Optimization Techniques

Lecture 2

1 . 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)

Optimization Techniques

2. Economic relationships

Economic relationships can be expressed in the form of tables, graphs, or equations.

2a. Table 2b. graph

Q TR

0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

TR

TR

Q

0 0 1 902 1603 2104 2405 2506 240

Economic Relationships

2c. Equation TR=100Q-10Q2

Two Approaches to Profit Maximization

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*

Marginal Approach: Profit is maximized when MR=MC or MR-MC=0 at Q* (Focus on this)

Illustrate graphically

Total Approach

Q

TC

TR

Q* maximizes profit

TRTC Profit is max at Q*

because TR-TC is Max

Marginal Approach

Q

MR

MC

MR

MC

Q*

3. Marginal analysis

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

4 . When the total value reaches maximum, the marginal (additional)value will be zero.

Given: π=100Q-10Q2

Illustrate

5 .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 ΔX0 = (a slope at a point).Example: If y=f(x-4), what is the limit of the function y as x5? 1 = 5-4.

6 . A review of the Rules of DifferentiationRule 1: Constant: The derivative of a constant is always zero.

Given: y= f(x)= 2000dy/dx= 0y=$2000 TFC= f(Q)= $2000 dTFC/dQ=0

0

1000

2000

3000

4000

0 1 2 3 4 5 6

Y

X

Y=2000

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!)

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 twoexpressions 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 let U= 2x-3 and V= 6x2

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

7. 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

7. Use of a derivative in the optimization process

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 z simultaneously to identify the maximum or minimum value

8b)Partial derivative Steps

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.

9.Types of Optimization

a)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

Constrained Optimization

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. Expenditures among TV, Radio, and Newspaper within a budget constraint of $1100; CTv = $300/ad; CR= $100/ad; CN= $200/ads.

• Budget constraint: $1100 =CTV TV+ CR R+ CN N

Constrained Optimization

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

Constrained Optimization

#of ads MBTv MBTv/CTv MBR MBR/CR MBNp MBNp/ 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 .090 10 .050

Constrained Optimization

Maximize Sales = f(TV, Radio, Newspaper)

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

Constrained Optimization

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 more Radio ad

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

Note: The marginal benefits are given.Refer to handout example # 3.

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

Solution: 2 TV ads, 3 Radio ads, and 1 Newspaper ad.

Constrained Optimization

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. Lagrangian Multiplier

a. Meaning: a mathematical technique for obtaining an optimal solution to constrained optimization problems.

The solutions are obtained by incorporating the zero value of the constraint equations(s) into the objective function.

10. Lagrangian Multiplier

(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.

10. Lagrangian Multiplier

(c) λ =-$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.