hessian using symbolic variables

2018 5 7
1 个回答
更新时间：2018 5 22
32 次查看（30 天）

0 个投票

Hi there,
I am struggling with obtaining the Hessian of my function. In the first case it works fine but in the second case I can not get it (still symbolic variables X are presented in H). So my question is why this happen? This is small portion of the code where I want to find hessian of my Rtot function with X being x1,x2,x3,x4.
Rtot= -sum(Rsub(:,i));
H=hessian(Rtot,X);
H1 = 0.5.*double(H);
with Rtot:
(6234858085755363*x1*((17865509505801117*x1)/14411518807585587200000 + (17865509505801117*x3)/14411518807585587200000 - 16863388614392257743519/1441151880758558720000))/140737488355328000 + (6069173180546921*x2*((17865509505801117*x1)/7205759403792793600000 + (17865509505801117*x2)/14411518807585587200000 + (17865509505801117*x3)/7205759403792793600000 + (17865509505801117*x4)/14411518807585587200000 - 16867211833426499182557/1441151880758558720000))/140737488355328000
and H1:
1.0e-07 *
0.5492 0.5346 0.2746 0
0.5346 0.5346 0.5346 0.2673
0.2746 0.5346 0 0
0 0.2673 0 0
This is fine. But then when I want to find hessian of different data set i get this. Rtot:
- (2587079436111693477*x2*(665101922855875194446501149/3153075739026468800000 - (4398046511104*x2)/5912017010674629 - (4398046511104*x3)/5912017010674629 - (4398046511104*x4)/5912017010674629 - (4398046511104*x1)/5912017010674629)^(1/3))/28147497671065600000000 - (4554081285294461427*x1*(1299027180826527113674752/6158351052786071875 - (4398046511104*x3)/5912017010674629 - (4398046511104*x1)/5912017010674629)^(1/3))/56294995342131200000000
and H still have symbolic variables inside which does not allow me to calculate H1. Difference is in Rtot but I still can not find out why it does not work.
Thank you a lot!!!

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Torsten
Torsten 2018-5-7
编辑：Torsten 2018-5-7
In the first case, your function is quadratic in the unknowns. Differentiation twice gives a constant Hessian that does not depend on x1, x2, x3 and x4.
In the second case, you have to substitute values for x1,x2,x3 and x4 to be able to get H1. Use MATLAB's "subs": (H1=double(subs(...))).
Best wishes
Torsten.
sensation
sensation 2018-5-7
Hi Torsten, Thanks for your insight. The code that comes before hessian is:
%
i=1;
Rsub(:,i) = subs(R(:,i),[c;s0(1,i);P(:,1);F(:,i);ro;eff;g;alpha(i);beta(i)],[C';init_storage(i);price1(1:N,1);inFlow1(1:N,i);ro1;eff1;gg;alpha_par1(i);b_par1(i)]);
Rtot= -sum(Rsub(:,i));
fsym= gradient(Rtot,X);
f= double(subs(fsym,X,zeros(size(X))));
H = hessian(Rtot,X);
H1=0.5.*double(H);
The substitute values youre talking about should be in H = hessian(Rtot,X) right? The X is unknown and the next step would be to do quadprog optimization knowing H and f.
However, if I try to do H1=double(subs(H,X)) I get ofc
"DOUBLE cannot convert the input expression into a double array" since x are unknown. Thanks for any tip!
M
Torsten
Torsten 2018-5-7
if you want to use the Hessian in quadprog, use
func_H = matlabFunction(H)
Best wishes
Torsten.
sensation
sensation 2018-5-22
Hi Torsten, Thanks for the tip. I created my Hessian as:
%
hessObj = hessian(Rtot(:,i),X(:,i));
H=matlabFunction(hessObj);
However, when I use H in my quadprog, H is still not double type. Any clue?
Thanks and cheers

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回答（1 个）

Alan Weiss
Alan Weiss 2018-5-7

0 个投票

There are some examples in the documentation that show how to do this:
Using Symbolic Math with Optimization Toolbox Solvers generates function handles.
Symbolic Math Toolbox Calculates Gradients and Hessians generates function files.
Alan Weiss
MATLAB mathematical toolbox documentation

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sensation
sensation 2018-5-8
Hi Alan, I had a view on this. My issue is following once I do:
%
gradf = jacobian(Rtot,X); % column gradf
hessf = jacobian(gradf,X);
qpoptions = optimset('Algorithm','interior-point-convex','Disp','iter');
tic
[x3(:,i),fval3(:,i)] = quadprog(hessf,f,A,b(:,i),Aeq,beq(:,i),LB(:,i),UB(:,i),[],qpoptions);
toc
hessf need to be of type H (double) but I can not convert it to double since there are still x unknowns in hessf. My hessf, instead of being a number able to be converted in double is:
(154325299586858956762382336*x2)/(13653103568166475135677228237359765625*(665101922855875194446501149/3153075739026468800000 - (4398046511104*x2)/5912017010674629 - (4398046511104*x3)/5912017010674629 - (4398046511104*x4)/5912017010674629 - (4398046511104*x1)/5912017010674629)^(5/3)) + (15092307240482528980631552*x1)/(1517011507574052792853025359706640625*(1299027180826527113674752/6158351052786071875 - (4398046511104*x3)/5912017010674629 - etc
thanks for any tip!
Alan Weiss
Alan Weiss 2018-5-8
Torsten gave you good advice: use matlabFunction exactly as he said.
Alan Weiss
MATLAB mathematical toolbox documentation

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