Why is strong duality holds even though the problem is not convex?
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I write the question $5.29$ in Convex Optimization ,and it said
minimize $-3x_1^2+x_2^2+2x_3^2+2(x_1+x_2+x_3)$
subject to $x_1^2+x_2^2+x_3^2=1$
so strong duality holds even though the problem is not convex
I wonder that if we have to prove it is strong duality and not convex,how should we prove it?
convex-analysis convex-optimization duality-theorems
asked Nov 26 at 13:09
shineele
377
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I write the question $5.29$ in Convex Optimization ,and it said
minimize $-3x_1^2+x_2^2+2x_3^2+2(x_1+x_2+x_3)$
subject to $x_1^2+x_2^2+x_3^2=1$
so strong duality holds even though the problem is not convex
I wonder that if we have to prove it is strong duality and not convex,how should we prove it?
convex-analysis convex-optimization duality-theorems
asked Nov 26 at 13:09
shineele
377
Perhaps you can read page 229 of the book (which is referred to from the question).
– LinAlg
Nov 26 at 14:28
The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
– A.Γ.
Dec 7 at 21:56
add a comment |
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up vote
0
down vote
favorite
I write the question $5.29$ in Convex Optimization ,and it said
minimize $-3x_1^2+x_2^2+2x_3^2+2(x_1+x_2+x_3)$
subject to $x_1^2+x_2^2+x_3^2=1$
so strong duality holds even though the problem is not convex
I wonder that if we have to prove it is strong duality and not convex,how should we prove it?
convex-analysis convex-optimization duality-theorems
asked Nov 26 at 13:09
shineele
377
I write the question $5.29$ in Convex Optimization ,and it said
minimize $-3x_1^2+x_2^2+2x_3^2+2(x_1+x_2+x_3)$
subject to $x_1^2+x_2^2+x_3^2=1$
so strong duality holds even though the problem is not convex
I wonder that if we have to prove it is strong duality and not convex,how should we prove it?
convex-analysis convex-optimization duality-theorems
convex-analysis convex-optimization duality-theorems
asked Nov 26 at 13:09
shineele
377
asked Nov 26 at 13:09
shineele
377
asked Nov 26 at 13:09
shineele
377
asked Nov 26 at 13:09
shineele
377
asked Nov 26 at 13:09
shineele
377
377
Perhaps you can read page 229 of the book (which is referred to from the question).
– LinAlg
Nov 26 at 14:28
The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
– A.Γ.
Dec 7 at 21:56
add a comment |
Perhaps you can read page 229 of the book (which is referred to from the question).
– LinAlg
Nov 26 at 14:28
The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
– A.Γ.
Dec 7 at 21:56
Perhaps you can read page 229 of the book (which is referred to from the question).
– LinAlg
Nov 26 at 14:28
Perhaps you can read page 229 of the book (which is referred to from the question).
– LinAlg
Nov 26 at 14:28
The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
– A.Γ.
Dec 7 at 21:56
The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
– A.Γ.
Dec 7 at 21:56
add a comment |
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Perhaps you can read page 229 of the book (which is referred to from the question).
– LinAlg
Nov 26 at 14:28
The problem actually becomes convex if we substitute $x_1^2=1-x_2^2-x_3^2$ and $x_1=-sqrt{1-x_2^2-x_3^2}$ into the objective function subject to $x_2^2+x_3^2le 1$.
– A.Γ.
Dec 7 at 21:56