he3001Notes_Word_2014.pdf

OPTIMIZATION CLASSICAL OPTIMIZATION Optimizing a function is defined as a rational action chosen consistently to get the decision maker nearto any desired goal as circumstances permit. In dealing with optimization problems we start with what is called an objective function, in which the left-hand side is the object of optimization (for example, profits) while in the right-hand side we have a set of choice variables and a set of coefficients or constants. The choice variables (or decision variables) are the ones that the economic agent (consumer, firm, government etc) can vary (decide on) in its attempt to optimize the objective function. We’ll examine 2 types of optimization problems: Unconstrained and Constrained. 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 or decreasing or with various increasing and decreasing parts. In optimization we are looking for an extreme value, i.e., 1 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, point C is a relative or local minimum, since it is an extreme point only in the immediate neighborhood of point C only. When we search for an extreme point the first thing to look at is the first derivative of the function. In the graph below: 2 When the function has a max. (at X=X*) the slope of the function changes from positive to 0 and then to negative. This means that this one-variable function has a max when the derivative is 0. However, in the graph below the same condition holds: 3 Another possible stationary point is a point of inflection. and the function has a max at x=5. This is the second order condition. This is neither a max nor a min. At this point we notice that in the case of a max the first derivative changes sign from positive to negative as X increases. A relative max exists if f’’(x) < 0 and a relative min exists if f’’(x) > 0. The first derivative 4 X* . This means that the first derivative test is a necessary condition for identifying both a max and a min. So. the second derivative is negative and the function has a max. Therefore. the condition f’(x)=0 is necessary to establish a max or a min. when the first derivative is decreasing as X increases. Example: Y = -2x2 + 20x Then. Summarizing. f’(x) = -4x + 20 and =0 when x=5. the second derivative is positive and the function has a min. but not sufficient. That is the first derivative identifies a stationary value (extreme point) of the function. the distinction between a max and a min relies on the direction of change of the first derivative. while in the case of a min the first derivative changes sign from negative to positive.The function has a minimum at point A where the first derivative is 0. The second derivative is f’’(x)= -4 < 0. Consider a stationary value characterized by f’(x)=0 and f’’(x)=0 at x=x*. while when the first derivative is increasing. which is measured by the second derivative. f(x) There is a stationary point at A (first derivative =0 in second diagram). Here. This is a stationary point of inflection. f’(x) is at min (f’’(x) = 0). Therefore. f’(x)=0 is not required for an inflection point. while f’’(x)=0 at C’’ as in the previous case.e. while for x > x*.. but remains positive as X increases. f’(x) increases in value (f’’(x) > 0). f(x) does not have a stationary point of inflection at x = x*. think of the short-run production function – in this case x stands for Labour and Y for output): 5 . For x< x*. the slope (f’(x)) is not horizontal at point C. f’(x) decreases (f’’(x) <0). So.does not change sign. f’’(x) f(x) X* f’(x) Another possibility is a non-stationary point of inflection. f’’(x) Example of non-stationary point of inflection (i. A necessary condition for a point of inflection is f’’(x) = 0. At x = x*. f’(x) > 0. f’(x) > 0. Inflection point where f’’(x) = 0. when x = 10 f’’’(x) = -4 < 0 (i) the non-stationary point of inflection is at x=10. (ii) f’(x) has a max at x = 10. f’’(x) (restricting f(x) to the non-negative quadrant).Y = -(2/3)x3 + 20x2 f(x) f’(x) Graph the functions f(x). Profit maximization 6 . (iv) Point (20) is a relative max. f’’(x) < 0 and f’(x) is concave to the x-axis. so f(x) is convex t o the x-axis. For x > 10. (iii) For x < 10. f’(x) = -2x2 + 40x f’’(x) = -4x + 40 = 0. f’(x). f’’(x) > 0. f’’(x) f’’’(x) Simple economic applications: 1. that is. the MC function must cut MR from below). So. or f’(L) = f(L)/L. C = C(q) = R(q) . and: R’’(q) < C’’(q) (2nd order condition) In economics this means that at the profit max output. 2.e. d(MPL)/dL < 0 (diminishing MPL) 7 . 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. and must be < 0 for max AP of labour. given that L>0. the second order condition for max AP of labour requires: f’’(L) < 0. or: MPL = APL then. the rate of change of MR must be less than the rate of change of MC (i.Choose the rate of output which maximizes profits: π = R – C.C(q) 1st order condition: π’(q) = R’(q) – C’(q) = 0. 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): d2(APL)/dL2 = f’’(L)/L – 2f’(L)/L2 + 2Lf’(L)/L3 = f’’(L)/L. R = R(q).. and at the same time the slope of the MP curve is negative (the MP curve cuts the AP curve from above). -------------------------------------APL = -(2/3)L2 + 10L ∂APL/∂L = -(4/3)L + 10 = 0 L = 7.5 MPL = -2L2 + 20L At L=7. MP=AP. the AP of labour equals the MP of labour.5: MPL = APL = 37. Practice exercise Q = -(2/3)L3 + 10L2.5 d2(APL)/dL2 = -(4/3) < 0 8 .At the max point of the AP of labour. Derive the AP of labour function and show that where AP is at max. Momentarily.Unconstrained optimization with more than one variable Two-variable case Consider the function: z = f(x. y) The first-order condition can be expressed in a similar fashion to the one variable case. For z to obtain an extreme value at point A (top of a hill). 9 . z neither increases nor decreases in value at this point: dz = f’(x)dx + f’(y)dy = 0 For any (non-zero) arbitrary variations dx and dy. the function must be stationary at this point. it is necessary that: f’(x) = f’(y) = 0 This is the first-order condition for a relative extreme point. y). fyx are defined as: fxy = ∂/∂y(∂z/∂x) = ∂2z/∂y∂x. Likewise. (because it also holds for minimum). Consider now the concept of the second-order differential. Given the function: 10 . The partial derivatives are: ∂z/∂x = f’(x) = fx. Given the function z = f(x. and fyx = ∂/∂x(∂z/∂y) = ∂2z/∂x∂y Cross-partial derivatives measure the rate of change of a first partial derivative with respect to the other variable. the tangent line drawn through point A // to the yz plane has a zero slope. The cross-partial derivatives fxy.From the graph. and fyy = ∂/∂y(∂z/∂y) = ∂2z/∂y2 The above measure the rate of change of the first derivative with respect to x and y. ∂z/∂y = f’(y) = fy The second partial derivative with respect to x (while y is held constant) are: fxx = ∂/∂x(∂z/∂x) = ∂2z/∂x2 . We need a sufficient condition or a second-order condition. this means that the tangent line through point A parallel to the xz plane has zero slope. since A is a candidate max. The first-order condition does not tell us whether the extreme point is a max or a min. In other words. The sufficient condition for max can be translated into a more convenient form: It can be shown that for any non-zero values of dx and dy. fyy <0 and fxxfyy > f2xy 11 . The above condition.Z = f(x. d2z will be negative (for max) iff: fxx <0. is not necessary because it is possible for d2z to be 0 at the max or min. while sufficient. we need dz to be decreasing (d2z < 0). given that dz at A equals zero. So the second order (sufficient) condition for max is d2z < 0 for non-zero arbitrary values of dx and dy. d2z > 0 is the sufficient condition for a minimum. It is going to be a max if a tiny movement away from A in any direction results in a decrease in the value of z (dz < 0) in the neighborhood of A. Likewise. 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. which can be < = > 0 Suppose we are looking for a max (at point A in the graph). Xn) The first order conditions for a stationary value are that all partial derivatives of Y must = 0 for all i = 1. Alternatively. if: fxxfyy = f2xy. X2.…n. That is the sign of d2z will be positive for some values of dx and dy and negative for others. For example. 12 .…. Functions with more than 2 variables Consider the following function for optimization: Y = f(X1. the second order conditions must be satisfied. the function will have a saddle point.Likewise for min: fxx > 0. wi are the prices of inputs and P is the price of output.) as we go away from point A in any direction (not only in the two basic direction). if: fxxfyy < f2xy.…. πi = Pfi – wi=0 (value of marginal products equal to input prices). this could be a max or a min but the test is inconclusive. fyy > 0 and fxxfyy > f2xy The second part of the sufficient condition will ensure that the surface decreases for max.∂(ΣwiXi)/∂Xi = 0 for all i. Or. where xi are the inputs. The first order conditions are: ∂π/∂Xi = P(∂f/∂Xi) . The above sufficiency condition can be inconclusive. In order to have a maximum. which can occur when fxx and fyy are of different sign.Xn) – Σwixi. X2. For example. in the case of profit max the objective function is: π = Pf(x1. (increases for min. In the 2 variables case. We can also for 3 principal minors of order n-2 by eliminating the 1st and 2nd. fyy <0 and fxxfyy > f2xy. we have called a principal minor of order k of this determinant. we can form 2 principal minors. With a 2x2 determinant. recall that these conditions required: fxx <0. 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 row and 1st column. Given some nth order determinant. what remains when any n-k rows and the corresponding number of columns are eliminated from the determinant. the 2nd row and column and the 3 rd row and column. the 1st and 3rd and the 2nd and 3rd rows and columns. The following theorem establishes the sufficiency condition: 13 . a22 and a11. or fxx fyx fxy >0 fyy To extend the second-order conditions to the n-variable case we make use of our knowledge of determinants. then it is not possible to determine whether there is a max or a min. X2. . H3 (=H) H1 = f11 f11 f12 H2 = f21 f22 H3 = H 14 . If all the principal minors are positive for all k at X = X* the function has a minimum at X = X*. with 3 variables: Max y = f(x1.: Construct the Hessian matrix of cross-partial derivatives: f11 f12 f13 H = f21 f22 f23 f31 f32 f33 Check the leading principal minors: H1. if some or all of the principal minors are 0 and the rest have the appropriate sign. if all the principal minors of the determinant: det(fij) of order k have sign (-1)k for all k = 1. then f(X1. x2.e. all we need to check are the signs of the leading principal minors] i.Theorem: Given a function y = f(X1. . which has a stationary value at X = X* (a vector of Xi). [in practice. X2. x3) Second order conditions for max and min.. H2. If the signs do not follow the pattern in any of the above cases..…n at X = X*. Xn). the function has a saddle point at X = X* Finally. and given the Hessian matrix of cross-partial derivatives of f (fij) then. Xn) has a maximum at X = X*. (1) The first order conditions (FOC) for max are: ∂π/∂Xi = P(∂f/∂Xi) – wi = 0 for all i. H2 >0.…n. H3>0 Going back to the profit max problem: π = Pf(X1. The (symmetric) Hessian determinant is: f11 f12 … f1n f21 f22 … f2n ……………… 15 . H3<0 [we start from negative because (-1)k = (-1)3=(-)] For min we need: H1>0.For max we need: H1<0. Given that P > 0. meaning that the FOC equations are independent) we can solve for the demand functions for inputs: Xi = Xi*(wi.Xn) – ΣwiXi. The second order conditions require that the successive principal minors of π ij = Pfij have sign (-1)k. for k = 1. we have to solve the system of equations (the First Order Condition equations (1)) for the Xs in terms of the parameters wi and P. P).…. we can express the matrix of second partial derivatives in terms of the production function. In order to find the candidate values of vector X for max. X2. If the system is solvable (that is the Jacobian determinant J = det(∂πi/∂Xj) is not zero. H2 >0. This in turn requires that all fij (diagonal elements) be < 0... i.fn1 fn2 … fnn and the successive principal minors are: f11 f12 H1 = f11. if satisfied.e. fiifjj – f2ij > 0. these conditions. starting with negative. By requiring that the signs alternate in sign starting with negative amounts to requiring that d2y is negative definite. In addition the 2nd order determinant must be > 0. make sure that the objective function is concave in the neighborhood of a maximum. Furthermore. which means diminishing marginal products. H2 = f21 f22 f11 f12 f13 H3 = f21 f22 f21 f31 f32 f33 Hn = H These principal minors have to alternate in sign. 16 . etc. f11 < 0. i. The 3rd order must be < 0.e. In the weak case of concavity. This is a necessary and sufficient condition. if the 2nd differential is everywhere negative semi-definite. that is we establish a relative maximum. When we use d2z to test (or alternatively the principal minor test). However... we identify an absolute minimum. establishing concavity or strict concavity can take care of this. The second order conditions we examined earlier. the function is (strictly) concave and this implies an absolute maximum. given a convex function and the FOC.e. from the first order conditions). The former are called concave functions. a stationary point can be identified as an absolute maximum (i.e. the hill may contain somewhere a flat section. d2z. Summarizing: (1) Given a concave function. i. Similarly. Establishing concavity and strict concavity can replace the 2 nd order conditions as a sufficient condition for maximum. as they apply to the 2 nd order differential. when d2z is negative definite (strict inequalities).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. the sign definiteness of the 2nd differential is evaluated only at the stationary point. A unique. can be distinguished in necessary conditions. absolute maximum requires strict concavity. Even if d2z happens to be zero at the stationary point. The strict case of concavity is associated with a global maximum. 17 (2) (3) . when d2z is negative-semi-definite (when we have weak inequalities in the relevant principal minor test) and in sufficient conditions. 18 . Q2) Prices of Q1 and Q2 are fixed at P1 and P2. Q2) = π(Q1. Q2) = 0 π2 = ∂π/∂Q2 = P2 – MC2(Q1. we can solve for the optimal values of Q1 and Q2.Application 1 Consider the case of profit maximization by a multiproduct firm in perfect competition: 2 products. the marginal costs depend on both goods: MC1 = ∂C/∂Q1 = c1(Q1.c(Q1. Q2) Profit maximization: Max π = π(Q1. Q2). Q2 Products can be technically interdependent (what does it mean?). Q2) MC2 = ∂C/∂Q2 = c2(Q1. Q2) π1 = ∂π/∂Q1 = P1 – MC1(Q1. Q1. Q2) = 0 Assuming the 2nd order conditions are satisfied. if: given the total cost function C = c(Q1. TR = P1Q1 + P2Q2 MR1 = ∂TR1/∂Q1 = P1 MR2 = ∂TR1/∂Q2 = P2 The profit function is: π = TR – C = P1Q1 + P2Q2 . π22 <0 and the second order determinant = 8 (> 0). π22 = -6. π = 4 The 2nd order conditions are: π11 = -2. Application 2 Price discrimination A single product firm (monopolist or firm with significant market power) is selling output in more than one market. (Q = Q1 + Q2) MC1 = ∂[c(Q)]/∂Q1 = (dC/dQ)*(dQ/dQ1) = dC/dQ = c’(Q) 19 .2Q1Q2 .(Q1)2 .3(Q2)2 . How much output should the firm sell in each market and what prices should it charge to each one to maximize profits? Q = Q1 + Q2 Total revenue: TR = TR1(Q1) + TR2(Q2) Total cost is: C = c(Q). Therefore the second order differential is negative definite and we have a maximum.2 π1 = 2Q1 + 2Q2 = 4 π2 = 2Q1 + 6Q2 = 8 → Q1 = Q2 = 1. π11 <0.Example: TR = 4Q1 + 8Q2 C(Q1. So. π12 = π21 = -2. Q2) = (Q1)2 + 2Q1Q2 + 3(Q2)2 + 2 [this cost function implies that the products are technically interdependent in production] then : π = 4Q1 + 8Q2 . Q2* and π* = π(Q1*. → MC1 = MC2 = MC Maximize profit: π = TR1(Q1) + TR2(Q2) – c(Q) First order conditions: π1 = ∂π/∂Q1 = ∂TR1/∂Q1 – (∂C/∂Q)(∂Q/∂Q1) = (MR1) – c’(Q) = 0 or (MR1) = c’(Q) π2 = ∂π/∂Q2 =∂TR2/∂Q2 – (∂C/∂Q)(∂Q/∂Q2) = (MR2) – c’(Q) = 0 or (MR2) = c’(Q) From the FOCs we can find Q1*. 