non linear optimization with fmincon
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Hi, I am solving a nonlinear optimization problem: Xop=fmincon(@ident1,[0.1;0.1],[],[],[],[],[1 1],[50 5]); problem.options =optimset('Display','iter-detailed'); I found Xop= 2.6147 1.0000 I don't understand why the second value of Xop don't change although i change the upper value. If upper value is 0.5 Xop(2)=0.5 Can you help me? thanks
11 个评论
Anoire BEN JDIDIA
2016-11-2
编辑:Walter Roberson
2016-11-2
function [eps] =ident1(x)
global Hilb
global tsim
x
eps=sum(abs(Hilb-(x(1)*13270.8617*(55.89*exp(-((tsim- 2.915)/7.182).^2) + 65.16*exp(-((tsim- 378.4)/224.1).^2) + 4.952 *exp(-((tsim-41.41)/9.817).^2) + 8.203*exp(-((tsim-228.6)/ 48.32).^2) + 6.925*exp(-((tsim- 166.4)/54.56).^2) + 2.585*exp(-((tsim-274.2 )/ 28.11).^2) +53.45*exp(-((tsim-50.93)/ 146.7).^2) + 1.155*exp(-((tsim-302)/16.94).^2)-0.29*(30.65*exp(-((tsim-1.463)/1.283).^2) -3.707*exp(-((tsim-12.16)/9.039).^2) + 158.4*exp(-((tsim-9.824e+004)/ 5.722e+004).^2) + 0.09067*exp(-((tsim- 156.9 )/ 0.2263 ).^2)+ 0.1171*exp(-((tsim- 220.8)/23.39).^2) -0.6953*exp(-((tsim- 182.4 )/396.7).^2) + 0.2898*exp(-((tsim- 158.8)/33.83).^2) -0.1958*exp(-((tsim- 183.8)/ 0.3628).^2)))+ x(2)*0.688*(55.89*exp(-((tsim- 2.915)/7.182).^2) + 65.16*exp(-((tsim- 378.4)/224.1).^2) + 4.952 *exp(-((tsim-41.41)/9.817).^2) + 8.203*exp(-((tsim-228.6)/ 48.32).^2) + 6.925*exp(-((tsim- 166.4)/54.56).^2) + 2.585*exp(-((tsim-274.2 )/ 28.11).^2) +53.45*exp(-((tsim-50.93)/ 146.7).^2) + 1.155*exp(-((tsim-302)/16.94).^2)-0.29*(30.65*exp(-((tsim-1.463)/1.283).^2) -3.707*exp(-((tsim-12.16)/9.039).^2) + 158.4*exp(-((tsim-9.824e+004)/ 5.722e+004).^2) + 0.09067*exp(-((tsim- 156.9 )/ 0.2263 ).^2)+ 0.1171*exp(-((tsim- 220.8)/23.39).^2) -0.6953*exp(-((tsim- 182.4 )/396.7).^2) + 0.2898*exp(-((tsim- 158.8)/33.83).^2) -0.1958*exp(-((tsim- 183.8)/ 0.3628).^2))).^2+ 0.87*( 30.65*exp(-((tsim-1.463)/1.283).^2) -3.707*exp(-((tsim-12.16)/9.039).^2) + 158.4*exp(-((tsim-9.824e+004)/ 5.722e+004).^2) + 0.09067*exp(-((tsim- 156.9 )/ 0.2263 ).^2)+ 0.1171*exp(-((tsim- 220.8)/23.39).^2) -0.6953*exp(-((tsim- 182.4 )/396.7).^2) + 0.2898*exp(-((tsim- 158.8)/33.83).^2) -0.1958*exp(-((tsim- 183.8)/ 0.3628).^2)).^2 + x(2)*6156180.1101 +1732750.255 + 567.84 - x(1)*13270.8617*( 39.78*exp(-((tsim- 62.36)/ 51.23 ).^2) + 39.99*exp(-((tsim+13.07)/ 49.11).^2) + 6.574*exp(-((tsim-95.15)/ 7.534).^2) + 7.534*exp(-((tsim-131)/ 20.23).^2) + 3.91e+013*exp(-((tsim- 1.36e+005)/ 2.603e+004).^2) + 15.31*exp(-((tsim-162.5)/49.37).^2) + 11.18*exp(-((tsim- 11.18)/ 41.93).^2) + 5.944*exp(-((tsim- 275.8)/ 15.14 ).^2)-0.29*( 4.629e+004*cos(0.01037*tsim) -8.526e+004*sin(0.01037*tsim ) -6.644e+004*cos(2*0.01037*tsim) + 3426 *sin(2*0.01037*tsim) + 3290*cos(3*0.01037*tsim) + 4.341e+004*sin(3*0.01037*tsim) + 2.338e+004 *cos(4*0.01037*tsim) -2284*sin(4*0.01037*tsim) -1166*cos(5*0.01037*tsim) -1.008e+004*sin(5*0.01037*tsim) -3297 *cos(6*0.01037*tsim) + 421.1*sin(6*0.01037*tsim) + 96.45 *cos(7*0.01037*tsim) + 738.4*sin(7*0.01037*tsim) +6.41*cos(8*0.01037*tsim) -10.68*sin(8*0.01037*tsim)))+ x(2)*0.688*(( 39.78*exp(-((tsim- 62.36)/ 51.23 ).^2) + 39.99*exp(-((tsim+13.07)/ 49.11).^2) + 6.574*exp(-((tsim-95.15)/ 7.534).^2) + 7.534*exp(-((tsim-131)/ 20.23).^2) + 3.91e+013*exp(-((tsim- 1.36e+005)/ 2.603e+004).^2) + 15.31*exp(-((tsim-162.5)/49.37).^2) + 11.18*exp(-((tsim- 11.18)/ 41.93).^2) + 5.944*exp(-((tsim- 275.8)/ 15.14 ).