程序的运行结果

P10-1. 写出下列程序的运行结果。

(2)P15-1. 写出下列字母的ASCII 码。

A____、B____、Y____、Z____、a____、b____、y____、z____ o____、l____、9____、+____、=____、$____、( ____、<____ (提示:如果记不起来,你可以让个程序来帮忙啊!) 2.写出下列程序的运行结果。

(2)3.找出下列程序中的错误，并改正。

(2)4.编一个程序，对你自己的姓名进行加密。

加密的方法是：取姓名每一个字母的ASCII 再加7，作为该字母对应的密码。

5.编一程序，求出微软公司名称“Microsoft ”中各个字符对应ASCII 码的和。

6.编个程序找出以下字符的前一个字符和后一个字符分别是什么？ B E G J要求每行输出一个字符的前一个字符、该字符和它的后继字符，如第7.假设一个储物柜S 中有很多个格子，每个各自都编上不同的编号。

已知编号为65、66、67和68的格子中分别保存着标记为“A ”、“B ”、“C ”和“D ”的宝物。

65 66 67 68如果用S[65]表示编号为65的格子中的宝物，你知道其余的几种宝物该如何表示吗？P20-1.比较下列字符串的大小，并在横线上填上“>”“<”或“=”。

MS-DOS___Windows IE___Excel FrontPage___Word 2000___Win excel___exercise Power___Page 2.写出下列程序的运行结果。

3.找出下列程序中的错误，并改正它。

4.遍个程序算一算：下列微软公司的产品名称长度分别是多少？总长度是多少？Access、Excel、FrontPage、Outlook5.编程序比较你和爸爸的姓名（用拼音表示），将大的显示在前面，小的显示在后面。

P24-1.指出下列哪些不能作变量名使用，并说说理由。

Ty age pred S11 V AR END3 writeln d:= p_12 ord succ2 String integer x+y w8n k22.指出下列语句的错误，并改正：(1)win+3:=50(2)x2:=‟years‟(变量x2已说明为integer类型)(3)ky:=‟windows‟-…dows‟(想取得除“dows”以外的那部分字符)(4)p1:=p2:=a×b （想把a×b的值存入p1和p2两个变量）3.写出下列程序的运行结果。

n=input('请输入节点数:n=');n1=input('请输入支路数:n1=');isb=input('请输入平衡母线节点号:isb=');pr=input('请输入误差精度:pr=');B1=input('请输入由各支路参数形成的矩阵:B1=');B2=input('请输入由各节点参数形成的矩阵:B2='); na=input('请输入PQ节点数na=');Y=zeros(n);YI=zeros(n);e=zeros(1,n);f=zeros(1,n);V=zeros(1,n);o=zeros(1,n);for i=1:n1if B1(i,6)==0p=B1(i,1);q=B1(i,2);else p=B1(i,2);q=B1(i,1);endY(p,q)=Y(p,q)-1./(B1(i,3)*B1(i,5));YI(p,q)=YI(p,q)-1./B1(i,3);Y(q,p)=Y(p,q);YI(q,p)=YI(p,q);Y(q,q)=Y(q,q)+1./(B1(i,3)*B1(i,5)^2)+B1(i,4)./2;YI(q,q)=YI(q,q)+1./B1(i,3);Y(p,p)=Y(p,p)+1./B1(i,3)+B1(i,4)./2;YI(p,p)=YI(p,p)+1./B1(i,3);end%求导纳矩阵disp('导纳矩阵Y=');disp(Y);G=real(Y);B=imag(YI);BI=imag(Y);for i=1:nS(i)=B2(i,1)-B2(i,2);BI(i,i)=BI(i,i)+B2(i,5);endP=real(S);Q=imag(S);for i=1:ne(i)=real(B2(i,3));f(i)=imag(B2(i,3));V(i)=B2(i,4);endfor i=1:nif B2(i,6)==2V(i)=sqrt(e(i)^2+f(i)^2);o(i)=atan(f(i)./e(i));endendfor i=2:nB(i,i)=1./B(i,i);else IC1=i+1;for j1=IC1:nB(i,j1)=B(i,j1)./B(i,i);endB(i,i)=1./B(i,i);for k=i+1:nfor j1=i+1:nB(k,j1)=B(k,j1)-B(k,i)*B(i,j1);endendendendp=0;q=0;for i=1:nif B2(i,6)==2p=p+1;k=0;for j1=1:nif B2(j1,6)==2k=k+1;A(p,k)=BI(i,j1);endendendendfor i=1:naif i==naA(i,i)=1./A(i,i);else k=i+1;for j1=k:naA(i,j1)=A(i,j1)./A(i,i);endA(i,i)=1./A(i,i);for k=i+1:nafor j1=i+1:naA(k,j1)=A(k,j1)-A(k,i)*A(i,j1);endendendendICT2=1;ICT1=0;kp=1;kq=1;K=1;DET=0;ICT3=1; while ICT2~=0|ICT3~=0ICT2=0;ICT3=0;if i~=isbC(i)=0;for k=1:nC(i)=C(i)+V(k)*(G(i,k)*cos(o(i)-o(k))+BI(i,k)*sin(o(i)-o(k)));endDP1(i)=P(i)-V(i)*C(i);DP(i)=DP1(i)./V(i);DET=abs(DP1(i));if DET>=prICT2=ICT2+1;endendendNp(K)=ICT2;if ICT2~=0for i=2:nDP(i)=B(i,i)*DP(i);if i~=nIC1=i+1;for k=IC1:nDP(k)=DP(k)-B(k,i)*DP(i);endelsefor LZ=3:iL=i+3-LZ;IC4=L-1;for MZ=2:IC4I=IC4+2-MZ;DP(I)=DP(I)-B(I,L)*DP(L);endendendendfor i=2:no(i)=o(i)-DP(i);endkq=1;L=0;for i=i:nif B2(i,6)==2C(i)=0;L=L+1;for k=1:nC(i)=C(i)+V(k)*(G(i,k)*sin(o(i)-o(k))-BI(i,k)*cos(o(i)-o(k)));endDQ1(i)=Q(i)-V(i)*C(i);DQ(L)=DQ1(i)./V(i);DET=abs(DQ1(i));if DET>=prICT3=ICT3+1;endendendelse kp=0;if kp~=0L=0;for i=1:nif B2(i,6)==2C(i)=0;L=L+1;for k=1:nC(i)=C(i)+V(k)*(G(i,k)*sin(o(i)-o(k))-BI(i,k)*cos(o(i)-o(k)));endDQ1(i)=Q(i)-V(i)*C(i);DQ(L)=DQ1(i)./V(i);DET=abs(DQ1(i));endendendendNq(K)=ICT3;if ICT3~=0L=0;for i=1:naDQ(i)=A(i,i)*DQ(i);if i==nafor LZ=2:iL=i+2-LZ;IC4=L-1;for MZ=1:IC4I=IC4+1-MZ;DQ(I)=DQ(I)-A(I,L)*DQ(L);endendelseIC1=i+1;for k=IC1:naDQ(k)=DQ(k)-A(k,i)*DQ(i);endendendL=0;for i=1:nif B2(i,6)==2L=L+1;V(i)=V(i)-DQ(L);endendkp=1;K=K+1;elsekq=0;if kp~=0;K=K+1;endendfor i=1:nDy(K-1,i)=V(i);endenddisp('迭代次数');disp(K);disp('没有达到精度要求的有功功率个数');disp(Np);disp('没有达到精度要求的无功功率个数');disp(Nq);for k=1:nE(k)=V(k)*cos(o(k))+V(k)*sin(o(k))*j;o(k)=o(k)*180/pi;enddisp('各节点的实际电压标么值E为(节点号从小到大排列):'); disp(E);disp('各节点的电压大小V为(节点号从小到大排列):');disp(V);disp('各节点的电压相角时o为(节点号从小到大排列):'); disp(o);for p=1:nC(p)=0;for q=1:nC(p)=C(p)+conj(Y(p,q))*conj(E(q));endS(p)=E(p)*C(p);enddisp('各节点的功率S为(节点号从小到大排列):');disp(S);disp('各条支路的首端功率Si为(顺序同您输入B1时一样):');for i=1:n1if B1(i,6)==0p=B1(i,1);q=B1(i,2);else p=B1(i,2);q=B1(i,1);endSi(p,q)=E(p)*(conj(E(p))*conj(B1(i,4)./2)+(conj(E(p)*B1(i,5))-conj(E(q)))*conj(1./(B1(i,3)*B1(i, 5))));disp(Si(p,q));enddisp('各条支路的末端功率Sj为(顺序同您输入B1时一样):');for i=1:n1if B1(i,6)==0p=B1(i,1);q=B1(i,2);else p=B1(i,2);q=B1(i,1);endSj(q,p)=E(q)*(conj(E(q))*conj(B1(i,4)./2)+(conj(E(q)./B1(i,5))-conj(E(p)))*conj(1./(B1(i,3)*B1(i, 5))));disp(Sj(q,p));enddisp('各条支路的功率损耗DS为(顺序同您输入B1时一样):');for i=1:n1if B1(i,6)==0p=B1(i,1);q=B1(i,2);else p=B1(i,2);q=B1(i,1);endDS(i)=Si(p,q)+Sj(q,p);disp(DS(i));endfor i=1:KCs(i)=i;for j=1:nDy(K,j)=Dy(K-1,j);endend请输入节点数:n=5请输入支路数:n1=5请输入平衡母线节点号:isb=1请输入误差精度:pr=0.00001请输入由各支路参数形成的矩阵:B1=[120.03i01.050;230.08+0.3i0.5i10;240.1+0.35i010;340.04+0.25i0.5i10;350.015i01.051]请输入由各节点参数形成的矩阵:B2=[001.051.0501;03.7+1.3i1.05002;02+1i1.0500 2;01.6+0.8i1.05002;501.051.0503]请输入PQ节点数na=3导纳矩阵Y=0-33.3333i0+31.7460i00 00+31.7460i 1.5846-35.7379i-0.8299+3.1120i-0.7547+2.6415i00-0.8299+3.1120i 1.4539-66.9808i-0.6240+3.9002i0 +63.4921i0-0.7547+2.6415i-0.6240+3.9002i 1.3787-6.2917i0000+63.4921i0 0-66.6667i迭代次数6没有达到精度要求的有功功率个数444330没有达到精度要求的无功功率个数000000各节点的实际电压标么值E为(节点号从小到大排列):1.0500 1.0473-0.0759i 1.0017+0.3149i 1.0465-0.0862i0.9766+ 0.3856i各节点的电压大小V为(节点号从小到大排列):1.0500 1.0500 1.0500 1.0500 1.0500各节点的电压相角时o为(节点号从小到大排列):0-4.145617.4510-4.711721.5470各节点的功率S为(节点号从小到大排列):2.5302+1.8416i-3.7000-1.2812i-2.0000-3.7435i-1.6000+0.3100i 5.0000+3.6788i各条支路的首端功率Si为(顺序同您输入B1时一样):2.5302+1.8416i-1.1986+0.3020i0.0288-0.0081i1.6729-0.2175i5.0000+3.6788i各条支路的末端功率Sj为(顺序同您输入B1时一样): -2.5302-1.5751i1.3271-0.3715i-0.0287+0.0084i-1.5713+0.3016i-5.0000-3.1545i各条支路的功率损耗DS为(顺序同您输入B1时一样): 0+0.2665i0.1285-0.0695i8.1234e-005+2.8432e-004i0.1017+0.0841i0.0000+0.5243i。

