1. 5. Nonlinear Functions of Several Variables
Jieun Yu
Kyungsun Kwon
Kyunghwi Kim
Sungjin Kim
Computer Networks Research Lab.
Dept. of Computer Science and Engineering
Korea University

2. Contents Introduction Newton’s Method for Nonlinear Systems
Fixed-Point Iteration for Nonlinear Systems
Minimum of a Nonlinear Function of Several Variables
MATLAB’s Methods
Case Study

3. Introduction
Chapter 1
Chapter 2
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
Newton’s method is applicable to this problem as well as that of a single variable

5. 5.1 Newton’s Method for Nonlinear Systems

6. Newton’s Method Tangent line function Iteration form
Tangent plane function
Iteration using r, s

7. Newton’s Method (cont’)
Step1
Step2

8. 5.1.1 Matrix-Vextor Notation
Step1
Step2

9. Ex5.1 Intersection of a Circle and a 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 Newton’s method
iteration x y
| |
0.5 1
0.875
0.625 2
The true solution is (x, y) = ( , )

13. 5.1.2 MATLAB Function for Newton’s Method for Nonlinear Systems
\ operation -> 100p

14. Example 5.2 a System of Three Equations

15. Example 5.2 a System of Three Equations

16. Example 5.3 Intersection of a Circle and an Ellipsey 1
0.5 x
-0.5 -1 -1
-0.5
0.5 1

17. Example 5.3 Intersection of a Circle and an Ellipse

18. Example 5.4 Positioning 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 β 4
(10,4)
(x1,y1) 2 α 2 4 6 8 10

19. Example 5.4 Positioning 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 (α , β) = (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
0.2
0.4
0.6
0.8 1
Iteration 1 2 3 4 5
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 = [ (-0.1*x(1)^ *x(2) + 0.5)
(0.1*x(1) + 0.1*x(2)^ )];
% Actually using function : Fixed_pt_sys
>> [ ], , 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
2.0000 1
3.7600
2.1000
4.0759 2
3.5729
1.6528
0.5518 3
3.6502
1.7621 4
3.6272
1.7232 5
3.6346
1.7350 6
3.6323
1.7312 7
3.6330
1.7324 8
3.6328
1.7320 9
1.7321 10
5.1098e-05

28. 5.2 Fixed-Point Iteration For Nonlinear Systems
Using MATLAB
function G = ex5_6(x)
G = [ (-0.02*x(1)^ *x(2)^ *x(3)^2 + 4)
(-0.05*x(1)^ *x(3)^ )
(0.025*x(1)^ *x(2)^ )];
% Actually using function : Fixed_pt_sys
>> [2 2 2], , 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 p=g(p)
If sequence of approximations to the fixed point
intial point
is sufficiently close to the fixed point p
If there is a constant K<1 such that for every x in D
For each i= 1,….,n and each j=1,…,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 ffmin Table 5.8 Approximate zeros at each step of iteration
Step x y
Change
0.5 1
0.875
0.625 2 3 4
0.6178 5
0.6181
e-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 [ ], 0, 5)

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 MATLAB’s Methods

40. MATLAB’s 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. MATLAB’s METHODS Examples a
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 DCF throughput

44. Nonlinear system: Case Study
The stationary probability τ
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 τ 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
Newton’s Method and fminsearch(fun,x0)
Problem1 by Newton’s Method (W=32, m = 3, N =3)
64τ p^4 + 34pτ - 33τ - 4p +2 =0 (1)
p = 1-(1- τ)^ (2)
*Equations (1) and (2) is nonlinear system

47. Nonlinear system: Case Study
function f = newton_case1(x)
f = [ (64*x(2)*x(1)^4 + 34*x(1)*x(2) -33*x(2) -4*x(1) +2)
(x(2)^2 - 2*x(2) + x(1)) ];
function df = newton_case1_j(x)
df = [ (256*x(2)*x(1)^3 + 34*x(2) - 4) (64*x(1)^4 + 34*x(1)-33)
( 2*x(2) -2) ];
τ =
p = aa

48. Nonlinear system: Case Study
Problem1 by fminsearch(fun,x0) (W=32, m = 3, N =3)
function h = fminsearch_case1(x)
h = (64*x(2)*x(1)^4 + 34*x(1)*x(2) -33*x(2) -4*x(1) +2)^ ((1-x(2))^2 -1 +x(1))^2;

49. Nonlinear system: Case Study
Problem2 by Newton’s Method (W=32, m = 3, N =10)
64τ p^4 + 34pτ - 33τ - 4p +2 =0 (1)
p = 1-(1- τ)^ (2)
*Equations (1) and (2) is nonlinear system
τ =
p = dd

50. Nonlinear system: Case Study
Problem2 by fminsearch(fun,x0) (W=32, m = 3, N =10)
function h = fminsearch_case1(x)
h = (64*x(2)*x(1)^4 + 34*x(1)*x(2) -33*x(2) -4*x(1) +2)^ ((1-x(2))^2 -1 +x(1))^9;

51. Nonlinear system: Case Study
Problem3 by Newton’s Method (W=32, m = 3, N =50)
64τ p^4 + 34pτ - 33τ - 4p +2 =0 (1)
p = 1-(1- τ)^ (2)
*Equations (1) and (2) is nonlinear system
τ =
p = dd

52. Nonlinear system: Case Study
Problem2 by fminsearch(fun,x0) (W=32, m = 3, N =50)
function h = fminsearch_case1(x)
h = (64*x(2)*x(1)^4 + 34*x(1)*x(2) -33*x(2) -4*x(1) +2)^ ((1-x(2))^2 -1 +x(1))^49;