极小极大优化- Gekko

Question

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我试图建立这个极小极大优化函数。我不是这个代码的哪个部分写得不对。

这项功能的目的是安排电动汽车的充电开始时间，以尽量减少需求的高峰。另一方面，需求的峰值是方程的最大值，它决定了每个EV在哪个时隙中接收电荷。

def testGEKKO():
        # set of all EVs
        n = np.linspace(1,10,10)
        
        # set of contract time for 10 hours
        t = np.linspace(1,10,10)
        
        
        arival_time = [0,2,1,5,2,3,6,0,6,4]
        
        duration = 2
        
        I = []
        L = []
        
        for ev in range(len(n)):
            all_intervals = []
            start = arival_time[ev]
            while start < 10 - duration:
                all_intervals.append([start, start+duration])
                L.append(start)
                start += 1
                
            I.append(all_intervals)
            
        L = list(set(L))
        
        h = np.array([50, 50, 50, 50, 50, 50, 50, 50, 50, 50])
        
        
        mdl = GEKKO(remote = False)
        idx = []
        
        H = mdl.Var(1,lb=0,integer=True)
        mdl.Minimize(H)
        
        for ev in range(len(n)):
            idx.append(mdl.Array(mdl.Var, (len(I[ev])), lb = 0, ub = 1, integer = True))
            
        
        for ev in range(len(n)):
            mdl.Equation(np.sum(idx[ev]) == 1)
        for time in L:
            evs = []
            all_idx_start = []
            for ev in range(len(n)):
                st = np.array(I[ev])[:,0]
                if time in st:
                    evs.append(ev)
                    all_idx_start.append(list(np.array(I[ev])[:,0]).index(time))
                
            equ = 0
            for i in range(len(all_idx_start)):
                equ += idx[evs[i]][all_idx_start[i]] * h[evs[i]]
                
            equations.append(equ  <= z)
            mdl.Equation(equ  <= z)

        #mdl.Equations([z>=equations[0],z>=equations[1],z>=equations[2], z>=equations[3],z>=equations[4],z>=equations[5],z>=equations[6],z>=equations[7]])
        #mdl.Equations([z[0]>=equations[0]])

        mdl.Minimize(z)

        mdl.options.MAX_ITER = 3000
                

        #mdl.open_folder()             
        mdl.options.SOLVER = 1    
        mdl.options.IMODE = 3
        mdl.solve(disp=True)
        
        return idx

我已经尝试过Gekko的建议，但它不起作用。

Answer 1

gekko中的Minimax要求变量(例如x1-x3)都是Gekko类型。

 min Z
 s.t. x1 + x2 + x3 = 15
      Z >= x1
      Z >= x2
      Z >= x3

以下是解决方案为x1=x2=x3=5的上述问题的一个示例。

from gekko import GEKKO
m = GEKKO(remote=False)
x1,x2,x3,Z = m.Array(m.Var,4)
m.Minimize(Z)
m.Equation(x1+x2+x3==15)
m.Equations([Z>=x1,Z>=x2,Z>=x3])
m.solve()
print(x1.value[0],x2.value[0],x3.value[0],Z.value[0])

对于特定的应用程序，不需要最小化变量H。最小化z是目标函数中的唯一一件事。

#(removed) mdl.Minimize(H)

这是完整的代码。原始代码缺少z的定义。对于未来的问题，请张贴一个完整和最小的问题的例子。

from gekko import GEKKO
import numpy as np

# set of all EVs
n = np.linspace(1,10,10)

# set of contract time for 10 hours
t = np.linspace(1,10,10)


arival_time = [0,2,1,5,2,3,6,0,6,4]

duration = 2

I = []
L = []

for ev in range(len(n)):
    all_intervals = []
    start = arival_time[ev]
    while start < 10 - duration:
        all_intervals.append([start, start+duration])
        L.append(start)
        start += 1
        
    I.append(all_intervals)
    
L = list(set(L))
h = np.array([50, 50, 50, 50, 50, 50, 50, 50, 50, 50])

mdl = GEKKO(remote = False)
idx = []
z = mdl.Var()
H = mdl.Var(1,lb=0,integer=True)
#mdl.Minimize(H)

for ev in range(len(n)):
    idx.append(mdl.Array(mdl.Var, (len(I[ev])), lb = 0, ub = 1, integer = True))
    
for ev in range(len(n)):
    mdl.Equation(np.sum(idx[ev]) == 1)
for time in L:
    evs = []
    all_idx_start = []
    for ev in range(len(n)):
        st = np.array(I[ev])[:,0]
        if time in st:
            evs.append(ev)
            all_idx_start.append(list(np.array(I[ev])[:,0]).index(time))
        
    equ = 0
    for i in range(len(all_idx_start)):
        equ += idx[evs[i]][all_idx_start[i]] * h[evs[i]]
        
    print(equ)
    mdl.Equation(equ  <= z)

mdl.Minimize(z)

mdl.options.MAX_ITER = 3000
        
#mdl.open_folder()             
mdl.options.SOLVER = 1    
mdl.options.IMODE = 3
mdl.solve(disp=True)

如果这仍然没有生成正确的答案，您可能需要增加MINLP迭代次数，然后它才能报告成功的解决方案，或者将间隙容忍降低到接近于零。

m.solver_options = ['minlp_gap_tol 1.0e-2',\
                    'minlp_maximum_iterations 10000',\
                    'minlp_max_iter_with_int_sol 1500']
m.options.solver = 1

另外，查看gk0_model.apm文件，该文件位于mdl.path (运行目录)中。您可以使用mdl.open_folder()打开run目录。这显示了脚本生成的模型方程。