Nonlinear Opt.: Gold section search 3: Unterschied zwischen den Versionen
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| − | + | After having computed λ<sub>k</sub> and µ<sub>0</sub> we insert, in succession, λ<sub>0</sub> and µ<sub>0</sub> in the start function f(x) to get a result for f(λ<sub>0</sub>) and f(µ<sub>0</sub>). | |
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<math>f(\lambda _{0})= -(1,91 ^{2})+ 4 * 1,91 + 2 </math> | <math>f(\lambda _{0})= -(1,91 ^{2})+ 4 * 1,91 + 2 </math> | ||
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<math>f(\mu _{0})= -(3,09 ^{2})+ 4 * 3,09 + 2</math> | <math>f(\mu _{0})= -(3,09 ^{2})+ 4 * 3,09 + 2</math> | ||
f(µ<sub>0</sub>) = 4,8119 | f(µ<sub>0</sub>) = 4,8119 | ||
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| + | ''Step 1:'' | ||
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| + | We compare the difference between our interval values and our termination tolerance. If the difference is smaller, we stop our computation since the optimal output is reached, otherwise we start with the first iteration. | ||
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| + | 5 – 0 > 0,8 => no stop | ||
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| + | ''Step 2:'' | ||
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| + | Iteration k = 0: | ||
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| + | Depending on whether f(λ<sub>0</sub>) < f(µ<sub>0</sub>) or f(λ<sub>0</sub>) > f(µ<sub>0</sub>), we use case a) or b) which are shown above in the mathematical formulation. In our example, we have f(λ<sub>0</sub>) > f(µ<sub>0</sub>), so we use case b). This means that our interval will be diminished from the right side. Our new b<sub>1</sub> := b<sub>0</sub> – λ<sub>0</sub> | ||
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| + | a<sub>1</sub> := 0 | ||
| + | b<sub>1</sub> := 3,09 | ||
| + | µ<sub>1</sub> := 1,91 | ||
| + | λ<sub>1</sub> := 0 + (1 – 0,618) * (3,09 – 0) | ||
| + | => λ<sub>1</sub> = 1,18038 | ||
| + | f(µ<sub>1</sub>) := 5,9919 | ||
| + | f(λ<sub>1</sub>) = 5,3282230556 | ||
| + | go to Step 1 | ||
Version vom 26. Juni 2013, 14:18 Uhr
Theory
Golden section search is a part of unconstrained nonlinear optimization for strictly unimodal functions. This method is used to find a minimum or maximum by narrowing values in a specific interval. The extrema exists in the range. Based on the unimodal function, we assume that the local extrema is simultaneously the global extrema. In this method we do not use derivation, but rather a "step-by-step" approximation to the extrema.
Mathematical formulation (for maximum)
The method of the golden section search has the function Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): f:\mathbb{R}^{n} \supseteq X \rightarrow \mathbb{R}^{n}
based on an interval (a0,b0), where “a” is smaller than ”b”. A termination tolerance of ε is given, with the objective ε to compare the distance between “a” and “b” in each iteration. Through each iteration, the interval will be diminished by the factor λ, which is given with the value 0,618. λ will be cut from the bigger part of the interval. k = n represents the iteration steps.
Step 0: Initialization
Set a0 := a, b0:= b, k := 0
Calculate λk := ak + (1 – δ) * (bk – ak)
and µk := ak + δ * (bk – ak)
as well as f(λk) and f(µk)
Step1: Compare the distance with the termination tolerance. If the distance is smaller than the termination tolerance, stop the iteration, because the maximum has been reached.
If bk – ak < ε, then STOP.
Step 2: Compare the distance with the termination tolerance. If the distance is greater than the termination tolerance, continue with the following two cases:
Case a)
If f(λk) < f(µk),
set ak+1 := λk,
bk+1 := bk,
λk+1 := µk,
µk+1 := ak+1 + δ * (bk+1 – ak+1),
f(λk+1) := f(µk),
and compute f(µk+1)
the interval will be diminished from the right side
Case b)
If f(λk) ≥ f(µk),
set ak+1 := ak,
bk+1 := µk,
µk+1 := λk,
λk+1 := ak+1 + (1 – δ) * (bk+1 – ak+1),
f(µk+1) := f(λk),
compute f(λk+1)
set k = k+1, go to Step 1.
the interval will be diminished from the left side
The result is:
figure 1: interval demonstration of the two possibilities
Example for maximum
Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): max f(x)= -(x^{2})+ 4x + 2
where x ϵ on interval [0, 5], termination tolerance ε = 0,8
Step 0:
a0 and b0 are the margins of the interval
a0 := 0 ; b0 := 5
k represents the iteration steps
k := 0
Compute λk with the formula:
λk := ak + (1 – δ) * (bk – ak)
λ0 := 0 + (1 – 0,618) * (5 – 0)
=> λ0 = 1,91
and then µ0 with the formula:
µk := ak + δ * (bk – ak)
µ0 := 0 + 0,618 * (5 – 0)
=> µ0 = 3,09
After having computed λk and µ0 we insert, in succession, λ0 and µ0 in the start function f(x) to get a result for f(λ0) and f(µ0).
Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): f(\lambda _{0})= -(1,91 ^{2})+ 4 * 1,91 + 2
f(λ0) = 5,9919
Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): f(\mu _{0})= -(3,09 ^{2})+ 4 * 3,09 + 2
f(µ0) = 4,8119
Step 1:
We compare the difference between our interval values and our termination tolerance. If the difference is smaller, we stop our computation since the optimal output is reached, otherwise we start with the first iteration.
5 – 0 > 0,8 => no stop
Step 2:
Iteration k = 0:
Depending on whether f(λ0) < f(µ0) or f(λ0) > f(µ0), we use case a) or b) which are shown above in the mathematical formulation. In our example, we have f(λ0) > f(µ0), so we use case b). This means that our interval will be diminished from the right side. Our new b1 := b0 – λ0
a1 := 0
b1 := 3,09
µ1 := 1,91
λ1 := 0 + (1 – 0,618) * (3,09 – 0)
=> λ1 = 1,18038
f(µ1) := 5,9919
f(λ1) = 5,3282230556
go to Step 1
