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# 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 con±ict 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 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 defne 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 suﬃcient 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 )= x 1 +2 x 2 ( x 2 1 + x 2 2 1) To fnd the infmum, solve x L =0For 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 .P lugthe s ein to 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 fnd the minimum oF a continuously di±erentiable Function f : R R given that the minimizer x Falls in some interval [ a , b ]suchthat 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 fnd 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 + xr 2 3. Repeat: 3.1 If f ° ( x m )=0or | x l x r | return x = x m 3.2 Otherwise, if f ° ( x l ) f ° ( x m ) > 0set x l = x m ;e lseset 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 14 / 45
Bisection Method Example 15 / 45

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Bisection Method Example 16 / 45
Bisection Method Example 17 / 45

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Bisection Method Example 18 / 45
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slide8_handout-1 - MS&E 211 Problem Session(Week 8 Allie...

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