Economics and Optimization Techniques

7/28/2019 Economics and Optimization Techniques

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CLASSICAL OPTIMIZATION

An optimizing function is defined as a rational action chosenconsistently to get the decision maker near to any desired goal as

circumstances permit.

In dealing with optimization problems we start with what is called anobjective function, in which the left-hand side is the object ofoptimization (for example, profits) while in the right-hand side we havea set of choice variables and a set of coefficients or constants. Thechoice variables (or decision variables) are the ones that the economicagent (consumer, firm, government etc) can vary (decide on) in its

attempt to optimize the objective function.

Well examine 2 types of optimization problems: Unconstrained andConstrained.

Unconstrained Optimization

Consider the function Y of variables Xi, and in the simplest case one

variable, X1:Y = f(X1)This function could be linear or non-linear, monotonically increasing ordecreasing or with various increasing and decreasing parts.

In optimization we are looking for an extreme value, i.e.,

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In the above graph on the left, suppose we are looking for a minimum,then point A represents the lowest point on the graph. Consequently, if

point A remains the lowest value of Y for all non-negative values of X,then A is called an absolute or global minimum. On the other hand, pointC is a relative orlocalminimum, since it is an extreme point only in theimmediate neighborhood of point C only.

When we search for an extreme point the first thing to look at is the firstderivative of the function. In the graph below:

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When the function has a max. (at X=X*) the slope of the functionchanges from positive to 0 and then to negative. This means that thisone-variable function has a max when the derivative is 0.However, in the graph below the same condition holds:

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The function has a minimum at point A where the first derivative is 0.This means that the first derivative test is a necessary condition foridentifying both a max and a min. That is the first derivative identifies a

stationary value (extreme point) of the function.

At this point we notice that in the case of a max the first derivativechanges sign from positive to negative as X increases, while in the caseof a min the first derivative changes sign from negative to positive.Therefore, the distinction between a max and a min relies on thedirection of change of the first derivative, which is measured by the

second derivative.

So, when the first derivative is decreasing as X increases, the secondderivative is negative and the function has a max, while when the firstderivative is increasing, the second derivative is positive and thefunction has a min.

Summarizing, the condition f(x)=0 is necessary to establish a max or amin, but not sufficient. A relative max exists if f(x) < 0 and a relative

min exists if f(x) > 0. This is thesecond order condition.Example:

Y = -2x2 + 20xThen, f(x) = -4x + 20 and =0 when x=5.The second derivative is f(x)= -4 < 0, and the function has a max atx=5.

Another possible stationary point is apoint of inflection.Consider a stationary value characterized by f(x)=0 and f(x)=0 atx=x*. This is neither a max nor a min.

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There is a stationary point at A (firstderivative =0 in second diagram). The first

derivative does not change sign, but remainspositive as X increases. For x< x*, f(x) decreases(f(x) x*, f(x) increases invalue (f(x) > 0). At x = x*, f(x) is at min (f(x)= 0). This is astationary point of inflection.

Another possibility is a non-stationary point ofinflection. Here, while f(x)=0 at C as in the

previous case, the slope (f(x)) is not horizontal atpoint C. Therefore, f(x) does not have a stationarypoint of inflection at x = x*. So, f(x)=0 is notrequired for an inflection point. A necessarycondition for a point of inflection is f(x) = 0.

5

f(x)

f(x)

f(x)

f(x)

f(x)

f(x)

*

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Example of non-stationary point of inflection (i.e., think of the short-runproduction function):

Y = -(2/3)x3 + 20x2

Graph the functions f(x), f(x), f(x) (restrictingf(x) to the non-negative quadrant).Inflection point where f(x) = 0.f(x) = -2x2 + 40xf(x) = -4x + 20 = 0, when x = 5

f(x) = -4 < 0(i) the non-stationary point of inflection is (5,500/3).

(ii) f(x) has a max at x = 5.(iii) For x < 5, f(x) > 0, f(x) > 0, so f(x) is

convex to the x-axis. For x > 5, f(x) > 0,f(x) < 0 and f(x) is concave to the x-axis.

(iv) Point (10, 1000/3) is a relative max.

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f(x)

f(x)

f(x)

f(x)


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Simple economic applications:

1. Profit maximization

Choose the rate of output which maximizes profits:

= R C, R = R(q), C = C(q)= R(q) - C(q)

1st order condition:

(q) = R(q) C(q) = 0, and:R(q) < C (q) (2nd order condition)

In economics this means that at the profit max output, the rate of changeof MR must be less than the rate of change of MC (i.e., the MC function

must cut MR from below).2. Relationship between AP of L (APL) and MP of L (MPL)

APL = Q/L = f(L)/L

To find the max point of the APL curve:d(APL)/dL = d[f(L)/L]/dL = [Lf(L) f(L)]/L2 = 0, or

f(L) = f(L)/L, or: MPL = APLthen,

d2(APL)/dL2 = { L2[Lf(L) + f(L) f(L)] [Lf(L) f(L)](2L)}/L4

= f(L)/L 2f(L)/L2 + 2f(L)/L3

Substituting the 1st order condition that Lf(L) = f(L):

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d2(APL)/dL2 = f(L)/L 2f(L)/L2 + 2Lf(L)/L3

= f(L)/L < 0 for max AP of labour.

So, given that L>0, the second order condition for max AP of labourrequires:

f(L) < 0, that is, d(MPL)/dL < 0 (diminishing MPL)

At the max point of the AP of labour, the AP of labour equals the MP oflabour, and at the same time the slope of the MP curve is negative (theMP curve cuts the AP curve from above).

Practice exercise

Q = -(2/3)L3 + 10L2,

Derive the AP of labour function and show that where AP is at max,MP=AP.--------------------------------------APL = -(2/3)L2 + 10L

APL/L = -(4/3)L + 10 = 0L = 7.5

MPL = -2L2 + 20LAt L=7.5: MPL = APL = 37.5

d2(APL)/dL2 = -(4/3) < 0

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Unconstrained optimization with more than one variable

Two-variable caseConsider the function:

z = f(x, y)

The first-order condition can be expressed in a similar fashion to the onevariable case. For z to obtain an extreme value at point A (top of a hill),the function must be stationary at this point. Momentarily, z neitherincreases nor decreases in value at this point:

dz = f(x)dx + f(y)dy = 0For any (non-zero) arbitrary variations dx and dy, it is necessary that:

f(x) = f(y) = 0

This is the first-order condition for a relative extreme point.

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From the graph, since A is a candidate max, this means that the tangentline through point A parallel to the xz plane has zero slope. Likewise,the tangent line drawn through point A // to the yz plane has a zeroslope.

The first-order condition does not tell us whether the extreme point is amax or a min. (because it also holds for minimum). We need a sufficientcondition or a second-order condition.

Given the function z = f(x, y),

The partial derivatives are:

z/x = f(x) = fx, z/y = f(y) = fy

The second partial derivatives with respect to x (while y is heldconstant) are:

fxx = /x(z/x) = 2z/x2 , and

fyy = /y(z/y) =

2

z/y

2

The above measure the rate of change of the first derivative with respectto x and y.

The cross-partial derivatives fxy, fyx are defined as:

fxy = /y(z/x) = 2z/yx, and

fyx = /x(z/y) = 2z/xy

Cross-partial derivatives measure the rate of change of a first partialderivative with respect to the other variable.

Consider now the concept of the second-order differential.

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Given the function:

Z = f(x, y)

The first order differential was:

dz = fxdx + fydy

The second order differential will be:

d2z = d(dz)

= (fxdx + fydy)/x(dx) + (fxdx + fydy)/y(dy)= (fxx dx + fyx dy)(dx) + (fxy dx + fyy dy)(dy)= fxx dx2 + fyx dydx + fxy dxdy + fyy dy2

= fxx dx2 + 2fxy dxdy + fyy dy2 < = > 0

Suppose we are looking for a max (at point A in the graph). It is going tobe a max if a tiny movement away from A in any direction results in adecrease in the value of z (dz < 0) in the neighborhood of A, given that

dz at A equals zero. In other words, we need dz to be decreasing (d

2

z 0 is the sufficient condition for a minimum.

The above condition, while sufficient, is not necessary because it ispossible for d2z to be 0 at the max or min.

The sufficient condition for max can be translated into a moreconvenient form:It can be shown that for any non-zero values of dx and dy, d2z will benegative (for max) iff:

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fxx 0, fyy > 0 and fxxfyy > f2xy

The second part of the sufficient condition will ensure that the surfacedecreases for max. (increases for min.) as we go away from point A inany direction (not only in the two basic direction).The above sufficiency condition can be inconclusive. For example, if:

fxxfyy = f2xy,this could be a max or a min but the test is inconclusive.

Alternatively, if: fxxfyy < f2xy, which can occur when fxx and fyy are ofdifferent sign, the function will have a saddle point. That is the sign ofd2z will be positive for some values of dx and dy and negative for others.

Functions with more than 2 variables

Consider the following function for optimization:

Y = f(X1, X2,.Xn)

The first order conditions for a stationary value are that all partialderivatives of Y must = 0 for all i = 1,n.

For example, in the case of profit max the objective function is:

= Pf(x1, X2,.Xn) wixi,

where xi are the inputs, wi are the prices of inputs and P is the price ofoutput.

The first order conditions are:

/Xi = P(f/Xi) - (wiXi)/Xi = 0 for all i.

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Or, i = Pfi wi=0 (value of marginal products equal to input prices).

In order to have a maximum, the second order conditions must besatisfied.

In the 2 variables case, recall that these conditions required:

fxx 0

f21 f22

To extend the second-order conditions to the n-variable case we makeuse of our knowledge of determinants.

Given some nth order determinant, we have called a principal minor oforder k of this determinant, what remains when any n-k rows and the

corresponding number of columns are eliminated from the determinant.With a 2x2 determinant, we can form 2 principal minors, a22 and a11.With a 3x3 determinant:

a11 a12 a13

a21 a22 a23

a31 a32 a33

we can form 3 principal minors of order n-1 by eliminating the 1st rowand 1st column, the 2nd row and column and the 3rd row and column.

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We can also for 3 principal minors of order n-2 by eliminating the 1 st

and 2nd, the 1st and 3rd and the 2nd and 3rd rows and columns.

The following theorem establishes the sufficiency condition:Theorem: Given a function y = f(X1, X2, .. Xn), which has a stationaryvalue at X = X* (a vector of Xi), and given the Hessian matrix of cross-

partial derivatives of f (fij) then, if all the principal minors of thedeterminant: det(fij) of order k have sign (-1)k for all k = 1,n at X =X*, then f(X1, X2, .. Xn) has a maximum at X = X*.

If all the principal minors are positive for all k at X = X* the function

has a minimum at X = X*.

