{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 1 - Linear Algebraic Equations"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa1.1 Page 24"
]
},
{
"cell_type": "code",
"execution_count": 54,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the solution of ex 1.1 by TDMA method is\n",
"317.5\n",
"395.0\n",
"432.5\n",
"430.0\n",
"387.5\n",
"305.0\n",
"182.5\n"
]
}
],
"source": [
"from numpy import zeros\n",
"from __future__ import division\n",
"a=[0];b=[];c=[]\n",
"for i in range(1,7):\n",
" a.append(1) #sub diagonal assignment\n",
"\n",
"for j in range(0,7):\n",
" b.append(-2) #main diagonal assignment\n",
"\n",
"for k in range(0,6):\n",
" c.append(1) #super diagonal assignment\n",
"\n",
"d=[-240] #given values assignment\n",
"for l in range(1,6):\n",
" d.append(-40) \n",
"\n",
"d.append(-60)\n",
"i=1#\n",
"n=7#\n",
"beta1=[b[i-1]]# #initial b is equal to beta since a1=0\n",
"gamma1=[d[i-1]/beta1[i-1]]# #since c7=0\n",
"m=i+1\n",
"for j in range(m,n+1):\n",
" beta1.append(b[j-1]-a[j-1]*c[j-1-1]/beta1[j-1-1])\n",
" gamma1.append((d[j-1]-a[j-1]*gamma1[j-1-1])/beta1[j-1])\n",
"\n",
"#x(n)=gamma1(n)# #since c7=0\n",
"x=zeros(n-1)\n",
"#x[n-1]=gamma1[n-1]\n",
"x=list(x)\n",
"x.append(gamma1[-1])\n",
"n1=n-i# \n",
"\n",
"for k in range(0,n1):\n",
" j=n-k-1\n",
" x[j-1]=gamma1[j-1]-c[j-1]*x[j+1-1]/beta1[j-1]\n",
"\n",
"\n",
"print \"the solution of ex 1.1 by TDMA method is\"\n",
"for i in range(0,7):\n",
" print x[i]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa1.2 Page 24"
]
},
{
"cell_type": "code",
"execution_count": 58,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the solution using gauss elimination method is 3, 4 and 5\n"
]
}
],
"source": [
"from __future__ import division\n",
"a1=10; a2=1; a3=2; #1st row\n",
"b1=2; b2=10; b3=1; #2nd row\n",
"c1=1; c2=2; c3=10; #3rd row \n",
"d1=44; d2=51; d3=61; #given values\n",
"\n",
"b3=b3-(b1/a1)*a3 # for making b1=0\n",
"b2=b2-(b1/a1)*a2\n",
"d2=d2-(b1/a1)*d1\n",
"b1=b1-(b1/a1)*a1\n",
"\n",
"c3=c3-(c1/a1)*a3 # for making c1=0\n",
"c2=c2-(c1/a1)*a2\n",
"d3=d3-(c1/a1)*d1\n",
"c1=c1-(c1/a1)*a1\n",
"\n",
"c3=c3-(c2/b2)*b3 # for making c2=0\n",
"d3=d3-(c2/b2)*d2\n",
"c2=c2-(c2/b2)*b2\n",
"\n",
"x3=d3/c3# # final values of x\n",
"x2=(d2-(b3*x3))/b2#\n",
"x1=(d1-(x3*a3)-(x2*a2))/a1#\n",
"print \"the solution using gauss elimination method is %d, %d and %d\"%(x1,x2,x3)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa1.3 Page 26"
]
},
{
"cell_type": "code",
"execution_count": 59,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the solution using gauss elimination method is 3, -2 and 0\n"
]
}
],
"source": [
"from __future__ import division\n",
"a1=3; a2=1; a3=-2; #1st row\n",
"b1=-1; b2=4; b3=-3; #2nd row\n",
"c1=1; c2=-1; c3=4; #3rd row \n",
"d1=9; d2=-8; d3=1; #given values\n",
"\n",
"b3=b3-(b1/a1)*a3 # for making b1=0\n",
"b2=b2-(b1/a1)*a2\n",
"d2=d2-(b1/a1)*d1\n",
"b1=b1-(b1/a1)*a1\n",
"\n",
"c3=c3-(c1/a1)*a3 # for making c1=0\n",
"c2=c2-(c1/a1)*a2\n",
"d3=d3-(c1/a1)*d1\n",
"c1=c1-(c1/a1)*a1\n",
"\n",
"c3=c3-(c2/b2)*b3 # for making c2=0\n",
"d3=d3-(c2/b2)*d2\n",
"c2=c2-(c2/b2)*b2\n",
"\n",
"x3=d3/c3# # final values of x\n",
"x2=(d2-(b3*x3))/b2#\n",
"x1=(d1-(x3*a3)-(x2*a2))/a1#\n",
"print \"the solution using gauss elimination method is %d, %d and %d\"%(x1,x2,x3)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa1.4 Page 27"
]
},
{
"cell_type": "code",
"execution_count": 61,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the values of MOLAR FLOW RATES of D1, B1, D2, B2 respectively are : 26, 18, 9 and 17\n",
