英文:
GEKKO (Python) giving incorrect solution
问题
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为什么
results.json文件与提示中显示的内容不匹配？这个问题可能是由于GEKKO求解器的设置或代码中的某些问题导致的。要解决这个问题，你可以检查以下几个方面：
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检查GEKKO模型的设置，包括求解器、参数和约束条件是否正确设置。确保模型的目标函数和约束条件与你的预期一致。
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检查代码中的随机数生成部分，确保随机数生成的方式和范围是正确的。
np.random.randint和np.random.random的使用应该是符合你的需求的。 -
检查结果文件
results.json是否包含了正确的输出数据。如果结果文件不正确，可能是代码中输出结果的部分存在问题。
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为什么我得到了错误的解决方案？
得到错误的解决方案可能是由于以下原因之一引起的：
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模型的约束条件可能不正确或不完整，导致求解器找到了一个满足约束条件的局部最优解，但这个解并不是全局最优解。你应该仔细检查模型的约束条件，确保它们涵盖了问题的所有限制。
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求解器选择和参数设置可能会影响结果。你可以尝试不同的求解器、不同的求解器选项或参数来寻找更好的解决方案。
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模型的目标函数可能不正确。确保你的目标函数正确地描述了问题的优化目标。
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代码中的数据生成部分可能存在问题，导致输入数据不符合预期。
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要解决这个问题，你可以逐步调试代码，检查模型设置、约束条件和目标函数是否正确，以及检查输入数据是否合理。还可以尝试不同的求解器和参数设置来查找更好的解决方案。
英文:
Consider the following GEKKO code:
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import numpy as npfrom gekko import GEKKOimport matplotlib.pyplot as plt# Build model#initialize GEKKO modelm = GEKKO()m.options.solver = 1# Seednp.random.seed(0)max_r = 5min_r = 2b = 2b_r = {}for i in range(b):b_r[i] = np.random.randint(low=min_r,high=max_r)u = 2br = np.random.random((b,u))lv = m.Array(m.Var,(b, u),lb=0,ub=1, integer=True)av = {}for i in range(b):av[i] = m.Array(m.Var,(b_r[i], u),lb=0,ub=1, integer=True)## Constraintsfor i in range(u):lv_sum = m.Const(value=0, name='Link Sum ' + str(i))for j in range(b):lv_sum += lv[j][i]m.Equation(lv_sum <= 1)for i in range(b):for j in range(b_r[i]):av_sum = m.Const(value=0, name='Alloc Sum ' + str((i,j)))for k in range(u):av_sum += av[i][j][k]m.Equation(av_sum <= 1)# Objective functionobj_u = m.Const(value=0, name='Final Objective')for i in range(u):obj_b = m.Const(value=0, name='Objective '+str(i))for j in range(b):for k in range(b_r[j]):obj_b += lv[j][i]*av[j][k][i]*br[j][i]obj_u += m.log(1+obj_b)# Maximize/Minimize objectivem.Maximize(obj_u)#Set global optionsm.options.IMODE = 3 #steady state optimization# m.options.IMODE = 6 #Dynamic optimizationprint(m.path)#Solve simulationm.solve()
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Quite obviously the solution to this piece of code will always be non-negative as we are maximizing log(1+x) where 0 <= x. Yet somehow GEKKO gives the following output:
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/tmp/tmpw15jz3zhgk_model0apm 203.192.204.172_gk_model0 <br><pre> ----------------------------------------------------------------APMonitor, Version 1.0.1APMonitor Optimization Suite----------------------------------------------------------------Warning: there is insufficient data in CSV file 203.192.204.172_gk_model0.csv--------- APM Model Size ------------Each time step containsObjects : 0Constants : 10Variables : 21Intermediates: 0Connections : 0Equations : 8Residuals : 8Number of state variables: 21Number of total equations: - 7Number of slack variables: - 7---------------------------------------Degrees of freedom : 7----------------------------------------------Steady State Optimization with APOPT Solver----------------------------------------------Iter: 1 I: 0 Tm: 0.00 NLPi: 7 Dpth: 0 Lvs: 0 Obj: -2.26E+00 Gap: 0.00E+00Successful solution---------------------------------------------------Solver : APOPT (v1.0)Solution time : 1.460000000952277E-002 secObjective : -2.26374259105544Successful solution---------------------------------------------------
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The model file is as follows:
