Optimization problem with lower and upper bounded constraints

5 次查看（过去 30 天）
Anthony Sirico
Anthony Sirico 2021-12-1
评论： Walter Roberson 2021-12-1
maximize:x0.063x4x7 −5.04x1 −0.035x2 −10x3 −3.36x5
subject to: x5 = 1.22x4 −x1
x6 = (98000 x3x4)/(x9 + 1000x3)
x8 = (x2 + x5)/x1
(99/100)x4 ≤x1(1.12 + 0.13167x8 −0.00667x28) ≤(100/99)x4
(99/100)x7 ≤86.35 + 1.098x8 −0.038x28 + 0.325(x6 −89) ≤(100/99)x7
(9/10)x9 ≤35.82 −0.222x10 ≤(10/9)x9
(99/100)x10 ≤−133 + 3x7 ≤(100/99)x10
[0,0,0,0,0,85,90,3,0.01,145] ≤x
x≤[2000,16000,120,5000,2000,93,95,12,4,162]
How do I represent constraints 4 through 6? This is what i have so far:
clc; clear
x = optimvar('x',10,'UpperBound',[2000;16000;120;5000;2000;93;95;12;4;162],"LowerBound",[0;0;0;0;0;85;90;3;0.01;145]);
prob = optimproblem("Objective",0.063*x(4)*x(7)-5.04*x(1)-0.035*x(2)-10*x(3)-3.36*x(5),"ObjectiveSense","maximize")
prob.Constraints.c1 = x(5) == 1.22*x(4)-x(1);
prob.Constraints.c2 = x(6) == 9800* x(3)/(x(4)*x(9)+1000*x(3));
prob.Constraints.c3 = x(8) == (x(2)+x(5))/x(1);
show(prob)

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2021-12-1
Ran in:
clc; clear
x = optimvar('x',10,'UpperBound',[2000;16000;120;5000;2000;93;95;12;4;162],"LowerBound",[0;0;0;0;0;85;90;3;0.01;145]);
prob = optimproblem("Objective",0.063*x(4)*x(7)-5.04*x(1)-0.035*x(2)-10*x(3)-3.36*x(5),"ObjectiveSense","maximize")
prob =
OptimizationProblem with properties: Description: '' ObjectiveSense: 'maximize' Variables: [1×1 struct] containing 1 OptimizationVariable Objective: [1×1 OptimizationExpression] Constraints: [0×0 struct] containing 0 OptimizationConstraints See problem formulation with show.
prob.Constraints.c1 = x(5) == 1.22*x(4)-x(1);
prob.Constraints.c2 = x(6) == 9800* x(3)/(x(4)*x(9)+1000*x(3));
prob.Constraints.c3 = x(8) == (x(2)+x(5))/x(1);
prob.Constraints.c4 = (99/100)*x(4) <= x(1)*(1.12 + 0.1367*x(8) - 0.00667*x(2)*x(8));
prob.Constraints.c5 = x(1)*(1.12 + 0.1367*x(8) - 0.00667*x(2)*x(8)) <= (100/99)*x(4);
prob.Constraints.c6 = (99/100)*x(7) <= 86.35 + 1.098*x(8) - 0.038*x(2)*x(8) + 0.325*(x(6)-89);
prob.Constraints.c7 = 86.35 + 1.098*x(8) - 0.038*x(2)*x(8) + 0.325*(x(6)-89) <= (100/99)*x(7);
show(prob)
OptimizationProblem : Solve for: x maximize : 0.063*x(4)*x(7) - 5.04*x(1) - 0.035*x(2) - 10*x(3) - 3.36*x(5) subject to c1: x(1) - 1.22*x(4) + x(5) == 0 subject to c2: x(6) == ((9800 .* x(3)) ./ ((x(4) .* x(9)) + (1000 .* x(3)))) subject to c3: x(8) == ((x(2) + x(5)) ./ x(1)) subject to c4: (0.99 .* x(4)) <= (x(1) .* ((1.12 + (0.1367 .* x(8))) - ((0.00667 .* x(2)) .* x(8)))) subject to c5: (x(1) .* ((1.12 + (0.1367 .* x(8))) - ((0.00667 .* x(2)) .* x(8)))) <= (1.0101 .* x(4)) subject to c6: (0.99 .* x(7)) <= (((86.35 + (1.098 .* x(8))) - ((0.038 .* x(2)) .* x(8))) + (0.325 .* (x(6) - 89))) subject to c7: (((86.35 + (1.098 .* x(8))) - ((0.038 .* x(2)) .* x(8))) + (0.325 .* (x(6) - 89))) <= (1.0101 .* x(7)) variable bounds: 0 <= x(1) <= 2000 0 <= x(2) <= 16000 0 <= x(3) <= 120 0 <= x(4) <= 5000 0 <= x(5) <= 2000 85 <= x(6) <= 93 90 <= x(7) <= 95 3 <= x(8) <= 12 0.01 <= x(9) <= 4 145 <= x(10) <= 162
and so on.
I had to guess about what x28 was; I coded it as x(2)*x(8)
  2 个评论
Anthony Sirico
Anthony Sirico 2021-12-1
Sorry its x^2...so did you break each of them up into 2 separate constraints?
Walter Roberson
Walter Roberson 2021-12-1
Yes, split the range inequality into two inequalities.
The x28 cannot be x^2 because x is a vector. Perhaps it is a clumsy x(8)^2

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Get Started with Optimization Toolbox 的更多信息

标签

产品


版本

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by