Question

我正在努力使用Python Pyomo中的for循环进行约束。我的代码概念如下：

model.Q = Set()
model.sbase=Param(model.Q, within = PositiveReals)

我的代码可以在一个Sbase上正常运行。但是如果我得到3个不同数量的Sbase。我想做的就是使用这3个数字输出3个不同的图像。

set Q := 1 2 3;
param sbase:=
1 800
2 1000 
3 1200;

到目前为止，我所做的是：（真的不知道该怎么做。。）

from pyomo.environ import *
import matplotlib.pyplot as plt
import numpy as np
import matplotlib.pyplot as plt
import random

# create a model
model = AbstractModel()

# declare decision variables
model.Q = Set()
model.J = Set()
model.A = Param(model.J, within = PositiveReals)
model.B = Param(model.J, within = PositiveReals)
model.C = Param(model.J, within = PositiveReals)
model.P_min = Param(model.J, within = PositiveReals)
model.P_max= Param(model.J, within = PositiveReals)
model.P = Var(model.J, within = NonNegativeReals)
model.sbase=Param(model.Q, within = PositiveReals)

# declare objective
def obj_expression(model):
    return sum(model.A[j] * model.P[j]**2 + model.B[j] * model.P[j] + model.C[j] for j in model.J)
model.profit = Objective(rule = obj_expression, sense=minimize)

# declare constraints
def lower_bound(model,j):
    return model.P_min[j] <= model.P[j]
model.laborA = Constraint(model.J, rule = lower_bound)

def upper_bound(model, j):
    return model.P[j] <= model.P_max[j]
model.laborB = Constraint(model.J, rule = upper_bound)

for q in range(2):    #from this line ，really confused about this
    def sum_labor(model,i,q):???
        if q==1:??
            return sum(model.P[j] for j in model.J) >= (model.sbase[Q] for q in model.Q)??
        elif q==2:??
            reture sum(model.P[j] for j in model.J) > = (model.sbase[Q] for q in model.Q)
        
model.laborC = Constraint(model.sbase,rule = sum_labor)

instance = model.create_instance("E:\pycharm_project\PD\pd.dat")
opt = SolverFactory('Ipopt')
results = opt.solve(instance)



plt.figure(figsize=(8,6))
plt.subplot(121)
for i in instance.J:
    plt.bar(i,value(instance.P[i]))
plt.xlabel('Generators')
plt.ylabel('Power')
plt.title('Power distribution') 

plt.subplot(122)
x=[0.1]
plt.bar(x,value(instance.profit),alpha=0.7,width=0.015)
plt.xlim(0,0.2)
plt.xlabel('')
plt.ylabel('Cost')
plt.title('minimum cost') 
plt.show()
print('\n\n---------------------------')
print('Cost: ',value(instance.profit))

#DATA 
set Q := 1 2 3;
set J := g1 g2 g3 g4 g5 g6 g7 g8 g9 g10  ;

param : A B C P_min P_max:=
g1   0.0148 12.1  82  80  200
g2   0.0289 12.6  49  120 320
g3   0.0135 13.2  100 50  150
g4   0.0127 13.9  105 250 520
g5   0.0261 13.5  72  80   280
g6   0.0212 15.4  29  50   150
g7   0.0382 14.0  32  30   120
g8   0.0393 13.5  40  30   110
g9   0.0396 15.0  25  20   80
g10  0.0510 14.3  15  20   60
;
param sbase:=
1 800
2 1000 
3 1200;

您对此有何建议？

谢谢！

Answer 1

这是一个玩具示例，我认为可以回答您的问题。您可以使用类似的方法，或者仍然从数据中读取sbase变量，然后仅对其进行索引以在循环内以类似方式设置值。

from pyomo.environ import *

m = ConcreteModel()

m.I = Set(initialize=[1,])      # mixed type set for demo only.

m.sbase = Param(mutable=True)   # for the purposes of each instantiation, sbase is FIXED, so no index reqd.

m.x = Var(m.I, domain=NonNegativeReals )

## constraint
m.C1 = Constraint(expr=sum(m.sbase*m.x[i] for i in m.I) <= 10 )

## objective
m.OBJ = Objective(expr=sum(m.sbase*m.x[i] for i in m.I), sense=maximize)

solver = SolverFactory('cbc')

sbase_values = {1, 2, 5}
for sbase in sbase_values:
    m.sbase = sbase
    results = solver.solve(m)
    print(f'\n ***** solved for sbase = {sbase} *****')
    m.display()

Pyomo：Pyomo约束中的For循环

1 个答案: