def durationBond(rate, couponRate, maturity):
"""Objective : estimte the durtion for a given bond
rate : discount rate
couponRate: coupon rate
maturity : number of years
Example 1: >>>discountRate=0.1
>>>couponRate=0.04
>>> n=4
>>>durationBond(rate,couponRate,n)
3.5616941835365492
Example #2>>>durationBond(0.1,0.04,4)
3.7465335177625576
"""
d = 0
n = maturity
for i in sp.arange(n):
d += (i + 1) * sp.pv(rate, i + 1, 0, -couponRate)
d += n * sp.pv(rate, n, 0, -1)
return d / sp.pv(rate, n, -couponRate, -1)
def myPV(r, n, fv):
"""
Objective: estimate present value of one future cash flow
r : period rate
n : number of periods
fv: fture value
formula used : pv=fv/(1+r)**n
Example 1: >>>myPV(0.1,1,100)
90.9090909090909
Example #2 >>>myPV(r=0.1,fv=100,n=1)
90.9090909090909
"""
import scipy as sp
return (-sp.pv(r, n, 0, fv))
def test_port(start_year,
h,
r=discount_r,
ws_bench=0.6,
c=1.8,
v_th=100,
eta=1.5,
lookback=50,
full_ret=full_ret):
v_init = sp.pv(r, h, c, -v_th)
v = v_init
v_bench = v_init
time_remain = h
i = 0
while time_remain >= 1:
train_ret = full_ret.loc[(start_year + i - lookback):(start_year + i -
1)]
policy = pd.Series()
for s in [t * 0.05 for t in range(0, 21)]:
policy[s] = calc_state_utility(time_remain,
train_ret,
s,
v_init=v,
c=c,
v_th=v_th,
eta=eta)
ws = policy.argmax()
port_ret = ws * full_ret.loc[start_year + i, 'stock'] + (
1 - ws) * full_ret.loc[start_year + i, 'bond']
bench_ret = ws_bench * full_ret.loc[start_year + i, 'stock'] + (
1 - ws_bench) * full_ret.loc[start_year + i, 'bond']
# update portfolio and benchmark value
v = v * (1 + port_ret) + c
v_bench = v_bench * (1 + bench_ret) + c
i = i + 1
time_remain = time_remain - 1
return (v, v_bench)
def myPV(r, n, fv):
import scipy as sp
return (-sp.pv(r, n, 0, fv))
# -*- coding: utf-8 -*- """ Created on Tue Jul 28 22:22:27 2020 @author: yomer """ import scipy as sp # 미래가치 구하기 : 연금리 10%, 100달러 예금 2년 후 print(sp.fv(0.1, 2, 0, 100)) # 현재가치 구하기 : 금리1.45%, 횟수5, 미래가치 234 print(sp.pv(0.0145, 5, 0, 234))
import scipy as sp ''' 每期利息率为：0.05 期数：3 每月等额收支：0 终值：100 求：现值 ''' pv1 = sp.pv(0.05, 3, 0, 100, when='begin') print(pv1) ''' 每期利息率为：0.05/12 期数：10*12 每月等额收支：+100 终值：0 求：现值 ''' pv2 = sp.pv(0.05 / 12, 10 * 12, 100, 0, when='begin') print(pv2) ''' 每期利息率为：0.05/12 期数：30*12 每月等额收支：+3000 终值：0 求：现值 60岁开始，每月领取3000元，共领取30年领完，60随时需要存入多少钱 ''' pv3 = sp.pv(0.05 / 12, 30 * 12, 3000, 0, when='begin') print(pv3)
x.sum()
x.sum(axis=0) ##sum by row
### simulation
ret = np.array(range(1, 100), float)/100 ##create a grid of return pct
rand_50 = np.random.rand(50) ##50 rand num btw 0 and 1
rnorm_100 = np.random.normal(size=100)
## 2. Scipy ------------
### Compute the NPV of a Cashflow
cf = [-100, 50, 40, 20, 10, 50]
sp.npv(rate=0.1, values=cf)
### Monthly payment: APR = 0.045, compounded monthly, mortgage = 250k, T=30yrs
sp.pmt(rate=0.045/12, nper=12*30, pv=250000)
### PV of one future cf
sp.pv(rate=0.1,nper=5,pmt=0,fv=100)
### PV of Annuity (series of payments)
sp.pv(rate=0.1, nper=5, pmt=100)
### Geometric mean returns
def geoMeanreturn(ret):
return pow(sp.prod(ret+1), 1./len(ret)) - 1
ret = np.array([0.1, 0.05, -0.02])
geoMeanreturn(ret)
### other fundamental stuff
sp.unique([1,1,2,3,4,1,1,1])
sp.median([1,1,2,3,4,1,1,1])
## 3. matplotlib ------------
### Series of Normal Distribution
plt.plot(rnorm_100)
""" Name : 4375OS_06_07_pv_opposite_sign.py Book : Python for Finance Publisher: Packt Publishing Ltd. Author : Yuxing Yan Date : 12/26/2013 email : [email protected] [email protected] """ import scipy as sp print round(sp.pv(0.1, 5, 0, 100), 2) print round(sp.pv(0.1, 5, 0, -100), 2)
def pv(*args):
rate, nper, pmt = args[0], args[1], args[2]
return scipy.pv(rate, nper, pmt)
PV_simple_1 = FV / (1 + r * n)
PV_simple_1
#%% m-기간 기준단리: FV=PV*(1+r/m*n*m) ==> PV=FV/(1+r/m*n*m)
# 년기준복리: FV=PV*(1+r)**n ==> PV=FV/(1+r)**n
# m-기간 기준복리: FV=PV*(1+r/m)*(n*m) ==> PV=FV/(1+r/m)*(n*m)
# 기존 모듈 이용
import scipy as sp
import numpy as np
help(sp.pv)
help(np.pv)
fv = 1276.2815625000003
r = 0.05
n = 5
m = 1
PV_sp = sp.pv(r, n, 0, fv) # pv(rate, nper, pmt, fv=0, when='end')
print(PV_sp)
PV_np = np.pv(r, n, 0, fv)
print(PV_np)
#%%[과제 43-2] 이자가 6개월에 한번 복리로 계산되고 년이자율이 5%인 계좌의 5년 후1000원의 현재가치는 얼마인가?
