Multiscale medalist learning algorithm and its application in engineering

Abstract

This paper presents the Multiscale Medalist Learning Algorithm (MMLA) as a novel heuristic approach for complex engineering optimization problems. By extending the Medalist Learning Algorithm, MMLA offers enhanced solution efficiency. The algorithm divides the learning process into successive periods with reduced search spaces, enabling focused search efforts in promising areas. Within each period, predefined learning stages are implemented. Top performers, known as medalists, engage in self-improvement operations through neighborhood searches, while common learners either learn from medalists or adapt based on the current state using neighborhood fluctuation. The MMLA balances exploration and exploitation capabilities through a natural growth curve that determines the learning efficiency. The MMLA's effectiveness and robustness are illustrated through the solution of a two-dimensional benchmark optimization problem and the successful resolution of ten well-known engineering design optimization problems. Comparative analysis demonstrates that the MMLA consistently outperforms other algorithms, providing competitive solutions with strict feasibility and minimal variation.

Appendix A. Constrained benchmark mechanical and engineering design problems

1.1 Multiple-disk clutch brake design

Consider variable $x=\left({r}_{i},{r}_{o},t,F,Z\right)$

Minimize $f\left(x\right)=\pi \left({r}_{o}^{2}-{r}_{i}^{2}\right)t\left(Z+1\right)\rho$

Subject to:

$$ g_{1} \left( x \right) = r_{o} - r_{i} - \Delta r \ge 0, g_{2} \left( x \right) = l_{\max } - \left( {Z + 1} \right)\left( {t + \delta } \right) \ge 0, g_{3} \left( x \right) = p_{\max } - p_{rz} \ge 0, $$

$$ g_{4} \left( x \right) = p_{\max } v_{sr\max } - p_{rz} v_{sr} \ge 0, g_{5} \left( x \right) = v_{sr\max } - v_{sr} \ge 0, g_{6} \left( x \right) = T_{\max } - T \ge 0, $$

$$ g_{7} \left( x \right) = M_{h} - sM_{s} \ge 0, g_{8} \left( x \right) = T \ge 0, $$

where ${M}_{h}=\frac{2}{3}\mu FZ\frac{{r}_{o}^{3}-{r}_{i}^{3}}{{r}_{o}^{2}-{r}_{i}^{2}}, {p}_{rz}=\frac{F}{\pi ({r}_{o}^{2}-{r}_{i}^{2})}, {M}_{h}=\frac{{2\pi n(r}_{o}^{3}-{r}_{i}^{3})}{90({r}_{o}^{2}-{r}_{i}^{2})}, T=\frac{{I}_{z}\pi n}{30({M}_{h}+{M}_{f})},$ $\Delta r=20\,\mathrm{mm}$, ${t}_{\mathrm{max}}$= 3 mm, ${t}_{\mathrm{min}}$= 1.5 mm, ${l}_{\mathrm{max}}=30\,\mathrm{mm}$,${Z}_{\mathrm{max}}$= 10,${v}_{\mathrm{srmax}}$= 10 m/s, $\upmu =0.5$, $\mathrm{s}=1.5, {M}_{s}$= 40 Nm$, {M}_{f}$= 3 Nm$, \mathrm{n}$= 250 rpm$, {p}_{\mathrm{max}}$= 1 MPa$, {I}_{z}$= 55 $\mathrm{kg }{\mathrm{mm}}^{2}, {T}_{\mathrm{max}}$= 15 s$, {F}_{\mathrm{max}}$= 1000 N, ${r}_{\mathrm{min}}$= 55 mm$, {r}_{o \mathrm{max}}$= 110 mm$,\uprho =7800\,\mathrm{kg}/{m}^{3}$.

Variable range ${r}_{i}\in \left\{60, 61, 62,\dots , 80\right\}{, r}_{o}\in \left\{90, 91, 92,\dots , 110\right\}, t\in \left\{1.0, 1.5, 2.0, 2.5, 3\right\}, F\in \left\{600, 610, 620, \dots , 1000\right\}, Z\in \left\{\mathrm{2,3},\mathrm{4,5},\mathrm{6,7},\mathrm{8,9}\right\}$.

1.2 Rolling element bearing

Maximize ${C}_{d}={f}_{c}{Z}^{2/3}{D}_{b}^{1.8}$ if ${D}_{b}\le 25.4$ mm.

${C}_{d}=3.64{f}_{c}{Z}^{2/3}{D}_{b}^{1.4}$ if ${D}_{b}>25.4$ mm.

