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alice919金蟲 (小有名氣)
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[求助]
幫我看看問題出在哪?
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options = Display: 'iter' MaxFunEvals: [] MaxIter: [] TolFun: [] TolX: [] FunValCheck: [] OutputFcn: [] PlotFcns: [] ActiveConstrTol: [] Algorithm: [] AlwaysHonorConstraints: [] BranchStrategy: [] DerivativeCheck: [] Diagnostics: [] DiffMaxChange: [] DiffMinChange: [] FinDiffType: 'central' GoalsExactAchieve: [] GradConstr: [] GradObj: [] HessFcn: [] Hessian: [] HessMult: [] HessPattern: [] HessUpdate: [] InitialHessType: [] InitialHessMatrix: [] InitBarrierParam: [] InitTrustRegionRadius: [] Jacobian: [] JacobMult: [] JacobPattern: [] LargeScale: 'off' LevenbergMarquardt: [] LineSearchType: [] MaxNodes: [] MaxPCGIter: [] MaxProjCGIter: [] MaxRLPIter: [] MaxSQPIter: [] MaxTime: [] MeritFunction: [] MinAbsMax: [] NodeDisplayInterval: [] NodeSearchStrategy: [] NonlEqnAlgorithm: [] NoStopIfFlatInfeas: [] ObjectiveLimit: [] PhaseOneTotalScaling: [] Preconditioner: [] PrecondBandWidth: [] RelLineSrchBnd: [] RelLineSrchBndDuration: [] ScaleProblem: [] Simplex: [] SubproblemAlgorithm: [] TolCon: [] TolConSQP: [] TolGradCon: [] TolPCG: [] TolProjCG: [] TolProjCGAbs: [] TolRLPFun: [] TolXInteger: [] TypicalX: [] UseParallel: [] Norm of First-order Trust-region Iteration Func-count f(x) step optimality radius 0 3 2.06697e+014 2.49e+014 1 1 6 1.03581e+012 0.92278 2.86e+013 1 2 9 4.38753e+007 0.0759773 5.33e+011 2.31 3 12 0.0824035 0.000500989 3.85e+007 2.31 4 15 1.18085e-013 2.17124e-008 3.77e+003 2.31 5 16 1.18085e-013 3.37394e-017 3.77e+003 2.31 Equation solved, fsolve stalled. fsolve stopped because the relative size of the current step is less than the default value of the step size tolerance squared and the vector of function values is near zero as measured by the default value of the function tolerance. <stopping criteria details> x = 1.0000 0.9993 fval = 1.0e-006 * -0.3436 0.0006 extiflag = 2 output = iterations: 5 funcCount: 16 algorithm: 'trust-region-dogleg' firstorderopt: 3.7653e+003 message: [1x774 char] JAC = 1.0e+010 * 1.0957 -0.0000 -0.0013 0.0013 幫我看看我的問題出在哪,我該怎么解決?這是一個用來計算化學(xué)平衡組成 [ 來自小組 化工工程工藝 ] |
金蟲 (小有名氣)
主管區(qū)長 (職業(yè)作家)
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專家經(jīng)驗: +1059 |

主管區(qū)長 (職業(yè)作家)
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專家經(jīng)驗: +1059 |

金蟲 (小有名氣)
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function ChemEquil clear all clc global P P=600; %壓力單位:kpa %m=50; %氫氣:草酸二甲酯摩爾比 %KP1=[23.4 21.8 20.3 18.9 17.6 16.4 15.3 14.2 13.3 12.4 11.5 10.8]; %KP2=[0.0393 0.0375 0.0358 0.0342 0.0328 0.0314 0.0301 0.0289 0.0277 0.0267... % 0.0257 0.0247]; %KP3=[4.4346E05 3.6476E05 3.0202E05 2.5168E05 2.1101E05 1.7796E05 1.5094E05... %1.2872E05 1.1036E05 0.9509E05 0.8233E05 0.7163E05]; %x0=0.001; %x=fsolve(@equil1,x0,[]) x0=[1 0 ]'; options=optimset('Display','iter', 'LargeScale','off','TolFun', 1.0e-22,'TolX',1.0e-22) [x,fval,extiflag,output,JAC]=fsolve(@equil1,x0,options) %................................................... %平衡組成關(guān)系式 function f=equil1(x) global P %f=x^2*(51-x)-21.8*((50-2*x)^2*(1-x)*P) tmp1=50-2*x(1)-2*x(2); tmp2=51-x(1)-x(2); f(1)=(x(1)-x(2))*(x(1)+x(2))*tmp2*100-8630*((1-x(1))*tmp1^2*P); f(2)=(x(1)+x(2))*x(2)*tmp2*100-10.4*((x(1)-x(2))*tmp1^2*P); 這是源程序,可以幫我看看錯在哪?是不是要改算法 |
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