slide 1 2002 south-western publishing web chapter a optimization techniques overview unconstrained...
TRANSCRIPT
Slide 12002 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
Slide 2
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!
Slide 3
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.
Slide 4
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).
Slide 5
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
Slide 6
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
Slide 7
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
Slide 8
• 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
Quick Differentiation Review
Slide 9
• 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
Quick Differentiation Review
Slide 10
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.
Slide 11
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.
Slide 12
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
Slide 13
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
Slide 14
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.
Slide 15
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
Slide 16
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
Slide 17
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.
Slide 18
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, .
Slide 19
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, .
Slide 20
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.
Slide 21
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?
Slide 22
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