5. Nonlinear Functions of Several Variables

1
5. Nonlinear Functions of Several Variables
Jieun Yu (jieunyu_at_korea.ac.kr) Kyungsun Kwon
(sadokwon_at_korea.ac.kr) Kyunghwi Kim
(kyunghwi_at_korea.ac.kr) Sungjin Kim
(sungsink_at_korea.ac.kr) Computer Networks Research
Lab. Dept. of Computer Science and
Engineering Korea University
2
Contents

• Introduction
• Newtons Method for Nonlinear Systems
• Fixed-Point Iteration for Nonlinear Systems
• Minimum of a Nonlinear Function of Several
  Variables
• MATLABs Methods
• Case Study

3
Introduction
Chapter 2
Chapter 1
Chapter 3,4
Chapter 5
4
Introduction

• A zero of a nonlinear function of a single
  variable
• Nonlinear functions of several variables
• The latter problem is much more difficult
• Newtons method is applicable to this problem as
  well as that of a single variable

5
5.1 Newtons Method for Nonlinear Systems
6
Newtons Method

• Tangent line function
• Iteration form

• Tangent plane function
• Iteration using r, s

7
Newtons Method (cont)

• Step1
• Step2

8
5.1.1 Matrix-Vextor Notation

• Step1
• Step2

9
Ex5.1 Intersection of a Circle and a Parabola
( Circle ) ( Parabola )
10
Ex5.1 Intersection of a Circle and a Parabola

• Step1
• Partial derivative
• Initial estimate (x0,y0) (1/2, 1/2)

11
Ex5.1 Intersection of a Circle and a Parabola

• Step2
• The second approximate solution (x1,y1)

12
Ex5.1 Intersection of a Circle and a Parabola

• Initial estimate and 2 iterations of Newtons
  method

iteration x y
0 0.5 0.5
1 0.875 0.625 0.39528
2 0.79067 0.61806 0.084611
The true solution is (x, y) (0.78615, 0.61803)
13
5.1.2 MATLAB Function for Newtons Method for
Nonlinear Systems
\ operation -gt 100p
14
Example 5.2 a System of Three Equations
15
Example 5.2 a System of Three Equations
16
Example 5.3Intersection of a Circle and an
Ellipse
y
1
0.5
x
0
-0.5
-1
-1
-0.5
0
0.5
1
17
Example 5.3Intersection of a Circle and an
Ellipse
18
Example 5.4Positioning a Robot Arm

• the end of the first link(x1,y1)
• the end of the second link(x2,y2)
• We need to solve

10
(x2,y2)
8
6
ß
(10,4)
4
(x1,y1)
2
a
0
0
2
4
6
8
10
19
Example 5.4Positioning a Robot Arm

• Find the angles so that the arm will move to the
  point(10,4)
• The length of link (d1,d2) (5, 6)
• initial angles(a , ß) (0.7,0.7)
• The system of equations

20
5.2 Fixed-Point Iteration For Nonlinear Systems
21
5.2 Fixed-Point Iteration For Nonlinear Systems

• Ex 5.5 Fixed-Point Iteration for a System of two
  Nonlinear Function
• Consider the problem of finding a zero of the
  system

• Coverting

form

• The graph of the equations and

22
5.2 Fixed-Point Iteration For Nonlinear Systems
Iteration
0 0.60000 0.60000
1 0.53840 0.18160 0.42291
2 0.50255 0.15444 0.044975
3 0.50275 0.15062 0.0038204
4 0.50235 0.15062 0.00039661
5 0.50238 0.15058 4.9381e-05
23
5.2 Fixed-Point Iteration For Nonlinear Systems
24
5.2 Fixed-Point Iteration For Nonlinear Systems

• Using MATLAB

function G ex5_5(x) G (-0.1x(1)3
0.1x(2) 0.5) (0.1x(1) 0.1x(2)3
0.1)
Actually using function Fixed_pt_sys gtgt
xFixed_pt_sys(_at_ex5_5, 0.6 0.6, 0.00001, 5)
25
5.2 Fixed-Point Iteration For Nonlinear Systems

