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.
2. Economic relationships can be expressed in the form of tables,graphs, or equations.
2a. Table 2b. graph
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*
Marginal Approach:Profit is
maximized when MR=MC or
MR-MC=0 at Q*
Profit is max at Q* because TR-TC is Max
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).
.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
= (a slope at a point).
Example: If y=f(x-4), what is the limit of the function y as x5? 1 = 5-4.
Rule 1:Constant: The derivative of a constant is always zero.
Given: y= f(x)= 2000
TFC= f(Q)= $2000
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
dy/dx=3.2x3-1=6x2 (We do it in our
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
dy/dx =18x+2 or dy/dx=18x-2
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.
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
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.
Given: y=u/v where u and v= f(x)
dy/dx = V(du/dx) – U(dv/dx)
If Y= (2x-3)/6x2, then
dy/dx = 6x2(2) - (2x-3) 12x
= [-12x2 +36x]/36x4]
Function (Chain Rule):the derivative of
such a function is found as follows:
Given: y=f(u) where u=g(x)
e.g. Y= 2U-U2; and U =2x3
Then dy/dx= (2-2U)6x2, or after
substituting U =2x3
Given y = lnx
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
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.
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.
Given: Budget =$1100, MCTv=$300, MCRadio =$100, MCNP =$200.
Determine the optimal unit of TV, Radio, Newspaper ads.
Choose the number of TV, Radio,
and Newspaper ads for which:
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
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
Sales = 2(300) + 1(100)+ 2(200)
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.
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 ?
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.
(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.