In [1]:

import numpy as np
import pandas as pd
import numpy.linalg as LA

Solve Matrix Equation : $\large{Ax = b}$ ¶

Example: 1 Modelling¶

In [2]:

df = pd.read_csv('data/data1.csv')
df

Out[2]:

	no.	f	x	y	z
0	1	0.720482	0.737451	4.10	0.8695
1	2	0.812391	0.884248	3.65	0.5750
2	3	0.813530	0.863871	3.55	0.5125
3	4	0.532572	0.872902	3.85	0.6520
4	5	0.697305	0.891665	4.70	0.5400
5	6	0.733407	0.880975	0.70	0.7745
6	7	0.654187	0.733362	2.25	0.7130
7	8	0.709219	0.839979	-1.10	0.9320
8	9	0.750032	0.917038	-1.25	0.7335
9	10	0.713197	0.915913	0.90	0.7345
10	11	0.521182	0.885547	-0.15	0.7685
11	12	0.824820	0.884369	0.60	0.7090
12	13	0.770785	0.873444	1.25	0.4890
13	14	0.772823	0.726831	3.40	0.5375
14	15	0.600803	0.638667	1.80	0.8985
15	16	0.792857	0.753639	1.10	0.5705
16	17	0.800259	0.819985	0.80	0.4965
17	18	0.537606	0.847556	-0.25	0.8340
18	19	0.575679	0.884614	2.85	0.7870
19	20	0.821320	0.863913	0.75	0.6245
20	21	0.325209	0.757005	-0.20	0.7630
21	22	0.790763	0.899136	5.30	0.4470
22	23	0.803798	0.897501	3.45	0.3915
23	24	0.785717	0.895834	4.40	0.4190
24	25	0.754932	0.808529	3.80	0.5265
25	26	0.793982	0.892413	5.10	0.4350
26	27	0.789723	0.881186	3.40	0.4320
27	28	0.797951	0.936163	3.20	0.5280
28	29	0.796906	0.934264	2.80	0.4445
29	30	0.800149	0.941757	3.00	0.4490

$\begin{equation} f=a+bx+cy+dz \end{equation}$

Normal equation are

$\sum_{}^{}{f}=na+b\sum{x}+c\sum{y}+d\sum{z}$

$\sum_{}^{}{xf}=a\sum{x}+b\sum{x^2}+c\sum{xy}+d\sum{xz}$

$\sum_{}^{}{yf}=a\sum{y}+b\sum{xy}+c\sum{y^2}+d\sum{yz}$

$\sum_{}^{}{zf}=a\sum{z}+b\sum{xz}+c\sum{zy}+d\sum{z^2}$

$\begin{equation} A= \begin{pmatrix} a11 & a12 & a13 &a14\\ a21 & a22 & a23 &a24\\ a31 & a32 & a33 &a34\\ a41 & a42 & a43 &a44\\ \end{pmatrix} \end{equation}$

$\begin{equation} b= \begin{pmatrix} b1 \\ b2 \\ b3 \\ b4\\ \end{pmatrix} \end{equation}$

In [3]:

n=30
a11=n
a12=np.sum(df['x'])     #sum of x
a13=np.sum(df['y'])     #sum of y
a14=np.sum(df['z'])     #sum of z
a21=np.sum(df['x'])     #sum of x
a22=np.sum(df['x']**2)  #sum of x2
a23=np.sum(np.dot(df['x'],df['y']))   #sum of xy
a24=np.sum(np.dot(df['x'],df['z']))   #sum of xz
a31=np.sum(df['y'])                   #sum of y
a32=np.sum(np.dot(df['y'],df['x']))   #sum of xy
a33=np.sum(df['y']**2)                #sum of y2   
a34=np.sum(np.dot(df['y'],df['z']))   #sum of yz
a41=np.sum(df['z'])                   #sum of z
a42=np.sum(np.dot(df['z'],df['x']))   #sum of xz
a43=np.sum(np.dot(df['z'],df['y']))   #sum of yz
a44=np.sum(df['z']**2)                #sum of z2
b1=np.sum(df['f'])                    #sum of f
b2=np.sum(np.dot(df['f'],df['x']))    #sum of fx
b3=np.sum(np.dot(df['f'],df['y']))    #sum of fy
b4=np.sum(np.dot(df['f'],df['z']))    #sum of fz

