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Optimisation The general problem: Want to minimise some function F(x) subject to constraints, a i (x) = 0, i=1,2,…,m 1 b i (x)  0, i=1,2,…,m 2 where x.

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Presentation on theme: "Optimisation The general problem: Want to minimise some function F(x) subject to constraints, a i (x) = 0, i=1,2,…,m 1 b i (x)  0, i=1,2,…,m 2 where x."— Presentation transcript:

1 Optimisation The general problem: Want to minimise some function F(x) subject to constraints, a i (x) = 0, i=1,2,…,m 1 b i (x)  0, i=1,2,…,m 2 where x is a vector of length n. F( ) is called the objective function. a i ( ) and b i ( ) are called the constraint functions.

2 Special Cases If n=1 there is just one variable, and we have the univariate case (as opposed to the multivariate case). If a i (x) and b i (x) are linear functions then we have linear constraints (as opposed to nonlinear constraints). If m 2 =0 we have equality constraints only. If m 1 =0 we have inequality constraints only. If m 1 =m 2 =0 we have the unconstrained case.

3 Techniques The techniques used to solve an optimisation problem depends on the properties of the functions F, a i, and b i. Important factors include: –Univariate or multivariate case? –Constrained or unconstrained problem? –Do we know the derivatives of F?

4 Example Linear Problem An oil refinery can buy light crude at £35/barrel and heavy crude at £30/barrel. Refining one barrel of oil produces petrol, heating oil, and jet fuel as follows: PetrolHeating oilJet fuel Light crude Heavy crude The refinery has contracts for 0.9M barrels of petrol, 0.8M barrels of heating oil and 0.5M barrels of jet fuel. How much light and heavy crude should the refinery buy to satisfy the contracts at least cost?

5 Problem Specification Let x 1 and x 2 be the number of barrels (in millions) of light and heavy crude that the refinery purchases. Cost (in millions of £): F(x) = 35x x 2 Constraints: 0.3x x 2  0.9 (petrol) 0.2x x 2  0.8 (heating oil) 0.3x x 2  0.5 (jet fuel) x 1  0, x 2  0 (non-negativity) This is called a “linear program”

6 Graphical Solution Minimum of F lies on boundary of feasible region. F varies linearly on each section of the boundary. Can get the solution by looking at the intersection points of the constraints forming the boundary. Feasible region x2x2 x1x (1, 2)

7 Solution (x 1,x 2 )F(x) (0,3)90 (2,1)100 (4,0)140 So minimum cost is for x 1 = 0 and x 2 = 3. Recall that: F(x) = 35x x 2

8 Unconstrained Univariate Case We seek to minimise f(x). If x* minimises f(x) then: i.f (x*) = 0 (first order condition) ii.f  (x*)  0 (second order condition) f(x) = (x-1)(x-1)+2

9 Example Minimise f(x) = x 2 + 4Cos(x) Solve: f (x) = 2x – 4Sin(x) = 0 y = Gives y =

10 Bisection Method Suppose we have already bracketed the zero in the interval [a,b]. Then: 1.Evaluate f at mid-point c=(a+b)/2. 2.If f(c) is zero then quit. 3.If f(a) and f(c) have the same sign then set a=c; else set b=c. 4.Go to Step 1. ab (a+b)/2

11 MATLAB Example >> >> a=1; fa=f(a); >> b=2; fb=f(b); >> c=(a+b)/2;fc=f(c);if fa*fc>0 a=c; else b=c; end;c Using the up arrow to repeat the last line we get values of c that converge to the solution of f(x)=0.

12 Convergence At each iteration the zero x* lies within the current interval from a to b. So the error |x*-x|

13 f(x k ) Newton’s Method Given an estimate x k of the zero a better estimate is obtained by approximating the function by the tangent line at x k. xkxk x k+1 f (x k ) = f(x k )/(x k -x k+1 ) x k+1 = x k – f(x k )/f (x k )

14 Convergence of Newton’s Method Error can be shown to be quadratic if initial estimate of zero is sufficiently close to x*. |x*-x k+1 | < M|x*-x k | 2 for some constant M. (Proof: Taylor series expansion of f(x*) about x k.)

15 Example Find real root of f(x)=x 3 +4x 2 -10=0. >> format long >> r=roots([ ]’); y=r(3);x=1; >> for i=1:8 fx=-10+x*x*(4+x);fxd=x*(8+3*x); err=y-x; a(i,1)=i;a(i,2)=x;a(i,3)=fx;a(i,4)=fxd;a(i,5)=err; x=x-fx/fxd; >> end; >> a

16 Problems with Newton’s Method Problems may arise if the initial estimate is not “sufficiently close” to the zero. Consider f(x)=ln(x). e1 If 0

17 Linear Interpolation Methods Newton method requires first derivative at each iteration. The bisection method doesn’t use the magnitudes of f at each end of the interval. Suppose we use f(a n ) and f(b n ) and finds a new estimate of the zero by approximating the function between a n and b n by a straight line. f(b n ) bnbn anan xnxn f(a n )

18 Secant Method The secant method is a linear interpolation method that generates approximations to the zero starting with x 0 and x 1 according to: x n-1 xnxn x n+1 x n+2 Problem with divergence!

19 Method of False Position To avoid possible divergence problem with the secant method we keep the zero bracketed in an interval (a,b), as in the bisection method. If f(c) = 0 we are finished. If f(a) and f(c) have the same sign we replace a by c; otherwise, we replace b by c.

20 Golden Section Method A function is unimodal on an interval [a,b] if it has a single local minimum on [a,b]. The Golden Section method can be used to find the minimum of function F on [a,b], where F is unimodal on [a,b]. This method is not based on solving F(x)=0. We seek to avoid unnecessary function evaluations.

21 If u>v then x* must lie in [a,x], and if u  v then x* must lie in [y,b]. Case 1: If u>v then new interval is [a,x] and length is x- a=  (b-a). At the next step we need to know F at: a +  (x-a) = a +  2 (b-a) a +  2 (x-a) = a +  3 (b-a) But we already know F at a +  2 (b-a) from the previous step so we can avoid this function evaluation. Golden Section Method Divide interval [a,b] at x and y as follows: x = a +  (b-a); u = F(x) y = a +  2 (b-a); v = F(y)

22 Golden Section Method Case 2: If u  v then new interval is [y,b] and length is b- y=  (b-a). At the next step we need to know F at: y +  (b-y) = a + 2  2 (b-a) y +  2 (b-y) = a +  2 (1+  )(b-a) = a +  (b-a) But we already know F at a +  (b-a) from the previous step so we can avoid this function evaluation. In both cases we get a new interval that is  times the length of the current interval, and each iteration requires only one function evaluation. After n iterations the error is bounded by (b-a)  n /2 Note:  2 +  - 1 = 0

23 MATLAB Code for Golden Section >> >> a=1; fa=f(a); b=2; fb=f(b);t=(sqrt(5)-1)/2; >> x=a+t*(b-a);y=a+t*t*(b-a);u=f(x);v=f(y); if u>v b=x;fb=u; else a=y;fa=v; end; c=(b+a)/2 Using the up arrow to repeat the last line we get values of c that converge to the minimum of F on [1,2].


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