Unformatted text preview: University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain ECO 204 Ajaz Hussain Basic Economics and Math Review ECO 100 review 1. Demand "curve" Assumes buyers are price takers Gives quantity demanded as a function of price: i.e. Q = f(P) Rigorously speaking, demand curves should have P on x axis and Q on y axis Historically, demand curves have P on y axis and Q on x axis Not all markets have a demand "curve"; indeed, some have a demand "point" 2. Supply "curve" Assumes sellers are price takers Gives quantity supplied as a function of price: i.e. Q = f(P) Rigorously speaking, supply curves should have P on x axis and Q on y axis Historically, supply curves have P on y axis and Q on x axis Not all markets have supply "curve"; indeed, some have a supply "point" 3. Elasticity Suppose y is a function of x; i.e. y = f(x) Elasticity is a measure of how y changes when x changes Defined as: E = % change in y / % change in x E can be positive, negative or zero. Example: Q = f(P). How does Q respond to changes in P? E = % change in Q / % change in P 1 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain The value of E is often compared to 1. There can be 3 cases: Case 1: E > 1 % change in y / % change in x > 1 % change in y > / % change in x That is, if x changes by a certain percentage, y will change by a greater percentage Case 2: E = 1 % change in y / % change in x = 1 % change in y = / % change in x That is, if x changes by a certain percentage, y will change by the same percentage Case 3: E < 1 % change in y / % change in x < 1 % change in y < / % change in x That is, if x changes by a certain percentage, y will change by a lesser percentage Example: ECO 100 tells you that if price E < 1, then raising P results in higher revenues. Let us see why. Now: R = PQ % change in R = % change in P + % change in Q [see calculus review below for why] Now, if P increases, Q decreases. If % change in P is greater than % change in Q, raising P will raise revenues. When is % change in P > % change in Q? E = % change in Q / % change in P Suppose E < 1. Then: % change in Q / % change in P < 1 % change in Q < / % change in P 2 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain Computing E. There are two methods: Arc E = % change in y / % change in x = (change in y / change in x)(average x / average y) Point E = % change in y / % change in x = (dy/y) / (dx/x) = (dy/dx)(x/y) Why the two methods? Arc E is used when you have two pairs of observations; for example, initial P, Q and final P, Q. Thus, Arc E is used to measure elasticity between two points. As shown in class, using the "traditional" formula for % change will give you different answers going from point A to point B versus point B to point A. Point E is used when you have a single pair of observations; for example, a single pair P, Q. Thus, Point E is used to measure elasticity at a point. To see where the Point E formula comes from look at Arc E formula for two points and bring the two points closer. From Math 133 this will be the: limit as change x goes to 0 of Arc E = (change in y / change in x)(average x / average y) limit as x 0 of Arc E = ( y / x)(average x / average y) Point E = (dy/dx)(x/y) 4. Opportunity cost. This is typically presented in ECO 100 as the "value of the next best alternative". This will take some explaining to do as many folks don't know why the value of the "next best alternative" is a cost, or that an opportunity cost doesn't always exist, or that it isn't always the "next highest alternative". We do this via an example. Suppose you've been blessed with $1m. What do you do with this? Let us consider different possibilities think of it as different people doing different things with their jackpot. Ajax is very happy to have $1m. He has always wanted to buy a business. Thus, without examining other opportunities, he uses the $1m to buy the business. Observe, Ajax is not torn between alternatives. He is very sure. Thus, what is the value of his decision to buy the business? Simply, the profits = R C. Observe we haven't said anything about opportunity cost. Britney is also very happy to have $1m. She thinks really hard about what she can do with the $1m. She can buy a business in fact, let's say this is the same business Ajax was thinking about or she could invest in real estate, gold, vineyard, brewery and so on. In contrast to Ajax, Britney is examining other opportunities. Which one should she pick? Of course, the opportunity with the highest value. Let's say that is the business. 