chin-shiuh shieh assignment 3.1, 文鶑 1. maximize f 1 (x,y) assignment - 20142 requirement: find...
TRANSCRIPT
Maximize F1(x,y)
Assignment - 2014 2
Requirement: Find maximize value of F1(x,y) used hill-climbing (gradient-ascent) method.
Step1: Initialize a value for: (x,y)(Choice one point:-5≤x,y<5)Example: x0 = 1; y0 = 1;
Assignment - 2014 3
If initialize value (x0,y0) is good, we can find global value. If not we can be trapped in local optima
The number of steps are change around: 100
Assignment - 2014 5
We initialize many points of (x0,y0) and used For loop to find optima value with each point. And then compare this to choice the best optima.
Assignment - 2014 6
The optimize when we you many initialize points (For loop) seem better than when we only used one point in some cases. But the number of steps size will be increase linear with number of initialize points.
Example: The number of steps size with 10 initialize points are change around: 1000
Assignment - 2014 7
But when we used Hill-climbing on this, some time I receive local optimize, some time is a strange value, have something are wrong in here? What happening?
I will point to you a case that we can’t find the optimize value if we only used Hill-climbing method.
Assignment - 2014 11
Now, you are standing in point a. You used Hill-climbing and you receive point a’. But it isn’t a optimize value in this location because the peak of mountains isn’t exit (out of range), and you are strapped on back of the mountains.
Assignment - 2014 12
We find maximum value of F(x,y) when (x,y) in line: y=-5*x/3 + 5
I replace y=-5*x/3 + 5 in function F(x,y). And then we can used For loop or Hill-Climbing to find this.
In here I used For loop with step size: 0.01.You can use Hill-Climbing to replace the For Loop to
have better result.
Assignment - 2014 13
After that we compare maximum value in case 1 and case 2 to find exactly global value
And we find out the optimize value is point b: Max = 4.1724
(x0, y0)= 1.8600 1.9000 Steps: 2009
Assignment - 2014 14
No. Optimize value in
Case1
(-5<x<3;-5<y<5
and y<-5*x/3 +
5)
Optimize value in
Case2
(y=-5*x/3 + 5)
Global optimize Total
numbe
r steps
1 Max = 2.3285
(x0, y0)=
-1.9660
1.9944
Steps: 730
Max = 4.1724
(x0, y0)=
1.8600 1.9000
Steps: 301
Max = 4.1724
(x0, y0)=
1.8600 1.9000
Steps:
1031
1031
2 Max = 3.2972
(x0, y0)=
1.9975 -1.9788
Steps: 1794
Max = 4.1724
(x0, y0)=
1.8600 1 .9000
Steps: 301
Max = 4.1724
(x0, y0)=
1.8600 1.9000
Steps: 2095
2095
3 Max = 2.3289
(x0, y0)=
1.3197 -1.8148
Steps: 556
Max = 4.1724
(x0, y0)= 1.8600
1.9000
Steps: 301
Max = 4.1724
(x0, y0)=
1.8600 1.9000
Steps: 857
857
4 Max = 4.1281
(x0, y0)=
1.7865 2.0232
Steps: 1708
Max = 4.1724
(x0, y0)= 1.8600
1.9000
Steps: 301
Max = 4.1724
(x0, y0)=
1.8600 1.9000
Steps: 2009
2009
Assignment - 2014 15