An enhanced moth flame optimization with mutualism scheme for function optimization

Abstract

Nature-inspired meta-heuristics have demonstrated superior efficiency in the solution of complicated nonlinear optimization problems than conventional techniques. In this article, an enhanced moth flame optimization (EMFO) is designed using the mutualism phase from the symbiotic organism search (SOS) algorithm. The suggested approach is examined on 36 classical benchmark functions taken from literature. The outputs of EMFO are compared with the latest meta-heuristic algorithms and variants of the MFO algorithm. The comparison results indicate that our proposed method is competitive from the compared methods. Also, the Friedman rank test is used to evaluate the new algorithm’s efficiency, and it is found that the rank of EMFO is superior. Finally, EMFO is being applied to solve seven real-world problems, and the outcomes of the proposed algorithm were found to be satisfactory.

Appendices

Appendix 1

See Tables

Table 16 Mathematical formulation of unimodal functions

16,

Table 17 Mathematical formulation of multimodal functions

17 and

Table 18 Mathematical formulation of multimodal functions

18.

Appendix 2

• a. Optimal capacity of gas production problem

  $$ {\text{Min }}f\left( x \right) = 61.8 + 5.72 \times x_{1} \times 0.2623 \times \left[ {\left( {40 - x_{1} } \right) \times \ln \frac{{x_{2} }}{200}} \right]^{ - 0.85} + 0.087 \times \left( {40 - x_{1} } \right) \times \ln \frac{{x_{2} }}{200} + 700.23 \times x_{2}^{ - 0.75} $$
  $$ {\text{s.t}}.\;x_{1} \ge 17.5, x_{2} \ge 200, 17.5 \le x_{1} \le 40, 300 \le x_{2} \le 600; $$

• b. Gear train design problem

  $$ {\text{Minimize}}\;f\left( {\vec{x}} \right) = \left[ {\frac{1}{6.931} - \frac{{x_{3} x_{2} }}{{x_{1} x_{4} }}} \right]^{2} , $$
  $$ {\text{Subjected to }}12\; \le x_{1} , x_{2} , x_{3} , x_{4} \le 60, $$
  $$ {\text{Where}}\;\vec{x} = \left[ {x_{1} x_{2} x_{3} x_{4} } \right] = \left[ {n\alpha \; n\beta \; n\gamma \; n\delta } \right]. $$

• c. Cantilever beam design problem

  $$ {\text{Minimize}}\;f\left( {\vec{x}} \right) = 0.06224\left( {x_{1} + x_{2} + x_{3} + x_{4} + x_{5} } \right), $$
  $$ {\text{Subjected to}}: \, h(\vec{x}) = \frac{61}{{x_{1}^{3} }} + \frac{37}{{x_{2}^{3} }} + \frac{19}{{x_{3}^{3} }} + \frac{7}{{x_{4}^{3} }} + \frac{1}{{x_{5}^{3} }} \le 1, $$
  $$ {\text{Where}}\;0.01 \le x_{1} , x_{2} , x_{3} , x_{4} , x_{5} \le 100. $$

• d. I- beam vertical deflection problem

  $$ {\text{Minimize}}\;f\left( {\vec{x}} \right) = \frac{5000}{{\frac{{2t_{f} \left( {h - t_{f} } \right)^{2} }}{2} + \frac{{t_{w} *\left( {h - 2t_{f} } \right)}}{12} + \frac{{bt_{f}^{3} }}{6}}} $$

  Subjected to the cross-sectional area (< 300 cm²).

  $$ x_{1} = t_{w} *\left( {h - 2t_{f} } \right) + 2*b*t_{w} \le 300, $$

  The stress constraint (where allowable bending stress of the beam is 56 KN/CM²)

  $$ x_{2} = \frac{{15*b*10^{3} }}{{\left( {h - 2t_{f} } \right)*t_{w}^{3} + 2t_{w} *b^{3} }} + \frac{{18*h*10^{4} }}{{\left( {h - t_{f} } \right)^{3} *t_{w} + 2b\left( {4t_{f}^{2} + 3h\left( {h - 2t_{f} } \right)} \right)t_{w} *b^{3} }} \le 56 $$
  $$ {\text{Provided}}\;0.9 \le t_{f} , t_{w} \le 5,{ }10 \le b \le 50\;{\text{ and }}\;10 \le h \le 80 $$

