Optimization. Thomas Dangl. March Thomas Dangl Optimization 1

Transcription

1 Optimization Thomas Dangl March 2010 Thomas Dangl Optimization 1

2 Introduction Course Preparation We use the program package R throughout the course. Please download the package from and install it on your system. Please also download and read the short introduction to the R language at and try to get started with the basics of R. For those using MS-Windows and MS-Excel: Also consider installing the package RExcel (with the load packages-function inside the R-Console) to perfectly integrate R with MS-Excel. Thomas Dangl Optimization 2

3 Introduction Literature The slides used to teach the course are still under construction! I will provide you with the most current version on a week-to-week basis. Please find the pdf of the slides at learn@wu. Additional literature for the course is Daniel Léonard and Ngo van Long (1992) Optimal Control Theory and Static Optimization in Economics, Cambridge University Press. Rangarajan K. Sundaram (1996) A First Course in Optimization Theory, Cambridge University Press. Morton I. Kamien and Nancy L. Schwartz (1991) Dynamic Optimization, Elsevier - North Holland. McGrawHill, some more applied texts Kenneth L. Judd (1998) Numerical Methods in Economics, MIT Press. Mario J. Miranda and Paul L. Fackler (2002) Applied Computational Economics and Finance, MIT Press. Thomas Dangl Optimization 3

4 Introduction Classes The classes will take place at March 1, March 8, March 15, March 22, April 19, April 26, April 28. There will be a mid-term exam on March 22 and a final exam on April 28 (approx. 90 minutes each). These exams will contribute 35% each to the final score. Preparation of the home exercises will contribute 30% to the final score. Active class participation, e.g., pointing out typos / errors in the slides can give up to an additional 10% score. A score of at least 50% is required to pass the course. Thomas Dangl Optimization 4

5 Introduction Course Outline 1 Introduction 2 Static Optimization Unconstrained Optimization Equality Constraints: The Method of Lagrange Inequality Constraints: The Method of Kuhn-Tucker Special Cases: LP, MILP, Porfolio Optimization 3 Dynamic Optimization The Bellman - Principle Theory of Optimal Control Theory of Optimal Control: Phase Diagram Analysis Thomas Dangl Optimization 5

6 Unconstrained Optimization Unconstrained Optimization (1) Deal with the choice of finitely many variables in order to maximize (minimize) some objective. sometimes, the values of these variables are unrestricted other times they are restricted by equality or inequality constraints The key concept of concavity (convexity) is introduced and will be associated with nice optimization problems. Assumptions, Theorems, Lemmas and Corollaries will be stated in the context we use and need them. I.e., there will be not section like Mathematical Preliminaries. Assumption: Differentiability If not stated otherwise, all functions we use are C 2, i.e., they have continuous second-order derivatives. Thomas Dangl Optimization 6

7 Unconstrained Optimization Unconstrained Optimization (2) Choose the values (x 1, x 2,..., x n ) R n in order to maximize the real-valued function f (x 1, x 2,..., x n ). Consider the example f (x 1, x 2 ) = 300 (10 x 1 ) 3 10x 1 2(10 x 2 ) 3 20x 2 Thomas Dangl Optimization 7

8 Unconstrained Optimization Unconstrained Optimization (3) Def. Convex Subsets of R n Consider a set D R n. If for D is called convex. x, y D [tx + (1 t)y] D, t [0, 1], Def. Open Subsets of R n The set D R n is called open subset if x D r > 0 such that x R n with x x < r x D. I.e., each x in D is surrounded by a neighborhood that is entirely contained in D. Or, each x D is an interior point. Thomas Dangl Optimization 8

9 Unconstrained Optimization Unconstrained Optimization (4) Assumption: Domains of Objective Functions We do only consider objective functions that are defined on domains that are convex and open subsets of R n. Def: Unconstrained Optimization Global Maximum Consider f : D R n R. Find x D such that x argmax x R n f (x) I.e., f (x ) f (x), x D. Then x is a global maximum. Sometimes, a maximum is only the maximum value in a certain neighborhood, i.e., the max. property is only valid locally Thomas Dangl Optimization 9

10 Unconstrained Optimization Unconstrained Optimization (5) Def: Local Maximum x is called a local maximum if r > 0 such that f (x ) f (x) x D with x x < r Def: Gradient The gradient vector f x R n is the column-vector created by f x = f x = f := The gradient f x evaluated at some point x and its components characterize the tangent plane to f at x and f x indicates the direction of the steepest increase in f. Thomas Dangl Optimization 10 f x 1 f x 2. f x n

11 Unconstrained Optimization Unconstrained Optimization (6) The Hesse matrix is the matrix that contains the second-order partial derivatives of f. Def. Hesse Matrix The Hesse matrix f xx is the symmetric n n matrix created by f xx = 2 f x 2 = H f := 2 f x1 2 2 f... 2 f x 1 x n 2 f x 2 x n x 2 x f x n x f x n x n According to our assumptions, f xx exists on the entire domain D of f and is continuous. Thomas Dangl Optimization 11

12 Unconstrained Optimization Unconstrained Optimization (7) To linearize a function at some x, i.e., to approximate it by the tangent-plane to f at the chosen x, we use Taylor s Theorem. First-Order Taylor-Series Expansion Consider x, x D R n, then we can write f (x) = f ( x) + (x x) f x ( x) + R 1 ( x, x) with R 1 ( x, x) = 1 2 (x x) f xx (x t )(x x). with x t = t x + (1 t)x for some t [0, 1]. Furthermore, ( ) R1 ( x, x) lim = 0. x x x x with. the Euclidean norm. I.e., R 1 ( x, x) = o( x x ). Thomas Dangl Optimization 12

