Scipy优化错误：遇到被零除的错误

Question

我很困惑，当我试图最大化一个函数时，我怎么会得到一个除以零的错误。在

solution = minimize(objective,x0, method='SLSQP', bounds = bnds, constraints = cons)
__main__:36: RuntimeWarning: invalid value encountered in double_scalars
C:\Users\loren\AppData\Roaming\Python\Python36\site-packages\scipy\stats\_distn_infrastructure.py:879: RuntimeWarning: invalid value encountered in greater
  return (self.a < x) & (x < self.b)
C:\Users\loren\AppData\Roaming\Python\Python36\site-packages\scipy\stats\_distn_infrastructure.py:879: RuntimeWarning: invalid value encountered in less
  return (self.a < x) & (x < self.b)
C:\Users\loren\AppData\Roaming\Python\Python36\site-packages\scipy\stats\_distn_infrastructure.py:1738: RuntimeWarning: invalid value encountered in greater_equal
  cond2 = (x >= self.b) & cond0
__main__:38: RuntimeWarning: invalid value encountered in double_scalars
__main__:40: RuntimeWarning: invalid value encountered in double_scalars
__main__:29: RuntimeWarning: invalid value encountered in double_scalars
__main__:31: RuntimeWarning: invalid value encountered in double_scalars
__main__:33: RuntimeWarning: invalid value encountered in double_scalars
__main__:36: RuntimeWarning: divide by zero encountered in double_scalars
__main__:29: RuntimeWarning: divide by zero encountered in double_scalars
__main__:38: RuntimeWarning: divide by zero encountered in double_scalars
__main__:31: RuntimeWarning: divide by zero encountered in double_scalars
__main__:40: RuntimeWarning: divide by zero encountered in double_scalars
__main__:33: RuntimeWarning: divide by zero encountered in double_scalars

代码仍在运行，但我没有得到任何有用的东西。在

^{pr2}$

这是我的目标函数：

def objective(x): #x <-> pl1 pl2 pl3 ph1 ph2 ph3 r prob1 prob2 prob3
    return -x[6]*(x[0]*NL1(x)+x[1]*NL2(x)+x[2]*NL3(x)+x[3]*NH1(x)+x[4]*NH2(x)+x[5]*NH3(x))

所以我最大化了10个变量。在

以下是定义函数的形式（不具有可笑的长目标函数）：

def CDF(x, parameter):
    return expon.cdf(x, scale = 1/parameter)
def NH1(x): #Number of high buyers at location 1
    return 1-CDF((x[3]-x[0])/(x[7]), lambdaB)
def NL1(x): #Number of low buyers at location 1
    return CDF((x[3]-x[0])/x[7], lambdaB) - CDF((s-x[0])/(1+x[7]), lambdaB)
def NS1(x): #Number of sellers at location 1
    return 1- CDF((NH1(x)*(x[3]-x[6])+NL1(x)*(x[0]-x[6]))/(NH1(x) + (1-x[7])*NL1(x)+2*x[7]*NL1(x)), lambdaS)
def constraintGenEq1(x):
    return NS1(x)- NH1(x) -(1+x[7])*NL1(x)
def constraintProbLow1(x):
    x[7] =  (x[3]-x[0])/(dist) 
    return x[7] - (x[3]-x[0])/(dist) 

conGen1= {'type': 'eq', 'fun': constraintGenEq1}
conGen2= {'type': 'eq', 'fun': constraintGenEq2}
conGen3= {'type': 'eq', 'fun': constraintGenEq3}
conProb1= {'type': 'eq', 'fun': constraintProbLow1}
conProb2= {'type': 'eq', 'fun': constraintProbLow2}
conProb3= {'type': 'eq', 'fun': constraintProbLow3}

upper = 150
cons = ([conGen1, conGen2, conGen3, conProb1, conProb2, conProb3])
bnds = ((0,upper),(0,upper),(0,upper),(0,upper),(0,upper),(0,upper),(0,1),(0,1),(0,1),(0,1))

n=len(bnds)
x0 = np.zeros(n)+0.1
solution = minimize(objective,x0, method='SLSQP', bounds = bnds, constraints = cons)

我明白如果这是一个密集的阅读。但是我试着利用边界和约束来让它返回一些有用的东西，但是我还没有运气。我想我在我的一些子功能中有除法运算的地方可能会遇到问题。最后，对参数进行了一定的假设。在

lambdaB = 1 #Buyer's exponential distribution parameter
lambdaS = 1 #Seller's exponential distribution parameter
dist = 1
s = 4