Matlab内部收益率

Question

你必须对我很轻松，我是 matlab 和 SO 的新手。我在使用 matlab 求解器计算内部收益率 (IRR) 时遇到问题。我看到 matlab 中的金融工具箱有这个功能，但是我不相信我已经安装它并且不想在他们的网站上获得试用版。

鉴于我的特定 IRR 计算的简单性质，我认为在 matlab 中简单地编写代码就足够简单了。每年的现金流是一样的，所以我在matlab中的输入如下：

syms x k;
IRR = solve(investment == yrSavings* symsum((1+x)^-k,1, nYears));

它没有失败，实际上给出了一个数字。唯一的问题是结果不正确！我手动输入内部收益率，它永远不会等于投资。使用 wolframalpha 我找到了实际的解决方案，返回并手动输入了 wolframalpha 的答案，symsum 函数返回了正确的结果。我不确定求解器是怎么回事！

Answer 1

按照您编写公式的方式，符号假设是您使用 x 作为迭代器变量。我相信你想使用k。试试这个：

syms x k;
IRR = solve(investment == yrSavings* symsum((1+x)^-k,k,1, nYears));

Answer 2

另一种方法是使用 MATLAB 函数 roots 计算折扣因子，然后将其转换为 IRR。前几天碰巧写了这样一个函数，所以我想我不妨把它贴在这里以供参考。它被大量注释，但实际代码只有三行。

% Compute the IRR of a stream of cashflows represented as a vector. For example:
%   > irr([-123400, 36200, 54800, 48100])
%   ans = 0.059616
%
% If the provided stream of cashflows starts with a negative cashflow and all
% other cashflows are positive then `irr` will return a single scalar result.
%
% If multiple IRRs exist, all possible IRRs are returned as a column vector. For
% example:
%   > irr([-1.0, 1.0, 1.1, 1.3, 1.0, -3.7])
%   ans =
%      0.050699
%      0.824254
%
% If no IRRs exist, or if infinitely many IRRs exist, an empty array is
% returned. For example:
%   > irr([1.0, 2.0, 3.0])
%   ans = [](0x1)
%
%   > irr([0.0])
%   ans = [](0x1)
%
% Unlike Excel's IRR function no initial guess can be provided because all
% possible IRRs are returned anyway
%
% This function is not vectorized and will fail if called with a matrix argument

function irrs = irr(cashflows)
  %% Overview
  % The IRR is defined as the rate, r, such that:
  %
  %   c0 + c1 / (1 + r) + c2 / (1 + r) ^ 2 + ... + cn / (1 + r) ^ n = 0
  %
  % where c0, c1, c2, ..., cn are the cashflows
  %
  % We define discount factors, d = 1 / (1 + r), so that the above becomes a
  % polynomial in terms of the discount factors:
  %
  %   c0 + c1 * d + c2 * d^2 + ... + cn * d^n = 0
  %
  % Such a polynomial will have n complex roots but between 0 and n real roots.
  % In the important special case that c0 < 0 and c1, c2, ..., cn > 0 there
  % will be exactly one real root.

  %% Check that input is a vector, not a matrix
  assert(isvector(cashflows), 'Input to irr must be a vector of cashflows');

  %% Calculation of IRRs
  % We use the built-in functions `roots` to compute all roots of the
  % polynomial, which are the discount factors `d`. The user will provide a
  % vector as [c0, c1, c2, ..., cn] but roots expects something of the form
  % [cn, ..., c2, c2, c0] so we reverse the order of cashflows using (end:-1:1).
  % At this stage `d` has n elements, most of which are likely complex.
  d = roots(cashflows(end:-1:1));

  % The function `roots` provides all roots, including complex ones. We are only
  % interested in the real roots, so we filter out the complex roots here. There
  % many also be spurious real roots less or equal to 0, which we also filter
  % out. Now `rd` could have between 0 and n elements but is likely to have a
  % single element
  rd = d(imag(d) == 0.0 & real(d) > 0);

  % We have solved everything in terms of discount factors, so we convert to
  % rates by inverting the defining formula. Since the discount factors are
  % real and positive, the IRRs are real and greater than -100%.
  irrs = 1.0 ./ rd - 1.0;

end