Golden section method

1. Golden section method Professor:Dr. SahandDaneshvar Presented by: Mahdibanar

2. Problem Statement

3. Basic Definitions • The functionf (x) is usually called the objective function. • gi (x) ≤ 0, I = 1;:::; n g, is called an inequality constraint. • hi(x) = 0, i = 1; : : : ; n h, is called an equality constraint.

4. Theorem • Let θ : R R be strictly quasi convex over the interval [a, b]. Let λ, με [a, b] be such that λ < μ. If θ (λ) > θ (μ), then θ (z) ≥ θ (μ) for all z ε [a, λ). If θ (λ) ≤ θ (μ), then θ (z) ≤ θ (λ) for all z ε (μ, b].

5. The Golden Section Method • The golden section search is a technique for finding the minimum or maximum of a strictly unimodal function by successively narrowing the range of values inside which the min or max is known to exist. The technique derives its name from the fact that the algorithm maintains the function values for triples of points whose distances form a golden ratio.

6. Cont. • At a general iteration k of the golden section method, let the interval of uncertainty be [ak, book].By Theorem the new interval of uncertainty [ak+1, bk+1] is given by [λk, bk] if θ (λ) > θ (μ) and by [ak,μk] if θ (λ) ≤ θ (μ). The points λk and μk are selected such that the following hold true. λk = ak+(1-α)(bk - ak) if α ε (0, 1) μk = ak +α (bk - ak)

7. Cont. • As λk+1 and μk+1 are selected for the purpose of a new iteration, either λk+1coincides with μk or coincides with λk . If this can be realized, then during iteration k + 1,only one extra observation is needed.

8. Golden section rule Case 1 : if θ(λk) > θ(μk) • ak+1= λk • bk+1 = bk • λk+1 =μk • μk+1 =?

9. Cont. Case 2: if θ (λk) ≤ θ (μk) • ak+1= ak • bk+1 = μk • μ k+1 =λk • λ k+1 =?

10. Golden Number (α) • Consider case 1 or 2 Case 1: μk= λk+1 = ak+1 + (1-α) (bk+1 - ak+1) =λk+ (1-α)(bk- λk) we know λk= ak+(1-α)(bk - ak) and μk = ak +α (bk - ak) (1) AfterSubstituting the expressions λk and μk from (1) into the above equation, we get The roots are α≈0. 618 and α≈-1.618 For α must be in the interval (0,1) then α≈0. 618 This number is golden.

11. Algorithm golden section method The golden section method for minimizing a strictly quasiconvex function over the interval [a1, b1]. Initialization Step: • Choose an allowable final length of uncertainty ℓ>0 • [a1, b1] is the initial interval of uncertainty: • We computed λ1 = a1 + (1-α) (b1 - a1) and • μ1= a1 +α (b1 - a1) let α≈0. 618 • Evaluate θ (λ1) and θ (μ1), let k=1 and go to main step

12. MAIN STEP • If bk – ak<ℓ stop. The optimal solution lies in the interval [ak, book]. Otherwise, if θ (λk) > θ (μk), go to step 2; and if θ (λk) ≤ θ (μk) go to step 3. 2. ak+1=λk, bk+1 = bk, λk+1 =μk , μk+1 = ? • ak+1=ak, bk+1 = μk ,μ k+1 =λk , λ k+1 = ? 4 . Replace k by k+1 and go to step 1

13. Example Consider the following problem: Solution: design a table with below components:

14. Cont. After eight iterations involving nine observations, the interval of uncertainty is [-1.112 , -0.9361],so that the minimum can be estimated to be the midpoint -1.024.Note that the true minimum is in fact -1.0.