20 . then: MR1 = d(TR1)/dQ1 = P1(dQ1/dQ1) + Q1(dP1/dQ1) = P1(dQ1/dQ1) + Q1(dP1/dQ1)(P1/P1) = P1 + (dP1/dQ1)Q1(P1/P1)) = P1[1 + (dP1/dQ1)(Q1/P1) MR1 = P1(1 – 1/Ed1) Similarly.MC2 = ∂[c(Q)]/∂Q2 = (dC/dQ)*(dQ/dQ2) = dC/dQ = c’(Q). MR2 = P2(1 – 1/ Ed2) Then. Q2*). If Ed1 < Ed2 at (optimal) Q1* and Q2*. from MR1 = MR2 = MC: P1(1 – 1/Ed1) = P2(1 – 1/ Ed2) We see that. then P1 > P2. Also observe that. since TR1 = P1Q1. The first-order conditions imply: MR1 = MR2 = MC. P1 = P2 only if price elasticities are equal for the 2 demand curves. 21 . are: πK = P∂f/∂K – r = PfK – r = 0 πL = P∂f/∂L – w = PfL – w = 0 We can solve the FOC for the optimal values of K and L (K* and L*). L) Total cost is given by: C = c(K.wL The first-order conditions for profit max. L) = rK + wL Total revenue is: TR = PQ = Pf(K. L) Profit is: π = TR. and optimal profit π*. The FOC state that: PfK (=VMPK)= r [Value of MP of K = unit price of capital] PfL (= VMPL)= w [Value of MP of L = wage rate] 22 .rK . K and L producing output Q: Q = f(K. L) .C = Pf(K.PROFIT MAXIMIZATION AND INPUT DECISIONS Consider a perfectly competitive firm using two inputs. 4(K-4/3L-1) = 4(K-4/3L-1) = 4/(K4/3L) > 0. maximum. Q* = 48 Check the Second Order Conditions: ΠKK = -8/3L1/2K-5/3 < 0 ΠLL = -3K1/3L-3/2 < 0 ΠKL = 2K-2/3L-1/2 ΠLK = 2K-2/3L-1/2 Given ΠKK and ΠLL both < 0. 23 . along with ΠKK. 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) . K* = 8. L* = 16.Example: Q =6K1/3L1/2 P=2. ΠLL < 0. r=4. w=3 Max π = 12K1/3L1/2 – 4K – 3L ΠK = 4K-2/3L1/2 – 4 = 0 ΠL = 6 K1/3L-1/2 – 3 = 0 Solving. ∂X2*/∂w2.Comparative Statics Exercises Starting from profit maximization: π = Pf(X1. X2 are 2 inputs) FOCs: Pf1(X1. f22 < 0. ∂X2*/∂P [FOCs are Equilibrium conditions] 24 . w2. P) The above are the factor demand curves. ∂X1*/∂w2. ∂X2*/∂w1. Comparative statics results What happens to the optimal/equilibrium (profit maximizing) values of X1 and X2.w1 = 0 (1) Pf2(X1. w1 and w2 we get: X1 = X1*(w1. in terms of the parameters P. X2) – w2 = 0 SOCs: f11 < 0. and f11 f22 – f212 > 0 Solving the FOCs (implicit functions) for the 2 unknowns. 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. w2. X2) – w1X1 + w2X2 (were X1. ∂X1*/∂P. X1 and X2. P) (2) X2 = X2*(w1. P)} – w2 = 0 (3b) That is. w2. X2* amounts of the 2 inputs to keep the values of the marginal products for each factor exactly equal to the price of each factor. the firm must employ X1*. w2. P). the equivalent of 4a and 4b will give: ∂X1*/∂w2 = -f12/P( f11 f22 – f212) ∂X2*/∂w2 = f11/P( f11 f22 – f212) (5c) (5d) 25 . for a change in w2. w2. w2. P). so: ∂X1*/∂w1 = f22/P( f11 f22 – f212) (5a) Note that the denominator is > 0 because of the SOCs. Suppose we multiply (4a) by f22 and (4b) by f12 and subtract (4b) from (4a) (and given that f12 = f21). X2*(w1. Now differentiate 3a and 3b partially with respect to w1 using the chain rule: P(∂f1/∂X1)(∂X1*/∂w1) + P(∂f1/∂X2)(∂X2*/∂w1) – 1 = 0 P(∂f2/∂X1)(∂X1*/∂w1) + P(∂f2/∂X2)(∂X2*/∂w1) =0 Or. X2*(w1. Then: ∂X2*/∂w1 = -f21/P( f11 f22 – f212) (5b) Likewise. then: P(f11 f22 – f212) (∂X1*/∂w1) = f22.First substitute equations (2) into equations (1): Pf1{X1*(w1. Pf11(∂X1*/∂w1) + Pf12(∂X2*/∂w1) = 1 Pf21(∂X1*/∂w1) + Pf22(∂X2*/∂w1) = 0 (4a) (4b) We need to solve for ∂X1*/∂w1 and ∂X2*/∂w1. P)} – w1 = 0 (3a) Pf2{X1*(w1. f2 f11 + f1 f21 / P(f11 f22 – f221) (6a) (6b) The signs of ∂X1*/∂P and ∂X2*/∂P cannot be established. w2.Conclusion: The factor demand curves (5a. However. However. Changes in the price of the product Differentiate (3a). w2. w2. it can be shown that not both ∂X1*/∂P and ∂X2*/∂P can be negative. as the signs of f12 and f21 are not known in the general case. X2*(w1. X2*(w1. P). P). then subtract second from first: P(f11 f22 – f212) (∂X1*/∂P) = . as the sign of the cross-derivative f12 (= f21) is unknown in the general case.f1 f22 + f2 f12 and. ∂X1*/∂P = . w2. P)} – w1 = 0 (3a) Pf2{X1*(w1. we cannot determine the sign of the cross-effects (∂X1*/∂w2 and ∂X2*/∂w1). 5d) are downward sloping. ∂X2*/∂P = . P)} – w2 = 0 (3b) with respect to P: → Pf11(∂X1*/∂P) + Pf12(∂X2*/∂P) = -f1 Pf21(∂X1*/∂P) + Pf22(∂X2*/∂P) = -f2 Multiply the first by f22 and the second by f12. (3b): Pf1{X1*(w1.f1 f22 + f2 f12 / P(f11 f22 – f212) then. Since output will increase in response to the P increase: [dQ = (∂Q/∂X1)(∂X1/∂P) + (∂Q/∂X2)(∂X2/∂P) > 0 ] Changes in Q 26 . Differentiate with respect to P to find how output changes when P changes (slope of the S-curve): ∂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). X2) and optimal output is: Q* = f(X1*. P). w2. it could be shown (and will be shown later on when we have a test to test for the concavity of a function).The production function is Q = f(X1. P) X2 = X2*(w1. (7) We could also show that: 27 . ∂Q*/∂P = . Q* = f{X1*(w1. w2. P) Then. P)} = Q*(w1.2f12 f1f2 + f21 f22 < 0 then. w2. w2. Conclusion: the sufficient condition for profit maximization (concavity of a function) also implies that the supply curve is upward-sloping. that the sufficient condition for concavity of a function can be written as: f22f11 . the numerator is positive and ∂Q*/∂P > 0. P) This is the supply function of the firm. X2*(w1.f21 f22 + 2f12 f1f2 –f22f11/ P(f11 f22 – f212) Although it is not immediately clear what sign the numerator has. X2*) Given: X1 = X1*(w1. w2. X2*(w1. ∂X1*/∂P = . Q*/∂w1 = f1f22/P(D) – f2f21/[P(f11 f22 – f212)] = .f1 f22 + f2 f12 / P(f11 f22 – f212)] Similarly for Q*/∂w2 28 . and the signs of these expressions are indeterminate: Q* = f{X1*(w1.∂X1*/∂P ∂Q*/∂w2 = . P). 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 – f212) Then.∂X1/∂P [Since. w2. w2.∂Q*/∂w1 = .∂X2*/∂P. Y = c(Y-T) + I0 + G0 (1) For a richer model we replace the exogenous investment function with: 29 . 1. We are going to begin with the basic closed economy Macro model and 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’ <1 Yd = Y-T I = I0 (exogenous) G = G0 (exogenous) So. 2. The output market The money market The bond market The labour market Because these markets are interdependent. Therefore. we can eliminate one of them (the bond market). We tend to aggregate the economy in up to 4 markets.COMPARATIVE STATICS IN MACROECONOMICS We are going to study the economy at a given state and observe how it is affected by changes in various events.. 3. i.e. the 4th one will necessarily be in equilibrium. once equilibrium is determined in 3 of them. 4. and money demand is: Md/P = L(Y.I(r) . Now combine the IS and the LM curves: The IS-LM model Y . Treating money supply. which is negative because dI/dr is < 0. Y and the interest rate.G = 0 (IS) (5) 30 . So. r).L1/L2 = -∂L/∂Y/∂L/∂r. r.c(Y-T) . To find the slope of the IS differentiate (3): dY = c’dY + I’dr. the LM curve is positively sloped. So. as exogenous. the IS curve is negatively sloped. r). and dr/dY = (1 – c’)/I’. The money market Money demand is a function of National Income.I = I(r) Then. L1 = ∂L/∂Y > 0 and L2 = ∂L/∂r < 0. (4) From (4): L1 dY + L2 dr = 0 dr/dY = . Y = c(Y-T) + I(r) + G (2) (3) This model gives rise to the IS curve. Equilibrium in the money market gives the LM curve: LM: Ms/P = L(Y. Ms. which is > 0. T. In particular. M. differentiate equations (5) with respect to P (which in the IS-LM model is taken as fixed): ∂Y/∂P – c’(∂Y/∂P) – I’(∂r/∂P) = 0 -L1∂Y/∂P – L2∂r/∂P = M/P2 Or. G. G. To find the slope of the AD curve. (1-c’) (∂Y/∂P) . M.M/P .L(Y. (1-c’) -L1 -I’ -L2 ∂Y/∂P = ∂r/∂P M/P2 0 0 -I’ 31 . P.I’(∂r/∂P) = 0 -L1(∂Y/∂P) – L2(∂r/∂P) = M/P2 Or. the solution: Y = Y(P. by allowing the price level to vary. r) = 0 (LM) Solve for the endogenous variables Y and r in terms of the exogenous variables. T) Is the aggregate demand curve (AD). The IS curve shifts to the right.L(Y.G = 0 M/P . r) = 0 with respect to G and solving for ∂Y/∂G and ∂r/∂G → ∂Y/∂G – (∂C/∂Y)(∂Y/∂G) – (∂I/∂r)(∂r/∂G) = ∂G/∂G = 1 (∂L/∂Y)(∂Y/∂G) + (∂L/∂r)(∂r/∂G) Or.I(r) . price of bonds decreases the interest rate (bond yield) increases (with money supply fixed). The effect of the increase in G can be obtained by differentiating (5): Y .∂Y/∂P = M/P2 -L2 ------------------(1-c’) -I’ -L1 -L2 I’M/P2 (-ve) = -----------------------L2(1 – c’) – L1I’ (+ve) <0 That is.c(Y-T) . Fiscal Policy An expansionary fiscal policy is characterized by an increase in G. This increase needs to be financed by: (1) (2) (3) Issuing bonds (debt financed increase) By raising taxes Printing money Assume case (1) Supply of bonds increases in the bond market. ∂Y/∂G – c’( ∂Y/∂G) – I’(∂r/∂G) = 1 L1(∂Y/∂G) +L2(∂r/∂G) =0 32 =0 . the AD curve is downward sloping. L1/ L2 which is the slope of the LM curve. L2 < 0: 1 ∂Y/∂G = ---------------------------. 33 .> 0 1-c’ + I’ (L1/ L2) (6) (7) The increase in the interest rate as a result of the increase in G represents the partial crowding out effect. Practice question: You can do the same for a change in T (taxes).(L1/ L2)(∂Y/∂G) Substituting in the first and solving for ∂Y/∂G and given that I’ < 0.L1/ L2 ∂r/∂G = ---------------------------. . in isolation or in combination with an increase in G (for combination see page 35).From the second. ∂r/∂G = . L1 > 0.> 0 1-c’ + I’ (L1/ L2) (-) (-) Then. ∂Y/∂G varies inversely with . = -------------------. (1-c’) L1 and.> 0 (1-c’) -I’ (1-c’)L2 + I’L1 (1-c’) + I’L1/L2 L2 (1 – c’)PL2 ∂r/∂M = ……….c(Y-T) . (1-c’) (∂Y/∂M) – I’(∂r/∂M) L1(∂Y/∂M) + L2(∂r/∂M) Or. 0 -I’ 1/P L2 I’/P I’/PL2 ∂Y/∂M = ------------------.G = 0 M/P . r) = 0 with respect to M and derive ∂Y/∂M and ∂r/∂M → ∂Y/∂M – (∂C/∂Y)(∂Y/∂M) .= -----------------.Monetary Policy Differentiate (5): Y .< 0 (1-c’) + I’L1/L2 34 = 1/P =0 = 1/P 0 = 1/P -I’ L2 ∂Y/∂M ∂r/∂M L1 .I(r) .∂I/∂r(∂r/∂M) = 0 ∂L/∂Y (∂Y/∂M) + ∂L/∂r(∂r/∂M) Or. = --------------------.L(Y. The relative effectiveness of fiscal and monetary policy is an empirical matter. The Balanced Budget Multiplier In this case dG = dT. Dividing ∂Y/∂G by ∂Y/∂M we can see the determinants of the relative effectiveness of the 2 policies: (∂Y/∂G)/(∂Y/∂M) = 1/(I’/PL2) = PL2/I’ The relative effectiveness depends upon the interest responsiveness of the demand for money (L2) and investment demand (I’).As I’ → 0 or L2 → ∞ monetary policy becomes inefficient in influencing national income. Using equations (5) : 35 . L(Y. the BBM will be equal to 1 only if investment is completely inelastic to changes in the interest rate.I(r) .< = 1 1 – c’ +I’(L1/ L2) The effect of the tax does not fully offset the effect of the increase in G.c(Y-T) . we need the aggregate supply curve (AS).(L1/ L2)(∂Y/∂G)] = 1 [1 – c’ +I’(L1/ L2](∂Y/∂G) = 1 – c’ (1 – c’) ∂Y/∂G = ---------------------. To be able to determine the price level. The Labour Market We have derived the IS. LM and AD curves. P.G = 0 M/P .Y .(L1/ L2)(∂Y/∂G) Then. or: ∂r/∂G = .e. pure Keynesian case). ∂Y/∂G – c’(∂Y/∂G) + c’ – I’(∂r/∂G) = 1 and. ∂Y/∂G – (∂C/∂Y)(∂Y/∂G) + (∂C/∂Y) – (∂I/∂r)(∂r/∂G) = ∂G/∂G = 1 [Why? Remember. r) = 0 Differentiate the model by allowing both G and T to change by the same amount. 36 . or L2 → ∞ (i. -L1(∂Y/∂G) – L2(∂r/∂G) = 0. (1-c’) (∂Y/∂G) + c’ – I’[. consumption depends on Yd = (Y-T)] Or. Ls. and φ’(L) < 0 (downward sloping demand for L). the demand for L will depend on the real wage rate. Assume a concave utility function which depends upon real income (Y) from work and leisure (T – Ls). demand for labour. (T – Ls)] ∂[(W/P)Ls] ∂(T – Ls) 37 . f’’ < 0 Competitive firms maximize profit: Pf’(L) = W → W/P = f’(L) [If on the other hand the firm has market power. where T is the fixed hours available to workers (24 hours) and Ls is hours of labour supplied. in the context of the worker’s work-leisure choice. φ(L) is a function of W/P (real wage rate). L). so that: Y = f(K. To determine the supply of labor. we need to perform utility maximization. In the short-run K is fixed. f’ >0 .This requires a labour market in the model. T-Ls) Subject to: Leisure = T – Ls Y = (W/P)Ls Substitute into the utility function: Max U = u[(W/P)Ls. So. Max U = u(Y. Assume a production function involving K and L. it could be shown that the demand for labour is derived from: P(1 + 1/Ed)f’(L) = W] In the general case. r) =0 Y – Y(L) =0 φ(L) = ψ(L) W/P = φ(L) This is a recursive model: We have 5 equations with 5 endogenous variables. since the assumption is that workers care about their real wage rate.G = 0 M/P – L(Y. Y. while nominal variables are: r. r.∂U/∂Ls = u1 ---------------. some function of Ls = real wage rate] So the labour supply function can be expressed as: W/P = ψ(Ls).. this model leads to the so called classical dichotomy.e. Equation 4 alone determines employment and then Y and W/P are determined from 38 . Equilibrium in the labour market requires: Ld = Ls = L.+ u2 ---------------. ψ’ > 0 This is a classical labour supply function. W/P = u2 (Ls) / u1 (Ls) [i.u2 = 0 ∂Ls ∂Ls and. [Note that: Real variables are Y.= u1 (W/P) . L (and W/P). P. or W/P = φ(L) = ψ(L) Add now the labour market equations. P. along with the production function into the model: (1) (2) (3) (4) (5) Y – c(Y-T) – I(r) . L. W] However. W. equations 3 and 5.I’(∂r/∂G) L1(∂Y/∂G) + L2(∂r/∂G) + M/P2 (∂P/∂G) (∂Y/∂G) . .Y’(∂L/∂G) (φ’-ψ’) (∂L/∂G) =1 =0 =0 =0 φ’(∂L/∂G) – (1/P)(∂W/∂G) + (W/P2)(∂P/∂G) = 0 then. . Y and W/P (the real variables) are determined exclusively in the labour market and are independent from G and M. So L. (1-c’) L1 1 0 0 . .I’ L2 0 0 0 -Y’ 0 0 0 0 M/P2 0 0 ∂Y/∂G ∂r/∂G ∂L/∂G ∂W/∂G ∂P/∂G = 1 0 0 0 0 0 (φ’-ψ’) 0 0 φ’ -1/P W/P2 1 . 4→ L → (5 → W/P) 3 → Y → 1 → r → 2 → P (and from W/P) → W Fiscal Policy Totally differentiate the system of equations allowing G to change: (1-c’)(∂Y/∂G) . So real activity is independent of fiscal and monetary policy. 39 . . .. ∂Y/∂G= --------------------------------. . . . . . . [i. The effect of the increase in G is to raise interest rates and reduce investment by an amount which exactly offsets the increase in G (complete crowding out). . . . . . With a higher r and fixed Y. Money market equilibrium requires higher P . 0 . . . . . .0 0 0 . ∂Y/∂G = 0. . Looking at the numerator first (the denominator = I’M(φ’-ψ’)/P3): You can verify that the numerator = 0. . and fiscal policy is ineffective in the classical model.e. . money wages (W) rise by the same proportion as the price level. So. . . . . . 40 . Md/P = Ms] In order to preserve the real wage rate. . . .given fixed M. . . demand for money decreases. . . Y’(∂L/∂M) (φ’-ψ’) (∂L/∂M) φ’(∂L/∂M) or.G = 0 M/P – L(Y. (1-c’) L1 1 0 0 + M/P2 (∂P/∂M) =0 = 1/P =0 =0 .Monetary Policy Y – c(Y-T) – I(r) . allowing for money supply to change: (1-c’)(∂Y/∂M) . differentiate the 5 equation model.(1/P)(∂W/∂M) + (W/ P2)(∂P/∂M) = 0 .I’(∂r/∂M) L1(∂Y/∂M) + L2(∂r/∂M) (∂Y/∂M) .I’ L2 0 0 0 -Y’ 0 0 0 0 M/P2 0 0 ∂Y/∂M ∂r/∂M ∂L/∂M ∂W/∂M ∂P/∂M = 0 1/P 0 0 0 41 0 (φ’-ψ’) 0 0 φ’ -1/P W/P2 . r) =0 Y – Y(L) =0 φ(L) = ψ(L) W/P = φ(L) (1) (2) (3) (4) (5) To determine the effect of monetary policy in the classical model. Look at the aggregate supply curve (AS). and from (2): (φ’ .∂Y/∂M = ------------------------ The numerator is again = 0.ψ’) (∂Y/∂P)/Y’ = 0 #0 #0 42 (1) (2) . from (1): ∂L/∂P = (∂Y/∂P)/Y’. the real activity is independent of both fiscal and monetary policy. so again ∂Y/∂M = 0. So. The above implies that the AS curve is vertical. Use the equations: Y – Y(L) = 0 φ(L) – ψ(L) = 0 The slope of the AS curve is ∂Y/∂P: ∂Y/∂P = Y’(∂L/∂P) φ’ (∂L/∂P) – ψ’(∂L/∂P) = 0 then. →∂Y/∂P = 0. 43 . this is hardly the case with labour supply because of money illusion on the part of the workers (lack of information. Differentiate the system with respect to G. W/P = φ(L) and W = ψ(L) → equilibrium: Pφ(L) = ψ(L)] We can see that it is no longer true that the labour market alone determines L. This suggests that fiscal and monetary policies may now have an effect. Instead.G = 0 M/P – L(Y. Consider the extreme case where labor supply depends exclusively on the money (nominal) wage rate [here it is not possible to work with an intermediate scenario]: W = ψ(Ls) (instead of W/P = ψ(Ls)) The system of equations now becomes: (1) (2) (3) (4) (5) Y – c(Y-T) – I(r) . lengthy and binding labour contracts. Denominator det: (1-c’)L2φY’/P + L1I’φY’/P . incorrect expectations about inflation. the entire system is interrelated. etc).I’M(Pφ’ – ψ’)/P3 44 . put it in matrix form and solve for ∂Y/∂G: [Differentiating (4): φ(L)*(dP/dG) + φ’P*(dL/dG) = ψ’(L)] Numerator determinant: L2Y’φ/P (verify).An alternative Labour Supply curve While it is reasonable to assume that the demand for labour depends on the real wage rate. r) =0 Y – Y(L) =0 Pφ(L) = ψ(L) (Ld = Ls) W/P = φ(L) (demand for L) [for 4. [in bracket sign is positive] 45 . we see that the value of the multiplier is reduced. compared to the pure classical model examined earlier. the interest rate does not rise sufficiently to lead to an equal reduction in investment. when prices were fixed. Differentiate the production function and the labor market equilibrium Y – Y(L) = 0 Pφ(L) – ψ(L) = 0 with respect to P: ∂Y/∂P – Y’(∂L/∂P) = 0 φ + Pφ’(∂L/∂P) – ψ’(∂L/∂P) = 0 then.Dividing by the numerator: 1 ∂Y/∂G = --------------------------------------------------------. ∂Y/∂P = Y’[– φ/(Pφ’– ψ’)].