quiz5 and ans

# quiz5 and ans - to the problem min x ( x 1-2) 4 + ( x 2 +...

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AMSC/CMSC 660 Quiz 5 , Fall 2006 1a. (5) Verify that the BFGS matrix B ( k +1) = B ( k ) - B ( k ) s ( k ) s ( k ) T B ( k ) s ( k ) T B ( k ) s ( k ) + y ( k ) y ( k ) T y ( k ) T s ( k ) . satis±es the secant condition: B ( k +1) s ( k ) = y ( k ) . 1b. (5) De±ne s ( k ) and y ( k ) . Why are quasi-Newton matrices designed to satisfy the secant condition? Answer: 1a. B ( k +1) s ( k ) = B ( k ) s ( k ) - B ( k ) s ( k ) s ( k ) T B ( k ) s ( k ) T B ( k ) s ( k ) s ( k ) + y ( k ) y ( k ) T y ( k ) T s ( k ) s ( k ) = B ( k ) s ( k ) - B ( k ) s ( k ) s ( k ) T B ( k ) s ( k ) s ( k ) T B ( k ) s ( k ) + y ( k ) y ( k ) T s ( k ) y ( k ) T s ( k ) = B ( k ) s ( k ) - B ( k ) s ( k ) s ( k ) T B ( k ) s ( k ) s ( k ) T B ( k ) s ( k ) + y ( k ) y ( k ) T s ( k ) y ( k ) T s ( k ) = B ( k ) s ( k ) - B ( k ) s ( k ) + y ( k ) = y ( k ) . 1b. For quadratic functions, we have Hs ( k ) = y ( k ) , where s ( k ) = x ( k +1) - x ( k ) (the change in x ) and y ( k ) = g ( k +1) - g ( k ) (the change in gradient). Therefore, we demand the same property from the Quasi-Newton matrix, since it forms a quadratic model for our function. 1

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2. (10) Write a Matlab program to apply 5 iterations of Newton’s method
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Unformatted text preview: to the problem min x ( x 1-2) 4 + ( x 2 + 1) 4-x 2 1 x 2 with a steplength of 1 (i.e, step in the Newton direction without a linesearch) and with an initial starting guess of x = [1 , 2] T . Answer: x = [1,2]; for k=1:5, g = [4*(x(1)-2)^3 - 2*x(1)*x(2); 4*(x(2)+1)^3 - x(1)^2]; H = [12*(x(1)-2)^2-2*x(2), -2*x(1)-2*x(1), 12*(x(2)+1)^2]; p = -H\g; x = x + p; end 2...
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## This homework help was uploaded on 02/05/2008 for the course CMSC 660 taught by Professor Oleary during the Fall '06 term at Maryland.

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quiz5 and ans - to the problem min x ( x 1-2) 4 + ( x 2 +...

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