In [4]:

A = np.array([[a11,a12,a13,a14],[a21,a22,a23,a24],[a31,a32,a33,a34],[a41,a42,a43,a44]])
b = np.array([b1,b2,b3,b4])
x = LA.solve(A,b)
x

Out[4]:

array([ 0.85507492,  0.14860913, -0.00131447, -0.41792768])

In [5]:

print('\n a=',"{:10.3}".format(x[0]))
print('\n b=',"{:.3}".format(x[1]))
print('\n c=',"{:.3}".format(x[2]))
print('\n d=',"{:.3}".format(x[3]))

 a=      0.855

 b= 0.149

 c= -0.00131

 d= -0.418

In [6]:

f_exp=[] # expected value of f
for i in range(n):
    f_exp.append(x[0]+x[1]*df.iloc[i,2]+x[2]*df.iloc[i,3]+x[3]*df.iloc[i,4])
f_exp    

Out[6]:

[0.5958894003599622,
 0.7413760360958024,
 0.7645997139268501,
 0.7072466193451008,
 0.7557255547611699,
 0.6613906839806531,
 0.6631192534951209,
 0.5918408249814666,
 0.68644830702124,
 0.6830369792734646,
 0.6656950521790324,
 0.6894008781179288,
 0.7788669964212885,
 0.7339832421031306,
 0.5721125629994265,
 0.7271988454495809,
 0.7683794858588099,
 0.6328064581873758,
 0.6538812940618572,
 0.7214786253986405,
 0.6489567737789436,
 0.7949143877809594,
 0.8202981096648657,
 0.8073086989485949,
 0.7501958066178496,
 0.7991933185714644,
 0.8010132155323733,
 0.7693252220065355,
 0.8044656816412954,
 0.8034356104392159]

$\begin{equation} RMSE=\sqrt{\frac{1}{n}\sum{(f-\hat{f})^2}}\\ R^2=1-\frac{\sum{(f-\hat{f})^2}}{\sum{(f-\bar{f})^2}} \end{equation}$

In [7]:

RMSE=np.sqrt(np.sum((df['f']-f_exp)**2)/n)
R_square=1-(np.sum(np.sum(np.sum((df['f']-f_exp)**2))/np.sum((df['f']-df['f'].mean())**2)))

In [12]:

print('\n RMSE=',"{:.2}".format(RMSE))
print('\n R square=',"{:.2}".format(R_square))

 RMSE= 0.093

 R square= 0.36

Example:2 Electric circuit¶

Calculate I1,I2,I3

Using Kirchhoff's laws in junction a,loop abdea,aegha

$I3=I1+I2$

$25.1I1+24+40.5I2=0$

$-18+40.5I2+78.25I3=0$

In [9]:

A = np.array([[1,1,-1],[25.1,40.5,0],[0,40.5,78.25]]) #arranging I1,I2,I3
b = np.array([0,-24,18])
x = LA.solve(A,b)
x

Out[9]:

array([ 18.9867374, -12.3596817,   6.6270557])

In [13]:

print('\n I1=',"{:.3}".format(x[0]))
print('\n I2=',"{:.3}".format(x[1]))
print('\n I3=',"{:.3}".format(x[2]))

 I1= 19.0

 I2= -12.4

 I3= 6.63

In [ ]:

Solve Matrix Equation : Ax=b\large{Ax = b}¶

Example: 1 Modelling¶

Example:2 Electric circuit¶

Solve Matrix Equation : $\large{Ax = b}$ ¶