ψ’)/L2Y’φP2] [verify why it is indeed > 0] {numer: +ve {denom: . which is > 0. so we get some increase in Y. The AS is upward sloping.> 0 (1-c’) + I’(L1/L2) – [I’M(Pφ’ . [ Ms/P ] However. thus raising the interest rate further. → (∂L/∂P) = – φ/(Pφ’– ψ’) and. The reason is that one of the effects of the increase in G is to raise prices.} Comparing the above multiplier with the one derived earlier from the IS-LM model (∂Y/∂G = 1/[(1-c’) + I’(L1/L2)]. (Pφ’– ψ’) (∂L/∂P) = – φ. thereby reducing real money supply (real money balances). 46 . we can derive an expression for ∂Y/∂M (>0).Monetary policy Similarly. Y = f(X1. whatever the value of λ. X2. The L-function will have a stationary value when the objective function has a stationary value.… Xn) s.… Xn)] The constraint must hold as equality. This could work nicely in this case but not in a case involving several choice variables and/or more than one constraint.Constrained Optimization We can introduce constrained optimization by considering the familiar utility maximization problem. the second term on the right-hand side disappears and we are left with the objective function. X2. so we have n+1variables to consider. X2. t : g = g(X1. X2.g(X1. when the constraint is satisfied.… Xn) + λ[g .… Xn) We confront this type of optimization problem using the Lagrangian method. Y) Subject to: PxX + PyY= M To deal with the above we can stick to the previous methodology by doing the following: Since in the constraint we have 2 arguments. We treat λ as an additional variable. Looking at the function. say. In the 2 variables case: Max. It involves setting up the following function: L = f(X1. The FOC require that: 47 . X and substitute into the objective function. X. which can then be optimized as usual. U=f(X. Y. Suppose we have: Max. we could solve the constraint for. .. X2.dividing the first 2 though: (∂u/∂X)/(∂u/∂Y) = MUx/MUy = Px/Py Second Order conditions (with one constraint) As in the case of optimization without constraints. fi/fj = gi/gj along with the contraint function allows you to solve the system of FOCs for the values of Xi and λ..… Xn) =0 (n+1 equations in n+1 unknowns) Take the pair-wise ratios of the FOC: f1/f2 = g1/g2. Going back to the utility maximization problem: Max. 48 . L=u(X. or negative definite for maximum and positive definite for minimum (sufficient conditions). . or fn = λgn ∂L/∂λ = g(X1. ∂L/∂Xn = ∂f/∂Xn – λ(∂g/∂Xn) = 0. the second order conditions require that the second order differential must be negative semi-definite for maximum and positive semi-definite for minimum (necessary conditions). or f1 = λg1 . Y) + λ(M – PxX – PyY) ∂L/∂X = ∂u/∂X – λPx = 0 ∂L/∂Y = ∂u/∂Y – λPy = 0 ∂L/∂λ = M – PxX – PyY = 0 Then .∂L/∂X1 = ∂f/∂X1 – λ(∂g/∂X1) = 0.. The difference when there are constraints is that, while before we were considering all non-zero values of dX and dY (small changes in X and Y away from the extreme point), now we consider those non-zero values of dX and dY which satisfy the constraint (dg = 0). The sufficient conditions translate into certain sign requirements of a bordered Hessian determinant: L11 L12 … L1n g1 ............................... Ln1 Ln2 … Lnn g2 H= g1 g2… gn 0 gn Ln1 Ln2 ... Lnn Find its successive border-preserving principal minors (note that H1 < 0): L11 H2 = L12 g1 g2 0 , H3 = L11 L21 L31 g1 .... with the last one being H itself. d2y is negative definite subject to dg = 0 iff: H2 > 0, H3 < 0, H4 > 0 ... for maximum (again notice that H1 is already negative), and: d2y is positive definite subject to dg = 0 iff: H2 , H3, H4 < 0 ... for minimum. 49 0 g1 g2 … gn g1 L11 L12 .... L1n or H= ................................ L12 L13 g1 L21 L22 g1 g2 L22 L23 g2 L32 L33 g3 g2 g3 0 In the 2 variable (and one constraint) case of utility maximization the bordered Hessian is: u11 u12 -P1 H= u21 u22 -P1 -P2 -P2 0 and H2 = H (while H1 = -(P1)2 < 0) Therefore, H must be >0 for maximum given that H1 is given to be negative. Case of more than one constraint: L = f(X1, X2,… Xn) + Σλj[gj – gj(X1, X2,… Xn)] L11 L12 … L1n g11 .. gr1 …………………………… Ln1 H= Ln2 … Lnn g1n .. grn g11 .................. g1n 0 ......0 ............................................. gr1 ............... ... grn 0 .......0 The sufficient conditions here state that for maximum the border preserving principal minors of order k > r alternate in sign, beginning with (-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 we observe the r x r matrix of zeros in the lower right side of the determinant; a 50 determinant involving fewer than r rows and columns from rows and columns 1 through n will be equal to 0. Establishing quasi-concavity and quasi-convexity in the constrained maximization case. [Note: this is not particular to constrained optimization– it is relevant to both constrained and unconstrained optimization, i.e., any time we want to establish the shape of a function]. If we establish the shape of the function we can disregard the 2nd order conditions. For twice-differentiable functions, the test for quasi-concavity and quasiconvexity we are going to use is as follows: From the following bordered determinant: f11 f12 … f1n f1 D= f21 f22 … f2n f2 ................................ fn1 fn2 ... fnn fn f1 f2 .... fn 0 , or 0 f1 f2 .... fn f1 f11 f12 … f1n ............................. fn ................. fnn The above differs from the bordered Hessian because the border consists of the first derivatives of the function, not the constraint. Then, we examine the sign of the principal minors: 0 D1 = f1 0 f1 f11 f2 , D3, … Dn. f1 or f1 0 f11 f1 D2 = f1 f11 f12 f2 f21 f22 51 = 8(x2 + y2) > 0 The function is quasi-concave. fy = -2y. (with Xi >=0) it is necessary that: D1 <= 0. fyy = -2. fxx = -2. Note that the non-positivity of D1 is already satisfied because: D1 = -f21 Example: Z = -x2 – y2 (x. Now. let’s go back to consumer’s maximization of utility. y. fxy = fyx = 0 -2 D= 0 0 -2 -2x -2y or 0 -2x -2y = -4x2 < 0 -2x -2y -2 0 0 -2 -2x -2y 0 0 D1 = -2x -2x -2 D2 = …….For Y = f(X1.…Xn) to be quasi-concave. >0) fx = -2x. X2. D2 <=0 … The sufficient conditions are the above with strict inequalities.. D2 >=0 … The necessary condition for quasi-convexity is: D1 <=0. 52 . 53 . a relaxation of the budget constraint): λ = ∂U*/∂M Thus. It can be considered as the gain (loss) in maximum utility resulting from a very small increase (decrease) in M. Interpretation of λ The Lagrange multiplier λ provides a measure of the sensitivity of U* to changes in the consumer’s budget (i. 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. it measures the change in maximum utility resulting from a change in money income. Y* and λ* yields the demand functions for the 2 inputs X and Y as functions of prices of goods and income. Y) + λ(M – PxX – PyY) FOCs: Lx = Ux – λPx = 0 Ly = Uy – λPy = 0 Lλ = M – PxX – PyY = 0 Solving the FOCs for X*.e. It can be interpreted as the Marginal Utility of Money (income) when the consumer’s utility is maximized..L = U(X. prices held constant. X1* = 12.4λ = 0 Lλ = 36 . 2X1 + 4X2 = 36 L = 2(ln X1) + ln X2 + λ(36 .4X2) LX1 = 2/X1 . λ measures what the marginal utility from saving a dollar for future consumption (not spending the last dollar) must be such that the consumer at the margin neither wants to incur debt nor to save)] Example: U = 2(ln X1) + ln X2 s.07 Checking the SOCs: L11 H= L21 g1 L12 L22 g2 g1 g2 0 then: 54 .4X2 = 0 → 2X2/X1 = ½.[Alternatively.2X1 .t.2λ = 0 LX2 = 1/X2 .2X1 . U* = ln(432) = 6. and X1 = 4X2 Substituting into the budget constraint: X2* = 3. λ* = 1/12. 2/X1 -2/(X12) 0 1/X2 0 -1/(X22) 0 D1 = 2/X1 2/X1 = -4/(X12) < 0 -2/(X12) 55 . we can check the shape of the function: f11 D = f21 f12 f22 f1 f1 = -2/(X12) 0 0 -1/(X22) 1/X2 2/X1 1/X2 0 f1 f2 0 2/X1 which can also be written as: 0 D= 2/X1 1/X2 Then.-2/(X12) H2 = H = 0 0 -1/(X22) 2 4 0 0 -0 2 4 +2 0 2 4 0 -1/(X22) 2 4 2 -1/(X2 ) 4 = -2/(X12) 4 0 = -2/(X12) (-16) + 2(2/(X22) = 32/(X12) + 4/(X22) > 0 Alternatively. L)) Let’s use an example: Q = 4K1/2L1/2.e.f(K. all along the isoquant. fK.2/X1 D2 = -2/X1 0 + 1/X2 2 2/X2 1/X2 -2/(X12) 0 1/X2 -1/(X2 ) = 4/(X12) (X22) + 2/(X12) (X22) > 0 The function is quasi-concave. > 0 C = rK + wL What quantities of K and L should the firm employ to produce a give output..] Now set up the Lagrangian function: L = rK + wL + λ(Q* . L). Q* = 32 56 . dQ = fK dK + fLdL = 0 MRTS = -dK/dL = fL/fK This means that to have convex isoquants. at minimum cost? [To obtain the slope of an isoquant we use the property that the change in Q along an isoquant is 0. fL. Q*. therefore the FOCs represent a maximum. we require diminishing MRTS (i. Cost Minimization subject to an output constraint Q = f(K. d(MRTS)/dL < 0). (which are derived from a concave production function). 2 λK1/2L-1/2 = 0 Lλ = 32 .2 λK-1/2L1/2 = 0 LL = 8 .λK-1/2L-1/2 .4K1/2L1/2 = 0 → L/K = ¼.4K1/2L1/2) FOC: LK = 2 . C* = 64. < 0 LKL LLL gL gK gL = 0 . λ* = 2 SOCs: LKK H= LLK gK λK-3/2L1/2 = .2λL-1/2K1/2 0 -λK-1/2L-1/2 λK1/2L-3/2 -2λL-1/2K1/2 57 . K* = 16.2λK-1/2L1/2 . K = 4L and using the constraint: L* = 4.C = 2K + 8L Then.2λK-1/2L1/2 = …. L = 2K + 8L + λ(32 . Comparative statics Certain comparative statics results can be obtained from the constrained maximization of utility problem. L = U(Q1. Q2) + λ(I – P1Q1 – P2Q2) For purposes of demonstration consider an example of a utility function such as: U = Q1Q2. Then: L = Q1Q2 + λ(I – P1Q1 – P2Q2) FOCs: Q2 – λP1 = 0 Q1 – λP2 = 0 I – P1Q1 – P2Q2 = 0 Take the total differential: dQ2 –P1dλ dQ1 -P1dQ1 – P2dQ2 In matrix form: = λdP1 –P2dλ = λdP2 = dI + Q1dP1 + Q2dP2 58 . 0 1 -P1 1 0 -P2 -P1 -P2 0 dQ1 dQ2 dλ = λdP1 λdP2 dI + Q1dP1 + Q2dP2 Suppose we consider only a change in P1 (so dP1 ≠ 0. dI = 0). Solve for dQ1: λdP1 0 1 0 -P1 -P2 dQ1 = Q1dP1 -P2 0 -(P2)2λdP1 – P2Q1dP1 ----------------------------. 59 .= -------------------------------0 1 -P1 2P1P2 1 -P1 0 -P2 -P2 0 Divide by dP1: dQ1/dP1 = (-λP2) /2P1 – Q1/2P1 Find the value of λ from the FOCs: Q2 = λP1 Q1 = λP2 I – P1Q1 – P2Q2 = 0 → λ = I/(2P1P2) So. dP2 = 0. ∂Q1/∂I. Differentiate the FOCs with respect to I: U11(∂Q1/∂I) + U12(∂Q2/∂I) – P1(∂λ/∂I) = 0 U21(∂Q1/∂I) + U22(∂Q2/∂I) – P2(∂λ/∂I) = 0 1 – P1(∂Q1/∂I) – P2(∂Q2/∂I) Or. P2 = 2 and I = 100 we can solve the FOCs for Q1* and Q2*: Q1* = 50. 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..∂Q1/∂P1 = -I/4(P1)2 – Q1/2P1 i. Q2) – λP2 = 0 I – P1Q1 – P2Q2 = 0 Suppose there is a change in consumer income. I. Q2) – λP1 = 0 U2(Q1.e. etc. Now: Let’s go back to the FOCs of the general model: U1(Q1. 60 (U1 = ∂U/∂Q1) (U2 = ∂U/∂Q2) =0 . Q2* = 25. Likewise we could derive ∂Q2/∂P2. if P1 = 1. = U11 U12 – P1 U21 U22 – P2 0 U22 – P2 --------------------. not both (∂Q1/∂I) and (∂Q2/∂I) can be negative (intuitively) and formally: 61 U12 .> = < 0 D (>0) – P1 – P2 Similarly.U11 U12 – P1 U21 U22 – P2 – P1 – P2 0 (∂Q1/∂I) 0 (∂Q2/∂I) = 0 (∂λ/∂I) – P1 – P2 U12 – P1 -1 0 0 U12 U22 -1 – P2 0 (∂Q1/∂I) = ------------------------.> = < 0 D (>0) So. However.> = < 0 D (>0) U11 U21 U22 (∂λ/∂I) = -----------------------. there is a possibility of inferior goods. U11 – P1 U21 – P2 (∂Q2/∂I) = ------------------. + Q1 ------------------D D (1) Only the sign of the first term on the right-hand side is known. P2. Let’s now differentiate the FOCs: U1(Q1. Q2) – λP1 = 0 (U1 = ∂U/∂Q1) U2(Q1. U11 U12 – P1 U21 U22 – P2 – P1 – P2 λ 0 =0 = Q1(dP1/dP1) (∂Q1/∂P1) (∂Q2/∂P1) = (∂λ/∂P1) λ 0 Q1 0 U12 – P1 U22 – P2 U22 – P2 U12 – P1 Q1 – P2 0 (∂Q1/∂P1) = --------------------. 62 .λ(dP1/dP1) = 0 U21(∂Q1/∂P1) + U22(∂Q2/∂P1) – P2(∂λ/∂P1) – P1(∂Q1/∂P1) – P2(∂Q2/∂P1) Or. say P1: U11(∂Q1/∂P1) + U12(∂Q2/∂P1) – P1(∂λ/∂P1) . > 0 not both effects can be negative.From: P1Q1 + P2Q2 = I: P1(∂Q1/∂I) + P2(∂Q2/∂I) = 1 (>0) And since P1.= D – P2 0 U22 – P2 λ -----------------. Q2) – λP2 = 0 (U2 = ∂U/∂Q2) I – P1Q1 – P2Q2 = 0 with respect to the price of one good. Likewise. ∂Qi/∂Pj > 0 means Qi and Qj are gross substitutes. and vice-versa for compliments.U21 – P2 U11 – P1 (2) – P1 0 U21 – P2 (∂Q2/∂P1) = …. D – P1 63 . Let’s attempt to interpret the results from (1) and (2): Consider the ∂Q1/∂I.+ Q1 --------------D D All 3 are of indeterminate sign.. We define goods as substitutes if an increase in the price of one good increases the demand for the other. Thus above. = -λ ------------------.Q1 -----------------D D [we see that the first term on the right-hand side is again known (+ve)] U21 (∂λ/∂P1) = …. = ---------------------U22 U11 U12 (3) – P1 -P2 U21 U22 λ ------------------. ∂Q2/∂I results: U12 U22 – P2 ∂Q1/∂I = -------------------. ∂Qu/∂P. The above are known as Slutsky equations. effect) (income effect) (∂Q2/∂P1) = ∂Q2u/∂P1 – Q1(∂Q2/∂I) (3) (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. (∂Qi/∂Pj) = ∂Qiu/∂Pj – Qj(∂Qi/∂I) 64 . On the other hand.e.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). so we can write equations (1) and (2) as: (∂Q1/∂P1) = ∂Q1u/∂P1 – Q1(∂Q1/∂I) (subst. the first parts in equations (1) and (2) are the pure substitution effects of a change in price (movement along the same indifference curve). Similarly: (∂Q1/∂P2) = (∂Q2/∂P2) = ∂Q1u/∂P2 – Q2(∂Q1/∂I) ∂Q2u/∂P2 – Q2(∂Q2/∂I) And in general.. i. and second. a pure substitution effect (or the response to a price change holding the consumer on the original indifference curve). where income (purchasing power of a given money income) is changed holding prices constant. 65 .The Slutsky equations show that the response of the consumer to the change in the price of a good can be split-up into 2 parts: first. a pure income effect. Q = f(X1.r1X1 + r2X2 = 0 and: f1 / f2 = r1/r2. Cost s.Comparative statics: Theory of the firm One can formulate a constrained optimization problem for the firm as: [earlier we had seen it also formulated as: min. Then (to explore what the economic meaning of λ is).r1X1 + r2X2) FOCs: ∂L/∂X1 = f1 – λr1 = 0 ∂L/∂X2 = f2 – λr2 = 0 ∂L/∂λ = C . or dC = 1/λ(f1dX1 + f2dX2) (1) By taking the differential of the production function: 66 .t. an output constraint. t. X2) s.] Max. X2) + λ(C . C = r1X1 + r2X2 Lagrangian function: L = f(X1. λ = f1 /r1 = f2 /r2 → λr1 = f1 → r1 = (1/λ) f1 and r2 = (1/λ) f2 By differentiating the constraint: dC = r1dX1 + r2dX2. X2) – r1X1 – r2X2. we return to that. Comparative statics results from the profit maximization problem can be derived by first deriving input demand functions using the FOCs: π1 = Pf1(X1. Recall the equivalent unconstrained profit maximization problem: Max: π = Pf(X1. X2) – r2 = 0 67 . dC/dQ = MC = 1/λ The SOCs require that the bordered Hessian determinant is positive: f11 f21 -r1 f12 f22 -r2 -r1 -r2 0 >0 This requires that the production function is quasi-concave. since the conventional formulation of the producer’s problem is as an (unconstrained) profit maximization problem. and recall the comparative statics results we have derived. However.dQ = f1dX1 + f2dX2 Dividing (2)/(1): (2) f1dX1 + f2dX2 dQ/dC = λ ( ---------------------) = λ = 1/MC f1dX1 + f2dX2 So. X2) – r1 = 0 π2 = Pf2(X1. r1 and r2): Pf11dX1 + Pf12dX2 = -f1dP + dr1 Pf21dX1 + Pf22dX2 = -f2dP + dr2 Or. taking the total differential for changing P. By totally differentiating these conditions (or equivalently. dr1 Pf12 0 Pf22 dr1Pf22 dX1 = -----------------. Pf11 Pf21 Pf12 Pf22 dX1 = dX2 -f2dP + dr2 -f1dP + dr1 Letting dr2 = 0 and dP = 0. the producer will change the use of inputs so that the FOCs continue to be satisfied.