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ID:44537437
大小:266.68 KB
页数:9页
时间:2019-10-23
《机械优化设计实例(附带程序和算法框图)》由会员上传分享,免费在线阅读,更多相关内容在工程资料-天天文库。
1、机械优化设计作业一、优化设计问题的提出预制一无盖水槽,现有一块长为4m,宽为3m的长方形铁板作为原材料,想在这块铁板的四个角处剪去相等的正方形以制成无盖水槽,问如何剪法使水槽的底面积最大?二、建立问题的数学模型为了建成此无盖水槽,可设在这块铁板的四个角处剪去相等的正方形的边长为X,所建造水槽的底面积为S,分析问题有次问题变成在约束条件:X204-2XN03-2XN0限制下,求目标函数:S(X)=(4-2X)(3-2X)=4X2-14X+12的最大值。由此可得此问题的数学模型为:、MinS(X)=4X2-14X+12T约束条件:(X)=XwOg2(X)=-(4-2X
2、)wO93(X)=-(3-2X)wO三.求解问题选择的优化算法此问题的求解属于一维优化问题,所选择的优化算法为黄金分割法。〈-用外推法确定函数s(X)=4X2-14X+12的初始搜索区间。设初始点%=0,初始步长h=loa0=0,S0=S(a0)=12;a1=ao+h=O+l=l,S1=S(a1)=2;比较So和Si,因为SiS1,此时已经形成So>Si3、.算法程序#inelude#ineludedoubleobfunc(doublex){doubleff;ff=4*X*X-14*X+12;return(ff);}voidjts(doublexO,doublehO,doubles[],intn,doublea[],doubleb[]){inti;doublex[3],h,f2,f2,f3;h=hO;for(i=0;ivn;i++)x[0]=x0;fl=obfunc(x[0]);for(i=0;i4、2>=fl){h=-hO;for(i=0;ivn;i++)x[2]=x[0];f3=fl;for(i=0;ivn;i++){x[0]=x[l];x[l]=x[2];}fl=f2;f2=f3;}for(;;){h=2.O*h;for(i=0;ivn;i++)x[2]=x[l]+h*s[i];f3=obfunc(x[2]);if(f25、=0;ivn;i++){a[i]=x[O];b[i]=x[2];}printf("%4d,,/n);}doublegold(doublea[],doubleb[],doubleeps,intn,doublexx)doublefL,f2,ff,q,w;doublex[3];for(i=0;if2){for(i=0;i6、]=x[l];}fl=f2;for(i=0;ieps);for(i=0;i7、b[i]);ff=obfunc(xx);printf("xx=ff=%5.2f////%5.2f"/xx,ff);return(ff);}voidmain(){intn=l;doublea[l],b[l],xx;doubles[]={l},x0=0;doubleepsl=0.001,h0=0.1;jts(xO,hO,s,nab);gold(a,b,epsl,n,xx);七、程序运行结果与分析(1)程序运行结果(截屏)uoidjts(doublexO,ttlllLlUUtMUdLII.II/Uincludedoubleobfunc(doublex)8、doubl
3、.算法程序#inelude#ineludedoubleobfunc(doublex){doubleff;ff=4*X*X-14*X+12;return(ff);}voidjts(doublexO,doublehO,doubles[],intn,doublea[],doubleb[]){inti;doublex[3],h,f2,f2,f3;h=hO;for(i=0;ivn;i++)x[0]=x0;fl=obfunc(x[0]);for(i=0;i4、2>=fl){h=-hO;for(i=0;ivn;i++)x[2]=x[0];f3=fl;for(i=0;ivn;i++){x[0]=x[l];x[l]=x[2];}fl=f2;f2=f3;}for(;;){h=2.O*h;for(i=0;ivn;i++)x[2]=x[l]+h*s[i];f3=obfunc(x[2]);if(f25、=0;ivn;i++){a[i]=x[O];b[i]=x[2];}printf("%4d,,/n);}doublegold(doublea[],doubleb[],doubleeps,intn,doublexx)doublefL,f2,ff,q,w;doublex[3];for(i=0;if2){for(i=0;i6、]=x[l];}fl=f2;for(i=0;ieps);for(i=0;i7、b[i]);ff=obfunc(xx);printf("xx=ff=%5.2f////%5.2f"/xx,ff);return(ff);}voidmain(){intn=l;doublea[l],b[l],xx;doubles[]={l},x0=0;doubleepsl=0.001,h0=0.1;jts(xO,hO,s,nab);gold(a,b,epsl,n,xx);七、程序运行结果与分析(1)程序运行结果(截屏)uoidjts(doublexO,ttlllLlUUtMUdLII.II/Uincludedoubleobfunc(doublex)8、doubl
