Topic 03: Optimal Decision Making Overview Unconstrained Optimization Univariate Calculus – review Constrained Optimization Multivariate Calculus - review 1

Overview Many economic decisions involve trying to decide what is the “best” decision to make. These problems involving trying to optimize some objective, such as maximize profits, or minimize risk exposure, etc. 2

Analytical Approach to Optimal Decision Making Express the relationship between the objective and various decision and exogenous variables. Dependent variable =f(independent variables) Example: Profit = f(quantity) or  = f(Q) Illustration: Find the quantity of output that maximizes profit: p(Q) = 16·Q - Q2 6

Unconstrained Optimization Unconstrained Optimization involves finding the optimum to some decision problem in which there are no constraints. 4

Average Profit = Profit / Q PROFITS Slope of ray from the origin Rise / Run Profit / Q = average profit Maximizing average profit doesn’t maximize total profit MAX C B profits quantity Q 4

Marginal Profits = /Q profits of the last unit produced maximum marginal profits occur at the inflection point (A) Decision Rule: produce where marginal profits = 0. profits max C B A Q average profits marginal profits Q 5

Optimization using calculus Calculus uses derivatives: dP/dQ = lim DP / DQ as DQ goes to 0. Decision rule - To maximize profits, find where dP/dQ = 0 -- first order condition 6

Review of Differentiation Rules Name Function Derivative Example Constant Y = c dY/dX = 0 Y = 5 dY/dX = 0 Line Y = c•X dY/dX = c Y = 5•X dY/dX = 5 Power Y = cXb dY/dX = b•c•X b-1 Y = 5•X2 dY/dX = 10•X 7

Review of Differentiation Rules Sum Rule: If Y = G(X) + H(X) , then dY/dX = dG/dX + dH/dX Example: Y = 5•X + 5•X2 dY/dX = 5 + 10•X Product Rule: If Y = G(X)•H(X), then dY/dX = (dG/dX)H + (dH/dX)G Example: Y = (5•X)(5•X2 ) dY/dX = 5(5•X2 ) + (10•X)(5•X) = 75•X2 8

Review of Differentiation Rules Quotient Rule: If Y = G(X) / H(X) , Then dY/dX = (dG/dX)•H - (dH/dX)•G H2 Example: Y = (5•X) / (5•X2) dY/dX = 5(5•X2) -(10•X)(5•X) (5•X2)2 = -25X2 / 25X4 = -X-2 9

Review of Differentiation Rules Chain Rule: If Y = G [ H(X) ] , then dY/dX = (dG/dH)•(dH/dX) Example: Y=(5+5X)2 , dY/dX = 2(5 + 5•X)1(5) = 50 + 50•X

Example #1: Maximization problem A profit function might look like an arch, rising to a peak and then declining at even larger outputs. A firm might sell huge amounts at very low prices, but discover that profits are low or negative. At the maximum, the slope of the profit function is zero. The first order condition for a maximum is that the derivative at that point is zero. 10

Example #1 (continued) Competitive Firm: Maximize Profits where P = TR - TC = P•Q - TC(Q) Use our first order condition: dP/dQ = P - dTC/dQ = 0 Decision Rule: P = MC a function of Q Max P = 100•Q - Q2 100 -2•Q = 0 implies Q = 50 and P = 2,500 12

Static Optimal Decisions The scale of a project should expand until MB = MC Example: screening for prostate or breat cancer Example of marginal reasoning MC MB frequency per decade 7

Capital Budgeting Example Investments Project A: $500m 23.0% Project B: $75m 18.0% Project C: $50m 21.0% Project D: $125m 16.0% Project E: $300m 14.0% Project F: $150m 13.0% Project G: $250m 19.0% Funds: First 250m 14.0% Next 250m 15.5% Next 100m 16.0% Next 250m 16.5% Next 200m 18.0% Next 200m 21.0%

How Much to Produce ? Graph of output and profit Possible Rule: Expand output until profits turn down But problem of local maxima vs. global maximum profit GLOBAL MAX MAX A quantity B 3

Optimum Can Be Highest or Lowest Finding the maximum flying range for the Stealth Bomber is an optimization problem. Calculus teaches that when the first derivative is zero, the solution is at an optimum. The original Stealth Bomber study showed that a controversial flying V-wing design optimized the bomber's range, but the original researchers failed to find that their solution in fact minimized the range. It is critical that managers make decision that maximize, not minimize, profit potential! 2

