我对Gurobi有点熟悉，但是由于Gekko似乎具有一些优势，因此可以过渡到Gekko。我遇到了一个问题，我将用我想象中的苹果园来说明。 5周的收获期（function selectItemIDFromMainGrid(element) {
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）临到我们了，我的-非常微薄的-农产品将是：
#horizon: T=5
我为自己保留一些苹果[3.0, 7.0, 9.0, 5.0, 4.0]，其余的农产品我将以下列价格在农贸市场出售：[2.0, 4.0, 2.0, 4.0, 2.0]。我的存储空间可容纳6个苹果，因此我可以提前计划并在最理想的时刻出售苹果，从而最大程度地提高收入。我尝试使用以下模型确定最佳计划：
[0.8, 0.9, 0.5, 1.2, 1.5]由于某种原因，这是不可行的（错误代码= 2）。有趣的是，如果设置m = GEKKO()
m.time = np.linspace(0,4,5)
orchard = m.Param([3.0, 7.0, 9.0, 5.0, 4.0])
demand = m.Param([2.0, 4.0, 2.0, 4.0, 2.0])
price = m.Param([0.8, 0.9, 0.5, 1.2, 1.5])
### manipulated variables
# selling on the market
sell = m.MV(lb=0)
sell.DCOST = 0
sell.STATUS = 1
# saving apples
storage_out = m.MV(value=0, lb=0)
storage_out.DCOST = 0
storage_out.STATUS = 1
storage_in = m.MV(lb=0)
storage_in.DCOST = 0
storage_in.STATUS = 1
### storage space
storage = m.Var(lb=0, ub=6)
### constraints
# storage change
m.Equation(storage.dt() == storage_in - storage_out)
# balance equation
m.Equation(sell + storage_in + demand == storage_out + orchard)
# Objective: argmax sum(sell[t]*price[t]) for t in [0,4]
m.Maximize(sell*price)
m.options.IMODE=6
m.options.NODES=3
m.options.SOLVER=3
m.options.MAX_ITER=1000
m.solve()
（即等于demand[0] to 3.0, instead of 2.0，则模型确实会产生成功的解决方案。
orchard[0]在最后一个时间步中没有得到适当的限制。显然，我没有正确地制定约束条件。我应该怎么做才能获得与gurobi输出相当的实际结果（请参见下面的代码）？storage_out用output = {'sell' : list(sell.VALUE),
's_out' : list(storage_out.VALUE),
's_in' : list(storage_in.VALUE),
'storage' : list(storage.VALUE)}
df_gekko = pd.DataFrame(output)
df_gekko.head()
> sell s_out s_in storage
0 0.0 0.000000 0.000000 0.0
1 3.0 0.719311 0.719311 0.0
2 7.0 0.859239 0.859239 0.0
3 1.0 1.095572 1.095572 0.0
4 26.0 24.124924 0.124923 0.0
解决了古罗比模型。请注意，gurobi还会使用demand = [3.0, 4.0, 2.0, 4.0, 2.0]产生一个解决方案。这只会对结果产生微不足道的影响：在时间t = 0出售的 n 苹果变成demand = [2.0, 4.0, 2.0, 4.0, 2.0]。
1输出：
T = 5
m = gp.Model()
### horizon (five weeks)
### supply, demand and price data
orchard = [3.0, 7.0, 9.0, 5.0, 4.0]
demand = [3.0, 4.0, 2.0, 4.0, 2.0]
price = [0.8, 0.9, 0.5, 1.2, 1.5]
### manipulated variables
# selling on the market
sell = m.addVars(T)
# saving apples
storage_out = m.addVars(T)
m.addConstr(storage_out[0] == 0)
storage_in = m.addVars(T)
# storage space
storage = m.addVars(T)
m.addConstrs((storage[t]<=6) for t in range(T))
m.addConstrs((storage[t]>=0) for t in range(T))
m.addConstr(storage[0] == 0)
# storage change
#m.addConstr(storage[0] == (0 - storage_out[0]*delta_t + storage_in[0]*delta_t))
m.addConstrs(storage[t] == (storage[t-1] - storage_out[t] + storage_in[t]) for t in range(1, T))
# balance equation
m.addConstrs(sell[t] + demand[t] + storage_in[t] == (storage_out[t] + orchard[t]) for t in range(T))
# Objective: argmax sum(a_sell[t]*a_price[t] - b_buy[t]*b_price[t])
obj = gp.quicksum((price[t]*sell[t]) for t in range(T))
m.setObjective(obj, gp.GRB.MAXIMIZE)
m.optimize()
答案 0 :(得分：2)
您可以通过以下方式获得成功的解决方案：
m.options.NODES=2
问题在于它正在用NODES=3求解主节点之间的平衡方程。您的微分方程具有线性解，因此NODES=2应该足够准确。
还有其他几种方法可以改善解决方案：
storage_in = storage_out找到较大的任意值。m.Minimize(1e-6*storage_in)和m.Minimize(1e-6*storage_out)。SOLVER=1的整数解决方案，则需要切换到APOPT求解器。 Successful solution
---------------------------------------------------
Solver : APOPT (v1.0)
Solution time : 0.058899999999999994 sec
Objective : -17.299986
Successful solution
---------------------------------------------------
Sell
[0.0, 0.0, 4.0, 1.0, 1.0, 8.0]
Storage Out
[0.0, 0.0, 1.0, 0.0, 0.0, 6.0]
Storage In
[0.0, 1.0, 0.0, 6.0, 0.0, 0.0]
Storage
[0.0, 1.0, 0.0, 6.0, 6.0, 0.0]
这是修改后的脚本。
from gekko import GEKKO
import numpy as np
m = GEKKO(remote=False)
m.time = np.linspace(0,5,6)
orchard = m.Param([0.0, 3.0, 7.0, 9.0, 5.0, 4.0])
demand = m.Param([0.0, 2.0, 4.0, 2.0, 4.0, 2.0])
price = m.Param([0.0, 0.8, 0.9, 0.5, 1.2, 1.5])
### manipulated variables
# selling on the market
sell = m.MV(lb=0, integer=True)
sell.DCOST = 0
sell.STATUS = 1
# saving apples
storage_out = m.MV(value=0, lb=0, integer=True)
storage_out.DCOST = 0
storage_out.STATUS = 1
storage_in = m.MV(lb=0, integer=True)
storage_in.DCOST = 0
storage_in.STATUS = 1
### storage space
storage = m.Var(lb=0, ub=6, integer=True)
### constraints
# storage change
m.Equation(storage.dt() == storage_in - storage_out)
# balance equation
m.Equation(sell + storage_in + demand == storage_out + orchard)
# Objective: argmax sum(sell[t]*price[t]) for t in [0,4]
m.Maximize(sell*price)
m.Minimize(1e-6 * storage_in)
m.Minimize(1e-6 * storage_out)
m.options.IMODE=6
m.options.NODES=2
m.options.SOLVER=1
m.options.MAX_ITER=1000
m.solve()
print('Sell')
print(sell.value)
print('Storage Out')
print(storage_out.value)
print('Storage In')
print(storage_in.value)
print('Storage')
print(storage.value)