Convex Optimization
- 4. Convex Optimization Problems
- Prof. Ying Cui
Department of Electrical Engineering Shanghai Jiao Tong University
2018
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Convex Optimization 4. Convex Optimization Problems Prof. Ying Cui - - PowerPoint PPT Presentation
Convex Optimization 4. Convex Optimization Problems Prof. Ying Cui Department of Electrical Engineering Shanghai Jiao Tong University 2018 SJTU Ying Cui 1 / 64 Outline Optimization problems Convex optimization Linear optimization
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i=0 domfi with m sublevel sets
i x = bi} (all convex)
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−∇f0(x) X x Figure 4.2 Geometric interpretation of the optimality condition (4.21). The feasible set X is shown shaded. Some level curves of f0 are shown as dashed lines. The point x is optimal: −∇f0(x) defines a supporting hyperplane (shown as a solid line) to X at x.
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(x, f(x)) Figure 4.3 A quasiconvex function f on R, with a locally optimal point x that is not globally optimal. This example shows that the simple optimality condition f ′(x) = 0, valid for convex functions, does not hold for quasiconvex functions.
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P x⋆ −c Figure 4.4 Geometric interpretation of an LP. The feasible set P, which is a polyhedron, is shaded. The objective cT x is linear, so its level curves are hyperplanes orthogonal to c (shown as dashed lines). The point x⋆ is
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xcheb xcheb Figure 8.5 Chebyshev center of a polyhedron C, in the Euclidean norm. The center xcheb is the deepest point inside C, in the sense that it is farthest from the exterior, or complement, of C. The center xcheb is also the center of the largest Euclidean ball (shown lightly shaded) that lies inside C.
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P x⋆ −∇f0(x⋆) Figure 4.5 Geometric illustration of QP. The feasible set P, which is a poly- hedron, is shown shaded. The contour lines of the objective function, which is convex quadratic, are shown as dashed curves. The point x⋆ is optimal.
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i x is Gaussian r.v. with mean ¯
i x and variance xTΣix:
i x ≤ bi) = Φ
i x
i
−∞ e−t2/2dt is CDF of N(0, 1)
x
i x ≤ bi) ≥ η,
x
i x + Φ−1(η)Σ1/2 i
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k y+bk
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F segment 4 segment 3 segment 2 segment 1 Figure 4.6 Segmented cantilever beam with 4 segments. Each segment has unit length and a rectangular profile. A vertical force F is applied at the right end of the beam.
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n
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pkgk K
i=1,i=k pigi+nk
K
i=1,i=k pigi+nk
K
i=1,i=k pigi + nk
K
i=1,i=k pigi + nk
k=1,...,K rk(p) ≈
k=1,...,K log2 pkgk K
i=1,i=k pigi+nk
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O f0(x⋆) Figure 4.7 The set O of achievable values for a vector optimization with
+, is shown shaded. In this case,
the point labeled f0(x⋆) is the optimal value of the problem, and x⋆ is an
achievable value f0(y), and is better than or equal to f0(y). (Here, ‘better than or equal to’ means ‘is below and to the left of’.) The lightly shaded region is f0(x⋆)+K, which is the set of all z ∈ R2 corresponding to objective values worse than (or equal to) f0(x⋆). O f0(xpo) Figure 4.8 The set O of achievable values for a vector optimization problem with objective values in R2, with cone K = R2
+, is shown shaded. This
problem does not have an optimal point or value, but it does have a set of Pareto optimal points, whose corresponding values are shown as the dark- ened curve on the lower left boundary of O. The point labeled f0(xpo) is a Pareto optimal value, and xpo is a Pareto optimal point. The lightly shaded region is f0(xpo) − K, which is the set of all z ∈ R2 corresponding to objective values better than (or equal to) f0(xpo).
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O f0(x1) λ1 f0(x2) λ2 f0(x3) Figure 4.9 Scalarization. The set O of achievable values for a vector opti- mization problem with cone K = R2
+. Three Pareto optimal values f0(x1),
f0(x2), f0(x3) are shown. The first two values can be obtained by scalar- ization: f0(x1) minimizes λT
1 u over all u ∈ O and f0(x2) minimizes λT 2 u,
where λ1, λ2 ≻ 0. The value f0(x3) is Pareto optimal, but cannot be found by scalarization.
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F1(x) = Ax − b2
2
F2(x) = x2
2
5 10 15 5 10 15 Figure 4.11 Optimal trade-off curve for a regularized least-squares problem. The shaded set is the set of achievable values (Ax−b2
2, x2 2). The optimal
trade-off curve, shown darker, is the lower left part of the boundary.
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mean return 0% 10% 20% 0% 5% 10% 15% standard deviation of return allocation x(1) x(2) x(3) x(4) 0% 10% 20% 0.5 1 Figure 4.12 Top. Optimal risk-return trade-off curve for a simple portfolio
The lefthand endpoint corresponds to putting all resources in the risk-free asset, and so has zero standard deviation. The righthand endpoint corresponds to putting all resources in asset 1, which has highest mean return. Bottom. Corresponding optimal allocations.
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