数学建模竞赛(化验结果的处理)解题论文.doc

2007年北京工业大学数学建模竞赛初赛试题B题：化验结果的处理 题解摘要：本文运用了距离判别和Fisher判别两种方法对问题进行分析求解，得出了我们想要的结论，即通过体内元素含量较准确的判别个体是否患有肾炎。1、 问题的提出人们到医院就诊时，通常要化验一些指标来协助医生的诊断。诊断就诊人员是否患肾炎时通常要化验人体内各种元素含量。表B.1是确诊病例的化验结果，其中130号病例是已经确诊为肾炎病人的化验结果；3160号病例是已经确定为健康人的结果。表B.2是就诊人员的化验结果。我们的问题是：1) 根据表B.1中的数据，提出一种或多种简便的判别方法，判别属于患者或健康人的方法，并检验你提出方法的正确性。2) 按照1提出的方法，判断表B.2中的30名就诊人员的化验结果进行判别，判定他（她）们是肾炎病人还是健康人。3) 能否根据表B.1的数据特征，确定哪些指标是影响人们患肾炎的关键或主要因素，以便减少化验的指标。4) 根据3的结果，重复2的工作。5) 对2和4的结果作进一步的分析。（表见附录）2、问题分析1) 题目中表.1中给出了已经确诊为肾炎病人和健康人的各30组数据；2) 每一组数据都有七个数，分别代表了Zn, Cu, Fe, Ca, Mg, K, Na在每个人体内的量；3) 第一问要求我们提出判别一个人属于患者还是健康人的方法，这就需要通过对60组数据的分析得出健康人和肾炎患者体中这些元素量之差异，这些差异的大小又同时是解决第三问的主要影响因素；4) 在寻找数据的差异时，我们用到的传统方法就是求数据的方差和均值，用excel列表分析，用matlab作直方图分析。5) 第二问最可靠的方法就是用判别分析来做，这就需要在R软件中进行一些必要的编程和处理；6) 第四问是建立在第三问的基础上的；当解决了第三问中到底是那些因素影响到了人们患肾炎的关键时，只需要在那些主要因素中进行判断就可以省去一些复杂繁琐的步骤；7) 将以上问题都解决之后，我们使用和步骤5）相同的方法，使用R软件帮助我们高效地对精简后的数据进行再次分析，并且把第二问和第四问的结果之间进行比较，观察差异和详细的分析。8) 为了进一步验证我们这种做法的合理性，我们又要用C语言编一个程序，把表B2中的数据与4）中所求出各元素的均值进行比较，进行了一下直观的分析。3符号约定后缀为1：患者体内元素的含量（例如：Zn1代表患者体内Zn的含量）；后缀为2：健康人体内元素的含量（例如：Zn2代表健康人体内Zn的含量）；1：患者；2：健康人；4模型假设1） 题中所给的内容和数据都是真实可信的；2） 除了表中列出的元素外，其他元素对是否会患肾炎的影响很小；3） 外界条件对肾炎患者的影响不计；4） 没病的个体都是健康体。5模型建立该问题的关键是如何判断一个人是有病的还是健康的，即这是个判别问题，可以采用统计方法中的判别分析法进行分析处理。题目中只有两类病体和健康体，所以可采用二类群体的判别方法。首先考虑用一种简单而直观的判别方法Mahalanobis距离判别。根据两个母体样本计算出他们的均值向量和协方差阵，求取待测样本x对两个样本的Mahalanobis距离，二者取差值，判断离那个母体近似。设x,y是从均值为，协方差阵为的总体A中抽取的样本，则总体A内两点x与y的Mahalanobis距离定义为.定义样本x与A的Mahalanobis距离为。在现实中，母体的均值向量和协方差阵由样本的均值向量和协方差阵来代替：设，是来自母体A的个样本，是来自母体A的个样本，则样本的均值与协方差为，对于待测样本x，如果两个母体样本的协方差相同，由得到判别函数为，其中，其判别准则是 。如果两个母体样本协方差不同，即，对于样本x判别函数定义为：，。其次考虑用另外一种方法求取解决办法Fisher判别法，即按类内方差尽量小，类间方差尽量大的准则来求判断函数。设两个总体A、B的均值和协方差阵分别是、和、，对任一测样本x，设它的判别函数为，并假设,使满足类内偏差平方和最小，而类间偏差平方和最大，其中。即要满足最大，若，则，否则。通过推导得出判别函数，其中，当，,否则。6模型求解利用模型求解时通过R软件将以上两种算法编写成程序代码，通过手动输入样本，利用计算机进行求解，程序清单如下：Mahalanobis距离判别：A-matrix(c(166,15.8,24.5,700,112,179,513,185,15.7,31.5,701,125,184,427,193,9.80,25.9,541,163,128,642,159,14.2,39.7,896,99.2,239,726,226,16.2,23.8,606,152,70.3,218,171,9.29,9.29,307,187,45.5,257,201,13.3,26.6,551,101,49.4,141,147,14.5,30.0,659,102,154,680,172,8.85,7.86,551,75.7,98.4,318,156,11.5,32.5,639,107,103,552,132,15.9,17.7,578,92.4,1314,1372,182,11.3,11.3,767,111,264,672,186,9.26,37.1,958,233,73.0,347,162,8.23,27.1,625,108,62.4,465,150,6.63,21.0,627,140,179,639,159,10.7,11.7,612,190,98.5,390,117,16.1,7.04,988,95.5,136,572,181,10.1,4.04,1437,184,