Department of Business Administration FALL 2007-08 Optimization Techniques by Asst. Prof. Sami...

Department of Business Administration

FALL 2007-08

Optimization TechniquesOptimization Techniques

by

Asst. Prof. Sami Fethi

2 Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved.

Ch 2: Optimisation Techniques

Optimization Techniques and New Optimization Techniques and New Management Tools Management Tools

The first step in presenting optimisation techniques is to examine ways to express economic relationships. Economic relationship can be expressed in the form of equation, tables, or graphs. When the relationship is simple, a table and/ or graph may be sufficient. However, if the relationship is complex, expressing the relationship in equational form may be necessary.



Optimization Techniques and New Optimization Techniques and New Management Tools Management Tools

Expressing an economic relationship in equational form is also useful because it allows us to use the powerful techniques of differential calculus in determining the optimal solution of the problem.

More importantly, in many cases calculus can be used to solve such problems more easily and with greater insight into the economic principles underlying the solution. This is the most efficient way for the firm or other organization to achieve its objectives or reach its goal.



Example 1Example 1

Suppose that the relationship between the total revenue (TR) of a firm and the quantity (Q) of the good and services that firm sells over a given period of time, say, one year, is given by

TR= 100Q-10Q2

(Recall: TR= The price per unit of commodity times the quantity sold; TR=f(Q), total revenue is a function of units sold; or TR= P x Q).



Example 1Example 1

By substituting into equation 1 various hypothetical values for the quantity sold, we generate the total revenue schedule of the firm, shown in Table 1. Plotting the TR schedule of table 1, we get the TR curve as in graph 1. In this graph, note that the TR curve rises up to Q=5 and declines thereafter.



0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

Q

TR

Example 1Example 1

Equation1: TR = 100Q - 10Q2

Table1:

Graph1:

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

Managerial Economics



Example 2Example 2

Suppose that we have a specific relationship between units sold and total revenue is precisely stated by the function: TR= $ 1.50 x Q. The relevant data are given in Table 2 and price is constant at $ 1.50 regardless of the quantity sold. This framework can be illustrated in graph 2.



Example 2Example 2

Unit Sold TR Price

1 1.5 1.5

2 3

3 4.5

4 6

5 7.5

6 9

Table2:

Graph of the relationship between total revenue and units sold

01.5

34.5

67.5

9

1 2 3 4 5 6

Unit sold for time period

Reve

nue

per

time

peri

odGraph2:



The relationship between total, average, and marginal concepts and measures is crucial in optimisation analysis. The definitions of totals and averages are too well known to warrant restating, but it is perhaps appropriate to define the term marginal.

Total, Average, and Marginal Cost

A marginal relationship is defined as the change in the dependent variable of a function associated with a unitary change in one of the independent variables.




In the total revenue function, marginal revenue is the change in total revenue associated with a one-unit change in units sold. Generally, we analyse an objective function by changing the various independent variables to see what effect these changes have on the dependent variables. In other words, we examine the marginal effect of changes in the independent variable. The purpose of this analysis is to determine that set of values for the independent or decision variables which optimises the decision maker’s objective function.




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

(Recall: Total cost: total fixed cost plus total variable costs; Marginal cost: the change in total costs or in total variable costs per unit change in output).

Table3:




The first two columns of Table 3 present a hypothetical total cost schedule of a firm, from which the average and marginal cost schedules are derived in columns 3 and 4 of the same table. Note that the total cost (TC) of the firm is $ 20 when output (Q) is zero and rises as output increases (see graph 3 to for the graphical presentation of TC). Average cost (AC) equals total cost divided by output. That is AC=TC/Q. Thus, at Q=1, AC=TC/1= $140/1= $140. At Q=2, AC=TC/Q =160/2= £80 and so on. Note that AC first falls and then rises.

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

Table3:



Total, Average, and Marginal Cost Marginal cost (MC), on the other

hand, equals the change in total cost per unit change in output. That is, MC= TC/Q where the delta () refers to “a change”. Since output increases by 1unit at a time in column 1 of table 3, the MC is obtained by subtracting successive values of TC shown in the second column of the same table. For instance, TC increases from $ 20 to $ 140 when the firm produces the first unit of output. Thus MC= $ 120 and so forth. Note that as for the case of the AC and MC also falls first and then rises (see graph 4 for the graphical presentation of both AC and MC). Also, note that at Q=3.5 MC=AC; this is the lowest AC point. At Q=2; that is the point of inflection whereas the point shows MC at the lowest point.

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

Table3:




0

60

120

180

240

0 1 2 3 4Q

TC ($)

0

60

120

0 1 2 3 4 Q

AC, MC ($)AC

MC

Graph3:

Graph4:



Profit MaximizationProfit Maximization

Table 4 indicates the relationship between TR, TC and Profit. In the top panel of graph 5, the TR curve and the TC curve are taken from the previous graphs. Total Profit () is the difference between total revenue and total cost. That is = TR-TC. The top panel of Table 4 and graph 5 shows that at Q=0, TR=0 but TC=$20. Therefore, = 0-$20= -$20. This means that the firm incurs a loss of $20 at zero output. At Q=1, TR=$90 and TC=$ 140. Therefore, = $90-$140= -$50. This is the largest loss. At Q=2, TR=TC=160. Therefore, = 0 and this means that firm breaks even. Between Q=2 and Q=4, TR exceeds TC and the firm earns a profit. The greatest profit is at Q=3 and equals $30.




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

Table 4:

Table 4 indicates the relationship between TR, TC and Profit. In the top panel of graph 5, the TR curve and the TC curve are taken from the previous graphs. Total Profit () is the difference between total revenue and total cost. That is = TR-TC. The top panel of Table 4 and graph 5 shows that at Q=0, TR=0 but TC=$20. Therefore, = 0-$20= -$20. This means that the firm incurs a loss of $20 at zero output. At Q=1, TR=$90 and TC=$ 140. Therefore, = $90-$140= -$50. This is the largest loss. At Q=2, TR=TC=160. Therefore, = 0 and this means that firm breaks even. Between Q=2 and Q=4, TR exceeds TC and the firm earns a profit. The greatest profit is at Q=3 and equals $30.




0

60

120

180

240

300

0 1 2 3 4 5Q

($)

MC

MR

TC

TR

-60

-30

0

30

60

Profit

Graph5:



Optimization by marginal Analysis

Marginal analysis is one of the most important concepts in managerial economics in general and in optimisation analysis in particular.

According to marginal analysis, the firm maximizes profits when marginal revenue equals marginal cost (i.e. MC=MR). Here, MC is given by the slope of TC curve and this tangential point is the point of inflection (i.e. at Q=2).



Optimization by marginal Analysis MR can be defined as the

change in total revenue per unit change in output or sales (i.e. MR=TR/Q) and is given by the slope of the TR curve. In graph 5, at Q=1 the slope of TR or MR is $80. At Q=2, the slope of TR or MR is $60. At Q=3 or 4, the slope of TR curve or MR is $40 and $20 respectively. At Q=5, the TR curve is highest or has zero slope so that MR=0. After that TR declines and MR is negative.

0

60

120

180

240

300

0 1 2 3 4 5Q

($)

MC

MR

TC

TR

-60

-30

0

30

60

Profit

Graph5:



Optimization by marginal Analysis Also At Q=3, the slope of the

TR curve or MR equals the slope of TC curve or MC, so that the TR curves are parallel and the vertical distance between them () is greatest. In the top panel of graph 5, at Q=3, MR=MC and is at a maximum. In the bottom panel of graph 5, the total loss of the firm is greatest when function faces up whereas the firm maximizes its total profit when function faces down.

0

60

120

180

240

300

0 1 2 3 4 5Q

($)

MC

MR

TC

TR

-60

-30

0

30

60

Profit

Graph5:



Concept of the DerivativeConcept of the Derivative The concept of derivative is

closely related to the concept of the margin. This concept can be explained in terms of the TR curve of graph1, reproduced with some modifications in graph6. Earlier, we defined the marginal revenue as the change in total revenue per unit change in output. For instance, when output increases from 2 to 3 units, total revenue from $160 to $ 210. Thus, MR= TR/ Q = $ 210-$ 160/3-2 =$ 50.

Graph 6:



Concept of the DerivativeConcept of the Derivative This is the slope of chord BC on the

total-revenue curve. However, when Q assumes values smaller than unity and as small as we want and even approaching zero in the limit, then MR is given by the slope of shorter chords, and it approaches the slope of the TR curve at a point in the limit. Thus, starting from point B, as the change in quantity approaches zero, the change in total revenue or marginal revenue approaches the slope of the TR curve at point B. That is MR= TR/ Q = $ 60- the slope of tangent BK to the TR curve at point B as change in output approaches zero in the limit.

Graph 6:



Concept of the DerivativeConcept of the Derivative

To summarize between points B and C on the total revenue curve of graph 6, the marginal revenue is given by the slope of chord BC ($ 50). This is average marginal revenue between 2 and 3 units of output. On the other hand, the marginal revenue at point B is given by the slope of line BK ($ 60), which is tangent to the total revenue curve at point B. For example, at point C, MR is $ 40. Similarly, at point D, MR= $20 whereas at point E, MR= $ 0- when total revenue curve reflect its concave shape its slope is always zero and then the shape indicates

declining slope.

Graph 6:




Graph 6:




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

In general, if we let TR=Y and Q=X, the derivative of Y with respect to X is given by the change in Y with respect to X, as the change in X approaches zero. So we define this concept in the following expression.



Concept of the Derivative-ExampleConcept of the Derivative-Example

Suppose we have y=x2

0limX

dY Y

dX X

0lim

X

dYdX f(x+dx)- f(x)

dX

lim (x+dx)2- x2

XdX

0lim

X

dY

dX

dX

2xdx-+ x2

limdY

dXX

(2xdx)

2x



Rules of DifferentiationRules 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 For example, Y=2 dY/dX=0

the slope of the line Y is zero.




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

( ) bY f X a X

1bdYb a X

dX

For example, Y=2xdY/dX=2




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

For example:

U=2x and V=x2

Y=U+V=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

For example:Y=2 x2 (3-2 x)and let U=2 x2 and V=3-2 xdY/dX=2x2(dV/dX)+(3-2x)(dU/dX)dY/dX=2 x2(-2)+ (3-2 x) (4x)dY/dX=-4x2+ 12x+8 x2

dY/dX= 12x-12 x2




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

For example:Y=3-2x/2x2

and let V=2 x2 and U=3-2 xdY/dX=(2 x2(dV/dX)+ (3-2 x) (dU/dX))/v2

dY/dX=2 x2(-2)+ (3-2 x) (4x)/ (2 x2)2

dY/dX=4x2-12/4x4= (4x)(x-3)/ (4x) (x3)=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 U

dY dY dU

dX dU dX

For example:Y=U3+10 and U=2X2

then dY/dU=3U2 and dU/dX=4XdY/dX=dY/dU.dU/dX=(3U2) 4XdY/dX=3(2X2)2(4X)=48X5



Optimization With CalculusOptimization With CalculusFind X such that dY/dX = 0 minimum or maximum.

First order is necessary not sufficient for min or max

Second derivative rules:

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

If d2Y/dX2 < 0, then X is a maximum.For example:TR=100-10Q2

d(TR)/dQ=100-20QSetting d(TR)/dQ=0, we get100-20Q=0Q=5 this means that its slope is zero and total revenue is maximum at the o/p level of 5 units.



Optimization With CalculusOptimization With Calculus

Distinguishing between a Maximum and a Minimum: The second derivative

For example:TR=100-10Q2

d(TR)/dQ=100-20Qd2(TR)/dQ2=-20

The rule is if the derivative is positive, we have a minimum, and if the second derivative is negative, we have a maximum. This means that TR function has zero slope at 5. Since d2(TR)/dQ2=-20, this TR function reaches a maximum at Q=5.



Maximizing a Multivariable FunctionMaximizing a Multivariable Function

To maximize or minimize a multivariable function, we must set each partial derivative equal to zero and solve the resulting set of simultaneous equations for the optimal value of independent or right-hand side variables.



ExampleExample =80X-2X2-XY-3Y2+100Y - total profit functionWe set d/dX and d/dY equal to zero and solve for X and Y.

d/dX=80-4X-Y=0d/dY=-X-6Y+100=0

Multiplying the first of the above expression by –6, rearranging the second and adding, we get-480+24X+6Y=0100-X-6Y=0-380=23X=0 X=16.52 Y=13.92

and substituting the values of x and y into the profit equation mentioned above, we have the max total profit of the firm is $ 1,356.52.



Constrained optimisation by substitutionConstrained optimisation by substitution

and Lagrangian Multiplier Methodsand Lagrangian Multiplier Methods Suppose that a firm seeks to maximize its total profit and the function as follows:

=80X-2X2-XY-3Y2+100Ybut faces the constrain that the o/p of commodity X plus the o/p of commodity Y must be 12. That is, X+Y=12First we can write X as a function of Y, such as X=12-YAnd substituting X=12-Y into the profit function in inspection.Finally, we get: =-4Y2+56Y+672



Solving y, we find the first derivative of: with respect to Y and then set it equal to zero,d/dY=-8Y+56=0 Y=7 and X=5 and the profit is =80X-2X2-XY-3Y2+100Y=$868.Example for lagrangian methodSuppose that we have a Lagrangian function as followsL=80X-2X2-XY-3Y2+100Y+(X+Y-12)



First we have to find the partial derivative of L with respect to X,Y, and and setting them equal to zero:

dL/dX=80-4X-Y+=0 (1)

dL/dY=-X-6Y+100+=0 (2)

dL/d=X+Y-12=0 (3)

First subtract eq2 from eq1 and get

–20-3X+5Y=0 (4)

Now, multiplying eq3 by 3 and adding with eq4 and get the followings

3X+3Y-36=0

-3X+5Y-20=0

8Y-56=Y=7 X=5 into eq2 to get the value of

-X-6Y+100+=0 =X+6Y-100 =-53 (economic interpretation?)

The total profit of the firm increase or decrease by about $ 53

In order to find the total profit of the firm, subs the relevant figures ($868)



New Management ToolsNew Management Tools

BenchmarkingTotal Quality ManagementReengineeringThe Learning Organization



Other Management ToolsOther Management Tools

BroadbandingDirect Business ModelNetworkingPricing PowerSmall-World ModelVirtual IntegrationVirtual Management



The EndThe End

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