slide8_handout-1

# slide8_handout-1 - MS&E 211 Problem Session(Week 8 Allie...

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MS&E 211 Problem Session (Week 8) Allie Leeper Department of Management Science and Engineering Stanford University November 15, 2013 1 / 45

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Announcements HW 5 due Thursday 11/21 Plan ahead: HW 6 will be assigned right before Thanksgiving break, due Thursday, Dec. 5 I will cover HW 6 during problem session next week And also I will bring snacks Extra credit auctions ongoing Final Wednesday, Dec. 11, 12:15-3:15pm in NVIDIA Auditorium Let us know if you have a conflict or require an exam accomodation 2 / 45
Agenda Nonlinear Duality and the Lagrange Function Bisection Method Golden Section Search Steepest Gradient Descent Newton’s Method for Unconstrained Problems Homework Hints 3 / 45

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Nonlinear Duality and the Lagrange Function Consider a generalized optimization problem: minimize f ( x ) subject to c i ( x ) 0 i = 1 , ..., m d i ( x ) = 0 i = 1 , ..., p The Lagrange Function is: L ( x , λ , y ) = f ( x ) m i =1 λ i c i ( x ) p i =1 y i d i ( x ) Further define φ ( λ , y ) = inf x L ( x , λ , y ) 4 / 45
Nonlinear Duality and the Lagrange Function Then the dual of the general optimization problem is: maximize φ ( λ , y ) subject to λ 0 The necessary conditions for optimality of the general primal and dual are just the KKT conditions. Recall that if the primal problem is additionally a convex optimization problem, then the KKT conditions are also su cient for optimality. 5 / 45

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Nonlinear Dual Example Find the dual of: minimize x 1 + 2 x 2 s.t. x 2 1 + x 2 2 1 6 / 45
Nonlinear Dual Example This gives us: L ( x , λ ) = x 1 + 2 x 2 ( x 2 1 + x 2 2 1) To find the infimum, solve x L = 0 for x 1 and x 2 x L = 1 2 x 1 λ 2 2 x 2 λ = 0 This yields x 1 = 1 / (2 λ ) and x 2 = 1 / λ . Plug these into L to get φ φ ( λ ) = λ + 5 / (4 λ ) 7 / 45

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Nonlinear Dual Example So we get that the dual of: minimize x 1 + 2 x 2 s.t. x 2 1 + x 2 2 1 is: maximize λ + 5 / (4 λ ) s.t. λ 0 8 / 45
Bisection Method We want to find the minimum of a continuously di ff erentiable function f : R R given that the minimizer x falls in some interval [ a , b ] such that f ( a ) f ( b ) < 0 Looking for point where f ( x ) = 0 Search by repeatedly bisecting the search interval At each step, check midpoint and keep half of interval which contains x End when either: we find an x such that f ( x ) = 0, or the length of the search interval | x x | < 9 / 45

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Bisection Method Given: f ( x ) Interval [ a , b ] containing the minimum Tolerance > 0 Algorithm: 1. Let x l = a and x r = b 2. Let x m = x l + x r 2 3. Repeat: 3.1 If f ( x m ) = 0 or | x l x r | < return x = x m 3.2 Otherwise, if f ( x l ) f ( x m ) > 0 set x l = x m ; else set x r = x m 3.3 Set x m = x l + x r 2 10 / 45
Bisection Method Example f ( x ) = 4 x 2 + 6 x 7 a = 10 , b = 10 = . 001 11 / 45

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Bisection Method Example 12 / 45
Bisection Method Example 13 / 45

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Bisection Method Example 15 / 45

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Bisection Method Example 18 / 45
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