Scipy 寻找方程组的一组非负根

Question

我有一个非线性方程组。我对初始值没有很好的猜测。我想要至少一组经济学中的所有积极根源，这些变量的负值没有多大意义。

# -*- coding: utf-8 -*-
"""
Created on Sat Oct 15 21:48:56 2016

@author: Nick
"""

import scipy as sp
from scipy.optimize import root, fsolve
import numpy as np

#from scipy.optimize import *



el          = 1.1  
eg          = el   
ej          = 10  
om          = 0.3  
omg         = 0.3  
rhog        = 0.8  
xi          = 0.9  
mun         = 2
pidss       = 0.02 
muc         = 0.001
ec          = 2.00 # sims obtains 2.47
beta        = 0.998
h           = 0.8   
kappa       = 4.00 
n           = 1/3.0  
alpha       = 1/3.0  
delta       = 0.025
egs         = eg   
oms         = 0.2  
omgs        = oms  
rhom        = 0.7  
psiygap     = 1.000
psipi       = 2.500
rhoicu      = 0.800
taudss      = 0.01    # steady state tax on domestic consumption (setting it as 0 would create algebraic difficulties)
taumss      = 0.01    # steady state tax on imported consumption for domestic country
taukss      = 0.01    # steady state tax on rental income from capital for domestic country block
taunss      = 0.01    # steady state tax on labor for domestic country
tauydss     = 0.05    
gss         = 0.23    # steady state government spending as a propostion of gdp for domestic country block    
gsss        = 0.23    # steady state government spending as a propostion of gdp for foreign country block    

taudsss     = 0.01    

taumsss     = 0.01    

tauksss     = 0.01    

taunsss     = 0.01    

tauydsss    = 0.01          # steady state tax rate on output for foreign country block
    tauss       = 1.0               # Steady state terms of trade

icu = ((1+pidss)/beta) - 1 
mc = ((ej - 1)/ej)
r = (1/taukss) * ((1/beta)  - (1-delta))
rs = (1-tauksss) * ((1/beta)  - (1-delta))
KN = (mc*alpha/r)**(1/(1-alpha))
KNs = (mc*alpha/rs)**(1/(1-alpha))
psigma = (1-xi) * (1/(1-tauydss) - xi)**(-1)
psigmas = (1-xi) * (1/(1-tauydsss) - xi)**(-1)
w =     (1-alpha) * mc * (KN)**(alpha)

z = np.zeros(16)

def fun(z):
    Yd = z[0]
    N = z[1]
    X = z[2]
    I = z[3]
    Cd = z[4]
    Cm = z[5]
    Gd = z[6]
    Gm = z[7]
    Yds = z[8]
    Ns = z[9]
    Xs = z[10]
    Is = z[11]
    Cds = z[12]
    Cms = z[13]
    Gds = z[14]
    Gms = z[15]
    print (z)
    f = np.zeros(16)
    f[0]  = N - ( (X - muc)**(-ec) * ((1-alpha)/(mun)) * (mc)**(1/(1-alpha)) * (alpha/r)** (1-taunss) )
    f[1]  = Yd - ( Cd + Gd + I + ((1-n)/n) *(Cms + Gms)       )
    f[2]  = Yd - ( (KN)**(alpha)  * (psigma/(1-tauydss)**(ej)) )
    f[3]  = Cd - ( X * ((1-om)/(1+taudss)**(el)) *((1-om)*(1+taudss)**(1-el) + om * (1+taumss)**(1-el) * tauss**(1-el))**(el/(1-el))  )
    f[4]  = Gd - ( ((gss*Yd * (1-omg))/(1+taudss)**(eg) ) *((1-omg)*(1+taudss)**(1-eg) + omg* (1+taumss)**(1-eg) * tauss**(1-eg))**(eg/(1-eg) ) )
    f[5]  = I - ( delta* KN * N )
    f[6]  = Cm -( (X * (1-om)/(1+tauydss)**(el) ) *((1-om)*(1+taudss)**(1-el) + om* (1+taumss)**(1-el) * tauss**(1-el))**(el/(1-el))   )
    f[7]  = Gm - ( ((gss*Yd * (omg))/(1+taumss)**(eg) ) *((1-omg)*(1+taudss)**(1-eg) + omg* (1+taumss)**(1-eg) * tauss**(1-eg))**(eg/(1-eg) ) )
    f[8]  = Ns - ( (Xs - muc)**(-ec) * ((1-alpha)/(mun)) * (mc)**(1/(1-alpha)) * (alpha/rs)** (1-taunsss) )
    f[9]  = Yds - ( Cds + Gds + Is + (n/(1-n)) *(Cm + Gm)       )
    f[10] = Yds - ( (KNs)**(alpha)  * (psigmas/(1-tauydsss)**(ej) ) )
    f[11] = Cds - ( Xs * ((1-oms)/(1+taudsss)**(el))* ((1-oms)*(1+taudsss)**(1-el) + oms* (1+taumsss)**(1-el) * tauss**(1-el))**(el/(1-el))  )
    f[12] = Gds - ( ((gsss*Yds * (1-omgs))/(1+taudsss)**(eg) ) *((1-omgs)*(1+taudsss)**(1-eg) + omgs* (1+taumsss)**(1-eg) * tauss**(1-eg))**(eg/(1-eg) ) )
    f[13] = Is - ( delta* KNs * Ns )
    f[14] = Cms -( (Xs * (1-oms)/(1+tauydsss)**(el) ) *((1-oms)*(1+taudsss)**(1-el) + oms* (1+taumsss)**(1-el) * tauss**(1-el))**(el/(1-el))   )
    f[15] = Gms - ( ((gsss*Yds * (omgs))/(1+taumsss)**(eg) ) *((1-omgs)*(1+taudsss)**(1-eg) + omgs* (1+taumsss)**(1-eg) * tauss**(1-eg))**(eg/(1-eg) ) )
    return f



