I have the following optimization problem:

\begin{alignat}2\min &\quad A(t)\cdot x(t)-B(t)\cdot y(t)+C(t)\cdot z(t)-D(t)\cdot k(t)\\\text{s.t.}&\quad z(t)+z_1(t)-y(t)-y_1(t)+x(t) = k(t);& \lambda(t)\\&\quad0 \le x(t) \le \bar{X};& \mu^1(t),\mu^2(t)\\&\quad0 \le k(t) \le \bar{K};& \mu^3(t),\mu^4(t)\\&\quad0 \le y(t) \le \bar{Y};& \mu^5(t),\mu^6(t)\\&\quad0 \le z(t) \le \bar{Z};& \mu^7(t),\mu^8(t)\\&\quad0 \le e(t) \le \bar{E};& \mu^9(t),\mu^{10}(t)\\&\quad e(t) = e(t-1)-(z+z_1)+\beta(y+y_1);&\mu^{11}(t)\\&\quad e(1) = 0;&\mu^{12}(t)\end{alignat}

The dual variables are listed against each equation. All the capital letters and letters under the bar are parameters and the rest are variables. Then, I derive the KKT optimality conditions for this problem by taking the Lagrangian. When I take the derivative of the Lagrangian w.r.t $z_1$ and $y_1$, I get the following primal feasibility conditions.

\begin{alignat}2-\lambda(t)+\mu^{11}(t) &= 0 \tag1\\\lambda(t)-\beta\cdot\mu^{11}(t) &= 0 \tag2\end{alignat}

where, $\beta = 0.85$. My question is - what does it mean to have one dual variable ($\lambda(t)$ in this case) equating to two different values as shown in $(1)$ and $(2)$?

  • 1
    $\begingroup$ Are $z_1(t)$ and $y_1(t)$ unbounded free variables? $\endgroup$
    – prubin
    Nov 25, 2019 at 19:07
  • $\begingroup$ Sorry, I forgot to include, they are non-negative variables. $\endgroup$
    – S_Scouse
    Nov 25, 2019 at 19:19
  • $\begingroup$ I interpret what you have here to mean that the optimal point has mu11 == 0. $\endgroup$
    – Brannon
    Oct 27, 2021 at 16:59


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