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响应曲面法与设计

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响应曲面法与设计,原理及实际的操作应用,对选用该法做实验的同学比较受用!7050等高当前运线图行条件最人值7D60504010040120x,=温度(C]160(10x2=压强si的162响应曲面的等高线图∑x,+∑Rnx+∑(16-2)几乎所有的RSM问题都用这些近似多项式中的一个或多个。当然,一个多项式模型在自变量的整个空间上是真实函数关系的合理的近似式是不可能的,但在-个相对小的区域内通常做的很好。第15章讨论的最小乘方法可用来估计近似多项式的参数然后在拟合曲面上儆嘀应曲面分析,如果拟合曲面是真实响应函数的个合适的近似式,则拟合曲面的分析就近似地等价于实际591系统的分析。如果能怜当地利用实验设计来收集数据,就能够最有效地估计模型参数。关于拟合响应曲面的设计叫做响应曲面设计。在16-4节中讨论这些设计RSM是一序贯方法。通常,当我们是在响应曲面相应的自变量区域内的某个点时,例如,像图162中当前运行条件那样的点,在此点处系统具有微小的弯曲,从而用一阶模型是恰当的。现在,我们的目的是要引导实验者快速而有效地到达最优点所在的邻近区域。一旦最优点的区域被找到,就可以用更精细的模型,例如阶模型并进行分析以便确定最优点的位置。由图162见出,响应曲面的分析法可以想像为“爬”一样,山顶代表响应的最大值点。如果真实的最优点是啊应的最小值点,则可设想为“落进山谷”。RSM的最终目的是确定系统的最优运行条件或确定因素空间中满足运行规范的区域。RSM主要不是用来了解系统的实际机制的,尽管RSM有助于得到这类知识。还有,RSM的“最优”是按特定的意义使用的。RSM的“爬山”方法只能保证收敛于局部的最优点162最速上升法系统最优运行条件的初步估计常常远离实际的最优点。在这种情况下实验者的目的是要快速地进入到最优点的附近区域。我们希望利用又简单又经济有效的实验方法。当远离最优点时,通常假定在x的一个小区域范围内一阶模型是真实曲面的合适近最速上升法是沿着最速上升的路径,郡响应有最大增量的方向逐步移动的方法。当然,如果求的是最小值,则叫做最速下降法。拟合的一阶模型是592·y=Rn+∑R;x与一阶响应曲面相应的y的等高线,是一系列平行的直线,如图l6-3所示。最速上升的方向就是y增加得最快的方面。这一方向一阶拟合响应最速上升路径曲面的区域=夕-20y-3图16-3--阶响应曲画的等高线与最速土升路径平行于合响应曲面等高线的法线方向。通常取通过所感兴趣的区城的中心并且垂直于拟合曲面等高线的直线为最速上升路径这样一来,沿着路径的步长就和回归系数{P}成正比。实际的步长大小是由实验者根据工序知识或其他的实际考虑来确定的实验是沿着最速上升的路径进行的直到观察到的响应不再593◆增加为止。然后,拟合一个新的一阶模型,确定-·条新的最速上升路径、继续按上述方法进行。最后,实验者到达最优点的附近区域。这一点,通常由一阶模型的拟合不足来指出。这时,进行如16-3节所述的添加的实验,会求得最优点的更为精确的估计例t6位化学工程帅要确定使化工产品收率最大的运行条件。影响收率的两个可控变量是:反应时和反应温度。工程师当前使用的运行条件是反应时同为35分钟,温度为155F,收率约为40%。因为此区域不大可能包含最优值←她拟合-阶模型并应用最速上升法。程师决定拟合一阶模型的探测区域应是反应时间为(30,40)分钟和(150,160)°F。为简化计算,将自变量规范在(-1,1)区间内。于是,如果记尔为自然时间变量,与:为自然温度变量,则规范变量是5155数据如表161所水。用来收集这些数据的设计是增加五个中心点的22析因设计,在中心点处的重复观察值是用来估计实验误差的,并可以用来检阶模型的适合性。还有,过程的当前运行条件也就在设计的中心点处用最小二乘法将一阶模型来拟合这些数据。用第15章的方法,求得以规范变量表示的下列模型y=49,44+0.775x1+0.325x2在沿着最速上升路径探测之前,应研究-阶模型的适合性。有中心点的22设计允许实验者去做1.求出误差的个估计量2.检测模型的交互作用(交叉乘积项3检测二次效应(弯曲性)。中心点处的重复观察值可月来计算误差的估计量如下:(40.3)2+(40.5)2+(49.7)2+(40,2)2+(40.6)2=(202.3)2/50.0430594表16-1拟會一阶模型的过程数据首然变量规范变量响应了1301539.3301604u.