Second Order Condition: One variable If the second derivative is negative, then it’s a maximum If the second derivative is positive, then it’s a minimum Max P = 50 + 5•X2 10•X = 0 second derivative is: 10 implies Q = 0 is a MIN Max P = 100•Q - Q2 100 -2•Q = 0 second derivative is: -2 implies Q =50 is a MAX 13

Multivariate Calculus Economic relationships usually involve several independent variables. A partial derivative is like a controlled experiment -- it holds the “other” variables constant For example, suppose that we are concerned with how demand for a product changes as its price changes, but assume that disposable income stays the same: Q = f (P, I ). Consequently, we are interested in: ¶Q/¶P holds income constant We can often find partial derivatives using the above differentiation rules. 14

Example #2 Sales are a function of advertising in newspapers and magazines ( X, Y) Max S = 200X + 100Y -10X2 -20Y2 +20XY Differentiate with respect to X and Y and set equal to zero. ¶S/¶X = 200 - 20X + 20Y= 0 ¶S/¶Y = 100 - 40Y + 20X = 0 solve for X & Y and Sales 15

Example #2 (continued) 200 - 20X + 20Y= 0 100 - 40Y + 20X = 0 Adding them, the -20X and +20X cancel, so we get 300 - 20Y = 0, or Y =15 Plug into one of them: 200 - 20X + 300 = 0, hence X = 25 To find Sales, plug into equation: S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250 16

Constrained Optimization Constrained Optimization involves finding the optimum to some decision problem in which the decision-maker faces constraints. Examples: constraints of money, time, capacity, or energy. Sometimes managers know that some constraints are binding, which means that they are equality constraints. Sometimes managers face optimization problems in which the constraints are not necessarily binding. For example, you do not have to spend all your income. These constraints are inequality constraints. Linear programming and other optimization techniques can be used in these cases. 5

Format of Constrained Optimization Economic problems require tradeoffs forced on us by the limits of our money, time, and energy. Optimization involves an objective function and one or more constraints. Maximize y = f(x1 , x2 , ..., xn ) Subject to g(x1 , x2 , ..., xn ) < b Minimize y = f(x1 , x2 , ..., xn ) Subject to g(x1 , x2 , ..., xn ) > b 3

Optimization with equality constraints An equality constraint is binding constraint that must be satisfied. An artificial variable is created for each constraint in the Lagrangian multiplier technique. This artificial variable is traditionally called lambda, l. Max L = (objective fct.) - l{constraint set to zero} Min L = (objective fct.) +l{constraint set to zero} A binding constraint tends to reduce the solution set of an optimal decision making problem, and thereby constrains the optimal decision. 18

Example #3 S(X,Y) = 200X + 100Y - 10X2 - 20Y2 + 20XY Bounds Inc has determined through regression analysis that its sales (S) are a function of the amount of advertising (measured in units) in two different media. This is given by the following relationship (X=newspapers, Y=magazines): S(X,Y) = 200X + 100Y - 10X2 - 20Y2 + 20XY Find the level of newspaper and magazine advertising that maximizes the firm’s sales. Calculate the firm’s sales at the optimal levels of newspaper and magazine advertising.

Example #3 (continued) Earlier we saw that: S/X = 200-20X+20Y =0 S/Y = 100-40X+20Y =0 You get X* = 15 units and Y* = 25 units. Plugging these optimal advertising levels back into the sales equation gives: S* = 200(25) + 100(15) – 10(25)2 – 20(15)2 + 20(25)(15) = 3250

Example #4 Suppose that Bounds Inc. (in example #3) faces a budget constraint such that it can only purchase 20 units of advertising. A. Determine the level of newspaper and magazine advertising that maximizes sales given this constraint. B. What does the Lagrange multipler represent in this case? C. Compare the constrained optimization solution to the unconstrained optimization solution.

Example #4 (continued) Max S = 200X + 100Y – 10X2 – 20Y2 +20XY subject to: X+Y=20. L= 200X + 100Y – 10X2 – 20Y2 +20XY – l(X+Y-20) Take partial derivatives wrt X, Y, and l: L/X = 200-20X+20Y – l L/Y = 100-40Y+20X – l L/l = -X - Y + 20 Set the partial derivatives equal to 0 and solve for X, Y, and l. You will get: Y*=7 units, X* = 13 units, and l* = 80. The Lagrange multiplier represents the amount that sales would increase if we increased our advertising budget by 1 unit. Unconstrained solutions: X*=25, Y*=15, S*=3250 Constrained solutions: X*=13, Y*=7, S*=2450