Engineering real-world test problems
Collection of real-world multi-objective optimization test problems. More details about
the test problems can be found in
RE-21: Four bar truss design problem
The two objectives of the problem are: \((f_1)\) minimize structural volume,
and \((f_2)\) minimize the joint displacement.
Definition
\[\begin{split}\min \; f_1(x) = & \; L \big( 2x_1 + \sqrt{2} x_2 + \sqrt{x_3} + x_4 \big) \\[2mm]
\min \; f_2(x) = & \; \frac{FL}{E} \Bigg( \frac{2}{x_1} + \frac{2\sqrt{2}}{x_2} - \frac{2\sqrt{2}}{x_3} + \frac{2}{x_4} \Bigg)\end{split}\]
The four variables determine the length of four bars, respectively.
Variable bounds are given as follows:
\[\begin{split}&a \leq x_1 \leq 3a \quad \quad \sqrt{2}a &\leq x_2 \leq 3a \quad \quad
\sqrt{2}a \leq x_3 \leq 3a \quad \quad a \leq x_4 \leq 3a \\
\\
&\text{where: }a = F/\sigma\end{split}\]
Parameters are given as follows:
\[\begin{split}F &= 10 \text{ kN} \quad & \quad
E &= 2 \times 10^5 \text{ kN/cm}^2 \\
L &= 200 \text{ cm} \quad & \quad
\sigma &= 10 \text{ kN/cm}^2\end{split}\]
RE-22 Reinforced concrete beam design problem
The two objectives of the problem are: \((f_1)\) minimize total cost of
concrete and reinforcing steel of the beam, and \((f_2)\) minimize the sum of constraint violations.
Definition
\[\begin{split}\min \; f_1(x) = & \; 29.4x_1 + 0.6x_2x_3\\[2mm]
\min \; f_2(x) = & \; \displaystyle\sum_{i=1}^{2} \max \{ g_i(x), 0 \}\end{split}\]
Constraints:
\[\begin{split}g_1(x) = & \; x_1x_3 - 7.735 \frac{x_1^2}{x_2} - 180 \geq 0 \\[2mm]
g_2(x) = & \; 4 - \frac{x_3}{x_2} \geq 0\end{split}\]
The three variables represent the area of reinforcement \((x_1)\), the width of the beam
\((x_2)\), and the depth of the beam \((x_3)\). \(x_1\) has a pre-defined discrete
value from 0.2 to 15.
Variable bounds are given as follows:
\[0.2 \leq x_1 \leq 15 \quad \quad 0 \leq x_2 \leq 20 \quad \quad
0 \leq x_3 \leq 40\]
RE-23 Pressure vessel design problem
The two objectives of the problem are: \((f_1)\) minimize total cost of a
cylindrical pressure vessel, and \((f_2)\) minimize the sum of constraint violations.
Definition
\[\begin{split}\min \; f_1(x) = & \; 0.6224x_1x_3x_4 + 1.7781x_2x_3^2\\[2mm]
& + 3.1661x_1^2x_4 + 19.84x_1^2x_3\\[2mm]
\min \; f_2(x) = & \; \displaystyle\sum_{i=1}^{3} \max \{ g_i(x), 0 \}\end{split}\]
Constraints:
\[\begin{split}g_1(x) = & \; x_1 - 0.0193x_3 \geq 0 \\[2mm]
g_2(x) = & \; x_2 - 0.00954x_3 \geq 0 \\[2mm]
g_3(x) = & \; \pi x_3^2 x_4 + \frac{4}{3} \pi x_3^3 - 1 \; 296 \; 000 \geq 0\end{split}\]
The four variables represent the thicknesses of the shell \((x_1)\),
the head of a pressure vessel \((x_2)\), the inner radius \((x_3)\), and
the length of the cylindrical section \((x_4)\).
Variable bounds are given as follows:
\[1 \leq x_1 \leq 100 \quad \quad 1 \leq x_2 \leq 100 \quad \quad
10 \leq x_3 \leq 200 \quad \quad 10 \leq x_4 \leq 240\]
RE-24 Hatch cover design problem
The two objectives of the problem are: \((f_1)\) minimize the weight of the hatch cover,
and \((f_2)\) minimize the sum of constraint violations.
Definition
\[\begin{split}\min \; f_1(x) = & \; x_1 + 120 x_2\\[2mm]
\min \; f_2(x) = & \; \displaystyle\sum_{i=1}^{4} \max \{ g_i(x), 0 \}\\[2mm]\end{split}\]
Constraints:
\[\begin{split}g_1(x) = & \; 1.0 - \frac{\sigma_b}{\sigma_{b,max}} \geq 0 \\[2mm]
g_2(x) = & \; 1.0 - \frac{\tau}{\tau_{max}} \geq 0 \\[2mm]
g_3(x) = & \; 1.0 - \frac{\delta}{\delta_{max}} \geq 0 \\[2mm]
g_4(x) = & \; 1.0 - \frac{\sigma_b}{\sigma_k} \geq 0\end{split}\]
The two variables represent the flange thickness \((x_1)\), and
the beam height of the hatch cover \((x_2)\).