Subject to:

$$ g_{1} \left( x \right) = \frac{{\emptyset_{o} }}{{2\sin^{ - 1} \left( {D_{b} /D_{m} } \right)}} - Z + 1 \ge 0, g_{2} \left( x \right) = 2D_{b} - K_{D\min } \left( {D - d} \right) \ge 0, $$

$$ g_{3} \left( x \right) = K_{D\max } \left( {D - d} \right) - 2D_{b} \ge 0, g_{4} \left( x \right) = \zeta B_{w} - D_{b} \le 0, $$

$$ g_{5} \left( x \right) = D_{m} - 0.5\left( {D + d} \right) \ge 0,g_{6} \left( x \right) = \left( {0.5 + e} \right)\left( {D + d} \right) - D_{m} \ge 0, $$

$$ g_{7} \left( x \right) = 0.5\left( {D - D_{m} - D_{b} } \right) - \varepsilon D_{b} \ge 0, g_{8} \left( x \right) = f_{i} \ge 0.515,\;\;g_{9} \left( x \right) = f_{o} \ge 0.515, $$

where

$$ f_{c} = 37.91\left\{ {1 + \left[ {1.04\left( {\frac{1 - \gamma }{{1 + \gamma }}} \right)^{1.72} \left( {\frac{{f_{i} \left( {2f_{o} - 1} \right)}}{{f_{o} \left( {2f_{i} - 1} \right)}}} \right)^{0.41} } \right]^{10/3} } \right\}^{ - 0.3} \left( {\frac{{\gamma^{0.3} \left( {1 - \gamma } \right)^{1.39} }}{{\left( {1 + \gamma } \right)^{1/3} }}} \right)\left( {\frac{{2f_{i} }}{{2f_{i} - 1}}} \right)^{0.41} , $$

$$ \phi_{o} = 2\pi - 2\cos^{ - 1} \frac{{\left[ {\frac{{\left( {D - d} \right)}}{2} - 3\left( \frac{T}{4} \right)} \right]^{2} + \left( {\frac{D}{2} - \frac{T}{4} - D_{b} } \right)^{2} - \left( {\frac{d}{2} + \frac{T}{4}} \right)^{2} }}{{2\left[ {\frac{{\left( {D - d} \right)}}{2} - 3\left( \frac{T}{4} \right)} \right]\left( {\frac{D}{2} - \frac{T}{4} - D_{b} } \right)}}, $$

$$ {\upgamma } = \frac{{D_{b} \cos \alpha }}{{D_{m} }}, f_{i} = \frac{{r_{i} }}{{D_{b} }},\;f_{o} = \frac{{r_{o} }}{{D_{b} }}, T = D - d - 2D_{b} , D = 160, d = 90,\;,B_{w} = 30,\;{\upalpha } = 0, $$

$$ 0.5\left( {D + d} \right) \le D_{m} \le 0.6\left( {D + d} \right), 0.15\left( {D - d} \right) \le D_{b} \le 0.45\left( {D - d} \right), Z \in \left\{ {4,5,6, \ldots ,50} \right\}, $$

$$ 0.515 \le f_{i} ,f_{o} \le 0.6, 0.4 \le K_{D\min } \le 0.5, 0.6 \le K_{D\max } \le 0.7, 0.3 \le \varepsilon \le 0.4, $$

$$ 0.02 \le {\text{e}} \le 0.1,\;0.6 \le {\upzeta } \le 0.85. $$

1.3 Hydrodynamic thrust bearing

Minimize $f(x)=\frac{Q{P}_{0}}{0.7}+{E}_{f}$

Subject to: ${\mathrm{g}}_{1}\left(x\right)=W-{W}_{s}\ge 0,{\text{g}}_{2}\left(x\right)={P}_{\mathrm{max}}-{P}_{0}\ge 0, {\text{g}}_{3}\left(x\right)=\Delta {T}_{\mathrm{max}}-\Delta T\ge 0, {\text{g}}_{4}\left(x\right)=h-{h}_{\mathrm{min}}\ge 0$, ${\text{g}}_{5}\left(x\right)=R-{R}_{0}\ge 0$, ${\text{g}}_{6}\left(x\right)=\text{0.001-}\frac{\gamma }{g{P}_{0}}{\left(\frac{Q}{2\pi Rh}\right)}^{2}\ge 0$, ${\mathrm{g}}_{7}\left(x\right)=5000-\frac{W}{\pi \left({R}^{2}-{R}_{0}^{2}\right)}\ge 0$,

Where $\upgamma =0.0307,$C = 0.5, n = − 3.55$, {C}_{1}$ = 10.04$, {W}_{s}$ = 101,000$, {P}_{\mathrm{max}}$ = 1000, $\Delta {T}_{\mathrm{max}}$ = 50$, {h}_{min}$ = 0.001, $\mathrm{g}=386.4,$N = 750$, 1\le \mathrm{R},{R}_{0},Q\le 16, 1\mathrm{e}-6\le\upmu \le 16\mathrm{e}-6$.

