Transcript Golden section method - Industrial Engineering Department

Professor:
Dr. Sahand Daneshvar
Presented by:
Mahdi banar
Problem Statement
Basic Definitions
 The function f (x) is usually called the objective function.
 gi (x) ≤ 0, I = 1;:::; n g, is called an inequality constraint.
 hi(x) = 0, i = 1; : : : ; n h, is called an equality constraint.
Theorem
 Let θ : R
R be strictly quasi convex over the interval
[a, b]. Let λ, μ ε [a, b] be such that λ < μ.
If θ (λ) > θ (μ), then θ (z) ≥ θ (μ) for all z ε [a, λ).
If θ (λ) ≤ θ (μ), then θ (z) ≤ θ (λ) for all z ε (μ, b].
The Golden Section Method
 The golden section search is a technique for
finding the minimum or maximum of a
strictly unimodal function by successively narrowing
the range of values inside which the min or max is
known to exist. The technique derives its name from
the fact that the algorithm maintains the function
values for triples of points whose distances form
a golden ratio.
Cont.
 At a general iteration k of the golden section method,
let the interval of uncertainty be [ak,
book].By
Theorem the new interval of uncertainty [ak+1, bk+1] is
given by [λk, bk] if θ (λ) > θ (μ) and by [ak, μk] if θ (λ) ≤
θ (μ). The points λk and μk are selected such that the
following hold true.
λk = ak +(1-α)(bk - ak) if α ε (0, 1)
μ = ak +α (bk - ak)
k
Cont.
 As λk+1 and μk+1 are selected for the purpose of a new
iteration, either λk+1 coincides with μk or coincides with
λk . If this can be realized, then during iteration
k+
1, only one extra observation is needed.
Golden section rule
Case 1
: if θ(λk) > θ(μk)
ak+1= λk
bk+1 = bk
λk+1 =μk
μ =?
k+1
Cont.
Case 2: if θ (λk) ≤ θ (μk)
 ak+1= ak
 bk+1 = μk
 μ k+1 =λk
 λ =?
k+1
Golden Number (α)
 Consider case 1 or 2
Case 1: μk = λk+1 = ak+1 + (1-α) (bk+1 - ak+1)
= λk + (1- α)(bk
- λk)
we know λk = ak +(1-α)(bk - ak) and μk = ak +α (bk - ak) (1)
After Substituting the expressions λk and μk from (1) into the
above equation, we get
The roots are α≈0. 618 and α≈-1.618
For α must be in the interval (0,1) then α≈0. 618
This number is golden.
Algorithm golden section method
The golden section method for minimizing a strictly
quasiconvex function over the interval [a1, b1].
Initialization Step:
 Choose an allowable final length of uncertainty ℓ>0
 [a1, b1] is the initial interval of uncertainty:
 We computed λ1 = a1 + (1-α) (b1 - a1) and
 μ1 = a1 +α (b1 - a1)
let α≈0. 618
 Evaluate θ (λ1) and θ (μ1), let k=1 and go to main step
MAIN STEP
1. If bk – ak <ℓ stop. The optimal solution lies in the
interval [ak, book]. Otherwise, if θ (λk) > θ (μk), go
to step 2; and if θ (λk) ≤ θ (μk) go to step 3.
2. ak+1= λk,
3.
bk+1 = bk ,
λk+1 =μk , μk+1 = ?
ak+1= ak , bk+1 = μk , μ k+1 =λk , λ k+1 = ?
4 . Replace k by k+1 and go to step 1
Example
Consider the following problem:
Solution: design a table with below components:
k
ak
bk
λk
μk
θ(λk)
θ(μk)
1
-3.000
5.000
0.056
1.944
0.115
7.667
Cont.
After eight iterations involving nine observations, the interval of
uncertainty is [-1.112 , -0.9361], so that the minimum can be
estimated to be the midpoint -1.024.Note that the true minimum
is in fact -1.0.