实验报告线性方程组的求解一．上机题目已知方程组为：x1-2*x2+2*x3=-2;2*x1-3*x2-3*x3=4;4*x1+x2+6*x3=3；分别用矩阵的三角分解法和高斯列主元消去法求方程组的解二．目的要求掌握用矩阵的三角分解法和高斯列主元消去法设计程序，从而实现对线性方程组的求解。

三．方法原理1．高斯消去法是通过逐步消元的方法把原方程组化为等价的上三角形方程组，然后回代的求解过程。

高斯列主元消去法是在高斯消去法第k步时，不取a[k][k]作为主元，而是取满足|a[r][k]|=max|a[i][k]|(k<=i<=n)的a[r][k]作为主元，若有多个r满足则取最小的，然后再去交换第r行与第k行；最后利用高斯消元法求解。

2. 矩阵的三角分解法是方程组的系数矩阵A可以分解为一个下三角阵L和一个上三角阵U的乘积，即A=LU，则AX=b为LUX=b,根据Ly=b和UX=y求出方程组的解。

其中求L,U的过程为：先计算U的第一行和L的第一列U[1][j]=a[1][j](1<=j<=n),L[i][1]=a[i][1]/U[1][1];然后根据U[k][j]=a[k][j]-∑L[k][r]*U[r][j](1<=r<=k-1;j=k,k+1,…n)和L[i][k]=(a[i][k]-∑L[i][r]*U[r][k])/U[k][k](1<=r<k-1;i=k+1,…n)计算出U的第k行和L的第k 列，从而求出L和U矩阵四．算法步骤（N-S流程图）1．矩阵的三角分解法算法步骤如下：Step1:计算U的第一行和L的第一列U的第一行：i=1时a[1][j]=L[1][1]*U[1][j]则U[1][j]=a[1][j] L[1][1]=1 (j=1,2,…n );L的第一列：j=1时a[i][1]=L[i][1]*U[1][1]则L[i][1]=a[i][1]/a[1][j] (i=2,3,…n);Step2:计算U的第二行和L的第二列U的第二行：i=2时a[2][j]=L[2][1]*U[1][j]+L[2][2]*U[2][j] L[2][2]=1则U[2][j]=a[2][j] –L[2][1]*U[1][j] (j=2,3,…n);L的第二列：j=2时 a[i][2]=L[i][1]*U[1][2]]+L[i][2]*U[2][2]则L[i][2]=(a[i][2]-L[i][1]*U[1][2])/U[2][2] (i=3,4,…n);Step3:假设已经进行了（k-1）步，得到了U的前k-1行和L的前k-1列，则计算U的第k行和L的第k列U的第k行：i=k时U[k][j]=a[k][j]-∑L[k][r]*U[r][j] (1<=r<=k-1;j=k,k+1,…n)L的第k列：j=k时L[i][k]=(a[i][k]-∑L[i][r]*U[r][k])/U[k][k] (1<=r<k-1;i=k+1,…n)Step4:利用L，U求解方程组的解根据A=L*U,A*X=y,从而得到 Ly=b和UX=y，然后根据高斯消元法的回代部分即可分别求出y和x2. 高斯列主元消去法算法步骤如下：Step1:选出列主元a[r][k]=max|a[i][k]| (k<=i<=n);Step2:判断方程组是否适合使用该方法求解若a[r][k]<e,则方程组不适合使用该方法求解，程序结束，否则进行下一步Step3:若r不等于k，交换r,k行Step4:高斯消元求解过程消元过程：当k=0,i=1时，令a[1][0]=a[1][0]/a[0][0];b[1]=b[1]-a[1][0]*b[1]； j=1时a[1][1]=a[1][1]-a[1][0]*a[0][1]当消元进行到第k次时，令a[k][k-1]=a[k][k-1]/a[k-1][k-1]; b[k]=b[k]-a[k][k-1]*b[k]； j=k时a[k][k]=a[k][k]-a[k][k-1]*a[k][k-1]回代求解过程：b[n-1]=b[n-1]/a[n-1][n-1];当k=n-1时，b[k]=(b[k]-∑a[k][j]*b[j])/a[k][k](k+1<=j<=n;k=n-1,n-2,...1);返回值b[k]即为所求解。

