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xizi19880211新蟲 (初入文壇)
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[求助]
1stopt求最小值 已有1人參與
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| 如果是非線性約束求最值,并且約束條件里面的量是變化,比如是隨時間變化的,可以用1stopt來實現么?MATlab的GA工具箱可以實現么?求實例代碼。 |
新蟲 (初入文壇)
榮譽版主 (文壇精英)
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新蟲 (初入文壇)
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也就是說把約束條件那一塊用循環(huán)來實現? 代碼在下面,幫我看看如何循環(huán),謝謝 Pw=[25.3300000000000 25.3300000000000 25.6700000000000 25.6700000000000 33.3300000000000 34.6700000000000 38.3300000000000 33 28 21 14.6700000000000 14.3300000000000 4.43000000000000 5.32000000000000 10.3300000000000 9.88000000000000 8.53000000000000 4.43000000000000 3.33000000000000 3.33000000000000 5.03000000000000 8.01000000000000 5 9.03000000000000 10.9900000000000 10.5300000000000 12.52 20 23.98 30.99 30 28.33 27.53 21.56 27.01 30.32 46.98 46.07 55.65 51.55 59.88 53.66 52.03 47.55 40 53.09 54.99 53.22 50.02]; Pl=[15 15 15 15 15 15 28.9900000000000 28.9900000000000 28.9900000000000 28.9900000000000 28.9900000000000 28.9900000000000 16.5500000000000 16.5500000000000 16.5500000000000 16.5500000000000 9.59000000000000 9.59000000000000 9.59000000000000 9.59000000000000 9.59000000000000 9.59000000000000 9.59000000000000 9.59000000000000 9.59000000000000 23.3300000000000 23.3300000000000 23.3300000000000 23.3300000000000 23.3300000000000 23.3300000000000 23.3300000000000 23.3300000000000 45.8800000000000 45.8800000000000 45.8800000000000 45.8800000000000 45.8800000000000 45.8800000000000 45.8800000000000 45.8800000000000 45.8800000000000 49.5500000000000 49.5500000000000 49.5500000000000 49.5500000000000 30.8900000000000 30.8900000000000 30.8900000000000]; Eb(0)=1.2e-3*x(1); Ec(0)=3.0375e-6*x(2); for k=1:49 if Pw(k)-Pl(k)>0; Eb(k)=Eb(k-1)+Pb(k)*k*0.7; Ec(k)=Ec(k-1)+Pc(k)*k*0.98; else Eb(k)=Eb(k-1)-Pb(k)*k*0.8; Ec(k)=Ec(k-1)-Pc(k)*k*0.98; end end for k=1:49 if dp=Pw(k)*0.95-Pl(k)>0; Elsp(k)=0; else Elsp(k)=[Pl(k)-Pw(k)]*0.95*k; end LPSP++=Elsp(k)/Pl(k); end; MinFunction 0.288*x(1)+7.35e-4*x(2); Eb(k)*1000/1.2<=x(1)<=Eb(k)*1000/0.72; Ec(k)*100000/0.3037<=x(2)<=Ec(k)*1000000/0.267;x(1)<=0.7*[Pb(k)+Pc(k)]*1000/1.2; |
新蟲 (初入文壇)
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