13 Unconstrained Optimization Unconstrained Optimization (8) Linearization Approx. the function f by ignoring terms of the order o( x x ) Take the above example: f (x) ˆ f (x) = f ( x) + (x x) f x ( x) f (x 1, x 2 = 300 (10 x 1 ) 3 10x 1 2(10 x 2 ) 3 20x 2 ( ) ( ) [( ) ( )] ( x1 x1 x1 x (10 x1 ) fˆ = f + 2 x 2 x 2 x (10 x 2 ) 2 x 2 ) See R implementation for an illustration. Thomas Dangl Optimization 13

14 Unconstrained Optimization Unconstrained Optimization (9) First-Order Necessary Conditions Suppose x is the desired maximum, i.e., f (x ) f (x) x D, then it follows f x = 0. That is, at a maximum all partial derivatives of f must be zero. Or in other words, the tangent plane at x must be flat. Suppose x is a maximum but f x (x ) 0. Then we can find some x in the neighborhood of x such that f ( x) > f (x ). (The proof is a simple application of Taylor s rule and will be presented in class.) Thomas Dangl Optimization 14

15 Unconstrained Optimization Unconstrained Optimization (10) In order to capture the curvature of the objective function, we use a second-order Taylor-series expansion. Second-Order Taylor-Series Expansion Consider x, x D R n, then we can write f (x) = f ( x) + (x x) f x ( x) (x x) f xx ( x)(x x) + R 2 ( x, x) with i.e., R 2 ( x, x) = o( x x 2 ). lim x x ( ) R2 ( x, x) x x 2 = 0. Second-order Taylor-series expansion of f at a (local) maximum results in f (x) = f (x ) (x x) f xx ( x)(x x) + o( x x 2 ) Thomas Dangl Optimization 15

16 Unconstrained Optimization Unconstrained Optimization (11) Sectional drawing of f from above at x 2 = (30 30)/3 + approximations at x 1 = first order approx. f x 1, second order approx. Thomas Dangl Optimization 16

17 Unconstrained Optimization Unconstrained Optimization (12) Sectional drawing of f from above at x 2 = (30 30)/3 + approximations at x 1 = (30 30)/3. 50 f x 1, first order approx second order approx. Thomas Dangl Optimization 17

18 Unconstrained Optimization Unconstrained Optimization (13) See Taylor.R for a 3D-illustration of a Taylor-series approximation of the above example, using the package rgl. Def. Definiteness Consider a symmetric matrix A such that x R n : x Ax 0 then A is called negative-semidefinite. If x R n : x Ax < 0 then A is called negative-definite. Thomas Dangl Optimization 18

19 Unconstrained Optimization Unconstrained Optimization (14) Th: Second-Order Neccesary Condition f (x) reaches a (local) maximum at x implies f xx is negative-semidefinite at x. Suppose there exists a x such that x f xx ( x)x > 0, then we can show that there exists some x in the neighborhood of x, such that f ( x) > f (x ). Can be shown using the second-order Taylor-series expansion of f at x. Thomas Dangl Optimization 19

20 Unconstrained Optimization Unconstrained Optimization (15) How to check the definiteness of a matix? Def: Leading Principal Minors Let A be a (n n) matrix, then the r-th leading principal minor B r is the determinant of the matrix that results from deleting the last n r rows and the last n r columns from A. Th: Definiteness A n n symmetric matrix A is negative-definite if and only if all its eigenvalues are negative. A is negative-semidefinite, if and only if all its eigenvalues are non-positive. A is real-valued, symmetric. The spectral theorem guarantees that all eigenvalues are real-valued. The matrix A is negative-definite if its leading principal minors alternate in sign beginning with a negative sign, or ( 1) r B r > 0. Thomas Dangl Optimization 20

21 Unconstrained Optimization Unconstrained Optimization (16) Some notes on minima: The first-order necessary condition for a (local) minimum at some x is identical to the first-order necessary condition for a maximum. The second-order necessary condition for a (local) minimum requires that the quadratic form y f xx y 0 for all y R n. Def. Definiteness Consider a symmetric matrix A such that x R n : x Ax 0 then A is called positive-semidefinite. If x R n : x Ax > 0 then A is called positive-definite. Thomas Dangl Optimization 21

22 Unconstrained Optimization Unconstrained Optimization (17) To check for the definiteness of a matrix: Th: Definiteness (cont.) A n n symmetric matrix A is positive-definite if and only if all its eigenvalues are positive. It is positive-semidefinite if and only if its eigenvalues are non-negative. The matrix A is positive-definite if and only if all its leading principal minors are positive. Th: Second-Order Neccesary Condition (cont.) f (x) reaches a (local) minimum at x implies f xx is positive-semidefinite at x. Thomas Dangl Optimization 22

23 Unconstrained Optimization Unconstrained Optimization (18) Sufficient conditions for a (local) maximum (minimum) are somewhat stricter than the necessary first- and second-order conditions, since semi-definiteness of the Hession matrix is not a sufficient condition to ensure that the the quadratic term in the Taylor-series expansion actually dominates the residual. Th: Sufficient Conditions If f x (x ) = 0 and f xx (x ) is negative-semidefinite, then f (x) reaches a local maximum at x. Note: It is not sufficient for a maximum that f is concave at x in each variable, i.e., that 2 f 0 for all i = 1,..., n. xi 2 Consider the example for various values of a 0. f (x 1, x 2 ) = (x 2 1 ) + ax 1x 2 (x 2 2 ) Thomas Dangl Optimization 23

24 Unconstrained Optimization Unconstrained Optimization (19) The gradient of f is f x ( x1 x 2 ) ( 2x1 + ax = 2 2x 2 + ax 1 ). Independent of a, the point x = (0, 0) satisfies the first-order conditions. The Hesse matrix of f is given by ( 2 a f xx (x) = a 2 thus, second-order partial derivatives with respect to x 1 and x 2 resp. are negative, independent of a. Leading principal minors are B 1 = 2 and B 2 = 4 a 2, hence, the Hesse matrix is negative-semidefinite for a 2. ), Thomas Dangl Optimization 24