101,542,146,20.7,23.8,1232,128,150,1092,42.3,10.3,9.70,629,93.7,439,888,28.2,12.4,53.1,370,44.1,454,852,154,13.8,53.3,621,105,160,723,179,12.2,17.9,1139,150,45.2,218,13.5,3.36,16.8,135,32.6,51.6,182,175,5.84,24.9,807,123,55.6,126,113,15.8,47.3,626,53.6,168,627,50.5,11.6,6.30,608,58.9,58.9,139,78.6,14.6,9.70,421,70.8,133,464,90.0,3.27,8.17,622,52.3,770,852,178,28.8,32.4,992,112,70.2,169),ncol=7, byrow=T)B-matrix(c(213,19.1,36.2,2220,249,40.0,168,170,13.9,29.8,1285,226,47.9,330,162,13.2,19.8,1521,166,36.2,133,203,13.0,90.8,1544,162,98.90,394,167,13.1,14.1,2278,212,46.3,134,164,12.9,18.6,2993,197,36.3,94.5,167,15.0,27.0,2056,260,64.6,237,158,14.4,37.0,1025,101,44.6,72.5,133,22.8,31.0,1633,401,180,899,156,135,322,6747,1090,228,810,169,8.00,308,1068,99.1,53.0,289,247,17.3,8.65,2554,241,77.9,373,166,8.10,62.8,1233,252,134,649,209,6.43,86.9,2157,288,74.0,219,182,6.49,61.7,3870,432,143,367,235,15.6,23.4,1806,166,68.8,188,173,19.1,17.0,2497,295,65.8,287,151,19.7,64.2,2031,403,182,874,191,65.4,35.0,5361,392,137,688,223,24.4,86.0,3603,353,97.7,479,221,20.1,155,3172,368,150,739,217,25.0,28.2,2343,373,110,494,164,22.2,35.5,2212,281,153,549,173,8.99,36.0,1624,216,103,257,202,18.6,17.7,3785,225,31.0,67.3,182,17.3,24.8,3073,246,50.7,109,211,24.0,17.0,3836,428,73.5,351,246,21.5,93.2,2112,354,71.7,195,164,16.1,38.0,2135,152,64.3,240,179,21.0,35.0,1560,226,47.9,330),ncol=7,byrow=T)X-matrix(c(58.2,5.42,29.7,323,138,179,513,106,1.87,40.5,542,177,184,427,152,0.80,12.5,1332,176,128,646,85.5,1.70,3.99,503,62.3,238,762.6,144,0.70,15.1,547,79.7,71.0,218.5,85.7,1.09,4.2,790,170,45.8,257.9,144,0.30,9.11,417,552,49.5,141.5,170,4.16,9.32,943,260,155,680.8,176,0.57,27.3,318,133,99.4,318.8,192,7.06,32.9,1969,343,103,553,188,8.28,22.6,1208,231,1314,1372,153,5.87,34.8,328,163,264,672.5,143,2.84,15.7,265,123,73.0,347.5,213,19.1,36.2,2220,249,62.0,465.8,192,20.1,23.8,1606,156,40.0,168,171,10.5,30.5,672,145,47.0,330.5,162,13.2,19.8,1521,166,36.2,133,203,13.0,90.8,1544,162,98.9,394.5,164,20.1,28.9,1062,161,47.3,134.5,167,13.1,14.1,2278,212,36.5,96.5,164,12.9,18.6,2993,197,65.5,237.8,167,15.0,27.0,2056,260,44.8,72.0,158,14.4,37.0,1025,101,180,899.5,133,22.8,31.3,1633,401,228,289,169,8.0,30.8,1068,99.1,53.0,817,247,17.3,8.65,2554,241,77.5,373.5,185,3.90,31.3,1211,190,134,649.8,209,6.43,86.9,2157,288,74.0,219.8,182,6.49,61.7,3870,432,143,367.5,235,15.6,23.4,1806,166,68.9,188),ncol=7, byrow=T)discri1-function(TrnA,TrnB,TstX=NULL,var.equal=FALSE) /*TrnA,TrnB,TstX分别为有病，健康和待测得样本， var.equal缺省值为FALSE，意思是以样本TrnA为参考if (is.null(TstX) = TRUE) TstX-rbind(TrnA,TrnB) nx-nrow(TstX); blong-array(0, c(nx) /*把待测样本数（30）给nx，建立一个向量值为0的1行30列的矩阵Ab-apply(TrnA,2,mean); Bb-apply(TrnB,2,mean) /Ab为1行7列的矩阵，列向量为TrnA矩阵对应列的所有数的均值，Bb同理if (var.equal = TRUE | var.equal = T) /两个样本的协方差相等时S-var(rbind(TrnA,TrnB); Xb-(Ab+Bb)/2 /计算，for (i in 1:nx)w0)blongi-1 /待测体有病elseblongi-2/待测体没病elseSa-var(TrnA); Sb-var(TrnB); /两个样本的协方差不等时for (i in 1:nx)y-TstXi,-Ab; z0)blongi-1elseblongi-2blongdiscri1(A,B,X)Fisher判别：A-matrix(c(166,15.8,24.5,700,112,179,513,185,15.7,31.5,701,125,184,427,193,9.80,25.9,541,163,128,642,159,14.2,39.7,896,99.2,239,726,226,16.2,23.8,606,152,70.3,218,171,9.29,9.29,307,187,45.5,257,201,13.3,26.6,551,101,49.4,141,147,14.5,30.0,659,102,154,680,172,8.85,7.86,551,75.7,98.4,318,156,11.5,32.5,639,107,103,552,132,15.9,17.7,578,92.4,1314,1372,182,11.3,11.3,767,111,264,672,186,9.26,37.1,958,233,73.0,347,162,8.23,27.1,625,108,62.4,465,150,6.63,21.0,627,140,179,639,159,10.7,11.7,612,190,98.5,390,117,16.1,7.04,988,95.5,136,572,181,10.1,4.04,1437,184,101,542,146,20.7,23.8,1232,128,150,1092,42.3,10.3,9.70,629,93.7,439,888,28.2,12.4,53.1,370,44.1,454,852,154,13.8,53.3,621,105,160,723,179,12.2,17.9,1139,150,45.2,218,13.5,3.36,16.8,135,32.6,51.6,182,175,5.84,24.9,807,123,55.6,126,113,15.8,47.3,626,53.6,168,627,50.5,11.6,6.30,608,58.9,58.9,139,78.6,14.6,9.70,421,70.8,133,464,90.0,3.27,8.17,622,52.3,770,852,178,28.8,32.4,992,112,70.2,169),ncol=7, byrow=T)B-matrix(c(213,19.1,36.2,2220,249,40.0,168,170,13.9,29.8,1285,226,47.9,330,162,13.2,19.8,1521,166,36.2,133,203,13.0,90.8,1544,162,98.90,394,167,13.1,14.1,2278,212,46.3,134,164,12.9,18.6,2993,197,36.3,94.5,167,15.0,27.0,2056,260,64.6,237,158,14.4,37.0,1025,101,44.6,72.5,133,22.8,31.0,1633,401,180,899,156,135,322,6747,1090,228,810,169,8.00,308,1068,99.1,53.0,289,247,17.3,8.65,2554,241,77.9,373,166,8.10,62.8,1233,252,134,649,209,6.43,86.9,2157,288,74.0,219,182,6.49,61.7,3870,432,143,367,235,15.6,23.4,1806,166,68.8,188,173,19.1,17.0,2497,295,65.8,287,151,19.7,64.2,2031,403,182,874,191,65.4,35.0,5361,392,137,688,223,24.4,86.0,3603,353,97.7,479,221,20.1,155,3172,368,150,739,217,25.0,28.2,2343,373,110,494,164,22.2,35.5,2212,281,153,549,173,8.99,36.0,1624,216,103,257,202,18.6,17.7,3785,225,31.0,67.