Part 04

Mech300 Numerical Methods, Hong Kong University of Science and Technology.

Part Four Optimization

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Mech300 Numerical Methods, Hong Kong University of Science and Technology.

Motivation Given a function f(x), find an x0 where f(x) is the maximum or minimum. Noncomputer method: derive the exact formula of the first derivative f’(x) using calculus and then solve the equation f’(x) = 0.

Question: How about find (x1, x2, x3, x4, x5) such that the following function is minimized?

sin log x1 + e  f ( x1 , x2 , x3 , x4 , x5 ) =

sin ( x 4 )

e − 3 x 2   cos −1 ( x1x5 )

cos ( x2 x3 − x52 )

− log x1 x5 + 5 x35

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Mech300 Numerical Methods, Hong Kong University of Science and Technology.

Example: Optimization of Parachute Cost For each chute: A = 2πr2

l = √2r

c = kcA

m = Mt/n

Cost = c0 + c1l + c2A2 Task:

v=

(

gm 1 − e −( c / m )t c

(

Given the total weight Mt, the impact speed limit vc, and the initial height z0, determine the size (r) and the number of chutes (n) such that the total cost n(c0 + c1l + c2A2) is the minimum while the impact speed of the chutes when reaching the ground is less than vc.

)

gm gm 2 z = z0 − t + 2 1 − e −( c / m )t c c

)

Minimize C = n(c0 + c1l + c2A2) Subject to v ≤ v0 and n ≥ 1 where A = 2πr2

l = √2r

c = kcA

m = Mt/n

t = root [ z0 – gmt/c + (gm2/c2)(1-e-(c/m)t)] v = (gm/c)(1-e-(c/m)t)

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Mech300 Numerical Methods, Hong Kong University of Science and Technology.

Mathematical Background Optimization or mathematical programming problem

Find x, which minimizes or maximizes f(x) Subject to di(x) ≤ ai

i = 1, 2, …, m

ci(x) = bi

i = 1, 2, …, p

where x:

an n-dimensional design vector

f(x): the objective function di(x): inequality constraints ci(x): equality constraints • If f(x) and the constraints are linear, it is linear programming. • If f(x) is quadratic and the constraints are linear, it is quadratic programming. • Otherwise, it is non-linear programming.

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Mech300 Numerical Methods, Hong Kong University of Science and Technology.

Overall Structure

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