• e. Welded beam design problem

  $$ {\text{Minimize}}\;f\left( {{\vec{\text{x}}}} \right) = f\left( {h, l, t, b} \right) = f\left( {{\text{x}}_{1} ,{\text{ x}}_{2} ,{\text{ x}}_{3} ,{\text{ x}}_{4} } \right) = {\text{A}}x_{2} x_{1}^{2} + Bx_{4} x_{3} \left( {C + x_{2} } \right). $$
  $$ {\text{Subjected to}}\;h_{1} \left( x \right) = \tau_{\max } - \tau \left( x \right) \ge 0 $$
  $$ h_{2} \left( x \right) = \sigma_{\max } - \sigma \left( x \right) \ge 0 $$
  $$ h_{3} \left( x \right) = \delta_{max} - \delta \left( x \right) \ge 0 $$
  $$ h_{4} \left( x \right) = x_{4} - x_{1} \ge 0 $$
  $$ h_{5} \left( x \right) = P_{c} \left( x \right) - P \ge 0 $$

  where A = 1.10471; B = 0.04811; C = 14.0

  $\tau_{\max } = 13600$ psi; $\sigma_{\max } =$ 30000psi; $\delta_{\max }$ = 0.25in;

  $$ \tau \left( x \right) = \sqrt {\left( {\tau^{\prime}\left( x \right)} \right)^{2} + \left( {\frac{{l\tau^{\prime\prime}\left( x \right)}}{{\sqrt {0.25\left( {x_{2}^{2} + \left( {x_{1} + x_{3} } \right)^{2} } \right)} }} + \left( {\tau^{\prime\prime}\left( x \right)} \right)^{2} } \right)} $$
  $$ \sigma \left( x \right) = \frac{50400}{{x_{4} x_{3}^{2} }} $$
  $$ P_{c} \left( x \right) = 64746x_{3} \left( {1 - 0.0282346x_{3} } \right)x_{4}^{3} $$
  $$ \delta \left( x \right) = \frac{2.1952}{{x_{4} x_{3}^{3} }} $$
  $$ \tau^{\prime}\left( x \right) = \frac{6000}{{\sqrt 2 x_{2} x_{1} }} $$
  $$ \tau^{\prime\prime}\left( x \right) = \frac{{6000\left( {0.5x_{2} + 14} \right)\sqrt {0.25\left( {x_{2}^{2} + \left( {x_{1} + x_{3} } \right)^{2} } \right)} }}{{2\left\{ {0.707x_{1} x_{2} \left( {\frac{{x_{2}^{2} }}{12} + 0.25\left( {x_{1} + x_{3} } \right)^{2} } \right)} \right\}}} $$
  $$ {\text{Range of variables}}\;0.125 \le x_{1} \le 5 $$
  $$ 0.1 \le x_{2} ,x_{3} \le 10 $$
  $$ 0.1 \le x_{4} \le 5 $$

  Parameters A = 1.10471; B = 0.04811; C = 14.0

  $\tau_{\max } = 13600$ psi; $\sigma_{\max } =$ 30000psi; $\delta_{\max }$ = 0.25in;

  E = 30 $\times 10^{6}$ psi G = 12 $\times 10^{6}$ psi.

  P = 6000 lb.

  L = 14in.

• f. Three-bar truss problem

  $$ \vec{\user2{k}} = \left\{ {k_{1} , k_{2} , } \right\} $$

  Objective function

  $$ {\text{Min}} \cdot f\left( k \right) = L\left\{ {k_{2} + 2\sqrt 2 k_{1} } \right\} $$

  Subject to

  $$ h_{1} \left( k \right) = \frac{{k_{2} }}{{2k_{2} k_{1} + \sqrt 2 k_{1}^{2} }} P - \sigma \le 0, $$
  $$ h_{2} \left( k \right) = \frac{{k_{2} + \sqrt 2 k_{1} }}{{2k_{2} k_{1} + \sqrt 2 k_{1}^{2} }} P - \sigma \le 0, $$
  $$ h_{3} \left( k \right) = \frac{1}{{k_{1} + \sqrt 2 k_{2} }}P - \sigma \le 0, $$

  where

  $$ 0 \le k_{1} ,k_{2} \le 1,\;{\text{and}} $$
  $$ P = 2, L = 100 \& \sigma = 2 $$