3,182,17.3,24.8,3073,246,50.7,109,211,24.0,17.0,3836,428,73.5,351,246,21.5,93.2,2112,354,71.7,195,164,16.1,38.0,2135,152,64.3,240,179,21.0,35.0,1560,226,47.9,330),ncol=7,byrow=T)X-matrix(c(58.2,5.42,29.7,323,138,179,513,106,1.87,40.5,542,177,184,427,152,0.80,12.5,1332,176,128,646,85.5,1.70,3.99,503,62.3,238,762.6,144,0.70,15.1,547,79.7,71.0,218.5,85.7,1.09,4.2,790,170,45.8,257.9,144,0.30,9.11,417,552,49.5,141.5,170,4.16,9.32,943,260,155,680.8,176,0.57,27.3,318,133,99.4,318.8,192,7.06,32.9,1969,343,103,553,188,8.28,22.6,1208,231,1314,1372,153,5.87,34.8,328,163,264,672.5,143,2.84,15.7,265,123,73.0,347.5,213,19.1,36.2,2220,249,62.0,465.8,192,20.1,23.8,1606,156,40.0,168,171,10.5,30.5,672,145,47.0,330.5,162,13.2,19.8,1521,166,36.2,133,203,13.0,90.8,1544,162,98.9,394.5,164,20.1,28.9,1062,161,47.3,134.5,167,13.1,14.1,2278,212,36.5,96.5,164,12.9,18.6,2993,197,65.5,237.8,167,15.0,27.0,2056,260,44.8,72.0,158,14.4,37.0,1025,101,180,899.5,133,22.8,31.3,1633,401,228,289,169,8.0,30.8,1068,99.1,53.0,817,247,17.3,8.65,2554,241,77.5,373.5,185,3.90,31.3,1211,190,134,649.8,209,6.43,86.9,2157,288,74.0,219.8,182,6.49,61.7,3870,432,143,367.5,235,15.6,23.4,1806,166,68.9,188),ncol=7, byrow=T)discri2-function(TrnA,TrnB,TstX=NULL) if(is.null(TstX)=TRUE)TstX-rbind(TrnA,TrnB) nx-nrow(TstX);blong-array(0,c(nx) na-nrow(TrnA);nb-nrow(TrnB) Ab-apply(TrnA,2,mean);Bb-apply(TrnB,2,mean)S-(na-1)*var(TrnA)+(nb-1)*var(TrnB)xb-na/(na+nb)*Ab+nb/(na+nb)*Bbfor(i in 1:nx)w-(TstXi,-xb)%*%solve(S,Bb-Ab);if(w=0)blongi-1elseblongi-2blongdiscri2(A,B,X)7模型结果及分析Mahalanobis距离判别法结果：1 1 2 1 2 1 2 2 2 2 1 1 2 2 2 1 2 2 2 2 2 2 1 2 1 2 2 2 2 2Fisher判别：1 1 2 1 1 1 2 1 1 2 1 1 1 2 1 1 2 2 1 2 2 2 1 2 1 2 2 2 2 2通过结果可以看出这两种方法得到的结果基本一致，但存在一定的误差，对结果有出入的个体应进行进一步的检查，以达到确诊的目的。期望值表示样本的平均值，通过健康者与患者的期望差的绝对值除以健康人的期望分析各元素对总体结果的影响程度，值越大说明该元素对诊断结果起的决定作用越大。首先确定Fe、Ca、Mg、K为主要元素，并再次通过程序对待测样本进行测试，再对Ca和K进行一次测试，之后单独对K进行测试，整理得到如下结果:Mahalanobis: 没去 discri11 1 1 2 1 2 1 2 2 2 2 1 1 2 2 2 1 2 2 2 2 2 2 1 2 1 2 2 2 2 24种discri1(A,B,X)1 1 1 2 1 1 1 2 2 1 2 1 1 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 22种discri1(A,B,X)1 1 1 2 1 1 1 1 1 1 2 1 1 1 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 21种discri1(A,B,X)1 1 1 1 1 2 1 1 1 2 2 1 1 2 1 1 1 1 2 1 1 2 1 1 1 1 2 1 2 1 2Fisher:没去 descri21 1 1 2 1 1 1 2 1 1 2 1 1 1 2 1 1 2 2 1 2 2 2 1 2 1 2 2 2 2 24种discri2(A,B,X)1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 2 1 2 2 1 2 2 2 1 1 1 2 1 2 2 22种discri2(A,B,X)1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 2 1 2 2 1 2 2 2 1 1 1 2 1 2 2 21种discri2(A,B,X)1 1 1 2 1 2 2 2 1 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2通过比较，发现用4种元素进行判断较准确，只用一种元素的方法误差较大，可以排除。