25 Unconstrained Optimization Unconstrained Optimization (20) Contour plots for a = 0, a = 1, a = 2, and a = 3. Thomas Dangl Optimization 25

26 Unconstrained Optimization Unconstrained Optimization (21) See SaddlePoint.R for a 3D-illustration. We next derive some global results for concave functions. Take a look at the exact form of Taylor s expansion: t [0, 1] such that f (x) = f (x ) + (x x ) f x (x ) (x x ) f xx (x t )(x x ) with x t = t x + (1 t)x, i.e., f xx is evaluated at some convex combination of x and x. Let us state three different definitions of concavity. Thomas Dangl Optimization 26

27 Unconstrained Optimization Unconstrained Optimization (22) Def. a: Concavity f C 2 defined on a convex set D is concave if and only if f xx is negative-semidefinite everywhere on D. Def. b: Concavity f C 1 defined on a convex set D is concave if and only if f ( x) f ( x) ( x x) f x ( x), x, x D. Def. c: Concavity f defined on a convex set D is concave if and only if f (t x + (1 t) x) t f ( x) + (1 t) f ( x), 0 t 1, and x, x D. Thomas Dangl Optimization 27

28 Unconstrained Optimization Unconstrained Optimization (23) With the help of concavity we can state global conditions Th. Sufficient Conditions for Concave Functions Let f (x) be a concave function, then it reaches a global max. at x if and only if f x (x ) = 0. How to find an extremum numerically. One possibility: Newtons Method Assume x is a local maximum and f xx is negative-definite in a neighborhood of x. Take start solution x 0 in the neighborhood of x and do an approximation of f (x ) by a second order Taylor-series of f at x 0, and take the derivative with respect to x. ˆ f x (x ) f x (x 0 ) + f xx (x 0 )(x x 0 ). Thomas Dangl Optimization 28

29 Unconstrained Optimization Unconstrained Optimization (24) Compute the next candidate for a solution to f x = 0 by solving the approximation for ˆ f x == 0. Obtain the iteration x i+1 = x i ( f xx (x i )) 1 f x (x i ) Convergence is generally not guaranteed, see Newton-Kantorovich Theorem for conditions. See simplenewton.r for a simple R implementation, see also the man pages for nlm. Newton s method does not distinguish between minima and maxima. Thomas Dangl Optimization 29

30 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (1) Typically, when searching for an optimum of the objective function we are restricted in the choice of x by constraints. consumers have budget constraints, society faces resource constraints, etc. The Classical Equality Constrained Problem (L.1) Find x = (x 1,..., x n) that maximizes f (x) subject to g 1 (x 1,..., x n ) = 0,. g m (x 1,..., x n ) = 0, m < n. Where f is the objective function and g j, j = 1,..., m, are the constraints. Thomas Dangl Optimization 30

31 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (2) Example: Maximize the objective function f (x 1, x 2 ) = 10 (10 x 1 ) 2 2(10 x 2 ) 2. subject to g(x 1, x 2 ) = 27 2x 1 x 2 = 0. g 0 f 0 f 9 f Thomas Dangl Optimization 31

32 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (3) Compare the gradient of the objective function, f x, to the gradient of the constraint, g x at a non-optimal choice of (x 1, x 2 ) g 0 fx gx 1 Thomas Dangl Optimization 32

33 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (4) Compare the gradient of the objective function, f x, to the gradient of the constraint, g x at a non-optimal choice of (x 1, x 2 ) g 0 fx 10.0 gx Thomas Dangl Optimization 33

34 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (5) In the optimum, the gradient of the objective function, f x, has to be parallel to the gradient of the constraint, g x, i.e., f x has to be some multiple of g x g 0 fx gx 9.0 Thomas Dangl Optimization 34 1

35 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (6) To formalize this idea, extend the dimensionality of the problem (introducing the before-mentioned multiples) and define the Lagrangian L as Def. The Lagrangian Define the Lagrangian L as L(λ 1,..., λ m, x 1,..., x n ) = f (x 1,..., x n ) + m λ j g j (x 1,..., x n ), j=1 or more compactly L(λ, x) = f (x) + λ g(x). Thomas Dangl Optimization 35

36 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (7) Define a matrix that contains the first-order derivatives of all the gs, where the rows correspond to the constraints and the columns to the variables, i.e., the matrix is (m n). Def. G Let the (m n) matrix G be defined as [ ] g j G = x i j,i = g 1 g 1 x n x g m x 1... g m x n Thomas Dangl Optimization 36

37 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (8) Th. First-Order Necessary Conditions Let x be a solution to problem (L.1) and let the (m n) matrix G have full rank m (this is known as the rank condition). Then there must exist a unique vector λ = (λ 1,..., λ m ) R m such that L λ = g(x ) = 0, L x = f x (x ) + G (x ) λ = 0. The first set of equations expresses that at the optimum the constraints must be satisfied. The second set of equations states that at the optimum the gradient of the objective function f x can be expressed as a linear combination of the gradients of the constraints. Thomas Dangl Optimization 37

38 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (9) In more detail, this set of equations is L λ 1 = g 1 (x ) = 0,. L = g m (x ) λ m = 0, L x 1 = f x1 (x ) + L x n = f xn (x ) + m λ j gx j 1 (x ) = 0, j=1. m λ j gx j n (x ) = 0. j=1 Thomas Dangl Optimization 38

39 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (10) Some more intuition Assume x is a constrained maximum. Any small change in x that is feasible must satisfy dg = 0 = G(x )dx i.e., dx is feasible if it dose not lead to the violation of any of the constraints. The first-order necessary condition of optimality requires that at x such a feasible change results in d f = f x(x )dx = 0 otherwise dx or dx leads to a marginal improvement. I.e., we do not require that f x (x ) = 0 (as in the unconstrained optimum), but only that f x (x ) = 0 in feasible directions. Thomas Dangl Optimization 39