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ModelConstantslink_sum_0 = 0link_sum_1 = 0alloc_sum_0_0_ = 0alloc_sum_0_1_ = 0alloc_sum_1_0_ = 0alloc_sum_1_1_ = 0alloc_sum_1_2_ = 0final_objective = 0objective_0 = 0objective_1 = 0End ConstantsVariablesint_v1 = 0, <= 1, >= 0int_v2 = 0, <= 1, >= 0int_v3 = 0, <= 1, >= 0int_v4 = 0, <= 1, >= 0int_v5 = 0, <= 1, >= 0int_v6 = 0, <= 1, >= 0int_v7 = 0, <= 1, >= 0int_v8 = 0, <= 1, >= 0int_v9 = 0, <= 1, >= 0int_v10 = 0, <= 1, >= 0int_v11 = 0, <= 1, >= 0int_v12 = 0, <= 1, >= 0int_v13 = 0, <= 1, >= 0int_v14 = 0, <= 1, >= 0End VariablesEquations((link_sum_0+int_v1)+int_v3)<=1((link_sum_1+int_v2)+int_v4)<=1((alloc_sum_0_0_+int_v5)+int_v6)<=1((alloc_sum_0_1_+int_v7)+int_v8)<=1((alloc_sum_1_0_+int_v9)+int_v10)<=1((alloc_sum_1_1_+int_v11)+int_v12)<=1((alloc_sum_1_2_+int_v13)+int_v14)<=1maximize ((final_objective+log((1+(((((objective_0+((((int_v1)*(int_v5)))*(0.8442657485810173)))+((((int_v1)*(int_v7)))*(0.8442657485810173)))+((((int_v3)*(int_v9)))*(0.8472517387841254)))+((((int_v3)*(int_v11)))*(0.8472517387841254)))+((((int_v3)*(int_v13)))*(0.8472517387841254))))))+log((1+(((((objective_1+((((int_v2)*(int_v6)))*(0.8579456176227568)))+((((int_v2)*(int_v8)))*(0.8579456176227568)))+((((int_v4)*(int_v10)))*(0.6235636967859723)))+((((int_v4)*(int_v12)))*(0.6235636967859723)))+((((int_v4)*(int_v14)))*(0.6235636967859723))))))End EquationsEnd Model
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And the results.json file is as follows:
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{"time" : [0.00],"link_sum_0" : [ 0.00 ],"link_sum_1" : [ 0.00 ],"alloc_sum_0_0_" : [ 0.00 ],"alloc_sum_0_1_" : [ 0.00 ],"alloc_sum_1_0_" : [ 0.00 ],"alloc_sum_1_1_" : [ 0.00 ],"alloc_sum_1_2_" : [ 0.00 ],"final_objective" : [ 0.00 ],"objective_0" : [ 0.00 ],"objective_1" : [ 0.00 ],"int_v1" : [ 0.00 ],"int_v2" : [ 1.0000000000E+00],"int_v3" : [ 1.0000000000E+00],"int_v4" : [ 0.00 ],"int_v5" : [ 0.00 ],"int_v6" : [ 1.0000000000E+00],"int_v7" : [ 0.00 ],"int_v8" : [ 1.0000000000E+00],"int_v9" : [ 1.0000000000E+00],"int_v10" : [ 0.00 ],"int_v11" : [ 1.0000000000E+00],"int_v12" : [ 0.00 ],"int_v13" : [ 1.0000000000E+00],"int_v14" : [ 0.00 ],"slk_1" : [ 0.00 ],"slk_2" : [ 0.00 ],"slk_3" : [ 0.00 ],"slk_4" : [ 0.00 ],"slk_5" : [ 0.00 ],"slk_6" : [ 0.00 ],"slk_7" : [ 0.00 ]}
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I am confused of the folllowing:
- Why does the
results.jsonnot match with what is being displayed in the prompt? - More importantly why am I getting an incorrect solution?
答案1
得分: 1
m.Const() 是一个常数声明，而不是一个约束。然后，像以下的循环会重新定义这个常数：
for i in range(u):lv_sum = m.Const(value=0, name='Link Sum ' + str(i))for j in range(b):lv_sum += lv[j][i]m.Equation(lv_sum <= 1)
相反，可以这样定义 lv_sum 和约束：
for i in range(u):lv_sum = m.Var(value=0, ub=1, name='Link Sum ' + str(i))m.Equation(lv_sum == sum(lv[1:b][i]))
使用 m.Var() 定义变量，使用 m.Equation() 定义方程，或者使用 m.Equations() 一次性定义多个方程。不要使用 +=，因为它会重新定义 gekko 变量。不要使用循环，可以尝试使用列表推导、矩阵操作和向量求和。这里有示例（参见 test_arrays.py、test_matrix.py、test_summations.py）。一旦修复了问题定义，它应该会给你一个正确的问题陈述，然后求解器可以解决。
英文:
The m.Const() is a constant declaration, not a constraint. Loops such as the following then redefine the constant:
for i in range(u):lv_sum = m.Const(value=0, name='Link Sum ' + str(i))for j in range(b):lv_sum += lv[j][i]m.Equation(lv_sum <= 1)
Instead, define lv_sum and the constraint this way:
for i in range(u):lv_sum = m.Var(value=0, ub=1, name='Link Sum ' + str(i))m.Equation(lv_sum = sum(lv[1:b][i]))
Use m.Var() to define the variable and m.Equation() to define the equation or m.Equations() to define multiple equations at once. Don't use += because it redefines the gekko variable. Instead of loops, try list comprehensions, matrix operations, and vector summations. Here are examples (see test_arrays.py, test_matrix.py, test_summations.py). Once you fix the problem definition, it should give you a correct problem statement that the solver can then solve.
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