ECON 186 Class Notes: Optimization Part 2

Transcription

1 ECON 186 Class Notes: Optimization Part 2 Jijian Fan Jijian Fan ECON / 26

2 Hessians The Hessian matrix is a matrix of all partial derivatives of a function. Given the function f (x 1,x 2,...,x n ), where all partial derivatives exist and are continuous, the Hessian of f is 2 f 2 f x 2 x 1 1 x f x 1 x n 2 f 2 f x H(f ) = 2 x f x2 2 x 2 x n f 2 f x n x 1 x n x f xn 2 Given the quadratic form d 2 z = f xx dx 2 + 2f xy dxdy + f yy dy 2, the Hessian determinant (sometimes called the Hessian) is H = f xx f xy f yx f yy Jijian Fan (UC Santa Cruz) ECON / 26

3 Examples Is q = 5u 2 + 3uv + 2v 2 either positive or negative definite? The discriminant of q is , with first leading principal minor 5 > 0 and second leading principal minor = = 7.75 > 0 So, q is positive definite. Given f xx = 2, f xy = 1, and f yy = 1 at a certain point on a function z = f (x,y), does d 2 z have a definite sign at that point? The discriminant of the quadratic form d 2 z is , which has leading principal minors 2 < 0 and = 1 > 0, so d 2 z is negative definite, which means the point in question is a local maximum. Jijian Fan (UC Santa Cruz) ECON / 26

4 Three-variable Quadratic Forms Similar conditions can analogously be obtained for a function of three or more variables. Consider a quadratic form q with three variables u 1, u 2, and u 3. Then: q(u 1,u 2,u 3 ) = d 11 (u 2 1) + d 12 (u 1 u 2 ) + d 13 (u 1 u 3 ) + d 21 (u 2 u 1 ) + d 22 (u 2 2) +d 23 (u 2 u 3 ) + d 31 (u 3 u 1 ) + d 32 (u 3 u 2 ) + d 33 (u 2 3) = = [ u 1 u 2 ] u 3 d 11 d 12 d 13 d 21 d 22 d 23 d 31 d 32 d 33 u 1 u 2 u i=1 j=1 u Du d ij u i u j Jijian Fan (UC Santa Cruz) ECON / 26

5 Three-variable Quadratic Forms Now, there are three leading principal minors: D 1 d 11 D 2 d 11 d 12 d 21 d 22 D 3 d 11 d 12 d 13 d 21 d 22 d 23 d 31 d 32 d 33 The sufficient condition for positive definiteness (local minimum) is that D 1 > 0, D 2 > 0, and D 3 > 0. The sufficient condition for negative definiteness (local maximum) is that D 1 < 0, D 2 > 0, and D 3 < 0. Jijian Fan (UC Santa Cruz) ECON / 26

6 Examples Find and classify the critical points of the function f (x,y,z) = x 2 + y 2 + 7z 2 + xy + 3yz. fx = 2x + y, f y = 2y + x + 3z, f z = 14z + 3y. It is easy to see that the only critical point is (0,0,0). f xx = 2, f yy = 2, f zz = 14, f xy = f yx = 1, f xz = f zx = 0, f yz = f zy = 3. We then compute the Hessian: f xx f yx f zx f yx f yy f yz f zx f zy f zz = D1 = 2 > 0, D 2 = = 4 1 = 3 > 0, D = = = 2(28 9) 14 = 24 > 0 So, since the Hessian is positive definite, the only critical point (0,0,0) is a local minimum. Jijian Fan (UC Santa Cruz) ECON / 26

7 Examples Find the extreme values of f (x 1,x 2,x 3 ) = z = x x 1x 3 + 2x 2 x 2 2 3x 2 3 f1 = 3x x 3, f 2 = 2 2x 2, f 3 = 3x 1 6x 3 So, we have a system of three equations: 3x x 3 = 0 x 3 = x 2 1 x 3 = ( 1 2 ) 2 = x 2 = 0 x 2 = 1 3x 1 6x 3 = 0 x 1 = 2x 3 x 1 = 2x 2 1 x 1 = 1 2 Additionally, since x 1 2x 3 = 0, (0,1,0) must also be a solution, so the two roots are (0,1,0) and ( 1 2,1, 1 4). Jijian Fan (UC Santa Cruz) ECON / 26