一、单选题(共25题，每题2分，共50分)1. 下列程序运行的结果是？（）s = 'hello'print(s+'world')A. sworldB. helloworldC. helloD. world标准答案：B试题难度：一般试题解析：s和'world'都属于字符串类型，加法运算表示的是字符串拼接的操作，所以最后得到的答案为helloworld，所以选择B选项。

2. 下列选项中不符合Python语言变量命名规则的是？（）A. ComputerB. PC. 3_1D. _WO1标准答案：C试题难度：一般3. 在Python中，运行9//2，输出的结果是？（）A. 3B. 4.5C. 4D. 4.0标准答案：C试题难度：一般4. 下面哪一行代码的输出结果不是World2021？（）A. print("World"+"2021")B. print("World"+"20"+"21")C. print("World"+2021)D. print("World2021")标准答案：C试题难度：一般5. 在Python中，输入3*4**2，运算结果是？（）A. 144B. 24C. 48D. 6标准答案：C试题难度：一般6. 关于比较运算符说法正确的是？（）①!=表示为不等于，如果两个操作数不相等，则为False②<=表示为小于等于，如果左边的数小于或等于右边的数，则为True③若a=2,b=5则a!=b为TrueA. ①②B. ②③C. ①③D. ①②③标准答案：B试题难度：一般7. Python中的乘法是用哪个符号表示的？（）A. *B. XC. xD. #标准答案：A试题难度：一般8. 以下哪个选项可以作为Python文件的后缀名？（）A. .pyB. .pngC. .docD. .pdf标准答案：A试题难度：容易9. 要给三个整型变量a、b、c赋值为5，下面Python程序正确的是？（）A. abc=5B. a=5,b=5,c=5C. a=b=c=5D. a=5 b=5 c=5标准答案：C试题难度：容易试题解析：此题考查对变量赋值的理解，根据Python中对变量的赋值语法故答案选择C选项10. 以下哪段程序能在画出三角形并隐藏turtle？（）A. import turtle turtle.circle(150,steps=3 )turtle.hideturtle() turtle.done()B. import turtleturtle.circle(150,3)turtle.hideturtle()turtle.done()C. import turtleturtle.circle(3)turtle.hideturtle()turtle.done()D. import turtleturtle.circle(150,3,3)turtle.hideturtle()标准答案：A试题难度：较难11. turtle.home() 的作用是下列哪一种？（）A. 移至初始坐标(0,0)B. 移至初始坐标(0,0)，并设置朝向为初始方向C. 移至屏幕左上角D. 设置朝向为初始方向标准答案：B试题难度：一般12. 关于Turtle绘图，下列说法错误的是？（）A. 色彩处理时，可以使用彩色画笔pencolor（），也可以直接由color( )方法更改目前画笔的颜色B. penup()指的是将笔提起，不会绘制任何图形C. 在选择画笔粗细时可以使用pensize（）D. 在海龟绘图中，画布中央是（0，0），往右X坐标值递减，往左X坐标值递增标准答案：D试题难度：一般13. 在Python中，输入18/6//3，输出结果为？（）A. 1B. 1.0C. 9D. 9.0标准答案：B试题难度：一般14. print(88-8)的运行结果是？（）A. 88B. 80C. 88-8D. 81标准答案：B试题难度：容易试题解析：print语句中是一个数学运算式，执行顺序是先执行88-8数学运算，再输出运算结果。

