Numerical Methods Golden Section Search Method - Theory
http://nm.mathforcollege.com http://nm.mathforcollege.com
Slide 2
For more details on this topic Go to
http://nm.mathforcollege.comhttp://nm.mathforcollege.com Click on
Keyword Click on Golden Section Search Method
Slide 3
You are free to Share to copy, distribute, display and perform
the work to Remix to make derivative works
Slide 4
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.
Slide 5
http://nm.mathforcollege.com5 Equal Interval Search Method
Figure 1 Equal interval search method. x f(x) a b Choose an
interval [a, b] over which the optima occurs. Compute and If then
the interval in which the maximum occurs is otherwise it occurs
in
Slide 6
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. X2X2 XLXL
X1X1 XuXu fufu f2f2 f1f1 fLfL Figure 2. Golden Section Search
method http://nm.mathforcollege.com
Slide 7
7 Golden Section Search Method- Selecting the Intermediate
Points ab XLXL X1X1 XuXu fufu f1f1 fLfL Determining the first
intermediate point a-b b X2X2 a XLXL X1X1 XuXu fufu f2f2 f1f1 fLfL
Determining the second intermediate point Golden Ratio=>
Let,hence b a http://nm.mathforcollege.com
Slide 8
Golden Section Search Method 8 Hence, after solving quadratic
equation, with initial guess = (0, 1.5708 rad) =Initial Iteration
Second Iteration Only 1 new inserted location need to be completed!
http://nm.mathforcollege.com
Slide 9
Golden Section Search- Determining the new search region Case1:
If then the new interval is Case2: If then the new interval is 9
X2X2 XLXL X1X1 XuXu fufu f2f2 f1f1 fLfL X2X2 XLXL X1X1 XuXu fufu
f2f2 f1f1 fLfL Case 2 Case 1 http://nm.mathforcollege.com
Slide 10
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
http://nm.mathforcollege.com
Slide 11
11 0 2.6185.2329.468 1.618 1.618 2 Min.g()=Max.[-g()] j th j-1
th j-2 th L = (1.618) v V = 0 j - 2 U = (1.618) v V = 0 j LL aa bb
UU 0.382( U - l ) 1 st 2 nd a = L + 0.382( U l ) = (1.618) v +
0.382(1.618) j-1 (1+1.618) V = 0 j - 2 3 rd a = (1.618) v +
1(1.618) j-1 = (1.618) v = already known ! V = 0 j - 2 V = 0 j - 1
4 th Figure 2.4 Bracketing the minimum point. Figure 2.5 Golden
section partition. = L _ Golden Section Search- (1-D) Line Search
Method http://nm.mathforcollege.com
Slide 12
12 If, Then the minimum will be between a & b. If as shown
in Figure 2.5, Then the minimum will be between & and. Notice
that : And Thus b ( wrt U & L ) plays same role as a ( wrt U
& L ) !! _ _ Golden Section Search- (1-D) Line Search Method
http://nm.mathforcollege.com
Slide 13
13 Step 1 : For a chosen small step size in ,say,let j be the
smallest integer such that. The upper and lower bound on i are and.
Step 2 : Compute g( b ),where a = L + 0.382( U - L ),and b = L +
0.618( U - L ). Note that, so g( a ) is already known. Step 3 :
Compare g( a ) and g( b ) and go to Step 4,5, or 6. Step 4 : If g(
a )
Step 5: If g( a ) > g( b ), then a i U. Similar to the
procedure in Step 4, putand. Compute,where and go to Step 7. Step
6: If g( a ) = g( b ) put L = a and u = b 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 http://nm.mathforcollege.com
Slide 15
THE END http://nm.mathforcollege.com
Slide 16
This instructional power point brought to you by Numerical
Methods for STEM undergraduate http://http://nm.mathforcollege.com
Committed to bringing numerical methods to the undergraduate
Acknowledgement
Slide 17
For instructional videos on other topics, go to
http://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.
Slide 18
The End - Really
Slide 19
Numerical Methods Golden Section Search Method - Example
http://nm.mathforcollege.com http://nm.mathforcollege.com
Slide 20
For more details on this topic Go to
http://nm.mathforcollege.comhttp://nm.mathforcollege.com Click on
Keyword Click on Golden Section Search Method
Slide 21
You are free to Share to copy, distribute, display and perform
the work to Remix to make derivative works
Slide 22
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.
Slide 23
23 Example The cross-sectional area A of a gutter with equal
base and edge length of 2 is given by (trapezoidal area):. 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 2 2
http://nm.mathforcollege.com
Slide 24
24 Solution The function to be maximized is Iteration 1: Given
the values for the boundaries of we can calculate the initial
intermediate points as follows: X2X2 XLXL X1X1 XuXu f2f2 f1f1 X L
=X 2 X 2 =X 1 XuXu X 1 =? http://nm.mathforcollege.com
Slide 25
25 Solution Cont To check the stopping criteria the difference
between and is calculated to be http://nm.mathforcollege.com
27 Theoretical Solution and Convergence Iterationxlxl xuxu x1x1
x2x2 f(x 1 )f(x 2 ) 10.00001.57140.97120.60025.16574.12381.5714
20.60021.5714 1.2005 0.97125.07845.16570.9712
30.60021.20050.97120.82955.16574.94260.6002
40.82951.20051.05880.97125.19555.16570.3710
50.97121.20051.11291.05885.17405.19550.2293
60.97121.11291.05881.02535.19555.19370.1417
71.02531.11291.07941.05885.19085.19550.0876
81.02531.07941.05881.04605.19555.19610.0541
91.02531.05881.04601.03815.19615.19570.0334 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. http://nm.mathforcollege.com
Slide 28
THE END http://nm.mathforcollege.com
Slide 29
This instructional power point brought to you by Numerical
Methods for STEM undergraduate http://http://nm.mathforcollege.com
Committed to bringing numerical methods to the undergraduate
Acknowledgement
Slide 30
For instructional videos on other topics, go to
http://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.