= -------------Pf11 Pf12 P2 |D| Pf21 and. 68 . dX1/dr1 = f22 /P |D| < 0 (|D| = f11f22 – f212 > 0) Pf22 The response to the change of an input price (other things constant) is always negative.As prices of inputs change. i. dr2 = 0 and allow P to change: -f1dP Pf12 -f2dP Pf22 -Pf1f22dP + Pf2f12dP dX1 = ----------------------. if f12 > 0. dX1/dP > 0 and an increase in output price will result in an increase in demand for X1. dP = 0. dX1/dr2 = . then: dX1/dr2 < 0 Finally. with dr1=0.. and: P |D| (f2f12 – f1f22) dX1/dP = ---------------------P |D| Here too.Similarly.e. if f12 > 0. Now consider a specific production technology (instead of a general production function) – for example the frequently used Cobb-Douglas PF: 69 .= ----------------------------P2 |D| P2 |D| dP(f2f12 – f1f22) = … = -----------------------. with dr1=0..f12 /P |D| If we assume that an increase in the use of one input increases the MP of the other. let dr1 = 0. dP = 0. dX2/dr2 = f11 /P |D| < 0 On the other hand. Assume a Cobb-Douglas production function.< 0 P2 |D| P2 |D| Similarly: dX1/dr2 < 0 dX1/dP > 0 Can you find these expressions if a + b =1? (see page 73).= -------------------------. with a and b > 0 and a + b <1 (decreasing returns to scale). 70 . Q = (X1)a(X2)b. Find the expression for dX1/dr1 and show that it is negative: Π=PX1aX2b – r1X1 – r2X2 π1 = PaX1a-1X2b – r1 = 0 π2 = PbX1aX2b-1 – r2 = 0 Take the differential of the first order conditions: 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 = ----------------------------------. Multiply (1) by a factor j = 1/L.e.…jxn) = jr f(x1. Homogeneous functions Given: f(x1. Linearly homogeneous functions (not linear and homogeneous) These are homogeneous functions of the first degree (r =1).e. x2. L) = f(K/L. k=K/L alone. L/L) = f(K/L.We are now going to explore certain properties of production functions (such as Homogeneity) and highlight those using Cobb-Douglas production functions. if K/L is kept constant. average product functions are homogeneous of degree 0). The above imply that as a result of linear homogeneity. Similarly for MPPL and MPPK (see book for proof) 71 . x2.…xn). as well as the MPPL and MPPK can be expressed as functions of the capital/labor ratio.…xn): A function is homogeneous of degree r. jx2. the average products will stay constant too (i. output will be multiplied by jQ = Q/L [since j=1/L]... 1) = f(k) APPL = Q/L = φ(k) [Q=Lφ(k)] then: APPK = Q/K = (Q/L)(L/K) = φ(k)/k. if: f(jx1. those exhibiting CRTS. L): (1) the APPL and APPK. so: f(K. i. Properties of CRTS production functions: Property (1): Given: Q = f(K. 1) = f(k. where j>0. Because of linear homogeneity (CRTS). Now, consider the Cobb-Douglas production function with constant returns to scale: Q = AKaLb = AKaL1-a MPPL = ∂Q/∂L = A (1-a)KaL-a = A(1-a)Ka/La = A(1-a)(K/L)a, and MPPK = ∂Q/∂K = … = Aa(K/L)a-1 In economic terms, this means that the MP of labor and capital are constant whenever the capital to labor ratio is constant. That is the MP (and the AP) are independent of the scale of production, and depend only on capital intensity (K/L). A puzzle Earlier we derived dX1/dr1< 0, dX1/dr2 > 0, dX1/dP > 0, etc., given a + b < 1. Now, use a production function exhibiting CRTS and perform comparative statics. Earlier we asked the question: Can you find the expressions dX1/dr1, dX2/dr2, dX1/dr2, etc, when if: =1? Pab X1a-1X2b-1 a+b 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 PabX1a-1X2b-1 Pab X1a-1X2b-1 Pb(b-1) X1aX2b-2 = (P2ab(a-1)(b-1)X12a-2X22b-2) - (P2a2b2 X12a-2X22b-2) 72 = (X12a-2X22b-2) [P2ab(a-1)(b-1) - P2a2b2] (and since b=1-a) = (X12a-2X22b-2) [P2ab(-b)(-a) - P2a2b2] = (X12a-2X22b-2) [P2ab(ba) - P2a2b2] =0 So these effects are indeterminate under CRTS. Note also that the SOCs are not satisfied in this case (since det D = 0). Why? Under CRTS, MC and AC are constant (independent of the scale of production). So with a horizontal long run average cost curve (instead of U-shaped), there is no unique combination of inputs for the firm; i.e., the firm is indifferent as long as the first order conditions are satisfied and the firm makes zero economic profits. Property (2) - Euler’s theorem If Q= f(K, L) is linearly homogeneous (CRTS), then: K(∂f/∂K) + L(∂f/∂L) ≡ Q (for proof see book). (1) One should distinguish this identity which holds for any values of K and L (but only for constant return to scale production functions) from the total differential of Q: dQ = (∂Q/∂K)dK + (∂Q/∂L)dL, which holds for any function, not only CRTS functions. Euler’s theorem says that under CRTS, if each input is paid according to its marginal product contribution, payments to inputs will exactly exhaust all 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/P Substituting into the Euler equation and multiplying both sides by P: 73 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 we consider increasing returns to scale the Euler theorem says: 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 at the existing market prices. So factors cannot receive rewards equal to their marginal products. Similarly, under decreasing returns to scale, total revenue is more than sufficient to reward factors according to their MP (rK + wL < PQ, i.e., there is a surplus). This leaves unanswered the question: what should be done with the surplus. Because of the above issues, neoclassical economic theory tends to use the assumption of CRTS. 74 subject to one or more constraints which are in general inequalities:.…Xn) of n variables.4)dX1 + 2(X2 – 3)dX2 – 0 = 0 2(X1. usually restricted to be non-negative (Xj ≥ 0).e.. g(X1. f* =2. ….Non-linear optimization with inequality constraints Introduction It refers to the problem of finding the optimal value of a given function f(X1. 75 . X2. 3X1 + 2X2 ≤ 12 -2X1 + 2X2 ≤ 3 2X1 – X2 ≤ 4 2X1 + 3X2 ≥ 6 X1. or ≤} bi. Min f = (X1 – 4)2 + (X2 -3)2 s.77. t. Xn) {≥ . To find the slope of the objective function. X2* = 27/13. The slope of g1 (the relevant constraint) is -3/2. To find this point we utilize the fact that at the tangency point (see graph). ≥ 0 (g1) (g2) (g3) (g4) We can solve this problem with the aid of a graph which allows us to see which constraint is binding. =. X1. the slopes are equal.4) + 2(X2 – 3)dX2/dX1= 0 But at tangency between f and g1 the slope is -3/2. we take the total differential: (X1 – 4)2 + (X2 -3)2 – f = 0 2(X1. The optimal point is X1* = 34/13. f and g are not both linear. X2. i. One can show that the optimal can occur inside the Feasible Set. we get the solution given above. Example: Min Z = (X1 – 9/5)2 + (X2 – 2)2 (same constraints) 76 .4) + 2(X2 – 3)(-3/2) = 0 → 2X1 – 3X2 = -1 Solving along with 3X1 + 2X2 = 12. 2(X1.So. Now the optimal value of Z occurs at an interior point (2. We can also show that a local optimum is not always a global one. Point A will provide a local maximum (Z = 169/100). example: X1X2 ≤ b1 X12 + X22 ≥ b2 X1. ≥ 0 77 . i.: Suppose we were looking for maximum instead of a minimum (same objective function and constraints). Finally. a further complication can result. but point B would give global maximum.e. X2. 9/5) and Z = 0 at minimum. that is Z at B > Z at A. O. Some N. Can we extend this approach to N. Necessary conditions for optimum (FOCs).O.This results in a feasible set that is not convex (it is disjoint). problems have a solution.L. some do not. problems (that is problems with inequality constraints)? Kuhn and Tucker have shown that: 78 . A solution may not exist if the constraints contain inconsistencies (as in the example above) or if the problem is unbounded.L. In handling classical constrained optimization problems we used calculus techniques. A convenient approach there was the Lagrangian method. This saddle point is a consequence of duality. or max-min relationship. when the Lagrangian function is optimized.…) that maximize the L-function and at the same time the values of the multipliers (λ*) which minimize the value of the Lfunction. Furthermore. X2*. then suppose we solve the problem. We say that it is an optimal solution only when we have found the values of the variables of the maximization problem (X1*. suppose we have a maximization problem: We set up the Lfunction and treat the L-multipliers as variables (as we did earlier). 79 .(1) (2) There is a wide class of such problems for which a Lagrangian approach can be followed. The duality concept specifies a primal-dual. This solution is called a saddle point. the same values that optimize the Lagrangian will also optimize the original function subject to the constraints. Let X* be the optimal set of the max. Given the problem: Max.Suppose we have a (primal) profit maximization problem. …Xn)] + … + λm[bm . t. X2. Then. …Xn)] 80 . will also maximize the L-function in which the L-multipliers (λ) take the dual optimal variable values. the Lagrangian multipliers in the primal problem are equal to the dual optimal variables. …Xn) + λ1[b1 . . X2. g1(X1. problem variables and Y* be the optimal set of the dual (minimization problem variables). Therefore. Z = f(X1. X*. L(Y*. X2. Z(X). . …Xn) s. The dual will be a cost minimization problem.Similarly. X2. X*) in which the multipliers are the primal optimal values. Y*. …Xn) ≤ b1 ……………………….gm(X1. To set up the L-expression in the case of NLO problems we have to follow a similar but slightly different approach from the case of classical constrained optimization. the values of the dual variables Y* that minimize the dual minimization problem will also minimize the L-function. and the dual multipliers are equal to the primal optimal solution values (More about duality in Linear Programming). X2.g1(X1. …Xn) ≤ bm Then.The values of the primal variables X* which maximize the primal objective function. X2. gm(X1. λ) = f(X1. L(X. and Xj[∂L(X. they are the First-Order-Conditions for NLO problems. λ)/∂Xj] = 0 ∂L(X. λ) = X13 + X1X2 + λ1(10 . λ)/∂λi ≥ 0. X2.X1 . λ)/∂λi] = 0 Xj.) ∂L(X. ∂L(X. Following are the K-T conditions side by side with the FOCs for classical constrained optimization: Classical optimization (Max. X1 + X25 ≤ 10 X12 + 2X2 ≥ 4 X1. λ)/∂Xj = 0 ∂L(X. and λi[∂L(X. λi. λ)/∂λi = 0 K-T Conditions for Max. In other words.X25) + λ2(-4 + X12 + 2X2) We entered the constraints in such a way so that when we take the derivative of L with respect to λ1 and λ2 we get the result that the K-T conditions ∂L/∂λi ≥ 0 are equivalent to the constraints: 81 . t. λ)/∂Xj ≤ 0.The Kuhn-Tucker conditions are necessary (but not sufficient) for optimization. ≥ 0 Then. ≥ 0 (for minimization the inequalities are reversed) Example: Max Z = X13 + X1X2 s. L(X. X1[3X12 + X2 – λ1 + 2λ2X1] = 0 X2[X1 .X25 ≥ 0 ∂L/∂λ2 = -4 + X12 + 2X2 ≥ 0 Along with.∂L/∂λ1 = 10 .X25] = 0 λ2[-4 + X12 + 2X2] = 0 X1. λ1[10 .5λ1X24 + 2λ2 ≤ 0 Along with. ∂L/∂λ1 = 10 .X1 . Having inequality constraints amounts to allowing for the possibility that the optimum occurs at a point where one or more of the solution values are = 0 (Xj 82 (1) (2) (1a) (2a) (3) (4) (3a) (4a) .X1 .5λ1X24 + 2λ2] = 0 and. X2 ≥ 0 The K-T conditions are in fact generalizations of the FOCs in classical constrained optimization.X25 ≥ 0 (equivalent to X1 + X25 ≤ 10) ∂L/∂λ2 = -4 + X12 + 2X2 ≥ 0 (equivalent to X12 + 2X2 ≥ 4) Now the K-T conditions are: ∂L/∂X1 = 3X12 + X2 – λ1 + 2λ2X1 ≤ 0 ∂L/∂X2 = X1 .X1 . for y = f(x). for max.. can occur at a corner point as well as an interior point. If ∂Y/∂X = 0 (point B).. The K-T conditions apply to both cases. Can this represent a maximum? Yes. In this respect. they describe a more general case and therefore contain the conditions for classical constrained optimization. for max. the possibility of a boundary or corner solution as opposed to an interior solution. i. because then. However.e.e. Putting the above together and generalizing for functions with more than one variables.= 0). At C we have ∂Y/∂X < 0 and X = 0. we may have a max. we have: 83 . we can increase the value of Y by increasing X. because the value of Y will decrease to the right of point C. which from the non-negativity constraints has to be at the positive orthand. A max. we cannot have ∂Y/∂X > 0 at X = 0. i. or ∂Y/∂Xi < 0 and Xi = 0 (corner max. λi>0 → ∂L/∂λi = 0.Either ∂Y/∂Xi = 0 and Xi > 0 (interior max. g1(X1. …Xn) ≤ b1 ………………………. or ∂L/∂λi ≠ 0 → λi = 0. either Xj = 0 or ∂L/∂Xj = 0 (or both) (point A) (point C) (point B) Similar conclusions are derived from λi[∂L/∂λ] = 0.) ∂Y/∂Xi = 0 and Xi = 0 is also possible Which one applies? From the K-T conditions: Xj[∂L/∂Xj] = 0 This means that if Xj > 0 → ∂L/∂Xj = 0 If ∂L/∂Xj ≠ 0 → Xj = 0 That is.. or both ∂L/∂λi = 0 and λi = 0] When λi>0. …Xn) s. X2.).e. the signs of the inequalities are reversed. this means that the corresponding constraint is binding. X2. gm(X1. ∂L/∂Xj ≥ 0 ∂L/∂λi ≤ 0 Intuitive explanation Max. t. X2. If we have minimization problem. Z = f(X1. …Xn) ≤ bm Consider the Lagrangian expression: 84 . [i. which is the ith constraint.gm(X1. …Xn)] = 0 for all i. Then. Therefore. …Xn)] + … + λm[bm .gi(X1. by the complimentary condition: λi (∂L/∂λi) = 0.g1(X1. That is. X2. …Xn) + λ1[b1 . Will make the value of L equal to that of the objective function. which is the objective function. …Xn) ≥ 0. …Xn)] Condition ∂L/∂λi ≥ 0 for maximum means: ∂L/∂λi = bi . X2. 85 . these terms on the right hand side vanish and we are left with: L(X*. λ) = f(X1. a solution which satisfies the K-T conditions for the L-function must also: (1) (2) Satisfy the constraints.L(X. or λi [bi . λ*) = f(X1. …Xn). X2.gi(X1. X2. X2. we know that the constraints are satisfied. X2. (3) The K-T conditions must be satisfied. concave can refer to a function or a region] Non-strictly concave Strictly concave Strictly convex 86 .e. the set of constraints form a Feasible Region which is convex (concave) – i. For a minimum it must be convex.. [Notice that the classification for convex vs. the constraint functions must be convex. (2) For maximum the objective function must be concave.Sufficient conditions The candidate point will be a global max (min) if: (1) For max (min). ≥ 0 Then. t.X1 . by examining the K-T conditions using examples. So. C = (X1 – 3)2 + (X2 – 4)2 s. X1 = 0. X2 > 0 (immediately rejected) 87 (=0 if X1 > 0) (=0 if X2 > 0) (=0 if λ > 0) .X1 . Consider the problem: Min.X2 ≤ 0 (3’) λ(4 . L = (X1 – 3)2 + (X2 – 4)2 + λ(4 . X2 = 0 2.X2) = 0 We have 4 cases to consider: 1. we use the determinant test to determine the shape of each function. X1 = 0. X2.X1 .X2) (1) ∂L/∂X1 = 2(X1 – 3) – λ ≥ 0 (1’) X1[2(X1 – 3) – λ] = 0 (2) ∂L/∂X2 = 2(X2 – 4) – λ ≥ 0 (2’) X2[2(X2 – 4) – λ] = 0 (3) ∂L/∂λ = 4 . We can approach this in an informal way. Computational aspects We are still looking for a method by which we can solve a NLO problem. X1 + X2 ≥ 4 X1.Note also that a linear function satisfies both convexity and concavity. Furthermore: From (2). X2 > 0. λ = -1. with λ < 0. we already have a violation. 88 (1) (2) (3) . C=4. X2 = 5/2. From (1): -6 – λ ≥ 0. with λ < 0: 2X2 – 8 – λ = 0 → 2X2 – 8 = λ (< 0) → X2 < 4. this is not the optimal solution. This solution satisfies the K-T conditions and the constraint. X2 = 0 4. From (3): 4 –X1 < 0 (contradiction/violation) Case 4: If X1 > 0. From (1).3.5. if permitted) 2(X1 – 3) – λ = 0 → 2X1 – 6 = λ 2(X2 – 4) – λ = 0 → 2X2 – 8 = λ And the constraint is equality. Then from (3): 4 – X2 < 0 (contradiction/violation). Case 3: If X1 > 0. or 6 + λ ≤ 0 → λ < 0 If λ is also restricted to be non-negative. with λ ≠ 0 (λ < 0. X2 = 0. X2 > 0. X1 > 0. X2 > 0 Case 2: If X1 = 0. From (2): -8 – λ ≥ 0 → 8 + λ ≤ 0 → λ < 0. From (1): 2X1 – 6 – λ = 0 (since X1 > 0). therefore: X1 + X2 = 4 Solve the 3 equations to get: X1 = 3/2. X1 > 0. however. 2X1 – 6 < 0 → X1 < 3. λ = 0. X1 + X2 ≤ 1 X1.Case 5: X1 > 0. ≥ 0 L = -3e-2X1 – 4e-5X2 + λ(1 – X1 – X2) ∂L/∂X1 = 6e-2X1 – λ ≤ 0 ∂L/∂X2 = 20e-5X2 – λ ≤ 0 ∂L/∂λ = 1 – X1 – X2 ≥ 0 Examine the 4 cases: (1) (2) X1 = 0.X1-X2=1–0-0= 0) X1 = 0. X2 > 0. Example 2 Max. X2 > 0. X2 = 4. The optimal value of the objective function is: C = 0. -3e-2X1 – 4e-5X2 s. X2 = 0 (rejected: it implies λ>0 → 1. with λ = 0 2X1 = 6 2X2 = 8 The point which minimizes the objective function and satisfies the K-T conditions and the constraint is: X1 = 3. then: (=0 if X1 > 0) (=0 if X2 > 0) (=0 if λ > 0) 6–λ≤0 Furthermore. 