0:00

The last topic that I'm going to talk about in

Â this module is that of optimization.

Â Now earlier on I had talked a little bit about

Â optimization when we have linear functions and linear constraints.

Â That was known as linear programming.

Â Now, I'm going to show you

Â some optimization in what I would term a more classical sense.

Â So, I'm going to use the mathematical technique calculus now

Â to help solve an optimization problem.

Â And remember, I had said previously that one of the key uses of quantitative models

Â is as inputs to optimization decisions.

Â So, businesses are always trying to optimize their performance in some way.

Â So optimization problems can sometimes be solved using a calculus approach,

Â and that's what I'm going to show you.

Â 0:52

So let's go back to the model that we had used to

Â understand the relationship between the price of a product and

Â the quantity demanded, and ask ourselves the question,

Â can we find the optimal price in order to maximize our profits?

Â Now, to do that, I've gotta put a little bit of a notation in place,

Â so that's going to happen on this slide.

Â So let's consider the model we had looked at before.

Â We call it demand model, where the quantity demanded of a product,

Â and here's the model is = 60,000 p to the minus 2.5.

Â So, I'm presenting you with that model.

Â Now, if you're sitting there thinking like where does he get a model like that from?

Â Well, that question actually isn't what I'm trying to do

Â right now in this particular module.

Â I'm going to talk about where do these models come from

Â when I talk about regression in one of the other modules.

Â So even though this looks like it's been pulled out of thin air,

Â there is a basis for creating these models that we'll discuss, but for right now,

Â I just want to show you this deterministic model.

Â So let's say our model for demand is quantity = 60,000 price to -2.5.

Â That shows me how quantity is related to price.

Â Further, I'm going to assume that the price of production is a constant,

Â act two, $2 for each unit.

Â 3:07

And revenue can be written as the price that you sell

Â the object at times the quantity that you sell.

Â So if you're selling candy bars, they cost $2 for someone to purchase,

Â I mean the price is $2 and you sell ten of them, then your revenue is $20, right?

Â Two times 10, so that's what's going on there.

Â We write that more generally as p x q, price times quantity.

Â Now, the profit is the revenue- cost.

Â So, the profit equals pq which is the revenue.

Â Now what is the cost of producing q units?

Â Each unit costs c dollars to produce, and I'm going to produce q of them.

Â So the total cost according to this model is c times q.

Â So I can simplify that equation into q(p-c).

Â So that's the profit that we're going to make.

Â But we've got a model for q in terms of price and

Â that model says quantity = 60,000 Price to the power -2.5.

Â So putting it together,

Â we can see that our profit is equal to 60,000 p to the negative 2.5,

Â that's the quantity, times (p- c) and in this particular example,

Â I'm taking the cost of production at $2.00 per unit so that's p- 2.

Â So now you're looking at an equation that has come out of

Â the quantitative model for quantity.

Â And I am now going to ask at what value of p is the profit maximized?

Â So choose profit p to maximize this equation.

Â So this is what we mean by an optimization.

Â As I said before,

Â optimization is one of the things that we tend to do with our quantitative models.

Â So how we going to do it?

Â Well there is a brute force approach to this.

Â We've got a function for profit.

Â 5:07

Let's just choose different values of price, which I'm writing as a little p,

Â and plug them into the function and see what the profit looks like.

Â And so in the table on this slide you can see I've

Â plugged in different values for price.

Â It's in the price column and I've used the equation,

Â the model to figure out what the profit is.

Â So if I charge $1.75 for

Â this product I actually don't make any profit off of it at all.

Â There's a negative profit, otherwise known as a loss, and of course, that makes

Â perfect sense, because 1.75 is less than the cost of production, which is $2.

Â If I were to price it $2, then I don't make any profit whatsoever because my

Â price is exactly equal to my cost.

Â So you get 0 for the second one and

Â then the subsequent numbers in there are just coming out of the profit equation.

Â Now if I look down through that table,

Â that optimization just corresponds to finding the biggest number in there.

Â And I've drawn a graph that shows you the profit as a function of price.

Â And you're already trying to figure out at which value of PDX axis

Â is the profit the highest.

Â Where's the top of that graph in other words?

Â So, this is a brute force approach because I haven't actually tried every value of p.

Â If I'm implementing this in a spreadsheet, spreadsheets have cells and

Â each cell you can only put in one number.

Â And so it's a discrete approach to solving this problem, and

Â it looks to me that the best value of price is somewhere sitting between 3 and

Â 4, but I don't know exactly where between 3 and 4 it is right?

