Linear Program with finite optimal value has strictly complemenetary solution
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In my lecture, the following statement was given without any proof:
Given a primal-dual linear problem
(P) $${min~ c'x mid Ax=b, x geq 0}$$
(D) $${max~ b'y mid A'y+s=c, s geq 0},$$
it holds that:
If (P), (D) has a finite optimal value, then there exists a strictly complementary solution $(x^*,y^*,s^*)$, i.e., $x_i^* s_i^*=0 forall i$ and $x_i^* + s_i^* >0 forall i$.
Do I miss something here that this is actually obvious? I am grateful for any comments or hints!
Thank you!
linear-programming karush-kuhn-tucker
asked Dec 5 '18 at 10:59
Andreas MuellerAndreas Mueller
11
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add a comment |
$begingroup$
In my lecture, the following statement was given without any proof:
Given a primal-dual linear problem
(P) $${min~ c'x mid Ax=b, x geq 0}$$
(D) $${max~ b'y mid A'y+s=c, s geq 0},$$
it holds that:
If (P), (D) has a finite optimal value, then there exists a strictly complementary solution $(x^*,y^*,s^*)$, i.e., $x_i^* s_i^*=0 forall i$ and $x_i^* + s_i^* >0 forall i$.
Do I miss something here that this is actually obvious? I am grateful for any comments or hints!
Thank you!
linear-programming karush-kuhn-tucker
asked Dec 5 '18 at 10:59
Andreas MuellerAndreas Mueller
11
$endgroup$
$begingroup$
Intuitively, if $x_i=0$, the $i^{th}$ constraint in the dual is not binding (so $s_i>0$).
$endgroup$
– LinAlg
Dec 5 '18 at 15:29
add a comment |
$begingroup$
In my lecture, the following statement was given without any proof:
Given a primal-dual linear problem
(P) $${min~ c'x mid Ax=b, x geq 0}$$
(D) $${max~ b'y mid A'y+s=c, s geq 0},$$
it holds that:
If (P), (D) has a finite optimal value, then there exists a strictly complementary solution $(x^*,y^*,s^*)$, i.e., $x_i^* s_i^*=0 forall i$ and $x_i^* + s_i^* >0 forall i$.
Do I miss something here that this is actually obvious? I am grateful for any comments or hints!
Thank you!
linear-programming karush-kuhn-tucker
asked Dec 5 '18 at 10:59
Andreas MuellerAndreas Mueller
11
$endgroup$
In my lecture, the following statement was given without any proof:
Given a primal-dual linear problem
(P) $${min~ c'x mid Ax=b, x geq 0}$$
(D) $${max~ b'y mid A'y+s=c, s geq 0},$$
it holds that:
If (P), (D) has a finite optimal value, then there exists a strictly complementary solution $(x^*,y^*,s^*)$, i.e., $x_i^* s_i^*=0 forall i$ and $x_i^* + s_i^* >0 forall i$.
Do I miss something here that this is actually obvious? I am grateful for any comments or hints!
Thank you!
linear-programming karush-kuhn-tucker
linear-programming karush-kuhn-tucker
asked Dec 5 '18 at 10:59
Andreas MuellerAndreas Mueller
11
asked Dec 5 '18 at 10:59
Andreas MuellerAndreas Mueller
11
asked Dec 5 '18 at 10:59
Andreas MuellerAndreas Mueller
11
asked Dec 5 '18 at 10:59
Andreas MuellerAndreas Mueller
11
asked Dec 5 '18 at 10:59
Andreas MuellerAndreas Mueller
11
11
$begingroup$
Intuitively, if $x_i=0$, the $i^{th}$ constraint in the dual is not binding (so $s_i>0$).
$endgroup$
– LinAlg
Dec 5 '18 at 15:29
add a comment |
$begingroup$
Intuitively, if $x_i=0$, the $i^{th}$ constraint in the dual is not binding (so $s_i>0$).
$endgroup$
– LinAlg
Dec 5 '18 at 15:29
$begingroup$
Intuitively, if $x_i=0$, the $i^{th}$ constraint in the dual is not binding (so $s_i>0$).
$endgroup$
– LinAlg
Dec 5 '18 at 15:29
$begingroup$
Intuitively, if $x_i=0$, the $i^{th}$ constraint in the dual is not binding (so $s_i>0$).
$endgroup$
– LinAlg
Dec 5 '18 at 15:29
add a comment |
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$begingroup$
Intuitively, if $x_i=0$, the $i^{th}$ constraint in the dual is not binding (so $s_i>0$).
$endgroup$
– LinAlg
Dec 5 '18 at 15:29