If the signs do not follow the pattern in any of the above cases, thefunction has a saddle point at X = X*

Finally, if some or all of the principal minors are 0 and the rest have theappropriate sign, then it is not possible to determine whether there is amax or a min.

Going back to the profit max problem:

= Pf(X1, X2,.Xn) wiXi, (1)

The first order conditions (FOC) for max are:

/Xi = P(f/Xi) wi = 0 for all i.

In order to find the candidate values of vector X for max, we have tosolve the system of equations (the First Order Condition equations (1))for the Xs in terms of the parameters wi and P.

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If the system is solvable (that is the Jacobian determinant J =det(i/Xj) is not zero, meaning that the FOC equations areindependent) we can solve for the demand functions for inputs:

Xi = Xi*(wi, P).

The second order conditions require that the successive principal minorsof ij = Pfij have sign (-1)k, for k = 1,n.

Given that P > 0, we can express the matrix of second partial derivativesin terms of the production function. The (symmetric) Hessiandeterminant is:

f11 f12 f1n

f21 f22 f2n

fn1 fn2 fnn

and the successive principal minors are:

f11 f12H1 = f11, H2 =

f21 f22

f11 f12 f13

H3 = f21 f22 f21

f31 f32 f33

Hn = H

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These principal minors have to alternate in sign, starting with negative,i.e., f11 < 0. This in turn requires that all fij (diagonal elements) be < 0,which means diminishing marginal products.

In addition the 2nd order determinant must be > 0, i.e.,

fiifjj f2ij > 0.

The 3rd order must be < 0, etc.

By requiring that the signs alternate in sign starting with negativeamounts to requiring that d2y is negative definite. Furthermore, theseconditions, if satisfied, make sure that the objective function is concavein the neighborhood of a maximum.

The relevance of concavity and convexity in optimization

Second order conditions test whether the stationary point is the top of a

hill shaped function or the bottom of a valley shaped function. Theformer are called concave functions. The strict case of concavity isassociated with a global maximum. In the weak case of concavity, thehill may contain somewhere a flat section. A unique, absolute maximumrequires strict concavity.

The second order conditions we examined earlier, as they apply to the2nd order differential, d2z, can be distinguished in necessary conditions,i.e., when d2z is negative-semi-definite (when we have weak inequalities

in the relevant principal minor test) and in sufficient conditions, whend2z is negative definite (strict inequalities).

When we use d2z to test (or alternatively the principal minor test), thesign definiteness of the 2nd differential is evaluated only at the stationary

point, that is we establish a relative maximum.

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However, if the 2nd differential is everywhere negative semi-definite, thefunction is (strictly) concave and this implies an absolute maximum.This is a necessary and sufficient condition.

The stronger property of everywhere negative semi-definiteness of d2z issufficient but not necessary for strict concavity, because strict concavityof z is compatible with zero value of d2z.

Summarizing:

(1) Given a concave function, a stationary point can be identified as

an absolute maximum (i.e., from the first order conditions).

(2) Establishing concavity and strict concavity can replace the 2nd

order conditions as a sufficient condition for maximum.

(3) Even if d2z happens to be zero at the stationary point,establishing concavity or strict concavity can take care of this.

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Application 1

Consider the case of profit maximization by a multiproduct firm inperfect competition:

2 products, Q1, Q2

Products are technically interdependent (what does it mean?), i.e.,

C = c(Q1, Q2)

MC1 = C/Q1 = c1(Q1, Q2)

MC2 = C/Q2 = c2(Q1, Q2)

Prices are fixed at P1 and P2

TR = P1Q1 + P2Q2MR1 = TR1/Q1 = P1MR2 = TR1/Q2 = P2

The profit function is: = TR C = P1Q1 + P2Q2 - c(Q1, Q2) = (Q1, Q2)

Max = (Q1, Q2)

1 = /Q1 = P1 MC1(Q1, Q2) = 02 = /Q2 = P2 MC2(Q1, Q2) = 0

Assuming the 2nd order conditions are satisfied, we can solve for theoptimal values of Q1 and Q2.

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Example:

TR = 4Q1 + 8Q2C(Q1, Q2) = (Q1)2 + 2Q1Q2 + 3(Q2)2 + 2, then :

= 4Q1 + 8Q2 - (Q1)2 - 2Q1Q2 - 3(Q2)2 - 2

1 = 2Q1 + 2Q2 = 42 = 2Q1 + 6Q2 = 8 Q1 = Q2 = 1, = 4

The 2nd order conditions are:

11 = -2, 22 = -6, 12 = 21 = -2.So, 11 0). Therefore thesecond order differential is negative definite and we have a maximum.

Application 2Price discrimination

A single product firm (monopolist or firm with significant market

power) is selling output in more than one market.How much output should the firm sell in each market and what pricesshould it charge to each one to maximize profits?

Q = Q1 + Q2

Total revenue:TR = TR1(Q1) + TR2(Q2)Total cost is:

C = c(Q), (Q = Q1 + Q2)

MC1 = [c(Q)]/Q1 = (dC/dQ)*(dQ/dQ1) = dC/dQ = c(Q)MC2 = [c(Q)]/Q2 = (dC/dQ)*(dQ/dQ2) = dC/dQ = c(Q),

MC1 = MC2

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= TR1(Q1) + TR2(Q2) c(Q)1 = TR1/Q1 (C/Q)(Q/Q1) = (MR1) c(Q) = 02 = TR2/Q2 (C/Q)(Q/Q2) = (MR2) c(Q) = 0

These are the first-order conditions. From these we can find Q1*, Q2*and * = (Q1*, Q2*).

The first-order conditions imply:

MR1 = MR2 = MC.

Also observe that, since TR1 = P1Q1, then:

MR1 = dTR1/dQ1 = P1(dQ1/dQ1) + Q1(dP1/dQ1)= P1(dQ1/dQ1) + (Q1dP1/dQ1)(P1/P1)= P1 + (dP1/dQ1)(Q1P1/P1)= P1[1 + (dP1/dQ1)(Q1/P1)= P1(1 1/Ed1)

Similarly, MR2 = P2(1 1/ Ed2)

Then, from MR1 = MR2 = MC:

P1(1 1/Ed1) = P2(1 1/ Ed2)

We see that, P1 = P2 only if price elasticities are equal for the 2 demandcurves. If Ed1 < Ed2 at (optimal) Q1* and Q2*, then P1 > P2.

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PROFIT MAXIMIZATION AND INPUT DECISIONS

Consider a perfectly competitive firm using two inputs, K and Lproducing output Q:

Q = f(K, L)

Total cost is given by:

C = c(K, L) = rK + wL

Total revenue is:

TR = PQ = Pf(K, L)

Profit is:

= TR- C = Pf(K, L) - rK - wL

The first-order conditions for profit max. are:

K= Pf/K r = PfK r = 0

L = Pf/L w = PfL w = 0

We can solve the FOC for the optimal values of K and L (K* and L*),and optimal profit *.

The FOC state that:

PfK (=VMPK)= r

PfL (= VMPL)= w

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Example:

Q =6K1/3L1/2

P=2, r=4, w=3

Max = 12K1/3L1/2 4K 3L

K = 4K-2/3L1/2 4 = 0L = 6 K1/3L-1/2 3 = 0

Solving,

K* = 8, L* = 16, Q* = 48

Check the Second Order Conditions:

KK= -8/3L1/2K-5/3 < 0

LL = -3K1/3L-3/2 < 0

KL = 2K

-2/3

L

-1/2

LK= 2K

-2/3

L

-1/2

The Hessian determinant is:

-8/3L1/2K-5/3 2K-2/3L-1/2

2K-2/3L-1/2 -3K1/3L-3/2

= 8(K-4/3L-1) - 4(K-4/3L-1) = 4(K-4/3L-1) = 4/(K4/3L) > 0, maximum, alongwith KK, LL < 0.

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Comparative Statics Exercises

Starting from profit maximization:

= Pf(X1, X2) w1X1 + w2X2 (were X1, X2 are 2 inputs)

FOCs:

Pf1(X1, X2) - w1 = 0(1) [FOCs are Equilibrium conditions]

Pf2(X1, X2) w2 = 0

SOCs:

f11 < 0, f22 < 0, and f11 f22 f212 > 0

Solving the FOCs (implicit functions) for the 2 unknowns, X1 and X2,in terms of the parameters P, w1 and w2 we get:

X1 = X1*(w1, w2, P)

(2)X2 = X2*(w1, w2, P)

The above are the factor demand curves.

Comparative statics results

What happens to the optimal (profit maximizing) values of X1 and X2,as well as Q and the profit of the firm if there is a change in a factor

price or the price of output?

Establish the sign of the following derivatives:

X1*/w1, X1*/w2, X1*/P, X2*/w1, X2*/w2, X2*/P

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First substitute equations (2) into equations (1):Pf1{X1*(w1, w2, P), X2*(w1, w2, P)} w1 = 0 (3a)Pf2{X1*(w1, w2, P), X2*(w1, w2, P)} w2 = 0 (3b)

That is, the firm must employ X1*, X2* amounts of the 2 inputs to keepthe values of the marginal products for each factor exactly equal to the

price of each factor.

Now differentiate 3a and 3b partially with respect to w1 using the chainrule:

P(f1/X1)(X1*/w1) + P(f1/X2)(X2*/w1) 1 = 0

P(f2/X1)(X1*/w1) + P(f2/X2)(X2*/w1) = 0

Or,

Pf11(X1*/w1) + Pf12(X2*/w1) = 1 (4a)Pf21(X1*/w1) + Pf22(X2*/w1) = 0 (4b)

We need to solve for X1*/w1 and X2*/w1.

Suppose we multiply (4a) by f22 and (4b) by f12 and subtract (4b) from(4a) (and given that f12 = f21), then:

P(f11 f22 f212) (X1*/w1) = f22, so:

X1*/ w1 = f22/P( f11 f22 f212) (5a)Note that the denominator is > 0 because of the SOCs.

X2*/ w1 = -f21/P( f11 f22 f212) (5b)

Likewise, for a change in w2:

X1*/ w2 = -f12/P( f11 f22 f212) (5c)X2*/ w2 = f11/P( f11 f22 f212) (5d)

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Conclusion:

The factor demand curves (5a, 5d) are downward sloping. However, we

cannot determine the sign of the cross-effects (X1*/w2 andX2*/w1), as the signs of f12 and f21 are not known in the general case.

