MHB How do we use dynamic programming?

[evinda]

Hello! (Wave)

Could you explain to me how we can use dynamic programming in order to solve a non linear programming problem?

What do we do for example if we are given the following problem?

$$\max (y_1^3-11 y_1^2+40 y_1+y_2^3-8y_2^2+21 y_2) \\ y_1+y_2 \leq 5.4 \\ y_1, y_2 \geq 0$$

 

[evinda]

If this is not possible, how else could we solve the problem using a method that is related to optimization?

 

[I like Serena]

Hello! (Wave)

Could you explain to me how we can use dynamic programming in order to solve a non linear programming problem?

What do we do for example if we are given the following problem?

$$\max (y_1^3-11 y_1^2+40 y_1+y_2^3-8y_2^2+21 y_2) \\ y_1+y_2 \leq 5.4 \\ y_1, y_2 \geq 0$$

If this is not possible, how else could we solve the problem using a method that is related to optimization?

Hey evinda! (Smile)

We can't do it with linear programming (the Simplex method), since it's not linear.
But there are many non-linear methods like:
- Graphing it, which is doable since there are only 2 dimensions,
- GRC non-linear, which is what Excel does
- Lagrange multipliers
- Powell's conjugate direction method

 

[evinda]

Hey evinda! (Smile)

We can't do it with linear programming (the Simplex method), since it's not linear.
But there are many non-linear methods like:
- GRC non-linear, which is what Excel does
- Lagrange multipliers
- Powell's conjugate direction method

We have done Lagrange multipliers in class... (Yes)
But how can we apply them in this case? (Thinking)

 

[I like Serena]

We have done Lagrange multipliers in class... (Yes)
But how can we apply them in this case? (Thinking)

There are 4 cases:

1. The objective function takes its maximum strictly inside the feasible region.
2. The maximum is where $y_1+y_2=5.4$.
3. The maximum is where $y_1=0$.
4. The maximum is where $y_2=0$.

If we assume that $y_1+y_2\le 5.4$ is a binding constraint (case 2), then we have:
$$\Lambda =(y_1^3-11 y_1^2+40 y_1+y_2^3-8y_2^2+21 y_2) + \lambda(y_1+y_2 - 5.4)$$
Solve for each of the derivatives being zero... (Thinking)

 

[evinda]

There are 4 cases:

1. The objective function takes its maximum strictly inside the feasible region.
2. The maximum is where $y_1+y_2=5.4$.
3. The maximum is where $y_1=0$.
4. The maximum is where $y_2=0$.

So do we have to check all of the cases to solve the problem? (Thinking)

If we assume that $y_1+y_2\le 5.4$ is a binding constraint (case 2), then we have:
$$\Lambda =(y_1^3-11 y_1^2+40 y_1+y_2^3-8y_2^2+21 y_2) + \lambda(y_1+y_2 - 5.4)$$
Solve for each of the derivatives being zero... (Thinking)

So now we check Case 1? Could you explain to me why we define $\Lambda$ like that?

You mean that we solve $\frac{\partial{\Lambda}}{\partial{y_1}}(y)=0$ and $\frac{\partial{\Lambda}}{\partial{y_2}}(y)=0$ ? (Thinking)

 

[I like Serena]

So do we have to check all of the cases to solve the problem? (Thinking)

From a graph, we can immediately see which case applies.
Otherwise, yes, we would have to check all cases.

So now we check Case 1? Could you explain to me why we define $\Lambda$ like that?

You mean that we solve $\frac{\partial{\Lambda}}{\partial{y_1}}(y)=0$ and $\frac{\partial{\Lambda}}{\partial{y_2}}(y)=0$ ? (Thinking)

My example was to check case 2.

That's just how I've learned Lagrange multipliers. How would you do it? (Wondering)

For case 1, that's indeed what we need to solve (where $\Lambda$ is without the $\lambda$ part).
For case 2 we need additionally that $\frac{\partial{\Lambda}}{\partial{\lambda}}(y_1, y_2; \lambda)=0$. (Nerd)

 

[evinda]

From a graph, we can immediately see which case applies.
Otherwise, yes, we would have to check all cases.

