Optimization Techniques Lecture 2 (Appendix C)

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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.

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### Optimization TechniquesLecture 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.
• Finding the cost minimizing units of product lines at Nissan(Trucks, Sentra, Altima)

TR

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

2a. Table 2b. graph

QTR

TR

Q

Economic Relationships

2c. Equation TR=100Q-10Q2

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

TR

TC

TC

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

TR

Q

Q* maximizes profit

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

.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.

.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

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

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

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.

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

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.

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]

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

Rule 7: Logarithmic function:

Given y = lnx

dy/dx= dlnx/dx=1/x

. 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)

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.

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

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.

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)

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.

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:

#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

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

  

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

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 ?

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.

(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.

(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.