内部收益计算函数
曾经看过一个帖子：有一个理财产品，每年年初存入10000元，每年年底得到利息1000元。持续5年，5年后返还本金50000元：问：利率是多少？
下面有个回复：每年存10000，利息1000，所以利率是10%。
那么问题来了，真实利率到底是多少？
office的excel有个公式叫xirr，参考 https://support.office.com/zh-cn/article/XIRR-%E5%87%BD%E6%95%B0-de1242ec-6477-445b-b11b-a303ad9adc9d
irr，参考 https://support.office.com/zh-cn/article/IRR-%E5%87%BD%E6%95%B0-64925eaa-9988-495b-b290-3ad0c163c1bc
实现如下：
require 'date' def simply_parse str
year, month, day = str.split('-')
return Date.new(year.to_i, month.to_i, day.to_i)
end def simply_sum arr
arr.inject(0) { |sum, element| sum + element }
end # guess : 从该值开始计算
# accuracy ： 计算精度，即每次递进的增量
# cal_times : 计算次数
def xirr(arr, guess = 0.05, accuracy = 0.0001, cal_times = 1000)
rate = guess
start_date = simply_parse(arr[0][1])
end_date = simply_parse(arr[-1][1])
__n = arr.map(&:first).map(&:to_f).select{|x| x < 0}
__p = arr.map(&:first).map(&:to_f).select{|x| x > 0}
t1 = ((simply_sum(__p))*accuracy).abs
t2 = ((simply_sum(__n))*accuracy).abs
tt = (t2+t1)/2
return 'number error' if __n.empty? && __p.empty?
last_sum = 0
sum = 0
flag = 0
cal_times.times do
sum = 0
arr.each do |item|
_date = simply_parse(item[1])
return 'date error' if _date < start_date || _date > end_date
profit = item[0].to_f * ((1+rate)**((end_date-_date)/365))
sum = sum + profit
end
if last_sum == 0
last_sum = sum
next
end
if last_sum.abs < sum.abs
accuracy = 0 - accuracy
flag = flag + 1
end
if sum > (0 + tt)
return rate if flag > 1
rate = rate + accuracy
last_sum = sum
elsif sum < (0 - tt)
return rate if flag > 1
rate = rate - accuracy
last_sum = sum
else
return rate
end
end
if sum > (0 - tt*2) && sum < (0 + tt*2)
return rate
end
return "guess failed, try guess = #{rate}"
end #test
arr = '10000 2010-01-01
-1000 2010-12-31
10000 2011-01-01
-1000 2011-12-31
10000 2012-01-01
-1000 2012-12-31
10000 2013-01-01
-1000 2013-12-31
10000 2014-01-01
-1000 2014-12-31
-50000 2014-12-31'.split("\n").map{|x| x.split}
xirr arr, 0.05, 0.0001, 2000 # => 0.03409999999999955 arr = '-10000 2008-1-1
2750 2008-3-1
4250 2008-10-30
3250 2009-2-15
2750 2009-4-1'.split("\n").map{|x| x.split}
xirr arr, 0.37 # => 0.37329999999999963 arr = '10000 2008-1-1
-2750 2008-3-1
-4250 2008-10-30
-3250 2009-2-15
-2750 2009-4-1'.split("\n").map{|x| x.split}
xirr arr, 0.37 # => 0.37329999999999963 arr = '-700000 2008-1-1
120000 2009-1-1
150000 2010-1-1
180000 2011-1-1
210000 2012-1-1'.split("\n").map{|x| x.split}
xirr arr, 0.0 # => -0.021199999999999927 arr = '-700000 2008-1-1
120000 2009-1-1
150000 2010-1-1
180000 2011-1-1
210000 2012-1-1
260000 2013-1-1'.split("\n").map{|x| x.split}
xirr arr, 0.07 # => 0.08670000000000049 arr = '-700000 2008-1-1
120000 2009-1-1
150000 2010-1-1'.split("\n").map{|x| x.split}
xirr arr, -0.47 # => -0.44300000000000295