8模型检验 Mahalanobis距离判别和Fisher的方法相互检验Mahalanobis: 1 1 2 1 2 1 2 2 2 2 1 1 2 2 2 1 2 2 2 2 2 2 1 2 1 2 2 2 2 2Fisher: 1 1 2 1 1 1 2 1 1 2 1 1 1 2 1 1 2 2 1 2 2 2 1 2 1 2 2 2 2 24种discri1(A,B,X)1 1 1 2 1 1 1 2 2 1 2 1 1 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 24种discri2(A,B,X)1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 2 1 2 2 1 2 2 2 1 1 1 2 1 2 2 22种discri1(A,B,X)1 1 1 2 1 1 1 1 1 1 2 1 1 1 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 22种discri2(A,B,X)1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 2 1 2 2 1 2 2 2 1 1 1 2 1 2 2 2只有四个样本个体的检验结果不同，因此判断这两种方法都可行。9模型应用这种判别在实际生活中很多，所以这些判别法在日常发挥着很多的作用，除了看病外，地质学中判断有矿无矿，工厂判断产品合格不合格，为新发现的物种分类都可以用这些方法进行分析。10总结曲折道路，我们一起走过刚刚拿到题目的时候，我们很简单的认为所给出的数据会比较有规律，然而细看发现样本的7种随即向量的数值之间有很大差异，离散程度非常的大，而且对于患病与非患病没有一个明确的指标，想要通过向量值轻松地建立判别函数比较困难，进行数据拟合也不太容易。我们进一步考虑均值和方差对于样本判别的作用，均值代表样本的平均值，我们通过待测样本的随即向量与有病体和健康体的均值做除法得到的结果进行判别，通过自己编写的C程序得到数据表如下：但发现数据依然十分的离散，无法得到正确的解。C程序清单如下：#includestdio.hvoid main() int i,j; char *counter7=Zn,Cu,Fe,Ca,Mg,K,Na; double imean7=143.1033,12.33433,23.06667,698.1667,113.3933,201.1333,526.8333; double hmean7=186.6,21.92367,62.01167,2511.133,295.1367,90.37,367.21; double array307=58.2,5.42,29.7,323,138,179,513, 106,1.87,40.5,542,177,184,427, 152,0.80,12.5,1332,176,128,646, 85.5,1.70,3.99,503,62.3,238,762.6, 144,0.70,15.1,547,79.7,71.0,218.5, 85.7,1.09,4.2,790,170,45.8,257.9, 144,0.30,9.11,417,552,49.5,141.5, 170,4.16,9.32,943,260,155,680.8, 176,0.57,27.3,318,133,99.4,318.8, 192,7.06,32.9,1969,343,103,553, 188,8.28,22.6,1208,231,1314,1372, 153,5.87,34.8,328,163,264,672.5, 143,2.84,15.7,265,123,73.0,347.5, 213,19.1,36.2,2220,249,62.0,465.8, 192,20.1,23.8,1606,156,40.0,168, 171,10.5,30.5,672,145,47.0,330.5, 162,13.2,19.8,1521,166,36.2,133, 203,13.0,90.8,1544,162,98.9,394.5, 164,20.1,28.9,1062,161,47.3,134.5, 167,13.1,14.1,2278,212,36.5,96.5, 164,12.9,18.6,2993,197,65.5,237.8, 167,15.0,27.0,2056,260,44.8,72.0, 158,14.4,37.0,1025,101,180,899.5, 133,22.8,31.3,1633,401,228,289, 169,8.0,30.8,1068,99.1,53.0,817, 247,17.3,8.65,2554,241,77.5,373.5, 185,3.90,31.3,1211,190,134,649.8, 