z = sp.optimize.root(fun, [100,100,70,30,50,20,50,20,100,100,100,100,100,100,100,100],  method='lm')
#z = fsolve(fun, [0,0,0.0,0,1,1,1,1,1,1,1,1,1,1,1,1])    
print(z)

解法有以下根

success: True
       x: array([  3.64725445e-01,   1.02848541e-06,  -1.86761721e+02,
         9.52089296e-10,  -1.30733205e+02,  -1.25265418e+02,
         5.87207967e-02,   2.51660557e-02,   3.36422990e+00,
         5.18324506e-04,   8.17060628e+01,   4.87111630e-04,
         6.53648502e+01,   6.53648502e+01,   6.19018302e-01,
         1.54754576e-01])

Answer 1

给定根的初始估计，数值求根算法在变量中沿特定方向移动-space，直到找到根。显然，使用这种方法，要求 returned 根在一定区间内是没有意义的——这完全取决于初始估计的好坏（以及算法使用的搜索方法）。

另一种可以 return 有界根的方法是将寻根问题作为优化（例如，最小化）问题，因为在优化问题中提供约束是有意义的。但是，您必须提供一个适当的 objective 函数，其最小值出现在原始函数的根部（此类函数有很多选项，通常选择是启发式的）。

其中一个函数是平方和 f[0]**2 + f[1]**2 + ... + f[15]**2。显然，此函数的最小值为零，这是在和的每一项都为零时（即在它们的根处）实现的。您可以使用 Scipy 的 least_squares 来执行此最小化，这也允许为优化变量提供界限。

对变量没有任何限制并使用相同的初始根估计，least_squares returns 与 root:

相同的解决方案

from scipy.optimize import least_squares

z_ls = least_squares(fun, [100,100,70,30,50,20,50,20,100,100,100,100,100,100,100,100])
print(z_ls.x)
print(z_ls.cost)

[  3.6473e-01   1.0285e-06  -1.8676e+02   9.5209e-10  -1.3073e+02  
  -1.2527e+02   5.8721e-02   2.5166e-02   3.3642e+00   5.1832e-04
   8.1706e+01   4.8711e-04   6.5365e+01   6.5365e+01   6.1902e-01
   1.5475e-01]
4.16527754459e-26

（请注意，z_ls.cost 是此时计算的平方和，（在数值精度内）为零。）

现在，使用 least_squares 将估计值限制为非负值：

z_ls = least_squares(fun, [100,100,70,30,50,20,50,20,100,100,100,100,100,100,100,100],bounds = (0,np.inf))
print(z_ls.x)
print(z_ls.cost)

[  5.9581e-01   2.1229e+01   4.2108e-02   1.1820e-37   2.0493e-33
   1.1914e-33   5.8857e-37   9.3812e-37   3.4508e+00   2.5054e+00
   1.1516e+00   2.2395e+00   8.0630e-01   4.2258e-01   5.1994e-01
   1.3867e-37]
0.237262813475

returned 估计确实有非负元素。但是，z_ls.cost（显着）大于零，表明此解不是根。这意味着两者之一：

初始点不够好，无法生成非负根。
这个问题不存在非负根。

如果你对以上没有任何见解，你唯一能做的就是尝试不同的初始化值，并希望得到一个想要的根 returned（直接通过 root 或如上所述的最小化公式）。

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