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Equal Interval Search Method
Figure 1 Equal interval search method.
x
f(x)
a b
2
2
Choose an interval [a, b] over which the optima occurs.
Compute and
22baf
If then the interval in which the maximum occurs is otherwise it occurs in
22baf
2222 bafbaf
bba ,
22
22
, baa
.Opt
2ba
2ba
6
Golden Section Search Method
The Equal Interval method is inefficient when is small. Also, we need to compute 2 interior points !
The Golden Section Search method divides the search more efficiently closing in on the optima in fewer iterations.
X2XL X1 Xu
fu
f2 f1
fL
Figure 2. Golden Section Search method
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7
Golden Section Search Method-Selecting the Intermediate Points
a bXL X1 Xu
fu
f1
fL
Determining the first intermediate point
a-b
b
X2
a
XL X1 Xu
fu
f2f1
fL
Determining the second intermediate point
)(*382.0),(*618.0
?);(618.0)(
lulu
lu
XXbandXXa
hencewhyab
XXbaa
bXaXX lu 2
Golden Ratio=> ...618.0ab
)(382.0 Lu xxb
ab
aba
ba
1
Let ,hence abR
)(618.0 Lu xxa
b
61803.02
)15(0111 2
RRRRRR
abXaXX ul 1
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Golden Section Search Method
8
3
2
1X2X
)(f
0]1cos2[4cos4
0)2cos(4cos4)()2sin(2sin4)(
)cos1(sin4)(
2
fff
Hence, after solving quadratic equation, with initial guess = (0, 1.5708 rad)
3. Opt
=Initial Iteration Second Iteration st1
3
Only 1 new inserted location need to be completed!
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Golden Section Search-Determining the new search region
Case1:If then the new interval is Case2:If then the new interval is
9
X2XL X1 Xu
fu
f2f1
fL
],,[ 12 xxxL
],,[ 12 uxxx
)()( 12 xfxf
)()( 12 xfxf
X2XL X1 Xu
fu
f2f1
fL
Case 2 Case 1
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At each new interval ,one needs to determine only 1(not 2) new inserted location (either compute the new ,or new )
Max. Min. It is desirable to have automated
procedure to compute and initially.
10
Golden Section Search-Determining the new search region
1x 2x )cos1(sin4)( f )cos1(sin4)( f
Lx ux
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11
0
δ
2.618δ 5.232δ 9.468δδ
1.618δ
1.6182δ
Min.g(α)=Max.[-g(α)]
α
jth
j-1thj-2th
αL = Σδ(1.618)vV = 0
j - 2
αU = Σδ(1.618)vV = 0
j • •
• •
αL αa αb αU
0.382(αU - αl)
1st
2nd
αa = αL + 0.382(αU – αl) = Σδ(1.618)v + 0.382δ(1.618)j-1(1+1.618)V = 0
j - 2
3rd
αa = Σδ(1.618)v + 1δ(1.618)j-1 = Σδ(1.618)v = already known ! V = 0
j - 2
V = 0
j - 1
4th
Figure 2.4 Bracketing the minimum point.
Figure 2.5 Golden section partition.= αL
_
Golden Section Search-(1-D) Line Search Method
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12
If , Then the minimum will be between αa & αb.
If as shown in Figure 2.5, Then the minimum will be between & and .
Notice that: And
Thus αb (wrt αU & αL ) plays same role as αa(wrt αU &
αL ) !!
_ _
Golden Section Search-(1-D) Line Search Method
jaULU )618.1(
)()( ba gg
)()( ba gg
a aLU UU
)(382.0]618.1[()382.0(618.1618.1)]618.1[(618.0])618.11[]618.1[)(236.0(
)]618.1[]618.1[)(236.0())(382.021(
11
1
LUj
Lb
jj
jjLUabLb
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13
Step 1 : For a chosen small step size δ in α,say ,let j be the smallest integer such that.
The upper and lower bound on αi are and .
Step 2: Compute g(αb) ,where αa= αL+ 0.382(αU- αL) ,and αb = αL+ 0.618(αU- αL).
Note that , so g(αa) is already known.
Step 3: Compare g(αa) and g(αb) and go to Step 4,5, or 6.