Changes in the price of the product

Differentiate (3a), (3b):Pf1{X1*(w1, w2, P), X2*(w1, w2, P)} w1 = 0 (3a)Pf2{X1*(w1, w2, P), X2*(w1, w2, P)} w2 = 0 (3b)

with respect to P:Pf11(X1*/P) + Pf12(X2*/P) = -f1Pf21(X1*/P) + Pf22(X2*/P) = -f2

Multiply the first by f22 and the second by f12, then subtract second fromfirst:

P(f11 f22 f

2

12) (X1*/P) = - f1 f22 + f2 f12and,

X1*/P = - f1 f22 + f2 f12 / P(f11 f22 f212) (6a)then,X2*/P = - f2 f11 + f1 f21 / P(f11 f22 f221) (6b)

As before, the signs of X1*/P and X2*/P cannot be established, asthe sign of the cross-derivative f12 (= f21) is unknown in the general case.

However, it can be shown that not both X1*/P and X2*/P can benegative.

[dQ = (Q/X1)(X1/P) + (Q/X2)(X2/P) > 0 ]

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Changes in Q

The production function is Q = f(X1, X2) and optimal output is:

Q* = f(X1*, X2*)

Given: X1 = X1*(w1, w2, P)X2 = X2*(w1, w2, P)

Then,Q* = f{X1*(w1, w2, P), X2*(w1, w2, P)} = Q*(w1, w2, P) (7)

This is the supply function of the firm.

Differentiate with respect to P to find how output changes when Pchanges:

Q*/P = (f/X1)(X1*/P) + (f/X2)(X2*/P)= f1(X1*/P) + f2/(X2*/P)

Substituting the expressions for (X1*/P) and (X2*/P),Q*/P = - f21 f22 + 2f12 f1f2 f22f11/ P(f11 f22 f212)

Although it is not immediately clear what sign the numerator has, itcould be shown that the sufficient condition for concavity of a functioncan be written as:

f22f11 - 2f12 f1f2 + f21 f22 < 0

then, the numerator is positive and Q*/P > 0.

Conclusion: the sufficient condition for profit maximization (concavityof a function) also implies that the supply curve is upward-sloping.

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We could also show that:Q*/w1 = - X1*/PQ*/w2 = - X2*/P,and the signs of these expressions are indeterminate.

Q* = f{X1*(w1, w2, P), X2*(w1, w2, P)}

Q*/w1 = (f/X1)(X1/w1) + (f/X2)(X2/w1)= f1(X1/w1) + f2(X2/w1)

X1*/w1 = f22/P( f11 f22 f212)

X2*/w1 = -f21/P( f11 f22 f2

12)

Then,

Q*/w1 = f1f22/P(D) f2f21/P(D) = - X1/P

[Since, X1*/P = - f1 f22 + f2 f12 / P(f11 f22 f212)]

Similarly for Q*/w2

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COMPARATIVE STATICS IN MACROECONOMICS

We are going to study the economy at a given state and observe how it isaffected by changes in various events.

We tend to aggregate the economy in up to 4 markets, i.e.,

1. The output market2. The money market3. the bond market4. The labour market

Because these markets are interdependent, once equilibrium isdetermined in 3 of them, the 4th one will necessarily be in equilibrium.Therefore, we can eliminate one of them (the bond market).

We are going to begin with the basic closed economy Macro modeland subsequently build on it.

The output market

Equilibrium in the product market:

Y = C + I + G (in real values)

In the simplest possible case:

C = c(Yd) 0 < c


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For a richer model we replace the exogenous investment functionwith:

I = I(r) (2)

Then,Y = c(Y-T) + I(r) + G (3)

This model gives rise to the IS curve.To find the slope of the IS differentiate (3):

dY = cdY + Idr, and

dr/dY = (1 c)/I,

which is negative because dI/dr is < 0. So, the IS curve is negativelysloped.

The money market

Money demand is a function of National Income, Y and the interest

rate, r.Treating money supply, Ms, as exogenous, and money demand is:

Md/P = L(Y, r), L1 = L/Y > 0 and L2 = L/r < 0.

Equilibrium in the money market gives the LM curve:LM: Ms/P = L(Y, r). (4)From (4):

L1 dY + L2 dr = 0

dr/dY = - L1/L2 = -L/Y/L/r, which is > 0.So, the LM curve is positively sloped.

Now combine the IS and the LM curves:

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The IS-LM model

Y - c(Y-T) - I(r) - G = 0 (IS)(5)

M/P - L(Y, r) = 0 (LM)

Solve for the endogenous variables Y and r in terms of the exogenousvariables, P, G, M, T.

In particular, by allowing the price level to vary, the solution:

Y = Y(P; G, M, T)

Is the aggregate demand curve (AD).

To find the slope of the AD curve, differentiate equations (5) withrespect to P (which in the IS-LM model is taken as fixed):

Y/P c(Y/P) I(r/P) = 0

-L1Y/P L2r/P = M/P

2

Or,

(1-c) (Y/P) - I(r/P) = 0

-L1(Y/P) L2(r/P) = M/P2

Or,

(1-c) -I Y/P 0=

-L1 -L2 r/P M/P2

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0 -I

M/P2 -L2 IM/P2 (-ve)Y/P = ------------------- = ----------------------- < 0

(1-c) -I -L2(1 c) L1I (+ve)

-L1 -L2

That is, the AD curve is downward sloping.

Fiscal Policy

An expansionary fiscal policy is characterized by an increase in G.This increase needs to be financed by:

(1) Issuing bonds (debt financed increase)(2) By raising taxes

(3) Printing moneyAssume case (1)Supply of bonds increases in the bond market, price of bonds decreasesthe interest rate (bond yield) increases (with money supply fixed). TheIS curve shifts to the right.

The effect of the increase in G can be obtained by differentiating (5):Y - c(Y-T) - I(r) - G = 0

M/P - L(Y, r) = 0with respect to G and solving for Y/G and r/GY/G (C/Y)(Y/G) (I/r)(r/G) = G/G = 1

(L/Y)( Y/G) + (L/r)(r/G) = 0

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Or,Y/G c( Y/G) I(r/G) = 1

L1(Y/G) +L2(r/G) = 0

From the second,

r/G = - (L1/ L2)(Y/G)

Substituting in the first and solving for Y/G and given that I < 0, L 1 >0, L2 < 0:

1Y/G = ---------------------------- > 0 (6)1-c + I (L1/ L2)

(-) (-)Then,

- L1/ L2r/G = ---------------------------- > 0 (7)

1-c + I (L1/ L2)The increase in the interest rate as a result of the increase in G representsthe partial crowding out effect.

Y/G varies inversely with - L1/ L2 which is the slope of the LM curve.

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Practice question:You can do the same for a change in T (taxes), in isolation or incombination with an increase in G (for combination see page 35).

Monetary Policy

Differentiate (5):Y - c(Y-T) - I(r) - G = 0M/P - L(Y, r) = 0with respect to M and derive Y/M and r/MY/M (C/Y)(Y/M) - I/r(r/M) = 0

L/Y (Y/M) + L/r(r/M) = 1/P

Or,

(1-c) (Y/M) I(r/M) = 0

L1(Y/M) + L2(r/M) = 1/P

Or,(1-c) -I Y/M 0

=L1 L2 r/M 1/P

and,0 -I

1/P L2 I/P I/PL2

Y/M = ------------------- = -------------------- = ------------------ > 0(1-c) -I (1-c)L2 + IL1 (1-c) + IL1/L2

L1 L2

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(1 c)PL2r/M = . = --------------------- < 0

(1-c) + IL1/L2

As I 0 or L2 monetary policy becomes inefficient in influencingnational income.

The relative effectiveness of fiscal and monetary policy is an empiricalmatter.

Dividing Y/G by Y/M we can see the determinants of the relativeeffectiveness of the 2 policies:

(Y/G)/(Y/M) = 1/(I/PL2) = PL2/I

The relative effectiveness depends upon the interest responsiveness ofthe demand for money (L2) and investment demand (I).

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The Balanced Budget Multiplier

In this case dG = dT.

Using equations (5) :

Y - c(Y-T) - I(r) - G = 0M/P - L(Y, r) = 0

Differentiate the model by allowing both G and T to change by thesame amount.

Y/G (C/Y)(Y/G) + (C/Y) (I/r)(r/G) = G/G = 1[Why? Remember, consumption depends on Yd = (Y-T)]Or,Y/G c(Y/G) + c I(r/G) = 1and,-L1(Y/G) L2(r/G) = 0, or:r/G = - (L1/ L2)(Y/G)

Then,(1-c) (Y/G) + c I[- (L1/ L2)(Y/G)] = 1

[1 c +I(L1/ L2](Y/G) = 1 c

(1 c)Y/G = ---------------------- < = 1

1 c +I(L1/ L2)

The effect of the tax does not fully offset the effect of the increase in G.the BBM will be equal to 1 only if investment is completely inelastic tochanges in the interest rate, or L2 (i.e, pure Keynesian case).

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The Labour Market

We have derived the IS, LM and AD curves. To be able to determine theprice level, P, we need the aggregate supply curve (AS).

This requires a labour market in the model.

Assume a production function involving K and L. In the short-run K isfixed, so that:

Y = f(K; L), f >0 , f < 0

Competitive firms maximize profit:

Pf(L) = W W/P = f(L)

[If on the other hand the firm has market power, it could be shown thatthe demand for labour is derived from: P(1 + 1/Ed)f(L) = W]

In the general case, the demand for L will depend on the real wage rate.

W/P = (L), with (L) < 0 (downward sloping demand for L).

To determine the supply of labor, Ls, we need to perform maximization,in the context of the workers work-leisure choice.

Assume a concave utility function which depends upon real income (Y)from work and leisure (T Ls), where T is the fixed hours available toworkers (24 hours) and Ls is hours of labour supplied.

Max U = u(Y, T-Ls)Subject to: Leisure = T Ls

Y = (W/P)Ls

Substitute into the utility function:

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Max U = u[(W/P)Ls, (T Ls)]

[(W/P)Ls] (T Ls)

U/Ls = u1 ---------------------- + u2 ---------------- = u1 (W/P) - u2 = 0Ls Ls

and,

W/P = u2 (Ls) / u1 (Ls)

So the labour supply function can be expressed as:

W/P = (Ls), > 0

This is a classical labour supply function, since the assumption is thatworkers care about their real wage rate.

Equilibrium in the labour market requires:

Ld = Ls = L, or

W/P = (L) = (L)

Add now the labour market functions into the model:

(1) Y c(Y-T) I(r) - G = 0(2) M/P L(Y, r) = 0(3) Y Y(L) = 0(4) (L) = (L)

(5) W/P = (L)

This is a recursive model:We have 5 equations with 5 endogenous variables, Y, r, P, L, W.[Note that: Real variables are Y, L (and W/P), while nominal variablesare: r, P, W]

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However, this model leads to the so called classical dichotomy. Equation4 alone determines employment and then Y and W/P are determinedfrom equations 3 and 5. So L, Y and W/P (the real variables) are

determined exclusively in the labour market and are independent from Gand M. So real activity is independent of fiscal and monetary policy.