close‎all %关闭打开了‎的所有图形‎窗口clc %清屏命令clear‎%清除工作空‎间中所有变‎量%定义时间范‎围t=[0:pi/10:8*pi];y=sin(t);plot(t,y,'b:squar‎e')close‎allclcclear‎%定义时间范‎围t=[0:pi/20:9*pi];grid onhold on %允许在同一‎坐标系下绘‎制不同的图‎形plot(t,sin(t),'r:*')plot(t,cos(t))plot(t,-cos(t),'k')%grid on %在所画出的‎图形坐标中‎添加栅格，注意用在p‎l ot之后‎4-1:close‎allclcclear‎%定义时间范‎围t=[0:pi/20:9*pi];hold on %允许在同一‎坐标系下绘‎制不同的图‎形plot(t,sin(t),'r:*')plot(t,cos(t))plot(t,-cos(t),'k')grid on %在所画出的‎图形坐标中‎添加栅格，注意用在p‎l ot之后‎hold off %覆盖旧图，自动把栅格‎去掉，且若要在加‎入栅格就%必须把gr‎i d on加在p‎l ot后面‎plot(t,-sin(t))grid on%主程序ex‎p2_10‎.mgloba‎l a %声明变量a‎为全局变量‎x=1:100;a=3;c=prods‎(x) %调用子程序‎p rods‎.m%子程序pr‎o d s.m% funct‎i on resul‎t=prods‎(x)% globa‎l a% resul‎t=a*sum(x);%声明了与主‎程序一样的‎全局变量a‎，以便在子程‎序中可以%使用主程序‎中定义的变‎量答案：15150‎exmdl‎2_1.mclear‎close‎allclct=[0:pi/20:5*pi];figur‎e(1)plot(t,out)grid onxlabe‎l('time')ylabe‎l('magni‎t ude')exp2_‎1.mclc %清屏clear‎%从内存中清‎除变量和函‎数more onecho on%求矩阵与矩‎阵的乘积，矩阵与向量‎的乘积A=[5 6 7;9 4 6;4 3 6]B=[3 4 5;5 7 9;7 3 1]C=A*BY=A*Xmore offecho off答案：%求矩阵与矩‎阵的乘积，矩阵与向量‎的乘积A=[5 6 7;9 4 6;4 3 6]A =5 6 79 4 64 3 6B=[3 4 5;5 7 9;7 3 1]B =3 4 55 7 97 3 1X=[5 ;7;8]X =578C=A*BC =94 83 8689 82 8769 55 53Y=A*XY =12189more offecho offexp2_‎2.mclcclear‎more onecho on%为便于理解‎，在程序等执‎行过程中显‎示程序的表‎达式a=16;b=12;c=3;d=4;e=a+b-c*df=e/2k=e\2h=c^3g=e+f+ ...2+1-9aa=sin(g)abs(aa)bb=2+3jcc=conj(bb)rbb=real(bb) log(rbb) sqrt(rbb) exp(rbb) echo off more offa=16;b=12;c=3;d=4;e=a+b-c*de =16f=e/2f =8k=e\2k =0.1250h=c^3h =27g=e+f+ ...2+1-9g =18aa=sin(g)aa =-0.7510abs(aa)ans =0.7510bb=2+3jbb =2.0000 +3.0000i‎cc=conj(bb)cc =2.0000 -3.0000i‎rbb=real(bb)rbb =2log(rbb)ans =0.6931sqrt(rbb)ans =1.4142exp(rbb)ans =7.3891echo offans =7.3891exp2_‎5.m%绘制单位圆‎clear‎close‎allclc%定义时间范‎围t=[0:0.01:2*pi];x=sin(t);y=cos(t);plot(x,y)axis([-1.5 1.5 -1.5 1.5])%限定x轴和‎y轴的显示‎范围grid onaxis('equal‎')%axis([xmin,xmax,ymin,ymax])函数来调整‎图轴的范围：答案：exp2_‎5_.mclear‎close‎allclct=[0:pi/20:5*pi];plot(t,sin(t),'r:*')axis([0 5*pi -1.5 1.5 ])%给x轴和y‎轴命名xlabe‎l('t(deg)')ylabe‎l('magni‎t ude')%给图形加标‎题title‎('sine wave from zero to 5\pi')%在指定位置‎创建说明性‎文字text(pi/2,sin(pi/2),'\bulle‎t\lefta‎r row The sin(t) at t=2')%图形文字标‎示命令的使‎用clear‎close‎allclct=[0:pi/20:5*pi];plot(t,sin(t),'r:*')axis([0 5*pi -1.5 1.5 ])%给x轴和y‎轴命名xlabe‎l('t(deg)')ylabe‎l('magni‎t ude')%给图形加标‎题title‎('sine wave from zero to 5\pi')%在指定位置‎创建说明性‎文字text(pi/2,sin(pi/2),'\bulle‎t\lefta‎r row The sin(t) at t=2') %输入特定的‎字符%\pi%\alpha‎%\lefta‎r row%\right‎a rrow‎%\bulle‎t(点号)hold onplot(t,cos(t))%区分图形上‎不同的曲线‎legen‎d('sin(t)','cos(t)')%用鼠标在特‎定位置输入‎文字gtext‎('文字标示命‎令举例') hold offexp2_‎6.m%图形分割命‎令的使用clear‎close‎allclct=[0:pi/20:5*pi];subpl‎o t(321)plot(t,sin(t))axis([0 16 -1.5 1.5]) xlabe‎l('t(deg)') ylabe‎l('magni‎t ude') grid ontitle‎('sin(t)')subpl‎o t(322)plot(t,-sin(t))axis([0 16 -1.5 1.5]) xlabe‎l('t(deg)') ylabe‎l('magni‎t ude') grid ontitle‎('-sin(t)') subpl‎o t(323)plot(t,cos(t))axis([0 16 -1.5 1.5]) xlabe‎l('t(deg)') ylabe‎l('magni‎t ude') grid ontitle‎('cos(t)')subpl‎o t(324)plot(t,-cos(t))axis([0 16 -1.5 1.5]) xlabe‎l('t(deg)') ylabe‎l('magni‎t ude') grid ontitle‎('-cos(t)') subpl‎o t(325) subpl‎o t(326)exp2_‎7.mclcclear‎%绘制对应于‎每个输入x‎的输出y的‎高度条形图‎subpl‎o t(221)x=[1 2 3 4 5 6 7 8 9 10];y=[5 6 3 4 8 1 10 3 5 6];bar(x,y)%绘制x1在‎以y1为中‎心的区间中‎分布的个数‎条形图subpl‎o t(222)x1=randn‎(1,1000);%生成100‎0个各随机‎数y1=-3:0.1:3;hist(x1,y1)%绘制y2对‎应于x2的‎梯形图subpl‎o t(223)x2=0:0.1:10;y2=1./(x2.^3-2.*x2+4);stair‎s(x2,y2)%绘制y3对‎应于x3的‎散点图subpl‎o t(224)x3=0:0.1:10;y3=1./(x2.^3-2.*x2+4);stem(x3,y3)exp2_‎8.mecho off % 不显示程序‎内容%clear‎allclear‎clca=4;b=6disp('暂停，请按任意键‎继续') % disp指‎令可以用来‎显示字符pause‎% 暂停，直到用户按‎任意键echo on% 显示程序内‎容，注意mat‎l ab默认‎是不显示c=a+b% 暂时把控制‎权交给键盘‎（在命令窗口‎中出现k提‎示符), % 输入ret‎u rn，回车后退出‎，继续执行下‎面的语句。

软件系统运行情况报告篇一：软件系统运行情况报告篇一：软件系统运行总结报告自2月份开始，我一直在跟进xx银行项目的测试工作，至此为止已近6个月时间，从公司内部系统测试、验收测试，再到uat测试，以及投产前的系统压力测试等等。

从开始到项目即将结束，一步步走过来。

本次项目中，我作为测试环节的主力人员之一，仅对此项目中测试工作进行总结。

一、项目测试进度控制。

项目的测试进度主要是按照项目计划进行的，完全按照项目组计划要求完成测试任务、提交测试类相关文档，包括测试案例的完善、制定测试计划、执行测试、缺陷跟踪以及bug回归测试等。

协调项目的内部测试工作，本此项目中测试小组一共组织了四轮次系统全面测试工作，认真配合项目工作，共同保证项目质量。

项目测试的问题跟踪及处理采用每日进行修改问题回归测试工作，每日同步更新问题跟踪单的模式，按照规划时间完成系统更新测试。

二、项目组内部成员关系处理。

在项目工作的这几个月里大家相处融洽，项目组内部共同探讨解决问题的方法，向各模块负责人学习模块功能处理方式，向业务人员了解系统中涉及的业务知识点，两者结合起来进行模块功能测试。

鉴于之前辖内对公交易系统和中行对公项目的经验，也向项目组提出了一些完善性意见。

三、协调用户测试方面。

用户验收测试是项目测试工作的重要组成部分之一，是项目验收阶段的最终把关阶段，业务人员结合日常业务处理情况对系统进行的尝试性使用过程。

本次项目客户测试方面也是我个人觉得不够安全感一个主要方面，客户测试介入力度太小，尽管我们已经很多次电话催促业务人员测试，每次联系相关业务人员进行测试，他们来到项目组开发现场测试，也仅仅一两个小时时间，简单的进行验证操作即可。

xx银行利用两批系统培训的时间安排了两次分行集中测试，也算给项目进行了一次全面的测试，从中也暴露出不少系统存在的问题，目前项目组均已解决。

[莲~山课件]四、测试成效方面。

中信系统测试中，共记录问题及客户新增需求825个，其中bug数量512个、系统完善类问题225个，新增需求类问题88个。

写出程序运行结果1、下面程序的运行结果是：________void swap1(int c[]){ int t;t=c[0];c[0]=c[1];c[1]=t;}void swap2(int c0,int c1){ int t;t=c0;c0=c1;c1=t;}main( ){ int a[2]={2,4},b[2]={3,5};swap1(a); swap2(b[0],b[1]);printf(“%d %d %d %d\n”,a[0],a[1],b[0],b[1]); }2、下面程序的运行结果是：_____________main(){int a=2,ifor(i=0;i<3;i++)printf("%4d",ff(a));}ff(int a){int b=0;static int c=3;b++;c++;return(a+b+c);}3、下面程序的运行结果是：________main(){int i,j,k=10;for(i=0;i<2;i++){k++;{int k=0;for(j=0;j<=3;j++){if(j%2)continue;k++;}}k++;}printf("k=%d\n",k);}4、下面程序的运行结果：_________。

