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

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

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 2 Optimization Techniques and New Management Tools 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 3 Optimization Techniques and New Management Tools 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 4 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-10Q 2 (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).

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 5 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.

© 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 7 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 9 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 10 Total, Average, and Marginal Cost 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 11 Total, Average, and Marginal Cost 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:

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 12 Total, Average, and Marginal Cost 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. Table3:

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 13 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. Table3:

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 15 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 16 Profit Maximization 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 18 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).

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 19 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. Graph5:

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 20 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. Graph5:

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 21 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:

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 22 Concept 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:

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 23 Concept 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:

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 25 Concept of the Derivative The derivative of Y with respect to X is equal to the limit of the ratio  Y/  X as  X approaches zero. 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 26 Concept of the Derivative-Example Suppose we have y=x 2 0 lim X dY dX   f(x+dx)- f(x) dX lim (x+dx )2- x 2  X  dX 0 lim X dY dX    dX 2xdx - + x 2 lim dY dX   X  (2xdx) 2x

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 27 Rules of Differentiation Constant Function Rule: The derivative of a constant, Y = f(X) = a, is zero for all values of a (the constant). For example, Y=2 dY/dX=0 the slope of the line Y is zero.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 28 Rules of Differentiation Power Function Rule: The derivative of a power function, where a and b are constants, is defined as follows. For example, Y=2x dY/dX=2

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 29 Rules of Differentiation Sum-and-Differences Rule: The derivative of the sum or difference of two functions U and V, is defined as follows. For example: U=2x and V=x 2 Y=U+V=2x+ x 2 dY/dX=2+2x

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 30 Rules of Differentiation Product Rule: The derivative of the product of two functions U and V, is defined as follows. For example:Y=2 x 2 (3-2 x) and let U=2 x 2 and V=3-2 x dY/dX=2x 2 (dV/dX)+(3-2x)(dU/dX) dY/dX=2 x 2 (-2)+ (3-2 x) (4x) dY/dX=-4x 2 + 12x+8 x 2 dY/dX= 12x-12 x 2

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 31 Rules of Differentiation Quotient Rule: The derivative of the ratio of two functions U and V, is defined as follows. For example: Y=3-2x/2x 2 and let V=2 x 2 and U=3-2 x dY/dX=(2 x 2 (dV/dX)+ (3-2 x) (dU/dX))/v 2 dY/dX=2 x 2 (-2)+ (3-2 x) (4x)/ (2 x 2 ) 2 dY/dX=4x 2 -12/4x 4 = (4x)(x-3)/ (4x) (x 3 )=x-3/x 3

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 32 Rules of Differentiation Chain Rule: The derivative of a function that is a function of X is defined as follows. For example: Y=U 3 +10 and U=2X 2 thendY/dU=3U 2 anddU/dX=4X dY/dX=dY/dU.dU/dX=(3U 2 ) 4X dY/dX=3(2X 2 ) 2 (4X)=48X 5

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 33 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 d 2 Y/dX 2 > 0, then X is a minimum. If d 2 Y/dX 2 < 0, then X is a maximum. For example: TR=100-10Q 2 d(TR)/dQ=100-20Q Setting d(TR)/dQ=0, we get 100-20Q=0 Q=5this means that its slope is zero and total revenue is maximum at the o/p level of 5 units.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 34 Optimization With Calculus Distinguishing between a Maximum and a Minimum: The second derivative For example: TR=100-10Q 2 d(TR)/dQ=100-20Q d 2 (TR)/dQ 2 =-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 d 2 (TR)/dQ 2 =-20, this TR function reaches a maximum at Q=5.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 35 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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 36 Example  =80X-2X 2 -XY-3Y 2 +100Y - total profit function We set d  /dX and d  /dY equal to zero and solve for X and Y. d  /dX=80-4X-Y=0 d  /dY=-X-6Y+100=0 Multiplying the first of the above expression by –6, rearranging the second and adding, we get -480+24X+6Y=0 100-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.

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 37 Constrained optimisation by substitution and Lagrangian Multiplier Methods Suppose that a firm seeks to maximize its total profit and the function as follows:  =80X-2X 2 -XY-3Y 2 +100Y but faces the constrain that the o/p of commodity X plus the o/p of commodity Y must be 12. That is, X+Y=12 First we can write X as a function of Y, such as X=12-Y And substituting X=12-Y into the profit function in inspection. Finally, we get:  =-4Y 2 +56Y+672

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 38 Solving y, we find the first derivative of:  with respect to Y and then set it equal to zero, d  /dY=-8Y+56=0Y=7 and X=5 and the profit is  =80X-2X 2 -XY- 3Y 2 +100Y=$868. Example for lagrangian method Suppose that we have a Lagrangian function as follows L  =80X-2X 2 -XY-3Y 2 +100Y+ (X+Y-12)

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 39 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=7X=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)

Managerial Economics © 2007/8, Sami Fethi, EMU, All Right Reserved. Ch 2: Optimisation Techniques 41 Other Management Tools Broadbanding Direct Business Model Networking Pricing Power Small-World Model Virtual Integration Virtual Management

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

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