Solve the following using steepest descent algorithm. Start with x = [1 1 1] and use stopping threshold E= 10-6. (a) Verify that the final solution satisfies the second order necessary conditions for a minimum. (b) Plot the value of the objective function with respect to the number of iterations and (c) Comment on the convergence speed of the algorithm. minimize f(x) = (x₁+ 5)² + (x₂+8)² + (x3 +7)² +2x²x² + 4x²x²

Solve the following using steepest descent algorithm. Start with x = [1 1 1] and use stopping threshold E= 10-6. (a) Verify that the final solution satisfies the second order necessary conditions for a minimum. (b) Plot the value of the objective function with respect to the number of iterations and (c) Comment on the convergence speed of the algorithm. minimize f(x) = (x₁+ 5)² + (x₂+8)² + (x3 +7)² +2x²x² + 4x²x²

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

Solve the following using steepest descent algorithm. Start with x =
threshold E= 10-6.
(a) Verify that the final solution satisfies the second order necessary conditions for a minimum.
(b) Plot the value of the objective function with respect to the number of iterations and
(c) Comment on the convergence speed of the algorithm.
[1 1 1] and use stopping
minimize f(x) = (x1 + 5)² + (x2 +8)² + (x3 +7)²
+2x²x² + 4x²x²

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Theorem 1 (Master Theorem). Let a ≥ 1 and b> 1 be constants, let f(n) be a function f: N→ R+, and let T(n) be defined on the nonnegative integers by the recurrence T(n)= aT(n/b) + f(n). Then T(n) has the following asymptotic bounds: 1. if f(n) = O(nlogs a-E) for some constant e > 0, then T(n) = (nossa), 2. if f(n) = (nlos a) then T(n) = (nloga log n), and 3. if f(n) = N(n¹og a+s) for some e > 0 and if af (n/b) ≤df (n) for some constant d < 1 and all sufficiently large n, then T(n) = (f(n)). Use the Master Theorem above to solve the following recurrences when possible (note that not every blank needs to be filled in in every problem):