1.4 Belleville spring

Minimize $f\left(x\right)=0.07075\pi \left({D}_{e}^{2}-{D}_{i}^{2}\right)t$

Subject to:

$${\text{g}}_{1}\text{(}x\text{)}=S-\frac{4E{\delta }_{\mathrm{max}}}{\left(1-{\mu }^{2}\right)\alpha {D}_{e}^{2}}\left[\beta \left(h-\frac{{\delta }_{\mathrm{max}}}{2}\right)+\gamma t\right]\ge 0,$$

$${\text{g}}_{2}\text{(}x\text{)}=\frac{4E{\delta }_{\mathrm{max}}}{\left(1-{\mu }^{2}\right)\alpha {D}_{e}^{2}}\left[\left(h-\frac{{\delta }_{\mathrm{max}}}{2}\right)\left(h-{\delta }_{\mathrm{max}}\right)t+{t}^{3}\right]-{P}_{\mathrm{max}}\ge 0,$$

$${\text{g}}_{3}\text{(}x\text{)}={\delta }_{l}-{\delta }_{\mathrm{max}}\ge 0, {\text{g}}_{4}\text{(}x\text{)=}H-h-t\ge 0, {\text{g}}_{5}\left(x\right)={D}_{\mathrm{max}}-{D}_{e}\ge 0,$$

${\text{g}}_{6}\left(x\right)={D}_{e}-{D}_{i}\ge 0, {\text{g}}_{7}\left(x\right)=0.3-\frac{1}{{D}_{e}-{D}_{i}}\ge 0,$ where $\mathrm{\alpha }=\frac{6}{\pi \mathrm{ln}K}{\left(\frac{K-1}{K}\right)}^{2}, \beta =\frac{6}{\pi \mathrm{ln}K}\left(\frac{K-1}{\mathrm{ln}K}-1\right),\upgamma =\frac{6}{\pi \mathrm{ln}K}\left(\frac{K-1}{2}\right)$, ${P}_{max}=5400\,\text{lb, }{\delta }_{\mathrm{max}}=0.2\,{\text{in}}., S$ = 200 kPsi, E = 30 e6psi, $\mu =0.3$, H = $0.2\,{\text{in}}.$, ${D}_{\mathrm{max}}$ = 12.01 in., $\mathrm{K}=\frac{{D}_{e}}{{D}_{i}}, {\delta }_{l}$ = $f\left(a\right)h, a=h/t$. $0.01\le t\le 6.0, 0.05\le h\le 0.5, 5.0\le {D}_{i},{D}_{e}\le 15.0$.

Values of $\mathrm{f}(\mathrm{a})$ vary as shown in Table

Table 13 Variation of $f(a)$ with $a$

13.

1.5 Step-cone pulley

Minimize $f\left(x\right)=\rho w\left\{{d}_{1}^{2}\left[1+{\left(\frac{{N}_{1}}{N}\right)}^{2}\right]+{d}_{2}^{2}\left[1+{\left(\frac{{N}_{2}}{N}\right)}^{2}\right]+{d}_{3}^{2}\left[1+{\left(\frac{{N}_{3}}{N}\right)}^{2}\right]+{d}_{4}^{2}\left[1+{\left(\frac{{N}_{4}}{N}\right)}^{2}\right]\right\}$

Subject to:

$${h}_{1}\left(x\right)={C}_{1}-{C}_{2}=0, {h}_{2}\left(x\right)={C}_{1}-{C}_{3}=0, {h}_{3}\left(x\right)={C}_{1}-{C}_{4}=0,$$

$${\text{g}}_{1,2,3,4}\left(x\right)={R}_{i}\ge 2, {\mathrm{g}}_{\mathrm{5,6},\mathrm{7,8}}\left(x\right)={P}_{i}\ge \left(0.75*745.6998\right),$$

where ${C}_{i}$ indicates the length of the belt to obtain speed ${N}_{i}$ and is given by.

