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Appendix F
A Sample Test Problem

The following example has two controls and one state,
and gradients are calculated.

par=[1, 2, 20, 0, 1, 0.25]; % Parameters
% nx, nu, nn, npa, grad, c

nnb=par(3); nx=par(1); nu=par(2);
% Initial value for the state

% Starting values for computing the control

% Lower bound (vector or matrix) for the control

% Upper bound (vector or matrix) for the control


% Calls constr package

[Objective,Constraint,State,Integral,Costate,Gradient] =
cqq(Control,par,subs,xinit) % Right hand side of differential

function yy=t3x(t,it,z,yin,hs,um,xm,lm,ps)

% Dependent variables on the right hand side
of a differential equation are coded as
z(1), z(2), etc.

function ff=t3j(t,it,z,yin,hs,um,xm,lm,ps)
% Integrand of objective function


function ff=t3f(xf,um,xm,ps)
% Endpoint term

ff(1)=0; % Control constraint

function gg=t3c(ii,hs,um,xm,lm,ps)
gg=um(ii,1) + um(ii,1) - 1;

function dg=t3k(ii,hs,um,xm,lm,ps)
% Gradient of constraint


function yy=t3g(t,hs,um,xm,lm,nn)
% Gradient of objective

temp= 0.5*(lm(t,1)+lm(t+1,1));
yy=[(1+temp)*li3(xm,hs,t2), 0.25+temp];

function yy=t3a(nn,xf,um,xm,ps)
%Boundary condition for adjoint equation (at t=1)


function yy=t3l(t,it,z,yin,hs,um,xm,lm,ps)
% Right hand side of adjoint equation


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control policy, 10, 29, 46, 54, 61, 67, 88, 101,
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103, 109
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control restraint, 95
analytical solution, xv, 1, 91, 104“108, 143
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approximate solution methods, 10
control system, 11, 17
approximation methods, 11
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CPET, 16, 19

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