• Result

26
5.2 Fixed-Point Iteration For Nonlinear Systems

• Ex 5.5 Fixed-Point Iteration for a System of
  Three Nonlinear Function

• Coverting this system to a fixed point iteration
  form

27
5.2 Fixed-Point Iteration For Nonlinear Systems

• Using the preceding MATLAB function and a
  starting estimate of (2, 2, 2)

Iteration
0 2.0000 2.0000 2.0000
1 3.7600 2.1000 -1.6750 4.0759
2 3.5729 1.6528 -1.4113 0.5518
3 3.6502 1.7621 -1.4876 0.15403
4 3.6272 1.7232 -1.4643 0.050903
5 3.6346 1.7350 -1.4719 0.0115899
6 3.6323 1.7312 -1.4695 0.0050696
7 3.6330 1.7324 -1.4702 0.0016031
8 3.6328 1.7320 -1.4700 0.00050866
9 3.6328 1.7321 -1.4701 0.00016117
10 3.6328 1.7321 -1.4701 5.1098e-05
28
5.2 Fixed-Point Iteration For Nonlinear Systems

• Using MATLAB

function G ex5_6(x) G (-0.02x(1)2 -
0.02x(2)2 - 0.02x(3)2 4) (-0.05x(1)2
- 0.05x(3)2 2.5) (0.025x(1)2
0.025x(2)2 - 1.875)
Actually using function Fixed_pt_sys gtgt
xFixed_pt_sys(_at_ex5_6, 2 2 2, 0.00001, 10)
29
5.2 Fixed-Point Iteration For Nonlinear Systems

• Result

30
5.2 Fixed-Point Iteration For Nonlinear Systems

• If g(x) maps D into D, then g has a fixed point
  in D. In other words, if g(x) is in D whenever x
  is in D, then there is some point p in D such
  that pg(p)
• If sequence of
  approximations to the fixed point

intial point
is sufficiently close to the fixed point p
If there is a constant Klt1 such that for every x
in D
For each i 1,.,n and each j1,,n,
A bound on the error at the mth step is given by
31
5.2 Fixed-Point Iteration For Nonlinear Systems

• Example 5.5

First check to make sure that g(x) maps the
retangle
That is, for
we have
In fact
and
Which are all less than 0.5 (for
), as is required for the corollary
32
5.3 Minimum of a Nonlinear Function of Several
Variables
33
Example 5.7

• and
  ,
• we define
• The minimum value of h(x,y) is 0, which occurs
  when f(x,y)0 and g(x,y)0.
• The derivatives for the gradient are

34
Table 5.8

• Result of ffminTable 5.8 Approximate zeros at
  each step of iteration

Step x y Change
0 0.5 0.5
1 0.875 0.625 -0.26831
2 0.74512 0.61133 -0.035987
3 0.79251 0.61902 -0.0079939
4 0.78446 0.6178 -0.00019414
5 0.78657 0.6181 -1.3631e-05
35
5.3.1 MATLAB Function for Minimization by
Gradient Descent
36
5.3.1 MATLAB Function for Minimization by
Gradient Descent
function h ex_min(x) h (x(1)2 x(2)2 -
1)2 (x(1)2 - x(2) )2
function dh ex_min_g(x) dh -(4(x(1)2
x(2)2 - 1)x(1) 4(x(1)2 - x(2))x(1))
-(4(x(1)2 x(2)2 - 1)x(2) - 2(x(1)2 -
x(2)))'
xmin ffmin(_at_ex_min, _at_ex_min_g, 0.5 0.5, 0,
5) 0 0.5000 0.5000 1.0000
0.8750 0.6250 -0.2683 2.0000 0.7451
0.6113 -0.0360 3.0000 0.7925
0.6190 -0.0080 4.0000 0.7845 0.6178
-0.0002 5.0000 0.7866 0.6181 -0.0000
37
Example 5.8

• Given three points in the plane, we wish to find
  the location of the point P(x,y) sp that the sum
  of the squares of the distances from P to the
  three given points, is as small as possible.