40 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (11) Every marginal feasible change dx that satisfies G dx = 0 must also satisfy f xdx = 0, i.e., every dx that satisfies the system of m linear equations in n variables g 1 x 1 dx g1 x n dx n = 0,... must also satisfy g m x 1 dx gm x n dx n = 0, f dx f dx n = 0. x 1 x n f x must be a linear combination of the g x, and since the rank of G equals m, the weights λ are unique. Thomas Dangl Optimization 40

41 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (12) What, if the rank condition is not satisfied? Then, multipliers might not be unique (system of equation is over-determined), it might be impossible to write f x as a linear combination of the g x (the case of cusps ). Will show an example of a cusp later. First, we return to our example subject to f (x 1, x 2 ) = 10 (10 x 1 ) 2 2(10 x 2 ) 2. g(x 1, x 2 ) = 27 2x 1 x 2 = 0. and solve the necessary conditions for a candidate solution x 1 = 26/3, x 2 = 29/3, f (x 1, x 2 ) = 8. Thomas Dangl Optimization 41

42 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (13) Th. Second-Order Necessary Conditions Let x be a local maximum for problem (L.1) and let (x, λ ) satisfy the first-order conditions. Then the matrix L = L xx is negative-semidefinite for all vectors z R n satisfying Gz = 0. For a minimum, negative-semidefinite has to be modified to positive-semidefinite. Th. Second-Order Sufficient Conditions Let (x, λ ) satisfy the first-order conditions and in addition let the matrix L = L xx be negative-definite for all vectors z R n satisfying Gz = 0, where L and G are evaluated at (x, λ ); then x is a local maximum for problem (L.1). For a minimum, negative-definite has to be modified to positive-definite. Thomas Dangl Optimization 42

43 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (14) While it is intuitively clear, that the definiteness does only matter in feasible directions, it is generally not easy to check for conditional definiteness. There is a theorem for sufficiency, that is easy to check (the variables and derivatives have to be done exactly in the manner stated here). Def. B is the Hesse Matrix of L Define B as the Hesse matrix of the Lagrangian L, such that we start with the derivatives with respect to the multipliers λ and continue with the derivatives with respect to the variables x, L λλ. L λx B = L xλ. L xx Thomas Dangl Optimization 43

44 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (15) Explicitly: 0 0 B =... g 1 x 1 g 1 x n g m x 1 g m x n g 1 x 1 g m x 1. f x1 x 1 + m j=1 λ j g j x 1 x 1 f x1 x n + m j=1 λ j g j x 1 x n..... g 1 x n g m x n. fx1 x n + m j=1 λ j gx j 1 x n f xn x n + m j=1 λ j gx j n x n Or more compactly 0. G B = G. L = f xx +, m j=1 λ j g j xx. Thomas Dangl Optimization 44

45 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (16) B is (m + n) (m + n) and the four submatrices 0, G, G and L are of order m m, m n, n m, and n n, respectively. Th. Sufficient Conditions Let (x, λ ) satisfy the first-order conditions and in addition let the last (n m) leading principal minors of B alternate in sign beginning with that of ( 1) m+1, where B is evaluated at (x, λ ), then x is a local maximum of problem (L.1). For a minimum, the last (n m) leading principal minors of B must be of the same sign as ( 1) m. Thomas Dangl Optimization 45

46 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (17) Some examples: Optimal consumption of an exhaustible resource. Socially optimal pricing of an exhaustible resource. Thomas Dangl Optimization 46

47 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (18) Continue our example. The Hesse matrix B is B = and since n = 2 and m = 1, only the last leading principal minor is relevant, i.e., B = 18 must have the same sign as ( 1) m+1 = 1, which is satisfied. Thus, the point (x 1 = 26/3), x 2 = 29/3, λ = 4/3 is a local maximum. Thomas Dangl Optimization 47

48 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (19) How does the optimized (maximum) value of the objective function change, if some parameter of the problem is altered? Suppose the objective function f as well as the constraints g j are affected by some vector of model parameters p. E.g., the total amount of the resource available. The natural question that arises: What is the total differential of f with respect to changes in p? The total differential considers that when p is altered also the choice of x changes! Th. The Envelope Theorem Let (x, λ ) solve problem (L.1) for some parameter vector p R k, i.e., f (x ; p) is a local maximum under the constraints g(x ; p) = 0. Then d f (x ; p) dp s = L(λ, x ; p) p s, s = 1,..., k. Thomas Dangl Optimization 48

49 Equality Constraints: The Method of Lagrange Optimization Under Equality Constraints (20) Back to our simple example: L = 10 (10 x 1 ) 2 2(10 x 2 ) 2 + λ(p 2x1 x2), with p = 27. How does the optimal value ( f (x ) = 8) change, if the parameter p of the constraint is altered by dp? Answer: d f (x ; p) = L(, λ, x ; p) = λ = 4/3. dp s p s If p moves to p + dp, then the optimized objective moves from f (x ; p) to f (x ; p) + λ dp Def. Shadow Price If the vector p characterizes the constraints in the form g j = p j h(x) = 0, then the associated Lagrange multipliers λ can be interpreted as shadow prices of the constraints. I.e., a rational agent maximizing the objective f is willing to pay up to λ j dp j in order to increase p j by a marginal dp j. Thomas Dangl Optimization 49

50 Inequality Constraints: The Method of Kuhn-Tucker Optimization Under Inequality Constraints (1) In many cases, constraints are not as strict as equality constraints. E.g., resource constraints do not imply that one has to fully exploit the resource. Def. The Standard Problem of Optimization Under Inequality Constraints (KT.1) subject to max x f (x) g(x) 0, x 0. Thomas Dangl Optimization 50