• g. Car-side crash design problem

  $$ \vec{\user2{a}} = \left\{ {a_{1} ,\user2{ }a_{2} ,a_{3} ,\user2{ }a_{4} ,\user2{ }a_{5} ,\user2{ }a_{6} ,\user2{ }a_{7} ,a_{8} ,a_{9} ,\user2{ }a_{10} ,a_{11} } \right\} $$

Objective function

$$ Min \; f\left( a \right) = 1.98 + 4.90 a_{1} + 6.67 a_{2} + 6.98 a_{3} + 4.01 a_{4} + 1.78 a_{5} + 2.73 a_{7} , $$

Subject to

$$ \begin{aligned} h_{1} \left( a \right) & = 1.16 - 0.3717 a_{2} a_{4} - 0.00931 a_{2} a_{10} - 0.484 a_{3} a_{9} \\ & + 0.01343 a_{6} a_{10 } \le 1, \\ \end{aligned} $$

$$ \begin{aligned} h_{2} \left( a \right) & = 0.261 - 0.0159 a_{1} a_{2} - 0.188 a_{1} a_{8} - 0.019 a_{2} a_{7} + 0.0144 a_{3} a_{5} + 0.0008757 a_{5} a_{10} \\ & + 0.080405 a_{6} a_{9} + 0.00139 a_{8} a_{11} + 0.00001575 a_{10} a_{11} \le 0.32, \\ \end{aligned} $$

$$ \begin{aligned} h_{3} \left( a \right) & = 0.214 + 0.00817 a_{5} - 0.131 a_{1} a_{8} - 0.0704 a_{1} a_{9} + 0.03099 a_{2} a_{6} - 0.018 a_{2} a_{7} + 0.0208 a_{3} a_{8} \\ & + 0.121 a_{3} a_{9} - 0.00364 a_{5} a_{6} + 0.0007715 a_{5} a_{10} - 0.0005354 a_{6} a_{10} + 0.00121 a_{8} a_{11} \le 0.32, \\ \end{aligned} $$

$$ h_{4} \left( a \right) = 0.074 - 0.061 a_{2} - 0.163 a_{3} a_{8} + 0.001232 a_{3} a_{10} - 0.166 a_{7 } a_{9} + 0.227 a_{2}^{2} \le 0.32, $$

$$ \begin{aligned} h_{5} \left( a \right) & = 28.98 + 3.818 a_{3} - 4.2 a_{1} a_{2} + 0.0207 a_{5} a_{10} + 6.63 a_{6} a_{9} \\ & - 7.7 a_{7} a_{8} + 0.32 a_{9} a_{10} \le 32, \\ \end{aligned} $$

$$ h_{6} \left( a \right) = 33.86 + 2.95 a_{3} + 0.1792 a_{10} - 5.05 a_{1} a_{2} - 11.0 a_{2} a_{8} - 0.0215 a_{5} a_{10} - 9.98 a_{7} a_{8} + 22.0 a_{8} a_{9} \le 32, $$

$$ \user2{ }h_{7} \left( a \right) = 46.36 - 9.9 a_{2} - 12.9 a_{1} a_{8} + 0.1107 a_{3} a_{10} \le 32, $$

$$ \begin{aligned} h_{8} \left( a \right) & = 4.72 - 0.5 a_{4} - 0.19 a_{2} a_{3} - 0.0122 a_{4} a_{10} + 0.009325 a_{6} a_{10} \\ & + 0.000191 a_{11}^{2} \le 4, \\ \end{aligned} $$

$$ \begin{gathered} h_{9} \left( a \right) = 10.58 - 0.674 a_{1} a_{2} - 1.95 a_{2} a_{8} + 0.02054 a_{3} a_{10} - \hfill \\ 0.0198 a_{4} a_{10} + 0.028 a\left( 6 \right) a\left( {10} \right) \le 9.9, \hfill \\ \end{gathered} $$

$$ h_{10} \left( a \right) = 16.45 - 0.489 a_{3} a_{7} - 0.843 a_{5} a_{6} + 0.0432 a_{9} a_{10} - 0.0556 a_{9} a_{11} - 0.000786 a_{11}^{2} \le 15.7, $$

where

$$ 0.5 \le a_{i} \le 1.5, i = 1, 2, 3, 4, 5, 6, 7 $$

$$ a_{8} ,a_{9} \in \left( {0.192, 0.345} \right), $$

$$ - 30 \le a_{10} , a_{11} \le 30. $$