我正在尝试解决上述优化问题。 我的代码如下。
它有效，但我遇到了负自由度问题。
而且客观值也是负数，这是我没想到的。我期待着积极的一面。
我不明白为什么会发生这种情况，也不知道如何解决这个问题。
有人可以给我一个建议吗？
代码
# Import package
from gekko import GEKKO
import numpy as np
# Define parameters
P_CO = 600 # $/tonCO
beta_CO2 = 1 # no unit
P_CO2 = 60 # $/tonCO2eq
E_ref = 3.1022616 # tonCO2eq/tonCO
E_dir = -1.600570692 # tonCO2eq/tonCO
E_indir_others = 0.3339226804 # tonCO2eq/tonCO
E_indir_elec_cons = 18.46607256 # GJ/tonCO
C1_CAPEX = 285695 # no unit
C2_CAPEX = 188.42 # no unit
C1_FOX = 82282 # no unit
C2_FOX = 24.094 # no unit
C1_ROX = 4471.5 # no unit
C2_ROX = 96.034 # no unit
C1_UOX = 1983.7 # no unit
C2_UOX = 249.79 # no unit
r = 0.08 # discount rate
N = 10 # number of scenarios
T = 30 # total time period
GWP_init = 0.338723235 # 2020 Electricity GWP in EU 27 countries
theta_max = 1600000 # Max capacity
# Function to make GWP_EU matrix (TxN matrix)
def Electricity_GWP(GWP_init, n_years, num_episodes):
GWP_mean = 0.36258224*np.exp(-0.16395611*np.arange(1, n_years+2)) + 0.03091272
GWP_mean = GWP_mean.reshape(-1,1)
GWP_Yearly = np.tile(GWP_mean, num_episodes)
noise = np.zeros((n_years+1, num_episodes))
stdev2050 = GWP_mean[-1] * 0.25
stdev = np.arange(0, stdev2050 * (1 + 1/n_years), stdev2050/n_years)
for i in range(n_years+1):
noise[i,:] = np.random.normal(0, stdev[i], num_episodes)
GWP_forecast = GWP_Yearly + noise
return GWP_forecast
GWP_EU = Electricity_GWP(GWP_init, T, N) # (T+1)*N matrix
GWP_EU = GWP_EU[1:,:] # T*N matrix
print(np.shape(GWP_EU))
# Build Gekko model
m = GEKKO(remote=False)
theta = m.Array(m.Var, N, lb=0, ub=theta_max)
demand = np.ones((T,1))
demand[0] = 8031887.589
for k in range(1,11):
demand[k] = demand[k-1] * 1.026
for k in range(11,21):
demand[k] = demand[k-1] * 1.016
for k in range(21,T):
demand[k] = demand[k-1] * 1.011
demand = 0.12 * demand
demand = np.tile(demand, N) # T*N matrix
print(np.shape(demand))
obj = m.sum([m.sum([((1/(1+r))**(t+1))*((P_CO*m.min3(demand[t,s], theta[s])) \
+ (beta_CO2*P_CO2*m.min3(demand[t,s], theta[s])*(E_ref-E_dir-E_indir_others-E_indir_elec_cons*GWP_EU[t,s])) \
- (C1_CAPEX+C2_CAPEX*theta[s]+C1_FOX+C2_FOX*theta[s])-(C1_ROX+C2_ROX*m.min3(demand[t,s], theta[s])+C1_UOX+C2_UOX*m.min3(demand[t,s], theta[s]))) for t in range(T)]) for s in range(N)])
m.Maximize(obj/N)
m.solve()
输出消息
(30, 10)
(30, 10)
----------------------------------------------------------------
APMonitor, Version 1.0.0
APMonitor Optimization Suite
----------------------------------------------------------------
--------- APM Model Size ------------
Each time step contains
Objects : 11
Constants : 0
Variables : 5121
Intermediates: 0
Connections : 321
Equations : 3901
Residuals : 3901
Number of state variables: 5121
Number of total equations: - 3911
Number of slack variables: - 2400
---------------------------------------
Degrees of freedom : -1190
* Warning: DOF <= 0
----------------------------------------------
Steady State Optimization with APOPT Solver
----------------------------------------------
Iter: 1 I: 0 Tm: 18.61 NLPi: 5 Dpth: 0 Lvs: 0 Obj: -1.87E+09 Gap: 0.00E+00
Successful solution
---------------------------------------------------
Solver : APOPT (v1.0)
Solution time : 18.619200000000003 sec
Objective : -1.8677021320161405E+9
Successful solution
---------------------------------------------------
最佳答案
负自由度警告是由于使用 min3() 函数时创建的松弛变量造成的。这只是一个警告，如果所有不等式都有效，那么这可能会导致方程组过度指定(方程多于变量)。如果有成功的解决方案，则可以忽略此警告。
负面目标是因为大多数求解器都需要目标的最小化。 Gekko 自动将 m.Maximize(obj) 转换为 m.Minimize(-obj)。这是一个同等的目标。如果您想报告最大化和积极目标，请在末尾使用以下内容:
print('Objective: ',-m.options.OBJFCNVAL)
关于python-3.x - 使用 GEKKO python 时的负自由度，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/74162557/