Table 8 Information of applied engineering problems.

Three-bar truss problem

\(f({\mathbf{x}}) = l\left( {x_{2} + 2\sqrt 2 x_{1} } \right)\)

\(g_{1} \left( {\mathbf{x}} \right) = P\left( {x_{2} + \sqrt 2 x_{1} } \right)\left( {2x_{1} x_{2} + \sqrt 2 x_{1}^{2} } \right)^{ - 1} - \sigma \le 0\)

\(g_{2} \left( {\mathbf{x}} \right) = P\left( {2x_{1} x_{2} + \sqrt 2 x_{1}^{2} } \right)^{ - 1} x_{2} - \sigma \le 0\)

\(g_{3} \left( {\mathbf{x}} \right) = P\left( {x_{1} + \sqrt 2 x_{2} } \right)^{ - 1} - \sigma \le 0\)

\(x_{1} \ge 0\)

\(x_{2} \le 1\)

Pressure vessel optimization problem

\(\begin{gathered} f({\mathbf{x}}) = 19.84x_{1}^{2} x_{3} + 3.1661x_{1}^{2} x_{4} + \hfill \\ \, 1.7781x_{2} x_{3}^{2} + 0.6224x_{1} x_{3} x_{4} \hfill \\ \end{gathered}\)

\(g_{1} \left( {\mathbf{x}} \right) = - x_{1} + 0.0193x_{3} \le 0\)

\(g_{2} \left( {\mathbf{x}} \right) = - x_{2} + 0.00954x_{3} \le 0\)

\(g_{3} \left( {\mathbf{x}} \right) = 1296000 - \pi x_{3}^{2} x_{4} - \frac{4}{3}\pi x_{3}^{3} \le 01296000\)\(- \pi x_{3}^{2} x_{4} - \frac{4}{3}\pi x_{3}^{3} \le 0\)

\(g_{4} \left( {\mathbf{x}} \right) = x_{4} - 240 \le 0\)

\(0 \le x_{i} \le 100{, }i = 1, \, 2\)

\(10 \le x_{i} \le 200{, }i = 3, \, 4\)

Tension/compression spring design

\(f({\mathbf{x}}) = x_{1}^{2} x_{2} \left( {2 + x_{3} } \right)\)

\(g_{1} \left( {\mathbf{x}} \right) = 1 - \frac{{x_{2}^{3} x_{3} }}{{71785x_{1}^{4} }} \le 0\)

\(g_{2} \left( {\mathbf{x}} \right) = - 1 + \frac{{4x_{2}^{2} - x_{1} x_{2} }}{{12566\left( {x_{1}^{3} x_{2} - x_{1}^{4} } \right)}} + \frac{1}{{5108x_{1}^{2} }} \le 0\)

\(g_{3} \left( {\mathbf{x}} \right) = - \frac{{140.45x_{1} }}{{x_{2}^{2} x_{3} }} + 1 \le 0\)

\(g_{4} \left( {\mathbf{x}} \right) = - 1 + \frac{{x_{1} + x_{2} }}{1.5} \le 0\)

\(0.05 \le x_{1} \le 2\)

\(0.25 \le x_{2} \le 1.3\)

\(2 \le x_{3} \le 15\)

Piston lever optimal design

\(f(H,B,X,D) = \frac{1}{4}\pi D^{2} \left( {L_{2} - L_{1} } \right)\)

\(g_{1} \left( {H,B,X,D} \right) = QL\cos \theta - RF \le 0 \, at \, \theta = 45^\circ\)

\(g_{2} \left( {H,B,X,D} \right) = Q\left( {L - X} \right) - M_{\max } \le 0\)

\(g_{3} \left( {H,B,X,D} \right) = 1.2\left( {L_{2} - L_{1} } \right) - L_{1} \le 0\)

\(g_{4} \left( {H,B,X,D} \right) = 0.5D - B \le 0\)

\(0.05 \le H,B,D \le 500\)

\(0.05 \le X \le 120\)

where

\(\begin{gathered} R = \left| {H\left( {B - X\cos \theta } \right) - X\left( {X\sin \theta + H} \right)} \right| \hfill \\ \left( {\sqrt {H^{2} + \left( {X - B} \right)^{2} } } \right)^{ - 1} \hfill \\ \end{gathered}\)

\(F = \frac{{\pi PD^{2} }}{4}\)

\(L_{1} = \sqrt {H^{2} + \left( {X - B} \right)^{2} }\)

\(L_{2} = \sqrt {\left( {B - X\cos \theta } \right)^{2} + \left( {X\sin \theta + H} \right)^{2} }\)

Tubular column design

\(f(d,t) = 9.8dt + 2d\)

\(g_{1} \left( {d,t} \right) = \frac{P}{{\pi dt\sigma_{y} }} - 1 \le 0\)

\(g_{2} \left( {d,t} \right) = \frac{8P}{{\pi^{3} Edt\left( {d^{2} + t^{2} } \right)}} - 1 \le 0\)

\(g_{3} \left( {d,t} \right) = \frac{2}{d} - 1 \le 0\)

\(g_{4} \left( {d,t} \right) = \frac{d}{14} - 1 \le 0\)

\(g_{5} \left( {d,t} \right) = \frac{1}{5t} - 1 \le 0\)

\(g_{6} \left( {d,t} \right) = \frac{5t}{4} - 1 \le 0\)