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Topic 03: Optimal Decision Making

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- Overview
- Unconstrained Optimization
- Univariate Calculus – review
- Constrained Optimization
- Multivariate Calculus - review
- Constrained Optimization

- 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.

- 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

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

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

profits

max

- profits of the last unit produced
- maximum marginal profits occur at the inflection point (A)
- Decision Rule: produce where marginal profits = 0.

C

B

A

Q

average

profits

marginal

profits

Q

- 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

Name Function DerivativeExample

- Constant Y = cdY/dX = 0Y = 5
dY/dX = 0

- Line Y = c•XdY/dX = cY = 5•X
dY/dX = 5

- Power Y = cXbdY/dX = b•c•X b-1 Y = 5•X2 dY/dX = 10•X

- SumRule: If Y = G(X) + H(X) , then
dY/dX = dG/dX + dH/dX

Example: Y = 5•X + 5•X2dY/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

- 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

- 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

- 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.

- 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

- 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

- Investments
- Project A:$500m23.0%
- Project B: $75m18.0%
- Project C:$50m21.0%
- Project D:$125m16.0%
- Project E: $300m14.0%
- Project F:$150m13.0%
- Project G:$250m19.0%

Funds:

First 250m14.0%

Next 250m15.5%

Next 100m16.0%

Next 250m16.5%

Next 200m18.0%

Next 200m21.0%

profit

- Graph of output and profit
- Possible Rule:
- Expand output until profits turn down
- But problem of local maxima vs. global maximum

GLOBAL

MAX

MAX

A

quantity B

- 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!

- 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

- 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.

- 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

- 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

- 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.

- 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

- 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.

- 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.

- 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

- 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.

- 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