Step 4: If g(αa)<g(αb),then αL ≤ αi ≤ αb. By the choice of αa and αb, the new points and have .
Compute , where and go to Step 7.
Golden Section Search-(1-D) Line Search Method
12 1010
j
V
j
VVV gg
0
1
0))618.1(())618.1((
J
V
VU
0
)618.1(
2
0
)618.1(j
V
VL
1
0
)618.1(j
V
Va
LL bu ab )( ag )(382.0 LuLa
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Step 5: If g(αa) > g(αb), then αa ≤ αi ≤ αU. Similar to the procedure in Step 4, put and .
Compute ,where and go to Step 7.
Step 6: If g(αa) = g(αb) put αL = αa and αu = αb and return to Step 2.
Step 7: If is suitably small, put and stop. Otherwise, delete the bar symbols on ,and and
return to Step 3.
14
Golden Section Search-(1-D) Line Search Method
aL )(618.0 LuLb
uu )( bg
Lu )(21
Lui
baL ,, u
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Acknowledgement
For instructional videos on other topics, go to
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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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Numerical Methods Golden Section Search Method - Examplehttp://nm.mathforcollege.com
For more details on this topic
Go to http://nm.mathforcollege.com Click on Keyword Click on Golden Section Search Method
You are free to Share – to copy, distribute,
display and perform the work to Remix – to make derivative works
Under the following conditions Attribution — You must attribute the
work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work).
Noncommercial — You may not use this work for commercial purposes.
Share Alike — If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one.
23
Example
The cross-sectional area A of a gutter with equal base and edge length of 2 is given by (trapezoidal area):
)2sin(2sin4)cos1(sin4)(. AfMax
05.0
.
Find the angle which maximizes the cross-sectional area of the gutter. Using an initial interval of find the solution after 2 iterations.Convergence achieved if “ interval length ” is within
]2
,0[
2
2
2
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24
Solution)cos1(sin4)( f
60000.0)5708.1(2
155708.1)(2
15
97080.0)5708.1(2
150)(2
15
2
1
Luu
LuL
xxxx
xxxx
The function to be maximized is
Iteration 1: Given the values for the boundaries of we can calculate the initial intermediate points as follows:
2/ 0 uL xandx
1654.5)97080.0( f
1227.4)60000.0( f
X2XLX1 Xu
f2f1
XL=X2X2=X1 Xu
X1=?
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25
Solution Cont2000.1)60000.05708.1(
21560000.0)(
215
1
LuL xxxx
To check the stopping criteria the difference between and is calculated to be
uxLx
97080.060000.05708.1 Lu xx
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Solution ContIteration 2
97080.02000.15708.160000.0
2
1
xxxx
u
L
0791.5)2000.1( f
1654.5)97080.0( f
)()( 21 xfxf
82918.0)6000.02000.1(2
152000.1)(2
152
Luu xxxx
XLX2 XuX1
97080.02000.160000.0
1
xxx
u
L
9000.06000.02000.12
Lu xx
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Theoretical Solution and Convergence
Iteration xl xu x1 x2 f(x1) f(x2) 1 0.0000 1.5714 0.9712 0.6002 5.1657 4.1238 1.57142 0.6002 1.5714 1.2005 0.9712 5.0784 5.1657 0.97123 0.6002 1.2005 0.9712 0.8295 5.1657 4.9426 0.60024 0.8295 1.2005 1.0588 0.9712 5.1955 5.1657 0.37105 0.9712 1.2005 1.1129 1.0588 5.1740 5.1955 0.22936 0.9712 1.1129 1.0588 1.0253 5.1955 5.1937 0.14177 1.0253 1.1129 1.0794 1.0588 5.1908 5.1955 0.08768 1.0253 1.0794 1.0588 1.0460 5.1955 5.1961 0.05419 1.0253 1.0588 1.0460 1.0381 5.1961 5.1957 0.0334
0420.12
0588.10253.12
Lu xx 1960.5)0420.1( f
The theoretically optimal solution to the problem happens at exactly 60 degrees which is 1.0472 radians and gives a maximum cross-
sectional area of 5.1962.
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Acknowledgement
For instructional videos on other topics, go to
http://nm.mathforcollege.com
This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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