optimization techniques lecture 2 appendix c n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Optimization Techniques Lecture 2 (Appendix C) PowerPoint Presentation
Download Presentation
Optimization Techniques Lecture 2 (Appendix C)

Loading in 2 Seconds...

play fullscreen
1 / 33

Optimization Techniques Lecture 2 (Appendix C) - PowerPoint PPT Presentation


  • 177 Views
  • Uploaded on

Optimization Techniques Lecture 2 (Appendix C). Optimization Techniques. . Optimization is: a process by which the maximum or minimum values of decision variables are determined. Examples Finding the profit maximizing PC sales units at Dell, or at COMPAQ, or at IBM.

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

PowerPoint Slideshow about 'Optimization Techniques Lecture 2 (Appendix C)' - duncan


An Image/Link below is provided (as is) to download presentation

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
slide2

Optimization Techniques

  • . Optimization is:
  • a process by which the maximum or minimum values of decision variables are determined.
  • Examples
  • Finding the profit maximizing PC sales units at Dell, or at COMPAQ, or at IBM.
  • Finding the cost minimizing units of product lines at Nissan(Trucks, Sentra, Altima)
slide4

TR

  • 0 0
  • 90
  • 160
  • 210
  • 240
  • 250
  • 240

2a. Table 2b. graph

QTR

TR

Q

economic relationships
Economic Relationships

2c. Equation TR=100Q-10Q2

slide6

3. An optimal sales value for profit maximization can be obtained by a total, or marginal approach.

Total Approach: Profit is maximized when TR-TC is maximum at Q*

Illustrate graphically

Marginal Approach:Profit is

maximized when MR=MC or

MR-MC=0 at Q*

total approach
Total Approach

TR

TC

TC

Profit is max at Q* because TR-TC is Max

TR

Q

Q* maximizes profit

slide9

Marginal analysis- a technique which postulates that an activity should be carried out until the marginal benefit (MB) equals the marginal cost (MC).

  • . When the total value reaches maximum, the marginal (additional)value will be zero(See page C-12 in the book)

Given: π=100Q-10Q2

Illustrate

slide10

.A derivative is simply a mathematical procedure for obtaining the marginal value (or slope)of a parent function at a point as the change in the explanatory variable approaches 0.

Given: y= f(x) as a parent function then,

dy/dx=lim(ΔY/ΔX):marginal value as

ΔX0

= (a slope at a point).

Example: If y=f(x-4), what is the limit of the function y as x5? 1 = 5-4.

slide11

.A review of the Rules of Differentiation

Rule 1:Constant: The derivative of a constant is always zero.

Given: y= f(x)= 2000

dy/dx= 0

y=$2000

TFC= f(Q)= $2000

dTFC/dQ=0

Y

Y=2000

X

slide12

Rule 2: Power Function:

The first derivative of a power function such as y=aXb where a & b are constants, is equal to the exponent b multiplied by a times the variable x raised to b-1 power

Given y=axb

dy/dx=b.axb-1

e.g. y=2x3

dy/dx=3.2x3-1=6x2 (We do it in our

heads!)

slide13

Rule 3: Sums and Differences

The derivative of a sum (difference) is equal to the derivative of the individual terms.

Given: y=u+v or y=u-v, where u=f(x) and v=g(x), then

dy/dx=du/dx + dv/dx

or

dy/dx=du/dx-dv/dx

y=9x2+2x+3 y=9x2-2x-3

dy/dx =18x+2 or dy/dx=18x-2

slide14

Rule 4: Products Rule

The derivative of the product of two

expressions is equal to the sum of

the first term multiplied by the

derivative of the second plus the

second term times the derivative of

the first term.

slide15

Rule 4: Products Rule

Given: y = U.V and U & V = f(x)

dy/dx= U dv/dx +V du/dx

y = 3x2(3-x);Let u = 3x2; v = 3-x

Then dy/dx =3x2(-1) + (3-x)(6x)

=-3x2+18x-6x2= 18x - 9x2

slide16

Rule 5: Quotient Rule:

The derivative of a quotient of two

expressions is equal to the

denominator multiplied by the

derivative of the numerator minus

the numerator times the derivative

of the denominator all divided

by the square of the denominator.

slide17

Rule 5: Quotient Rule

Given: y=u/v where u and v= f(x)

dy/dx = V(du/dx) – U(dv/dx)

V2

If Y= (2x-3)/6x2, then

dy/dx = 6x2(2) - (2x-3) 12x

(6x2)2

= [-12x2 +36x]/36x4]

slide18

Rule 6: Derivative of A Function of a

Function (Chain Rule):the derivative of

such a function is found as follows:

Given: y=f(u) where u=g(x)

Then dy/dx=(dy/du)(du/dx)

e.g. Y= 2U-U2; and U =2x3

Then dy/dx= (2-2U)6x2, or after

substituting U =2x3

=[2-2(2x3)]6x2 =12x2-24x5

slide19

Rule 7: Logarithmic function:

Given y = lnx

dy/dx= dlnx/dx=1/x

slide20

. Use of a derivative in the optimization process.

