slide 1 2002 south-western publishing web chapter a optimization techniques overview unconstrained...

South-Western Publishing

Web Chapter AOptimization Techniques

Overview

• Unconstrained & Constrained Optimization

• Calculus of one variable

• Partial Differentiation in Economics

• Appendix to Web Chapter A: » Lagrangians and Constrained Optimization

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 derivativeis 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!

Unconstrained Optimization • Unconstrained Optimization is a relatively

simple calculus problem that can be solved using differentiation, such as finding the quantity that maximizes profit in the function:

(Q) = 16·Q - Q2. • The answer is Q = 8 as we will see.

Where d/dQ = 0.

Constrained Optimization• Constrained Optimization involves one or more

constraints of money, time, capacity, or energy.

• When there are inequality constraints (as when you must spend less than or equal to your total income), linear programming can be used.

• Most often, managers know that some constraints are binding, which means that they are equality constraints. » Lagrangian multipliers are used to solve these problems (which

appears in the Appendix to Web Chapter A).

Optimization Format• 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 , b.

Maximize y = f(x1 , x2 , ..., xn )

Subject to g(x1 , x2 , ..., xn ) < b

or: Minimize y = f(x1 , x2 , ..., xn )

Subject to g(x1 , x2 , ..., xn ) > b

Using Equations• profit = f(quantity) or = f(Q)

» dependent variable & independent variable(s)» average profit =Q» marginal profit = / Q

• Calculus uses derivatives » d/dQ = lim / Q Q

0

» SLOPE = MARGINAL = DERIVATIVE» NEW DECISION RULE: To maximize profits, find where

d/dQ = 0 -- first order condition

Quick Differentiation Review

• 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

Name Function Derivative Example

• Sum Rule Y = G(X) + H(X) dY/dX = dG/dX + dH/dX

example Y = 5•X + 5•X2 dY/dX = 5 + 10•X

• Product Rule Y = G(X)•H(X)

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 Y = G(X) / H(X)

dY/dX = (dG/dX)•H - (dH/dX)•G H2

Y = (5•X) / (5•X2) dY/dX = 5(5•X2) -(10•X)(5•X) (5•X2)2

= -25X2 / 25•X4 = - X-2

• Chain Rule Y = G [ H(X) ]

dY/dX = (dG/dH)•(dH/dX) Y = (5 + 5•X)2

dY/dX = 2(5 + 5•X)1(5) = 50 + 50•X


Applications of Calculus in Managerial Economics

• 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.

• If = 50·Q - Q2, then d/dQ = 50 - 2·Q, using the rules of differentiation.

• Hence, Q = 25 will maximize profits where 50 - 2•Q = 0.

More Applications of Calculus

• minimization problem: Cost minimization supposes that there is a least cost point to produce. An average cost curve might have a U-shape. At the least cost point, the slope of the cost function is zero.

• The first order condition for a minimum is that the derivative at that point is zero.

• If C = 5·Q2 - 60·Q, then dC/dQ = 10·Q - 60.

• Hence, Q = 6 will minimize cost where 10•Q - 60 = 0.

More Examples

• Competitive Firm: Maximize Profits » where = TR - TC = P•Q - TC(Q)» Use our first order condition:

d/dQ = P - dTC/dQ = 0. » Decision Rule: P = MC.

a function of Q

Max = 100•Q - Q2

100 -2•Q = 0 implies Q = 50 and = 2,500

Max= 50 + 5•X2

So, 10•X = 0 implies Q = 0 and= 50

Problem 1 Problem 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 = 100•Q - Q2

100 -2•Q = 0

second derivative is: -2 implies Q =50 is a MAX

Max= 50 + 5•X2

10•X = 0

second derivative is: 10 implies Q = 0 is a MIN

Problem 1 Problem 2

Partial Differentiation• Economic relationships usually involve several

independent variables.

• A partial derivative is like a controlled experiment -- it holds the “other” variables constant

• Suppose price is increased, holding the disposable income of the economy constant as in Q = f (P, I ), then Q/P holds income constant.

Problem:• 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

Solution: 2 equations & 2 unknowns

• 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

International Import Restraints

• Import quotas of Japanese automobiles are inequality constraints. The added constraint will affect decisions.

• A Japanese manufacturer will shift more production to U.S. assembly facilities and increase the price of cars exported to the U.S.

• We may also expect that the exported cars will be "top of the line" models, and we expect U.S. manufacturers to raise domestic car prices.

Web Chapter A -- Appendix

Objective functions are often constrained by one or more “constraints” (time, capacity, or money)

Max L = (objective fct.) -{constraint set to zero}

Min L = (objective fct.) +{constraint set to zero}

An artificial variable is created for each constraint in the Lagrangian multiplier technique. This artificial variable is traditionally called lambda, .

Maximize Utility Exampleexample:

Max Utility subject to a money constraint

Max U = X•Y2 subject to a $12 total budget with the prices of X as $1, the price of Y as $4 (suppose X represents soda and Y movie tickets).

Max L = X•Y2 -{ X + 4Y - 12}• differentiate with respect to X, Y and lambda, .

L/X = Y2 - = 0 Y2 = L/Y = 2XY - 4= 0 2XY = 4L/= X + 4Y- 12 = 0

Three equations and three unknowns

Solve: Ratio of first two equations is:

Y/2X = 1/4 or Y = .5 X. Substitute into the third equation: We get:

X = 4; Y = 2; and= 4• Lambda is the marginal (objective function) of the

(constraint). In the parentheses, substitute the words used for the objective function and constraint.

• Here, the marginal utility of money.

ProblemMinimize crime in your town

• Police, P, costs $15,000 each.

• Jail, J, costs $10,000 each.

• Budget is $900,000.

• Crime function is estimated: C = 5600 - 4PJ

» Set up the problem as a Lagrangian

» Solve for optimal P and J, and C

» What is economic meaning of lambda?

Answer• Min L= 5600 - 4PJ + {15,000•P + 10,000•J -900,000 }• To Solve, differentiate

L/P: - 4•J +15,000•L/J: - 4•P +10,000•L/ : 15,000•P +10,000•J -900,000 =0

J/P = 1.5 so J = 1.5•P & substitute into (3.)

15,000•P +10,000•[1.5•P] - 900,000 = 0

solution: P = 30, J = 45, C = 200 and = -.012

• Lambda is the marginal crime (reduction) for a dollar of additional budget spent

slide 1 2002 south-western publishing web chapter a optimization techniques overview unconstrained...

Documents