$${C}_{i}=\frac{\pi {d}_{i}}{2}\left(1+\frac{{N}_{i}}{N}\right)+\frac{{\left(\frac{{N}_{i}}{N}-1\right)}^{2}{d}_{i}^{2}}{4a}+2a, {i}=\left(\mathrm{1,2},\mathrm{3,4}\right).$$

${R}_{i}$ Is the tension ratio and is given by.

$${R}_{i}=\mathrm{exp}\left(\mu \left[\pi -2{\mathrm{sin}}^{-1}\left(\left(\frac{{N}_{i}}{N}-1\right)\frac{{d}_{i}}{2a}\right)\right]\right),{ i}=\left(\mathrm{1,2},\mathrm{3,4}\right).$$

$$ P_{i} = {\text{stw}}\left[ {1 - \exp \left[ { - \mu \left\{ {\pi - 2\sin^{ - 1} \left( {\left( {\frac{{N_{i} }}{N} - 1} \right)\frac{{d_{i} }}{2a}} \right)} \right\}} \right]} \right]\frac{{\pi d_{i} N_{i} }}{60}, i = \left( {1,2,3,4} \right). $$

$$\uprho =\frac{7200\,\mathrm{kg}}{{m}^{3}},a=3\,\mathrm{m},\upmu =0.35,\mathrm{ s}=1.75\,\mathrm{MPa},\mathrm{ t}=8\,\mathrm{mm}.$$

$$1\,\mathrm{mm}\le {d}_{1},{d}_{2},{d}_{3},{d}_{4},w\le 100\,{\text{mm}}\text{.}$$

1.6 Speed reducer design

Minimize $f\left(x\right)=0.7854{x}_{1}{x}_{2}^{2}\left(3.3333{x}_{3}^{2}+14.9334{x}_{3}-43.0934\right)-1.508{x}_{1}\left({x}_{6}^{2}+{x}_{7}^{2}\right) +7.4777\left({x}_{6}^{3}+{x}_{7}^{3}\right)+0.7854\left({x}_{4}{x}_{6}^{2}+{x}_{5}{x}_{7}^{2}\right)$

Subject to:

$${\mathrm{g}}_{1}\left(x\right)=\frac{27}{{x}_{1}{x}_{2}^{2}{x}_{3}}-1\le 0, {\mathrm{g}}_{2}\left(x\right)=\frac{397.5}{{x}_{1}{x}_{2}^{2}{x}_{3}^{2}}-1\le 0, {\mathrm{g}}_{3}\left(x\right)=\frac{1.93{x}_{4}^{3}}{{x}_{2}{x}_{3}{x}_{6}^{4}}-1\le 0, {\mathrm{g}}_{4}\left(x\right)=\frac{1.93{x}_{5}^{3}}{{x}_{2}{x}_{3}{x}_{7}^{4}}-1\le 0,$$

$${\text{g}}_{5}\left(x\right)=\frac{\sqrt{{\left[745{x}_{4}/\left({x}_{2}{x}_{3}\right)\right]}^{2}+16.9e6}}{110{x}_{6}^{3}}-1\le 0, {\text{g}}_{6}\left(x\right)=\frac{\sqrt{{\left[745{x}_{5}/\left({x}_{2}{x}_{3}\right)\right]}^{2}+157.5e6}}{85{x}_{7}^{3}}-1\le 0,$$

$$ g_{7} \left( x \right) = \frac{{x_{2} x_{3} }}{40} - 1 \le 0, g_{8} \left( x \right) = \frac{{5x_{2} }}{{x_{1} }} - 1 \le 0,\,g_{9} \left( x \right) = \frac{{x_{1} }}{{12x_{2} }} - 1 \le 0, $$

$${\mathrm{g}}_{10}\left(x\right)=\frac{1.5{x}_{6}+1.9}{{x}_{4}}-1\le 0, {\mathrm{g}}_{11}\left(x\right)=\frac{1.1{x}_{7}+1.9}{{x}_{5}}-1\le 0,$$

where $2.6\le {x}_{1}\le 3.6; 0.7\le {x}_{2}\le 0.8;$ $17\le {x}_{3}\le 28;$ $7.3\le {x}_{4}\le 8.3; 7.8\le {x}_{5}\le 8.3$; $2.9\le {x}_{6}\le 3.9$; $5.0\le {x}_{7}\le 5.5$.