20e-5X2 = λ (since X2 > 0) → λ > 0 89 . t. The constraint is satisfied as a strict inequality. X2. Thus, 1 – X2 = 0 → X2 = 1 Then, 20e-5 = 0.134 which violates 6 – λ ≤ 0. (3) X1 > 0, X2 = 0, then: 6e-2X1 – λ = 0 20 - λ ≤ 0 → λ > 0, and: 1– X1 = 0 → X1 = 1 So, 6e-2 = 0.81 = λ, which violates 20 – λ ≤ 0. (4) X1 > 0, X2 > 0, then: 6e-2X1 = λ 20e-5X2 = λ →λ>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 90 One more example to highlight how a graph can help by seing which constraint/s are binding. The firm produces and sells products x1 and x2 with demand curves: P1 = 10 – x1 and P2 = 20 – x2 The problem is: Max R = x1(10 – x1) + x2(20 – x2) Subject to: 5x1 + 3x2 ≤ 40 x1 ≤ 5 x2 ≤ 10 x1, x2 ≥ 0 (1) Use a Lagrangian function and the Kuhn-Tucker conditions to find the revenue maximizing values of x1, x2, as well as the values of the λs. (2) Find the effect on maximum revenue if the firm had one more unit of the first resource; then find the effect on maximum revenue if the firm had one more unit of the second resource. Check that K-T sufficiency conditions for maximum hold. Answer L = x1(10 – x1) + x2(20 – x2) + λ1(40 – 5x1 – 3x2) + λ2(5 – x1) + λ3(10 – x2) ∂L/∂x1 = 10 – 2x1 - 5λ1 – λ2 ≤ 0 [x1*(10 – 2x1 - 5λ1 – λ2)=0] ∂L/∂x2 = 20 – 2x2 - 3λ1 – λ3 ≤ 0 [x2*(20 – 2x2 - 3λ1 – λ3)=0] ∂L/∂λ1 = 40 – 5x1 – 3x2 ≥ 0 [λ1*(40 – 5x1 – 3x2)=0] ∂L/∂λ2 = 5 – x1 ≥ 0 [λ2*(5 – x1)=0] ∂L/∂λ3 = 10 – x2 ≥ 0 [λ3*(10 – x2)=0] Consider point A: 5λ1 + λ2 = 10-4 = 6 3λ1 + λ3= 20-20 = 0 Given that λ2=0 (the corresponding constraint is not binding): The values associated with point A are x1 = 2, x2 = 10, λ1=6/5, λ2 = 0, λ3 = -18/5. 91 This solution satisfies all K-T conditions and the constraints: It also gives a better value than point B. However, you can check that the optimal solution is somewhere along segment AB. To find: set λ1>0 (since the first constraint is binding) and λ2 = λ3 = 0 (since the other two constraints are not binding). Then: 10 – 2x1 - 5λ1 = 0 20 – 2x2 - 3λ1 = 0 40 – 5x1 – 3x2 = 0 5 – x1 > 0 10 – x2 > 0 The solution is: x1* = 2.8, x2* = 8.66, λ1* = 0.88 and R* = 118.3 > 116 at point A. 92 P1X1 . U(X1. X2). λ.P1X1 . X2) s. X2. compared to the usual treatment. X2) + λ(I . is that now the consumer is not necessarily spending all income and doesn’t need to consume positive amounts of both goods. X2.P2X2) K-T conditions: ∂L/∂X1 = U1 – λP1 ≤ 0 ∂L/∂X2 = U2 – λP2 ≤ 0 ∂L/∂λ = I . subject to a budget constraint can be given a treatment using K-T analysis. Consider as a first case: X1. ≥ 0 L = U(X1. then: U1 = λP1 U2 = λP2 → λ (MU of money income) = U1 / P1 = U2 / P2 (>0) This represents the MU per $ spent on X1 and X2. 93 . Max: U(X1. I.P2X2 ≥ 0 (=0 if X1 > 0) (=0 if X2 > 0) (=0 if λ > 0) Recall that the Lagrange multiplier. The qualifying difference here. has been interpreted as the MU of money income. t: P1X1 + P2X2 ≤ I X1. > 0.Example: Maximization of Utility The model of a consumer maximizing utility. If the constraint is not binding (i.e. This means that the consumer is satiated in all goods.P2X2 = 0 (consumer exhausts all income). In this case. U1/P1 = U2/P2 = λ = 0.. the consumer does not exhaust income). this can be written as: P1X1 ..e.In this case (interior point maximization). then. i. The remaining possibility is that the consumer is maximizing at a corner: 94 . Going to the alternative case (if consumer not necessarily spends all income): ∂L/∂λ = I . ∂L/∂λ = I . To see this: U1 = λP1 = 0 U2 = λP2 = 0. the consumer is maximizing at some point inside the feasible region. That is the consumer will not consume more of the 2 goods even if they are free.P2X2 < I.P1X1 .P1X1 .P2X2 ≥ 0. the consumer equates the MU per $ spent on each good to the MU of money income. that is MU of income is 0 (utility from spending one more $ of income = 0). λP1 < 0 [and X1* (U1 . At the same time. his MU of money exceeds the MU of consuming X1 per $ spent.λP2 = 0. 95 . or U1 < λP1 → λ > U1 / P1 That is. since X2 > 0. → U2 . → U1 .λP1) = 0]. i. the consumer is consuming X2 so that he equates the MU per $ spent on X2 to the MU of money income. or U2 = λP2 → λ = U2 / P2.e..Since X1 = 0. y* = y*(φ). The optimum of a direct objective function is given for given values of parameters. The indirect objective function traces out all the maximum values of the objective function as these parameters (φ) vary. Given the FOCs holding: dV/dφ = fφ So. Illustration using the Profit function 96 . or an indirect objective function (IOF). i. the derivative dV/dφ depends only on the direct effect of the change in φ. 2 indirect and 1 direct effect of the change in φ). φ) We can call this a maximum value function. the derivative gives the same result as if x* and y* are treated as constants. y*(φ). Substitute into the objective function: V(φ) = f(x*(φ).e.. with φ changing and x*. We can therefore state that the IOF is an “envelope” of all the optimized objective functions generated from varying the parameters.e. y. Therefore. y* allowed to adjust.Maximum value functions and the Envelope theorem (Unconstrained Optimization) Suppose you have found the optimal values of an objective function (which are themselves functions of exogenous variables and parameters): From: Max U = f(x. Now differentiate V(φ) with respect to φ: dV/dφ = fx*(∂x*/∂φ) + fy*(∂y*/∂φ) + fφ (i. the envelope theorem says that only the direct effect of the change in the exogenous variable needs to be considered. φ) From the FOCs and the SOCs satisfied: x* = x*(φ).. P) = Pf(K*.π = Pf(K. Similarly. we can show that: ∂π*/∂r = – K*(w. L*). (i. r. so: ∂π*/∂w = – L*(w. supply function) Equations 3-5 are known as Hotelling’s lemma. P) ∂π*/∂P = f(K*.e. r. L*) – wL* .e. for labour when the wage rate changes.rK* (2) The effect of a change in w (holding everything else constant) on the firm’s profits usinf equation (1) is: ∂π/∂w = -L. r. (3) The above means that at the profit maximizing position. But from the FOCs the two terms in parentheses vanish. the change in profits arising from a change in the wage rate is the same independently of whether the inputs are held constant or not as input prices change. and the profit function is the IOF (max profit given the values of the parameters): π*(w. equals the negative of the labour demand function). which does not take into account the possibility of substituting capita.. L) – wL – rK (1) Solve the FOCs for the demand for input functions: L* = L*(w. P). which takes all capital-labour substitutions into account: ∂π*/∂w = (PfL – w) ∂L*/∂w + (PfK – r) ∂K*/∂w – L*. r. r. Then using equation (2). P) K* = K*(w.. (4) (5) 97 . P) (i. B). The dual problem is: Min. y) = U* Using the Lagrangian approach and solving for the (Hicksian – compensated. B) y = y(Px. and solving for the (Marshalian uncompensated) demand functions and λ: x = x(Px. P) while holding K* and L* constant. y(Px. The indirect utility function is: U* = U*(x(Px. So Hotelling’s lemma is another manifestation of the Envelope theorem (dV/dφ = fφ). U*) μ = μ(Px. Primal and dual Consider the utility maximization problem as an example. y) s.. Py. real income held constant) demand functions and μ: x = x(Px. U*) 98 . (maximum utility given the values of the parameters). Py. r. U*) y = y(Px. Py. U(x. Py. B).e. Py. B) λ = λ(Px. i. t. Py.One can also show that the same results emerge if we differentiate equation (2) with respect to each parameter (w. E = Pxx + Pyy s.t. Py. Py. The primal problem is: Max U = U(x. Py. Pxx + PyY = B Using the Lagrangian approach. B)) = V(Px. B). Which is the minimum value function (min expenditure function). The minimized expenditures in the dual equal the given budget B in the primal: E(Px.e. the solutions for the optimal consumptions are the same in the two problems. However. Duality The tangency condition is reflected in: Ux/Uy = Px/Py The tangency condition is identical for both problems. U*). B) = y(Px. U*) = B. Py. Py. U*) + Pyy(Px. U*) y(Px. Py. U*) = E(Px. Py.. B) = x(Px. Py. U*) i. since the solutions are functions of different exogenous variables. given U*. Hence: x(Px. Py. comparative-static exercises will generally produce different results. Py. 99 .Substituting x and y into the objective function: Pxx(Px. Py. and the maximized value of utility in the primal equals the target level of utility in the dual. The above results are due to duality conditions. λx Then. Taking the ration of these two partial derivatives: ∂V/∂Px/∂V/∂B = -x These results.Note however. ∂V/∂B = λ. hence the solutions values for these two are reciprocal to each other: λ = 1/μ Roy’s Identity (application of Envelope theorem) In the primal utility maximization problem. gives the maximum value function: V(Px.λx Using the FOCs: ∂V/∂Px = . that λ ≠ μ: λ = Ux/Px. 100 . Similarly for y. known as Roy’s identity. y) + λ(B – Pxx – Pyy) Differentiate with respect to Px: ∂V/∂Px = (Ux–λPx)∂x/∂Px + (Uy–λPy) )∂y/∂Px + (B–Pxx–Pyy) ∂λ/∂Px . while μ = Px/Ux. Py. B) = U(x. says that the Marshallian demand for good x is the negative of the ratio of two partial derivatives of the maximum value function V with respect to Px and B. respectively. differentiate with respect to B: ∂V/∂B = (Ux–λPx)∂x/∂B + (Uy–λPy) )∂y/∂B + (B–Pxx–Pyy) ∂λ/∂B + λ Or. substituting the optimal values into the Lagrangian. Example (Example 1.Pxx . Shephard’s Lemma One can show that the partial derivatives of the consumer’s expenditure function E(Px. Py. page 439): Utility maximization: Max. rearranging and solving for B: B = (4PxPyU*)1/2 = 2Px1/2Py1/2U1/2* The consumer’s dual problem is to minimize expenditure: E(Px. ∂V/∂B = B/2 PxPy. Py. Pxx + Pyy = B Lagrangian: Z = xy + λ(B . and: -(∂V/∂Px/∂V/∂B) = … = B/2Px = x (Roy’s identity). U*) with respect to Px and Py are the consumer’s (Hicksian) demand functions. B) = B2/4PxPy Setting V=U* (the maximized utility). U= xy s. 101 . U*) = B = 2Px1/2Py1/2U1/2* Differentiating the IUF with respect to Px and then B: ∂V/∂Px = -(B/4Px2Py).t.Pyy) Deriving and then solving the FOCs for the demand functions gives: x = B/2Px y = B/2Py λ = B/2PxPy (2hr SOCs are satisfied) The IUF is: V(Px. Py. xy = U* Lagrangian: Z = Pxx + Pyy + μ(U* . results in the minimum-value function: E = Pxx + Pyy = Px(PyU*/Px)1/2 + Py(PxU*/Py)1/2 = … = 2Px1/2Py1/2U1/2* (also found in Example 1). U*)/∂U* = Px1/2Py1/2/U*1/2 = μ 102 . U*)/∂Py = Px1/2U*1/2/Py1/2 = y ∂ E(Px. Py. To demonstrate Shephard’s lemma: ∂ E(Px. U*)/∂Px = Py1/2U*1/2/Px1/2 = x ∂ E(Px.Using an example: (Example 2. Pxx + Pyy s. t.xy) Solving the FOCs (and given that the SOCs hold): x = (PyU*/Px)1/2 y = (PxU*/Py)1/2 μ = (PxPy/U*)1/2 Substituting the values of the x and y into the objective function. Py. Py. page 440) Corresponding to Example 1: Min. for any other value of x.e.. in which variables at different points in time are related in an essential way. we have also y = a. i. the value of the function will be different from a. if its behaviour over time is determined by functional equations. Continuous vs. If we put x = (a b)/a. It is easy to confirm that the function we are looking for is: y(x) = Aex. A system is dynamical. 103 . we are concerned with growth/change of key economic variables and stability over time. Dynamics: Continuous time Example: Consider the functional equation: y’(x) – y(x) = 0. To solve a functional equation means to find an unknown function which satisfies the functional equation identically.DYNAMIC SYSTEMS In dynamics. y’ – y = 0. discrete time. since y’(x) = Aex for any x. To solve an equation in the usual sense is to find the value (or values) of the unknown which satisfies the equation. we need to make an assumption: Continuous vs. Now consider the function: y = ax + b. Applications of continuous time are mostly found in natural sciences. What is a functional equation: It is an equation in which the unknown is a function. which gives y’ = a. therefore. However. discrete time Since we are interested in the behaviour of economic variables over-time. the function y = ax + b does not satisfy out functional equation identically. we know that y(0) = 10. In general. y(t). This is the essence of integration. We are looking for the function y = y(t). Indeed. We need to find the primitive function corresponding to the derivative (1).. usually at time 0. Use of integration in dynamics (continuous time) Consider a relation such as the one given above. If. i. y(t) = 2t1/2 + c.Now if we acknowledge that the symbol x stands for time. 104 . This is called an initial condition. Suppose that a variable (for example. where c is an arbitrary constant. given that we know its derivative. in most cases we have such information. then: y(0) = 2(0)1/2 + c = 10 → c = 10. In order to be able to solve for the constant. we need additional information. The function whose derivative is t-1/2 is 2t1/2. Using the earlier example: y’(t) – y(t) = 0 In fact. population) is known to grow over time at the rate: dy/dt = y’(t) = 1/√t ( or t-1/2) (1) We want to find the time path of this population. and: y(t) = 2t1/2 + 10.e. in particular we know the value of the function at some point in time. y’(t) = y(t) can be considered as a relation which involves the value of y at any point in time and the value it takes at an arbitrarily close point determined by y’(t). for example. we are ready to understand the definition of dynamic systems. given that this population is growing at the rate given by (1). y(t) = 2t1/2 + y(0). which implies: F’(x) = f(x) Rules: (1) Integral of a constant: ∫kdx = kx + c Integral of a power function: ∫xndx = (xn+1/n+1) + c (n ≠ -1) Integral of 1/x: ∫(1/x)dx = ln(x) + c (also a sub-case of (2)) Integral of a natural exponential function: ∫ekxdx = ekx/k + c (special case: ∫exdx = ex + c) Integral of a constant times a function: ∫kf(x)dx = k∫f(x)dx Integral of a sum of 2 or more functions: ∫[f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx 105 (2) (3) (4) (5) (6) . we can refresh the rules of integration: Rules of integration Notation: ∫f(x)dx The outcome of integration is: ∫f(x)dx = F(x) + c. Before we proceed with more examples. ∫12x-1dx = 12∫(1/x)dx = 12 ln|x| + c 5.∫xdx + ∫dx = 2(x3/3) .½(x)2 + x + c 4. ∫2x2 – x + 1)dx = 2∫x2 . 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: . ∫x4dx = x4+1/4+1 = x5/5 + c 3.(7) Integral of the negative of a function: ∫-f(x)dx = . ∫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 use the simple rules.∫20x4(x5 + 7)dx = ∫20x4u(du/5x4) = ∫4udu = 4∫udu Integrate with respect to u: 106 .½(x)2 + x + c = 2/3(x)3 .∫f(x)dx Examples 1. ∫9dx = 9x + c 2. ∫f’(x)g(x)dx Now. then du/dx = 1 → dx = du ∫3x(x + 6)2 dx = ∫3xu2du = 3∫xu2du. And we stop there because integration by substitution will not work. let’s go back to the example: ∫3x(x + 6)2 dx 107 .4∫udu = 4(1/2)u2 = 2u2 + c 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. 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) . However. another method will work. Usually we consider first the simpler function for f(x) and the more complex for g’(x)..1/4(x + 6)4 + c Note that c1 does not appear in the final solution.e. Let f(x) = 3x. what is to be integrated) into 2 parts as in the formula. g(x) = ∫(x + 6)2dx = 1/3(x + 6)3 + c1 . then: f’(x) = 3. then: f’(x) = 5.Separate the integrand (i. f’(x) and g(x) in the formula (note that 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 . g(x) = ∫ex-9dx = ex-9 108 . . g’(x) = (x + 6)2.1/4(x + 6)4 + c So: ∫3x(x + 6)2 dx = x(x + 6)3 .Substitute the values for f(x).∫[(x + 6)3 + 3c1]dx = x(x + 6)3 + 3xc1 – 1/4(x + 6)4 – 3xc1 + c = x(x + 6)3 . We can always assume that c1 is always = 0..Check the answer by differentiating Another example: ∫5xex-9dx Let f(x) = 5x and g’(x) = ex-9. ∫f’(x)g(x)dx = 5xex-9 .5ex-9 + c Definite integrals The definite integral is a number which can be evaluated using the following: The numerical value of a definite integral of a continuous function f(x) over the interval [a to b]: b ∫f(x)dx a is given by the primitive function F(x) + c. a.[2(1)3 + 5(1)] = 69 – 7 = 62 Properties of definite integrals 109 . minus the same primitive evaluated at the lower limit.∫ex-9 5dx = 5xex-9 . b b ∫f(x)dx = [F(x)] = F(b) – F(a) a a 3 3 2 3 Example: ∫(6x + 5)dx = [2x + 5x] 1 1 = [2(3)3 + 5(3)] . ∫5xex-9dx = 5xex-9 .5ex-9 dx. So.