51 Inequality Constraints: The Method of Kuhn-Tucker Optimization Under Inequality Constraints (2) Example Consider a firm that produces a non-durable good in two consecutive time periods. Production cost is a constant EUR 10, per unit produced. Let x 1 and x 2 denote the quantity produced in periods 1 and 2. Inverse demand in the first period is given by p 1 = x 1. In period 2, the attained price is p 2 = 50 2x 2. In the course of production, emissions are unavoidable and legal restrictions admit a maximum of total 33 units of production (sum of x 1 and x 2 ). Capacity constraints restrict the production in each of the periods to be less than 28 units. The time value of money is 0. a) Determine the optimal production decision x 1 and x 1 that maximizes the total profit of the firm. b) Give an interpretation of the result. What can be derived from the magnitude of the Lagrangian multipliers? c) Assume that the emission constraint is relaxed and substituted by an emission tax of EUR 12 per unit produced. What is then the optimal production decision x 1, x 2? Thomas Dangl Optimization 51

52 Inequality Constraints: The Method of Kuhn-Tucker Optimization Under Inequality Constraints (3) Def. Binding Constraint A constraint g j 0 is called binding at some point x if g j ( x) = 0. Regularity Condition The set of constraints g meets the regularity condition if one of the following is satisfied (i) g j is linear for all j = 1,..., m. (ii) g j is concave and there exists x > 0 such that g j > 0 for all j = 1,..., m. (iii) The feasible set (g j 0, x 0) is convex and has a nonempty interior, and g j x 0 if constraint j is binding. (iv) Renumber the constraints so that the first k m constraints are binding. Then the matrix [ g j / x] ji, i = 1,..., n, j = 1,..., k has full rank k. Thomas Dangl Optimization 52

53 Inequality Constraints: The Method of Kuhn-Tucker Optimization Under Inequality Constraints (4) Th. Kuhn-Tucker Let x be a solution to problem (KT.1) and assume that the regulatory condition applies (see above). Then there must exist a set of multipliers λ 1,..., λ m such that g j (x ) 0, λ j 0, and λ j g j (x ) = 0, j = 1,..., m, m f xi (x ) + λ j gx j i (x ) 0, xi 0, x i j=1 f x i (x ) + m λ j gx j i (x ) = 0, i = 1,..., n. j=1 Thomas Dangl Optimization 53

54 Inequality Constraints: The Method of Kuhn-Tucker Optimization Under Inequality Constraints (5) Th. Kuhn-Tucker (cont) More compactly: L λ (λ, x ) 0, λ 0, and λ L λ (λ, x ) = 0, L x (λ, x ) 0, x 0, and x L x (λ, x ) = 0. The subset of the Kuhn-Tucker conditions λ L λ (λ, x ) = 0, x L x (λ, x ) = 0. is called the complementary slackness conditions. In words: Only binding constraints can have non-zero shadow prices. Sloppy speaking: Kuhn-Tucker is Lagrange Method for an arbitrary chosen subset of binding constraints + a criterion for deciding whether the chosen subset is consistent with the problem (KT.1). Thomas Dangl Optimization 54

55 Inequality Constraints: The Method of Kuhn-Tucker Optimization Under Inequality Constraints (6) Contra: There is no guideline for finding the proper selection of binding constraints. I.e., we need to have some (economic) intuition and / or have to solve up to 2 m Lagrange problems. Th. Sufficient Condition Suppose (x, λ ) satisfies the Kuhn-Tucker conditions and the gs satisfy the regularity condition (at x ). Suppose further that the first k m constraints are binding at x and the set of non-binding constraints is strictly non-binding in an entire neighborhood of x. If (x, λ j ), j = 1,..., k, together with the binding constraints g j = 0, j = 1,..., k, satisfy the sufficient optimality condition for the implied (L.1) Lagrange problem, then x is a local maximum of (KT.1). Thomas Dangl Optimization 55

56 Inequality Constraints: The Method of Kuhn-Tucker Optimization Under Inequality Constraints (7) Example (Cusp): Maximize f (x 1, x 2 ) = x 1 subject to g 1 : x 2 (x 1 2) 3 + 1, g 2 : x g x fx g x 2 Thomas Dangl Optimization 56

57 Inequality Constraints: The Method of Kuhn-Tucker Optimization Under Inequality Constraints (8) Some notes: Let gs depend on the selection of some parameters p. Let U {} be the feasible range of x. If a change of p p leads to a feasible range Ũ with Ũ U we call it a reinforcement of constraint g. If a change of p p leads to a feasible range Ũ with U Ũ we call it a relaxation of constraint g. Let f (x (p)) be a (local) maximum. A change p p that reinforces constraint(s) implies f (x ( p)) f (x (p)) Let f (x (p)) be a (local) maximum. A change p p that relaxes constraint(s) implies f (x ( p)) f (x (p)) Especially the introduction of an additional constraint is always a reinforcement of the set of constraints. Thomas Dangl Optimization 57

58 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (1) Linear Programming is the special case where the objective function as well as all constraints are linear functions. The regularity condition is met in any case (see above). The set of feasible solutions (i.e., x that meet the boundary conditions) is either empty, or a convex, n-dimensional polyhedron. Due to the linearity of the problem, we know: If there exists a maximum, one of the vertices of the polyhedron is always part of the optimal solution. The Simplex method was developed by Georg Dantzig in the 1940-ies. It is a method that systematically searches the vertices of the polyhedron created by the restrictions along a path that follows the steepest improvement of the objective function. Simplex-method and Simplex-Tableau are not covered here. Thomas Dangl Optimization 58

59 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (2) Solve a simple example and learn how to use the R-package lpsolve. Example: Resource Allocation A cabinetmaker is planning the coming year. She wants to invest a maximum of working hours in the production of chairs and tables. There are only 282 units of wood available per year. (i) Due to space constraints not more than 320 tables per year can be produced. (ii) The production of one table and one chair takes 2 hours and 4 hours, respectively. (iii) For producing a table and a chair, 0.8 and 0.4 units of wood is needed, resp. The profit margin of a table is EUR 90., the profit margin of a chair is 60.. What is the maximum attainable profit margin? What is the optimal allocation of resources over the year? Thomas Dangl Optimization 59

