Question

我尝试将启动成本添加到工作调度问题（无需启动成本即可工作）。应该在一个动态时期内优化时间表，即一天中的所有96个季度。我有96个价格的清单，我将其转换为矩阵：

prices = np.matrix([prices.as_matrix()])
prices = prices.T
length = prices.size
Range = range(np.size(prices))

以及以下变量：

c_turb = -100
min_turb = 0
max_turb = 8
c_pump = -50
min_pump = -10
max_pump = 0

这是我的客观功能：

 prob = pulp.LpProblem("The PSP",LpMaximize)
 prob +=lpSum([(price[i])*pump[i] for i in RANGE]
             + [price[i]*turbine[i] for i in RANGE]
             + [z_turb[i]*c_turb for i in RANGE]
             + [z_pump[i]*c_pump for i in RANGE])

我的变量定义如下：

pump = LpVariable.dicts("pump", RANGE,upBound=0,  cat="Integer")
turbine = LpVariable.dicts("turbine", RANGE, lowBound=0, cat="Integer")
pump_status=LpVariable.dicts("pump_status", RANGE,  cat="Binary")
turbine_status=LpVariable.dicts("turbine_status", RANGE,  cat="Binary")
z_pump = LpVariable.dicts("pump_startup", RANGE,  cat="Binary")
z_turb= LpVariable.dicts("turbine_startup", RANGE, cat="Binary")

for i in pump:
     pump[i].lowBound = -10 #bounds can vary for each i but irrelevant for problem

for i in turbine:
     turbine[i].upBound = 8

现在我添加了以下约束：

for i in range(1,length+1):
    n=range(0,i)
    prob += lpSum((-0.8*pump[i]-turbine[i])*0.25 for i in n ) <=50        prob += lpSum((-0.8*pump[i]-turbine[i])*0.25 for i in n) >=0
    prob += pump[i]>=min_pump*pump_status[i]
    prob += pump[i]<= max_pump*pump_status[i]
    prob += turbine[i] >= min_turb * turbine_status[i]
    prob += turbine[i] <= max_turb * turbine_status[i]

    if i == 1:
        prob += z_pump[i] == pump_status[i]
    else:
        prob+= z_pump[i]>=pump_status[i]-pump_status[i-1]
        prob+= z_pump[i]<=pump_status[i]
        prob+= z_pump[i]<=(1-pump_status[i-1])

    if i == 1:
        prob += z_turb[i] == pump_status[i]
    else:
        prob+= z_turb[i]>=turbine_status[i]-turbine_status[i-1]
        prob+= z_turb[i]<=turbine_status[i]
        prob+= z_turb[i]<=(1-turbine_status[i-1])

其中turbine_status和pump_status应该是此特定季度i中涡轮机/泵的开/关状态。 z_pump和z_turb应存储启动次数，即从0到1的状态切换。

但是，我收到了错误：prob += pump[i]>=min_pump*pump_status[i]。不幸的是error info: 96并不是真的很冗长....

有人能指出我正确的方向吗？

谢谢!!

线性优化python

0 个答案: