ch2optech1011

Department of Business Administration

FALL 2010-11

Optimization Techniques

byAssoc. Prof. Sami Fethi

2

Ch 2: Optimisation Techniques

© Dominick Salvatore; ed. 2007, 2010/11, Sami Fethi, EMU, All Right Reserved. Managerial Economics in a Global Economy

Optimization Techniques and New 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.

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Optimization Techniques and New 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.

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Example 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).

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Example 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.

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0

50

100

150

200

250

300

0 1 2 3 4 5 6 7

Q

TR

Example 1

Equation1: TR = 100Q - 10Q2

Table1:

Graph1:

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

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Example 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.

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Example 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

od

Graph2:

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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.

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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.

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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:

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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:

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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:

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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:

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Profit 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.

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Profit Maximization

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.

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Profit Maximization

0

60

120

180

240

300

0 1 2 3 4 5Q

($)

MC

MR

TC

TR

-60

-30

0

30

60

Profit

Graph5:

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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).

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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:

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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:

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Example-TP

Given the following total product (TP) schedule, (a) drive the average product (AP) and marginal product (MP) schedules. (b) On the same set of axes plot the total, average, and marginal product schedules of part a. (c) Using the figure you drew for part b, briefly explain the relationship among the total, average, and marginal product curves.

Table-TP

Q 0 1 2 3 4 5 6 7

TP 0 3 8 12 15 17 17 16

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Answer-TP-(a)

Q TP AP MP

0 0 - -

1 3 3 3

2 8 4 5

3 12 4 4

4 15 3.75 3

5 17 3.4 2

6 17 2.8333333 0

7 16 2.2857143 -1

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Answer-TP-(b)

Schedule

-5

0

5

10

15

20

0 1 2 3 4 5 6 7

quantity

tota

l a

ve

rag

e m

arg

ina

l

pro

du

ct

TP

AP

MP

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Answer-TP-(c)

The slope of a ray from the origin to the TP curve or the average product rises to a point between 2 and 3. then after 5 start to fall but it remains positive as long as TP is positive. Thus the AP curve rises to a point between 2 and 3 and then declines. At the same time, the slope of the TP curve (i.e. The marginal product) rises to the point 1.5 (i.e. The point of inflation of the TP curve and falls thereafter. Thus the MP curve rises to the intersection point of TP and MP and then declines. When TP is at its maximum, the slope of the TP curve is zero (i.e. top point of TP) and so is MP intersection point on horizontal axis. Past point (i.e. top point of TP) , TP curve declines and MP is negative. It is important to mention that when the AP curve rises, the MP curve is above it and when the AP curve declines and MP curve is below it. The MP curve intersects the AP curve at the highest point of AP so that AP=MP at the level of ouput.

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Example-TR

Given Px=8-Qdx (a) Drive (calculate) TR, AR, MR. (b) Plot the schedules of part a. (c) Using the figure you drew for part b, briefly explain the

relationship among the total, average, and marginal revenue curves.

Table-TR

P 8 7 6 5 4 3 2 1 0

Q

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Answer-TR-(a)

P Q TR AR MR

8 0 0

7 1 7 7 7

6 2 12 6 5

5 3 15 5 3

4 4 16 4 1

3 5 15 3 -1

2 6 12 2 -3

1 7 7 1 -5

0 8 0 0 -7

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Answer-TR-(b)

Plot

-10

-5

0

5

10

15

20

0 1 2 3 4 5 6 7 8

quantity

MR

-TR

-AR TR

AR

MR

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Answer-TR-(c)

The slope of a ray from the origin to the TR curve or the average revenue rises to a point between 1 and 3. then after 4 start to fall but it remains positive as long as TR is positive. Thus the AR curve declines from 1.5 to 7.5. At the same time, The marginal revenue curve decreases and intersect the horizontal axis at 5. When TR is at its maximum, the slope of the TR curve is zero (i.e. top point of TR) and so is MR intersection point on horizontal axis. Past point (i.e. top point of TR) , TR curve declines and MR is negative. It is important to mention that when the AR curve declines, the MR curve is below it. The MR curve intersects the AR curve at the highest point of AR so that AR=MR at the level of ouput.

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Concept 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:

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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:

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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:

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Graph 6:

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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.

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Concept of the Derivative-Example

Suppose we have y=x2

0limX

dY Y

dX X

0limX

dYdX

f(x+dx)- f(x)dX

lim (x+dx)2- x2

XdX

0lim

X

dY

dX

dX

2xdx-+ x2 +dx2 - x2

limdY

dX

X(2xdx)

2x

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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 For example, Y=2 dY/dX=0

the slope of the line Y is zero.

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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

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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

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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

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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

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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

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Optimization With Calculus

Find 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.

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Optimization 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.

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Maximizing 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.

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Example-Profit

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

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.

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Constrained optimisation

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

Example-Example-substitutionsubstitution and Lagrangian Multiplier Methodsand Lagrangian Multiplier Methods

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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 follows:Lagrangian=profit fuction +(constraint function is set to equal to zero)L=80X-2X2-XY-3Y2+100Y+(X+Y-12)


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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)


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For the following total profit function of a firm:2y2-120y+xy = 144x --3x2 -35

Determine (a) the level of output of each commodity at which the firm maximizes its profit. (b) the value of maximum amount of the total profit of the firm.

Example-Profit function

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Answer-Profit function

For the following total profit function of a firm:2y2-120y+xy = 144x --3x2 -35

(a) d/dx=144-6x-y=0, d/dy=-x-4y+120=0

x= 19.82 and y=25.04

(b) 2(25.04)2-120 (25.04)+(19.82)(25.049 = 144 (19.82)--3 (19.82)2 -35

=$ 2,895.09

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For the following total revenue and cost functions:TR=22Q-0.5Q2 and TC=(1/3) Q3- 8.5Q2 +50Q+90Determine (a) the level of output of Q commodity at which the firm maximizes its profit. (b) the value of maximum amount of the total profit of the firm. (c) Explain briefly part a and b

Example-TR/TC

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For the following total revenue and cost functions:TR=22Q-0.5Q2 and TC=(1/3) Q3- 8.5Q2 +50Q+90a=TR-TC

= 22Q-0.5Q2-((1/3) Q3- 8.5Q2 +50Q+90)= -1/3 Q3 + 8 Q2 -28Q-90 d/dQ= - Q2 + 16 Q2 -28Q

Q1 = 14 Q2=2 (b) = -1/3 (14)3 + 8 (14)2 -28 (14)-90=$ 171.4

(c) profit is max as Q=14 and min as Q=2.

d2/dQ2= -2 Q +16=0 (14) for -12 Max; (2) for 12 Min.

Answer-TR/TC

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New Management Tools

Benchmarking Total Quality Management Reengineering The Learning Organization

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Other Management Tools

Broadbanding Direct Business

Model Networking Pricing Power Small-World Model Virtual Integration Virtual Management

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The EndThe End

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