3 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain Britney makes the same decision as Ajax. Observe we haven't said anything about opportunity cost. Cindy is ecstatic to have $1m. She thinks really hard about what she can do with the $1m. She can buy a business in fact, let's say this is the same business Ajax was thinking about or she could invest in real estate, gold, vineyard, brewery and so on. Suppose Cindy is particularly intrigued by the business, especially since her good chums Ajax and Britney have plunked their $1m into one. Cindy wonders, "should I put my $1m into the business and, by definition, not put the $1m into other opportunities?". Note that Cindy is examining the business against all other alternatives. The answer is, she should compare the value of the business against the value of all other alternatives. That is, she could make a pair wise comparison of the business against each of the alternatives. If she had, say 100 other alternatives, this would take a long time. In fact, to make sure the business is a good idea, she need only compare the value of the business against the best of all other alternatives. The value of the next best alternative is labeled opportunity cost. Why the nomenclature opportunity cost? Here's why. Cindy will invest in the business if: Value of business > Value of the next best alternative R C > Value of the next best alternative R C Value of the next best alternative > 0 R [C + Value of the next best alternative] > 0 R "Cost" > 0 That is, Cindy will invest in the business if revenue minus cost minus the value of the next best alternative is positive. Observe how from a decision making perspective, the value of the next best alternative is a "cost". It's not an actual cost rather, it is only for decision making. This is why the value of the next best alternative is called "opportunity cost". In Cindy's case, she should buy the business if: R [C + Opportunity Cost] > 0 Let's dispel another myth that an opportunity cost is the value of next best alternative. In the Ajax, Britney and Cindy case, it is the value of the next best alternative. In more complex cases, you have to think twice. 4 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain A "twisted" example: Suppose a store orders from its suppliers a hot selling item. We will examine the following 3 cases: 1. How many items should store order if it has ample shelf capacity? 2. How many items should store order if it has limited shelf capacity? 3. Suppose the store has ordered items from part 2. However, before shelving, demand turns out to be weaker than expected. If the supplier has a buyback program, how any many items at the loading dock should be shelved? Assume limited shelf space. Math Review 5. Calculus From Math 133 you know that if y = f(x), then dy/dx is the slope: the change in y over the change in x. Let's review some rules. Suppose y = xn . Then: dy/dx = n xn1 You should recognize that this can also be written as: dy = n xn1 dx Suppose y = log x. Here "log" is base e. Now: dy/dx = 1/x dy = dx/x Observe that the right hand side "almost" looks like a % change. This is useful for Elasticity calculations. For example, take the Point price E: E = (dQ/dP)(P/Q) E = (dQ/Q)/(dP/P) E = (d log Q)/(d log P) This is very useful because we will do nonlinear demand functions. For example, one could have: Q = 3P2 If you're asked to compute point E via E = (dQ/dP)(P/Q), you could do it, but it'd be 5 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain messy. Here's a simpler way: Q = 3P2 log Q = log (3P2) log Q = log 3 + log P2 log Q = log 3 2 log P Now, note that E = (d log Q)/(d log P). Treating log Q as the "y" variable, log P as the "x" variable, perform dy/dx or dlog Q/dlogP to get: d log Q/d log P = 2 E = 2 No matter what the price, the elasticity is always equal to 2. This is why these demand functions are known as constant elasticity demand functions. You can use this to also see why we said that if: R = PQ % change in R = % change in P + % change in Q Start with R = PQ log R = log (PQ) log R = log P + log Q d log R =d log P + d log Q Read this as "change in log R = change in log P + change in log Q". Now, recall if: y = log x dy/dx = 1/x dy = dx/x But y = log x. Simply rewrite as: 6 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain d log x = dx/x Returning to: d log R =d log P + d log Q