"the composition of stream B is 0.077, 0.247, 0.467 & 0.210\n",
"the composition of stream D is 0.274 0.492 0.12 0.114\n"
]
}
],
"source": [
"from __future__ import division\n",
"\n",
"a1=.35; a2=.16; a3=.21; a4=.01 #1st row \n",
"b1=.54; b2=.42; b3=.54; b4=.1 #2nd row\n",
"c1=.04; c2=.24; c3=.1; c4=.65 #3rd row\n",
"d1=.07; d2=.18; d3=.15; d4=.24 #4th row \n",
"r1=14; r2=28; r3=17.5; r4=10.5 #given values\n",
"\n",
"b4=b4-(b1/a1)*a4 # for making b1=0\n",
"b3=b3-(b1/a1)*a3\n",
"b2=b2-(b1/a1)*a2\n",
"r2=r2-(b1/a1)*r1\n",
"b1=b1-(b1/a1)*a1\n",
"\n",
"c4=c4-(c1/a1)*a4 # for making c1=0\n",
"c3=c3-(c1/a1)*a3\n",
"c2=c2-(c1/a1)*a2\n",
"r3=r3-(c1/a1)*r1\n",
"c1=c1-(c1/a1)*a1\n",
"\n",
"d4=d4-(d1/a1)*a4 # for making d1=0\n",
"d3=d3-(d1/a1)*a3\n",
"d2=d2-(d1/a1)*a2\n",
"r4=r4-(d1/a1)*r1\n",
"d1=d1-(d1/a1)*a1\n",
"\n",
"c4=c4-(c2/b2)*b4 # for making c2=0\n",
"c3=c3-(c2/b2)*b3\n",
"r3=r3-(c2/b2)*r2\n",
"c2=c2-(c2/b2)*b2\n",
"\n",
"d4=d4-(d2/b2)*b4 # for making d2=0\n",
"d3=d3-(d2/b2)*b3\n",
"r4=r4-(d2/b2)*r2\n",
"d2=d2-(d2/b2)*b2\n",
"\n",
"d4=d4-(d3/c3)*c4 #for making d3=0\n",
"r4=r4-(d3/c3)*r3\n",
"d3=d3-(d3/c3)*c3\n",
"\n",
"B2=r4/d4#\n",
"D2=(r3-(c4*B2))/c3#\n",
"B1=(r2-(D2*b3)-(B2*b4))/b2#\n",
"D1=(r1-(B2*a4)-(D2*a3)-(B1*a2))/a1#\n",
"print \"the values of MOLAR FLOW RATES of D1, B1, D2, B2 respectively are : %.f, %.f, %.f and %.f\"%(D1,B1,D2,B2)\n",
"\n",
"B=D2+B2#\n",
"x1B=(.21*D2 + .01*B2)/B#\n",
"x2B=(.54*D2 + .1*B2)/B#\n",
"x3B=(.1*D2 + .65*B2)/B#\n",
"x4B=(.15*D2 + .24*B2)/B#\n",
"print \"the composition of stream B is %.3f, %.3f, %.3f & %.3f\"%(x1B,x2B,x3B,x4B)\n",
"\n",
"D=D1+B1#\n",
"x1D=(.35*D1 + .16*B1)/D#\n",
"x2D=(.54*D1 + .42*B1)/D#\n",
"x3D=(.04*D1 + .24*B1)/D#\n",
"x4D=(.07*D1 + .18*B1)/D#\n",
"print \"the composition of stream D is\",x1D,x2D,x3D,x4D"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa1.5 Page 28"
]
},
{
"cell_type": "code",
"execution_count": 69,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the values of x1,x2,x3 respectively is\n",
"3\n",
"4\n",
"5\n"
]
}
],
"source": [
"from __future__ import division\n",
"xnew=[];e=[]\n",
"for i in range(0,3):\n",
" xnew.append(2)\n",
" e.append(1)\n",
"\n",
"x=1e-6\n",
"while e[0]>x and e[1]>x and e[2]>x:\n",
" xold=[]\n",
" for i in range(0,3):\n",
" xold.append(xnew[i])\n",
" \n",
" xnew[0]=(44-xold[1]-2*xold[2])/10\n",
" xnew[1]=(-2*xnew[0]+51-xold[2])/10\n",
" xnew[2]=(-2*xnew[1]-xnew[0]+61)/10\n",
" e=[]\n",
" for i in range(0,3):\n",
" e.append(abs(xnew[i]-xold[i]))\n",
" \n",
"print \"the values of x1,x2,x3 respectively is\"\n",
"for i in range(0,3):\n",
" print '%.f'%xnew[i]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa1.6 Page 28"
]
},
{
"cell_type": "code",
"execution_count": 75,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"the values of x1,x2,x3 respectively is\n",
"3\n",
"-2\n",
"-1\n"
]
}
],
"source": [
"from __future__ import division\n",
"xnew=[];e=[];\n",
"for i in range(0,3):\n",
" xnew.append(2)\n",
" e.append(1)\n",
"\n",
"x=1e-6\n",
"while e[0]>x and e[1]>x and e[2]>x:\n",
" xold=[]\n",
" for i in range(0,3):\n",
" xold.append(xnew[i])\n",
" \n",
" xnew[0]=(9-xold[1]+2*xold[2])/3\n",
" xnew[1]=(xnew[0]-8+3*xold[2])/4\n",
" xnew[2]=(xnew[1]-xnew[0]+1)/4\n",
" e=[]\n",
" for i in range(0,3):\n",
" e.append(abs(xnew[i]-xold[i]))\n",
" \n",
"print \"the values of x1,x2,x3 respectively is\"\n",
"for i in range(0,3):\n",
" print '%.f'%xnew[i]"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.9"
}
},
"nbformat": 4,
"nbformat_minor": 0
}