4 L (5 W/P) 3 Y 1 r 2 P (and W/P) W

Fiscal PolicyTotally differentiate the system of equations allowing G to change:

(1-c)(Y/G) - I(r/G) = 1

L1(Y/G) + L2(r/G) + M/P2 (P/G) = 0

(Y/G) - Y(L/G) = 0

(-) (L/G) = 0

(L/G) (1/P)(W/G) + (W/P

2

)(P/G) = 0then,(1-c) - I 0 0 0 Y/G 1

L1 L2 0 0 M/P2 r/G 0

1 0 -Y 0 0 L/G = 0

0 0 (-) 0 0 W/G 0

0 0 -1/P W/P2 P/G 0

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1 . . . .

0 . . . .

0 . . . .

0 . . . .

0 . . . .Y/G= ---------------------------------

. . . . .

. . . . .

. . . . .

. . . . .

Looking at the numerator first (the denominator = IM(-)/P

3

):You can verify that the numerator = 0.

So, Y/G = 0, and fiscal policy is ineffective in the classical model.

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Monetary Policy

(1) Y c(Y-T) I(r) - G = 0

(2) M/P L(Y, r) = 0(3) Y Y(L) = 0(4) (L) = (L)(5) W/P = (L)

To determine the effect of monetary policy in the classical model,differentiate the 5 equation model, allowing for money supply to change:

(1-c)(Y/M) - I(r/M) = 0

L1(Y/M) + L2(r/M) + M/P2 (P/M) = 1/P

(Y/M) - Y(L/M) = 0

(-) (L/M) = 0

(L/M) - (1/P)(W/M) + (W/ P

2

)(P/M) = 0or,

(1-c) -I 0 0 0 Y/M 0

L1 L2 0 0 M/P2 r/M 1/P

1 0 -Y 0 0 L/M = 0

0 0 (-) 0 0 W/M 0

0 0 -1/P W/P2 P/M 0

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Y/M = ------------------------

The numerator is again = 0, so again Y/M = 0.

So, the real activity is independent of both fiscal and monetary policy.

The effect of the increase in G is to raise interest rates and reduceinvestment by an amount which exactly offsets the increase in G

(complete crowding out).With a higher r and fixed Y, money market equilibrium requires higherP. In order to preserve the real wage rate, money wages rise by the same

proportion as the price level.

Look at the aggregate supply curve (AS). Use the equations:Y Y(L) = 0(L) (L) = 0

The slope of the AS curve is Y/P:

Y/P = Y(L/P)

(L/P) (L/P) = 0

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then,

L/P = (Y/P)/Y, and

( - ) (Y/P)/Y = 0#0 #0

Y/P = 0.

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An alternative Labour Supply curve

While it is reasonable to assume that the demand for labour depends onthe real wage rate, this is hardly the case with labour supply because of

money illusion on the part of the workers (lack of information, incorrectexpectations about inflation, etc).

Consider the extreme case where labor supply depends exclusively onthe money (nominal) wage rate:

W = (Ls) (instead of W/P = (Ls))

The system of equations now becomes:

(1) Y c(Y-T) I(r) - G = 0(2) M/P L(Y, r) = 0(3) Y Y(L) = 0(4) P(L) = (L) (Ld = Ls)(5) W/P = (L) (demand for L)

[for 4, W/P = (L) and W = (L) equilibrium: P(L) = (L)]We can see that it is no longer true that the labour market alonedetermines L. Instead, the entire system is interrelated.This suggests that fiscal and monetary policies may now have aneffect.

Differentiate the system with respect to G and derive Y/G:[Differentiating (4): (L)*(dP/dG) + P*(dL/dG) = (L)]

Numerator determinant: L2Y/P (verify).

Denominator det: (1-c)L2Y/P + L1IY/P - IM(P )/P2

Dividing by the numerator:

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1Y/G = ----------------------------------------------------- > 0

(1-c) + I(L1/L2) - IM(P - )/L2YP2

[verify why it is indeed > 0]

Comparing the above multiplier with the one derived earlier from the IS-LM model (Y/G = 1/[(1-c) + I(L1/L2)], when prices were fixed, wesee that the value of the multiplier is reduced. The reason is that one ofthe effects of the increase in G is to raise prices, thereby reducing realmoney supply (real money balances), thus raising the interest rate

further.

However, compared to the pure classical model examined earlier, theinterest rate does not rise sufficiently to lead to an equal reduction ininvestment; so we get some increase in Y. The AS is upward sloping.

Y/P Y(L/P) = 0

+ P(L/P) (L/P) = 0 [i.e., differentiation of P(L) = (L)]then,

(P ) (L/P) = , (L/P) = /(P )and,

Y/P = Y[ /(P )], which is > 0.

Monetary policy

Similarly, we can derive an expression for Y/M (>0).

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Constrained Optimization

We can introduce constrained optimization by considering the familiarutility maximization problem. In the 2 variables case:

Max. U=f(X, Y)Subject to: PxX + PyY= M

To deal with the above we can stick to the previous methodology bydoing the following: Since in the constraint we have 2 arguments, X, Y,we could solve the constraint for, say, X and substitute into the objectivefunction, which can then be optimized as usual.

This could work nicely in this case but not in a case involving severalchoice variables and/or more than one constraint.

Suppose we have:

Max. Y = f(X1, X2, Xn)s. t : g = g(X1, X2, Xn)

We confront this type of optimization problem using the Lagrangianmethod.

It involves setting up the following function:

L = f(X1, X2, Xn) + (g - g(X1, X2, Xn))

The constraint must hold as equality. Looking at the function, whatever

the value of , when the constraint is satisfied, the second term on theright-hand side disappears and we are left with the objective function.

We treat as an additional variable, so we have n+1variables toconsider.

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The L-function will have a stationary value when the objective functionhas a stationary value. The FOC require that:

L/X1 = f/X1 (g/X1) = 0

.

.L/Xn = f/Xn (g/Xn) = 0L/ = g(X1, X2, Xn) = 0(n+1 equations in n+1 unknowns)

Take the pair-wise ratios of the FOC:

f1/f2 = g1/g2, ot fi/fj = gi/gj

Going back to the utility maximization problem:

Max. L=u(X, Y) + (M PxX PyY)

L/X = u/X Px = 0

L/Y = u/Y Py = 0L/ = M PxX PyY = 0Then,

(u/X)/(u/Y) = MUx/MUy = Px/Py

Second Order conditions

As in the case of optimization without constraints, the second orderconditions require that the second order differential must be negativesemi-definite for maximum and positive semi-definite for minimum(necessary conditions), or negative definite for maximum and positivedefinite for minimum (sufficient conditions).

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The difference when there are constraints is that, while before we wereconsidering all non-zero values of dX and dY, now we consider thosenon-zero values of dX and dY which satisfy the constraint (dg = 0).

The sufficient conditions translate into certain sign requirements of abordered Hessian determinant:

L11 L12 L1n g1 0 g1 g2 gn

L21 L22 L2n g2 g1 L11 L12 .... L1nH = .. or H = ................................

g1 g1 gn 0 gn Ln1 Ln2 ... LnnFind its successive border-preserving principal minors:

L11 L12 g1 L11 L12 L13 g1

H2 = L21 L22 g2 , H3 = L21 L22 L23 g2

g1 g2 0 L31 L32 L33 g3

g1 g2 g3 0....with the last one being H itself.

d2y is negative definite subject to dg = 0 iff:

H2 > 0, H3 < 0, H4 > 0 ... for maximum, and:

d2y is positive definite subject to dg = 0 iff:

H2 , H3, H4 < 0 ... for minimum.

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In the 2 variable case of utility maximization the bordered Hessian is:

u11 u12 -P1

H = u21 u22 -P2 and H2 = H

-P1 -P2 0

Therefore, H must be >0 for maximum.

Case of more than one constraints

L = f(X1, X2, Xn) + j[gj gj

(X1, X2, Xn)]

L11 L12 L1n g11 .. gr1

Ln1 Ln2 Lnn g1n .. grn

H = g11 ..................g1n 0 ......0

.............................................gr1 .................. grn 0 .......0

The sufficient conditions here state that for maximum the borderpreserving principal minors of order k > r alternate in sign, beginningwith (-1)r+1 , the second of opposite sign, etc.[Note that the order k is for rows and columns from 1 to n]For minimum, the border preserving principal minors of order k > r have

sign (-1)r.

The principal minors must be of order k > r because, by inspecting H weobserve the r x r matrix of zeros in the lower right side of thedeterminant; a determinant involving fewer than r rows and columnsfrom rows and columns 1 through n will be equal to 0.

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Establishing quasi-concavity and quasi-convexity in the constrainedmaximization case.

If we establish the shape of the function we can disregard the 2nd orderconditions.

For twice-differentiable functions, the test for quasi-concavity and quasi-convexity we are going to use is as follows:

From the following bordered determinant:

f11 f12 f1n f1 0 f1 f2 .... fn

f21 f22 f2n f2 f1 f11 f12 f1nD = ................................ , or .............................

fn1 fn2 ... fnn fnfn ................. fnn

f1 f2 .... fn 0

The above differs from the bordered Hessian because the border consistsof the first derivatives of the function, not the constraint. Then, weexamine the sign of the principal minors:

0 f1 f11 f1D1 = or

f1 f11 f1 0

0 f1 f2

D2 = f1 f11 f12 , D3, Dn.

f2 f21 f22

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For Y = f(X1, X2,Xn) to be quasi-concave, (with Xi >=0) it isnecessary that:

D1 =0

The necessary condition for quasi-convexity is:

D1 0

The function is quasi-concave.

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Now, lets go back to consumers maximization of utility.

L = U(X, Y) + (M PxX PyY)

FOCs:

Lx = Ux Px = 0

Ly = Uy Py = 0

L = M PxX PyY = 0

Solving the FOCs for X*, Y* and * yields the demand functions for the2 inputs X and Y as functions of prices of inputs and income.

Divide the first 2 condition through to get:

Ux / Uy = Px / Py

Solve for X as a function of Y (or vice-versa) and then use the budget

constraint to solve for the other.Interpretation of

The Lagrange multiplier provides a measure of the sensitivity of U* tochanges in the consumers budget (i.e., a relaxation of the budgetconstraint):

= U*/M

Thus, it measures the change in maximum utility resulting from achange in money income. It can be considered as the gain (loss) inmaximum utility resulting from a very small increase (decrease) in M,

prices held constant. It can be interpreted as the Marginal Utility ofMoney (income) when the consumers utility is maximized.

