decade was on average, 10 or more grows in GDP, industry value added and manufacturing value added. When we see the entry process, you can see in 80, there were only 351 number of establishments that employee 10 or more employees that go down at the end of the DERG regime to 75. After that, it improved and in 2010, it reached 2172. This is like more than tripled in the last 10 years. But the public share has declined as you see. Let me give you the ideas. I'm in the present industrial development strategy, major principles. Well, the primary principle is the linkage between industry and agriculture because this is based on the philosophy of agricultural developmental-led industrialization. Other principles, export-oriented, labor-intensive and public-private partnership. Mechanism of engagement, well, creating conducive environment and also direct support for selected sectors. Here are the main ways the government sees to create conducive business environment. When we look at the macro stability, well, government envisaged single-digit inflation, but there was a high record of inflation since 2005-2006. It was also envisaged low interest rate, but virtually it was negative because of high inflation but controlled interest rate. The real effective exchange rate also appreciated due to high domestic inflation. So there was unexpected developments in terms of macroeconomic. There was also institutional regulatory reforms in this period. Well, competition policy, business registration, investment code, et cetera, et cetera, but the main initiative at this period was the civil service reform program and that has improved service delivery substantially, particularly for the private sector. Well, if you see this doing business indicator, they have improvement up to 2010, but declined from 2011. What happened? There was some reversal of reforms and ranking. In the post-2010, went for an exchange shortage, uphearest government shut down 90 for warehouse of coffee exporters when inflation source makes some futile efforts using price cap to control prices and also it revised the business registration to reduce the monopoly power of wholesalers, particularly importers, et cetera, et cetera. And starting 2011, private banks are forced to purchase NB bonds, national bank bonds to finance the government huge investment, public investment. So these are reduced in some way the confidence and policy predictability. That's why we see some declining here. Sectoral policies, well, textile, meat, other agro-industry construction and the micro-industrial enterprise are the priority sectors, but the list of priority sectors has been updated through time. Well, there are some economic incentives that are also direct supports. I don't have time to go to all these details, but I have here compared the three sectors, three industries particularly, the textile and the leather industry and the flour industry. The flour industry was a sector that emerged by itself, but the leather and the textile industry was in the priority list from the beginning and their performance you can see. The flour industry, it emerges itself, it reaches like 189 USD in 2011 for an exchange, but textile, it's only 71, but it was envisaged to generate 500 million USD. But of course, this is like seven-fold in comparison to the base year in 2004, 2005, but in terms of target, it is shortfall. It is the same also in the leather industry. Let me go through to the emerging and remaining issues. Where are the goals yet achieved, that is mixed outcome, GDP and all other sub-sectors grew 10% or more, but industry contribution to GDP remains stagnant still less than 14%. The country more than double its country export earnings, but diversification mainly outside manufacturing. High entry of firms into manufacturing sector, but new entrants are mostly small, there is dual structure continuing. The new entrants are also mostly domestic oriented, export sectors are underperforming, so manufacturing contribution to merchandise export remain low, even by sub-saharan Africa standard. There is increasing import dependency and weak domestic linkage, so exports increasingly constrained by lack of quality of inputs in domestic markets, despite inferences to address them. So a regular review of policies and instruments need to be instituted with the aim of identifying emerging bottlenecks, but of course the policies have to look the whole value chain and at the same time horizontal linkage. Second, there is an issue of choice of champion or activities or products. Why has the flower industry that has been successful but not the textile and laser industry that received the most attention long before the flower industry? There are three alternative views regarding identification of potential products whether government should choose or let it for the market whether it's combined. Well, data and experience supports the third view, but one need to introduce mechanism to elicit valuable information from the private sector on the potential industry through continuous consultation. This also involves continuous experimentation. Lastly, on the nature of public-private partnership, the European Industrial Policy made distinction between developmental and rent-seeking private sector to promote the former and to curtail the latter. Government provides generous incentives and support programs to build the private sector capacity, the carots. At times, particularly recently, it has introduced a number of measures, sticks, alleging to discipline the reg private sector, increasing tension and policy uncertainties around the private sector. Yet a number issues arise regarding the effectiveness of instruments. How much rent and how long should the private sector in the select sectors be given to bear fruits? What form of relationship should be instituted between the government and the private sector? How do you create an environment that maximizes the social benefits and limit rent-seeking? There are critics that the instruments, that means the carots and sticks are not transparent and policymakers tend to patronize the private sector instead of encouraging competition and innovation. There is also emerging concerns that the public investment expansion is dwarfing the private sector. For example, credit for an exchange availability, et cetera. So the public investment rate, I mean, a recent World Bank report that maybe last month says that the public investment rate in Ethiopia is the third highest in the world, while the private investment rate is the sixth lowest in the world. So a lot of money or finance is going to the public sector like drying the source of finance for the private sector. So vibrant private sector is critical for effectiveness of industrial policy. Therefore, the government need to address the above and other emerging issue. Thank you.