209,6.43,86.9,2157,288,74.0,219.8, 182,6.49,61.7,3870,432,143,367.5, 235,15.6,23.4,1806,166,68.9,188; double iresult307=0.0; double hresult307=0.0; int judgement30=1,1,2,1,2,1,2,2,2,2,1,1,2,2,2,1,2,2,2,2,2,2,1,2,1,2,2,2,2,2; FILE *fp=NULL; for(i=0;i30;i+) for(j=0;j7;j+) iresultij=arrayij/imeanj; hresultij=arrayij/hmeanj; fp=fopen(math.txt,w); fprintf(fp, ); for(i=0;i7;i+) fprintf(fp, %5s,counteri); fprintf(fp, ); for(i=0;i7;i+) fprintf(fp, %5s,counteri); fprintf(fp,n); for(i=0;i30;i+) fprintf(fp,%d ,judgementi); fprintf(fp,%2d,i+1); for(j=0;j7;j+) fprintf(fp, %5.2lf,iresultij); fprintf(fp, %2d ,i+1); for(j=0;j7;j+) fprintf(fp, %5.2lf,hresultij); fprintf(fp,n); getch();其中的初始数据imean和hmean为R计算得到的两个样本的均值。我们认真地学习了一下数学建模基础，发现这是一个典型的二类判别问题，可以用文中所述的两种分析方法通过数学模型进行分析，经过模型计算，我们得到了两组结果。这两组结果相互检验。然而这两种方法对于某些个体存在一定的差异，我们希望得到一个更准确的结果，但从正面下手很难进行，所以我们考虑用假设的方法检验是否存在矛盾，即将某个存在差异的个体划入健康体，即健康体的样本中多了一组数据，再将这个差异个体对两个样本进行距离判别，如果出现属于病体的情况，则出现了矛盾，则他一定是病体，否则他是健康体，然而在实际解决问题过程中，只要进行运算所选择的模型不变已经得到的结果再进行一次运算也还是同样的结果，不会发生改变，因此这个方法被否定了。而且对于这个问题我们暂时并没有进一步的突破，其实在实际生活中医学诊断结果也未必能准确，是存在一定的误差的。在考虑哪些是较重要的元素时我们产生了分歧，最简单最直观的方法是对每个元素进行测试，但误差会很大，有人提出只测一种不如去掉一种再测试，经过商量我们认为这样也不好，对于这七种元素，它的幂集有27种，如果一一测量实在是一个巨大的工作量，绝不是一个好的方法。之后有人提出用病体与健康体的样本均值差的绝对值除以健康样本的均值，通过比值的大小判断该元素对判断起的作用大小。通过这种方法我们解决了这个问题。通过这道题目，我们体会到了学习数学的快乐，思考的快乐，以前只是枯燥为了学习数学而学习数学，其实实际中的数学更有意思，而且通过数学建模，我们回顾了以前的知识，并学到了新的东西，感觉参加这个活动是一个明智的选择，非常荣幸。11参考文献数学建模基础 薛毅 主编北京工业大学出版社精通MATLAB6.5版 张志勇 等编著北京航空航天大学出版社概率论与数理统计 王松桂 等编著科学出版社附录：表B.1 确诊病例的化验结果病例号ZnCuFeCaMgKNa116615.824.5700112179513218515.731.570112518442731939.8025.9541163128642415914.239.789699.2239726522616.223.860615270.321861719.299.2930718745.5257720113.326.655110149.4141814714.530.065910215468091728.857.8655175.798.43181015611.532.56391071035521113215.917.757892.4131413721218211.311.3767111264672131869.2637.195823373.0347141628.2327.162510862.4465151506.6321.06271401796391615910.711.761219098.53901711716.17.0498895.51365721818110.14.0414371841015421914620.723.8123212815010922042.310.39.7062993.74398882128.212.453.137044.14548522215413.853.36211051607232317912.217.9113915045.22182413.53.3616.813532.651.6182251755.8424.980712355.61262611315.847.362653.61686272750.511.66.3060858.958.91392878.614.69.7042170.81334642990.03.278.1762252.37708523017828.832.499211270.21693121319.136.2222024940.01683217013.929.8128522647.93303316213.219.8152116636.21333420313.090.8154416298.903