8 Examples f 11 = 6x 1, f 22 = 2, f 33 = 6, f 12 = f 21 = 0, f 13 = f 31 = 3, f 23 = f 32 = 0 So, the Hessian is 6x D 1 (0,1,0) = 0, so we already know that the point (0,1,0) is indefinite, and in fact is not an extremum at all. D 1 ( 1 2,1, 1 4 ) = 3 < 0, D 2 ( 1 2,1, 1 4 ) = = 6 > 0, D 3 ( 1 2,1, 1 4 ) = = = = 18 < 0 The Hessian is negative definite, so the point ( 1 2,1, 1 4) is a maximum. Jijian Fan (UC Santa Cruz) ECON / 26

9 Profit Maximization Example Consider a competitive firm with the following profit function: π = R C = PQ wl rk where P is price, Q is output, L is labor, K is capital, w is wage, r is the rental rate of capital. Since the firm is in a competitive market, P, w, and r are exogenous, while L, K, and Q are endogenous. However, Q is also function of K and L via the Cobb-Douglas production function Q = Q(K,L) = L α K β Assume that there are decreasing returns to scale where α = β < 1 2. Substituting in, the objective function becomes π(k,l) = PL α K α wl rk Jijian Fan (UC Santa Cruz) ECON / 26

10 Profit Maximization Example First order conditions: π ( w 1 L = PαLα 1 K α w = 0 K = Pα L1 α) α π K = PαLα K α 1 r = 0 Jijian Fan (UC Santa Cruz) ECON / 26

11 Profit Maximization Example Before we continue, let s make sure that these equations for L and K do actually give us a maximum. H = π LL π LK = Pα(α 1)L α 2 K α Pα 2 L α 1 K α 1 Pα 2 L α 1 K α 1 Pα(α 1)L α K α 2 π KL π KK = P 2 α 2 (α 1) 2 L 2α 2 K 2α 2 P 2 α 4 L 2α 2 K 2α 2 = P 2 α 2 (α 2 2α + 1)L 2α 2 K 2α 2 P 2 α 4 L 2α 2 K 2α 2 = P 2 α 2 L 2α 2 K 2α 2 (1 2α) + P 2 α 4 L 2α 2 K 2α 2 P 2 α 4 L 2α 2 K 2α 2 = P 2 α 2 L 2α 2 K 2α 2 (1 2α) H 1 = Pα(α 1)L α 2 K α < 0 and H > 0, so the hessian is negative definite, so L and K as defined by the FOC s represents the optimal quantities that will maximize profit. Jijian Fan (UC Santa Cruz) ECON / 26

12 Profit Maximization Example Plugging in the FOC for L into the FOC for K, we get [ ( ] w 1 α 1 PαL α K α 1 r = PαL α Pα L1 α) α r = 0 PαL α [ ( w Pα ) 1 ] α 1 α ( L 1 α w ) α 1 α α = Pα Pα = P α 1 α +1 α α 1 = (Pα) α 1 L 2α 1 α w α 1 α α +1 L α2 +2α 1+α 2 α w α 1 α L (1 α)(α 1) α +α r r r = 0 (Pα) α 1 L 2α 1 α w α 1 α = r (Pα) α 1 w α 1 α r 1 = L 2α 1 α = L 1 2α α L = ( Pαw α 1 r α) 1 2α 1 Jijian Fan (UC Santa Cruz) ECON / 26

