Consider the Primal-Dual problem,

(P)min $c^Tx$

s.t. Ax = b, x $\geq 0$

(D) max $b^Ty$

s.t. $A^Ty + s = c$, s $\geq 0$

The log-barrier function for (P) is :

min $c^Tx - \mu \sum_{i=1}^n ln(x_i)$

s.t. Ax = b, x > 0

How to prove that if $\mu > \mu^{\prime}$, then $c^Tx(\mu) > c^Tx(\mu^{\prime})$, where $x(\mu)$ is the optimal solution of log-barrier function with parameter $\mu$.

  • 1
    $\begingroup$ can you prove it by contradiction? $\endgroup$ – LinAlg Nov 10 '19 at 22:40
  • $\begingroup$ I think constradiction is a little difficult. $\endgroup$ – Icy Nov 12 '19 at 6:31

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