dR/R = dP/P + dQ/Q (dR/R)*100 = (dP/P)*100 + (dQ/Q)*100 % change in R = % change in P + % change in Q 6. Unconstrained optimization (maximization/minimization without constraints) We'll see lots of unconstrained optimization problems in this course (and future economics and finance courses). Examples: revenue maximization, profit maximization, total cost minimization, cost per unit minimization, etc. Back in Math 133, it was drummed into you that if you want to find the "maximum" or "minimum" of y = f(x) then you should set dy/dx = 0 ("first order condition") and see whether you have a max or min by checking the "second order conditions". This is slightly misleading as this procedure doesn't really find a max or min; rather, it identified the stationary points where the slope is 0 and checks to see if these stationary points are a local max or min. Put simply, you may not find the global max or min. In fact, you should complete another step. Look at the boundaries of the function and evaluate y = f(x) at the boundaries. Then compare these to the value of y at the stationary points. Example: a firm faces demand given by P = 100 10Q. Suppose it has unlimited capacity what is the revenue maximizing output? Now suppose it has limited capacity of 6 units what is the revenue maximizing output? Suppose it has limited capacity of 4 units what is the revenue maximizing output? 7. Envelope Theorem In any optimization problem, there will be variables and parameters. The optimal "x" will always be a function of the parameters. If we plug the optimal x into the objective, we will get the optimized objective. We may then ask, how does the optimized objective change when a parameter changes? The answer is the envelope theorem: simply differentiate the original objective with respect to the parameter. Example: suppose a firm has unlimited capacity and seeks to maximize revenues with demand function P = 100 10Q. How will the maximized revenues change, if the intercept of the demand curve increases? 7 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain I will do the problem without the envelope theorem, then with the envelope theorem. First, the objective is R = PQ: R = PQ R = (100 10Q)Q The variables are R and Q and the parameters are 100 (the intercept of the demand curve) and 10 (the slope of the demand curve). Now: R = 100Q 10Q2 To maximize set dR/dQ = 0: dR/dQ = 100 20Q = 0 Solve for optimal "x": Q = 100/20 = 5 From which: R = 100Q 10Q2 = 100(5) 10(25) = 500 250 = $250. The optimized objective is $250 and the optimal Q is 5. Note that the Q = 5 is implicitly a function of the intercept and slope. You can see this by redoing the problem and using letters for the parameters. Denote the demand curve by: P = a bQ R = PQ R = (a bQ)Q The variables are R and Q and the parameters are a (the intercept of the demand curve) and b (the slope of the demand curve). Now: R = a Q b Q2 To maximize set dR/dQ = 0: dR/dQ = a 2b Q = 0 8 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain Solve for optimal "x": Q = a/2b Now, what happens if the intercept increases, say, by 1 unit: how will the optimized revenues currently at $250 change? To do this, we could repeat the problem using the demand curve P = 101 10Q, solving for the new optimal Q, and, the new optimized revenues. But there is a faster way. Recall we expressed R as: R = a Q b Q2 Use the envelope theorem. To see how the optimized revenues will change when the intercept increases, differentiate the objective with respect to the parameter, which in this case is a. dR/da = Q The optimal Q was 5. Thus: dR/da = 5 Put simply, with the initial demand curve P = 100 10Q, the optimal Q was 5, and optimal revenues were $250. If the intercept increased, then the optimal revenues will increase by 5: the new revenues will be (approximately) $255. You can see why this is useful for business. Should a company focus on changing the intercept or the slope? We know that a slight increase in the intercept results in extra $5 of revenues. What about the slope? Easy: dR/db = Q2 or dR/db = 25. Recalling that the slope of the demand curve is negative, this says: a flatter slope lowers revenues by $25. Hence, a steeper slope raises revenues by $25. If it has the choice of either changing the intercept or the slope by 1 unit, the company should focus on making demand curve steeper. 8. Constrained optimization (maximization/minimization with constraints) We'll see lots of constrained optimization problems in this course (and future economics and finance courses). 