Transcript Document

Golden Section Method
Prepared by Shahrzad Kavianirad
Submitted to Prof. Dr. Sahand Daneshvar
Outline
 What is the Golden Section Search
 intermediate points
 Golden Ratio
 Algorithm
 example
Golden Section Method
• What is the Golden Section Search method used
for and how does it work? 1. 𝑎𝑘 < λ𝑘 < 𝑏𝑘
2. 𝜃(𝑎)𝑘 < 𝜃(λ)𝑘
 Unimodal function
3. 𝜃(λ) < 𝜃(𝑏)
𝜽𝒌
𝑘
𝜽(𝝁)𝑘
𝜽(𝒂)𝒌
𝒂𝒌 𝝁𝑘
3. New point
4. 𝑎𝑘 < 𝜇𝑘 < λ𝑘
𝛉(λ)𝒌
5. If𝜃(𝑎)𝑘 < 𝜃(λ)𝑘
new points : 𝑎𝑘 < 𝜇𝑘 < λ𝑘
6. If else new points:
𝜇𝑘 < λ𝑘 < 𝑏𝑘
𝜽(𝒃)𝒌
λ𝒌
𝑘
𝒃𝒌
𝑘
How are the intermediate points in the
Golden Section Search determined?
• Note that 𝑎 + 𝑏 is equal to the distance between
the lower and upper boundary points 𝑎𝑘 and 𝑏𝑘 .
𝑎
𝑏
=
𝑎+𝑏 𝑎
𝜃
𝜃
𝜽(𝝁)𝑘
𝛉(λ)𝒌
𝜽(𝒂)𝒌
𝑏 𝑎−𝑏
=
𝑎
𝑏
𝜽(𝒂)𝒌
𝜽(𝒃)𝒌
𝛉(λ)𝒌
𝜽(𝒃)𝒌
a-b
b
𝒂𝒌
𝒃𝒌
λ𝒌
a
b
𝑘
𝒂𝒌 𝝁𝑘
a
λ𝒌
𝒃𝒌
𝑘
Does the Golden Section Search have
anything to do with the Golden Ratio?
• The Golden Ratio has been used since ancient times in
various fields such as architecture, design, art and
engineering. To determine the value of the Golden Ratio :
𝑎
𝜑=
𝑏
1
1 + 𝜑 = 𝑜𝑟 𝜑2 + 𝜑 − 1 = 0
𝜑
−1 + 1 − (4 ∗ −1)
5−1
𝜑 =
=
= 0.61803
2
2
Does the Golden Section Search have
anything to do with the Golden Ratio?
• In other words, the intermediate points λ𝑘 and 𝜇𝑘 are chosen such that,
the ratio of the distance from these points to the boundaries of the
search region is equal to the golden ratio.
𝜃
𝜽(𝝁)𝑘
𝜽(λ𝒌 )
𝜽(𝒂)𝒌
b
𝒂𝒌 𝝁𝑘
a
𝜽(𝒃)𝒌
a
λ𝒌
𝒃𝒌
b
𝑘
The Golden Section Search Algorithm
1. Initialization: Determine 𝑎𝑘 and 𝑏𝑘 which is known to contain
the maximum of the function 𝜃
2. Determine two intermediate points λ𝑘 and 𝜇𝑘 such that: λ𝑘 =
𝑎𝑘 + 𝑑 𝑎𝑛𝑑𝜇𝑘 = 𝑏𝑘 − 𝑑 Where 𝑑 =
5−1
(𝑏𝑘
2
− 𝑎𝑘 )
• Evaluate 𝜃(λ𝑘 )and𝜃(𝜇𝑘 )
1. If 𝜃(λ𝑘 ) > 𝜃(𝜇𝑘 ) then :
 𝑎𝑘 = 𝜇𝑘
 𝜇𝑘 = λ𝑘
 𝑏𝑘 = 𝑏𝑘
λ𝑘 = 𝑎𝑘 +
5−1
(𝑏𝑘
2
− 𝑎𝑘 )
The Golden Section Search Algorithm
5. If 𝜃(λ𝑘 ) < 𝜃(𝜇𝑘 ) then :
 𝑎𝑘 = 𝑎𝑘
 𝑏𝑘 = λ𝑘
 λ𝑘 = 𝜇𝑘
𝜇𝑘 = 𝑏𝑘 −
5−1
(𝑏𝑘
2
− 𝑎𝑘 )
6. If 𝑏𝑘 − 𝑎𝑘 < 𝜀 (a sufficiently small number), then
𝑏𝑘 +𝑎𝑘
the maximum occurs at
and stop iterating, else
2
go to Step 2.