∫5xex-9dx = f(x)g(x) . evaluated at the upper limit b. TC= ∫(1/2)qdq = (1/2)[q2/2] = q2/4 + c We can definitize the constant. ∫f(x)dx = ∫f(x)dx + ∫f(x)dx a a b (a <b < c) A few simple economic applications Suppose we are given the slope of a cost function of a firm. You are essentially given the marginal cost (MC) curve. Find the equation of the total cost curve. TC = 7. ∫f(x)dx = 0 a c b c 3. 110 . Then. which is at any point equal to (1/2)q and the information that the total cost curve passes through the point (2.7). Given that when q =2. → TC = (2)2/4 + c = 7 → c = 6. ∫f(x)dx = . and: TC = 6 + q2/4 (=TFC + TVC) c. we utilize the additional information. if Calculation of capital stock This is done on the basis of a known pattern of change of investment.∫f(x)dx a b a 2.b a 1. On the other hand. of a cash flow (a series of revenues or costs receivable at various times in the future). i. which is the time path of capital stock. 111 . and K(t) = 8t3/2 + 25. is invested at the rate of interest. we know that the 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. where I is the present value of R. we have: K(t) = ∫12t1/2 dt = 12(t3/2/3/2) + c = 8t3/2 + c K(0) = 25 = 8(0) + c → c = 25. 3) in years: 3 3 1/2 3/2 K(3) – K(1) = ∫12t dt = 8[t ] 1 1 Present value of a cash flow Another application is finding the P. if you are asked to find the amount of capital accumulated over a particular time interval. R.V. This refers to a single future value. I. to receive R. say (1. Suppose an amount.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). t years later) This is the continuous case counterpart of: I = R/(1 + i)t . he has to invest: 10 I = ∫100e-itdt = (100/0. Investment will increase productive capacity at the same time it increases total effective demand (aggregate demand).But if a person expects to earn R at the end of the first period and subsequent periods. if i = 0. and the investor wishes to receive $100 for 10 years. will change the objective situation on which the current equilibrium is based. Keynesian analysis is different. the very fact new investment is undertaken.e. then he has to invest: t I = ∫Re-itdt 1 where I is the present value of a series of annuities given by the return expected from an investment outlay for a unit of a particular asset.. i.04)e-0. Before Keynes. The explanation stopped there.4 = $726 1 Economic dynamics We may introduce economic dynamics by considering pre-Keynesian and Keynesian notions of equilibrium. If the achievement of full employment equilibrium requires a certain amount of new investment to be undertaken to bring total effective demand to the level of full capacity utilization.04 .04. investment plays a special role in Keynesian economic dynamics and acts trough 2 different channels: 112 .(100/0. So.04)e-0. the question asked was whether the equilibrium exists and if it is unique and stable. however. K and L are combined in fixed proportion. By differentiating (1): (full employment output over time): dk/dt = ρ(dK/dt) = ρI along with: (from the demand side): dY/dt = 1/s(dI/dt) [= m(dI/dt)] There is no reason why these two effects should necessarily be compatible with the maintenance of full capacity utilization.1. We can explore. since labour is indispensable. First of all. because only then we can consider capital alone. taken to be a multiple of capital. Application: Domar’s growth model The model assumes a rather rigid production function which involves only capital: k = ρK (1) where k is capacity output (full employment output). One implication here is that. Total new investment is an addition to existing productive capacity. total effective demand must be equal to productive capacity to begin with: Y(0) = k(0) 113 . As a result. ρ = k/K. the conditions to be satisfied in order that effective demand and productive capacity may expand in a compatible way over time. Change in effective demand through the multiplier: dy/dt = m(dI/dt) = 1/s(dI/dt) 2. e. or eln(I) = e(ρst + c) → It = eρstec. i... in order to maintain equilibrium in the long-run. new investment must expand over time according to the above exponential function (i. dk/dt = dY/dt must always hold over time. or: dI/I = ρsdt → ∫(1/I)dI = ∫ρsdt → ln(I) = ρst + c. 114 . An economy might satisfy or not satisfy this condition.then: It = Aeρst So: (I(0) = Ae0 = A) It = I0eρst The above says that. expand at a rate g= ρ*s over time). The above discussion of Domar’s growth model and the application of integration will take us to the concept and use of a differential equation in economic dynamics.Secondly.e. ρI = 1/s(dI/dt). In the case where in place of f(x) on the right hand side we have a constant instead. the equation contains a second derivative of the function we are looking for. are all constants not involving x... The differential equation is linear with constant coefficients. dy/dt.+ an-1(dy/dx) + any = f(x) where f(x) is an function of x. If in addition. dny/dxn) = 0 (1) The order of the equation is that of the highest derivative of it.. the unknown function y and its first derivative. the general form of a linear differential equation with constant coefficients and order n is: a0(dny/dxn) + a1(dn-1y/dxn-1) + . 115 . So. The general expressions of an ordinary differential equation (the unknown is a function of only one variable) in a function y of an independent variable x. dy/dx. is called a differential equation of the first order. can be written as: F(y. x.Differential equations Basic concepts and definitions (Functional) equations in which the unknowns are functions and which contain not only the unknown function but also their derivatives. are called differential equations. An equation relating the independent variable x.. if the coefficients of y. then such an equation is of the second order. the equation is called linear with constant coefficients and a constant term.. it is to be found by solving the equation. The form of the function is not given.. . d2y/dx2. . Accordingly we can define deferential equations of higher order.. i. emphasis will be given to linear equations of the 1st and 2nd order. In discussing differential equations. a0(dny/dxn) + a1(dn-1y/dxn-1) + .The first-order case is written as: dy/dx + a1y = f(x) The second order: (d2y/dx2) + a1(dy/dx) + a2y = f(x) The particular case where f(x) is absent altogether is called the homogeneous form.. 116 ..+ an-1(dy/dx) + any = 0 Summarizing: Differential equations No –linear (1 or higher order) st Linear (1 or higher order) st Constant coefficients Variable coefficients Homogenous Non-homogeneous Homogeneous Non-homog.e.. At the end of the year the principal will be: P = X(1 + i) If instead the interest is compounded semi-annually: P = X(1 + i/2)2 Generalizing: P = X(1 + i/n)n 117 .e. it is clear that the equation tells us something about the behaviour of y over time.. and since d2y/dt2 describes how dy/dt is changing over time. since dy/dt describes how y changes over time. use t instead of x): (d2y/dt2) + a1(dy/dt) + a2y = 0 If we take y to be an economic variable and t stands for time. compounded annually. The number e The number e plays a role in our analysis using differential equations (continuous time).Now suppose we have an equation like the following (i. Solving linear differential equations By solution we mean an expression (”formula”) giving y at all points in time such that it satisfies the equation. Say that we invest X$ at a rate of interest i. so. it is about economic dynamics. then. In particular. it will enter the solution of the differential equations. if X=$1 and the rate of interest i = 100% ( i = 1): P = (1 + 1/n)n The limit approached by P as n → ∞ is equal to e (the principal of one dollar compounded continuously for 1 year at 100% interest). money is not compounded continuously.ay → dy/y = -adt. then: ∫dy/y = -a∫dt → ln(y) + c1 = -at + c2 → ln(y) = -at + c.If interest was compounded continuously. Of course. e ≈ 2. then: 118 . the solution can be found as follows: dy/dt = . we would have a principal given by the value approached by P as n → ∞. In particular. but one could think of events which behave very much like a sum on which interest is compounded continuously.718…. Solution of Linear Differential Equations of the 1st order These can be written in the general form: dy/dt + ay = f(t) if f(t) = 0 (homogeneous form): dy/dt + ay = 0 Since the above equation is separable. plus the particular solution. and: λceλt + aceλt = 0 → ceλt (λ + a) = 0 → λ = -a This is a general way of solving the homogeneous equation and there is an arbitrary constant. f(t). To remove the unknown constant requires additional information – i. we need to take into account the non-homogeneous part. From the general solution: y(t) = ce-at.e. The general solution will be the solution to the homogeneous form (call it complimentary solution. 119 . so: y(t) = y(0)e-at. For t=0: y(0) = ce0 = c = y(0). c. Besides the solution to the homogenous equation there is also a particular solution. yp.. yc). appearing. The particular solution will be any particular solution to the complete equation. an initial condition. we can find the solution in a slightly different (and more general) way: Given: dy/dt + ay = 0 Assume that the solution: y(t) = ceλt. then: dy/dt = λceλt.elny = e(-at + c) → y(t) = e-atec = Ae-at Verify by substituting into the differential equation: dy/dt = -a Ae-at and: -a Ae-at + a(Ae-at) = 0 Alternatively. try yp = k. try yp = ekt and so on. Then: dy/dt = 0 and we have: 0 – 3k = -3 → k = 1.5t (check solution) To find the particular solution. y(0) = 1 Using the standard solution. Using the general way of solving the homogeneous form: yc =ce3t [complimentary solution] To find the particular solution: let yp = k. i. y(0)= c + 1 = 2 → c =1 and yt = e3t + 1 [definate solution] Verify: dy/dt – 3y = 3e3t – 3(e3t + 1) = -3 Example 2: y – 2(dy/dt) + 2t = 6.e. Then: yc = ce-(-1/2)t = ce0.One way to find the particular solution is by setting the solution to be a general expression of the same form as f(t). dy/dt – 3y = -3. set yp = ko + k1t. Example 1: Solve. y(0) = 2. then: 120 .. The complete solution is: yt = ce3t + 1 [general solution] Using the initial condition. we first need to re-write the equation: 2(dy/dt) – y = 2t – 6. if f(t) is a constant k (such as f(t) = 3). or dy/dt – ½y = t – 3. If f(t) = e3t. k1) = -3 → (ko /2 .k1) = 3 Solving. and: yp = 2 – 2t. and the initial equation is: y – 2(dy/dt) + 2t = 6: -e0.5e0. y(0) = 1: y(0)= c + 2 = 1 → c = -1 Then: yt = -e0.5t + -e0.(ko /2 .5t.k1) = t – 3 Then: -k1/2 = 1 and .5t + 4 = 6 121 .5t + 2 – 2t Using the initial condition.5e0. we get k1 = -2 and ko = 2. so: yt = ce0.5t + 2 – 2t – 2(-2 – 0.5t + 2 – 2t Check by substituting into y – 2(dy/dt) + 2t = 6.dy/dt = k1 and: k1 – ½( ko + k1t) = t – 3 re-write as: k1 – 1/2 k1t – 1/2 ko = t – 3 → (-k1/2)t – (ko /2 . given that: dy/dt = -2 – 0.5t) + 2t = 2 – 2t + 2t + e0. However. s. our solution will converge to infinity as t → ∞ if g > 0. of g < 0 the solution converges to 0 (!!). of income at time t: St = sYt Investment is proportional to the rate of change of income over time: (1) It = g(dy/dt) Given the equilibrium condition. s/g). On the other hand.Economic Applications Consider the following case: A community’s savings (= investment at equilibrium) at any point in time t. we derive a non-homogeneous differential equation: 122 . or: dy/dt – (s/g)Yt = 0 (1st order homogeneous equation. growth of output accelerates the expansion of productive capacity and investment. This is an unreasonable implication. when we introduce a slight modification to the above model. Solution: Yt = A e(s/g)t = Y0e(s/g)t [A = Y0] Implications: Given that s is positive. output will finally decline to 0. which suggests that if output does not grow enough. is a constant proportion. with constant coefficient. that is. since if g < 0: lim{ e(s/g)t } = 0. with investment falling and unused capacity appearing. I = S: sYt = g(dy/dt). and: It = I1t + I0t From the definition of National Income: Yt = Ct + It . a + s = 1.I0t/g The solution to the homogeneous equation is: Yc = Y0e(s/g)t 123 . The simple multiplier is: m = 1/s. The accelerator expresses planned investment as: I1t = g(dY/dt) Putting everything together: Yt = aYt + g(dY/dt) + I0t (= consumption + induced investment + autonomous investment). s = dS/dY (MPS). or s = 1 – a. → g(dY/dt) = [(1-a)]Yt .Harrod’s growth model I1t = g(dY/dt) (induced investment) I0t: autonomous investment.I0t (dY/dt) = [(1-a)/g]Yt . Ct = aYt [a = dC/dY (MPC)] St = sYt .I0t/g dY/dt – (s/g) Yt = . income would rise exponentially. equilibrium income will finally reach I0t/s. So. for convergence we need: s/g < 0 so that lim{ e(s/g)t} = 0 as t →∞. which may either converge or diverge from the equilibrium value of I 0t/s. If s/g < 0. income grows at a steady exponential rate s/g. 124 . then: Yt = Y0e(s/g)t + I0t/s Since I0t/s is a constant.I0t/g → Yp = I0t/s. but if (as one would expect) s/g > 0.To find the particular solution: Yp = k and: -(s/g)Y = . depending on the sign of s/g. In particular. price tends to rise. if we ever reach it. our objective is to see what happens when the system is not in equilibrium. The first thing we have to do is adopt an assumption about the behaviour of the variables in the system (in this case.Dynamic models of market price – Stability of equilibrium In a perfectly competitive market. if there is excess demand. Stability conditions will depend on this assumption. Starting from the simple case in which D and S of a commodity are assumed to depend only on price (other things unchanged). then the question concerns the static stability of the system. the question is whether the system will tend to equilibrium or not. the usual behavioural assumption is the Walrasian assumption: According to it. we cannot exclude the case where equilibrium is overtaken. So the question is that of dynamic stability. equilibrium is determined at the intersection of D and S. Equilibrium is statically stable if a P increase caused by positive excess demand will decrease excess demand: 125 . if there is excess supply. price tends to fall. without being concerned about the time path of price (existence and stability of equilibrium). But even if economic forces push the system towards equilibrium. When the question is simply whether the economic forces will finally bring the system to equilibrium. This would give rise to oscillations before approaching equilibrium. P). In the case of dynamic models of market price. b1)P + c(a – a1). S = a1 + b1P. from which we can solve for Pt.Differentiate excess demand. In the case of linear demand and supply: D = a + bP. which is then solved to derive the time path of the variable: dP/dt = P’ = c[D(P) – S(P)]. so: Pt = Aec(b – b1)t + Pe 126 .c(b . E(P)= D(P)-S(P).b1)P + c(a – a1) Equilibrium price is derived by setting P’ = 0: Pe = . c].(a – a1)/(b – b1) To find the time path of price: dP/dt = c(b . with an adjustment positive coefficient.(a – a1)/(b – b1) = Pe. with respect to P: dE(P)/dP = d[D(P) – S(P)]/dP < 0 or. [This is the behavioural assumption which says that price responds to excess demand or supply.b1)P = c(a – a1). The solution of the homogeneous equation (complimentary solution) is: Pc = Aec(b – b1)t (complimentary solution) For the particular solution we let: Pp = k → -c(b – b1)k = c(a – a1) → k = . dD/dP – dS/dP < 0 (slope of supply > slope of demand) In analysing dynamic stability we express the behavioural assumption in the form of a functional equation. or P’ . dP/dt = P’ = c[a + bP – a1 – b1P] = c(b . . static stability..Looking at the exponent in the solution. note that this condition is the same as: dD/dP – dS/dP < 0.e. When they are different. we need: c(b – b1) to be negative for stability (since c >0). we are more interested in the dynamic stability condition. Qe = 20 Using the Walrasian (static) stability assumption: E(P) = 80 – 4P – (-10 + 2P) = 90 – 6P dE/dP = -6 < 0. P0 = 18 Find the equilibrium point and examine static stability. so in this case the static and dynamic stability conditions are the same. So. S = -10 + 2P. At equilibrium: 80 – 4P = -10 + 2P → Pe = 15.e. This is not always the case though. This is the dynamic stability condition. the slope of the supply must exceed that of the demand. Example: Static stability D = 80 – 4P. the stability condition is: b – b1 < 0 or b < b1 i. however. 