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Web Chapter A Optimization Techniques

Web Chapter A Optimization Techniques. Overview Unconstrained & Constrained Optimization Calculus of one variable Partial Differentiation in Economics Appendix to Web Chapter A: Lagrangians and Constrained Optimization.  2002 South-Western Publishing .

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Web Chapter A Optimization Techniques

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  1. Web Chapter AOptimization Techniques Overview • Unconstrained & ConstrainedOptimization • Calculus of one variable • Partial Differentiation in Economics • Appendix to Web Chapter A: • Lagrangians and Constrained Optimization 2002 South-Western Publishing

  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 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!

  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 dp/dQ = 0.

  4. ConstrainedOptimization • 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).

  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

  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

  7. Quick Differentiation Review 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 = cXbdY/dX = b•c•X b-1Y = 5•X2 dY/dX = 10•X

  8. Quick Differentiation Review • SumRuleY = G(X) + H(X) dY/dX = dG/dX + dH/dX exampleY = 5•X + 5•X2 dY/dX = 5 + 10•X • Product RuleY= G(X)•H(X) dY/dX = (dG/dX)H + (dH/dX)G exampleY = (5•X)(5•X2 ) dY/dX = 5(5•X2 ) + (10•X)(5•X) = 75•X2

  9. Quick Differentiation Review • 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 RuleY = 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

  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.

  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.

  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 Problem 1 Problem 2 • 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

  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= 50 + 5•X2 • 10•X = 0 • second derivative is: 10 implies Q = 0 is a MIN Problem 1 Problem 2 • Max = 100•Q - Q2 • 100 -2•Q = 0 • second derivative is: -2 implies Q =50 is a MAX

  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/Pholds income constant.

  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

  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

  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.

  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, .

  19. Maximize Utility Example example: Max Utility subject to a money constraint Max U = X•Y2subject 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, .

  20. L/X = Y2 - = 0 Y2 =  L/Y = 2XY - 4= 0 2XY = 4 L/= 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.

  21. Problem Minimize 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?

  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

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