Optimization Techniques and New Management Tools

Prepared by Robert F. Brooker, Ph.D. Copyright ©2004 by South-Western, a division of Thomson Learning. All rights reserved. Slide 1

Optimization Techniques


Tujuan perusahaan adalah: memaksimumkan profit memaksimumkan value of the firm atau Meminimumkan cost berdasarkan constraint yang ada


• The objective of a business firm is to maximize profits or the value of the firm or to minimize costs, subject to some constraints.

• Techniques or methods for maximizing or minimizing the objective function of the firm are very important to be described.

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

Economic Optimization Process

• Optimal Decisions– Best decision helps achieve objectives

most efficiently.

• Maximizing the Value of the Firm– Value maximization requires serving

customers efficiently.• What do customers want?• How can customers best be served?

PowerPoint Slides Prepared by Robert F. Brooker, Ph.D. Slide 5

Steps in Optimization

• Define an objective mathematically as a function of one or more choice variables

• Define one or more constraints on the values of the objective function and/or the choice variables

• Determine the values of the choice variables that maximize or minimize the objective function while satisfying all of the constraints


Optimization Techniques

• Methods for maximizing or minimizing an objective function

• Examples– Consumers maximize utility by purchasing

an optimal combination of goods– Firms maximize profit by producing and

selling an optimal quantity of goods– Firms minimize their cost of production by

using an optimal combination of inputs

Maximizing the Value of the Firm

•The value of the firm is impacted by ▫Total Revenue.... which is a function of

marketing strategies, pricing and distribution policies, nature of competition....

▫Total Cost .... which is a function of the price and availability of inputs, alternative production methods...

Review of Basic Microeconomics

• What is the objective of the firm?

PROFIT MAXIMIZATION• Profit = Total Revenue – Total Cost• To understand profit, you need to understand

both revenue and cost. Understanding total revenue begins with the Law of Demand. Understanding total cost begins with the Law of Diminishing Returns.

Review of Basic Microeconomics

• The Law of Diminishing Returns – As a variable input increases, holding all else constant, the rate of increase in output will eventually diminish.

• OR..... As a firm increases its employment of labor, holding capital and all other factors constant, the productivity of each worker hired will eventually diminish.

• WHY? Each additional worker has less and less capital to work with.

• The Law of Diminishing Returns is the foundation from which we build the relationship between output and total cost, marginal cost, and average cost.


0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

Q

TR

Expressing Economic Relationships

Equations: TR = 100Q - 10Q2

Tables:

Graphs:

Q 0 1 2 3 4 5 6TR 0 90 160 210 240 250 240


Total, Average, and Marginal Revenue

TR = PQ

AR = TR/Q

MR = TR/Q

Q TR AR MR0 0 - -1 90 90 902 160 80 703 210 70 504 240 60 305 250 50 106 240 40 -10


0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

Q

TR

-40

-20

0

20

40

60

80

100

120

0 1 2 3 4 5 6 7

Q

AR, MR

Total Revenue

Average andMarginal Revenue


Total, Average, andMarginal Cost

Q TC AC MC0 20 - -1 140 140 1202 160 80 203 180 60 204 240 60 605 480 96 240

AC = TC/Q

MC = TC/Q


Profit Maximization

Q TR TC Profit0 0 20 -201 90 140 -502 160 160 03 210 180 304 240 240 05 250 480 -230


Profit Maximization

0

60

120

180

240

300

0 1 2 3 4 5Q

($)

MC

MR

TC

TR

-60

-30

0

30

60

Profit

The firm maximizes total profit at Q=3, where thhe positive difference between TR and TC is greatest, MR=MC, and the п function is at its highest point


Geometric Relationships

• The slope of a tangent to a total curve at a point is equal to the marginal value at that point

• The slope of a ray from the origin to a point on a total curve is equal to the average value at that point


Geometric Relationships

• A marginal value is positive, zero, and negative, respectively, when a total curve slopes upward, is horizontal, and slopes downward

• A marginal value is above, equal to, and below an average value, respectively, when the slope of the average curve is positive, zero, and negative


Concept of the Derivative

The concept of derivative is closely related to the concept of the margin. For example when output increase from 2 to 3 unit and TR increases from $160 t0 $210 then MR = $50/1 = $ 50

The derivative of Y with respect to X is equal to the limit of the ratio Y/X as X approaches zero.