main(){int a,b,k=4,m=6,*p1=&k,*p2=&m;a=p1==&m;b=(-*p1)/(*p2)+7;printf("a=%d,",a);printf("b=%d\n",b);}5、以下程序运行结果：________。

funa(int a){int b=0;static int c=0;a=c++,b++;return (a);main(){int a=2,i,k;for(i=0;i<2;i++)k=fun(a++);printf("%d\n",k);}6、下面程序的运行结果：_____________struct stu{int num;char name[10];int age;}void fun(struct stu *p){printf("%s\n",(*p).name);}main(){struct stu students[3]={{2010,"zhang",20},{2011,"wang",19},{2012,"zhao",18};fun(students+2);}7、下面程序的运行结果是：_________。

Stata程序及运行结果(前面整理合并数据部分省略，考试会直接给合并好的数据)（由于是copy的，所以会缺少表格线）use "C:\Documents and Settings\Administrator\桌面\合并数据.dta", clear drop if q==. leverage==. profit==. fratio==. lna==.(416 observations deleted). sum q, dqPercentiles Smallest1% .601852 -7.6692855% .849397 .25189310% .938017 .283239 Obs 10397 25% 1.12204 .283675 Sum of Wgt. 1039750% 1.4441 Mean 4.98767 Largest Std. Dev. 184.905575% 2.066348 955.415890% 3.189917 1736.165 Variance 34190.03 95% 4.42306 11665.94 Skewness 72.16768 99% 11.19809 14665.54 Kurtosis 5324.15. gsort - q. drop if q<0 q>50(25 observations deleted). sum leverage, dleveragePercentiles Smallest1% .0378113 .00172535% .087829 .007079910% .1376331 .0075213 Obs 10372 25% .2841151 .0108272 Sum of Wgt. 1037250% .4686987 Mean .5390607 Largest Std. Dev. 1.67691175% .6367149 41.9393890% .7588252 41.93938 Variance 2.81203195% .8406378 96.95931 Skewness 42.38926 99% 1.6956 96.95931 Kurtosis 2225.629. drop if leverage>1 leverage<0(232 observations deleted). sum profit, dprofitPercentiles Smallest1% -.1462 -2.6817675% -.0283393 -1.14023210% .0102038 -1.094207 Obs 10140 25% .0322641 -.6430355 Sum of Wgt. 1014050% .0566138 Mean .0653552 Largest Std. Dev. .268733375% .086125 7.18842390% .1256777 7.188423 Variance .0722176 95% .1595877 8.599852 Skewness 60.59119 99% .2545736 22.00289 Kurtosis 4581.194. drop if profit<-1 profit>20(4 observations deleted). sum fratio, dfratioPercentiles Smallest1% .0277692 05% .1028079 010% .1602922 0 Obs 10136 25% .2636119 0 Sum of Wgt. 1013650% .402213 Mean .4088882 Largest Std. Dev. .192008375% .5483563 .955230890% .6696964 .9709213 Variance .0368672 95% .7325709 .9745661 Skewness .1536334 99% .8396031 .9745661 Kurtosis 2.416593. drop if fratio<=0(14 observations deleted). sum lna, dlnaPercentiles Smallest1% 19.19192 15.376355% 20.00134 17.4672910% 20.31043 17.49501 Obs 1012225% 20.82361 17.60426 Sum of Wgt. 1012250% 21.5525 Mean 21.72332 Largest Std. Dev. 1.2946575% 22.43455 28.003190% 23.44376 28.13565 Variance 1.67611995% 24.13284 28.28206 Skewness .780972499% 25.51874 28.40521 Kurtosis 4.300583. *sum q leverage profit fratio lna, d. reg leverage qSource SS df MS Number of obs = 10122 F( 1, 10120) = 146.29Model 7.03261354 1 7.03261354 Prob > F = 0.0000 Residual 486.489226 10120 .048072058 R-squared = 0.0142 Adj R-squared = 0.0142Total 493.52184 10121 .048762162 Root MSE = .21925 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q -.0171031 .001414 -12.10 0.000 -.0198749 -.0143313_cons .4828087 .0033785 142.91 0.000 .4761862 .4894312 . reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479 lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. test q( 1) q = 0F( 1, 10117) = 68.26Prob > F = 0.0000. test q profit( 1) q = 0( 2) profit = 0F( 2, 10117) = 250.68Prob > F = 0.0000. *test q=1. *test q=profit=1. *test q+profit=1. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. ereturn listscalars:e(N) = 10122e(df_m) = 4e(df_r) = 10117e(F) = 1241.0995********e(r2) = .329173605634022e(rmse) = .1808974246986086e(mss) = 162.4543634900055e(rss) = 331.0674763826105e(r2_a) = .3289083782368228e(ll) = 2946.854676865168e(ll_0) = 926.2762267636834e(rank) = 5macros:e(cmdline) : "regress leverage q profit fratio lna"e(title) : "Linear regression"e(marginsok) : "XB default"e(vce) : "ols"e(depvar) : "leverage"e(cmd) : "regress"e(properties) : "b V"e(predict) : "regres_p"e(model) : "ols"e(estat_cmd) : "regress_estat"matrices:e(b) : 1 x 5e(V) : 5 x 5functions:e(sample). gen r2ur=e(r2). reg leverage fratio lnaSource SS df MS Number of obs = 10122F( 2, 10119) = 2126.58Model 146.048038 2 73.0240189 Prob > F = 0.0000 Residual 347.473802 10119 .034338749 R-squared = 0.2959 Adj R-squared = 0.2958Total 493.52184 10121 .048762162 Root MSE = .18531 leverage Coef. Std. Err. t P>t [95% Conf. Interval]fratio .4065295 .0098893 41.11 0.000 .3871446 .4259144 lna .0580092 .0014631 39.65 0.000 .0551413 .0608771_cons -.9750239 .0311435 -31.31 0.000 -1.036071 -.9139764. ereturn listscalars:e(N) = 10122e(df_m) = 2e(df_r) = 10119e(F) = 2126.577725823851e(r2) = .29593024275006e(rmse) = .1853071749435181e(mss) = 146.0480378759594e(rss) = 347.4738019966565e(r2_a) = .2957910847784719e(ll) = 2702.068978698881e(ll_0) = 926.2762267636834e(rank) = 3macros:e(cmdline) : "regress leverage fratio lna"e(title) : "Linear regression"e(marginsok) : "XB default"e(vce) : "ols"e(depvar) : "leverage"e(cmd) : "regress"e(properties) : "b V"e(predict) : "regres_p"e(model) : "ols"e(estat_cmd) : "regress_estat"matrices:e(b) : 1 x 3e(V) : 3 x 3functions:e(sample). gen r2r=e(r2). gen F=((r2ur-r2r)/2)/((1-r2ur)/(10122-4-1)).. capture drop r2ur r2r F. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479 lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. ereturn listscalars:e(N) = 10122e(df_m) = 4e(df_r) = 10117e(F) = 1241.0995********e(r2) = .329173605634022e(rmse) = .1808974246986086e(mss) = 162.4543634900055e(rss) = 331.0674763826105e(r2_a) = .3289083782368228e(ll) = 2946.854676865168e(ll_0) = 926.2762267636834e(rank) = 5macros:e(cmdline) : "regress leverage q profit fratio lna"e(title) : "Linear regression"e(marginsok) : "XB default"e(vce) : "ols"e(depvar) : "leverage"e(cmd) : "regress"e(properties) : "b V"e(predict) : "regres_p"e(model) : "ols"e(estat_cmd) : "regress_estat"matrices:e(b) : 1 x 5e(V) : 5 x 5functions:e(sample). gen r2ur=e(r2). reg leverage lnaSource SS df MS Number of obs = 10122 F( 1, 10120) = 2196.68Model 88.0196216 1 88.0196216 Prob > F = 0.0000 Residual 405.502218 10120 .040069389 R-squared = 0.1784 Adj R-squared = 0.1783Total 493.52184 10121 .048762162 Root MSE = .20017 leverage Coef. Std. Err. t P>t [95% Conf. Interval]lna .072032 .0015369 46.87 0.000 .0690194 .0750446_cons -1.113191 .0334455 -33.28 0.000 -1.178751 -1.047631. ereturn listscalars:e(N) = 10122e(df_m) = 1e(df_r) = 10120e(F) = 2196.679894874584e(r2) = .1783500028923137e(rmse) = .2001733977342292e(mss) = 88.01962156870104e(rss) = 405.5022183039149e(r2_a) = .1782688121811371e(ll) = 1920.46295803989e(ll_0) = 926.2762267636834e(rank) = 2macros:e(cmdline) : "regress leverage lna"e(title) : "Linear regression"e(marginsok) : "XB default"e(vce) : "ols"e(depvar) : "leverage"e(cmd) : "regress"e(properties) : "b V"e(predict) : "regres_p"e(model) : "ols"e(estat_cmd) : "regress_estat"matrices:e(b) : 1 x 2e(V) : 2 x 2functions:e(sample). gen r2r=e(r2). gen F=((r2ur-r2r)/3)/((1-r2ur)/(10122-4-1)).. drop r2ur r2r F. gen lnq=ln(q). gen lnleverage=ln(leverage). gen lnprofit=ln(profit)(770 missing values generated). gen