How can we make the graph? For $y_1=0$ the function gets for example $y_2^3-8y_2^2+21 y_2$... (Sweating)

For case 1, that's indeed what we need to solve (where $\Lambda$ is without the $\lambda$ part).
For case 2 we need additionally that $\frac{\partial{\Lambda}}{\partial{\lambda}}(y_1, y_2; \lambda)=0$. (Nerd)

What does $\frac{\partial{\Lambda}}{\partial{\lambda}}(y_1, y_2; \lambda)$ represent? Do we consider $\lambda$ to be a constant? (Thinking)

 

[I like Serena]

How can we make the graph? For $y_1=0$ the function gets for example $y_2^3-8y_2^2+21 y_2$... (Sweating)

Take a look at this contour plot on Wolfram.
Which clue can we get from it? (Wondering)

What does $\frac{\partial{\Lambda}}{\partial{\lambda}}(y_1, y_2; \lambda)$ represent? Do we consider $\lambda$ to be a constant? (Thinking)

No, $\lambda$ is a dummy variable.
If we have a problem of the form $\max f(\mathbf x)$ where $g(\mathbf x)=0$ and $h(\mathbf x)=0$, we can apply the Lagrange multipliers method.
That means that we first define:
$$\Lambda(\mathbf x; \lambda, \mu) = f(\mathbf x) + \lambda g(\mathbf x) + \mu h(\mathbf x)$$
and then solve:
$$\pd \Lambda {x_1} = 0 \\
\vdots \\
\pd \Lambda {x_n} = 0 \\
\pd \Lambda \lambda = 0 \\
\pd \Lambda \mu = 0
$$

 

[evinda]

No, $\lambda$ is a dummy variable.
If we have a problem of the form $\max f(\mathbf x)$ where $g(\mathbf x)=0$ and $h(\mathbf x)=0$, we can apply the Lagrange multipliers method.
That means that we first define:
$$\Lambda(\mathbf x; \lambda, \mu) = f(\mathbf x) + \lambda g(\mathbf x) + \mu h(\mathbf x)$$
and then solve:
$$\pd \Lambda {x_1} = 0 \\
\vdots \\
\pd \Lambda {x_n} = 0 \\
\pd \Lambda \lambda = 0 \\
\pd \Lambda \mu = 0
$$

In this case do we pick $h(x) \equiv 0$ ?

We have $\frac{\partial{\Lambda}}{\partial{y_1}}=3y_1^2-22y_1+40+ \lambda \\ \frac{\partial{\Lambda}}{\partial{y_2}}=3y_2^2-16y_2+21+ \lambda $

Right?

 

[I like Serena]

In this case do we pick $h(x) \equiv 0$ ?

For instance, or we just leave the whole $h(\mathbf x)$ and $\mu$ part out. (Nerd)

We have $\frac{\partial{\Lambda}}{\partial{y_1}}=3y_1^2-22y_1+40+ \lambda \\ \frac{\partial{\Lambda}}{\partial{y_2}}=3y_2^2-16y_2+21+ \lambda $

Right?

And additionally $\pd \Lambda \lambda = y_1+y_2-5.4$.
Hey! That simply is the constraint! (Happy)

 

[evinda]

And additionally $\pd \Lambda \lambda = y_1+y_2-5.4$.
Hey! That simply is the constraint! (Happy)

Oh yes... And what do we get from that? (Thinking)

 

[I like Serena]

Oh yes... And what do we get from that? (Thinking)

That we need to solve the system:
\begin{cases}3y_1^2-22y_1+40+\lambda &= 0 \\
3y_2^2-16y_2+21+\lambda &= 0\\
y_1+y_2-5.4 &= 0
\end{cases}
That is, 3 equations with 3 unknowns.
Btw, we're not really interested in $\lambda$ itself, so it makes sense to eliminate $\lambda$ first. (Mmm)

 

[evinda]

That we need to solve the system:
\begin{cases}3y_1^2-22y_1+40+\lambda &= 0 \\
3y_2^2-16y_2+21+\lambda &= 0\\
y_1+y_2-5.4 &= 0
\end{cases}
That is, 3 equations with 3 unknowns.
Btw, we're not really interested in $\lambda$ itself, so it makes sense to eliminate $\lambda$ first. (Mmm)

Then we get $y_1=3.2$ and $y_2=2.2$. Or am I wrong? (Thinking)

 

[I like Serena]

Then we get $y_1=3.2$ and $y_2=2.2$. Or am I wrong? (Thinking)

That is correct. (Nod)

 

[evinda]

That is correct. (Nod)

Then the function is equal to $66.256$, right?

So now do we check the other cases?