0limX

dY Y

dX X

• Optimization analysis can be conducted much more efficiently and precisely with differential calculus,

with relies on concept derivative.


Rules of Differentiation

Constant Function Rule: The derivative of a constant, Y = f(X) = a, is zero for all values of a (the constant).

( )Y f X a

0dY

dX If Y = 2, dX/dY =0



Power Function Rule: The derivative of a power function, where a and b are constants, is defined as follows.

( ) bY f X aX

1bdYb aX

dX

Y=4X², dY/dX = 8X



Sum-and-Differences Rule: The derivative of the sum or difference of two functions U and V, is defined as follows.

( )U g X ( )V h X

dY dU dV

dX dX dX

Y U V

Y = 2X + X², dY/dX = 2 + 2X



Product Rule: The derivative of the product of two functions U and V, is defined as follows.

( )U g X ( )V h X

dY dV dUU V

dX dX dX

Y U V

Y=2X²(3-2X), U=2X², V=3-2X

dY/dX=2x²(dV/dX)+(3-2X)(dU/dX)

=2X²(-2)+(3-2X)(4x)

=-4X²+12X-8X² = 12X-12X²



Quotient Rule: The derivative of the ratio of two functions U and V, is defined as follows.

( )U g X ( )V h X UY

V

2

dU dVV UdY dX dXdX V

Y=(3-2X)/2X², U=3-2X, V=2X²

dY/dX=(4X²-12X)/(2X²)²= (4X²-12X)/4X(X³) = 4X(x-3)/4X(X³)

dY/dX= (X-3)/X³



Chain Rule: The derivative of a function that is a function of X is defined as follows.

( )U g X( )Y f UdY dY dU

dX dU dX Y=(3X²+10)³, U=3X²+10, Y =U³

dY/dU = 3U² and dU/dX = 6X

dY/dX = 3(3X³+10)²(6X)

=3(9X4+60X²+100)(6X)


Optimization With Calculus

Optimization often requires finding the maximum or the minimum value of the function.

Find X such that dY/dX = 0

Second derivative rules:

If d2Y/dX2 > 0, then X is a minimum.

If d2Y/dX2 < 0, then X is a maximum.


Example 1

• Given the following total revenue (TR) function, determine the quantity of output (Q) that will maximize total revenue:

• TR = 100Q – 10Q2

• dTR/dQ = 100 – 20Q = 0

• Q* = 5 and d2TR/dQ2 = -20 < 0


Example 2

• Given the following total revenue (TR) function, determine the quantity of output (Q) that will maximize total revenue:

• TR = 45Q – 0.5Q2

• dTR/dQ = 45 – Q = 0

• Q* = 45 and d2TR/dQ2 = -1 < 0

For example, for total-revenue function:• TR = 100Q – 10Q²

d(TR)/dQ = 100 – 20 Q, setting d(TR)/dQ = 0

100 – 20 Q = 0 20 Q = 100, Q = 5

d²(TR)/d²(Q) = - 20, d²(TR)/d²(Q) < 0 , maka TR mencapai maximum pada Q=5

**Another example: TR = 45Q -0.5Q², d(TR)/d(Q)= 45 – Q, d²(TR)/d²Q = - 1

TR mencapai maximum pada Q = 45


Example 3

• Given the following marginal cost function (MC), determine the quantity of output that will minimize MC:

• MC = 3Q2 – 16Q + 57

• dMC/dQ = 6Q - 16 = 0

• Q*= 2.67 and d2MC/dQ2 = 6 > 0

****TR = 45Q - 0.5Q² TC = Q³ - 8Q² + 57Q + 2 п = TR – TC 45 Q – 0.5 Q² - (Q³ - 8Q² + 57Q + 2) = - Q³ + 7.5Q² - 12Q – 2

dп/dQ = - 3Q² + 15 Q – 12 = 0, (- 3Q + 3)(Q – 4) = 0, Q1 = 1, Q2 = 4

d²п/d²Q = - 6 Q + 15 at Q =1 d²п/d²Q = 9, at Q = 4 d²п/d²Q = - 9,

jadi п pada Q = 4 п = - (4)³ + 7.5 (4)² - 12(4) – 2 = $ 6


****TR = 45Q - 0.5Q² TC = Q³ - 8Q² + 57Q + 2

Max Profit MR=MC MR = - Q + 45 MC = 3Q2 – 16 Q + 57

MR=MC -Q+45 = 3Q 2-16Q+57 -3Q2 + 15 Q – 12 = 0

(-3Q + 3) (Q - 4) = 0 Q1 = 1, Q2 = 4 d2 profit/d2 Q = -6 Q + 15 at Q=1 d2profit/d2Q = 9 minimum at Q =4 d2profit/d2Q = - 9

Profit => TR – TC = 45Q – 0.5Q2 – (Q3 – 8Q2 + 57Q + 2) = - Q3 + 7.5Q2 – 12Q -2 = 6

• Demonstration Problem (1) An engineering firm recently conducted a study to determine

its profit and cost structure. The results of the study are as follows:

TR = 300Q – 6Q² and TC = 4Q² so that MR = 300 -12Q and MC = 8Q , The manager has been

asked to determine that maximum level of net profits and the level of Q that will yield that result.

Answer: Equating MR and MC yield 300 -12Q = 8Q Solving this equation for Q reveals that the optimal level of Q is Q = 15 Plugging Q = 15 into the net profit relation yields the maximum level of

net profits:

п = 300 (15) – (6)(15)² - 4 (15)² = $ 2,250

• Demonstration Problem (2) Suppose TR = 10 Q – 2 Q² and TC = 2 + Q² What value of the managerial control variable, Q, maximize net

profit?

Answer: Net profits are

п = TR – TC = 10Q – 2Q² - 2 - Q² taking the derivative of п with respect to Q and setting it equal to

zero gives dп/dQ = 10 -4Q - 2Q = 0 and Q = 10/6 to verify that this indeed a maximum, we must check that

the second derivative of п is negative:

d²п/d²Q = - 4 – 2 = - 6 < 0

therefore, Q = 10/6 is indeed a maximum


Multivariate Optimization

• Objective function Y = f(X1, X2, ...,Xk)

• Find all Xi such that ∂Y/∂Xi = 0

• Partial derivative:– ∂Y/∂Xi = dY/dXi while all Xj (where j ≠ i) are

held constant

• Multivariate Optimization - the process of determining the maximum and minimum point of a

function of two or more variables.

Partial DerivativesSuppose that the total profit (п) function of the the firm depend on sales of

commodities X and Y as follows: п = f(X,Y) = 80X – 2X² - XY – 3Y² + 100Y

to find the partial derivative of п with respect to X, dп/dX, we hold Y constant

dп/dX = 80 – 4X – Y and to find derivative of п with respect to Y

dп/dY = - X - 6Y + 100,

Multiply the first equation by – 6, rearranged - 480 + 24 X – 6Y = 0 100 - X - 6Y = 0 adding - 380 + 23 X = 0 X = 380/23 = 16.52 Y = 80 – 4 (16.52) = 13.92

п = 80(16.52) – 2 (16.52)² - (16.52)(13.92) – 3(13.92)² + 100(13.92) = $1,356.52


Constrained Optimization

• Substitution Method– Substitute constraints into the objective

function and then maximize the objective function

• Lagrangian Method– Form the Lagrangian function by adding

the Lagrangian variables and constraints to the objective function and then maximize the Lagrangian function

• Constrained Optimization Most of the time, managers face constraints in their optimization decisions.

For example, a firm may face a limitation on its production capacity or on the availability of skilled personnel and raw material constraint optimization problem.