127 . i. the term 3e-6ct → 0 and equilibrium is dynamically stable. 128 . 6ck = 90c → k = 15) Given that P0 = 18 → A = P0 – Pe = 18 – 15 = 3 Pt = 3e-6ct + 15 Given c> 0.Example: Dynamic stability dP/dt = P’ = c[80 – 4P – (-10 + 2P)] = c(90 – 6P) = -6cP + 90c or: dP/dt + 6cP = 90c which has the solution: Pt = Ae-6ct + 15 (Pp = k. b1 > 0 b1 < b. stable 129 . b1 < 0 |b1| < |b|.Some abnormal cases 1. b> 0. b1 > 0 b1 > b. unstable 3. b> 0. stable (note that P is on vertical axis) 2. b < 0. b < 0. b1 < 0 |b1| > |b|. unstable 4. Consider the following demand and supply for a product: Dt = St Dt = D(Pt. Using an alternative behavioural assumption. dP/dt) St = S(Pt) Assuming linear functions: QD = a1 + a2P + a3(dP/dt) QS = b1 + b2P → Q = a1 + a2P + a3(dP/dt) = b1 + b2P → a1 + a2P + a3(dP/dt) = b1 + b2P → a3(dP/dt) + (a2 – b2)P = b1 – a1 → dP/dt + [(a2 – b2)/a3]P = (b1 – a1)/a3. or Pt = Ae[(b2 – a2)/a3]t + P* Pt = (P0 – P*)e[(b2 – a2)/a3]t + P* (since P0 = A + P* → A= P0 – P*) (1) For price to converge to P* (stability) we need: [(b2 – a2)/a3] < 0.A dynamic model of market price Stability in a single market Assume small shocks in market price. the market clears at any point in time (instead of: price adjusts according to excess demand or supply). i. and Pt → P* 130 .. then as t → ∞: lim (e[(b2 – a2)/a3]t) = 0. and the time path of P: Pt = Ae-[(a2 – b2)/a3]t + P*.e. (+ .(+))/ . stability if b2 > a2 (.= + ve (unstable) (.= -ve (stable) (+ . stable if |a2| > |b2| 131 ..(-))/ .= (±).We can highlight 4 cases. 4. 3.(-))/ . assuming a3<0: 1.= (±). 2..(+))/ . y’. It can be shown that for any two initial conditions: y0 = y(x0). y’) (2) is known as the equation resolved with respect to the 2 nd derivative. y’0 = y’(x0) (3) there exists only one solution. and y’ = y’(x). k) if k = 0. Homogeneous linear equations with constant coefficients y’’ + by’ + cy = k (2nd order with a constant term. y’’ + by’ + cy = 0 (homogeneous) 132 . y’’=y’’(x) are its 1st and 2nd derivatives. y. y’’) = 0 (1) where y=y(x) is the unknown function we are looking for. The function φ(x) will be called a solution of (1) if it has derivatives φ’(x). y = y(x) of equation (2) which satisfies the initial conditions (3). φ(x). φ’’(x)) = 0 A differential equation of the form: y’’ = f(x. φ’’(x) such that for any x the following equality holds: F(x. y.Second-order differential equations with constant coefficients In the general case a 2nd order differential equation can be written in the following general form: F(x. This constitutes the basic contents of Cauchy’s theorem. φ’(x). Substitute this into (1) and since: y’ = λeλx.4eλx = 0 → (λ2 – 4) eλx therefore. Therefore the solution is: 133 . the function: y = c1ex + c2e-x is also a solution to (1). The roots that satisfy the equation are λ1=2 and λ2=-2.Example 1 y’’ – y = 0 It is easy to verify that the function: y = ex is a solution. y’’ = λ2eλx. then: λ2eλx . y’’ – y = 0 From Cauchy’s theorem there is no other solution to the homogeneous equation. where λ is an unknown number. c2. and.c2e-x y’’ = c1ex + c2e-x = y so. a function of the form eλx satisfied equation (1) if an only if: (λ2 – 4) = 0. Example 2 y’’ – 4y = 0 (1) We can find the solution in the form: y = eλx. since: y’ = c1ex . We can show that for any two constants c1. Given a homogeneous linear equation with constant coefficients: y’’ + by’ + cy = 0 (1) substitute eλx as well as its first and second derivatives into (1): y’’ + by’ + cy = 0 → λ2 eλx + bλeλx + c eλx = 0 → eλx(λ2 + bλ + c) = 0 the function eλx will be a solution of (1) if λ satisfies: (λ2 + bλ + c) = 0 (2) Equation (2) is called the characteristic equation of (1). use time as the independent variable (t instead of x). 134 . which is the general solution. Now.y = c1e2x + c2e-2x Characteristic equation The above examples suggest an idea to use a solution of the form eλx in the general case as well. so y’ is dy/dt. etc. then so is the function φ = c1φ1 + c2φ2. Then: y = c1eλ1*x + c2eλ2*x Proposition: If φ1 and φ2 are both solutions of (1). y’’ = d2y/dt2. y(0) = 4. and λ = (-b ± √Δ)/2 = (3 ± 1)/2 = λ12 = 2. The characteristic equation is: λ2 . 1.3λ + 2 = 0. y0 = 4 or: y’’ – 3y’ + 2y = 0. dy0/dt = 6. λ2 = -3. Example 2 (with initial conditions): d2y/dy2 – 3(dy/dt) + 2y = 0. to satisfy the initial conditions: y(0) = c1e0 + c2e0 = c1 + c2 = 4 y’(0) = 2c1e2t + c2et = 2c1e0 + c2e0 = 2c1 + c2 = 6 135 . and: [discriminant: Δ = b2 – 4c = 9 – 8 = 1. and: λ1 = -1. then: y = c1e-x + c2e-3x Consider now an example where in addition to the equation we are given initial conditions. y’(0) = 6.Case of two distinct real roots Example 1 (no initial conditions): y’’ + 4y’ + 3y = 0 The characteristic equation is: λ2 + 4λ + 3 = 0.] The general solution is: yt = c1e2t + c2et Then. b = -2λ. Solution: eλt is a solution to equation (1). y’’ = λeλt + λeλt + λ2teλt = 2λeλt + λ2teλt (1) 136 . c = λ2 [since b2 – 4c = 0. We can show that in this case (case of multiple /repeated) root. c1 = 2. also. λ = [-b ± √0]/2 → 2λ = -b. and again since b2 – 4c = 0→ c = b2/4 → b2 = 4c. teλt is also a solution. c2 = 2 and the solution of the homogeneous equation (definite solution) is: yt = 2e2t + 2et (definite solution) Case of multiple roots y’’ + by’ + cy = 0 Characteristic equation: λ2 + bλ + c root. and b2=4λ2]. y’ = eλt + λteλt. We have: y = teλt .solving. 2λ2 + λ2) = 0 Therefore.2λ) + teλt (λ2 . the solution is: yt = c1eλt + c2 teλt Example: y’’ – 2y’ + y = 0 λ12 = -1 (repeated root) solution: yt = c1e-t + c2te-t Case of complex roots y’’ + by’ + cy = 0 λ2 + bλ + c (1) (2) When: b2 – 4c < 0. If we find the conjugate complex roots: λ1.so.2 = h ± iv (where h is the real part of the root). then the general solution will be: yt = c1e(h + iv)t + c2e(h . y’’ + by’ + cy = 2λeλt + λ2teλt + beλt + btλeλt + cteλt = eλt (2λ + b) + teλt (λ2 + bλ + c) = eλt (2λ .iv)t 137 . the roots are complex. -1 yp = k → 0 – 0 – 3k = 10 → k = -10/3 then. and k = d/c = yp Example (real roots): y’’ – 2y’ – 3y = 10. then: y’ = 0. y(0) = 4.which can be written as: yt = eht(c1eivt + c2e-ivt) [we’ll come back to this later] Particular solution (for any case of roots) Suppose the right hand side is a constant. c2 = 4 The definite solution is: yt = (10/3)e3t + 4e-t – 10/3 138 . y’’ + by’ + cy = d (1) Consider a particular solution of the form: yp = k. y’’ = 0. λ12 = 3. general solution: yt = c1e3t + c2e-t – 10/3 y(0) = c1e0 + c2e0 – 10/3 = c1 + c2 – 10/3 = 4 y’(0) = 3c1 – c2 = 6. and c1 = 10/3. y’(0) = 6. then eλt = 1 and the solution remains constant over time. and since the denominator will grow at ever-increasing rates 1/(et)η will approach 0. Since e > 1. then (et)λ = 1/(et)η. or c(et)λ.λ is > 0. The solution is convergent. If λ = 0. Since c can be positive or negative. the solution is unstable (explosive solution). First-order differential equations The solution is of the form: ceλt. If λ < 0 so that η = . if λ > 0.Stability of equilibrium 1. Thus. we identify 6 possibilities for the time path of y (complimentary solution): 139 . et will grow at an everincreasing rate if λ>0. The amplitude of fluctuations will depend largely on the magnitude of e ht where h is the real part of the complex root. eventually the solution will approximate: yt = ci(eλi*t). When we have a pair of complex roots there will be a term in the solution of the form: yt = eht(c1eivt + c2e-ivt) The expression in parenthesis will repeat itself (trigonometric fluctuations).e. stable solution (why?) Stability in the case of complex roots. i. immediately). c1 and c2. both positive or negative.2. what will happen will depend on the relative values of the constants. Stability requires h < 0. 140 . But “eventually” may be a long time and we may be more interested what happens before that (i.. When this is not the case. since the other term becomes relatively negligible. We can answer this if the roots are of the same sort.e. The time path is unstable when there is at least one positive root. Second-order differential equations We have 2 terms in the solution: yt = c1eλ1*t + c2eλ2*t + yp If the largest root is λi. If multiple roots (still case of real roots): yt = c1eλt + c2 teλt If λ < 0. the real part of the complex root.b2 Hence. If Qd and Qs depend on d2P/dt2 in addition to dP/dt then: Qd = D[P(t). if the roots are complex. δ > 0) Equating D to S (that is assuming that the market always clears): 141 . by inspecting the equation itself. On the other hand. these will be damped (convergent) or explosive.The distinction between +ve and –ve real part of the root (h) is of particular importance because it determines whether. d2P/dt2] Qs = S[P(t). The derivative dP/dt relates to whether price is rising or falling over time. whatever fluctuations may occur. β > 0) (γ. the second derivative d2P/dt2 indicates whether price increases at an increasing rate (or decreasing at a decreasing rate). λ2 + bλ + c = 0 The roots are: -b/2 ± √b2 – 4c = -b/2 ± i√4c . dP/dt. There is an easy way to determine whether. the real part is + ve or – ve. h = –b/2 which is – ve if b is + ve and vice versa. dP/dt. d2P/dt2] Using linear D and S equations and the alternative notation: Qd = α – βP + mP’ + nP’’ Qs = -γ + δP + uP’ + wP’’ (α. Economic applications Dynamic model of market price Consider a market model where the market behaviour of individuals is affected by expectations about the future price. with h = ½ and v = 3/2. The particular solution is found by setting yp = k.P’’ + bP’ + cP = d Where b. P’(0) = 4 Equating Qd and Qs and simplifying: P’’ – P’ -2. d are related to the original coefficients. c. then: k = d/c The solution in the case of 2 roots is: Pt = c1eλ1t +c2eλ2t + d/c Example (from textbook) Qd = 3P’’ + P’ – P + 9 Qs = 5P’’ – P’ + 4P – 1 P(0) = 4.5P = -5 → P =2) The characteristic equations is: λ2 – λ – 5/2 = 0 and has complex roots: ½ ± (3/2)i. The general solution will be: Pt = e1/2t( ) 142 .5P = -5 Particular solution: Pp = 2 ( -2. so –b/2 is positive). 143 .The path is non-convergent and the fluctuations in price are explosive (b = -1. this implies that monetary policy cannot affect unemployment. The "short-run Phillips curve" is also called the "expectations-augmented Phillips curve". In the long run unemployment will converge to the “natural rate” and will be no trade off. policies can result in short run declines in unemployment by increasing permanent inflation.Philips (1958) published: The relation between unemployment and the rate of change of money wages in the UK.background . . In the long run.However. .Friedman argued this relationship was only a short run phenomenon (in the long run workers would take inflation into account resulting in employment contracts with higher wage rates according to anticipated inflation).What is today still influential is a modified form of the Phillips Curve that takes inflationary expectations into account and distinguishes between a short-run and a long-run PC. but not permanently. Found a statistically negative relationship between (wage) inflation and unemployment (see graph). .In its modern version. Most of this criticism was initiated during the 1970s when the world observed high rates of both unemployment and inflation. since 1974. . During those years economists believed that there was a permanent/stable relationship which could be exploited for policy purposes. also called the "NAIRU" (long-run PC is vertical).Similar patterns found in other countries. and in 1960 Samuelson and Solow formalized this relationship between unemployment and inflation. since it shifts up when inflationary expectations rise.An example from Macroeconomics The dynamic Philips curve . which adjusts back to its "natural rate". Inflation can lower unemployment temporarily. The dynamic Philips curve – model (1) Original Phillips Relation 144 . seven Nobel prizes have been given for work critical of the PC. . 1861-1957. Now we add one more equation for unemployment. β > 0) (2) Introducing expectations w = f(U) + gπ (0 < g < 1) (π is the expected inflation rate) p = α – T – βU + gπ (1) Assuming adaptive expectations. i. T is labour productivity.e. The time path of π: Equations (1) – (3): 145 . induced by technology). U: rate of Using a linear function: w = α – βU. dπ/dt > 0. Assume (for simplicity) that the only government policy that matters is monetary policy..w = f(U) (f’(U) < 0) unemployment] [w: rate of change of nominal wage. and denoting the rate of growth of nominal money supply by m: dU/dt= -k(m-p) (where k>0 and m-p is the growth of real M-supply) [note that real M-supply is M/P and the growth rate of M/P is m-p] (3) According to (3). they revise their expectations up. then the rate of change of expected inflation over time is: dπ/dt = j(p – π) (0 < j < 1) (2) If individuals under-predict the rate of inflation. the change in the unemployment rate over time is determined by the rate of growth or real money supply. So: p = α – T – βU (α. Also: p = w – T (p is rate of change in price level. From (2): p = (1/j)(dπ/dt) + π Substituting into (5)’: d2π/dt2 + [βk + j(1-g)]dπ/dt + (jβk)π = jβkm [2nd order linear differential eqn with constant coefficients and constant term] The particular solution is: πp = (jβkm)/ (jβk) = m Which means that the inter-temporal equilibrium value (equilibrium over time) of expected inflation depends only on (i. converges to) the rate of change of money supply.e..p = α – T – βU + gπ dπ/dt = j(p – π) dU/dt= -k(m-p) (1) (2) (3) constitute a model with 3 variables. We can choose to collapse the model into (any) one equation. i. for example if we choose π. differentiate (4) with respect to time: d2π/dt2 = -jβ(dU/dt) – j(1-g) (dπ/dt) Put (3) into (5) to replace dU/dt with: -k(m-p): d2π/dt2 = jβkm – jβkp – j(1-g) (dπ/dt) We still need to eliminate p. p and U. in π. or U.e. putting (1) into (2): dπ/dt = j(α – T – βU + gπ – π) dπ/dt = j(α – T – βU) – j(1-g)π (4) To create a dU/dt term.. p. (5)’ (5) 146 . π. all expected inflation is transferred to wages and prices). 147 . the particular solution (where the unemployment rate converges in the long-run) will be a number independent of the policy variable m (rate of growth of money supply) and regardless of the equilibrium rate of inflation.5. and depending on the type of roots we’ll know the time path (convergent or divergent. Similarly. On the other hand.For the complimentary solution (which will tell us whether or not we’ll reach this new equilibrium and if yes in what manner) we have two roots. as long as the coefficient g=1 (i. the rate of change of the price level (inflation) will equal to the rate of growth of nominal money supply. Tutorial question 16.3 deals with what happens if g < 1.e. if we derive the time path of the unemployment rate. highlighting the result that in the long-run Philips curve is vertical. that is in the long-run (give it enough time). we can derive the time path of p and find that the particular solution (inter-temporal equilibrium price) is again: m. monotonic or oscillatory). Δ2yt. t: Δyt = f(t + 1) – f(t) = yt+1 – yt Δyt+1 = f(t + 2) – f(t+1) = yt+2 – yt+1 ……………………………… Likewise we can compute second differences: Δ2yt = Δyt+1 . Δyt. its first difference is defined as the difference between the value of the function when the argument takes the value t + h and the value of the function at time t. Since the differences of any order can be expressed. i. Δy = f(t + 1) – f(t) Consider unit increments of the independent variable..Difference equations (discrete time) General principles Consider the function y = f(t). as we have seen above.e.Δyt = (yt+2 – yt+1) – (yt+1 – yt) = yt+2 – 2yt+1 + yt Δ2yt+1 = Δyt+2 . etc. The order of the difference equation is that of the highest difference appearing. in terms of various values of the function. Then. of an unknown function of time. a difference equation may also be 148 .