Why I get index out of bounds error even though I debug the loop and its Ok

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Muna Shehan
Muna Shehan 2016-7-15
评论： Muna Shehan 2016-7-16
Hi all I try to optimize three variables using fmincon. Inside the objective function there is a loop which the error message point on it. I debug the loop and its work and its reach the end and go to fmincon to complete the check after that I get the error Kindly, can any on explain why I get this error. Regards
??? Attempted to access t02(206); index out of bounds because numel(t02)=205. % t02 is a time vector (206x1)
Error in ==> ObjFun_Zdd_F at 52
ttt=t02(i+1)-t02(i);
Error in ==> @(Fkc)ObjFun_Zdd_F(Fkc)
Error in ==> E:\Program
Files\MATLAB\R2011a\toolbox\optim\optim\private\evalObjAndConstr.p>evalObjAndConstr
at 135
Error in ==> E:\Program
Files\MATLAB\R2011a\toolbox\optim\optim\sqpLineSearch.p>sqpLineSearch at 265
Error in ==> fmincon at 832
[X,FVAL,EXITFLAG,OUTPUT,LAMBDA,GRAD,HESSIAN] =
sqpLineSearch(funfcn,X,full(A),full(B),full(Aeq),full(Beq), ...
Error in ==> Fmain at 9
[Fkcopt,fval,exitflag,output] = fmincon(@(Fkc) ObjFun_Zdd_F(Fkc),Fkco,...
and this is fmincon
Fkco=[100,1000,500];
lb=[0,100,50];
ub=[3000,160000,50000];
options = optimset('Display','iter','TolFun',1e-8,'algorithm','sqp');
[Fkcopt,fval,exitflag,output] = fmincon(@(Fkc) ObjFun_Zdd_F(Fkc),Fkco,...
[],[],[],[],lb,ub,[],options);
and the objective function is:
function [val] = ObjFun_Zdd_F(Fkc)
F=Fkc(1);
kc(1)=Fkc(2);
kc(2)=Fkc(3);
load IRI_737b
zdot = calc_zdot(road_x, road_z, 10);
zt=zdot.u;
x00=[0,0,0,0];
input=@(t)lookup_u(zdot,t);
opt = odeset('RelTol', 1e-2, 'AbsTol', 1e-3);
[t02,y] = ode23(@(t,x)f(t,x,F,kc, @(t)lookup_u(zdot,t)), [0 2], x00,opt);
xx=size(t02,1);
x_frame=y';
FF=F * ones(1, xx);
xu_frame=[x_frame;FF];
val = 0;
r1 = 1e+5
r2 = 0.5;
q = 1e-5;
R = diag([r1, 0, 0, 0, q,r2]);
xu_frame_zs = zeros(6,length(t02));
for i=1:205
ddx=ff(i,xu_frame(1:4,i),F,kc);
Acc=ddx(4);
xu_frame_zs(:,i) = [xu_frame(:,i);Acc];
ttt=t02(i+1)-t02(i); % here the error pointed
val=val+ttt*xu_frame_zs(:,i)'*R*xu_frame_zs(:,i);
end
end
  8 个评论
显示 6更早的评论
Walter Roberson
Walter Roberson 2016-7-15
At the command line, give the command
dbstop if error
and run the program. When it stops with that error message, show size(t02) and size(y) and look at xx, and between all of those figure out whether the ode routine returned 205 or 206 values.
Muna Shehan
Muna Shehan 2016-7-16
I followed your instruction, and what you mention at the beginning is exactly what occur in my case. It is true size(t02) is 206 for the first fmincon iteration and since the ode23 is inside the objective function ObjFun_Zdd_F(Fkc) so size(t02) is changed based on what you explain before "the ode will decide by itself how many points to use, based upon what it decides it needs to meet the integration tolerances" which what happened in the next fmincon iteration. I have to put ode23 inside the objective function ObjFun_Zdd_F(Fkc) and even if I use length(t02)-1 as the loop upper limit I need to determine the output of lookup_u(zdot,t) which I did it before manually by using a breakpoint and debug the change in both t and u inside lookup_u(zdot,t) and re-write t and u in a vectors to filter it later based on the t02 values. After that I write filtered u values in a vector and use it later in the appropriate location inside ff(i,xu_frame(1:4,i),F,kc). Now I use length(t02)-1 as the upper limit and inside the loop I call lookup_u(zdot, t02(i)) and save its value in Zot02 then send this value to ff(i,xu_frame(1:4,i),F,kc,Zot02) as a passing parameter
Zot02= lookup_u(zdot, t02(i));
