# How to prove this convex-optimization problem?

I am struggling with the following optimization problems.

Problem 1

\begin{align}\max_{\alpha, s_1, s_2}&\quad s_1 + s_2 - \gamma (s_1 (K_1 +c_1 + s_1) + s_2 (K_2+ c_2 + s_2) + 2\alpha K) +C\\\text{s.t.}&\quad s_1 \geq 0, s_2 \geq 0, \alpha \geq 0, \alpha \geq A-s_1 - \beta s_2\end{align}

Problem 2

\begin{align}\max_{\alpha, s_1, s_2}&\quad s_1 + s_2 - \gamma (s_1 (K_1 + s_1) + s_2 (K_2 + s_2) + \alpha K)\\\text{s.t.}&\quad s_1 \geq 0, s_2 \geq 0, \alpha \geq 0, B-s_1 - \beta s_2 \leq \alpha \leq A-s_1 - \beta s_2\end{align}

where $$K_1 >0, K_2 >0, C>0, 0<\beta<1, 00, c_1, c_2 >0$$ are constants.

If for a given $$\gamma = \gamma'$$, the optimal objective value of Problem 1 is greater than that of Problem 2, is the optimal objective value of Problem 1 greater than that of Problem 2 for all $$0< \gamma < \gamma'$$?

I can prove this when the constraints $$s_1, s_2, \alpha \geq 0$$ do not exist. I proved it by determining a closed form solution for each problem and just compared the two (derivative with respect to $$\gamma$$ yields lower value for Problem 1 for all $$\gamma >0$$). How can this be solved when the non-negativity constraints are introduced?

If I understood everything correctly this should be false. The following is a counter example.

Let $$\epsilon >0$$ be small. Choose $$K_1=K_2=2$$, $$c_1 = c_2 = 3$$, $$K=\epsilon$$, $$B=\epsilon, A=2\epsilon$$, $$\beta = \epsilon$$ and $$C = 4\epsilon^2$$.

First consider $$\gamma=\gamma'=1$$:

Then the first objective becomes: $$-(s_1^2+4s_1+s_2^2+4s_2+2\epsilon\alpha) + 4\epsilon^2$$ This is optimized by $$s_1=s_2=0$$ and $$\alpha=A=2\epsilon$$ giving an objective of $$0$$.

Then the second objective we find: $$-(s_1^2+s_1+s_2^2+s_2+\epsilon\alpha)$$ This is optimized by $$s_1=s_2=0$$ and $$\alpha=B=\epsilon$$ giving an objective of $$-\epsilon^2$$.

Thus the optimal solution for the first problem is bigger than the one for the second.

Now consider $$\gamma=\frac{1}{4}<\gamma'$$:

Now the first objective becomes: $$-\frac{1}{4}(s_1^2+s_1+s_2^2+s_2+2\epsilon\alpha) + 4\epsilon^2$$ This is again optimized by $$s_1=s_2=0$$ and $$\alpha=A=2\epsilon$$ giving an objective of $$3\epsilon^2$$.

And for the second objective we find: $$\frac{1}{2}(s_1+s_2)-\frac{1}{4}(s_1^2+s_2^2+\epsilon\alpha))$$ Here we consider the (not necessarily optimal) solution $$s_1=0, s_2=1$$ and $$\alpha=0$$ giving an objective of $$\frac{1}{4}>3\epsilon^2$$ for sufficient $$\epsilon$$.

• First of all, thank you so much for your comment. But in Problem 2, we have the constraint $\alpha \leq A - s_1 - \beta s_2$. If we use $A = 2 \epsilon$, $(\alpha, s_1, s_2) = (0, 1, 1)$ is infeasible. Nov 25 '20 at 14:44
• That's a good point, but we should be able to fix it by adjusting $\beta$ accordingly and using $s_1=0, s_2=1$. I have adjusted my answer! Nov 25 '20 at 15:23