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Example:

U = 2(ln X1) + ln X2

s.t. 2X1 + 4X2 = 36

L = 2(ln X1) + ln X2 + (36 - 2X1 - 4X2)

LX1 = 2/X1 - 2 = 0

LX2 = 1/X2 - 4 = 0

L = 36 - 2X1 - 4X2 = 0

2X2/X1 = , and X1 = 4X2

Substituting into the budget constraint:X2* = 3, X1* = 12, * = 1/12, U* = ln(432) = 6.07

Checking the SOCs:L11 L12 g1

H = L21 L22 g2 then:

g1 g2 0

-2/(X12) 0 2

H2 = H = 0 -1/(X22) 4

2 4 0

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-1/(X22) 4 0 -1/(X22)= -2/(X12) - 0 + 2

4 0 2 4

= -2/(X12) (-16) + 2(2/(X22) = 32/(X12) + 4/(X22) > 0

Alternatively, we can check the shape of the function:

f11 f12 f1 -2/(X12) 0 2/X1

D = f21 f22 f1 = 0 -1/(X22) 1/X2

f1 f2 0 2/X1 1/X2 0which can also be written as:

0 2/X1 1/X2

D = 2/X1 -2/(X12) 0

1/X2 0 -1/(X2

2

)Then,

0 2/X1D1 = = -4/(X12) < 0

2/X1 -2/(X12)

2/X1 0 2/X2 -2/(X12)

D2 = -2/X1 + 1/X21/X2 -1/(X22) 1/X2 0

= 4/(X12) (X22) + 2/(X12) (X22) > 0The function is quasi-concave, therefore the FOCs represent amaximum.

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Cost Minimization subject to an output constraint

Q = f(K, L), fK, fL, > 0

C = rK + wL

What quantities of K and L should the firm employ to produce a giveoutput, Q*, at minimum cost?

[To obtain the slope of an isoquant we use the property that the changein Q along an isoquant is 0.

dQ = fKdK + fLdL = 0

MRTS = -dK/dL = fL/fK

This means that to have convex isoquants, (which are derived from aconcave production function), we require diminishing MRTS (i.e.,d(MRTS)/dL < 0), all along the isoquant.]

Now set up the Lagrangian function:L = rK + wL + (Q* - f(K, L))

Lets use an example:

Q = 4K1/2L1/2, Q* = 32

C = 2K + 8L

Then,

L = 2K + 8L + (32 - 4K1/2L1/2)

FOC:LK = 2 - 2 K-1/2L1/2 = 0

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LL = 8 - 2 K1/2L-1/2 = 0

L = 32 - 4K1/2L1/2 = 0

L/K = , K = 4L and using the constraint:

L* = 4, K* = 16, C* = 64, * = 2

SOCs:

LKK LKL gK

H = LLK LLL gL =

gK gL 0

K-3/2L1/2 -K-1/2L-1/2 - 2K-1/2L1/2

= - K

-1/2

L

-1/2

K

1/2

L

-3/2

- 2L

-1/2

K

1/2

- 2K-1/2L1/2 -2L-1/2K1/2 0

= . < 0

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Comparative statics

Certain comparative statics results can be obtained from the constrainedmaximization of utility problem.

L = U(Q1, Q2) + (I P1Q1 P2Q2)

For purposes of demonstration consider an example of a utility functionsuch as:

U = Q1Q2. Then:

L = Q1Q2 + (I P1Q1 P2Q2)

FOCs:

Q2 P1 = 0

Q1 P2 = 0

I

P1Q1

P2Q2 = 0

Take the total differential:

dQ2 P1d = dP1

dQ1 P2d = dP2

-P1dQ1 P2dQ2 = dI + Q1dP1 + Q2dP2

In matrix form:

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0 1 -P1 dQ1 dP1

1 0 -P2 dQ2 = dP2

-P1 -P2 0 d dI + Q1dP1 + Q2dP2

Suppose we consider only a change in P1 (so dP1 0, dP2 = 0, dI = 0).

Solve for dQ1:

dP1 1 -P1

0 0 -P2

Q1dP1 -P2 0 -(P2)2dP1 P2Q1dP1dQ1 = ----------------------------- = --------------------------------

0 1 -P1 2P1P2

1 0 -P2-P1 -P2 0

Divide by dP1:

dQ1/dP1 = (-P2) /2P1 Q1/2P1

Find the value of from the FOCs:

Q2 = P1Q1 = P2I P1Q1 P2Q2 = 0 = I/(2P1P2)So,

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Q1/P1 = -I/4(P1)2 Q1/2P1

i.e., if P1 = 1, P2 = 2 and I = 100

we can solve the FOCs for Q1* and Q2*:

Q1* = 50, Q2* = 25, and

Q1/P1 = -100/4 50/2 = -25 25 = -50

For every $ change in the price of good 1 the consumer will change his

purchases of Q1 by 50 units in the opposite direction.

Likewise we could derive Q2/P2, Q1/I, etc.

Lets go back to the FOCs of the general model:

U1(Q1, Q2) P1 = 0 (U1 = U/Q1)

U2(Q1, Q2) P2 = 0 (U2 = U/Q2)I P1Q1 P2Q2 = 0

Suppose there is a change in consumer income, I.

Differentiate the FOCs with respect to I:

U11(Q1/I) + U12(Q2/I) P1(/I) = 0

U21(Q1/I) + U22(Q2/I) P2(/I) = 0

1 P1(Q1/I) P2(Q2/I) = 0

Or,

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U11 U12 P1 (Q1/I) 0

U21 U22 P2 (Q2/I) = 0

P1 P2 0 (/I) -1

0 U12 P1

0 U22 P2 U12 P1-

-1 P2 0 U22 P2(Q1/I) = ------------------------- = --------------------- > = < 0U11 U12 P1 D (>0)

U21 U22 P2

P1 P2 0

Similarly,U11 P1

U21 P2(Q2/I) = ------------------- > = < 0

D (>0)

U11 U12-

U21 U22(/I) = ------------------------ > = < 0

D (>0)

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So, there is a possibility of inferior goods. However, not both (Q1/I)and (Q2/I) can be negative (intuitively) and formally:From: P1Q1 + P2Q2 = I:P1(Q1/I) + P2(Q2/I) = 1 (>0)

And since P1, P2, > 0 not both effects can be negative.

Lets now differentiate the FOCs:U1(Q1, Q2) P1 = 0 (U1 = U/Q1)U2(Q1, Q2) P2 = 0 (U2 = U/Q2)I P1Q1 P2Q2 = 0

with respect to the price of one input, say P1:

U11(Q1/P1) + U12(Q2/P1) P1(/P1) - (dP1/dP1) = 0

U21(Q1/P1) + U22(Q2/P1) P2(/P1) = 0

P1(Q1/P1) P2(Q2/P1) = Q1(dP1/dP1)

Or,U11 U12 P1 (Q1/P1)

U21 U22 P2 (Q2/P1) = 0

P1 P2 0 (/P1) Q1

U12 P1

0 U22 P2 U22 P2 U12 P1

Q1 P2 0 P2 0 U22 P2(Q1/P1) = --------------------- = ------------------ + Q1 ------------------- (1)

D D D

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Likewise,

U21 P2 U11 P1

P1 0 U21 P2(Q2/P1) = . = - ------------------- - Q1 ------------------ (2)

D D

U21 U22 U11 U12

P1 -P2 U21 U22

(/P1) = . = ------------------- + Q1 --------------- (3)D D

All 3 are of indeterminate sign.

We define goods as substitutes if an increase in the price of one goodincreases the demand for the other. And vice-versa for compliments.

Thus above, Qi/Pj > 0 means Qi and Qj are gross substitutes.

Lets attempt to interpret the results from (1) and (2):

Consider the Q1/I, Q2/I results:

U12 P1-

U22 P2Q1/I = -------------------,

D

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U11 P1

U21 P2

Q2/I = -------------------D

The above expressions are exactly the second terms of equations (1) and(2) multiplied by Q1 (income effect).

On the other hand, the first parts in equations (1) and (2) are the puresubstitution effects of a change in price (movement along the same

indifference curve), i.e., Qu

/P, so we can write equations (1) and (2) as:

(Q1/P1) = Q1u/P1 Q1(Q1/I) (3) (subst. effect) (income effect)

(Q2/P1) = Q2u/P1 Q1(Q2/I) (4)

In (3) the substitution effect is clearly negative while the income effect is

of indeterminate sign because the sign of Q1/I is unknown.The above are known as Slutsky equations.

Similarly:

(Q1/P2) = Q1u/P2 Q2(Q1/I)

(Q2/P2) = Q2u/P2 Q2(Q2/I)

And in general,

(Qi/Pj) = Qiu/Pj Qj(Qi/I)

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The Slutsky equations show that the response of the consumer to thechange in the price can be split-up into 2 parts: first, a pure substitutioneffect (or the response to a price change holding the consumer on theoriginal indifference curve), and second, a pure income effect, where

income is changed (purchasing power of a given money income) holdingprices constant.

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Comparative statics: Theory of the firm

Max. Q = f(X1, X2)

s. t. C = r1X1 + r2X2

Lagrangian function:

L = f(X1, X2) + (C - r1X1 + r2X2)

FOCs:

L/X1 = f1 r1 = 0

L/X2 = f2 r2 = 0

L/ = C - r1X1 + r2X2 = 0

Then,

f1 / f2 = r1/r2, = f1 /r1 = f2 /r2 r1 = f1 r1 = (1/) f1

and r2 = (1/) f2

By differentiating the constraint:

dC = r1dX1 + r2dX2, or

dC = 1/(f1dX1 + f2dX2) (1)

By taking the differential of the production function:

dQ = f1dX1 + f2dX2 (2)

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Dividing (2)/(1):

f1dX1 + f2dX2dQ/dC = ( ---------------------) = = 1/MC

f1dX1 + f2dX2

The SOCs require that the bordered Hessian determinant is positive:

f11 f12 -r1

f21 f22 -r2 > 0

-r1 -r2 0

This requires that the production function is quasi-concave.

Lets go back now at the equivalent unconstrained profit maximization

problem: = Pf(X1, X2) r1X1 r2X2, and derive some comparative staticsresults.

Comparative statics: Profit maximizationMax: = Pf(X1, X2) r1X1 r2X2

Comparative statics results from the profit maximization problem can bederived by first deriving input demand functions using the FOCs:1 = Pf1(X1, X2) r1 = 02 = Pf2(X1, X2) r2 = 0

As prices of inputs change, the producer will change the use of inputs sothat the FOCs continue to be satisfied.

