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ϵ2 - Le Cermics - ENPC

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30 Chapitre 2 : Analyse numérique dans un cas simplié<br />

Proof. Without loss of generality, we assume that the diagonal elements of H i<br />

are strictly separated :<br />

ɛ i1 < ɛ i2 < . . . < ɛ in<br />

for i = 1, 2. If some of the diagonal elements were equal, then we could establish the<br />

corollary by considering a sequence of perturbed problems and letting the perturbation<br />

go to zero. By the rst-order optimality conditions (2.8) and by the diagonal<br />

structure of the H i , we have<br />

(ɛ 1j − λ 1 )y 1j = µy 2j and (ɛ 2j − λ 2 )y 2j = µy 1j .<br />

We combine these equations to obtain<br />

]<br />

]<br />

[(ɛ 1j − λ 1 )(ɛ 2j − λ 2 ) − µ 2 y 1j = 0 =<br />

[(ɛ 1j − λ 1 )(ɛ 2j − λ 2 ) − µ 2 y 2j . (2.36)<br />

By Corollary 1, the multipliers λ 1 and λ 2 satisfy λ i ∈ [ɛ i1 , ɛ i2 ]. Hence, the coecient<br />

of y 1j and y 2j in (2.36) are strictly increasing functions of j ∈ [2, n]. It<br />

follows that these coecients can vanish for at most one j ∈ [2, n] and possibly for<br />

j = 1. When the coecients of y 1j and y 2j do not vanish in (2.36), we must have<br />

y 1j = y 2j = 0. In summary, at the global optimum, all the components of y ij vanish<br />

except possibly y 11 , y 21 , y 1j , and y 2j for some j ∈ [2, n]. We focus on the case j = 2<br />

since j > 2 leads to a larger cost.<br />

Dene x 2 1j = v j and x 2 2j = w j for j = 1, 2. The optimization problem (2.1) with<br />

H i diagonal and x ij = 0 for j > 2 reduces to<br />

min v 1 ɛ 11 + v 2 ɛ 12 + w 1 ɛ 21 + w 2 ɛ 22<br />

subject to v 1 + v 2 = 1 = w 1 + w 2 ,<br />

v 1 w 1 = v 2 w 2 , v 1 , v 2 , w 1 , w 2 ≥ 0.<br />

The equation v 1 w 1 = v 2 w 2 is the orthogonality condition x 11 x 21 = −x 12 x 22 squared.<br />

We substitute v 1 = 1 − v 2 and w 1 = 1 − w 2 to reduce the optimization problem to<br />

min ɛ 11 + ɛ 21 + v 2 (ɛ 12 − ɛ 11 ) + w 2 (ɛ 22 − ɛ 21 ) (2.37)<br />

subject to v 2 + w 2 = 1, v 2 ≥ 0, w 2 ≥ 0.<br />

We substitute v 2 = 1−w 2 in the objective function to further reduce the optimization<br />

problem to<br />

min ɛ 21 + ɛ 12 + w 2 (ɛ 11 + ɛ 22 − ɛ 21 − ɛ 12 )<br />

subject to 0 ≤ w 2 ≤ 1.<br />

Since the cost function is linear in w 2 , the minimum is achieved at either w 2 = 0<br />

(w 1 = 1, v 1 = 0, v 2 = 1) with objective function value ɛ 21 + ɛ 12 or w 1 = 1 (w 2 = 0,<br />

v 1 = 1, v 2 = 0) with objective function value ɛ 11 + ɛ 22 .

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