나는 비선형 방정식 시스템을 가지고있다. 나는 초기 값에 대해 좋은 추측을하지 못했다. 그리고 경제에서 적어도 하나의 긍정적 인 뿌리가 필요하기를 바랍니다. 이러한 변수에 대한 음수 값은별로 의미가 없습니다.Sci Py 비 방정식의 집합을 방정식 시스템으로 찾는 것
# -*- 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)
이 솔루션은
루트의 초기 추정을 감안할 때success: 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])
코드를 실행하면 어떻게됩니까?BTW는'n'과'alpha'에 대해 파이썬에서 정수로 간주되는'1/3'을 가지고 있습니다. 플로트가되도록하려면 '1/3.0'으로 변경할 수 있습니다. – Paul
폴 감사합니다. :) 내가 그 질문을 편집하면서 뿌리를 내리고있다. 일부는 부정적입니다. 그래서 나는 부정적이지 않은 뿌리를 찾을 수있는 특별한 방법이 있는지 궁금해하고있었습니다. – Nck