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By totally differentiating these conditions:

Pf11dX1 + Pf12dX2 = -f1dP + dr1

Pf21dX1 + Pf22dX2 = -f2dP + dr2

Or,

Pf11 Pf12 dX1 -f1dP + dr1=

Pf21 Pf22 dX2 -f2dP + dr2

Letting dr2 = 0 and dP = 0,

dr1 Pf12

0 Pf22 dr1Pf22dX1 = ------------------ = --------------

Pf11 Pf12 P

2

|D|Pf21 Pf22

and,

dX1/dr1 = f22 /P |D| < 0 (|D| = f11f22 f212 > 0)

The response to the change of an input price (other things constant) isalways negative.

Similarly, with dr1=0, dP = 0,

dX2/dr2 = f11 /P |D| < 0

On the other hand, with dr1=0, dP = 0,

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dX1/dr2 = - f12 /P |D|

If we assume that an increase in the use of one input increases the MP ofthe other, i.e., if f12 > 0, then:

dX1/dr2 < 0

Finally, let dr1 = 0, dr2 = 0 and allow P to change:

-f1dP Pf12

-f2dP Pf22 -Pf1f22dP + Pf2f12dPdX1 = ----------------------- = -----------------------------P2 |D| P2 |D|

dP(f2f12 f1f22)= = ------------------------ , and:

P |D|

(f2f12 f1f22)dX1/dP = ----------------------P |D|

Here too, if f12 > 0, dX1/dP > 0 and an increase in output price will resultin an increase in input demand.

Further question:

Assume a Cobb-Douglas production function, Q = (X1)a(X2)b, with a andb > 0 and a + b


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=PX1aX2b r1X1 r2X2

1 = PaX1a-1X2b r1 = 02 = PbX1aX2b-1 r2 = 0

Pa(a-1)X1a-2X2b(dX1) + Pab X1a-1X2b-1(dX2) = -aX1a-1X2b (dP) + dr1

PabX1a-1X2b-1(dX1) + Pb(b-1) X1aX2b-2(dX2) = -bX1aX2b-1 (dP) + dr2

1 Pab X1a-1X2b-1

0 Pb(b-1) X1aX2b-2 Pb(b-1) X1aX2b-2

dX1/dr1 = ----------------------------------- = -------------------------- < 0P2 |D| P2 |D|

Similarly:

dX1/dr2 > 0dX1/dP > 0

Can you find these expressions ifa + b =1? (see page 70).

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Homogeneous functions

Given: f(x1, x2,xn),A function is homogeneous of degree r, if:

f(jx1, jx2,jxn) = jrf(x1, x2,xn), where j>0.

Linearly homogeneous functions

These are homogeneous functions of the first degree, i.e., those exhibitingCRTS.

Properties:

(1)Given: Q = f(K, L): (1)the APPL and APPK, as well as the MPPL and MPPK can be expressed asfunctions of the capital/labor ratio, k=K/L alone.

Multiply (1) by a factor j = 1/L. Because of linear homogeneity (CRTS),output will be multiplied by jQ = Q/L, So:

f(K, L) = f(K/L, L/L) = f(K/L, 1) = f(k, 1) = f(k)APPL = Q/L = (k) [Q=L(k)]then:

APPK = Q/K = (Q/L)(L/K) = (k)/k.The above imply that as a result of linear homogeneity, if K/L is keptconstant, the average products will stay constant too (i.e., average productfunctions are homogeneous of degree 0).

Similarly for MPPL and MPPK (se book for proof)

Consider the Cobb-Douglas production function with constant returns toscale: Q = AKaLb = AKaL1-a

MPPL = Q/L = A (1-a)KaL-a = A(1-a)Ka/La = A(1-a)(K/L)a, andMPPK = Q/K = = Aa(K/L)a-1

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In economic terms, this means that the MP of labor and capital areconstant whenever the capital to labor ratio is constant. That is the MP areindependent of the scale of production, and depend only on capitalintensity (K/L).

A puzzle

Use a production function exhibiting CRTS and perform comparativestatics. Earlier we asked the question:

Can you find the expressions dX1/dr1, dX2/dr2, dX1/dr2, etc, when ifa+ b =1?

1 Pab X1a-1X2b-1

0 Pb(b-1) X1aX2b-2 Pb(b-1) X1aX2b-2

dX1/dr1 = ----------------------------------- = --------------------------P2 |D| P2 |D|

The determinant |D| is:

Pa(a-1)X1a-2X2b Pab X1a-1X2b-1

PabX1a-1X2b-1 Pb(b-1) X1aX2b-2

= (P2ab(a-1)(b-1)X12a-2X22b-2) - (P2a2b2 X12a-1X22b-1)

= (X12a-2X22b-2) [P2ab(a-1)(b-1) - P2a2b2]= (X12a-2X22b-2) [P2ab(-b)(-a) - P2a2b2]= 0So these effects are indeterminate under CRTS.

Why?

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Under CRTS, MC and AC are constant (independent of the scale ofproduction). So with a horizontal long run average cost curve (instead ofU-shaped), there is no unique combination of inputs for the firm; i.e., thefirm is indifferent as long as the first order conditions are satisfied and the

firm makes zero economic profits.

Eulers theorem

Property (2)If Q= f(K, L) is linearly homogeneous:

K(f/K) + L(f/L) Q (for proof see book). (1)

One should distinguish this identity which holds for any values of K andL (but only for constant return to scale production functions) from thetotal differential of Q: dQ = (Q/K)dK + (Q/L)dL, which holds forany function, not only CRTS functions.

Eulers theorem says that under CRTS, if each input is paid according toits marginal product contribution, payments to inputs will exactly exhaustall output (product) i.e., the pure economic profit will be 0.

Note that from the first order conditions:

(f/K) = r/P and (f/L) = w/PSubstituting into the Euler equation and multiplying both sides by P:

rK + wL = PQ (2)(under CRTS payments to owners of inputs exhaust revenue).

However, the above equation is subject to the adding up problem. If weconsider increasing returns to scale the Euler theorem says:

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K(f/K) + L(f/L) mQ, with mQ>Q, and form equation (2):rK + wL > PQ,That is total revenue will not be enough to reward factors of production atthe existing market prices. So factors cannot receive rewards equal to

their marginal products.

Similarly, under decreasing returns to scale, total revenue is more thansufficient to reward factors according to their MP (there is a surplus).This leave unanswered the question: what should be done with thesurplus.Because of the above issues, neoclassical economic theory tends to usethe assumption of CRTS.

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Non-linear optimization with inequality constraints

Introduction

It refers to the problem of finding the optimal value of a given functionf(X1, X2,Xn) of n variables, usually restricted to be non-negative (Xj 0), subject to one or more constraints which are in generalinequalities:, g(X1, X2, , Xn) { , =, or } bi.f and g are not both linear.

i.e.,

Min f = (X1 4)2

+ (X2 -3)2

s. t. 3X1 + 2X2 12 (g1)-2X1 + 2X2 3 (g2)2X1 X2 4 (g3)2X1 + 3X2 6 (g4)

X1, X1, 0

The optimal point is X1 = 34/13, X2 = 27/13, f =2.77.To find this point we utilize the fact that at the tangency point (seegraph), the slopes are equal. The slope of g1 is -3/2. To find the slope ofthe objective function, we take the total differential:

(X1 4)2 + (X2 -3)2 f = 0

2(X1- 4)dX1 + 2(X2 3)dX2 0 = 0

2(X1- 4) + 2(X2 3)dX2/dX1= 0

But at tangency between f and g1 the slope is -3/2.

So, 2(X1- 4) + 2(X2 3)(-3/2) = 0 2X1 3X2 = -1

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Solving along with 3X1 + 2X2 = 12, we get the solution given above.

One can show that the optimal can occur inside the Feasible Set.Example: Min Z = (X1 9/5)2 + X2 2)2 (same constraints)

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Now the optimal value of Z occurs at an interior point (2, 9/5) and Z = 0at minimum.

We can also show that a local optimum is not always a global one, i.e.:

Suppose we were looking for maximum instead of a minimum (sameobjective function and constraints). Point A will provide a localmaximum (Z = 169/100), but point B would give global maximum, thatis Z at B > Z at A.

Finally, a further complication can result; example:

X1X2 b1

X12 + X22 b2

X1, X2, 0

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This results in a feasible set that is not convex (it is disjoint).

Some N.L.O. problems have a solution, some do not. A solution may notexist if the constraints contain inconsistencies (as in the example above)

or if the problem is unbounded.Necessary conditions for optimum.

In handling classical optimization problems we used calculustechniques. A convenient approach there was the Lagrangian method.Can we extend this approach to N.L.O. problems (that is problems withinequality constraints)?

Kuhn and Tucker have shown that:

(1) There is a wide class of problems for which a Lagrangian approachcan be followed; when the Lagrangian function is optimized, thesame values that optimize the Lagrangian will also optimize theoriginal function subject to the constraints.

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(2) Furthermore, suppose we have a maximization problem: We set upthe L-function and treat the L-multipliers as variables (as we didearlier); then suppose we solve the problem. We say that it is anoptimal solution only when we have found the values of the

variables of the maximization problem (X1, X2,) that maximizethe L-function and at the same time the values of the multipliers() which minimize the value of the L-function. This solution iscalled a saddle point.

**

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This saddle point is a consequence of duality. The duality conceptspecifies a primal-dual, or max-min relationship.

Suppose we have a profit maximization problem. The dual will be a costminimization problem. Let X* be the optimal set of the max. problemvariables and Y* be the optimal set of the dual variables.

Then,

- The values of the primal variables X* which maximize the primalobjective function, Z(X), will also maximize the L-function inwhich the L-multipliers take the dual optimal variable values, Y*.

- Similarly, the value of the dual variables Y* that minimize the dualminimization problem will also minimize the L-function, L(Y*,X*) in which the multipliers are the primal optimal values, X*.

Therefore, the Lagrangian multipliers in the primal problem are equal tothe dual optimal variables, and the dual multipliers are equal to the

primal optimal solution values (More about duality in Linear

Programming).

To set up the L-expression in the case of NLO problems we have tofollow a similar but slightly different approach from the case of classicalconstrained optimization.

Given the problem:

Max. Z = f(X1, X2, Xn)s. t. g1(X1, X2, Xn) b1.gm(X1, X2, Xn) bm

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Then,

L(X, ) = f(X1, X2, Xn) + 1[b1 - g1(X1, X2, Xn)] + +m[bm - gm(X1, X2, Xn)]

The Kuhn-Tucker conditions are necessary (but not sufficient) foroptimization. In other words, they are the First-Order-Conditions for

NLO problems.

