I need help for solving quadratic programming, on how to avoid solution(x) of the objective function to be zero ?

x0=10*[1;1;1;1;1;1;1;1;1;1;1;1;1]; lb=-10^8*ones(13,1); ub=100000000*ones(13,1); f=zeros(1,13); Y=eye(13,13); H=Y*2; A1=[-0.048042 -0.049226 -0.233376 -0.205312 -0.036378 -0.060762 -0.0812 -0.138498 -0.14592 -0.061104 -0.02225 -0.006138 -0.12585; 0.048048 0.049296 0.24528 0.202608 0.034882 0.066138 0.09165 0.169776 0.167552 0.06072 0.022826 0.045762 0.0252;-0.022896 -0.288798 0.20091 0.159904 -0.050274 -0.033066 0.008216 0.06513 0.075168 -0.030576 -0.079928 -0.021504 0.361248; 0.014448 0.26 -0.018262 -0.010864 -0.0372 -0.16687 -0.485306 -0.027456 -0.017936 -0.0588 -0.052098 -0.04221 0.030528;0.003842 0.01435 -0.043898 -0.01955 -0.09216 0.172966 0.488186 0.02784 0.01817 -0.00528 -0.091014 -0.058086 -0.005776; 0.004576 0.014378 -0.149152 -0.13632 0.18048 0.02541 -0.01457 -0.123 -0.124352 0.088506 0.22287 0.130088 -0.238502;-0.037506 -0.078336 -0.46295 -0.47376 -0.13867 -0.164328 -0.200046 -0.28741 -0.305592 -0.172666 -0.113046 -0.180072 -0.430122;-0.118144 -0.073944 0.714432 0.68579 0.028942 0.086592 0.186648 0.40369 0.424974 0.09792 -0.039064 0.097384 0.670704; 0.154882 0.138574 -0.488394 -0.374976 0.081168 0.01536 -0.072922 -0.224472 -0.231054 0.02584 0.135424 0.03675 -0.400582;0.00075 0.00917 0.10229 0.00889 0.00681 0.030336 0.03753 0.063 0.05787 0.01953 0.005678 0.013376 0.067158;1.8E-05 0.004446 0.12555 0.1568 0.018944 0.02772 0.069192 0.1344 0.13904 0.040086 -0.001232 0.03952 0.212842;-0.00624 -0.26775 -0.271478 -0.198522 -0.0368 -0.074226 -0.125 -0.2254 -0.22523 -0.073486 -0.01351 -0.068208 -0.318682; -0.03813 -0.226106 0.543368 0.41245 0.02519 0.06693 -0.068442 0.245752 0.249504 0.053728 -0.002808 0.03465 0.440794;-0.029326 0.3176 0.0184 0.015198 -0.581094 -0.197064 0.07241 -0.047104 -0.063714 -0.3498 -0.455328 -0.423168 -0.035754; 0.073728 0.176418 -0.269346 -0.219648 0.59607 0.209814 0.102982 -0.009386 -0.033136 0.361862 0.482976 0.453606 -0.14863;-2E-06 -8E-06 -0.00739 0.002672 -0.000638 -0.004224 -0.0084 -0.02511 -0.00095 -0.015824 -0.003848 0.002562 -0.011178;-0.00024 -0.00064 -0.03213 -0.01352 0.000728 -0.002838 -0.0048 -0.007104 0.008406 0.017264 0.003648 -0.00325 -0.01935;0.000242 0.000646 0.035098 0.012186 -0.000128 0.004598 0.008778 0.01728 0.009834 0.00616 -0.00201 0.002482 0.0256; -1.8E-09 1.99181E-06 0.002820976 -4.67625E-05 0.000199784 0.001267978 0.00071502 0.007641212 0.01315048 -0.002705872 0.000590152 -0.001887964 -0.001825876;-0.000721998 -0.002887992 -0.080142976 -0.040379238 -0.129031784 -0.024387978 -0.03741102 -0.080691212 -0.11655248 -0.093756128 -0.093150152 -0.140192036 -0.234712124;0.000718339 0.002864398 0.087019892 0.043901594 0.1285613 0.026382439 0.043309201 0.094963676 0.13103198 0.095754939 0.090999358 0.142984658 0.261219522;-0.000446339 -0.006938398 -0.015649892 -0.010205594 0.0003987 -0.002668439 -0.005773201 -0.016243676 -0.01745398 -0.002826939 0.003354642 -0.002378658 -0.027169522];

A=-A1; b=zeros(22,1); [x,fval,exitflag,output,lambda]=quadprog(H,f,A,b,[],[],lb,ub,x0) % [x,fval,exitflag,output]=quadprog(H,f,A,b,[],[],[],[],x0)

I have tried varying ub lb and x0, but sometimes it gives me vector of zero as solution, which i dont wish, kindly tell me as to how to avoid this, by making any changes in objective function or something? Thank you, your reply will be appreciated.

Answer 1

[Torsten]

Use b=-ones(22,1),e.g.

Best wishes

Torsten.

[Megha dwivedi]

hey Torsten, thank you for the reply!! Actually, the inequality equation is Ax<=b , and in my constraints, b =0.

[Torsten]

If all the settings from above are fixed, then x=0 is the solution for your optimization problem.

So if you wish or not - quadprog will always return the zero solution.

Best wishes

Torsten.