(小01504U.9404J.53543.335010.5351534(.了35l5544.235l554〔.6阶模型假定变量r2和x2对响应有可加的效应。变量间的交互作用可用增加于模型的交叉乘积项x2的系数12来度量。此系数的最小二乘估计恰好是按普通22析因设计算得的交作用效应的二分之一,或B=1[(1×3.93)÷(1×41-5)+(-1×40.0)+(-1×40.9-0。1)0.025单自由度的交可作用平方和是SS交互作甲〔.02比较SS炊作用和。给出下刘拟合个足统计量:交五卡0.0250.0430=0058与F…进行比较。显然.交可作用是不显著的对直线模型适合性的另一个检测是比较设计的析因部分的四个点处的平均响应,即y=40,425、和在设计的中心点处的平均响应,即兴=40.46如果设计于弯曲的曲面上·则yr-y是曲面的总弯曲性的度量。如果月1与A2“纯二次”项x与x的系数,则y-y是A1+R的一个估计量。在我们的例中,纯二次项的个估计量是B1:+B40.425—40.460.35与零假设H:1+P2=0有关的单自由度的平方和是tf(÷)(5)(-优35)24+027其中7利n分别是析因部分的点数和中心点数。因F0,0027。063〔.0430将它与F、,比较。没有显示出纯二次项的影响。此模型的方差分析概括在表15-2中。交互作用和弯曲性的检测都是不显菩的,前总回归的F检验是显著的。还有,月和P2的标准差是MS0.94300.10=1,24问归系数月和B2相对于它们的标准差都较大。在这一点上.我们没有理由怀疑阶模型的适合性要离开设计中心·点(x:=0,x2=0)—沿最速上升跸径移动,就要对沿x2方向每移动0.325个单位.我们将沿x1方向移动0.775个单位。于是,最速上升路径经过点(x1-0,xz=0)且斜率为0.325/0.775。工程师决定用5分钟反应时间作为基本步长。用与x1之间的关系式,知道5分钟反应时问等价于规范变量x1的步长为4x=1。因此,沿最速上升路径的步长是△x1-1.00和4x2=(0.325/0.775)△x;=042。L程师计算了沿此路径的点并观察了在这些点处的收率直至响应下降为止。其结果见表16-3,表中既列出了规范变量也列出了自然变量。虽然规范变量在数学上容易计算,但在过程运有中必须用自然变量。图16-4画出了沿最速上升路径的每一步处的收率图。直到第十步所观察到的响应都是增加的;但是,这以后的每·步收率都是减少的。因此,另一一个一阶模型应该在点(41=85,2=175)的附近区域进行拟合。596·衰L42一阶模型的方差分析变差来源平方和自由度均方西归(月1,A2)825:214412547.83残差0.1772(交互作用(0.自025)0.4025).058〔纯二次)U.0027)0.00270.053纯误差)0.⊥7200.0430总和3.002281%的显着性表16-3例16-1的最谅上升实验规范变量自然变量响应步长_巴原点351550.42原点+△1.0,42401574且,原点+2△2.000.8445ig42原点十343.001.2650原点+444.[0685563原点+5▲5.2.106016553.8原点+646-供2.526516759.9原点十7▲7.002.9470l6965.0原点+88.03.:6751710.4原点+9△78173原点+10419.420L75原点+11411.004.6290ITs76原点十12412.00549575.上个新的一阶模型在点〔51=85,52=175)附近拟合。探测的区域对与是[80,90],对2是[170,180],于是。规范变量是5979F0Z了456785t112步长图16-4例16-1中沿最速上升路径的收率对步长的图形35,-175再次用五个中心点的2设计。数据见表6-在。拟合表16-4的规范数据的一阶模型是y=:78,97+1.00x1+0.50x2此模型的方差分析,包括交作用和纯次项的检测,如表16-5所示。交可作用和纯次项的检测表明、阶模型不是合适的近似。真实曲的弯曲性指岄了我们已接近最优点。为更精确地确定最优点,在该点必须做进步的分忻2由例16-1见出,最速上升路径是和拟合的一阶模型598

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