Example
 The cross-sectional area A of a gutter with equal base and edge
length of 2 is given by : 𝐴 = 4 sin 𝜃(1 + cos 𝜃). Find the angle
θ which maximizes the cross sectional area of the gutter. Using
an initial interval of [0,π/2], find the solution after 2 iterations.
Use an initial ε=0.05
2
2
θ
2
θ
Example
Iteration 1:
Given the values for the boundaries of 𝑎𝑘 = 0 and 𝑏𝑘 = π/2, we
can calculate the initial intermediate points as follows:
• λ𝑘 = 𝑎𝑘 +
5−1
2
𝑏𝑘 − 𝑎𝑘 = 0 +
5−1
2
• λ𝑘 = 𝑏𝑘 −
5−1
2
𝑏𝑘 − 𝑎𝑘 = 1.5708 −
1.5708 = 0.97080
5−1
2
1.5708 = 0.6
The function is evaluated at the intermediate points as
𝑓(0.97080 ) = 5.1654 and 𝑓(0.6 ) = 4.1227. since 𝜃(λ𝑘 ) > 𝜃(𝜇𝑘 )
we eliminate the region to the left of 𝜇𝑘 and update the lower
boundary point as 𝑎𝑘 =𝜇𝑘 .
Example
• The upper boundary point 𝑏𝑘 remains unchanged. The second
intermediate point 𝜇𝑘 is updated to assume the value of λ𝑘 and
finally the first intermediate point λ𝑘 is re-calculated as
follows:
5−1
2
5−1
2
• λ𝑘 = 𝑎𝑘 +
𝑏𝑘 − 𝑎𝑘 = 0.6 +
1.5708 − 0.6 =
1.2
It is greater
To check the stopping criteria :
then ε
𝑏𝑘 − 𝑎𝑘 = 1.5708 − 0.6 = 0.97080
The process is repeated in the
second iteration.
Example
Iteration 2
𝑎𝑘 =0.6 , 𝑏𝑘 =1.5708 , λ𝑘 =1.2 , 𝜇𝑘 =0.9708
5−1
5−1
𝜇𝑘 = 𝑏𝑘 −
𝑏𝑘 − 𝑎𝑘 = 1.2 −
1.2 − 0.6
2
2
= 0.82918
0.6
𝑏𝑘 − 𝑎𝑘 1.2 + 0.6
=
= 0.9
Max area
2
2
when θ=0.9 or
51.6 degree
Example
Iteration
𝒂𝒌
𝒃𝒌
λ𝒌
𝝁𝑘
𝜽(λ𝒌 ) 𝜽(𝝁𝒌 )
𝜀
1
0.00000
1.5708
0.97081
0.59999
5.1654
4.1226
1.5708
2
0.59999
1.5708
1.2000
0.97081
5.0791
5.1654
0.97081
3
0.59999
1.2000
0.97081
0.82917
5.1654
4.9418
0.59999
4
0.82917
1.2000
1.0583
0.97081
5.1955
5.1654
0.37081
5
0.97081
1.2000
1.1124
1.0583
5.1743
5.1955
0.22918
6
0.97081
1.1124
1.0583
1.0249
5.1955
5.1936
0.14164
7
1.0249
1.1124
1.0790
1.0583
5.1909
5.1955
0.08754
8
1.0249
1.0790
1.0583
1.0456
5.1955
5.1961
0.05410
9
1.0249
1.0583
1.0456
1.0377
5.1961
5.1957
0.03344
Golden Section Method