lnfratio=ln(fratio). reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479 lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. display .0104224*1.541244.01606346. reg lnleverage lnq lnprofit lnfratio lnaSource SS df MS Number of obs = 9352 F( 4, 9347) = 1165.37Model 1549.75886 4 387.439715 Prob > F = 0.0000 Residual 3107.51249 9347 .332460949 R-squared = 0.3328 Adj R-squared = 0.3325Total 4657.27136 9351 .498050621 Root MSE = .57659 lnleverage Coef. Std. Err. t P>t [95% Conf. Interval]lnq .1182021 .0137478 8.60 0.000 .0912534 .1451508 lnprofit -.1480528 .0081074 -18.26 0.000 -.163945 -.1321606 lnfratio .3743693 .0091822 40.77 0.000 .3563702 .3923683 lna .2064762 .005279 39.11 0.000 .1961283 .2168242_cons -5.572398 .1249991 -44.58 0.000 -5.817423 -5.327372.. gen profit_2=profit^2. reg leverage q profit profit_2 fratio lnaSource SS df MS Number of obs = 10122 F( 5, 10116) = 1123.89Model 176.246892 5 35.2493785 Prob > F = 0.0000 Residual 317.274948 10116 .031363676 R-squared = 0.3571 Adj R-squared = 0.3568Total 493.52184 10121 .048762162 Root MSE = .1771 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0097995 .0012353 7.93 0.000 .007378 .012221profit -.8495927 .0280139 -30.33 0.000 -.9045055 -.7946799 profit_2 .4402733 .0209949 20.97 0.000 .3991192 .4814275 fratio .3651988 .0095655 38.18 0.000 .3464485 .3839492 lna .0673593 .0015058 44.73 0.000 .0644078 .0703109_cons -1.131486 .0328249 -34.47 0.000 -1.19583 -1.067143. sum profitVariable Obs Mean Std. Dev. Min Maxprofit 10122 .0613722 .0806199 -.6430355 2.645649. display (-.8495927+2*.4402733*.0613722)*.0806199-.06413729. display -1*(-.8495927)/(2*.4402733).96484695.. gen q_profit=q*profit. reg leverage q profit q_profit fratio lnaSource SS df MS Number of obs = 10122 F( 5, 10116) = 1024.71Model 165.922728 5 33.1845455 Prob > F = 0.0000 Residual 327.599112 10116 .032384254 R-squared = 0.3362Adj R-squared = 0.3359Total 493.52184 10121 .048762162 Root MSE = .17996 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0086458 .0012666 6.83 0.000 .0061631 .0111286profit -.7199138 .0315488 -22.82 0.000 -.7817558 -.6580718q_profit .0531124 .0051322 10.35 0.000 .0430523 .0631725 fratio .3745416 .0097085 38.58 0.000 .3555111 .3935721 lna .0654486 .0015264 42.88 0.000 .0624566 .0684405_cons -1.101763 .033317 -33.07 0.000 -1.167071 -1.036455. sum profitVariable Obs Mean Std. Dev. Min Maxprofit 10122 .0613722 .0806199 -.6430355 2.645649. sum qVariable Obs Mean Std. Dev. Min Maxq 10122 1.825711 1.541244 .283239 44.53. display (.0086458+.0531124*.0613722)*1.541244.01834916.. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102fratio .3803813 .0097427 39.04 0.000 .3612837 .399479lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. predict pre_leverage, xb. gen pre_leverage_2=pre_leverage^2. reg leverage q profit fratio lna pre_leverage_2Source SS df MS Number of obs = 10122F( 5, 10116) = 1014.85Model 164.859349 5 32.9718698 Prob > F = 0.0000 Residual 328.662491 10116 .032489372 R-squared = 0.3340 Adj R-squared = 0.3337Total 493.52184 10121 .048762162 Root MSE = .18025leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0056752 .0013727 4.13 0.000 .0029844 .008366profit -.3891551 .0255773 -15.21 0.000 -.4392918 -.3390185fratio .217856 .0212386 10.26 0.000 .1762241 .2594879lna .0359964 .0036657 9.82 0.000 .0288108 .0431819pre_levera~2 .4584789 .0532885 8.60 0.000 .3540228 .562935 _cons -.5069118 .0763578 -6.64 0.000 -.6565883 -.3572353.. *第六章. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479 lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. reg leverage q profit fratioSource SS df MS Number of obs = 10122 F( 3, 10118) = 902.47Model 104.181387 3 34.7271289 Prob > F = 0.0000 Residual 389.340453 10118 .038479982 R-squared = 0.2111 Adj R-squared = 0.2109Total 493.52184 10121 .048762162 Root MSE = .19616leverage Coef. Std. Err. t P>t [95% Conf. Interval]q -.0085362 .0012782 -6.68 0.000 -.0110418 -.0060306profit -.3857851 .0244635 -15.77 0.000 -.4337385 -.3378317 fratio .4707303 .0103067 45.67 0.000 .4505272 .4909334 _cons .2981022 .0056612 52.66 0.000 .2870051 .3091993. corr q profit fratio lna(obs=10122)q profit fratio lnaq 1.0000profit 0.1024 1.0000fratio -0.1111 -0.1206 1.0000lna -0.3615 0.0415 0.2332 1.0000.. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479 lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. estat vifVariable VIF 1/VIFlna 1.22 0.821395q 1.17 0.855360fratio 1.08 0.928478profit 1.04 0.965439Mean VIF 1.12. gen lna_1=lna. reg leverage q profit fratio lna lna_1note: lna_1 omitted because of collinearitySource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479 lna .0646686 .0015325 42.20 0.000 .0616646 .0676725lna_1 (omitted)_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. replace lna_1=0 in 1/5(5 real changes made). reg leverage q profit fratio lna lna_1Source SS df MS Number of obs = 10122 F( 5, 10116) = 992.93Model 162.471085 5 32.494217 Prob > F = 0.0000 Residual 331.050755 10116 .03272546 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0099378 .0014321 6.94 0.000 .0071306 .0127451profit -.490468 .022789 -21.52 0.000 -.5351389 -.4457971 fratio .3805559 .009746 39.05 0.000 .3614518 .3996601 lna .0683764 .0054087 12.64 0.000 .0577743 .0789784lna_1 -.0038382 .0053695 -0.71 0.475 -.0143634 .006687_cons -1.094295 .0338511 -32.33 0.000 -1.16065 -1.02794. estat vifVariable VIF 1/VIFlna_1 16.93 0.059075lna 15.16 0.065944q 1.51 0.663687fratio 1.08 0.927895profit 1.04 0.957918Mean VIF 7.14.. reg lna_1 q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) =40285.00Model 18079.0202 4 4519.75505 Prob > F = 0.0000 Residual 1135.0716 10117 .112194485 R-squared = 0.9409 Adj R-squared = 0.9409Total 19214.0918 10121 1.89843808 Root MSE = .33495lna_1 Coef. Std. Err. t P>t [95% Conf. Interval]q -.1262561 .0023358 -54.05 0.000 -.1308347 -.1216776 profit .3745886 .042031 8.91 0.000 .2921996 .4569777 fratio .0454821 .0180399 2.52 0.012 .0101203 .080844 lna .9660203 .0028376 340.44 0.000 .9604581 .9715825 _cons .9185765 .0620092 14.81 0.000 .7970262 1.040127. display 1/(1-0.9409)16.920474. ereturn listscalars:e(N) = 10122e(df_m) = 4e(df_r) = 10117e(F) = 40285.00211816753e(r2) = .9409250454026557e(rmse) = .3349544514557118e(mss) = 18079.020********e(rss) = 1135.0716********e(r2_a) = .9409016886943045e(ll) = -3288.949185267316e(ll_0) = -17606.256149305e(rank) = 5macros:e(cmdline) : "regress lna_1 q profit fratio lna"e(title) : "Linear regression"e(marginsok) : "XB default"e(vce) : "ols"e(depvar) : "lna_1"e(cmd) : "regress"e(properties) : "b V"e(predict) : "regres_p"e(model) : "ols"e(estat_cmd) : "regress_estat"matrices:e(b) : 1 x 5e(V) : 5 x 5functions:e(sample). display 1/(1-e(r2))16.927647.. *排除其中一个变量、主成分分析和因子分析、增加观测值（增加横截面的观测值、搜集? > 姘迨 荩?.. *第七章. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000Residual 331.067476 10117 .032723878 R-squared = 0.3292Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102fratio .3803813 .0097427 39.04 0.000 .3612837 .399479lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. rvpplot q. *graph save Graph "C:\Documents and Settings\Administrator\桌面\graph1.gph", r> eplace.. estat hettest qBreusch-Pagan / Cook-Weisberg test for heteroskedasticityHo: Constant varianceVariables: qchi2(1) = 231.02Prob > chi2 = 0.0000. estat hettest fratioBreusch-Pagan / Cook-Weisberg test for heteroskedasticityHo: Constant varianceVariables: fratiochi2(1) = 86.45Prob > chi2 = 0.0000. estat hettest q profit fratio lnaBreusch-Pagan / Cook-Weisberg test for heteroskedasticityHo: Constant varianceVariables: q profit fratio lnachi2(4) = 2777.73Prob > chi2 = 0.0000.. *自己构造BP统计量. capture drop pre_leverage pre_leverage_2. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479 lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. predict resid, residuals. gen resid_2=resid^2. reg resid_2 qSource SS df MS Number of obs = 10122 F( 1, 10120) = 118.51Model .494284609 1 .494284609 Prob > F = 0.0000 Residual 42.2098625 10120 .004170935 R-squared = 0.0116 Adj R-squared = 0.0115Total 42.7041471 10121 .00421936 Root MSE = .06458 resid_2 Coef. Std. Err. t P>t [95% Conf. Interval]q .0045342 .0004165 10.89 0.000 .0037178 .0053507_cons .0244295 .0009952 24.55 0.000 .0224788 .0263802 . reg resid_2 fratioSource SS df MS Number of obs = 10122 F( 1, 10120) = 44.02Model .184961456 1 .184961456 Prob > F = 0.0000 Residual 42.5191857 10120 .004201501 R-squared = 0.0043 Adj R-squared = 0.0042Total 42.7041471 10121 .00421936 Root MSE = .06482 resid_2 Coef. Std. Err. t P>t [95% Conf. Interval]fratio -.022319 .0033639 -6.63 0.000 -.0289128 -.0157252_cons .0418463 .0015206 27.52 0.000 .0388657 .044827 . reg resid_2 q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 408.91Model 5.94319536 4 1.48579884 Prob > F = 0.0000 Residual 36.7609518 10117 .003633582 R-squared = 0.1392 Adj R-squared = 0.1388Total 42.7041471 10121 .00421936 Root MSE = .06028resid_2 Coef. Std. Err. t P>t [95% Conf. Interval]q .0014776 .0004203 3.52 0.000 .0006536 .0023015profit .2861667 .007564 37.83 0.000 .2713397 .3009936 fratio .0015508 .0032465 0.48 0.633 -.004813 .0079146 lna -.0050894 .0005107 -9.97 0.000 -.0060903 -.0040884_cons .12237 .0111593 10.97 0.000 .1004955 .1442445.. *自己构造White统计量. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102 fratio .3803813 .0097427 39.04 0.000 .3612837 .399479 lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. predict pre_leverage, xb. gen pre_leverage_2=pre_leverage^2. reg resid_2 pre_leverage pre_leverage_2Source SS df MS Number of obs = 10122 F( 2, 10119) = 2812.26Model 15.2564632 2 7.6282316 Prob > F = 0.0000 Residual 27.4476839 10119 .00271249 R-squared = 0.3573 Adj R-squared = 0.3571Total 42.7041471 10121 .00421936 Root MSE = .05208resid_2 Coef. Std. Err. t P>t [95% Conf. Interval]pre_leverage -.9542024 .0127245 -74.99 0.000 -.979145 -.9292599 pre_levera~2 .9501041 .0134433 70.68 0.000 .9237526 .9764556 _cons .2546084 .0031379 81.14 0.000 .2484576 .2607593.. reg leverage q profit fratio lnaSource SS df MS Number of obs = 10122F( 4, 10117) = 1241.10Model 162.454363 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0012615 8.26 0.000 .0079497 .0128951profit -.4919057 .0226995 -21.67 0.000 -.5364012 -.4474102fratio .3803813 .0097427 39.04 0.000 .3612837 .399479lna .0646686 .0015325 42.20 0.000 .0616646 .0676725_cons -1.097821 .033489 -32.78 0.000 -1.163466 -1.032175. reg leverage q profit fratio lna, robustLinear regression Number of obs = 10122 F( 4, 10117) = 1006.08Prob > F = 0.0000R-squared = 0.3292Root MSE = .1809Robustleverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0104224 .0017306 6.02 0.000 .0070302 .0138147profit -.4919057 .1054406 -4.67 0.000 -.6985902 -.2852213fratio .3803813 .011546 32.94 0.000 .3577489 .4030138lna .0646686 .0018246 35.44 0.000 .061092 .0682452_cons -1.097821 .037007 -29.67 0.000 -1.170362 -1.02528.. *findit wls0. *wls0 leverage q profit fratio lna, wvar(q profit fratio lna) type(abse).. *第九章. *为了不引入新的数据，我们假定内生变量为lna，工具变量为fratio. ivregress 2sls leverage q profit (lna= fratio)Instrumental variables (2SLS) regression Number of obs = 10122 Wald chi2(3) = 773.13Prob > chi2 = 0.0000R-squared = .Root MSE = .36709leverage Coef. Std. Err. z P>z [95% Conf. Interval]lna .3369321 .0138051 24.41 0.000 .3098746 .3639896q .0902409 .0049029 18.41 0.000 .0806314 .0998503profit -.938688 .048783 -19.24 0.000 -1.034301 -.8430751_cons -6.974845 .3066999 -22.74 0.000 -7.575965 -6.373724Instrumented: lnaInstruments: q profit fratio.. *人工的两阶段最小二乘法. reg lna q profit fratioSource SS df MS Number of obs = 10122 F( 3, 10118) = 733.36Model 3029.8545 3 1009.9515 Prob > F = 0.0000 Residual 13934.1472 10118 1.37716419 R-squared = 0.1786 Adj R-squared = 0.1784Total 16964.0017 10121 1.67611913 Root MSE = 1.1735 lna Coef. Std. Err. t P>t [95% Conf. Interval]q -.2931661 .0076468 -38.34 0.000 -.3081554 -.2781768profit 1.640992 .1463507 11.21 0.000 1.354115 1.927868 fratio 1.397107 .0616585 22.66 0.000 1.276244 1.51797 _cons 21.58579 .0338677 637.36 0.000 21.5194 21.65218. predict pre_lna, xb. reg leverage q profit pre_lnaSource SS df MS Number of obs = 10122 F( 3, 10118) = 902.47Model 104.181393 3 34.7271309 Prob > F = 0.0000 Residual 389.340447 10118 .038479981 R-squared = 0.2111 Adj R-squared = 0.2109Total 493.52184 10121 .048762162 Root MSE = .19616leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0902409 .00262 34.44 0.000 .0851052 .0953765profit -.9386881 .0260685 -36.01 0.000 -.9897875 -.8875886pre_lna .3369322 .0073771 45.67 0.000 .3224715 .3513928 _cons -6.974845 .1638935 -42.56 0.000 -7.296109 -6.653581.. *Durbin-Wu-Hausman检验. reg lna q profit fratioSource SS df MS Number of obs = 10122 F( 3, 10118) = 733.36Model 3029.8545 3 1009.9515 Prob > F = 0.0000 Residual 13934.1472 10118 1.37716419 R-squared = 0.1786 Adj R-squared = 0.1784Total 16964.0017 10121 1.67611913 Root MSE = 1.1735lna Coef. Std. Err. t P>t [95% Conf. Interval]q -.2931661 .0076468 -38.34 0.000 -.3081554 -.2781768profit 1.640992 .1463507 11.21 0.000 1.354115 1.927868 fratio 1.397107 .0616585 22.66 0.000 1.276244 1.51797 _cons 21.58579 .0338677 637.36 0.000 21.5194 21.65218. predict res, residuals. label var res "第一阶段的残差". reg leverage q profit lna resSource SS df MS Number of obs = 10122 F( 4, 10117) = 1241.10Model 162.454364 4 40.6135909 Prob > F = 0.0000 Residual 331.067476 10117 .032723878 R-squared = 0.3292 Adj R-squared = 0.3289Total 493.52184 10121 .048762162 Root MSE = .1809 leverage Coef. Std. Err. t P>t [95% Conf. Interval]q .0902409 .0024161 37.35 0.000 .0855049 .0949769profit -.938688 .0240398 -39.05 0.000 -.9858108 -.8915652lna .3369321 .006803 49.53 0.000 .3235968 .3502674res -.2722635 .0069735 -39.04 0.000 -.285933 -.2585941_cons -6.974845 .1511391 -46.15 0.000 -7.271107 -6.678582. test res( 1) res = 0F( 1, 10117) = 1524.32Prob > F = 0.0000.. ivregress 2sls leverage q profit (lna= fratio)Instrumental variables (2SLS) regression Number of obs = 10122 Wald chi2(3) = 773.13Prob > chi2 = 0.0000R-squared = .Root MSE = .36709leverage Coef. Std. Err. z P>z [95% Conf. Interval]lna .3369321 .0138051 24.41 0.000 .3098746 .3639896。

