Introductory example - Mines Nancy

Problem statement¶

To motivate first definitions, let us consider the following practical shape optimization problem.

A can producer wants to optimize his consumption of raw material.

They formulate one explicit constraint: the volume of the can must be $V_0 = 1$ L.

We assume objects of constant thickness $e$ . Therefore, the quantity of raw material is simply given by

\text{quantity of raw material } = \text{ area of cylinder } \times e

(1)

Since the goal is to optimize (i.e., in this case, minimize) the consumption of raw material, the optimization problems can be formulated in layman’s terms as follows:

Mathematical formulation¶

Expression of the area $A$ (objective function or criterion)

A(r, h) = A_{\text{cylinder}}(r, h) + A_{\text{lids}}(r, h) = 2\pi r h + 2 \pi r^2

(2)

The dimensions $r$ and $h$ are the variables of the optimization problem.

Constraints We have explicit and implicit constraints:

Constant volume $V_0 = 1$ L. This one is explicitly given in the statement of the problem, and reads
$V(r, h) = \pi r^2 h = V_0$
(3)
Non-negativity: the variables $r$ and $h$ must be non-negative owing to their interpretation as physical dimensions
$r\geq 0,\qquad h\geq 0$
(4)

The constrained optimization problem is finally written as

\text{Find } (r^\star, h^\star) \text{ such that } A(r^\star, h^\star) \text{ is minimal and } \begin{cases}r\geq 0 \text{ and } h\geq 0\\ \pi r^2 h = V_0\end{cases}

(5)

Simplification¶

Notice that $h$ and $r$ are related by the constraint $V_0 = \pi r^2 h$ . Therefore, the area $A(r, h)$ can be rewritten as a function of $r$ only. Using

h(r) = \frac{V_0}{\pi r^2} \quad (h\geq 0 \quad \text{OK!})

(6)

and substituting this into the expression for $A(r, h)$ , one obtains

\begin{align*} A(r, h) &= 2\pi r h + 2\pi r^2\\ & = 2 \pi r \frac{V_0}{\pi r^2} + 2 \pi r^2\\ &= \frac{2V_0}{r} + 2\pi r^2\\ &:= \tilde{A}(r) \end{align*}

(7)

The volume constraint is directly taken into account in the new objective function $\tilde{A}(r)$ . We consider the constraint $r > 0$ when searching for the minimum of $\tilde{A}(r)$ over $\mathbb{R}_+^*$ ;

Visualization using Python¶

Let us plot the objective function using basic Python commands.

# import libraries
import numpy as np # for manipulation of arrays
import matplotlib.pyplot as plt # for plotting

# function definitions
def h(r, V0=1000):
    return V0 / (np.pi * r**2)

def A_lids(r):
    return 2 * np.pi * r**2

def A_cylinder(r):
    return 2 * np.pi * r * h(r)

def A(r):
    return A_lids(r) + A_cylinder(r)

# define variables and constraints
V0 = 1000  # volume in cm^3 (=1 L)
r = np.linspace(1, 15, 100)  # vector of 100 values linearly spaced between 1 and 15

# do the plotting
plt.plot(r, A_lids(r), label="lids area", linewidth=2, linestyle='--')
plt.plot(r, A_cylinder(r), label="cylinder area", linewidth=2, linestyle=':')
plt.plot(r, A_lids(r) + A_cylinder(r), label="total area", linewidth=2)
plt.xlabel("radius r [cm]")
plt.ylabel("objective value [cm^2]")
plt.legend()
plt.show()

The plot suggests a minimizer exists, somewhere between $r=4$ cm and $r = 6$ cm. Quite naturally, the minimum area value is obtained when the derivative of $\tilde{A}(r)$ vanishes. Let us do the explicit computation.

Explicit solutions¶

The derivative of the objective function vanishes at its minimum.

\frac{\mathrm{d}\tilde{A}(r)}{\mathrm{d}r} = 4\pi r - \frac{2 V_0}{r^2}

(8)

We solve

\left.\frac{\mathrm{d}\tilde{A}(r)}{\mathrm{d}r}\right|_{r = r^\star} = 0 \iff 4\pi r^\star = \frac{2V_0}{(r^\star)^2} \iff r^\star = \left[\frac{V_0}{2\pi}\right]^{1/3}

(9)

Since $\tilde{A}''(r) = \frac{4V_0}{r^3} + 4\pi > 0$ for all $r > 0$ , $\tilde{A}$ is strictly convex on $\mathbb{R}_+^*$ , so this stationary point is the unique global minimizer.

In summary, the solutions to the optimization problem are

r^\star = \left[\frac{V_0}{2\pi}\right]^{1/3} \qquad \hat{h} = \frac{V_0}{\pi (r^\star)^2}= 2r^\star

(10)

We can also check that our computations are correct using Python, by modifying the previous code.

# optimal value
r_star= (V0/(2*np.pi))**(1/3)

# do the plotting
plt.plot(r, A_lids(r), label="lids area", linewidth=2, linestyle='--')
plt.plot(r, A_cylinder(r), label="cylinder area", linewidth=2, linestyle=':')
plt.plot(r, A_lids(r) + A_cylinder(r), label="total area", linewidth=2)
plt.axvline(r_star, label="optimal radius", linewidth=2, color="black")
plt.axhline(A_lids(r_star) + A_cylinder(r_star), label="optimal area", linewidth=2, color="black", linestyle='--')
plt.xlabel("radius r [cm]")
plt.ylabel("objective value [cm^2]")
plt.legend()
plt.show()

print(f"Optimal value of r minimizing A(r): r_opt = {r_star:.2f} cm")
print(f"Minimal area A(r_opt): A_min = {A_lids(r_star) + A_cylinder(r_star):.2f} cm^2")
print(f"Associated optimal height value h(r_opt): h_opt = {h(r_star):.2f} cm")

Optimal value of r minimizing A(r): r_opt = 5.42 cm
Minimal area A(r_opt): A_min = 553.58 cm^2
Associated optimal height value h(r_opt): h_opt = 10.84 cm

You can play around with this code snippets and see how the solution changes as we modify the value of $V_0$ .

1.1 Introductory example

Problem statement¶

Mathematical formulation¶

Simplification¶

Visualization using Python¶

Explicit solutions¶