38
Figure 5.7
39
5.4 MATLABs Methods
40
MATLABs METHODS

• FMINS
• finds the minimum of a scalar function of several
  variables, starting at an initial estimate
• Note
• The fmins function was replaced by fminsearch in
  Release 11 (MATLAB 5.3).
• In Release 12 (MATLAB 6.0), fmins displays a
  warning message and calls fminsearch
• Syntax
• x fmins('fun',x0)
• x fminsearch(fun,x0)
• starts at the point x0 and finds a local minimum
  x of the function described in fun.
• x0 can be a scalar, vector, or matrix.

41
MATLABs METHODS

• Examples
• A classic test example for multidimensional
  minimization is the Rosenbrock banana function
• The traditional starting point is (-1.2,1). The
  M-file banana.m defines the function.
• The minimum is at (1,1) and has the value 0.
• a

42
5.5 Nonlinear system Case Study
43
Nonlinear system Case Study

• The analytical model to compute the 802.11 DCF
  throughput

44
Nonlinear system Case Study

• The stationary probability t
• The station transmits a packet in a generic slot
  time
• The conditional collision probability p
• aa

45
Nonlinear system Case Study

• Two Equations represent a nonlinear system in the
  two unknowns t and p
• W Cwmin
• m Maximum backoff stage
• N The number of stations

46
Nonlinear system Case Study

• Solve the nonlinear problem of two variables
• Assumptions
• W 32
• m 3
• N 3, 10, 50
• Newtons Method and fminsearch(fun,x0)
• Problem1 by Newtons Method (W32, m 3, N 3)
• 64t p4 34pt - 33t - 4p 2 0 (1)
• p 1-(1- t)2
  (2)
• Equations (1) and (2) is nonlinear system

47
Nonlinear system Case Study

• function f newton_case1(x)
• f (64x(2)x(1)4 34x(1)x(2) -33x(2)
  -4x(1) 2)
• (x(2)2 - 2x(2) x(1))
  
• function df newton_case1_j(x)
• df (256x(2)x(1)3 34x(2) - 4)
  (64x(1)4 34x(1)-33)
• 1
  ( 2x(2) -2)
• t 0.4165
• 
  p 0.6595

48
Nonlinear system Case Study

• Problem1 by fminsearch(fun,x0) (W32, m 3, N
  3)
• function h fminsearch_case1(x)
• h (64x(2)x(1)4 34x(1)x(2) -33x(2)
  -4x(1) 2)2 ((1-x(2))2 -1 x(1))2

49
Nonlinear system Case Study

• Problem2 by Newtons Method (W32, m 3, N 10)
• 64t p4 34pt - 33t - 4p 2 0 (1)
• p 1-(1- t)9
  (2)
• Equations (1) and (2) is nonlinear system
• t 0.1259
• 
  p 0.7021

50
Nonlinear system Case Study

• Problem2 by fminsearch(fun,x0) (W32, m 3, N
  10)
• function h fminsearch_case1(x)
• h (64x(2)x(1)4 34x(1)x(2) -33x(2)
  -4x(1) 2)2 ((1-x(2))2 -1 x(1))9

51
Nonlinear system Case Study

• Problem3 by Newtons Method (W32, m 3, N 50)
• 64t p4 34pt - 33t - 4p 2 0 (1)
• p 1-(1- t)49
  (2)
• Equations (1) and (2) is nonlinear system
• 
  t 0.0427
• 
  p 0.8823

52
Nonlinear system Case Study

• Problem2 by fminsearch(fun,x0) (W32, m 3, N
  50)
• function h fminsearch_case1(x)
• h (64x(2)x(1)4 34x(1)x(2) -33x(2)
  -4x(1) 2)2 ((1-x(2))2 -1 x(1))49