Step 1: Helps to Identify: the maximum or minimum values of decision variables (Q, Ad units)

Given: y=f(x)=> Nonlinear function

Get dy/dx=0 and solve for x =>

First Order Condition (FOC),

or Necessary condition)

slide21

Step 2: Helps to distinguish the maximum values from minimum values  second order condition (SOC)

If d2y/dx2 <0, then a maximum value of the decision variable(X) is obtained.

If d2y/dx2 >0, then a minimum value of the decision variable(X) is obtained.

Example: Given  = -100 + 400Q - 2Q2

Question: What level of output(Q) will maximize Profit? Illustrate.

slide22

8a)Partial derivative helps us to find the maximum or minimum values of decision variables from an equation with three, or more variables.

(8b) Yes. Given: y=f(x, z)

Step 1: δy/δx=0 and δy/δz=0 and solve for x and zsimultaneously to identify the maximum or minimum value

slide23

Step 2: If δ2y/δx2 and δ2y/δz2 <0,then

the value maximizing units of x and z are obtained.

If δ2y/δx2 and δ2y/δx2 >0, then the value minimizing units x and z are obtained.

(c) Example: y= f(x,z)

= 2x + z -x2 + xz -z2

Find x and z which maximize y.

slide24

9a)Unconstrained optimization- a process of choosing a level of some activity by comparing the marginal benefits and marginal costs of an activity (MB=MC).

b)Constrained Optimization-In the real world, optimization often involves maximization or minimization of some objective function subject to a series of constraints (Resources, output quantity and quality, legal constraints)

slide25

Rule: An objective function is maximized or minimized s.t. a constraint if for all of the variables in the objective function, the ratios of MBs to MCs are equal for all activities.

MB1/MC1=MB2/MC2 =......=MBn/MCn

Example 1: Optimal Allocation of Ad. Exp. among TV, Radio, and Newspaper within a budget constraint of $1100; CTv = $300/ad; CR= $100/ad; CN= $200/ad.

slide26

The optimal Allocation of Advertising Budget

Given: Budget =$1100, MCTv=$300, MCRadio =$100, MCNP =$200.

Determine the optimal unit of TV, Radio, Newspaper ads.

Decision Rule:

Choose the number of TV, Radio,

and Newspaper ads for which:

MBTv/MCTv=MBRadio/MCRadio=MBNP/MCNP

slide27

#of ads MBTvMBTv/CTvMBR MBR/CR MBNpMBNp/CNp

1 40 .133 15 .151 20 .100

2 30 .100 13 .131 15 .075

3 22 .073 10 .100 12 .060

4 18 .060 9 .09 10 .050

5 14 .047 6 .06 8 .040

6 10 .033 4 .04 6 .030

7 7 .023 3 .03 5 .025

slide28

Maximize Sales = f(TV, Radio, Newspaper)

S.t. 300TV +100R + 200N = $1100

Solution: 2 TV Ads; 3 Radio Ads; 1 Newspaper Ad will maximize total sales.

What is the total sales for this combination?

Sales=40+30+15+13+10 +20 = 128

  

TV Radio NP

slide29

What is the total sales for 2 Tv+1R+2Np?

Sales = 2(300) + 1(100)+ 2(200)

= $1100

Is the above combination optimal? Yes or No. Why? No! Total benefit=118 (=40+30+15+10+20)instead of 128

MBR/CR > MBTV/CTV => Use Radio

MBR/CR < MBNP/CNP => Use more NP ad

The marginal benefits are given.

Refer to handout example # 3.

Rule: MBTV/CTV= MBR/CR=MBN/CN

Solution: 2 TV ads, 3 Radio Ads,

1 Newspaper ad.

slide30

Applications--optimal combination of inputs, optimal allocation of time etc.

Given: Q= f(k,L)

MPK/Pk =MPL/PL ; (LCC Rule)

What if MPK/PK > MPL/PL ?

slide31

10(a) Lagrangian Multiplier

  • a mathematical technique for obtaining an optimal solution to constrained optimization problems.
  • These solutions are obtained by incorporating the zero value of the constraint equations(s) into the objective function.
slide32

(b) Example

Minimize: TC= 3x2+6y2-xy:

s.t: x+y =20 :constraint eq.

L= 3x2+6y2-xy+ λ(x+y-20)

Where L stands for the Lagrange function

λ= tells us the marginal change in the objective function associated with a one unit change in the binding constraint.

Solution: X=13; Y= 7 will minimize TC.

Illustrate.

slide33

(c) =-$71 means that a reduction in the binding constraint of 20 by one unit (say to 19) will reduce the total cost by $71 or an increase in the binding constraint of 20 by one unit (say to 21)will increase TC by $71.