Transcript Lecture 5 Chapter 3 Improving Search

Lecture 5
Chapter 3 Improving Search
Local Improvement
Hill Climbing
Local Search
Neighborhood Search
Def 3.3 P 83 Improving searches are algorithms that
begin at a feasible point and move along a search
path of feasible points with improving objective
value.
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Hill Climbing
Feasible
Region
300
400
200
100
Maximum
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Neighborhood
The neighborhood of a point is all nearby points.
I haven’t
defined
nearby yet.
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Local & Global Optimum
Def 3.5 P 85 Local Optimum – point must be feasible
and in a small neighborhood no feasible point is
better
Def 3.7 P 85 Global Optimum – point must be
feasible and no other feasible point is better
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Example
Point
X
Y
Z
Elevation
10K
15.3K
12.8K
X
Y
Z
Any Local Optima?
Any Global Optima?
What is W?
W
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Improving Search
Def 3.12 P 88 Improving searches move from point 1
to point 2 by taking a step size of lambda in the
direction delta x.
X2 = X1 + (X)
where  is the step size and X is the direction
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Simplex Algorithm
Obj Value
Improves at Each
Step
4
3
1
2
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Interior Point Algorithm
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Obj Value
Improves at Each
Step
3
2
1
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Improving Direction
Def 3.13 P 90 Improving Direction – objective
function must improve for a sufficiently small step
size
C
B
What are the
improving
directions at
A, B, and C?
A
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Improving Direction
Def 3.13 P 90 Improving Direction – objective
function must improve for a sufficiently small step
size
C
B
What are the
improving
directions at
A, B, and C?
A
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Feasible Direction
Feasible Direction – for sufficiently small , the new
point is feasible.
At the point A is
the direction (0,1)
feasible?
Is (1,0) feasible?
(1,0) =
A
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Algorithm 3A: Continuous
Improving Search
0. Initialization. Let X0 be an feasible point and set
t to 0.
1. Check for local optimum. If no improving feasible
direction exists, then stop with Xt as a local
optimum.
2. Move direction. Let d be an improving feasible
direction.
3. Step size. Let s be the largest possible step size
that leads to a better feasible solution.
4. Make your move. Set Xt+1 = Xt+(s)(d), set t to t+1
and go to 1.
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Gradient
Def 3.19 P 99 The gradient of a function evaluated
at a point is the vector of partial derivatives
evaluated at the point.
Def 3.20 P 101 The gradient of a function evaluated
at a point is the direction of maximum increase.
Let f(X,Y) = X+2Y. The gradient of f at the point
(4,0) is [1,2]. Why?
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Linear Functions
For linear functions the gradient is constant at all
points.
f(X,Y) = 50X+75Y
What is the
direction of max
increase? Looks like
(50,75) or (10,15) or
(2,3) or (1,1.5).
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Nonlinear Function
For f(X,Y) = (x-2)2 + (Y-4)2
This is a circle with a center at (2,4). f increases
the further you move away from the center.
Gradient of f at (4,4) = [2(x-2), 2(y-4)] evaluated at
(4,4) is [4,0].
Draw it.
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Important Result
Def 3.21 P 102 Let G(z) (1xn) be the gradient of f(x)
evaluated at the point z. Let d (nx1) be a direction.
G(z)d < 0 implies f decreases in the direction d
= 0 implies f stays the same in the direction d
> 0 implies f increases in the direction d
G(z)d is (1xn)(nx1) = 1x1 (a constant)
This allows us to determine if a direction is an
improving direction.
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Example f(x,y) = (x-2)2+(y-2)2
G(0,0) = [2(x-2), 2(y-2)] at (0,0) = [-4,-4]
D
G.d
Change in f
[1,1]
-8
Decreases
[1,0]
-4
Decreases
[1,-1]
0
No Change
[-1,-1]
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Increases
f(2,2) = 0
(2,2)
f(0,0) = 8
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Improving Directions
Def 3.23 P 104. Let G(z) (1xn) be the gradient of
f(x) at the point z. If G(z) not 0, then d = G(z)
(nx1) is an improving direction for the
maximization problem. -G(z) is an improving
direction for the minimization problem.
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Example f(x,y) = (x-2)2+(y-2)2
z
[0,0]
G(z)=
[2(x-2),2(y-2)]
[-4,-4]
[2,0]
[0,-4]
[0,2]
[-4,0]
[3,3]
[2,2]
[3,0]
[2,-4]
(0,2)
(2,2)
(2,0)
(0,0)
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Feasible Directions
The question:
Is a given direction d feasible at the point z?
To answer this we must examine the constraints:
Three terms imply the same thing:
active at
tight at z
binding at z
Equality constraints are always tight!
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What are the feasible directions
at z=[0.0]
Minimize f(x,y)=(x-2)2+(y-2)2
Subject to x-y > 0
x > 0, 0 < y < 4
Point
Y/N
[0,1]
[-1,-1]
[1,0]
[1,1]
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What are the feasible directions
at z=[0.0]
Minimize f(x,y)=(x-2)2+(y-2)2
Subject to x-y > 0
x > 0, y > 0
Point
Y/N
[0,1]
NO
[-1,-1]
NO
[1,0]
YES
[1,1]
YES
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Binding Constraints
x-y > 0, x > 0, and y > 0 are all binding
If d=[d1,d2] is a feasible direction at [0,0], then
there must be some  > 0 such that the point
[0,0]+[d1,d2] satisfies all binding constraints.
1st constraint:
2nd constraint:
3rd constraint:
d1-d2 > 0
d1 > 0
d2 > 0
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Feasible Directions
1st constraint:
3rd constraint:
d1-d2 > 0
d2 > 0
2nd constraint:
d1 > 0
Try d = [1,1].  > 0,  > 0,  > 0 implies  can be infinite
Try d = [1,0].  > 0,  > 0, 0 > 0 implies  can be infinite
Try d = [0,1]. - > 0, 0 > 0,  > 0 implies  = 0, hence
direction not feasible
Try d = [1,-1]. 2 > 0,  > 0, - > 0 implies =0, hence
direction not feasible
Try d = [-1,0]
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To Obtain Local Optimum
For improving search, you need improving feasible
directions to get a local optimum.
Tractable Models (The Best Models For Applications)
Local Optimum = Global Optimum (see page 109)
Def 3.26 Unimodal Objective Function
x1 is feasible and x2 is feasible
f(x2) is better than f(x1), then d = x2 – x1 is an
improving direction
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Unimodal
Unimodal
Not Unimodal
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Linear Functions Are Unimodal
Def 3.28 P 112 If the objective function is unimodal,
then every unconstrained local optimum is an
unconstrained global optimum.
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Convex Sets
Def 3.29 If x and y are in the set, then all points on
the line segment between x and y are also in the
set.
Convex Set
Convex Set
Not
Convex
Not
Convex
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Discrete Sets
Def 3.30 Discrete Sets are not convex.
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Linear Constraints
Def 3.32 If all constraints are linear, then the
feasible set of points is a convex set.
Convex
Set
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Local Optimum = Global Optimum
Def 3.34
Objective function is unimodal
Constraint set is convex
Local = Global
Improving search works for this type of problem.
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