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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, 0<B<A, \gamma >0, 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?

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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 $\epsilon$.

Thus we have a contradiction to your conjecture.

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  • $\begingroup$ 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. $\endgroup$ – OR_student Nov 25 '20 at 14:44
  • $\begingroup$ 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! $\endgroup$ – SimonT Nov 25 '20 at 15:23

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