multipliers and their use for sensitivity of optimal solutions Constrained optimization x 1 x 2 Infeasible regions Feasible region Optimum Decreasing fx g1x g2x Inequality constraints ID: 347953
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Slide1
Optimality conditions for constrained local optima, Lagrange multipliers and their use for sensitivity of optimal solutions Slide2
Constrained optimization
x
1
x
2
Infeasible regions
Feasible region
Optimum
Decreasing
f(x)
g1(x)
g2(x
)
Inequality constraintsSlide3
Equality constraintsWe will develop the optimality conditions for equality constraints and then generalize them for inequality constraints
Give an example of an engineering equality constraint.Slide4
Lagrangian function
where j are unknown Lagrange multipliersStationary point conditions for equality constraints:
Lagrangian
and
stationaritySlide5
ExampleQuadratic objective and constraint
LagrangianStationarity conditionsFour stationary points Slide6
Problem Lagrange multipliersSolve the problem of minimizing the surface area of a cylinder of given value V. The two design variables are the radius and height. The equality constraint is the volume constraint
. SolutionSlide7
Inequality constraints require transformation to equality constraints:
This yields the following Lagrangian:Why is the slack variable squared?Inequality constraintsSlide8
Karush-Kuhn-Tucker conditions
Conditions for stationary points are then:
If inequality constraint is inactive (t
≠ 0
) then Lagrange multiplier = 0
For minimum, non-negative multipliersSlide9
Convex optimization problem hasconvex objective function
convex feasible domain if the line segment connecting any two feasible points is entirely feasible.All inequality constraints are convex (or gj = convex)All equality constraints are linearonly one optimumKarush-Kuhn-Tucker
conditions necessary and will also
be
sufficient for global
minimum
Why do the equality constraints have to be linear?Convex problemsSlide10
Example extended to inequality constraints
Minimize quadratic objective in a ringIs feasible domain convex?Example solved with fmincon using two functions: quad2 for the objective and ring for constraints (see note page)Slide11
Message and solution
Warning: The default trust-region-reflective algorithm does not solve …. FMINCON will use the active-set algorithm instead. Local minimum found ….Optimization completed because the objective function is non-decreasing in feasible directions, to within the default value of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance.
x =10.0000 -0.0000
fval
=100.0000
lambda =
lower: [2x1 double] upper: [2x1 double] eqlin: [0x1 double] eqnonlin: [0x1 double] ineqlin: [0x1 double]
ineqnonlin: [2x1 double]lambda.ineqnonlin’=1.0000 0What assumption Matlab likely makes in selecting the default value of the constraint tolerance?Slide12
Problem inequalitySolve the problem of minimizing the surface area of the cylinder subject to a minimum value constraint as an inequality constraint. Do also with
Matlab by defining non-dimensional radius and height using the cubic root of the volume. SolutionSlide13
Sensitivity of optimum solution to problem parameters
Assuming problem objective and constraintsdepend on parameter pThe optimum solution is x*(p)The corresponding function value
f
*(p)=f(x*(p),p) Slide14
Sensitivity of optimum solution to problem parameters (contd.)
We would like to obtain derivatives of f* w.r.t. pAfter manipulating governing equations we obtain
Lagrange multipliers called “shadow prices” because they provide the price of imposing constraints
Why do we have ordinary derivative on the left side and partial on the right side?Slide15
ExampleA simpler version of ring problem
For p=100 we found Here it is easy to see that solution isWhich agrees withSlide16
Problems sensitivity of optima
For find the minimum for p=0, estimate the derivative df*/dp there, and check by solving again for p=0.1 and comparing to finite difference derivative. SolutionCalculate the derivative of the cylinder surface
area with respect to change in volume using the Lagrange multiplier and compare to the derivative obtained by differentiating the exact solution.
Solution