^2)-0.29*( 4.629e+004*cos(0.01037*tsim) -8.526e+004*sin(0.01037*tsim ) -6.644e+004*cos(2*0.01037*tsim) + 3426 *sin(2*0.01037*tsim) + 3290*cos(3*0.01037*tsim) + 4.341e+004*sin(3*0.01037*tsim) + 2.338e+004 *cos(4*0.01037*tsim) -2284*sin(4*0.01037*tsim) -1166*cos(5*0.01037*tsim) -1.008e+004*sin(5*0.01037*tsim) -3297 *cos(6*0.01037*tsim) + 421.1*sin(6*0.01037*tsim) + 96.45 *cos(7*0.01037*tsim) + 738.4*sin(7*0.01037*tsim) +6.41*cos(8*0.01037*tsim) -10.68*sin(8*0.01037*tsim)))).^2+ 0.87*( 4.629e+004*cos(0.01037*tsim) -8.526e+004*sin(0.01037*tsim ) -6.644e+004*cos(2*0.01037*tsim) + 3426 *sin(2*0.01037*tsim) + 3290*cos(3*0.01037*tsim) + 4.341e+004*sin(3*0.01037*tsim) + 2.338e+004 *cos(4*0.01037*tsim) -2284*sin(4*0.01037*tsim) -1166*cos(5*0.01037*tsim) -1.008e+004*sin(5*0.01037*tsim) -3297 *cos(6*0.01037*tsim) + 421.1*sin(6*0.01037*tsim) + 96.45 *cos(7*0.01037*tsim) + 738.4*sin(7*0.01037*tsim) +6.41*cos(8*0.01037*tsim) -10.68*sin(8*0.01037*tsim)).^2)));
end
Alexandra Harkai
2016-11-2
Why should the optimum be at a different Xop(2) value? (Didn't decode the long function but it seems to be the an absolute value of a polynomial, so there could be bits where even though the range is wider, it will not move away from the local optimum.)
Can you plot the ident1 function for a range of [x y] values where x is fixed?
When the upper bound is 0.5, then the lower bound can't be 1, are you changing the problem config?
Alexandra Harkai
2016-11-2
编辑:Alexandra Harkai
2016-11-2
In addition to plotting parts of the [x y] surface, the 'iter-detailed' display carries lots of info why the second coordinate doesn't change. Using these optimalisation options it shows where the function argument moves during the iterations. With the 'PlotFcn' defined this would also be displayed:
opts = optimoptions(@fmincon, 'Display','iter-detailed', 'PlotFcn', @optimplotx); % this has to be defined first
Xop = fmincon(@ident1,[0.1;0.1],[],[],[],[],[1 1],[50 5], opts); % then calling the optimisation using those options
Anoire BEN JDIDIA
2016-11-2
not work. error msg: ??? Undefined function or method 'optimoptions' for input arguments of type 'function_handle'.
Error in ==> simultest2 at 18 opts = optimoptions('Display','iter-detailed', 'PlotFcn', @optimplotx); % this has to be defined first
Anoire BEN JDIDIA
2016-11-2
problem.options = optimset('Display','iter','Algorithm','sqp','PlotFcn',@optimplotx); xop=fmincon(@ident1,[0.1;0.1],[],[],[],[],[1 1],[100 100],[],problem.options); its work but i have bad result. xop =
2.6147
1.0000
if lwer value of X(2)=2 we will have xop(2)=2
Alexandra Harkai
2016-11-2
编辑:Alexandra Harkai
2016-11-2
What does the iterative display tell you? What reason does it give for finishing the optimisation? You can try to find the exit flag:
[xop, ~, exitflag] = fmincon(...)
How do these compare to when you use a different lower/upper boundary?
These give information for you about why the algorithm stopped and help to think about how you might want to change your optimisation according to What Can Be Wrong If The Solver Succeeds?
Anoire BEN JDIDIA
2016-11-2
if i try : problem.options = optimset('Display','iter-detailed','Algorithm','SQP','PlotFcn',@optimplotx);
[xop, ~, exitflag] = fmincon(@ident1,[3;3],[],[],[],[],[1 2.5],[100 100],[],problem.options); i found xop =
3.1273
2.5000
Alexandra Harkai
2016-11-2
I don't know all the parameters you are using for your optimisation, but regardless, the best advice I can give is to look at all the previous questions above and try to find out how your optimisation behaves. Check the iteration process, not just the final result. Check the linked MATLAB Documentation here too.
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