9 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain Examples: consumer maximizes utility given budget constraint; firm minimizes total cost given target output; investor maximizes expected returns given risk tolerance. Most but not all problems will have an equality constraint. For example: Consumer's expenditure = consumer income (true if preferences are monotone) Firm's production = target output There are two ways of solving such problems. One is to simply substitute the constraint into the objective and the second is to use the Lagrangian. Let's illustrate by an example. Ajax Airlines operates flights between London and Toronto. Ajax serves two types of consumers with demand functions: P1 = 330  Q1 P2 = 250  Q2 Suppose all flights have a total of 180 seats capacity and must fly full. Solve for the revenue maximizing number of type 1 and type 2 passengers. First, let's do the direct substitution method. The problem is: Choose Q1 and Q2 to maximize R = R1 + R2 subject to Q1 + Q2 = 180 From the constraint: Q1 + Q2 = 180 Q2 = 180 Q1 Substitute in objective: R = R1 + R2 R = P1 Q1 + P2 Q2 R = (330  Q1) Q1 + (250  Q2)Q2 R = 330 Q1  Q21 + 250 Q2  Q22 R = 330 Q1  Q21 + 250 (180 Q1)  (180 Q1)2 R = 330 Q1  Q21 + 45,000 250Q1  (32,400 360 Q1 + Q21 ) 10 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain R = 330 Q1  Q21 + 45,000 250Q1  32,400 + 360 Q1 Q21 R = 440 Q1  2Q21 + 12,600 First order condition: dR/dQ = 440  4Q 1 = 0 4Q 1 = 440 Q 1 = 110 And therefore, from the constraint we have that Q 2 = 70. Let's solve the same problem with the Lagrangian approach. The problem was: Choose Q1 and Q2 to maximize R = R1 + R2 subject to Q1 + Q2 = 180 With the Lagrangian, the problem is: Choose Q1 and Q2 to maximize L = R1 + R2 [Q1 + Q2 180] Note we are maximizing the Lagrangian L where is the Lagrangian multiplier and the constraint Q1 + Q2 = 180 has been rewritten as: Q1 + Q2 180 Now: L = P1 Q1 + P2 Q2 [Q1 + Q2 180] L = P1 Q1 + P2 Q2 [Q1 + Q2 180] Skipping a few steps: L = 330 Q1  Q21 + 250 Q2  Q22 [Q1 + Q2 180] There are 3 variables: Q1, Q2 and and therefore 3 first order conditions (FOCs): (i) dL /dQ1 = 0 330 2Q1 = 0 11 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain 330 2Q1 = (ii) dL /dQ2 = 0 250 2Q2 = 0 250 2Q2 = (iii) dL/d = 0 Q1 + Q2 180 = 0 Observe how the third FOC is simply the constraint again. Now, the 1st and 2nd FOCs imply: 330 2Q1 = and 250 2Q2 = Which implies that: 330 2Q1 = 250 2Q2 80 = 2Q1 2Q2 40 = Q1 Q2 This equation can be solved simultaneously with Q1 + Q2 = 180 to get (what else?): Q 1 = 110 Q 2 = 70 It's nice to know the two approaches give the same answer. So is there an advantage of doing the Lagrangian? Yes, otherwise I wouldn't be writing this! The direct substitution method is fine but it won't answer one critical question: "will Ajax Air benefit from operating larger planes?" Currently, aircrafts have 180 seats: should Ajax expand capacity? To answer this question, we'd have to compare the impact on revenues of adding another seat versus the cost of adding seats. Suppose adding a seat costs $10, regardless of how many seats are added. (As a test question, you may want to think about how you'd do the calculations below if the cost 12 University of Toronto, Department of Economics, ECO 204 20082009 S. Ajaz Hussain of adding a seat depended on the number of seats). Now, what is the value of adding another seat? To see what it is, look at: L = R1 + R2 [Q1 + Q2 180] Now, we are asking: will Ajax make more money if has > 180 seats? Sound familiar? It should we are asking the impact of changing the capacity on optimized revenues. This is of course the envelope theorem! Write capacity as "c". The Lagrangian becomes: L = R1 + R2 [Q1 + Q2 c] By the envelope theorem, the value of changing c is simply: dL/dc = . So the multiplier is the value of adding another seat. To find , we can use any of the two equations: 330 2Q1 = and 250 2Q2 = Use the first: we know Q 1 = 110 and thus: 330 2(110) = = $110 Adding another seat raises optimized revenues by $110. Given that it cost $10 to add a seat that sounds like a good deal. 13 ...
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This note was uploaded on 05/02/2011 for the course ECO 204 taught by Professor Hussein during the Fall '08 term at University of Toronto.
 Fall '08
 HUSSEIN
 Economics, Microeconomics

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