是否有用于非线性和非标量目标函数的 Python 优化包

Question

I am looking to solve problems with the following characteristics:

max {x * max (f(var_1),f(var_2)) + y * min (f(var_3),f(var_4)) + ...}
s.t.
b_0 <= x <= b_1
c_0 <= y <= c_1
d_0 <= var_1 <= d_1
d_2 <= var_2 <= d_3
d_4 <= var_3 <= d_5
d_6 <= var_4 <= d_7

etc...

using python.

Thanks

Answer 1

I can solve this kind of problems using CasADi software, please follow this instruction ( https://web.casadi.org/get/ ) to install the package into your computer

Here, I provide you to how to solve your problem, you may change the functions (f1,..f4) and values as you want.

    from casadi import *

    x = SX.sym('x')
    y = SX.sym('y')
    var_1 = SX.sym('var_1')
    var_2 = SX.sym('var_2')
    var_3 = SX.sym('var_3')
    var_4 = SX.sym('var_4')

    f1 = x + var_1 -y
    f2 = x + var_2 -y
    f3 = var_3*x
    f4 = var_4-y

    obj = fmax(f1, f2) + y*fmin(f3,f4)

    opt_val = vertcat(x, y, var_1, var_2, var_3, var_4)
    nlp_prob = {'f':obj, 'x':opt_val, 'p':[]}

    opts = {}
    opts["expand"] = True
    opts["ipopt.max_iter"] = 100
    opts["ipopt.tol"] = 1e-4
    opts["ipopt.print_level"] = 0
    opts["print_time"] = 1
    opts["ipopt.acceptable_tol"] = 1e-8

    solver = nlpsol('solver', 'ipopt', nlp_prob, opts)

    lbx = DM(opt_val.size1(),1)
    ubx = DM(opt_val.size1(),1)

    lbx[0] = -100 #b_0
    lbx[1] = -100 #c_0
    lbx[2] = -100 #d_0
    lbx[3] = -100 #d_2
    lbx[4] = -100 #d_4
    lbx[5] = -100 #d_6

    ubx[0] = 500 #b_1
    ubx[1] = 500 #c_1
    ubx[2] = 500 #d_1
    ubx[3] = 500 #d_3
    ubx[4] = 500 #d_5
    ubx[5] = 500 #d_7


    # initial values for x,y, val_1, val_2, val_3, val_4
    x0 = DM([[3],[6],[34],[3],[23],[25]])

    args = {'lbx':lbx, 'ubx':ubx, 'lbg':-inf, 'ubg':inf, 'p':[], 'x0':x0}
    sol = solver(**args)

    print("======Solution=====")
    sol = sol['x'].full().flatten()
    print("x: ", sol[0])
    print("y: ", sol[1])
    print("val_1: ", sol[2])
    print("val_2: ", sol[3])
    print("val_3: ", sol[4])
    print("val_4: ", sol[5])

Expected output

======Solution=====
('x: ', 3.0756501068719735)
('y: ', -99.999974650383308)
('val_1: ', -98.442843531628299)
('val_2: ', 108.68842844611186)
('val_3: ', 195.0797522765082)
('val_4: ', 499.99687090550719)

Let me know anything is not clear