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by: Ms. Hipolito Willms


Ms. Hipolito Willms
Texas A&M
GPA 3.56

Sergiy Butenko

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Sergiy Butenko
Study Guide
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This 4 page Study Guide was uploaded by Ms. Hipolito Willms on Wednesday October 21, 2015. The Study Guide belongs to ISEN 629 at Texas A&M University taught by Sergiy Butenko in Fall. Since its upload, it has received 41 views. For similar materials see /class/226197/isen-629-texas-a-m-university in Industrial Engineering at Texas A&M University.

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Date Created: 10/21/15
ISEN 629 Engineering Optimization Lecture 9 Sergiy Buten k0 Industrial and Systems Engineering Texas AampM University Fall 2007 11n Optimal methods Next we need to make sure that the condition of Lemma 221 is satisfied Assume that we already have Xk such that lt15 2 KW Then by the above lemma we have a2 1 04kfgtltk IMOk mllWJkhlz mflykTVk Jk i1 2 Since fXk 2 fyk f ykTXk 7 yk we get 2 2 fyk72 k llf ykll2 1 akV UkVQ fEUk yk Xk 7M We want to have 15 2 HM 21n Optimal methods From fX 7 fy g f yTX 7 y llx7 yllz we can ensure the inequality 1 WM illf OOll2 2 fgtltk1 by eg taking the gradient step Xk1 yk 7 InfH with hk and using the inequality fy 7 fX 7 f XTy 7X g LllX 7 yll2 with y Xk1 and X yk et us define 04k as fol ows L04 n1 1 040 04 Then 24 TIE and we have a g 2 fgtltk1 1 Mylo0T 77104 yk Xk in Since we are free to choose M We can find it from the equation ak7 Yk1 7 ak kak l Vki Xk yquot 7 Yk l ak l 39 3 Vk7ykgtltk7yk 07 yielding General scheme of optimal method 0 Choose X0 6 Rquot and 70 gt 0 Set v0 X0 1 kth iteration k 2 0 3 Compute ark 6 01 from equation Lari 1 010 04W Set Yk1 1 i 010 WM b Choose 0mm 7k1Xk 7k l 0444 and compute fyk and fyk c Find Xk1 such that Vk rm M 7 21 iifmir 7 17 lkaak yk7akfl k d Set vk1 7 mm mm Optimal methods Theorem 221 The above scheme generates a sequence X1lt k 2 0 such that rm 7 n g mm 7 n llxo 7 we lt71 Where A0 1 and Ak H 1 7 L1 I 0 Proof Choose 0X fXo 7 vollz Then t39X0 153 and we get t39Xk g 45 for any k Therefore we can apply Lemma 221 B 51n Optimal methods To estimate the rate of convergence of fXk k 2 0 we can use the rate of convergence of Ak Lemma 224 lf39yo 2 u then k Akgminlt1igt 4 L nin Optimal methods Theorem 222 Choose 70 L Then our scheme generates a sequence Xlt k 2 0 such that k fXk7f Lminlt em k 2V This means that our scheme is optimal for unconstrained minimization of functions from 5101 u 2 0 HM Xll2 71n Optimal methods Proof We will use a property of f E fi IURquot L TU gt0 f XTy X S EllXJll27 VXVJ E R7 Since for y X tquotX 0 we have L 2 f Xof gillx07x Hence from the previous theorem and lemma MWXo f yg llxo Xllzl rxk7r k g minlt17 Liixocxwz k Lminlt17 llxo 7M2 Next we will show that our scheme is indeed optimal for W Optimal methods Recall the lower complexity bound for 51Rquot Qf71 2 y 4k rx 7r2 lt gt R227 7 R2 k 2 MM 2 exp M71 where Qf Lu R llXo 7Xll Hence if fXk 7 V g e we have M 4k 2 0771 1 M 7 lt gt a 2explt M71gtR76 k7 4 ln ln22lnR For our scheme wk 7 P lt LR2lt17 x1Qfk lt LR2 exp lt7Lgt a 7 xQ f To get an upper bound on k for our method we have LRZexp 7L gte klt xQ lnlln 2lnR m e 2 Etn Optimal methods Thus we have k2 VQquot1linlIn 2ian 4 e 2 and kSQflnlln2lnR 6 1 and the main term In the upper bound x Qf ln 2 Is proportional to the lower boun So the proposed method is indeed optimal for Sii RU D 101n Optimal methods If we use the constantstep gradient iteration to find Xk1 in our general scheme we obtain the following scheme 0 Choose X0 6 Rquot and 70 gt 0 Set v0 X0 1 kth iteration k 2 0 3 Compute 01k 6 01 from equation Lari 1 7 010 0W Set Yk1 7 1 7 arm 0w b Choose BMWk 7k1Xk Yk WM and compute fyk and fyk c Set Xk1 yr 7 ifyk 7 17 1kvkakgyk7aw yk 1 Set Vk1 7 7k Vk 111n Optimal methods Note that 7 7 g 7 i yk 7 W aWWWkik 7k1Xk 7 Vk 7 W 2 W W Xk1 yk fYk7 Vk1 1 i 0407ka 1ka 7 04ktquotykl lli fkl w WWW Yk1Xk 1ka 7 LINMN 7 1 170 39y 17 7 W l afl kyk Wk Tfl Xk flitM Xk alkOk 7 Xk 7 f Ok me m 70 7am W Lmil Xk aikXk1 7 Xk Hence n1 Wak17kl Vk1 Yk2Xk1 Xk1 kXIlt1 Xk 7 7 0w 1 1 170 where W 7 121n Optimal methods Since 2 akL 1 i 040 Mak Yk17 we have k W M 9 w1akw Qk Yk1Qi1L 1 k1quotllt1 7 W11 1 05k 7 05k I Qk T 05k w1ak1L T a kg Note that in 1 0414004 ak17 so we can completely eliminate the sequence yk k 2 0 131n Optimal methods We obtain the following scheme 0 Choose X0 6 Rquot and a0 6 01 Sety0X0 and qu L 1 kth iteration k 2 0 3 Compute fyk and fyk Set 1 Xk1 yr 7 If yr b Compute Dtk1 6 071 from equation a 11 MOW 404k 17m set w akak1 39 YK1 Xk1 WW M um Optimal methods Theorem 223 If in the above scheme 0402E7 my r g min 17 XWXO f yg llxo Xll217 then Where 70 1 Proof The condition 10 2 is equivalent to yo 2 u Therefore the statement of this theorem follows from Theorem 221 and Lemma 224 151n Optimal methods L lfwe choose 10 WML then 70 u 04k xuL k C and we obtain the following scheme 0 Choose yo X0 6 1Rquot 1 kth iteration k 2 0 Xk1 Jk HM L Yk1 Xk1 t CXkl Xkl 1n1n


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