hw2sol_fa09

# hw2sol_fa09 - UC Berkeley Department of Electrical...

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UC Berkeley Department of Electrical Engineering and Computer Science EECS 227A Nonlinear and Convex Optimization Solutions 2 Fall 2009 Reading: Sections 1.2—1.3 of Nonlinear programming by Bertsekas. ( Note: Be sure to use the scanned PDFs on webpage to obtain the correct numbering of the problems in this assignment.) Solution 2.1 The problem is to perform an unconstrained minimization of the function f ( x,y ) = 3 x 2 + y 4 using various methods. Using the starting point x 0 = (1 , - 2), we have f ( x 0 ) = 19 , f ( x 0 ) = ± 6 x 4 y 3 ² = ± 6 - 32 ² (a) Steepest descent with s = 1, σ = 0 . 1 and β = 0 . 5 takes the form: x 1 = x 0 + β m sd 0 where d 0 = -∇ f ( x 0 ). Here are the results of determining the appropriate m according to the Armijo rule: m β m s f ( x 0 ) - f ( x 0 + β m sd 0 ) - σβ m s f ( x 0 ) T d 0 0 1 -810056 106 1 .5 -38409 53 2 .25 -1277.75 26.5 3 .125 2.8125 13.25 4 .0625 17.828125 6.625 So one iteration of steepest descent requires 5 internal iterations to determine the value of m ; it yields the new point x 1 = (0 . 625 , 0) where f ( x 1 ) = 1 . 17. (b) Steepest descent with s = 1 = 0 . 1 and β = 0 . 1. m β m s f ( x 0 ) - f ( x 0 + β m sd 0 ) - σβ m s f ( x 0 ) T d 0 0 1 -810056 106 1 .1 -16.4464 10.6 Now only 2 internal iterations are required, and yield the new point x 1 = (0 . 4 , 1 . 2) where f ( x 1 ) = 2 . 55. The smaller value of β reduced the amount of time to ﬁnd the stepsize, but yielded a new point x 1 with higher cost. (c) Newton’s method with s = 1 = 0 . 1 and β = 0 . 5. We now use the descent direction d 0 = - ( 2 f ( x 0 )) - 1 f ( x 0 ). Calculations yield d 0 = - ³ 6 0 0 48 ´ - 1 ± 6 32 ² = ± - 1 2 / 3 ² . 1

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m β m s f ( x 0 ) - f ( x 0 + β m sd 0 ) - σβ m s f ( x 0 ) T d 0 0 1 15.8395 2.733 So one iteration of Newton’s method requires only 1 internal iteration to determine m , and yields x 1 = (0 , - 4 / 3) 0 with f ( x 1 ) = 3 . 1605. Although fewer internal iterations were required, the cost of the resulting x 1 is higher. In this case, determining the descent direction (involving the inverse of the Hessian) was straightforward, but this calculation is potentially expensive. Solution 2.2
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## This note was uploaded on 09/03/2011 for the course EE 227A taught by Professor Staff during the Spring '11 term at Berkeley.

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hw2sol_fa09 - UC Berkeley Department of Electrical...

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