# KKT conditions validation- one dual variable equating to two values

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 $$(2)$$?

• Are $z_1(t)$ and $y_1(t)$ unbounded free variables? Nov 25 '19 at 19:07
• Sorry, I forgot to include, they are non-negative variables. Nov 25 '19 at 19:19
• I interpret what you have here to mean that the optimal point has mu11 == 0. Oct 27 at 16:59