60 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (3) The linear program is: max x t,x c f (x t, x c ) = 90x t + 60x c subject to g 1 (x t, x c ) = 320 x t 0, g 2 (x t, x c ) = x t 4x c 0, g 3 (x t, x c ) = x t 0.4x c 0, Thomas Dangl Optimization 60

61 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (4) The region of feasible production schedules together with iso-value lines of the objective function f f 0 f 100 Thomas Dangl Optimization 61

62 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (5) At a non-optimal vertex: The gradient of the objective function can be composed by a linear combination of the gradients of the active boundaries. But f const g 2 g fx g x g x 2 Thomas Dangl Optimization 62

63 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (6)... the sign of the Lagrange-multipliers is not consistent with the Kuhn-Tucker conditions. 220 f const g 2 g Note: Λ Λ 1 g x 1 fx Λ 2 g x Thomas Dangl Optimization 63

64 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (7) Implementation: see LP.R. library(lpsolve) # coefficients of the objective function obj <- c(90,60) # constraints # space constraint lhs <- matrix(c(1,0),nrow=1) rhs <- 320 dir <- "<=" # time constraint lhs <- rbind(lhs,c(2,4)) rhs <- c(rhs,1200) dir <- c(dir,"<=") Thomas Dangl Optimization 64

65 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (8) LP.R cont.: # wood constraint lhs <- rbind(lhs,c(0.8,0.4)) rhs <- c(rhs,282) dir <- c(dir,"<=") # compute the solution sol <- lp("max",obj,lhs,dir,rhs,compute.sens=t) The components of sol can be seen with names(sol). The most important are sol$objval sol$solution sol$duals (first the duals for the constraints, then for the variables) Thomas Dangl Optimization 65

66 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (9) Result: optimal production plan number tables 270 chairs 165 max. objective: dual variables of the constraints (shadow prices): lambda g1 0 g2 5 g3 100 Thomas Dangl Optimization 66

67 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (10) Result (cont): duals of the variables (relative profit margins): rel. margin x1 0 x2 0 The λs say that the cabinetmaker values one additional hour at EUR 5, and one additional unit of wood at EUR 100. Additional space for producing tables has a value of EUR 0, since the space constraint is not binding. The relative profit margins constitute the marginal opportunity costs when deviating from the optimal production plan. Both variables are strictly positive, thus, the relative profit margins are 0. Thomas Dangl Optimization 67

68 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (11) Now change the profit margins such that only one of the products is in the production plan: max x t,x c f (x t, x c ) = 60x t + 130x c New result: optimal production plan number tables 0 chairs 300 max. objective: Thomas Dangl Optimization 68

69 Special Cases: LP, MILP, Porfolio Optimization Linear Programming (12) New results (cont.): dual variables of the constraints (shadow prices): lambda g1 0.0 g g3 0.0 duals of the variables (relative profit margins): rel. margin x1-5 x2 0 Now tables are not in the production plan. The opportunity cost of producing a (marginal) table is EUR 5 per unit. Thomas Dangl Optimization 69

70 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (1) While large LPs can be solved very efficiently and fast, things change, if some or all of the free variables x are restricted to integer values. Since differentiability is lost in this case, the Lagrange and Kuhn-Tucker conditions cannot be applied. As long as there exists an efficient algorithm for solving the related non-integer problem, the branch-and-bound algorithm is a suitable method to solve the integer problem. The simplest approach to integer problems is the complete enumeration, i.e., the evaluation of the objective function on each of the possible combinations of the integer variables. Thomas Dangl Optimization 70

71 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (2) Assume a problem with n binary variables (integers that can either be 0 of 1. Suppose that a computer is able to calculate (and compare) the objective function at 1 billion locations per second. Then the time to completely enumerate the problem is for n = 10 it takes 1.024µs, for n = ms, for n = s, for n = min, for n = d, for n = y. for n = y. I.e., we must use a method that avoids complete enumeration! Thomas Dangl Optimization 71

72 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (3) Branch-and-bound algorithm for mixed integer linear programming: Consider a linear maximization program where the first k n variables in x are required to be integers. Some considerations: First ignore all integer conditions and solve the linear program. Call this initial problem P0 and the solution of the problem f 0. If f 0 satisfies all integer constraints, then f 0 is also a solution to the overall problem. If not, take one of the variables for which the integer condition is not satisfied, say x i. Let x 0,i be the value of x i in the solution f 0 and x 0,i the largest integer with x 0,i x 0,i. Furthermore, x 0,i is the smallest integer with x 0,i x 0,i. Create two new models (branches) and add them to the candidate list: P1 is P0 with the additional constraint x i x 0,i. P2 is P0 with the additional constraint x i x 0,i Thomas Dangl Optimization 72

73 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (4) We can now solve P1 and P2 and in the case there are still variables that do not satisfy the required integer conditions create new branches. So far, the branching approach is only a systematic approach for the total enumeration. Fortunately, we can calculate upper limits for the maximum attainable objective value in a branch (bounds): Consider a problem Ph with maximum LP-solution f h. All problems that follow in the branch rooted at Ph have additional constraints, i.e., the are reinforcements to Ph. Thus, f l f h for all successor problems Pl. Now we are ready to formulate the branch-and-bound algorithm. Thomas Dangl Optimization 73

74 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (5) Branch-and-Bound: 1 Start with P0 by ignoring all integer conditions and add it as the first problem to the candidate list. The model status is non-evaluated and the parent objective is set to -. 2 Select among the non-evaluated candidate problems one with maximum parent objective, say Ph, and compute the LP-solution f h. 3 If none of the following stop criteria is met, set status to evaluated, branch into two models at one of the variables for which the LP-solution f h does not meet the integer condition, pass f h as parent objective to these two child problems, status non-evaluated. 4 Continue at step 2 as long as there are non-evaluated models in the candidate list. 5 After no non-evaluated model is left If none of the evaluated models satisfies all integer conditions, the overall model is infeasible. If one or more candidate models satisfy all integer conditions, report the max. as overall solution. Thomas Dangl Optimization 74