Â So, this gives me a sense of where the answer is.

Â And it might be fit for you, say it might be enough for you to say, I just want to

Â set the price between 3 and 4, but optimization does give us the potential to

Â be a bit more precise about it, so that's what I'm going to do now.

Â So, the calculus approached to these problems involves

Â the mathematical technique of differentiation and

Â what we need to be able to do is to find the derivative, which means the rate

Â of change of a function, with of the profit with respect to price,

Â and we need to see where that derivative equals to zero.

Â So optimization, the actual mathematics of optimization,

Â is not the goal of this course.

Â The goal of this course is to talk about modeling, and

Â this is one of the places that models are used.

Â And so I'm not actually going to do that, I'm going to present you with the results.

Â If you're interested in calculus one you can, and

Â its use in business you can certainly find other courses that will address that.

Â So I'm just going to skip to the answer here, it turns out that

Â by applying calculus to this problem you can obtain the optimal price.

Â And the optimal price which I'll write as Popt, opt for

Â optimal, is qual to c b / 1 + b,

Â where c is the production cost and b is the exponent in the power function.

Â So with this neat little mathematical model that we had for

Â quantity demanded as a function of price I'm able to leverage that equation,

Â leverage that model, and come up with an answer to the question.

Â What's the best price to set in order to maximize my profits?

Â Now, going back to this example, c was equal to 2,

Â that was the cost of production, and b was equal to negative 2.5.

Â If I plug in those numbers to the equation, you can convince yourself that

Â the optimal value for p, for the profit, is about equal to 3.33,

Â it's already three and a third is the best value for price.

Â So that is solution to the problem.

Â And by creating, or using a simple model for the quantity for

Â the demand I'm able to end up with a simple model,

Â you can even call it a rule of thumb if you want,

Â a simple formula for pricing.

Â 9:17

Now, in terms of interpretation again, well, we know what c is,

Â that was the cost, that coefficient the minus 2.5 in the power function model

Â that we're looking at, remember this model for demand is a power function model.

Â It was 60,000 times price to the power negative 2.5.

Â That's a special quantity, the exponent b in this situation and

Â it gets called the price elasticity of demand.

Â And so, oftentimes economists will put a negative in front of that because

Â the coefficient is negative 2.5.

Â And one might say the price elasticity of demand is 2.5 for this particular product.

Â What that negative 2.5 means in terms of the business process is that as you

Â increase price by 1%,

Â you can anticipate a fall in quantity demanded of 2.5%.

Â So the coefficient relates percent change in x to percent change in y,

Â and the negative 2.5 means that as x is going up, y is going down,

Â so a one percent increase in price is associated with a 2.5%

Â fall in quantity demanded and that proportionate relationship,

Â proportional change in x to proportional change in y is true for any value of x.

Â That's what's very special about the power functions that that proportionate

Â change between x and y is a constant and in this case, it's negative 2.5.

Â And, as I say, people might call the price elasticity of demand here, 2.5.

Â So that's the calculus approach.

Â And I'll finish this off with a slide that shows you

Â what the calculus approach is doing.

Â The blue curve is the demand equation.

Â That is the curve 60,000 times price to the negative 2.5.

Â So that shows how price and quantity demanded are related.

Â Now for any value of the price, so

Â fix the price, stop the box moving, means fix the price.

Â 11:25

For any value of the price you can go up to the curve and

Â record what that point is.

Â That will define a box.

Â And that box that's the light grey shaded area

Â in the graph here actually is the profit that's associated with that price.

Â And you know it's the profit because the width of the box is p minus z,

Â that was price minus cost and the height of the box is q, the quantity demanded.

Â Remember, that was how we were able to write our profit here

Â as q times p minus c.

Â So q is the height of the box, p minus c is the width of the box,

Â the product of those two numbers is the profit, and

Â the product of those two numbers is the area of the box.

Â So what the calculus approach does is take your quantitative model,

Â which is the blue curve here, that's what we contribute with the modeling, and

Â then we do the optimization which is simply to find the value of p

Â at which the area of the grey shaded box is largest.

Â If I can find that value of p, I've solved the optimization problem.

Â And so again, one of the nice things about these quantitative models is they can help

Â you visualize the solution to a problem.

Â And without this visual here it's hard to kind of

Â see in the same way as to what we're really trying to achieve, and as I say,

Â what we're trying to achieve when we maximize the profit is essentially to find

Â the value of price of which this grey shaded box is maximized.

Â And we know from having gone through the calculus approach that it's at p

Â equal to three and a third, 3.33.

Â