How to maximize an expression based on a piecewised function with respect to inequality constraints? Background
I have the input/output transfer characteristics of an inverter as shown in the graph below:

I want to define the static discipline of this inverter, that is define the boundaries a sender can use to produce a logical "0" or a logical "1" and the boundaries a receiver can expect to interpret the signal as a logical "0" and a logical "1". I also want to maximize the noise margin in order to reduce the sensibility of the inverter to the noise.
Mathematical issue
I have four dependents variables $V_{IL}$, $V_{OL}$, $V_{IH}$ and $V_{OH}$ (who correspond to the boundaries I mentioned above) and a piecewised function $f$ (who corresponds to the transfer characteristics of the inverter) defined as below:

*

*$f(x) = \begin{cases} 
      5-0.5x & 0 \leq x\leq 1 \\
      \frac{35-8x}{6} & 1\leq x\leq 4 \\
      2.5 - 0.5x & 4\leq x \leq 5
   \end{cases}$ (P1)

*$V_{IL} - V_{OL} = V_{OH} - V_{IH}$ (P2)

I have to maximize $NM = V_{IL} - V_{OL}$ with respect to these constraints:

*

*$V_{OL} \leq V_{IL} \leq V_{IH} \leq V_{OH}$ (C1)

*$0 \leq f(V_{IH}) \leq V_{OL}$ (C2)

*$V_{OH} \leq f(V_{IL}) \leq 5$ (C3)

My solution
Because I had no idea about the way to solve this, I assumed from the context:

*

*$V_{IL} = 5 - V_{IH}$ (H1)

*$V_{OL} = 5 - V_{OH}$ (H2)

I also arrange (C2) and (C3) like that:

*

*$V_{OL} = f(V_{IH})$ (C2')

*$V_{OH} = f(V_{IL})$ (C3')

From that point, I can write all my variables in function of $V_{IL}$:

*

*$V_{IH} = 5 - V_{IL}$

*$V_{OH} = f(V_{IL})$

*$V_{OL} = 5 - f(V_{IL})$
Then, I have to maximize $NM = V_{IL} - 5 + f(V_{IL})$. Because $0 \leq V_{IL} \leq 2.5$, I imagine two scenarios:
When $0 \leq V_{IL} \leq 1$:
$NM = V_{IL} - 5 + 5 - 0.5 V_{IL} = 0.5 V_{IL}$
If I want to maximize $NM$, $V_{IL}$ should be maximum so $V_{IL} \geq 1$.
When $1 \leq V_{IL} \leq 2.5$:
$NM = V_{IL} - 5 + \frac{35 - 8 V_{IL}}{6} = \frac{5 - 2V_{IL}}{6}$
If I want to maximize $NM$, $V_{IL}$ should be minimum so $V_{IL} \leq 1$.
Answer
I can now conclude that to maximize $NM$, $V_{IL} = 1$. I can now deduce the other variables: $V_{OL} = 0.5$, $V_{IH} = 4$ and $V_{OH} = 4.5$.

My questions are:

*

*is it possible to solve my problem without modifying the constraints (C2) and (C3)?

*is it possible to solve my problem without make the assumptions (H1) and (H2)?

Also, is my solution and my reasoning correct?
 A: I find the answer to my second question: yes, it is possible to solve my problem without making the assumptions (H1) and (H2) because we can deduce them from the premises.
From the definition of $f$, we can deduce:

*

*if $f(x) = y$ and $x \leq y$, then $x \leq 2.5$ and $y \geq 2.5$;

*if $f(x) = y$ and $x \geq y$, then $x \geq 2.5$ and $y \leq 2.5$.

Then, from (C1) and the modified constraints (C2') and (C3'): $0 \leq V_{OL} \leq V_{IL} \leq 2.5 \leq V_{OH} \leq V_{IH} \leq 5$ (C1').
Now, we can rewrite (P2) this way:
$V_{IH} - V_{OL} = V_{OH} - V_{IL} \iff V_{IH} - f(V_{IH}) = f(V_{IL}) - V_{IL}$ (P2')
To deduce $V_{IH} = 5 - V_{IL}$, we have to analyze the four following cases. Keep in mind (C1').
Case 1
$4 \leq V_{IH} \leq 5$ and $0 \leq V_{IL} \leq 1$
From (P1) and (P2'):
$V_{IH}-2.5+0.5V_{IH}=5-0.5V_{IL}-V_{IL} \iff V_{IH}+V_{IL}=5$
Case 2
$2.5 \leq V_{IH} \leq 4$ and $1 \leq V_{IL} \leq 2.5$
From (P1) and (P2'):
$V_{IH}-\frac{35-8V_{IH}}{6}=\frac{35-8V_{IL}}{6}-V_{IL} \iff V_{IH}+V_{IL}=5$
Case 3
$4 \leq V_{IH} \leq 5$ and $1 \leq V_{IL} \leq 2.5$
From (P1) and (P2'):
$V_{IH}-2.5+0.5V_{IH}=\frac{35-8V_{IL}}{6}-V_{IL} \iff 9V_{IH}+14V_{IL}=50$
This is true only for $V_{IH}=4$ and $V_{IL}=1$, so $V_{IH}+V_{IL}=5$.
Case 4
$2.5 \leq V_{IH} \leq 4$ and $0 \leq V_{IL} \leq 1$
From (P1) and (P2'):
$V_{IH}-\frac{35-8V_{IH}}{6}=5-0.5V_{IL}-V_{IL} \iff 14V_{IH}+9V_{IL}=65$
This is true only for $V_{IH}=4$ and $V_{IL}=1$, so $V_{IH}+V_{IL}=5$.

Then, we prove (H1) from the premises and (C2') and (C3'). We can now deduce (H2):
$V_{OH}=V_{IH}+V_{IL}-V_{OL} \iff V_{OH}=5-V_{OL}$.
It's now possible to find the answer of this problem with the solution I already mentioned in my question.
