we have E Y x t ( ( xl)) 1 ( xl), e a in I( Λ ) are as follows:
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1 APPENDICES Aendx : the roof of Equaton (6 For j m n we have Smary from Equaton ( note that j '( ( ( j E Y x t ( ( x ( x a V ( ( x a ( ( x ( x b V ( ( x b V x e d ( abx ( ( x e a a bx ( x xe b a bx By usng the best asymtotc normaty (BAN roerty we have ˆ ~ N I ( Λ Λ Λ where the eements I j n I( Λ are as foows: n ( d ( abx I E N e t a n ( d ( abx I E N x e t b n ( d ( abx I I E N xe t ab n I I E 0 a n I I E 0 b By the saddeont aroxmaton (Jørgensen 997 t can be shown that c( y t ~ j j d d ( tj ( yj
2 Therefore I E E n n m n c( yj t j Nm j By the deta s method we have ˆ x h I Λ h T AVar( q = ( ( ( f ( q ˆ ( f( q N m ˆ qq a( d bd x h e e x t h bd x uxv e xv xu tutv u v uv Aendx : the roof of Resut When { x } are refxed the agrange functon of the constraned-otmzaton robem stated n Equatons (7-8 wth = can be exressed as foows: z x zu zv ( xv xu u v uv M (A where b d x z = te ( are caed KKT muters et zu zv ( xv xu u v uv A x z x z z x z Then the KKT condtons are: M A 0 (A M 0 (A 0 (A4 0 (A5 0 (A6
3 Eght combnatons satsfyng Equaton (A6 s aso nown as the comementary sacness condtons These combnatons can be further cassfed nto four grous as shown n Tabe A Tabe A: The grous of condtons Grou Grou >0 0 Grou 0 >0 0 0 > >0 >0 0 0 Grou >0 0 >0 0 0 >0 > Grou 4 >0 >0 > (the symbo denotes a non-zero vaue Among these combnatons f a are non-zeros then 0 whch contradcts to Smary f ony one of the vaues of s equa to 0 then j 0 j Ths stuaton reduces the robem to a snge-eve robem whch s not an ADT Therefore a the combnatons for { } n Grous 4 are nfeasbe Hence ony the combnatons n Grou Grou are needed to be consdered In Grou note that f 0 then from Equaton (A we have A A A
4 Hence the soutons { } for the above smutaneous equatons together wth Equaton (A are: z z x x ( x x ( ( ( zz xx xx z zxx x x z zxx x x z z x x ( x x ( ( ( zz xx x x z zxx x x z zxx x x z z x x ( x x ( ( ( zz xx xx z zxx x x z zxx x x Under the condton x x x we have 0 0 Obvousy t w contradct to 0 Therefore t s mossbe to have a non-trva otmum aocaton under In Grou note that t mes that the test unts ony need to aocate to any two out of three stress eves Ths s exacty the same as the otmum aocaton for Therefore the otmum aocaton may beong to one of the foowng three ossbe aocatons: ( 0 ( 0 or (0 For the case of ( 0 substtute ( (( r r ( r ( (00 nto the constrants stated n Equaton (A5 then sove as foows: z z x x g g g g g g zx zx Note that ( g g g 0 Therefore to ensure 0 the condton that
5 g g g s requred Smary g g g g g g are the condtons for ( 0 whe g g g s the condton for (0 Hence we comete the roof of Resut Aendx : the roof of Resut When d t t t then bdx bd x bd x e e e Hence g g g t t ( x x e t t ( x x e t t ( x x e b d ( x x b d ( x x b d ( x x t t e ( x x x x t t ( x x e b d ( x x b d ( x x t e ( x x ( t e t e b d x b d x b d x 0 Ths mes that g g s aways ess than g From Resut t s mossbe to have the settng of ( 0 for d Smary when d t s mossbe to have the settng of (0 Aendx 4: the roof of Resut 4 For the case ( of Resut 4 snce xh H the otmum soutons of x can be obtaned by the foowng equatons: G( x ( x ( x 0 (A7 G( x x ( x ( x 0 (A8 From Equaton (A7 we have ( r where H
6 r x ( t t ex{ b( d ( x / } H H Settng bd ( substtutng nto Equaton (A8 then we have x x e t +e ( ( x (e e t x 0 (A9 t t ( x H From Equaton (A9 then we have ether e x ( ( x t t (A0 ( H or e ( x t x (A By settng u x t ( 0 then Equaton (A0 H t can be rewrtten as e u u (A By the ambert W functon the souton for u n Equaton (A can be exressed as foows: W e u W e t t b( d Therefore f d W e t b then x u W b( d H ( e t 0 Smary another souton for x can be obtaned from Equaton (A: (A x b( d W e t t b( d b ( d H (A4 Note that ony x n Equatons (A s the otmum souton due to the fact that
7 when evauatng the Hessan matrx of G( x at ( x ( x the determnant of the Hessan matrx s t t t H 5 t e t 0 6t W e b( d W b( d b ( d e W e b( d W e f d W e t b Note that f d W e t b G( x we have 5 x 0 Therefore the otmum owest stress s x 0 The case ( of Resut 4 can be roved smary
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