Variable bounds are given as follows:
\[0.5 \leq x_1 \leq 4 \quad \quad 4\leq x_2 \leq 40\]
Parameters are given as follows:
\[\begin{split}\sigma_{b,max} &= 700 \text{ kg/cm}^2 \quad & \quad
\tau_{max} &= 450 \text{ kg/cm}^2 \\
\delta_{}max &= 1.5 \text{ cm} \quad & \quad
\sigma_k &= Ex_1^2/100 \text{ kg/cm}^2 \\
\sigma_b &= 4500/(x_1x_2)\text{ kg/cm}^2 \quad & \quad
\tau &= 1800/x_2\text{ kg/cm}^2 \\
\delta &= 56.2 \times 10^4 /(Ex_1x_2^2) \quad & \quad
E &= 700 \; 000 \text{ kg/cm}^2\end{split}\]
RE-25 Coil compression spring design problem
The two objectives of the problem are: \((f_1)\) minimize the volume of spring steel
wire which is used to manufacture the spring, and \((f_2)\) minimize the sum of constraint violations.
Definition
\[\begin{split}\min \; f_1(x) = & \; \frac{\pi^2 x_2 x_3^2 (x_1 + 2)}{4} \\[2mm]
\min \; f_2(x) = & \; \displaystyle\sum_{i=1}^{6} \max \{ g_i(x), 0 \} \\[2mm]
C_f = & \; \frac{4(x_2/x_3) - 1}{4(x_2/x_3) - 4} + \frac{0.615x_3}{x_2} \\[2mm]
K = & \; \frac{Gx_3^4}{8x_1x_2^3} \\[2mm]
\sigma_p = & \; \frac{F_p}{K} \\[2mm]
l_f = & \; \frac{F_{max}}{K} + 1.05(x_1 + 2) x_3\end{split}\]
Constraints:
\[\begin{split}g_1(x) = & \; - \frac{8C_f F_{max} x_2}{\pi x_3^3} + S \geq 0 \\[2mm]
g_2(x) = & \; -l_f + l_{max} \geq 0 \\[2mm]
g_3(x) = & \; -3 + \frac{x_2}{x_3} \geq 0 \\[2mm]
g_4(x) = & \; - \sigma_p + \sigma_{pm} \geq 0 \\[2mm]
g_5(x) = & \; - \sigma_p - \frac{F_{max} - F_p}{K} \\ & - 1.05 (x_1 + 2) x_3 + l_f \geq 0 \\[2mm]
g_6(x) = & \; - \sigma_w + \frac{F_{max} - F_p}{K} \geq 0 \\[2mm]\end{split}\]
The three variables represent the number of of spring coils \((x_1)\),
the outside diameter of the spring \((x_2)\), and the spring wire diameter \((x_3)\).
\(x_3\) has a pre-defined discrete value from 0.009 to 0.5.
Variable bounds are given as follows:
\[1 \leq x_1 \leq 70 \quad \quad 0.6 \leq x_2 \leq 30 \quad \quad 0.009 \leq x_3 \leq 0.5\]
Parameters are given as follows:
\[\begin{split}F_{max} &= 1000 \text{ lb} \quad & \quad
S &= 189 \; 000 \text{ psi} \\
l_{max} &= 14 \text{ inch} \quad & \quad
d_{min} &= 0.2 \text{ inch} \\
D_{max} &= 3 \text{ inch} \quad & \quad
F_p &= 300 \text{ lb} \\
\sigma_{pm} &= 6 \text{ inch} \quad & \quad
\sigma_w &= 1.25 \text{ inch} \\
G &= 11.5 \times 10^6\end{split}\]
RE-31 Two bar truss design problem
The three objectives of the problem are: \((f_1)\) minimize the structural weight,
\((f_2)\) minimize the resultant displacement of joint, and \((f_3)\) minimize
the sum of constraint violations.
Definition
\[\begin{split}\min \; f_1(x) = & \; x_1 \sqrt{16 + x_3^2} + x_2 \sqrt{1 + x_3^2} \\[2mm]
\min \; f_2(x) = & \; \frac{20 \sqrt{16 + x_3^2}}{x_3x_1} \\[2mm]
\min \; f_3(x) = & \; \displaystyle\sum_{i=1}^{3} \max \{ g_i(x), 0 \} \\[2mm]\end{split}\]
Constraints:
\[\begin{split}g_1(x) = & \; 0.1 - f_1(x) \geq 0 \\[2mm]
g_2(x) = & \; 10^5 - f_2(x) \geq 0 \\[2mm]
g_3(x) = & \; 10^5 - \frac{80 \sqrt{1 + x_3^2}}{x_3x_2} \geq 0\end{split}\]
\(x_1\) and \(x_2\) represent the length of the two bars, and \(x_3\)
represents the vertical distance from the second bar.