75 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (6) Stopping condition: Do not further branch the tree after evaluation of problem Ph if one of the following criteria is met i The LP-solution f h satisfies all integer conditions. Then f h is a feasible solution to the overall problem. Compare f h to the best overall solution found so far. If it exceeds the max. found so far, remember Ph and f h. Search for non-evaluated problems in the candidate list with parent objectives that are less than f h. Change the status of these models from non-evaluated to dominated. I.e., the total branch rooted at such a model is dominated and we need not care about these successor models. ii Ph is infeasible. Set its status to infeasible. Thomas Dangl Optimization 75

76 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (7) Consider the following integer linear program: max f (x 1, x 2 ) = x 1 + 2x 2 subject to 6x 1 +5x x 1 +6x 2 21 x j 0 and integer, j = 1, 2 Thomas Dangl Optimization 76

77 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (8) Here you see the initial problem P0 together with the grid of feasible integer pairs (x 1, x 2 ) x1 5x2 27 2x1 6x2 21 f 0 2 P The LP-solution to P0 is: f 0 = 7.73, x 0,1 = 2.19,x 0,2 = Thomas Dangl Optimization 77

78 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (9) Branching at x 0,1 leads to further candidates P1, P2. 6 6x1 5x x1 6x2 21 f 0 2 P1 P Thomas Dangl Optimization 78

79 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (10) The candidate list model status parent objective boundaries P0 evaluated P0 P1 non-evaluated 7.73 P0 & x 1 2 P2 non-evaluated 7.73 P0 & x 1 3 Continue with P1 The LP-solution is: f 1 = 7.67, x 1,1 = 2.0,x 1,2 = New candidate list: model status parent objective boundaries P0 evaluated P0 P1 evaluated 7.73 P0 & x 1 2 P2 non-evaluated 7.73 P0 & x 1 3 P3 non-evaluated 7.67 P0 & x 1 2 & x 2 2 P4 non-evaluated 7.67 P0 & x 1 2 & x 2 3 Thomas Dangl Optimization 79

80 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (11) Illustration of the problems P3 and P4: 6 6x1 5x P4 f 0 2 P3 2x1 6x Thomas Dangl Optimization 80

81 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (12) We continue with problem P2. The LP-solution is: f 2 = 6.6, x 2,1 = 3.0,x 2,2 = 1.8. Branch again at x 2,2, the new candidate list: model status parent objective boundaries P0 evaluated P0 P1 evaluated 7.73 P0 & x 1 2 P2 evaluated 7.73 P0 & x 1 3 P3 non-evaluated 7.67 P0 & x 1 2 & x 2 2 P4 non-evaluated 7.67 P0 & x 1 2 & x 2 3 P5 non-evaluated 6.60 P0 & x 1 3 & x 2 1 P6 non-evaluated 6.60 P0 & x 1 3 & x 2 2 Next evaluate P3. The LP-solution is: f 3 = 6, x 3,1 = 2,x 3,2 = 2. Solution satisfies the integer conditions! Do not branch further. This is the best overall solution found so far. There are no branches dominated. Thomas Dangl Optimization 81

82 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (13) Continue with P4. LP-solution is: f 4 = 7.5, x 4,1 = 1.5, x 4,2 = 3. Branching creates a new list: model status parent objective boundaries P0 evaluated P0 P1 evaluated 7.73 P0 & x 1 2 P2 evaluated 7.73 P0 & x 1 3 P3 solution 7.67 P0 & x 1 2 & x 2 2 P4 evaluated 7.67 P0 & x 1 2 & x 2 3 P5 non-evaluated 6.60 P0 & x 1 3 & x 2 1 P6 non-evaluated 6.60 P0 & x 1 3 & x 2 2 P7 non-evaluated 7.50 P0 & x 1 1 & x 2 3 P8 non-evaluated 7.50 P0 & x 1 = 2 & x 2 3 Continue with problem P7: f 7 = 7.33, x 7,1 = 1, x 7,2 = Branch again at x 7,2 = Thomas Dangl Optimization 82

83 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (14) The new candidate list (skip evaluated problems): model status parent objective boundaries P3 solution 7.67 P0 & x 1 2 & x 2 2 P5 non-evaluated 6.60 P0 & x 1 3 & x 2 1 P6 non-evaluated 6.60 P0 & x 1 3 & x 2 2 P8 non-evaluated 7.50 P0 & x 1 = 2 & x 2 3 P9 non-evaluated 7.33 P0 & x 1 1 & x 2 = 3 P10 non-evaluated 7.33 P0 & x 1 1 & x 2 4 P8 is infeasible, stop criterion is met. Next P9: f 9 = 7, x 9,1 = 1, x 9,2 = 3. This is an integer solution. It is better than the solution of P4( f 4 = 6). Problems P5 and P6 are dominated by P9 Thomas Dangl Optimization 83

84 Special Cases: LP, MILP, Porfolio Optimization Mixed Integer Linear Programming (15) The only remaining problem P10 is infeasible. The overall maximum is given by the solution to problem P9: x1 = 1 x2 = 3 f (x1, x 2 ) = 7 Thomas Dangl Optimization 84

85 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (1) Suppose there is an investment universe consisting of n assets. Some notation: r... return in excess to some riskless asset Σ... variance-covariance matrix of returns x... portfolio weights µ... return expectations, E(r) 1... a (n 1) column vector consisting of 1s r p... portfolio return, r p = x r Assume that the riskiness of an investment is measured by the return variance (one period). So our first objective in portfolio selection is: Minimize the return variance for a desired expected portfolio return. Thomas Dangl Optimization 85

86 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (2) The problem subject to: 1 min x 2 Var[r p,t+1] = 1 2 x Σx E(r p,t+1 ) = x µ = µ x 1 = 1 The Lagrangian of this equality constrained problem is L = 1 2 x Σx + λ(1 x 1) + δ( µ x µ) Thomas Dangl Optimization 86

