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Finance 30210: Managerial Economics Optimization

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Optimization deals with functions. A function is simply a mapping from one space to another. (that is, a set of instructions describing how to get from one location to another) is a function Is the range Is the domain

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For example For Domain Range Function

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For 05 Domain 5 20 Range X =3 Y =14

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05 Domain 5 20 Range Optimization involves finding the maximum value for y over an allowable domain. Here, the optimum occurs at x = 5 (y = 20)

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What is the solution to this optimization problem? 105 There is no optimum because f(x) is discontinuous at x = 5

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06 12 What is the solution to this optimization problem? There is no optimum because the domain is open (that is, the maximum occurs at x = 6, but x = 6 is NOT in the domain!)

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0 12 What is the solution to this optimization problem? There is no optimum because the domain is unbounded (x is allowed to become arbitrarily large)

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The Weierstrass theorem provides sufficient conditions for an optimum to exist, the conditions are as follows: The domain foris closed and bounded is continuous over the domain of

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Formally, the derivative ofis defined as follows: All you need to remember is the derivative represents a slope (a rate of change) Actually, to be more accurate, the derivative represents a trajectory Finding maxima/minima involves taking derivatives….

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Graphically… Now, let the change in x get arbitrarily small

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First Order Necessary Conditions Ifis a solution to the optimization problem then or

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Useful derivatives Exponents Logarithms Linear Functions Example:

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An Example Suppose that your company owns a corporate jet. Your annual expenses are as follows: You pay your flight crew (pilot, co-pilot, and navigator a combined annual salary of $500,000. Annual insurance costs on the jet are $250,000 Fuel/Supplies cost $1,500 per flight hour Per hour maintenance costs on the jet are proportional to the number of hours flown per year. Maintenance costs (per flight hour) = 1.5(Annual Flight Hours) If you would like to minimize the hourly cost of your jet, how many hours should you use it per year?

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Let x = Number of Flight Hours First Order Necessary Condition

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An Example Hourly Cost ($) Annual Flight Hours

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Slope is increasingSlope is decreasing How can we be sure we are at a maximum/minimum? For a maximization…For a minimization…

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Let x = Number of Flight Hours First Order Necessary Conditions Second Order Necessary Conditions For X>0

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Suppose you know that demand for your product depends on the price that you set and the level of advertising expenditures. Choose the level of advertising AND price to maximize sales

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When you have functions of multiple variables, a partial derivative is the derivative with respect to one variable, holding everything else constant First Order Necessary Conditions

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With our two first order conditions, we have two variables and two unknowns

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its generally sufficient to see if all the second derivatives are negative… How can we be sure we are at a maximum?

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Practice Questions 1) Suppose that profits are a function of quantity produced and can be written as Find the quantity that maximizes profits 2) Suppose that costs are a function of two inputs and can be written as Find the quantities of the two inputs to minimize costs

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Constrained optimizations attempt to maximize/minimize a function subject to a series of restrictions on the allowable domain To solve these types of problems, we set up the lagrangian Function to be maximized Constraint(s) Multiplier

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To solve these types of problems, we set up the lagrangian We know that at the maximum…

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Once you have set up the lagrangian, take the derivatives and set them equal to zero First Order Necessary Conditions Now, we have the Multiplier conditions…

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Example: Suppose you sell two products ( X and Y ). Your profits as a function of sales of X and Y are as follows: Your production capacity is equal to 100 total units. Choose X and Y to maximize profits subject to your capacity constraints.

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The key is to get the problem in the right format The first step is to create a Lagrangian Objective Function Constraint Multiplier

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Now, take the derivative with respect to x and y First Order Necessary Conditions Multiplier conditions

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First, lets consider the possibility that lambda equals zero Nope! This cant work!

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The other possibility is that lambda is positive

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Lambda indicates the marginal value of relaxing the constraint. In this case, suppose that our capacity increased to 101 units of total production. Assuming we respond optimally, our profits increase by $5

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Example: Postal regulations require that a package whose length plus girth exceeds 108 inches must be mailed at an oversize rate. What size package will maximize the volume while staying within the 108 inch limit? X Y Z Girth = 2x +2y Volume = x*y*z

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First set up the lagrangian… Now, take derivatives…

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Lets assume lambda is positive

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Suppose that you are able to produce output using capital (k) and labor (L) according to the following process: Labor costs $10 per hour and capital costs $40 per unit. You want to minimize the cost of producing 100 units of output.

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Minimizations need a minor adjustment… A negative sign instead of a positive sign!! So, we set up the lagrangian again…now with a negative sign Take derivatives…

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Lets again assume lambda is positive

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Suppose that you are choosing purchases of apples and bananas. Your total satisfaction as a function of your consumption of apples and bananas can be written as Apples cost $4 each and bananas cost $5 each. You want to maximize your satisfaction given that you have $100 to spend

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First set up the lagrangian… Now, take derivatives…

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Lets again assume lambda is positive

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Suppose that you are able to produce output using capital (k) and labor (l) according to the following process: The prices of capital and labor areandrespectively. Union agreements obligate you to use at least one unit of labor. Assuming you need to produce units of output, how would you choose capital and labor to minimize costs?

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Just as in the previous problem, we set up the lagrangian. This time we have two constraints. Doesnt necessarily hold with equality Will hold with equality Non-Binding Constraints

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First Order Necessary Conditions

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Case #1: Constraint is non-binding

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First Order Necessary Conditions Case #2: Constraint is binding

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Constraint is Non-Binding Constraint is Binding

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Try this one… You have the choice between buying apples and bananas. You utility (enjoyment) from eating apples and bananas can be written as: The prices of Apples and Bananas are given byand Maximize your utility assuming that you have $100 available to spend

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(Income Constraint) (Objective) (You cant eat negative apples/bananas!!) Objective Non-Negative Consumption Constraint Income Constraint

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First Order Necessary Conditions We can eliminate some of the multiplier conditions with a little reasoning… 1.You will always spend all your income 2.You will always consume a positive amount of apples

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Case #1: Constraint is non-binding First Order Necessary Conditions

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Case #1: Constraint is binding First Order Necessary Conditions

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Constraint is Non-Binding Constraint is Binding

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Linear Programming. Linear programming A technique that allows decision makers to solve maximization and minimization problems where there are certain.

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