Variable bounds are given as follows:
\[10^{-5} \leq x_1 \leq 100 \quad \quad 10^{-5} \leq x_2 \leq 100
\quad \quad 1 \leq x_3 \leq 3\]
RE-32 Welded beam design problem
The three objectives of the problem are: minimize the cost \((f_1)\) and end deflection
\((f_2)\) of a welded beam, and \((f_3)\) minimize the sum of constraint violations.
Definition
\[\begin{split}\min \; f_1(x) = & \; 1.10471x_1^2x_2 + 0.04811x_3x_4 (14 + x_2) \\[2mm]
\min \; f_2(x) = & \; \frac{4PL^3}{Ex_4x_3^3} \\[2mm]
\min \; f_3(x) = & \; \displaystyle\sum_{i=1}^{4} \max \{ g_i(x), 0 \} \\[2mm]
\tau(x) = & \; \sqrt{(\tau')^2 + \frac{2\tau'\tau''x_2}{2R} + (\tau'')^2} \\[2mm]
\tau' = & \; \frac{P}{\sqrt{2}x_1x_2} \\[2mm]
\tau'' = & \; \frac{MR}{J} \\[2mm]
M = & \; P \Big( L + \frac{x_2}{2} \Big) \\[2mm]
R = & \; \sqrt{\frac{x_2^2}{4} + \bigg( \frac{x_1 + x_3}{2} \bigg)^2 } \\[2mm]
J = & \; 2 \Bigg( \sqrt{2} x_1x_2 \bigg( \frac{x_2^2}{12} + \Big( \frac{x_1 + x_3}{2} \Big)^2 \bigg) \Bigg) \\[2mm]
\sigma(x) = & \; \frac{6PL}{x_4x_3^2} \\[2mm]
P_C(x) = & \; \frac{4.013E \sqrt{x_3^2x_4^6 / 36}}{L^2} \Bigg( 1 - \frac{x_3}{2L} \sqrt{\frac{E}{4G}} \Bigg)\end{split}\]
Constraints:
\[\begin{split}g_1(x) = & \; \tau_{max} - \tau(x) \geq 0 \\[2mm]
g_2(x) = & \; \sigma_{max} - \sigma(x) \geq 0 \\[2mm]
g_3(x) = & \; x_4 - x_1 \geq 0 \\[2mm]
g_4(x) = & \; P_C(x) - P \geq 0\end{split}\]
The four variables adjust the size of the beam and the variable bounds are given as follows:
\[0.125 \leq x_1 \leq 5 \quad \quad 0.1 \leq x_2 \leq 10 \quad \quad 0.1 \leq x_3 \leq 10
\quad \quad 0.125 \leq x_4 \leq 5\]
Parameters are given as follows:
\[\begin{split}P &= 6000 \text{ lb} \quad & \quad
L &= 14 \text{ in} \\
E &= 30 \times 10^6 \text{ psi} \quad & \quad
G &= 12 \times 10^6 \text{ psi} \\
\tau_{max} &= 13 \; 600 \text{ psi} \quad & \quad
\sigma_{max} &= 30 \; 000 \text{ psi}\end{split}\]
RE-33 Disc brake design problem
The three objectives of the problem are: minimize the the mass of the brake \((f_1)\)
and the minimum stopping time \((f_2)\) of a disc brake, and \((f_3)\)
minimize the sum of constraint violations.
Definition
\[\begin{split}\min \; f_1(x) = & \; 4.9 \times 10^{-5}(x_2^2 - x_1^2)(x_4 - 1) \\[2mm]
\min \; f_2(x) = & \; 9.82 \times 10^6 \bigg(\frac{x_2^2 - x_1^2}{x_3x_4(x_2^3 - x_1^3)} \bigg) \\[2mm]
\min \; f_3(x) = & \; \displaystyle\sum_{i=1}^{4} \max \{ g_i(x), 0 \}\end{split}\]
Constraints:
\[\begin{split}g_1(x) = & \; (x_2 - x_1) - 20 \geq 0 \\[2mm]
g_2(x) = & \; 0.4 - \frac{x_3}{3.14(x_2^2 - x_1^2)} \geq 0 \\[2mm]
g_3(x) = & \; 1 - \frac{2.22 \times 10^{-3}x_3 (x_2^3 - x_1^3)}{(x_2^2 - x_1^2)^2} \geq 0 \\[2mm]
g_4(x) = & \; \frac{2.66 \times 10^{-2}x_3 x_4 (x_2^3-x_1^3)}{(x_2^2-x_1^2)} - 900 \geq 0\end{split}\]
The four variables represent the inner radius of the discs \((x_1)\),
the outer radius of the discs \((x_2)\), the engaging force \((x_3)\),
and the number of friction surfaces \((x_4)\).
Variable bounds are given as follows:
\[55 \leq x_1 \leq 80 \quad \quad 75 \leq x_2 \leq 110 \quad \quad 1000 \leq x_3 \leq 3000
\quad \quad 11 \leq x_4 \leq 20\]
