NUMERICAL ON
BISECTION METHOD
BY Sumita Das
Bisection Method
 It is a Derivative Based Method for Optimization
 Requirements for Bisection Method
 f -> c’ i.e. f is continuous for the first derivative.
 There exists a minima in the level of uncertainty [a
b]
 Function must be unimodal.
PowerPoint Presentation by Sumita Das, GHRCE
Algorithm
 Initialize Level of uncertainty [a b]
k=1
ak =a
bk =b
ϵ > 0
l : Allowable level of uncertainty such that
(1/2) n <= (1/(b`-a`))
PowerPoint Presentation by Sumita Das, GHRCE
While k<= n
ck= (ak +bk )/2
if f(ck)=0
Stop with ck as the
solution
if f(ck)>0
ak+1 = ak
bk+1 = ck
else
ak+1 = ck
bk+1 = bk
end if
k=k+1
end while
Find midpoint
c is now b.
a remains same
c is now a.
b remains same
Midpoint is the minima
PowerPoint Presentation by Sumita Das, GHRCE
In Simple words
Midpoint
m
a b
Example: f(x)=3x2 – 2x
f’(x)=6x-2
Put midpoint value in
derivative.
f’(x)=6*50-2=298
1. if f(m)=0, Midpoint is
minima
2. if f(m)>0, Level of
uncertainty will be [a, m]
3. if f(m)<0, Level of
uncertainty will be [m, b]
50 9010
So, 298>0, level of uncertainty will be
[10, 50] ,Follow the procedure.
PowerPoint Presentation by Sumita Das, GHRCE
Example
Que: Find Minima f(x)=(x-2)2
[0 6]
Solution: f ‘(x)=2x-4
k ak bk ck f’(ck )
1 0 6 3 2
2 0 3 1.5 -1
3 1.5 3 2.25 0.5
4 1.5 2.25 1.875 -0.25
5 1.875 2.25 2.062 0.123
6 1.875 2.062 1.9685 -0.063
7 1.9685 2.062 2.015 0.0305
8 1.9685 2.015 1.99175 -0.016
9 1.99175 2.015 2.003 0.006
PowerPoint Presentation by Sumita Das, GHRCE
References
[1] Singiresu S. Rao, “Engineering Optimization, Chapter 5:
Nonlinear Programming I: One-Dimensional Minimization
Methods”, 4th Edition
PowerPoint Presentation by Sumita Das, GHRCE

Numerical on bisection method

  • 1.
  • 2.
    Bisection Method  Itis a Derivative Based Method for Optimization  Requirements for Bisection Method  f -> c’ i.e. f is continuous for the first derivative.  There exists a minima in the level of uncertainty [a b]  Function must be unimodal. PowerPoint Presentation by Sumita Das, GHRCE
  • 3.
    Algorithm  Initialize Levelof uncertainty [a b] k=1 ak =a bk =b ϵ > 0 l : Allowable level of uncertainty such that (1/2) n <= (1/(b`-a`)) PowerPoint Presentation by Sumita Das, GHRCE
  • 4.
    While k<= n ck=(ak +bk )/2 if f(ck)=0 Stop with ck as the solution if f(ck)>0 ak+1 = ak bk+1 = ck else ak+1 = ck bk+1 = bk end if k=k+1 end while Find midpoint c is now b. a remains same c is now a. b remains same Midpoint is the minima PowerPoint Presentation by Sumita Das, GHRCE
  • 5.
    In Simple words Midpoint m ab Example: f(x)=3x2 – 2x f’(x)=6x-2 Put midpoint value in derivative. f’(x)=6*50-2=298 1. if f(m)=0, Midpoint is minima 2. if f(m)>0, Level of uncertainty will be [a, m] 3. if f(m)<0, Level of uncertainty will be [m, b] 50 9010 So, 298>0, level of uncertainty will be [10, 50] ,Follow the procedure. PowerPoint Presentation by Sumita Das, GHRCE
  • 6.
    Example Que: Find Minimaf(x)=(x-2)2 [0 6] Solution: f ‘(x)=2x-4 k ak bk ck f’(ck ) 1 0 6 3 2 2 0 3 1.5 -1 3 1.5 3 2.25 0.5 4 1.5 2.25 1.875 -0.25 5 1.875 2.25 2.062 0.123 6 1.875 2.062 1.9685 -0.063 7 1.9685 2.062 2.015 0.0305 8 1.9685 2.015 1.99175 -0.016 9 1.99175 2.015 2.003 0.006 PowerPoint Presentation by Sumita Das, GHRCE
  • 7.
    References [1] Singiresu S.Rao, “Engineering Optimization, Chapter 5: Nonlinear Programming I: One-Dimensional Minimization Methods”, 4th Edition PowerPoint Presentation by Sumita Das, GHRCE