Mathematics Review For GSB 420. Instructor: Tim Opiela

Transcription

1 Mathematics Review For GSB 40 Instructor: Tim Opiela

2 I. lgebra Review. Solving Simultaneous Equations Two equations with two unknowns Supply: Q S = 75 +3P Demand: Q D = 5 P Solve for Equilibrium P and Q P = 5 -P 5P = 50 P = 0 Q = (0) = 05 or Q = 5 (0) = 05 B. Simplifying lgebraic Epressions 0. =. a = a 3. a b = a b 4. ( ) = a b a* b 5. ( ) * ( ) = a b a +b C. Eponential and Logarithmic Functions. Eponential Functions a. Simple Eponential Function: t y = b, has base b and eponent t and where b > [graph] b. Generalized Eponential Function

3 3 ct y = ab, where a and c are compressing and etending parameters. a pivots the function vertically and c etends it horizontally. [graph] c. Eponential Functions with base e (where e is the irrational number ) t y = e rt y = e, or more generally, rt pplications of y = e : () Interest Compounding () Instantaneous Rate of Growth (when we get to derivatives). Logarithmic Functions Natural Log t y = e t = loge y (or t = ln y) pplications with Logs: () Calculating Growth Rates () Linearizing non-linear functions Rules of Logarithms () Log of a Product: ln(uv) = ln(u) + ln(v) () Log of a Quotient: ln(u/v) = ln(u) ln(v) (3) Log of a Power: ln(u a ) = aln(u) [graph of y vs. t and t vs. y and lny vs. t]

4 4 II. Calculus The Derivative Concept: We wish to find the slope of a function. Knowledge of a slope is useful in providing a numerical measure of a relationship between two variables. This is a basis of causality and decision making. The slope is defined as the rise over the run: Given the function, Y = f ( ), Slope = rise run Y = Y For a linear function the slope is a constant throughout the function and.easy to calculate. rise = Y run = Y However, with a non-linear function the slope varies with. However, we cannot measure the slop at a particular because we need a change in. If we try measuring the slope for a discrete change in, we end up measuring it over an arc of the function. To measure the slope for any particular we must make the change in infinitesimally small. Y Y We define the slope as: = lim o Y That is; the derivative of Y with respect to is defined as the limit of Y as approaches zero. This is as close to a particular as we can achieve.

5 5 Rules for Differentiation: Constants: If : Y = a, where a is a constant. = 0 Linear Functions: If : Y = a + b Y = b Power Functions: If: Y = a b (note how this corresponds to the slope of a linear function) = ba b Sums and Differences: If: Y = u + w If: Products: If: Quotients: u w = + Y = u - w u w = Y=u*w u w w = * + * u If: Y = u w = u w w * * u w

6 6 Eponential base e: If: If: t Y = e rt Y = e = t t e Natural Logarithms: If: Y = lnt = t t If: Y = ln at Partial Derivatives = t = t t rt re Y = f (, z) = a z = f = az, which assumes that z is constant = z f z = a, which assumes that is constant The Differential dy = d f '( ) (derivative) dy = f '( ) d (differential) dy = f, z) d f (, z) dz (total differential of a multivariate function) ( +

7 7 III. Optimization. Unconstrained Optimization We are interested in finding the value of a control variable (i.e., the one that a decision maker can manipulate below) to maimize or minimize the dependent variable (i.e., the objective function). For eample, a firm may want to know how many workers to hire to minimize the cost of producing a given output, or a firm may want to know the level of output that maimizes profits. When the dependent variable (objective function) is maimized or minimized, the slope of the function is equal to zero. Thus, taking the derivative of the function (i.e., the slope) and setting it equal to zero will allow us to find the optimal. To know whether we have found a maimum or minimum requires us to look and the direction in which the slope is moving as increases. This latter information is gleaned from the second derivative of the function. Y ma min Y 0

8 8 Eample: S= (this function was taken from plotting past advertising vs. sales and fitting a function) where: S: total sales (millions of $) : advertising ependitures (millions of $) Objective: Find the $ amount of advertising ependitures MXIMIZES sales revenue? Solution: ) Take the first derivative (or slope of the objective function). S = 5 3 ) Set the slope equal to zero and solve for (the control variable). We call this the first-order condition (FOC) 5 3 = 0 5= 3 5 = 3 = $. 67million How do we know we are maimizing the objective function? We could plot the function or we could take the second derivative of the function. We call this the second-order condition (SOC). Y Y (5 3) = = 3 < 0 This tells us that as increases from its value determined in the FOC, the slope decreases. Therefore, we must have found a maimum when = $.67 million. Note: The Second Derivative is the Slope of the Slope direction the slope is moving as increases. We write this as Y or f or f.

9 9 The second derivative is defined as: Y Y = Y If: < 0 you have a maimum If: Y > 0 you have a minimum Optimizing Multivariate Functions: Often, there are more than one control variable that determines the objective function. For eample, assume where: Π=f( Q, Q) Π is total profits Q is output of good is output of good Q The objective function is given by, FOC: Π= Q Q Q 0Q 5Q Q Q 00 0Q 5Q = 0 () Q 80 0Q 5Q = 0 () Solve them for Q and Q ( equations with unknowns): from () 5Q 00 0Q Q = = 0 4Q (3) 5Q = 80 0Q from () (4) Q = 6 4Q

10 0 Substituting (3) into (4) we obtain: Q = 6 4( 0 4Q ) = Q 5Q = 64 Q Substituting from Q into (3) we obtain, Q = 0 4 * Q B. Constrained Optimization 933. In most applications, optimizing an objective function involves one or more constraints. Eample: Maimize total sales across two regions subject to the constraint of an advertising budget. Your company hired some hotshot consultant that estimated the following regions specific relationships between sales revenue in regions and advertising ependitures: S S = = where: S refers to sales revenue in region i (millions of $) i refers to advertising ependitures in region i (millions of $) i dditionally, you are provided with an advertising budget of $750,000 that cannot be eceeded.. Objective: To allocate advertising ependitures so as to maimize sales. Stated precisely, i.e., MX( S + S ) Subject to (s.t.) + = 075. MX s.t. + = 075. There are two general approaches: i) Substitution approach ii) Lagrangian approach

11 The substitution approach: MX s.t. + = 075. Note: = 075. (from the constraint) This allows the objective function to be rewritten as: + 5*( ) 5. *( ) MX *( ) wrt FOC 5. 4 = 0 = 035. From the constraint it follows that: = 075. = = Thus, you allocate advertising dollars over the two regions in the following manner: = $437, 500 = $3, 500 NOTE: general rule is that the MRGINL BENEFITS (MB refers to the sales associated with the last dollar spent) must be equal for optimization. i.e., S S = = = 0.35 We have: S = = 5 3 = 5 3*( ) = S = 0.35 = = = 4

12 Lagrangian pproach: Here we specify the constraint as part of the optimization function, called a Lagrangian function: L= λ( ) We then differentiate wrt each of the control variables and the constraint. FOC 5 3 λ = 0 () 4 λ = 0 () λ 075. = 0 (3) From () and () we solve for in terms of, 5 3 = 4 3 = + = From (3) and (4) we solve for, = = (4) nd from (3) we solve for we solve for, 035. = = The value added of this approach is that we have identified something called the shadow price ( λ ) of the constraint. From () we know that: λ= 4 = = This value is the marginal benefit (in terms of sales) of another dollar of advertising, in the case that constraint is loosened.