1.7 Cantilever beam design

Minimize $\mathrm{f}\left(\mathrm{x}\right)=0.0624\sum_{i=1}^{5}{x}_{i}$

Subject to: ${\mathrm{g}}_{1}\left(x\right)=\frac{61}{{x}_{1}^{3}}+\frac{37}{{x}_{2}^{3}}+\frac{19}{{x}_{3}^{3}}+\frac{7}{{x}_{4}^{3}}+\frac{1}{{x}_{5}^{3}}-1\le 0$,

Variable range $0.01\le {x}_{i}\le 100,i=\mathrm{1,2},\mathrm{3,4},5.$

1.8 Three-bar truss design

Minimize $\mathrm{f}\left(\mathrm{x}\right)=\left(2\sqrt{2}{x}_{1}+{x}_{2}\right)l$

Subject to:

$${\mathrm{g}}_{1}\left(x\right)=\frac{\sqrt{2}{x}_{1}+{x}_{2}}{\sqrt{2}{x}_{1}^{2}+2{x}_{1}{x}_{2}}P-\sigma \le 0, {\mathrm{g}}_{2}\left(x\right)=\frac{{x}_{2}}{\sqrt{2}{x}_{1}^{2}+2{x}_{1}{x}_{2}}P-\sigma \le 0, {\mathrm{g}}_{3}\left(x\right)=\frac{{x}_{2}}{{x}_{1}+\sqrt{2}{x}_{2}}P-\sigma \le 0,$$

where $l$ = 10 cm, $P=2$ KN/${\text{cm}}^{2},\sigma =2$ KN/${\text{cm}}^{2}$, variable range $0\le {x}_{1},{x}_{2}\le 1$.

1.9 Welded beam design

Consider $x=\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right)=\left(w,l,d,h\right)$,

minimize $f\left(x\right)=1.10471{w}^{2}l+0.04811dh(14.0+l)$,

subject to.

$${g}_{1}\left(x\right)=\tau \left(x\right)-13600\le 0,$$

$${g}_{2}\left(x\right)=\sigma \left(x\right)-30000\le 0,$$

$${g}_{3}\left(x\right)=w-h\le 0,$$

$${g}_{4}\left(x\right)=0.10471{w}^{2}+0.04811hd\left(14+l\right)-5.0\le 0,$$

$${g}_{5}\left(x\right)=0.125-w\le 0,$$

$${g}_{6}\left(x\right)=\delta \left(x\right)-0.25\le 0,$$

$${g}_{7}\left(x\right)=6000-P\left(x\right)\le 0,$$

where

$$\upsigma \left(x\right)=\frac{504000}{h{d}^{2}}, \delta \left(x\right)=\frac{65856}{30000h{d}^{3}}, Q=6000\left(14+\frac{l}{2}\right), D=\frac{1}{2}\sqrt{{l}^{2}+{\left(w+d\right)}^{2}},$$

$$J=\sqrt{2}wl\left[\frac{{l}^{2}}{6}+\frac{{\left(w+d\right)}^{2}}{2}\right], \beta =\frac{QD}{J}, \alpha =\frac{6000}{\sqrt{2}wl}, \tau \left(x\right)=\sqrt{{\alpha }^{2}+\frac{\alpha \beta l}{D}+{\beta }^{2}},$$

$$P\left(x\right)=0.61423\times {10}^{6}\frac{d{h}^{3}}{6}\left(1-\frac{d\sqrt{30/48}}{28}\right),$$

With the variable range.

$$ 0.1 \le l,d \le 10,\,0.1 \le w,h \le 2.0. $$

1.10 The pressure vessel design

Consider $x=\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right)=\left({T}_{s},{T}_{h},R,L\right)$,

Minimize $f\left(x\right)=0.6224{x}_{1}{x}_{3}{x}_{4}+1.7781{x}_{2}{x}_{3}^{2}+3.1661{x}_{1}^{2}{x}_{4}+19.84{x}_{1}^{2}{x}_{3}$,

Subject to ${g}_{1}\left(x\right)=-{x}_{1}+0.0193{x}_{3}\le 0$

$${g}_{2}\left(x\right)=-{x}_{2}+0.00954{x}_{3}\le 0,$$

$${g}_{3}\left(x\right)=-\pi {x}_{3}^{2}{x}_{4}-\frac{4}{3}\pi {x}_{3}^{3}+1296000\le 0$$

$${g}_{4}\left(x\right)={x}_{4}-240\le 0,$$

$$0\le {x}_{1},{x}_{2}\le 99, 10\le {x}_{3},{x}_{4}\le 200.$$