Example: the firm seeks to max. п function п = 80X - 2X² -XY - 3Y² + 100Y

but faces the constraint that the output of commodity X plus the output commodity Y must be 12, X + Y = 12

solving the constraint for X, then X = 12 – Y, substituting into п п = 80 (12-Y) – 2(12-Y)² - (12-Y) Y – 3Y² + 100Y

= 960 – 80Y- 2(144-24Y + Y²) – 12Y + Y² - 3Y² + 100Y = - 4Y² + 56Y + 672,

dп/dY = - 8Y + 56 = 0 therefore Y = 7, X = 12 -7 = 5 and

п with constraint = 80 (5) – 2(5)² - (5)(7) -3(7)² = 100(7) = $ 868

п without constraint = $ 1356.52

• Langrangian Multiplier Method Set the constraint function (X + Y = 12) equal to zero X + Y – 12 = 0

then this Constraint function multiply by λ and add to п function:

Lп = 80X – 2X² - XY – 3Y² + 100Y + λ (X + Y – 12) with three unknown variables: X, Y and λ

to maximize Lλ, we set the partial derivative of Lλ with respect to X, Y and λ equal to zero,

1. dLλ/dX = 80 – 4X – Y + λ = 0

2. dLλ/dY = - X – 6Y + 100 + λ = 0

3. dLλ/dλ = X + Y – 12 = 0

to find the value of X , Y and λ that maximizes Lλ and п ,we simultaneously 80 – 4X – Y + λ = 0

100 - X - 6Y + λ = 0 substract - 20 -3X + 5Y = 0 (4)

Multiplying equation (3) dLп/dλ = X + Y – 12 = 0 by 3 3X + 3Y – 36 = 0

adding to equation (4) we obtain: 3X + 3Y – 36 = 0

-3X + 5Y – 20 = 0 8Y - 56 = 0

Y = 7, X = 5 substituting into equation (2) -5 – 42 + 100 = -λ

λ = - 53

• λ = - 53 the value of λ has an important economic interpretation. It is a

marginal effect on the objective-function solution associated with a 1-unit change in the constraint. In the above problem, this means that a decrease in the output capacity constraint from 12 to 11 unit or an increase to 13 units will reduce or increase, respectively, the total profit of the firm (п ) by about $ 53.


Interpretation of the Lagrangian Multiplier,

• Lambda, , is the derivative of the optimal value of the objective function with respect to the constraint– In Example , = -53, so a one-unit

increase in the value of the constraint (from -12 to -11) will cause profit to decrease by approximately 53 units

• New Management Tools for OptimizationSelama 2 dekade terakhir ini banyak management tools baru

diperkenalkan yang merubah the way firms are managed.

Management tools yang baru ini al:

a. Benchmarking

refers to the finding out how other firms may be doing something better (cheaper) so that your firm can copy and possibly improve on its technique, usually accomplished by field trip.

b.Total Quality Management

refers to constantly improving the quality of products and the firm’s processes so as to consistently deliver value to customers. TQM constantly asks, “How can we do this cheaper, faster, or better?” This involves maximizing quality and minimizing costs.

c. Reengineering Its asks “If this were an entirely new firm,

how would you organize it? Or if you were able to start all over again, how would you do it?” Reengineering involves the radical redesign of all the firm’s processes to achieve major gains in speed, quality, service and profitability.

There are 2 major reasons to reengineer:1. Fear that competitors may come up with new

products, services, or ways of doing business that might destroy your firm

2. Desire if you believe that by reengineering, your company can erase competition.

Reengineering is extremely difficult to carry it out in the real world, and not all firms are capable or need to reengineer.

d. The Leaning Organization is one that value continuing learning, both individual and collective, and believes that competitive advantage derives from and requires continuous learning. A learning organization is based on five basic ingredient:

1. to develop a new mental model –putting aside old ways of thinking and being willing to change. 2. to achieve personal mastery- by learning to be open with others and listen to them rather than telling them what to do. 3. to develop system thinking – an understanding of how the firm really works. 4. to develop a shared vision - a strategy for the firm that all its employee share 5. to strive/make an effort for team learning – seeing how all the

firm’s employees can be made to work and learn together to realize the shared vision and carry out the strategy of the firm.

• Problems

1. Given the following total revenue and total cost functions of a firm:

TR = 22 Q – 0.5Q²

TC = 1/3 Q³ - 8.5 Q² + 50 Q + 90

determine (a) the level of output at which the firm maximizes its total profit; (b) the maximum profit the firm could earn.

2. for the following total profit function of a firm:

п = 144X – 3X² - XY – 2Y² + 120Y – 35 determine (a) the level of output of each commodity at which the

firm maximize it total profit; (b) the value of maximum amount of the total profit of the firm.

Optimization Techniques and New Management Tools

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