4、2>=fl){h=-hO;for(i=0;ivn;i++)x[2]=x[0];f3=fl;for(i=0;ivn;i++){x[0]=x[l];x[l]=x[2];}fl=f2;f2=f3;}for(;;){h=2.O*h;for(i=0;ivn;i++)x[2]=x[l]+h*s[i];f3=obfunc(x[2]);if(f25、=0;ivn;i++){a[i]=x[O];b[i]=x[2];}printf("%4d,,/n);}doublegold(doublea[],doubleb[],doubleeps,intn,doublexx)doublefL,f2,ff,q,w;doublex[3];for(i=0;if2){for(i=0;i6、]=x[l];}fl=f2;for(i=0;ieps);for(i=0;i7、b[i]);ff=obfunc(xx);printf("xx=ff=%5.2f////%5.2f"/xx,ff);return(ff);}voidmain(){intn=l;doublea[l],b[l],xx;doubles[]={l},x0=0;doubleepsl=0.001,h0=0.1;jts(xO,hO,s,nab);gold(a,b,epsl,n,xx);七、程序运行结果与分析(1)程序运行结果(截屏)uoidjts(doublexO,ttlllLlUUtMUdLII.II/Uincludedoubleobfunc(doublex)8、doubl
5、=0;ivn;i++){a[i]=x[O];b[i]=x[2];}printf("%4d,,/n);}doublegold(doublea[],doubleb[],doubleeps,intn,doublexx)doublefL,f2,ff,q,w;doublex[3];for(i=0;if2){for(i=0;i6、]=x[l];}fl=f2;for(i=0;ieps);for(i=0;i7、b[i]);ff=obfunc(xx);printf("xx=ff=%5.2f////%5.2f"/xx,ff);return(ff);}voidmain(){intn=l;doublea[l],b[l],xx;doubles[]={l},x0=0;doubleepsl=0.001,h0=0.1;jts(xO,hO,s,nab);gold(a,b,epsl,n,xx);七、程序运行结果与分析(1)程序运行结果(截屏)uoidjts(doublexO,ttlllLlUUtMUdLII.II/Uincludedoubleobfunc(doublex)8、doubl
6、]=x[l];}fl=f2;for(i=0;ieps);for(i=0;i7、b[i]);ff=obfunc(xx);printf("xx=ff=%5.2f////%5.2f"/xx,ff);return(ff);}voidmain(){intn=l;doublea[l],b[l],xx;doubles[]={l},x0=0;doubleepsl=0.001,h0=0.1;jts(xO,hO,s,nab);gold(a,b,epsl,n,xx);七、程序运行结果与分析(1)程序运行结果(截屏)uoidjts(doublexO,ttlllLlUUtMUdLII.II/Uincludedoubleobfunc(doublex)8、doubl
7、b[i]);ff=obfunc(xx);printf("xx=ff=%5.2f////%5.2f"/xx,ff);return(ff);}voidmain(){intn=l;doublea[l],b[l],xx;doubles[]={l},x0=0;doubleepsl=0.001,h0=0.1;jts(xO,hO,s,nab);gold(a,b,epsl,n,xx);七、程序运行结果与分析(1)程序运行结果(截屏)uoidjts(doublexO,ttlllLlUUtMUdLII.II/Uincludedoubleobfunc(doublex)
8、doubl
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