Nonlinear Optimisation Seminar Exercises II-6 Solutions PDF

Solutions to Seminar Exercises II-6 on Nonlinear Optimisation 1. De…ne f (x; y) = (2x y 2 ) (x y2) : (a) Is f a convex function? We …rst construct the Hessian. x y2 2x y 2 = 2x2 3xy 2 + y 4 fx = 4x 3y 2 ; fxx = 4; f...

Solutions to Seminar Exercises II-6 on Nonlinear Optimisation 1. De…ne f (x; y) = (2x y 2 ) (x y2) : (a) Is f a convex function? We …rst construct the Hessian. x y2 2x y 2 = 2x2 3xy 2 + y 4 fx = 4x 3y 2 ; fxx = 4; fxy = 6y fy = 4y 3 6xy; fyx = 6y 4 6y f 00 (x; y) = H = 6y 12y 2 6x Since the second principal leading minor is 12y 2 24x; which is not always 0, f is not convex. (It is convex on the set x 0). 1  1000 800 600 z 400 200 0 -4 -4 -2 -2 00 2 2 4 4 y x Plot of (2x y 2 ) (x y2) b) Show that f (x; y) has a local minimum at P = (0; 0) when restricted to any line x = at; y = bt; (a; b not both 0)through P: de…ne g (t) = f (at; bt) = b2 t2 at b2 t2 2at g (t) = b4 t4 + 3ab2 t3 + 2a2 t2 g 0 (t) = 4a2 t 9ab2 t2 + 4b4 t3 = 0 at t = 0: g 00 (t) = 4a2 18ab2 t + 12b4 t2 = 4a2 > 0; t = 0; a 6= 0: g (t) = b4 t4 if a = 0: 2  30 y 20 10 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 x Plot of g (t) ; (a; b) = (1; 1) red (2; 1) green, (1; 2) blue. 3  c) Show that f (x; y) does not have a local minimum at (0; 0) : Consider the curve x = (3=4) y 2 , which goes through (0; 0) : On this curve 1 4 f (x; y) = f (3=4) y 2 ; y = y ; 8 which is negative (so < f (0; 0) = 0) for y 6= 0: d) If we set fx = fy = 0; we get fx = 4x 3y 2 = 0 fy = 4y 3 6xy = 0 x = 3y 2 =4 4y 3 6 3y 2 =4 y = 0 1 3 y = 0 2 y = 0; so also x = 0: But we have shown that (0; 0) is not a local minimum. New question. How could we get this curve where (0; 0) is not a local minimum? f (x; y) = 2x y2 x y2 To make this negative want one factor positive, one negative. So 2x > y 2 but x < y 2 : Take x = cy 2 where 2c > 1 and c < 1; or 1=2 < c < 1: For example c = 3=4: 4  2. a. Compute all local min, max h (x; y; z) = x2 + 2y 2 + 5z 2 2xy 4yz 2z hx = 2x 2y = 0 hy = 4y 2x 4z = 0 hz = 10z (4y + 2) = 0 Only solution is Q = (2; 2; 1) : 2x 2y = 0 4y 2x 4z = 0 10z (4y + 2) = 0 0 1 2 2 0 Hessian is @ 2 4 4 A : Principle leading minors are 2,12,8. Since 0 4 10 all are positive, Hessian is positive de…nite and the function is convex. So Q = (2; 2; 1) must be the global minimum. 5  2b. Show that P = (0; 0) is a local minimum of f (x; y) = x2 2xy 2 + y 4 y5; on any line x = at; y = bt through P (a; b not both 0). Note that f (at; bt) = g (t) = a2 t2 2ab2 t3 b5 t5 + b4 t4 g 0 (t) = 2a2 t 6ab2 t2 5b5 t4 + 4b4 t3 g 00 (t) = 2a2 12ab2 t 20b5 t3 + 12b4 t2 So if a 6= 0; g 00 (0) = 2a2 > 0 and we have a local minimum. If a = 0; then since b 6= 0 we have g (t) = f (0; bt) = b4 t4 b5 t5 = b4 t4 (1 bt) : Since g (0) = 0 and for t close to 0 we have 1 bt > 0; 0 is a local minimum of g: 2c. If P = (0; 0) a local minimum of f (x; y)? No. Consider f (x; y) on the curve x = y 2 ; where f (x; y) = f y 2 ; y = 2 = y2 2 y2 y2 + y4 y5 = y5 < 0 6  3. Identify all local minimum points of the function f (x; y) = x3 + x2 + 2y 2 xy + y fx = 3x2 + 2x y = 0; fy = 4y x + 1 = 0 6x + 2 1 , Hessian is 1 4 3x2 + 2x y = 0 4y x + 1 = 0 1 5 1 1 6 ( 1=4) + 2 1 Solutions P = 4 ; 16 ,Q=( 3 ; 3 ): At P; H = = 1 4 1 2 1 is positive de…nite, so P is a local minimum. At Q; 1 4 6 ( 1=3) + 2 1 H = 1 4 0 1 = ; 1 4 which is not positive semide…nite so Q is not a local minimum. Clearly as x goes to 1 the function also goes to 1 so there can be no global minimum. 7  1 2 4. Maximize f (x; y) = 8x + 2y x2 2 y , Gradient is for (x; y) 2 R2 : The …rst order conditions are fx = 8 2x = 0 fy = 2 y = 0; with unique solution P = (4; 2) : 2 0 Hessian is is negative de…nite, so the function is concave 0 1 and has a unique global maximum, which must be at P: 8  20 -4 -4 -2 -2 0 0 0 2 y -20 x2 4 4 z -40 -60 -80 1 2 Plot of 8x + 2y x2 2 y 9