87 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (3) At the optimal selection the partial derivatives of the Lagrangian with respect to its variables must vanish (i) (ii) (iii) L = Σx λ1 δµ = 0 x L λ = x 1 1 = 0 L δ = x µ µ = 0 From (i) and the fact that λ and δ are real numbers it follows that x = λσ δσ 1 µ Using (ii) we get x 1 = 1 Thomas Dangl Optimization 87

88 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (4) Together with the expression for x this yields From (iii) it follows that and thus λ1 Σ δµ Σ 1 1 = 1 x µ = µ λ1 Σ 1 µ + δµ Σ 1 µ = µ And compactly written ( 1 Σ 1 1 µ Σ Σ 1 µ µ Σ 1 µ ) ( λ δ ) = ( 1 µ ) Thomas Dangl Optimization 88

89 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (5) And with the following notation A = 1 Σ 1 1 B = 1 Σ 1 µ C = µ Σ 1 µ we can write this as ( ) ( A B λ B C δ } {{ } M ) = ( 1 µ ) We can show that from Σ positive definite it follows that M is invertible. Thomas Dangl Optimization 89

90 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (6) The Lagrangian multipliers are given by ( ) ( ) λ = M 1 = δ 1 µ 1 ( C B M B A ) ( 1 µ The Lagrangian multiplier is associated with the condition x 1 = 1 (portfolio weights sum to 1), λ, is given by λ = C µb M Lagrangian multiplier associated with the condition x µ = µ (given expected return), δ, is given by δ = µa B M ) Thomas Dangl Optimization 90

91 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (7) Optimal portfolio weights: x = C µb Σ µa B Σ 1 µ M M 1 = M (CΣ 1 1 BΣ 1 µ) + 1 M (AΣ 1 µ BΣ 1 1) µ } {{ }} {{ } Λ 1 Λ 2 = Λ 1 + Λ 2 µ That is, the optimal portfolio weights are a linear affine function of µ. Thomas Dangl Optimization 91

92 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (8) The portfolio variance at the optimum is given by Var(r p ) = σ 2 p = x Σx = x Σ(λΣ δσ 1 µ) = λx ΣΣ δx ΣΣ 1 µ = λx 1 + δx µ = λ + δ µ 1 = M (C µb + µ2 A µb) 1 = M ( µ2 A 2 µb + C) Thomas Dangl Optimization 92

93 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (9) These minimum variance portfolios (for different values of µ) form the so called efficient portfolio frontier. The case of three assets: Thomas Dangl Optimization 93

94 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (10) The location of the global minimum variance portfolio can be computed in two ways. and, thus, dσ 2 µ = 1 (2 µa 2B) = 0 M µ = B A The second possibility to derive the location of the minimum variance portfolio: At the min. var. portfolio, the Lagrangian multiplier associated with x µ = µ must be zero, because this boundary condition is not binding in that case: δ = 0 1 M ( µa B) = 0 µ = B A Thomas Dangl Optimization 94

95 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (11) Now we have characterized the diversification potential inherent in the portfolio. I.e., we know the return-variance that an investor must accept if he / she demands an expected portfolio return of µ. Consider a representative mean-variance optimizing investor, i.e., an investor with utility u(r) = E(r) 1 2 γvar(r) = µ 1 2 γσ2 The representative investor must hold the market. Therefore u = µ 1 2 γ 1 M ( µ2 A 2 µb + C) The first order condition for the maximum utility choice is du d µ = γ 1 (2A µ 2B) = 0 M Thomas Dangl Optimization 95

96 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (12) Using the expression for δ leads to γ A µ B M = γδ = 1 i.e., the mean-variance optimizing investor with a relative risk aversion of γ maximizes her utility by choosing the efficient portfolio with expected return µ given by µ = M γa + B A. For this utility maximizing portfolio it holds that δ = 1 γ. Thomas Dangl Optimization 96

97 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (13) Utility maximization: The investor selects the portfolio which lies on the highest attainable iso-utility line Thomas Dangl Optimization 97

98 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (14) Utility maximization: Higher risk aversion means the choice of a portfolio with lower variance Thomas Dangl Optimization 98

99 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (15) Now consider the riskless asset (which is in zero net supply). The riskless asset has an excess return of 0 (as defined) and variance of 0. An individual investor can now mix a risky portfolio and the riskless asset and has, thus, two parameters of choice: a portfolio of risky assets (characterized by µ) the weight w of the risky portfolio in the overall investment. For a given µ get a min. var portfolio x ( µ) with expected return µ and variance σ 2 ( µ) = x Σx. If the investor choses a weight of w for the risky part of her investment, it is characterized by µ p = w µ, σ 2 p = w 2 x Σx = w 2 σ 2 ( µ) Thomas Dangl Optimization 99

100 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (16) What is the optimal choice of w (for a given µ)? For different w the utility of the mean-variance optimizer is u = w µ 1 2 γw2 σ 2 ( µ) Then the first order condition for the optimal choice of w is du dw = µ γwσ2 ( µ) = 0, which yields: w = µ γσ 2 ( µ). Thomas Dangl Optimization 100

101 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (17) Optimal weight w for a non-optimally chosen portfolio µ Thomas Dangl Optimization 101

102 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (18) What is the optimal choice of µ? Select µ such that the investor can reach the highest attainable iso-utility line. This is achieved by selecting the so called tangency portfolio. I.e., independent of the risk aversion of the individual investor, all investors will choose to invest a certain fraction of their wealth in the tangency portfolio and the rest in the riskless asset.! The weight w, however, depends on the investor s risk aversion! Thomas Dangl Optimization 102

103 Special Cases: LP, MILP, Porfolio Optimization Portfolio Optimization (19) The tangency portfolio and two different investors choice of the weight w of the risky sub-portfolio Thomas Dangl Optimization 103