13 Profit Maximization Example Similarly, we can find that K = ( Pαr α 1 w α) 1 1 2α Then, we can find the optimal quantity expressed only as a function of the exogenous parameters: Q = (L ) α (K ) α = ( Pαw α 1 r α) α 1 2α ( Pαr α 1 w α) α 1 2α = ( α 2 P 2 = wr ) α 1 2α Jijian Fan (UC Santa Cruz) ECON / 26

14 Constrained Optimization Up to this point, we have considered only problems of unconstrained optimization, that is, where an economic entity chooses the values of some variables to optimize a dependent variable with no restriction. However, consider a firm which seeks to maximizes profits with the production of two goods, but faces a production quota where Q 1 + Q 2 = 950. In this case, the choice variables are not only simultaneous, but also dependent. The solving of this problem is called constrained optimization. As another example, consider that a consumer wants to maximize their utility, given by U = x 1 x 2 + 2x 1 However, the consumer does not have an infinite amount of money, so they cannot buy an infinite amount of goods as would maximize their utility. Instead, the individual only has $60 to spend and x 1 costs $4 and x 2 costs $2, so their budget constraint is 4x 1 + 2x 2 = 60 Jijian Fan (UC Santa Cruz) ECON / 26

15 Constrained Optimization So, the individual s optimization problem can be stated as max U = x 1 x 2 + 2x 1 subject to 4x 1 + 2x 2 = 60 We call this constraint a budget constraint and it restricts the domain of the utility function, and as a result, the range of the objective function. In an unconstrained setting, x 1 and x 2 could take any value 0, but now the pair (x 1,x 2 ) must lie on the budget line. Jijian Fan (UC Santa Cruz) ECON / 26

16 Constrained Optimization The first method of solving constrained optimization is that of substitution. In the above example, we can take the budget constraint and find: x 2 = 60 4x 1 = 30 2x 1 2 Plug into the utility function to get: Also, we can easily see that du2 dx1 2 constrained maximum of U. U = x 1 (30 2x 1 ) + 2x 1 = 32x 1 2x 2 1 U x 1 = 32 4x 1 = 0 x 1 = 8 x 2 = 30 2(8) = 14 U = 8(14) + 2(8) = 128 = 4 < 0, so x 1 = 8 represents a Jijian Fan (UC Santa Cruz) ECON / 26

17 Constrained Optimization Another method, which is generally much more useful, especially for more complex and more than one constraint is called the Lagrange-multiplier method. The Lagrangian function for the previous example is: L = x 1 x 2 + 2x 1 + λ (60 4x 1 2x 2 ) λ is called the Lagrange multiplier (which we will discuss later). To solve the Lagrangian, we treat λ as a choice variable, so that the derivative with respect to λ will automatically satisfy the constraint. The first order conditions are: L = x λ = 0 λ = x x 1 4 L = x 1 2λ = 0 λ = x 1 x 2 2 L λ = 60 4x 1 2x 2 = 0 Jijian Fan (UC Santa Cruz) ECON / 26

18 Constrained Optimization λ = λ x ( x = x 1 2 Marginal rate of substitution x 1 = x ) 2x 2 = x 2 4 = 0 4x 2 = 56 x 2 = x 1 28 = 0 4x 1 = 32 x 1 = 8 Jijian Fan (UC Santa Cruz) ECON / 26

19 Lagrangian Method Given an objective function z = f (x,y) subject to g(x,y) = c we can write the Lagrangian function as L = f (x,y) + λ [c g(x,y)] Then, the first order conditions are L λ : c g(x,y) = 0 L x : f x λg x = 0 L y : f y λg y = 0 Jijian Fan (UC Santa Cruz) ECON / 26

20 Lagrangian Example A firm s production function is y = x + z and input prices are w x and w z. Find the quantities of x and z that minimize cost subject to the production function. L = w x x + w z z λ( x + z y) L x : w x 1 2 λx 1 2 = 0 wx = 1 2 λx 1 2 λ = 2wx x 1 2 L z : w z 1 2 λz 1 2 = 0 wz = 1 2 λz 1 2 λ = 2wz z 1 2 2w x x 1 2 = 2wz z 1 2 x 1 2 = z 1 2 w z w x x = z w 2 z w 2 x Jijian Fan (UC Santa Cruz) ECON / 26