Δyt+1 = yt+3 – 2yt+2 + yt+1 We define an ordinary difference equation as a functional equation involving one or more of the differences. Ct-1 = 100 + 0.7Yt-1 implies.defined as a functional equation involving two or more of the values yt. or: ayt+1 + (b-a)yt = 0 It makes no difference whether the equally spaced values of t are computed forwards or backwards. from (3): y1 = -by0 = -bA y2 = -by1 = -b2A …………………. t = 0. of an unknown function of time. the equation: ayt+1 + (b-a)yt = 0 is identical to: ayt + (b-a)yt-1 = 0 Example: Consumption function: Ct = 100 + 0. Then.7Yt-2 First-order difference equations The general form is: c1yt + c0yt-1 = g(t) The homogeneous form is: c1yt + c0yt-1 = 0 or: yt + byt-1 = 0 where b = c0/c1. the function y takes on an arbitrary value A. i. The solution appears to be: yt = A(-b)t Check if: yt + byt-1 = 0 149 .e. i.e.. the difference equation: aΔy t + byt = 0 can be transformed into: a (yt+1 –yt) + byt = 0. yt+1 etc.. so long as the structure of the lags remains the same. (1) (2) (3) Suppose that in the initial period. This need derives from the fact that the solution yt = A(-b)t gives only the form of the function. when g(t)= 0 we have a homogeneous equation. then Ay1(t). Linear difference equations with constant coefficients The general nth order form is: cnyt+n + cn-1yt+n-1 + … + c1yy+1 + c0yt = g(t) where the ci are constants and g(t) is a known function. the following theorem will be used: 1.A(-b)t + b A(-b)t-1 = A(-b)t – (-) b A(-b)t-1 = A(-b)t – (-) b A(-b)t(-b)-1 = A(-b)t – A(-b)t = 0 Determining the arbitrary constant (A). substituting into the equation: cnAy1(t+n) + cn-1 Ay1(t+n-1) + … + c1 Ay1(y+1) + c0y1(t) = 0. Since y 1(t) is a solution. therefore: A[ brackets vanishes. When looking for a solution. where A is an arbitrary constant. Both cn and c0 must be ≠ 0 if the equation is of order n. We need to use an initial condition. is also a solution. [Proof: If y1(t) satisfies the equation. If y1(t) is a solution of a homogeneous equation.] ] = 0. the expression in 150 . with c1 = -c0 = c cμt . so: yp = (a/c)t 151 (1) . yp = a/(c1 + c0). for any 2 constants A1 and A2. but with undetermined coefficients. y2(t) are 2 distinct solutions of the homogeneous equation (with n > 1). Particular solution Case 1: g(t) = constant c1yt + c0yt-1 = a try yt = yt-1 = μ. then A1y1(t) + A2y2(t) is also a solution. So. The approach is to find the particular solution by trying a function having the same form as g(t). then: (c1 + c0)μ = a → μ = a/(c1 + c0).2. as in the case of differential equations. [Proof similar] 3.cμ(t-1) = a → μt – μt + μ = a/c → μ = a/c. If yp(t) is any particular solution of the non-homogeneous equation. the general solution is obtained by adding the particular solution to the general solution of the homogeneous equation. The particular solution will depend on the form of g(t). with (c1 + c0) ≠ 0 If (c1 + c0) = 0. If y1(t). try yt = yt-1 = μt Substituting into (1): c1μt + c0μ(t-1) = a. (c1 + c0)α – c0β + (c1 + c0)βt = α0 + α1t → (c1 + c0)α – c0β = α0 (c1 + c0)β = α1t → solve for α. Case 2: g(t) is a polynomial of degree n. β Example: yt-1 . then: from: c1yt + c0yt-1 = α0 + α1t. i.this suggests the following generalization: If the function you try as the particular solution does not work. c1(α + βt) + c0[α + β(t-1)] = α0 + α1t therefore. y(0) = A – ¼ = 7/4 → A = 2. g(t) = α0 + α1t.. try the same multiplied by t.5yt = 1 (y0 = 7/4) The complimentary solution (solution to the homogeneous equation) is: yc = A[(5)t] = A(5)t yt = yt-1 = k k – 5k = 1 → k = -1/4 Solution: yt = A(5)t – ¼. and: yt = 2(5)t – ¼ Applications The Cobweb model 152 .e. D has a negative slope (b < 0).This is a dynamic demand-supply model. That is. on the other hand. bPt . These oscillations are explosive if |b1| > |b|.. while b1 > 0.. it is the same price obtained from the solution of the static model: D = a + bP S = a1 + b1P D=S Usually. etc).e.e. agricultural products. At the end of each period. 153 . if the slope of the supply > slope of the demand. while demand depends only on current price. Producers think (assumption) that this price will hold also in the next period.b1Pt-1 = a1 – a Pt – (b1/b) Pt-1 = (a1 – a)/b And the solution (time path of price) is: Pt = A(b1/b)t + Pe. and price will exhibit an oscillatory movement around the equilibrium price. Then. When market clears: Dt = St or. Dt = a + bPt St = a1 + b1Pt-1 This model could apply to goods whose production is not instantaneous or continuous. raising pigs. b1/b < 0. they will converge to the equilibrium price if |b1| < |b|. but requires a period of time (i. the output started at the beginning of the period reaches the market. P0 = A(b1/b)0 + Pe = A + Pe) Pe is the static equilibrium price. i. Assume that supply reacts to price with a lag of one period. or Pt = (P0 – Pe)(b1/b)t + Pe (since. so the quantity of their new production is based on current price. 154 . 155 . the movement is monotonically stable. In this case. because the sign of b1/b is positive and |b1/b| < 1. 156 .Two abnormal cases Positively sloped demand curve and |b1| < |b|. 157 . This case leads to monotonic divergence (unstable solution).Positively sloped demand curve and |b1| > |b|. Pt-1) (0 < c < 1) [if c = 0. they expect that the former will change towards the latter..Pt-1)] = a1 + b1Pt-1 + b1cPe – b1c Pt-1 bPt . We can consider the case where producers have a price in mind (call it “normal” price) that they think is going to prevail sooner or later. which is a rather naive assumption. think will prevail when the output reaches the market. i. If current price is different from the “normal” price.e. cobweb case] But how we determine what the “normal” price is? One way is to assume that PN is a constant and equals Pe.The cobweb model can be considered as a special case of a more general model: Dt = a + bPt St = a1 + b1Pet Dt = St where Pe is an expected price.(b1 + b1c) Pt-1 = a1 – a + c Pe → Pt . This would be realistic if the system was in equilibrium for a long time before the shock. Pet = Pt-1 + c(PN . the price that producers. Substitute the equation for Pet into the model: a + bPt = a1 + b1[Pt-1 + c(PN ..[(b1 + b1c)/b] Pt-1 = a1 – a + c Pe And the homogeneous equation is: 158 . In the Cobweb model the assumption is that Pe = Pt-1. at the time production begins.e. i. the static equilibrium price. Pt . A price movement which was convergent before converges faster now. A divergent movement may become convergent if the parameter c is sufficiently close to 1. 159 . So. 2. the introduction of expectations based on the above assumption makes the model more stable. A constant oscillation before becomes a convergent oscillation. 3.[b1(1 – c)/b] Pt-1 = 0 Solution: Pt = A[b1(1 – c)/b]t + Pe (Pe is the particular solution) The stability condition is: | b1(1 – c)| < |b| If we compare the new model with the original Cobweb model we obtain the following observations: 1. 160 . To make the model dynamic we must introduce lags: Ct = a + bYt-1 It = I(0) + ΔI Yt = Ct + It Then: Yt .Dynamic multipliers The simple Keynesian static multiplier in macroeconomics is: m = ΔY/∆I = [1/(1-b)] (where I represents any autonomous expenditure) This reveals nothing about the transition from the old to the new equilibrium. hence no fluctuations). the term A(b)t → 0 and income moves monotonically towards its equilibrium value (a + I(0) + ΔI)/(1 b).bYt-1 = a + I(0) + ΔI Solution (time path): Yt = A(b)t + (a + I(0) + ΔI)/(1-b). or [Note that: Yp = k → k – bk = a + I(0) + ΔI →k= (a + I(0) + ΔI)/(1-b)] Since |b| < 1 (and positive . 161 . 6Yt-1 = 50 Solution: Yt = A(0. Is equilibrium stable? The total increment in income is given from the static multiplier: 2. C(0) = 60.6Yt-1 + 50 Yt .5 → ΔY = 25.6Yt-1 and investment is autonomous. The new equilibrium is 125.Example: In a certain period income is in equilibrium at level Y(0) = 100. Yt tends to 125 and the equilibrium is stable.(b + h) Yt-1 = a + I(0) + ΔI 162 1/(1-0. Compute the new equilibrium income.6)t → monotonically to 0.6)t + 125 A = Y(0) – 125 = -25. I(0) = 40.6) = .6)t + 125 Since (0. so Yt = -25(0. Assume investment increases from 40 to 50. Concerning stability: Yt = Ct + It Yt = 0.0. The consumption function is: Ct = 0. Now suppose that investment is not entirely autonomous: It = I(0) + hYt-1 Then: Ct = a + bYt-1 It = I(0) + hYt-1 + ΔI and: Yt . . This is called the “warranted” rate of growth. To confirm this.The time path of income is: Yt = A(b + h)t + (a + I(0) + ΔI)/(1-b-h) Stability requires: b + h < 1 or h < 1 – b. Then: St = sYt-1 = s A[(k + s)/k]t-1 163 . This means that. if for any reason income deviates from the equilibrium path (given by the solution). suppose that in period t income is not A[(k + s)/k]t but say. St = sYt-1 It = k(yt – yt-1) In equilibrium: sYt-1 = k(yt – yt-1) kyt – (k + s) yt-1 = 0 yt . It is a rate such that when income grows according to it. dynamic equilibrium). Harrod’s growth model Assume. there is a continuous equality between savings and investment (i. The “warranted” rate of growth has the peculiarity of being unstable. with B > 0. A[(k + s)/k]t + B. it will go on deviating further and further away from the path.e.[(k + s)/k] yt-1 = 0 Solution: yt = A[(k + s)/k]t = A[1 + s/k]t (1) [in continuous time: dI/dt = k(dY/dt)] Equation (1) tells us that income increases over time at a constant rate of growth s/k. It = k(yt – yt-1) = k[A((k + s)/k)t + B .A((k + s)/k)t-1]. So. It > St which will give an additional impact to income. 164 .And. As t →∞. a2 = c0/c2) The general solution will involve a function of the form: λt where λ is determined by means of the coefficients of the model. Furthermore.2.Second-order difference equations General form: c2yt + c1yt-1 + c0yt-2 = g(t) (c2. c0 ≠ 0) Consider the homogeneous equation written as: yt + a1yt-1 + a2yt-2 = 0 (a1 = c1/c2. We can substitute y t = λt to obtain: λt + a1λt-1 + a2λt-2 = 0. or λt-2 (λ2 + a1λ + a2) = 0. The movement will be convergent if and only if both roots are in absolute value < 1. when one or both roots are negative there will be stepped fluctuations. where. Case of 2 distinct real roots Solution: yt = A1(λ1)t + A2(λ2)t The kind of movement of y over time depends on the sign as well as the values of λ1 and λ2. λ2 + a1λ + a2 is the characteristic equation with roots λ1. 165 . the movement of y over time will be dominated by the root which is numerically larger (dominant root). The general solution is: yt = A1(λ)t + A2t(λ)t = (A1 + A2t)λt when |λ| < 1 the solution is convergent. 166 . since the conversing effect due to λt dominates the diverging effect of the multiplicative t. λ = (-a1 ± 0)/2] [also.Case of multiple (repeated) roots yt + a1yt-1 + a2yt-2 = 0 λ1 = λ2 = λ = -(1/2)a1 [a12 – 4a2 = 0. we can show that tλt is also a solution. As shown in the case of differential equations. since a12 – 4a2 = 0 → a2 = (¼)a12] Solution involves λt.t(1/2)a12 + (1/2)a12 + a2t – 2a2] = [(-1/2)a1)t-2 [t(-1/4)a12 + (1/2)a12 + a2t – 2a2] = 0 = [(-1/2)a1)t-2 [t(-a2) + (1/2)a12 + a2t – 2(1/4a12)] = 0 Which proves that tλt is also a solution. Substitute into yt + a1yt-1 + a2yt-2 : tλt + a1(t-1) tλt-1 + a2(t-2)λt-2 λt-2 [tλ2 +a1(t-1)λ + a2(t-2)] from which: [(-1/2)a1)t-2 [t(1/4)a12 . i. the solution can be written in a more suitable form: yt = A1[r(cosw + isinw)]t + A2[r(cosw . h ± iv. Now. r(sinw) = v. The solution is: yt = A1(h + iv)t + A2(h – iv)t Mathematical Note Any complex number h ± iv can be written in its equivalent trigonometric form: r(cosw ± isinw) by transformation from the Cartesian to Polar coordinates. The expression in the brackets produces trigonometric oscillations (stepped trigonometric oscillations). Whether they are explosive or damped. depends on 167 .e. let’s leave it at this. yt = rt[A1(cosw + isinw)t + A2(cosw .Case of complex roots Two complex conjugate roots. r(cosw) = h. with: r2 = h2 + v2 The positive number r = √ h2 + v2 is called the modulus or absolute value of the complex number..isinw)t] Although this can be manipulated further. after manipulations.isinw)]t or. 8yt-2 then: y2 = -1. The stability condition is a2 < 1..8yt-1 + 0... Since √a2 > = < 1 as a2 > = < 1...8y1 . y0 = 0 One could compute the successive values of y recursively. by writing the equation in the form: yt = -1...0.the magnitude of r.88 y4 = .6) . There is a simple formula connecting r to the coefficients of the characteristic equation: r = √a2 (given without proof) where the square root is taken with positive sign (because r is the absolute value of the complex number h ± iv)...........8yt-1 . constant.8y0 = 3.9 y5 = . it follows that the oscillations will have an increasing.. y1 = -2.8yt-2 = 0.0.. = -6....0....8(3... or decreasing amplitude according to whether a2 > = < 1...7 etc. = 5.8(-2) = -4..6 y3 = -1.. 168 .. Example: Consider the equation: yt + 1. We might be tempted to conclude that the movement is oscillatory and divergent. Let’s check this by solving the equation. λ2 + 1.8λ + 0.8 = 0 λ1,2 = -1, -0.8, then: yt = A1(-1)t + A2(-0.8)t y0 = A1(-1)0 + A2(-0.8)0 = A1 + A2 = 0 y1 = A1(-1)1 + A2(-0.8)1 = -A1 – 0.8A2 = 2 → A1 = 10, A2 = -10, so: yt = 10(-1)t -10(-0.8)t from which we can see that, given enough time, the term -10(-0.8)t will approach 0, so we are left with long-term oscillations of constant amplitude. The oscillations are actually of increasing amplitude, but they do not diverge, but tend to a limit cycle of constant amplitude. 169 The inference we were tempted to make on the basis of the recursive solution, therefore, is wrong. Economic application Samuelson’s multiplier-accelerator model Ct = bYt-1 It = I0 + I1t (0< b < 1) (autonomous + induced investment) [I0t = I (constant)] I1t = k(Ct – Ct-1) Yt = Ct + It Substitute: Yt = bYt-1 + k(bYt-1 - bYt-2) + I0 Yt – b(1 + k) Yt-1 + bk Yt-2 = I0 Particular solution: Yp = I0/(1-b) Characteristic equation: λ2 – b(1 + k)λ + bk = 0 Example: In a certain period an economic system is in equilibrium with (for the sake of simplicity), Y(0) = 0, C(0) = 0, I(0) = 0, and assume that in period 1 autonomous investment increases to 100, afterwards remaining at the same level. Given: Ct = 0.8Yt-1, I1t = 3(Ct – Ct-1), examine the behaviour of Y over time. 170 Yt = 0.8Yt-1 + 3(Ct – Ct-1) + 100 Yt – 3.2Yt-1 + 2.4 Yt-2 = 100 (1) (2) If we want to compute the values of Y recursively, we use equation (1) because it allows for better economic understanding of the process. t 0 1 2 3 4 etc I 0 100 100 100 100 Ct = 0.8Yt-1 0 0 80 336 963 Ct – Ct-1 0 0 80 256 627 3(Ct – Ct-1) Yt = Ct + It + I 0 0 0 100 240 420 768 1204 1882 2945 which exhibits a tendency for Y to increase exponentially. This is confirmed by solving the difference equation: λ2 – 3.4λ + 2.4 = 0 → λ1,2 = 1.2, 2 Yp = 100/(1 – 3.2 + 2.4) = 500, and: Yt = A1(1.2)t + A2(2)t + 500 Y0 = ... = A1 + A2 + 500 = 0 Y1 = ... = 1.2A1 + 2A2 + 500 = 100 → A1 = -750, A2 = 250, so: Yt = -750(1.2)t + 250(2)t + 500 which confirms the monotonically explosive pattern of Y. 171 Pt-2).Pt-2).Pt-2)] Dt = St Then. with c positive (negative) according the weather price is expected to continue moving in the same direction (or to reverse direction). Suppose that expectations are assumed to be formed according to the relation: Pet = Pt-1 + c(Pt-1 . bPt – b1(1+c)Pt-1 + b1cPt-2 = a1 – a Pt – [b1(1+c)]/bPt-1 + (b1c/b)Pt-2 = (a1 – a)/b λ2 – [b1(1+c)/b]λ + (b1c)/b = 0 Pp = Pe = (a1 – a)/(b – b1) Example 1: Dt = 80 . c = 1 172 .4Pt St = -10 + Pet Pet = Pt-1 + c(Pt-1 . Dt = a + bPt St = a1 + b1Pet = a1 + b1[Pt-1 + c(Pt-1 .The Cobweb model with expectations We have seen earlier that the expectations assumption underlying the Cobweb model is unrealistic. 81)t + 18 P0 = A1 – A2 + 18 = 18 P1 = 0.5 – 0.4Pt = -10 + Pt-1 + Pt-1 .Pt-2 Pt + 0.5Pt-1 – 0.81A2 + 18 = 20 → A1 = A2 = -4. Check if price converges to the equilibrium value or not and how. and: Pt = -4(0. price rises to 20. 18 Example 2: 173 . given enough time.31A1 – 0.5λ . Because of the negative second root.25) = 18.81.2 = 0.5/(1 + 0. because of a disturbance.31)t – 4(-0.31)t + A2(-0. while in period 1.81)t + 18 Because the 2 roots are < 1 in absolute value the solution is convergent and price tends to the equilibrium value of 18. so: Pt = A1(0.In period 0 price is in equilibrium. -0.25 = 0 → λ1.5 Pp = Pe = 22.25Pt-2 = 22. 80 .31. the time path is oscillatory.0. λ2 +0. 5. no real part) Pt = rt( ) r = √h2 + v2 = √0. The stability condition is: √a2 < 1or a2 < 1. so: Pt = 0.5 λ2 + 0. since r = √a2 = √0.25Pt-2 = 22. 174 .25 = 0. 2 = 0 ± 0.25 = 0.25 = 0.5 < 1.Same but with c = -1 80 . Alternatively.4Pt = -10 + [Pt-1 – (Pt-1 . λ1.5i (purely imaginary roots.5.Pt-2)] Pt + 0.5t( ) We have convergence since 0.