my equation has 3 variables but i want to integrate with respect to only one variable ,so that i can optimize the ramaining two from the resulting equation

조회 수: 2(최근 30일)
bhanu kiran vandrangi
bhanu kiran vandrangi 2021년 7월 27일
댓글: bhanu kiran vandrangi 2021년 7월 27일
my equation(looks big) is as follows , which has three variables (d,e,t) I want to integrate with t from 0 to 2000 (by keeping d,e constant ) so that i can optimize those two using any optimization method . i used int(err,0,2000) command but gettting no result .so how to get past this error?
the eqaution at its basic term is in form 1483/(k-1) , where K=function(d,e,t)
err =
1483/(200*(0.99966444607127868948737159371376*d - 99966.44460712373256683349609375*d*(0.0000000061953699799058132798891540549135*exp(-0.031830333683444479980079178105257*t) + 0.000010273947474151098879779908656928*exp(-9.1132612523704400742108333588476*t) - 0.00000028014284413259036773946597520535*exp(-0.21190841394611541767881157660725*t)) + 15567.003559999167919158935546875*e*(0.0000000061953699799058132798891540549135*exp(-0.031830333683444479980079178105257*t) + 0.000010273947474151098879779908656928*exp(-9.1132612523704400742108333588476*t) - 0.00000028014284413259036773946597520535*exp(-0.21190841394611541767881157660725*t)) + 919819.0186288356781005859375*d*(0.00000019463729288918805472585749072323*exp(-0.031830333683444479980079178105257*t) + 0.0000011273623338164517199144754044937*exp(-9.1132612523704400742108333588476*t) - 0.0000013219996267057210898032693080495*exp(-0.21190841394611541767881157660725*t)) - 143208.7303999960422515869140625*e*(0.00000019463729288918805472585749072323*exp(-0.031830333683444479980079178105257*t) + 0.0000011273623338164517199144754044937*exp(-9.1132612523704400742108333588476*t) - 0.0000013219996267057210898032693080495*exp(-0.21190841394611541767881157660725*t)) - 79516.5081846714019775390625*d*(0.0000061148367096900205219789370403305*exp(-0.031830333683444479980079178105257*t) + 0.00000012370569685174297169805157636802*exp(-9.1132612523704400742108333588476*t) - 0.0000062385424065411129723734973140381*exp(-0.21190841394611541767881157660725*t)) + 6144.9373500002548098564147949219*e*(0.0000061148367096900205219789370403305*exp(-0.031830333683444479980079178105257*t) + 0.00000012370569685174297169805157636802*exp(-9.1132612523704400742108333588476*t) - 0.0000062385424065411129723734973140381*exp(-0.21190841394611541767881157660725*t)) - 1))

채택된 답변

Alan Weiss
Alan Weiss 2021년 7월 27일
This seems like a solution. Of course, I don't know what your real bounds are on d and e, so I just guessed.
Notice that I used ./ in the first line of the function myfun. That is because the integral function requires vectorized inputs.
lb = [1/2 1/2];
ub = [3/4 3/4];
[sol,fval,eflag,output] = fmincon(@minfn,[1/2 1/2],[],[],[],[],lb,ub)
Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.
sol = 1×2
0.7500 0.7500
fval = -5.8613e+04
eflag = 1
output = struct with fields:
iterations: 4 funcCount: 15 constrviolation: 0 stepsize: 2.2605e-04 algorithm: 'interior-point' firstorderopt: 0.0757 cgiterations: 0 message: '↵Local minimum found that satisfies the constraints.↵↵Optimization completed because the objective function is non-decreasing in ↵feasible directions, to within the value of the optimality tolerance,↵and constraints are satisfied to within the value of the constraint tolerance.↵↵<stopping criteria details>↵↵Optimization completed: The relative first-order optimality measure, 3.264963e-07,↵is less than options.OptimalityTolerance = 1.000000e-06, and the relative maximum constraint↵violation, 0.000000e+00, is less than options.ConstraintTolerance = 1.000000e-06.↵↵' bestfeasible: [1×1 struct]
function minit = minfn(x)
d = x(1);
e = x(2);
minit = intval(d,e);
end
function r = intval(d,e)
r = integral(@(t)myfun(d,e,t),0,2000);
end
function val = myfun(d,e,t)
val = 1483./(200*(0.99966444607127868948737159371376*d - ...
99966.44460712373256683349609375*d*(0.0000000061953699799058132798891540549135*exp(-0.031830333683444479980079178105257*t) +...
0.000010273947474151098879779908656928*exp(-9.1132612523704400742108333588476*t) -...
0.00000028014284413259036773946597520535*exp(-0.21190841394611541767881157660725*t)) + ...
15567.003559999167919158935546875*e*(0.0000000061953699799058132798891540549135*exp(-0.031830333683444479980079178105257*t) +...
0.000010273947474151098879779908656928*exp(-9.1132612523704400742108333588476*t) -...
0.00000028014284413259036773946597520535*exp(-0.21190841394611541767881157660725*t)) +...
919819.0186288356781005859375*d*(0.00000019463729288918805472585749072323*exp(-0.031830333683444479980079178105257*t) +...
0.0000011273623338164517199144754044937*exp(-9.1132612523704400742108333588476*t) -...
0.0000013219996267057210898032693080495*exp(-0.21190841394611541767881157660725*t)) -...
143208.7303999960422515869140625*e*(0.00000019463729288918805472585749072323*exp(-0.031830333683444479980079178105257*t) +...
0.0000011273623338164517199144754044937*exp(-9.1132612523704400742108333588476*t) -...
0.0000013219996267057210898032693080495*exp(-0.21190841394611541767881157660725*t)) -...
79516.5081846714019775390625*d*(0.0000061148367096900205219789370403305*exp(-0.031830333683444479980079178105257*t) +...
0.00000012370569685174297169805157636802*exp(-9.1132612523704400742108333588476*t) -...
0.0000062385424065411129723734973140381*exp(-0.21190841394611541767881157660725*t)) +...
6144.9373500002548098564147949219*e*(0.0000061148367096900205219789370403305*exp(-0.031830333683444479980079178105257*t) +...
0.00000012370569685174297169805157636802*exp(-9.1132612523704400742108333588476*t) -...
0.0000062385424065411129723734973140381*exp(-0.21190841394611541767881157660725*t)) - 1));
end
Alan Weiss
MATLAB mathematical toolbox documentation
  댓글 수: 1
bhanu kiran vandrangi
bhanu kiran vandrangi 2021년 7월 27일
thank you for the solution
i want the equation after integration so that i can use that as a cost function in particleswarm command(PSO) with 0,1 as bounds for bothd,e

댓글을 달려면 로그인하십시오.

추가 답변(0개)

제품


릴리스

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by