1 / 53

# Finance 510: Microeconomic Analysis - PowerPoint PPT Presentation

Finance 510: Microeconomic Analysis. Optimization . Don't Panic!. Functions. 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 the range . Is a function.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Finance 510: Microeconomic Analysis' - giacinto

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Finance 510: Microeconomic Analysis

Optimization

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 the range

Is a function

Is the domain

For any

and

Note: A function maps each value of x to one and only one value for y

For

Range

Domain

For

Range

Y =14

5

Domain

0

5

X =3

Here, the optimum occurs at x = 5 (y = 20)

Range

5

Domain

0

5

Optimization involves finding the maximum value for y over an allowable range.

5

10

There is no optimum because f(x) is discontinuous at x = 5

12

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!)

0

6

12

There is no optimum because the domain is unbounded (x is allowed to become arbitrarily large)

0

Sufficient conditions guarantee a solution, but are not required

Necessary conditions are required for a solution to exist

Gas is a necessary condition to drive a car

A gun is a sufficient condition to kill an ant

The Weierstrass Theorem provides sufficient conditions for an optimum to exist, the conditions are as follows:

is continuous

over the domain of

The domain for

is closed and bounded

Formally, the derivative of

is defined as follows:

All you need to remember is the derivative represents aslope (a rate of change)

Linear Functions

Exponents

Logarithms

Products

Composites

Strictly speaking, no problem is truly unconstrained. However, sometimes the constraints don’t “bite” (the constraints don’t influence the maximum)

First Order Necessary Conditions

If

is a solution to the optimization problem

or

then

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?

Let x = Number of Flight Hours

First Order Necessary Conditions

Hourly Cost (\$)

Annual Flight Hours

Secondary Order Necessary Conditions

If

is a solution to the maximization problem

then

If

is a solution to the minimization problem

then

Slope is decreasing

Slope is increasing

An Example derivative

Let x = Number of Flight Hours

First Order Necessary Conditions

Second Order Necessary Conditions

For X>0

Multiple Variables derivative

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

Partial Derivatives derivative

When you have functions of multiple variables, a partial derivativeis the derivative with respect to one variable, holding everything else constant

Example (One you will see a lot!!)

Multiple Variables derivative

First Order Necessary Conditions

Multiple Variables derivative

(2)

(1)

(1)

(2)

40

50

For a function of more than one variable, it’s a bit more complicated…

Constrained optimizations derivatives are negative… 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 thelagrangian

Function to be maximized

Constraint(s)

Multiplier

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…

Constrained Optimization and set them equal to zero

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.

Constrained Optimization and set them equal to zero

Multiplier

The first step is to create a Lagrangian

Constraint

Objective Function

Constrained Optimization and set them equal to zero

First Order Necessary Conditions

“Multiplier” conditions

Note that this will always hold with equality

Constrained Optimization and set them equal to zero

The Multiplier and set them equal to zero

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

Another Example and set them equal to zero

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 are

and

respectively.

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?

Minimizations need a minor adjustment… and set them equal to zero

To solve these types of problems, we set up thelagrangian

Inequality Constraints and set them equal to zero

Just as in the previous problem, we set up the lagrangian. This time we have two constraints.

Holds with equality

Doesn’t necessarily hold with equality

First Order Necessary Conditions and set them equal to zero

Case #1: and set them equal to zero

Constraint is non-binding

First Order Necessary Conditions

Case #2: and set them equal to zero

Constraint is binding

First Order Necessary Conditions

Constraint is Binding and set them equal to zero

Constraint is Non-Binding

Try this one… and set them equal to zero

You have the choice between buying apples and oranges. You utility (enjoyment) from eating apples and bananas can be written as:

The prices of Apples and Bananas are given by

and

Maximize your utility assuming that you have \$100 available to spend

(Objective) and set them equal to zero

(Income Constraint)

(You can’t eat negative apples/oranges!!)

Objective

Non-Negative Consumption Constraint

Income Constraint

First Order Necessary Conditions and set them equal to zero

• We can eliminate some of the multiplier conditions with a little reasoning…

• You will always spend all your income

• You will always consume a positive amount of apples

Case #1: Constraint is non-binding and set them equal to zero

First Order Necessary Conditions

Case #1: Constraint is binding and set them equal to zero

First Order Necessary Conditions