程序运行现状分析报告1. 引言本报告旨在对程序的运行现状进行全面分析，以便了解程序在当前环境下的性能表现和存在的问题。

通过分析现状，我们可以对程序的优化方向和改进方案进行合理规划。

2. 程序概述本程序是一个基于Java开发的企业资源管理系统，用于管理企业的各类资源、业务和人员。

程序使用了Spring框架作为开发基础，并集成了数据库等多种技术实现各项功能。

3. 程序运行环境程序目前运行在一台配置较高的服务器上，服务器配备了8核处理器、16GB内存和500GB硬盘，并安装了Ubuntu Server操作系统和MySQL 数据库。

程序使用Tomcat作为应用服务器，并通过Nginx进行反向代理和负载均衡。

4. 程序的性能评估4.1 响应时间通过对程序进行压力测试，得到了一组响应时间数据，并绘制成如下的折线图：从图中可以看出，程序的响应时间整体较稳定，但在高峰期间会有轻微的升高。

响应时间主要受到网络延迟、数据库查询和业务逻辑处理等因素的影响。

4.2 并发处理能力通过模拟多个用户同时访问程序，并观察服务器的系统负载情况，得到了如下的负载曲线图：从图中可以看出，程序在并发请求较高时，服务器的负载会呈现线性增长的趋势。

当服务器负载达到一定程度时，会出现性能下降、响应时间延长的情况。

4.3 内存占用情况通过监测程序运行过程中的内存使用情况，得到了如下的内存占用曲线图：从图中可以看出，程序的内存占用较为稳定，在正常范围内波动。

在程序执行大量数据库查询时，内存占用会略微增加，但仍然在可接受范围内。

5. 程序存在的问题与改进方案根据以上的性能评估结果，我们总结了程序存在的主要问题，并提出相应的改进方案：5.1 响应时间略高虽然程序的响应时间整体较为稳定，但仍然存在高峰期间响应时间略高的情况。