21 Lagrangian Example y = x + z = z 2 1 w 1 z + z 1 z 2 w z + z 1 2 w x 2 = w x w x z 1 2 = w x y z = w x 2 y 2 w x + w z (w x + w z ) 2 = z 1 2 (w x + w z ) w x Plugging back into the marginal rate of technical substitution, ( w x 2 = x y 2 ) w 2 z (w x + w z ) 2 wx 2 = w 2 z y 2 (w x + w z ) 2 Jijian Fan (UC Santa Cruz) ECON / 26

22 Lagrangian Multiplier Interpretation The optimal value of L depends on λ (c), x (c), y (c), so L = f (x,y ) + λ [c g (x,y )] dl dc = (f x λ g x ) dx dc + (f y λ g y ) dy dc + [c g (x,y )] dλ dc + λ However, the first order conditions tell us that c = g (x,y ), f x = λ g x, and f y = λ g y, so the first three terms on the right hand side drop out and we are left with dl dc = λ So, the value of the Lagrange multiplier at the solution of the problem is a measure of the effect of a change in the constraint via the parameter c on the optimal value of the objective function. Jijian Fan (UC Santa Cruz) ECON / 26

23 Lagrangian-Method with Multiple Constraints The Lagrange-multiplier method is equally applicable when there is more than one constraint, we just need a Lagrange-multiplier for each constraint. Consider the function f (x 1,x 2,...,x n ) subject to two constraints: g (x 1,x 2,...,x n ) = c and h (x 1,x 2,...,x n ) = d. Then, the Lagrangian function can be written as: L = f (x 1,x 2,...,x n ) + λ [c g (x 1,x 2,...,x n )] + µ [d h (x 1,x 2,...,x n )] Then, the first-order conditions will consist of the following (n + 2) simultaneous equations: L λ = c g (x 1,x 2,...,x n ) = 0 L µ = d h (x 1,x 2,...,x n ) = 0 L i = f i λg i µh i = 0 (i = 1,2,...,n) Jijian Fan (UC Santa Cruz) ECON / 26

24 Multi-Constraint Lagrangian Example Find the maximum and minimum of f (x,y,z) = 4y 2z subject to 2x y z = 2 and x 2 + y 2 = 1. The Lagrangian function is: L = 4y 2z + λ (2 2x + y + z) + µ ( 1 x 2 y 2) The first order conditions are: L λ : 2 2x + y + z = 0 (1) L µ : 1 x 2 y 2 = 0 (2) L x : 2λ 2x µ = 0 (3) L y : 4 + λ 2y µ = 0 (4) L z : 2 + λ = 0 (5) Jijian Fan (UC Santa Cruz) ECON / 26

25 Multi-Constraint Lagrangian Example Plug in λ = 2 to (3) and (4): (5) λ = 2 ( 3) 2(2) 2x µ = 0 2x µ = 4 x = 2 µ (6) (4) y µ = 0 6 = 2y µ y = 3 µ (7) Plug in (6) and (7) to (2): ( 1 2 ) 2 ( ) 3 2 = 0 1 = 13 µ µ µ 2 µ = ± 13 Jijian Fan (UC Santa Cruz) ECON / 26

26 Multi-Constraint Lagrangian Example So there are two possible solutions, where µ = 13 and µ = 13. Case 1: µ = 13 Plugging back in to (6), (7), and then (2), we get x = 2 13, y = 3 13, and 0 = 2 + 2( 2 13 ) z z = Case 2: µ = 13 Plugging back in to (6), (7), and then (2), we get x = 2 13, y = 3 13, and 0 = 2 2( 2 13 ) z z = These are both potential optimum. To confirm, we must use the second order conditions we will learn in the next lecture. Jijian Fan (UC Santa Cruz) ECON / 26