ddx=ff(i,xu_frame(1:4,i),F,kc,Zot02);
and lookup_u(tu, t) is
function u = lookup_u(tu, t)
i = find(tu.t <= t, 1, 'last');
[m,n] = size(tu.u);
if (m==1) || (n==1)
if i<length(tu.t)
u = (tu.u(i)*(tu.t(i+1) -t) + tu.u(i+1)*(t-tu.t(i)))/(tu.t(i+1)-tu.t(i));
else
u = tu.u(i);
end
else
if i < length(tu.t)
u = (tu.u(:,i)*(tu.t(i+1) -t) + ...
tu.u(:,i+1)*(t-tu.t(i)))/(tu.t(i+1)-tu.t(i));
else
u = tu.u(:,i);
end
end
end
This modification is allowed to calculate u corresponding to the t02 index, the code is run but it seem I still need to enhance fmincon results. Thanks Walter for all your replays because your comments are the reason which highlights my way to understand what happen and lead me to this conclusion.
Norm of First-order
Iter F-count f(x) Feasibility Steplength step optimality
0 4 4.771200e+002 0.000e+000 4.350e+002
1 8 3.767602e+002 0.000e+000 1.000e+000 2.081e+002 5.935e+003
2 12 2.864263e+002 0.000e+000 1.000e+000 2.656e+002 3.344e+002
3 16 2.097120e+002 0.000e+000 1.000e+000 4.654e+002 4.192e+003
4 30 1.709489e+002 0.000e+000 2.825e-002 7.196e+002 1.239e+004
5 58 1.696533e+002 0.000e+000 1.916e-004 4.805e+000 1.103e+002
6 73 1.696308e+002 0.000e+000 1.977e-002 1.265e-001 1.106e+002
7 77 1.696229e+002 0.000e+000 1.000e+000 3.172e-002 2.103e-001
8 81 1.696210e+002 0.000e+000 1.000e+000 1.031e-002 1.301e-001
9 86 1.696206e+002 0.000e+000 7.000e-001 1.277e-002 3.906e-002
10 90 1.696206e+002 0.000e+000 1.000e+000 4.313e-003 3.354e-002
11 94 1.696205e+002 0.000e+000 1.000e+000 1.991e-002 2.437e-002
12 98 1.696203e+002 0.000e+000 1.000e+000 6.413e-002 3.515e-002
13 104 1.696198e+002 0.000e+000 4.900e-001 1.593e-001 3.342e-001
14 110 1.696197e+002 0.000e+000 4.900e-001 5.265e-002 6.015e-002
15 114 1.696196e+002 0.000e+000 1.000e+000 2.605e-002 5.945e-002
16 120 1.696193e+002 0.000e+000 4.900e-001 6.390e-002 3.182e-001
17 129 1.696193e+002 0.000e+000 1.681e-001 4.015e-003 5.966e-002
18 133 1.696192e+002 0.000e+000 1.000e+000 2.304e-002 9.673e-002
19 137 1.696190e+002 0.000e+000 1.000e+000 1.701e-001 3.285e-001
20 141 1.696186e+002 0.000e+000 1.000e+000 1.593e-001 3.340e-001
21 145 1.696171e+002 0.000e+000 1.000e+000 8.108e-001 4.778e-001
22 149 1.696056e+002 0.000e+000 1.000e+000 4.002e+000 4.248e-001
23 153 1.687674e+002 0.000e+000 1.000e+000 4.070e+001 1.103e+001
24 171 1.685538e+002 0.000e+000 6.782e-003 2.743e-001 2.253e+000
25 176 1.685358e+002 0.000e+000 7.000e-001 5.257e-002 4.522e+000
26 180 1.685354e+002 0.000e+000 1.000e+000 3.341e-002 1.945e+001
27 184 1.685342e+002 0.000e+000 1.000e+000 2.767e-002 1.488e+000
28 188 1.685340e+002 0.000e+000 1.000e+000 3.975e-003 1.894e-001
29 192 1.685340e+002 0.000e+000 1.000e+000 2.070e-004 1.683e-001
Local minimum possible. Constraints satisfied.
fmincon stopped because the size of the current step is less than
the default value of the step size tolerance and constraints are
satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
>>

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采纳的回答

Walter Roberson
Walter Roberson 2016-7-16
Poster had hard-coded the expected number of points returned by ode23s, but ode23s decides dynamically how many points to use and was not always using the same number of points. User has changed the code to not hard-code the number of points, solving the original problem.
  1 个评论
Muna Shehan
Muna Shehan 2016-7-16
what you say "'ode23s decides dynamically how many points to use and was not always using the same number of points" is true.The ode23s change the number of points at every optimization iteration which I did not know this information before this time. Thanks again

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