#i) 사용자 정의 함수를 사용하여 계산하여 보세요.
import numpy as np
FV = 1000
r = 0.05
n = 5
m = 2
def get_pv(Fv, r, n, m):
return FV / (1 + r / m * n * m)
""" Name : 4375OS_06_07_pv_opposite_sign.py Book : Python for Finance Publisher: Packt Publishing Ltd. Author : Yuxing Yan Date : 12/26/2013 email : [email protected] [email protected] """ import scipy as sp print round(sp.pv(0.1,5,0,100),2) print round(sp.pv(0.1,5,0,-100),2)
import scipy as sp ''' 每期利息率为：0.05 期数：3 每月等额收支：0 终值：100 求：现值 ''' pv1 = sp.pv(0.05,3,0,100,when='begin') print(pv1) ''' 每期利息率为：0.05/12 期数：10*12 每月等额收支：+100 终值：0 求：现值 ''' pv2 = sp.pv(0.05/12, 10*12, 100, 0,when = 'begin') print(pv2) ''' 每期利息率为：0.05/12 期数：30*12 每月等额收支：+3000 终值：0 求：现值 60岁开始，每月领取3000元，共领取30年领完，60随时需要存入多少钱 ''' pv3 = sp.pv(0.05/12,30*12,3000,0,when = 'begin') print(pv3)
import scipy as sp
n = float(raw_input('N = '))
iy = float(raw_input('I/Y = '))
pv = float(raw_input('PV = '))
pmt = float(raw_input('PMT = '))
fv = float(raw_input('FV = '))
sol = raw_input('Solve for (n,iy,pv,pmt,fv): ')
if sol == "n":
print("N = %f" % sp.nper(iy, pmt, pv, fv))
if sol == "iy":
print("I/Y = %f" % sp.rate(n, pmt, pv, fv))
if sol == "pv":
print("PV = %f" % sp.pv(iy, n, pmt, fv))
if sol == "pmt":
print("PMT = %f" % sp.pmt(iy, n, pv, fv))
if sol == "fv":
print("FV = %f" % sp.fv(iy, n, pmt, pv))
'''
print 'N = %f, ' % n
print 'I/Y = %f, ' % iy
print 'PV = %f, ' % pv
print 'PMT = %f, ' % pmt
print 'FV = %f, ' % fv
'''
Python 3.7.3 (v3.7.3:ef4ec6ed12, Mar 25 2019, 21:26:53) [MSC v.1916 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license()" for more information. >>> import scipy as sp >>> sp.fv(.0025,25,0,5000) -5322.057184363162 >>> import scipy as sp >>> sp.pv(.0541/2,20,55,0,1) -863.7816463604273 >>> import scipy as sp >>> sp.pmt(.0312,5,2400) -525.8478442817194 >>>
#!/usr/bin/env python
#
# Classes and functions given as examples from Chapter 5
# in the book.
#
# Stock valuation using the dividend discount model
#
import scipy as sp
if __name__ == "__main__":
"""
Go go super awesome stock valutation DDM
"""
"""
Present Value demo:
Given R=12%; D=1; and FV=50, where
R is appropriate cost of equity
D is the dividend
and FV is the future value
"""
sp.pv(0.12, 1, 1 + 50)
"""
Grammatical growth demo:
"""
dividends = [1.80, 2.07, 2.48193, 2.680, 2.7877]
R = 0.182
g = 0.03
print(float(sp.npv(R, dividends[:-1]) * (1 + R)))