Following are the K-T conditions side by side with the FOCs forclassical constrained optimization:

Classical optimization (Max.) K-T Conditions forMax.L(X, )/Xj = 0 L(X, )/Xj 0, and

Xj[L(X, )/Xj] = 0

L(X, )/i = 0 L(X, )/i 0, andi[L(X, )/i] = 0

Xj, i, 0

(for minimization the inequalities are reversed)

Example:

Max Z = X13 + X1X2s. t. X1 + X25 10

X1

2

+ 2X2 4X1, X2, 0

Then,

L(X, ) = X13 + X1X2 + 1(10 - X1 - X25) + 2(-4 + X12 + 2X2)

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We entered the constraints in such a way so that when we take thederivative of L with respect to 1 and 2 we get the result that the K-Tconditions L/i 0 are equivalent to the constraints:

L/1 = 10 - X1 - X25 0 (equivalent to X1 + X25 10)

L/2 = -4 + X12 + 2X2 0 (equivalent to X12 + 2X2 4)

Now the K-T conditions are:

L/X1 = 3X12 + X2 1 + 22X1 0 (1)

L/X2 = X1 - 51X24 + 22 0 (2)

Along with,

X1[3X12 + X2 1 + 22X1] = 0 (1a)

X2[X1 - 51X24 + 22] = 0 (2a)

and,

L/1 = 10 - X1 - X25 0 (3)

L/2 = -4 + X12 + 2X2 0 (4)

Along with,

1[10 - X1 - X25 = 0] (3a)

2[-4 + X12 + 2X2 = 0] (4a)

X1, X2, 1, 2 0

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The K-T conditions are in fact generalizations of the FOCs in classicalconstrained optimization.

Having inequality constraints amounts to allowing for the possibility thatthe optimum occurs at a point where one or more of the solution valuesare = 0 (Xj = 0), i.e., the possibility of a boundary or corner solution asopposed to an interior solution. The K-T conditions apply to both cases.In this respect, they describe a more general case and therefore containthe conditions for classical optimization.

i.e.,

for y = f(x),

A max. can occur at a corner point as well as an interior point. If Y/X= 0 (point B), we may have a max.

At C we have Y/X < 0 and X = 0. Can this represent a maximum?Yes, because the value of Y will decrease to the right of point C, whichfrom the non-negativity constraints has to be at the positive orthand.

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However, for max. we cannot have Y/X > 0 at X = 0, because then,we can increase the value of Y by increasing X.Putting the above together and generalizing for functions with more thanone variables, for max. we have:

Either Y/Xi = 0 and Xi > 0 (interior max.), or (point A)Y/Xi < 0 and Xi = 0 (corner max.) (point C)

Y/Xi = 0 and Xi = 0 is also possible (point B)

Which one applies?

From the K-T conditions: Xj[L/Xj] = 0

This means that if Xj > 0 L/Xj = 0If L/Xj 0 Xj = 0

That is, either Xj = 0 or L/Xj = 0 (or both)

Similar conclusions are derived from i[L/] = 0.

If we have minimization problem, the signs of the inequalities are

reversed.L/Xj 0

L/i 0

Intuitive explanation

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Max. Z = f(X1, X2, Xn)s. t. g1(X1, X2, Xn) b1

.gm(X1, X2, Xn) bm

Consider the Lagrangian expression:

L(X, ) = f(X1, X2, Xn) + 1[b1 - g1(X1, X2, Xn)] + +m[bm - gm(X1, X2, Xn)]

Condition L/i 0 for maximum means:

L/i = bi - gi(X1, X2, Xn) 0, which is the ith constraint. That is,we know that the constraints are satisfied.

Then, by the complimentary condition: i (L/i) = 0, ori [bi - gi(X1, X2, Xn)] = 0 for all i, these terms on the right hand sidevanish and we are left with:

L(X*, *) = f(X1, X2, Xn), which is the objective function.Therefore, a solution which satisfies the K-T conditions for the L-function must also:

(1) Satisfy the constraints.(2) Will make the value of L equal to that of the objective function.

Sufficient conditions

The candidate point will be a global max (min) if:

(1) For max (min), the set of constraints form a Feasible Region whichis convex (concave) i.e., the constraint functions must be convex.

(2) For maximum the objective function must be concave. For aminimum it must be convex.

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(3) The K-T conditions must be satisfied.[Notice that the classification for convex vs. concave can refer to afunction or a region]

84

Non-strictlyconcave

Strictl concave

Strictl convex


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Note also that a linear function satisfies both convexity and concavity.

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Computational aspects

We are still looking for a method by which we can solve a NLOproblem.

We can approach this in an informal way, by examining the K-Tconditions using examples.

Consider the problem:

Min. C = (X1 3)2 + (X2 4)2s. t. X1 + X2 4

X1, X2, 0

Then,

L = (X1 3)2 + (X2 4)2 + (4 - X1 - X2)

(1) L/X1 = 2(X1 3) 0 (=0 if X1 > 0)(1) X1[2(X1 3) ] = 0

(2) L/X2 = 2(X2 4) 0 (=0 if X2 > 0)(2) X2[2(X2 4) ] = 0

(3) L/ = 4 - X1 - X2 0 (=0 if > 0)(3) (4 - X1 - X2) = 0

We have 4 cases to consider:

1. X1 = 0, X2 = 0 (immediately rejected)2. X1 = 0, X2 > 03. X1 > 0, X2 = 04. X1 > 0, X2 > 0

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Case 2: If X1 = 0, X2 > 0,

From (1): -6 0, or 6 + 0 < 0If is also restricted to be non-negative, we already have a violation.

Furthermore:

From (2), with < 0: 2X2 8 = 0 2X2 8 = (< 0) X2 < 4.Then from (3): 4 X2 < 0 (contradiction/violation).

Case 3: If X1 > 0, X2 = 0,

From (1): 2X1 6 = 0 (since X1 > 0).

From (2): -8 0 8 + 0 < 0.From (1), with < 0, 2X1 6 < 0 X1 < 3.From (3): 4 X1 < 0 (contradiction/violation)

Case 4: If X1 > 0, X2 > 0, >0

2(X1 3) = 0 2X1 6 = (1)2(X2 4) = 0 2X2 8 = (2)

And the constraint is equality, therefore:X1 + X2 = 4 (3)

Solve the 3 equations to get:

X1 = 3/2, X2 = 5/2, C=4.5.This solution satisfies the K-T conditions and the constraint; however,this is not the optimal solution.

Case 5: X1 > 0, X2 > 0, = 0

The point which minimizes the objective function and satisfies the K-Tconditions and the constraint is: X1 = 3, X2 = 4, = 0.[2X1-6 = 0 X1 = 3, and 2X2 8 = 0 X2 = 4]

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The optimal value of the objective function is: C = 0. The constraint issatisfied as a strict inequality.

Example 2

Max. -3e-2X1 4e-5X2

s. t. X1 + X2 1X1, X2, 0

L = -3e-2X1 4e-5X2 + (1 X1 X2)

L/X1 = 6e-2X1

0 (=0 if X1 > 0)

L/X2 = 20e-5X2 0 (=0 if X2 > 0)

L/ = 1 X1 X2 0 (=0 if > 0)

Examine the 4 cases:

(1) X1 = 0, X2 = 0 (rejected: it implies >0 1- X1-X2 = 0)(2) X1 = 0, X2 > 0, then:

6 0Furthermore,

20e-5X2 = (since X2 > 0) > 0

Thus, 1 X2 = 0 X2 = 1

Then, 20e-5 = 0.134 which violates 6 0.

(3) X1 > 0, X2 = 0, then:

6e-2X1 = 020 - 0 > 0, and:

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1 X1 = 0 X1 = 1So, 6e-2 = 0.81 = , which violates 20 0.

(4) X1 > 0, X2 > 0, then:

6e-2X1 =

20e-5X2 =

> 0

Hence, 1 X1 X2 = 0 X2 = 1 X1.

Substitute:

20e-5(1-X1) = 6e-2X1

(20/6) e-5(1 X1) = e-2X1

(20/6) = e-2X1 + 5(1-X1)

Or, 3.33 = e-7X1 + 5

Or, ln(3.33) = -7X1 + 5 = 1.2

X1 = 0.54, X2 = 0.46

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Example: Maximization of Utility

The model of a consumer maximizing utility, U(X1, X2), subject to abudget constraint can be given a treatment using K-T analysis.

The qualifying difference here, compared to the usual treatment, is thatnow the consumer is not necessarily spending all his income and doesntneed to consume positive amounts of both goods.

Max: U(X1, X2)

s. t: P1X1 + P2X2 I

X1, X2, 0

L = U(X1, X2) + (I - P1X1 - P2X2)

K-T conditions:

L/X1 = U1 P1 0 (=0 if X1 > 0)

L/X2 = U2 P2 0 (=0 if X2 > 0)L/ = I - P1X1 - P2X2 0 (=0 if > 0)

Recall that the Lagrange multiplier, , has been interpreted as the MU ofmoney income, I. Consider as a first case:

X1, X2, > 0, then:

U1 = P1U2 = P2

(MU of money income) = U1 / P1 = U2 / P2

This represents the MU per $ spent on X1 and X2.

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In this case (interior point maximization), the consumer equates the MUper $ spent on each good to the MU of money income. In this case,L/ = I - P1X1 - P2X2 = 0 (consumer exhausts all income).

Going to the alternative case (if consumer not necessarily spends allincime): L/ = I - P1X1 - P2X2 0, this can be written as: P1X1 -P2X2 I. If the constraint is not binding (i.e., the consumer does notexhaust income), then, U1/P1 = U2/P2 = = 0, that is MU of income is 0(utility from spending one more $ of income = 0). This means that theconsumer is satiated in all goods. To see this:

U1 = P1 = 0

U2 = P2 = 0,

That is the consumer will not consume more of the 2 goods even if theyare free.

The remaining possibility is that the consumer is maximizing at a corner:

Since X1 = 0, U1 - P1 < 0, or U1 < P1 > U1 / P1That is, his MU of money exceeds the MU of consuming X1 per $ spent.

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At the same time, since X2 > 0, U2 - P2 = 0, or U2 = P2 = U2 / P2, i.e., the consumer is consuming X2 so that he equates theMU per $ spent on X2 to the MU of money income.

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DYNAMIC SYSTEMS

In dynamics, we are concerned with growth (change of key economicvariables, such as Y, P, etc) and stability over time.

A system is dynamical, if its behaviour over time is determined byfunctional equations, in which variables at different points in time arerelated in an essential way.

What is a functional equation: It is an equation in which the unknown isa function.

To solve an equation in the usual sense is to find the value (or values) ofthe unknown which satisfies the equation. To solve a functional equationmeans to find an unknown function which satisfies the functionalequation identically.

Example:

Consider the functional equation: y(x) y(x) = 0. It is easy to confirm

that the function we are looking for is: y(x) = Ae

x

, since y(x) = Ae

x

forany x.

