Introduction to Optimization Problems and Methodsmillerpd/docs/501_Fall09/... · 2009-12-11 ·...

Linear OptimizationNonlinear Optimization

Integer OptimizationDynamic Optimization

Introduction to Optimization Problems andMethods

Jingchen [email protected]

December 10, 2009

Jingchen Wu Introduction to Optimization Problems and Methods



Outline

1 Linear OptimizationProblemSimplex Method

2 Nonlinear OptimizationOne-Dimensional OptimizationMulti-Dimensional Unconstrained OptimizationMulti-Dimensional Constrained Optimization

3 Integer OptimizationCutting Plane Method

4 Dynamic OptimizationDiscrete Dynamic Programming




ProblemSimplex Method

General and Standard form of Linear Programming problem

General Form

min cT xs.t. aT

i x ≥ bi i ∈ M1

aTi x ≤ bi i ∈ M2

aTi x = bi i ∈ M3

xj ≥ 0 j ∈ N1

xj ≤ 0 j ∈ N2

Standard Form

min cT xs.t. Ax = b

x ≥ 0





Simplex Method

Polyhedron Geometric Interpretation ofSimplex Method

If optimal solution exists, it canbe achieved on one of thevertices. The algorithm travelsalong edges of the polyhedron tovertices with higher functionvalue until it achieves optimalvalue.





Pros and Cons of Simplex Method

Pros

Remarkably efficient in practice, especially when the scale ofthe problem is small.

Cons

Worst-case complexity is exponential time.

For almost every pivoting rule, there is an exponentialworst-case complexity example.




One-Dimensional OptimizationMulti-Dimensional Unconstrained OptimizationMulti-Dimensional Constrained Optimization

One-Dimensional Nonlinear Problem

Problem

minx∈[a,b] f (x)

Assumption

There is a unique local minimum in the interval considered.

f (x) may not be differentiable.





Bisection Method

5b

X5

5a

In each iteration, the lengthof the interval becomes halfof the previous one.

Calculate the value at twomore points in eachiteration.





Bisection Method

5b

X5

5a In each iteration, the lengthof the interval becomes halfof the previous one.

Calculate the value at twomore points in eachiteration.





Golden Section Method

b

a=

a

c=

b − c

c= 1.618

In each iteration, the lengthof the interval becomes0.618 of the previous one.

Only need to calculate onemore point in each iteration.





Multi-Dimensional Unconstrained Problem

Problem

min f (x), x ∈ Rn

Assumption

f (x) is 2 times differentiable.

Questions:

How to choose direction?

How to choose step length?





Steepest Descent Method

Choose di at xi as di = −∇f (xi ).

Since steepest descent direction is a local property, themethod is not effective.





Conjugate Gradient Method (1)

Definition

Nonzero vectors u and v are conjugate with respect to positivedefinite matrix A if uTAv = 0

Property

If any two of nonzero vectors v1, v2, ...vn are conjugate in Rn, thenv1, v2, ..., vn form a base in Rn.






Consider f (x) = 12xTAx − bT x , where A is positive definite.

rk = b − Axk is called residual, which is gradient descent.

Let

pk+1 = rk −∑i≤k

pTi Ark

pTi Api

pi .

xk+1 = xk + αk+1pk+1, where αk+1 =pTk+1rk

pTk+1Apk+1

.






Avoid repeating directions usedbefore.

Convergence in finite steps forquadratic objective functions.





Newton Method (1)

Consider f (x) = 12(x − xk)TA(x − xk)− bT (x − xk), where A

is positive definite.

The minimizer is xk+1 = xk + A−1b.

The idea of Newton Method is approximating a functionlocally at xk by its first three terms of Taylor expansion, andset the next iterate be the minimizer of the approximation.





Newton Method (2)Pros and Cons

Pros

Newton’s method can often converge remarkably quickly,especially if the iteration begins “sufficiently near” the optimalsolution.

A modification Quasi-Newton methods saves computation.

Cons

If f (x) is not convex function, the Newton’s method maysometimes diverge or converge to saddle point and localminimum.





Problem

Problem

min f (x) s.t. gi (x) ≤ 0, j = 1, 2, ...,m

Assumption

gi (x), j = 1, 2, ...,m are convex function. So feasible region isconvex.





Interior Penalty Function Method

Choose penalty function as1

gj(x), and let φ(x , rk) =

f (x)− rk

m∑j=1

1

gj(x).

Suppose we have an initialfeasible solution. Dounconstrained optimizationfor φ(x , rk) in each step.

let rk → 0 as k →∞. When rk is very small, φ approximatelyequal to the original constrained problem.

More often, penalty function log(gj(x)) is used.




Cutting Plane Method

Integer Programming-Cutting Plane Method

Blue lines are the boundary;

Red dots are feasible solutions;

Assume cost vector is verticallyupward;

Green line is the cutting plane.




Discrete Dynamic Programming

A Discrete Dynamic Programming Example

There are T + 1 periods, t = 0, 1, ...,T .kt : capital in period t; ct : consumption in period t; u(ct)utility of consuming ct .

kt+1 = Akat − ct , A > 0, 0 < a < 1

Problem

maxT∑

t=0

λtu(ct), subject to kt+1 = Akat − ct ≥ 0





Discrete Bellman Equation

Define value function VT+1(k) = 0;

Bellman Equation

Vt(kt) = max(u(ct) + λVt+1(kt+1))subject to kt+1 = Aka

t − ct ≥ 0 for t = 0, 1, ...,T

Bellman Equation can be solved by backward induction.





A Continuous Dynamic Programming Example

Problem

min

{∫ T

0C [x(t), a(t)] + D[x(T )]

}where C (·) is the cost rate function and D(·) is the utility at thefinal state, x(t) is the system state vector, x(0) is given, and a(t)is control vector.The system is also subject to x(t) = F [x(t), u(t)].





HJB Equation

Define value function V (x ,T ) = D(x);

V (x(t), t) = mina{C (x(t), a)dt + V (x(t + dt), t + dt)}

= mina{C (x(t), a)dt + V (x , t)dt +

∇V (x , t)xdt + o(dt2)}

HJB Equation

V (x , t) + mina{∇V (x , t)F (x , a) + C (x , a)} = 0

V (x ,T ) = D(x)





Thank you very much for your attention!


Introduction to Optimization Problems and Methodsmillerpd/docs/501_Fall09/... · 2009-12-11 ·...

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Transcript of Introduction to Optimization Problems and Methodsmillerpd/docs/501_Fall09/... · 2009-12-11 ·...