Now consider the function: y = ax + b, which gives y = a. If we put x =(a-b)/a, we have also y = a, i.e., y y = 0. However, for any other valueof x, the value of the function will be different from a, therefore, thefunction y = ax + b does not satisfy out functional equation identically.

Now if we suppose that the symbol x stands for time , we are ready to

understand the definition of dynamic systems:

y(t) y(t) = 0 [example]

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In fact, y(t) = y(t) can be considered as a relation which involves thevalue of y at any point in time and the value it takes at an arbitrarilyclose point determined by y(t).

Use of integration in dynamics (continuous time)

Consider a relation such as the one given above. Suppose that a variable(for example, population) is known to grow over time at the rate:

dy/dt = 1/t ( or t-1/2) (1)

We want to find the time path of this population given that this

population is growing at the rate given by (1).

We are looking for the function y = y(t), given that we know itsderivative. We need to find the primitive function corresponding to thederivative (1). This is the essence of integration.

The function whose derivative is t-1/2 is 2t1/2, i.e., y(t) = 2t1/2 + c, where cis an arbitrary constant. In order to be able to solve for the constant, we

need additional information. Indeed, in most cases we have suchinformation; in particular we know the value of the function at somepoint in time, usually at time 0. This is called an initial condition. If, forexample, we know that y(0) = 10, then:

y(0) = 2(0)1/2 + c = 10 c = 10, and:

y(t) = 2t1/2 + 10.In general,

y(t) = 2t1/2 + y(0).

Before we proceed with more examples, we can refresh the rules ofintegration:

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Rules of integration

Notation: f(x)dxThe outcome of integration is:

f(x)dx = F(x) + c, which implies:

F(x) = f(x)

Rules:

(1) Integral of a constant:kdx = kx + c

(2) Integral of a power function:xndx = xn+1/n+1 + c (n -1)

(3) Integral of 1/x:(1/x)dx = ln(x) + c (n > 0)(also a sub-case of (2))

(4) Integral of a natural exponential function:ekxdx = ekx/k + c(special case: exdx = ex + c)

(5) Integral of a constant times a function:kf(x)dx = kf(x)dx

(6) Integral of a sum of 2 or more functions:[f(x) g(x)]dx = f(x)dx g(x)dx

(7) Integral of the negative of a function:-f(x)dx = - f(x)dx

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Examples

1. 9dx = 9x + c

2. x4dx = x4+1/4+1 = x5/5 + c

3. 2x2 x + 1)dx = 2x2 - xdx + dx= 2(x3/3) - (x)2 + x + c= 2/3(x)3 - (x)2 + x + c

4. 12x-1dx = 12(1/x)dx = 12 ln|x| + c

5. 20e-4x

= 20[(-1/4)e-4x

] = -5e-4x

+ c

Integration by substitution

Its use becomes clear in cases where it is difficult or impossible to usethe simple rules.

Example (here we could still have used the simple rules):20x4(x5 + 7)dx

- Let another function, u, equal (x5 + 7)- Differentiate u: du/dx = 5x4

- Solve for dx: dx = du/5x4

- Substitute u for (x5 + 7) and du/5x4 for dx:

- 20x4(x5 + 7)dx = 20x4u(du/5x4)= 4udu = 4udu

Integrate with respect to u:

4udu = 4(1/2)u2 = 2u2 + c

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Convert back in terms of the original problem:

20x4(x5 + 7)dx = 2(x5 + 7)2 + c

Now consider the following example:

3x(x + 6)2 dx

Let u=x +6, then du/dx = 1 dx = du

3x(x + 6)2 dx = 3xu2du = 3xu2du,

And we stop there because integration by substitution will not work.However, another method will work.

Integration by parts

It is derived by reversing the process of differentiation of a product:

d/dx[f(x)g(x)] = f(x)g(x) + f(x) g(x)

take the integral of the derivative:

f(x)g(x) = f(x)g(x)dx + f(x)g(x)dx

then solve algebraically for the first integral on the right hand side:

f(x)g(x)dx = f(x)g(x) - f(x)g(x)dx

Now, lets go back to the example:

3x(x + 6)2 dx

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- Separate the integrand (i.e., what is to be integrated) into 2 parts asin the formula. Usually we consider first the simpler function forf(x) and the more complex for g(X).

Let f(x) = 3x, g(x) = (x + 6)2, then:

f(x) = 3, g(x) = (x + 6)2dx = 1/3(x + 6)3 + c1

- Substitute the values for f(x), f(x) and g(x) in the formula (notethat g(x) is not used):

3x(x + 6)2 dx = f(x)g(x) - f(x)g(x)dx

= 3x[1/3(x + 6)3

+ c1] 3[1/3(x + 6)3

+ c1]dx= x(x + 6)3 + 3xc1 - [(x + 6)3 + 3c1]dx= x(x + 6)3 + 3xc1 1/4(x + 6)4 3xc1 + c= x(x + 6)3 - 1/4(x + 6)4 + c

So:

3x(x + 6)2 dx = x(x + 6)3 - 1/4(x + 6)4 + c

Note that c1 does not appear in the final solution. We can always assumethat c1 is always = 0.

- Check the answer by differentiating

Another example:

5xex-9dx

Let f(x) = 5x and g(x) = ex-9, then:

f(x) = 5, g(x) = ex-9dx = ex-9

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5xex-9dx = f(x)g(x) - f(x)g(x)dx= 5xex-9 - ex-9 5dx= 5xex-9 - 5ex-9 dx,

So, 5xex-9dx = 5xex-9 - 5ex-9 + c

Definite integrals

The definite integral is a number which can be evaluated using thefollowing:

The numerical value of a definite integral of a continuous function f(x)

over the interval [a to b]:

bf(x)dxa

is given by the primitive function F(x) + c, evaluated at the upper limitb, minus the same primitive evaluated at the lower limit, a:

b bf(x)dx = [F(x)] = F(b) F(a)a a

3 3Example: (6x2 + 5)dx = [2x3 + 5x]

1 1

= [2(3)3 + 5(3)] - [2(1)3 + 5(1)]

= 69 7 = 62

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Properties of definite integrals

b a

1. f(x)dx = - f(x)dxa b

a2. f(x)dx = 0

a

c b c3. f(x)dx = f(x)dx + f(x)dx (a


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Calculation of capital stock

This is done on the basis of a known pattern of change of investment.

If we are given that investment at time t is given by the function:

I(t) = 12t1/2

and K(0) = 25 (this is an example from Chiang and Wainwright), wehave:

K(t) = 12t1/2 dt = 12(t3/2/3/2)+ c = 8t3/2 + c

K(0) = 25 = 8(0) + c c = 25, and K(t) = 8t3/2 + 25,

which is the time path of capital stock.

On the other hand, if you are asked to find the amount of capitalaccumulated over a particular time interval, say (1, 3) in years:

3 3K(3) K(1) = 12t1/2 dt = 8[t3/2 ]1 1

Present value of a cash flow

Another application is finding the P.V. of a cash flow (a series ofrevenues or costs receivable at various times in the future).

Suppose an amount, I, is invested at the rate of interest, i; we know thatthe total amount expected t years later is:

R(t) = Ioeit

We can rearrange to get: Io = R(t)/eit = Re-it

(how much to invest now, to receive R, t years later)

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This is the continuous case counterpart of: I = R/(1 + i)t , where I is thepresent value of R. This refers to a single future value, R.

But if a person expects to earn R at the end of the first period andsubsequent periods, then he has to invest:

tI = Re-itdt

1where I is the present value of a series of annuities given by the returnexpected from an investment outlay for a unit of a particular asset; i.e., if

i = 0.04, compounded annually and the investor wishes to receive $100for 10 years, he has to invest:

10I = 100e-itdt = (100/0.04)e-0.04 - (100/0.04)e-0.4 = $726

1

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Economic dynamics

We may introduce economic dynamics by considering pre-Keynesianand Keynesian notions of equilibrium.

Before Keynes, the question asked was whether the equilibrium existsand if it is unique and stable. The explanation stopped there.

Keynesian analysis is different. If the achievement of full employmentequilibrium requires a certain amount of new investment to beundertaken to bring total effective demand to the level of full capacityutilization, the very fact new investment is undertaken, will change the

objective situation on which the current equilibrium is based. Investmentwill increase productive capacity at the same time it increases totaleffective demand (aggregate demand).

So, investment plays a special role in Keynesian economic dynamics andacts trough 2 different channels:

1. Change in effective demand through the multiplier:dy/dt = m(dI/dt) = 1/s(dI/dt)

2. Total new investment is an addition to existing productivecapacity.

Application: Domars growth model

The model assumes a rather rigid production function which involves

only capital:

k = K (1)

where k is capacity output (full employment output), taken to be amultiple of capital. As a result, = k/K.

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One implication here is that, since labour is indispensable, K and L arecombined in fixed proportion, because only then we can consider capitalalone.

By differentiating (1):

(full employment output over time): dk/dt = (dK/dt) = I

along with:

(from the demand side): dY/dt = 1/s(dI/dt)

There is no reason why these two effects should necessarily becompatible with the maintenance of full capacity utilization. We canexplore, however, the conditions to be satisfied in order that effectivedemand and productive capacity may expand in a compatible way overtime.

First of all, total effective demand must be equal to productive capacityto begin with:

Y(0) = k(0)

Secondly, dk/dt = dY/dt must always hold over time, i.e.,

1/s(dI/dt) = I, or:

dI/I = sdt (1/I)dI = sdt ln(I) = st + c, oreln(I) = e(st + c) It = estec,then:

It = Aest (I(0) = Ae0 = A)So:

It = I0est

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The above says that, in order to maintain equilibrium in the long-run,new investment must expand over time according to the aboveexponential function (i.e., expand at a rate g= *s over time). Aneconomy might satisfy or not satisfy this condition.

We could arrive at the same condition for It using definite integralsinstead [take this as a practice question]

The above discussion of Domars growth model and the application ofintegration will take us to the concept and use of a differential equationin economic dynamics.

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Differential equations

Basic concepts and definitions

Equations in which the unknowns are functions and which contain notonly the unknown function but also their derivatives, are calleddifferential equations.

The form of the function is not given; it is to be found by solving theequation.

The general expressions of an ordinary differential equation in a

function y of an independent variable x, can be written as:

F(y, x, dy/dx, d2y/dx2, ..., dny/dxn) = 0 (1)

The order of the equation is that of the highest derivative of it.

An equation relating the independent variable x, the unknown function yand its first derivative, is called a differential equation of the first order.

If in addition, the equation contains a second derivative of the functionwe are looking for, then such an equation is of the second order.Accordingly we c

Economics and Optimization Techniques

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