IRR Function

IRR.js

// Copyright (c) 2012 Sutoiku, Inc. (MIT License)
 
// Some algorithms have been ported from Apache OpenOffice:
 
/**************************************************************
 * 
 * Licensed to the Apache Software Foundation (ASF) under one
 * or more contributor license agreements.  See the NOTICE file
 * distributed with this work for additional information
 * regarding copyright ownership.  The ASF licenses this file
 * to you under the Apache License, Version 2.0 (the
 * "License"); you may not use this file except in compliance
 * with the License.  You may obtain a copy of the License at
 * 
 *   http://www.apache.org/licenses/LICENSE-2.0
 * 
 * Unless required by applicable law or agreed to in writing,
 * software distributed under the License is distributed on an
 * "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
 * KIND, either express or implied.  See the License for the
 * specific language governing permissions and limitations
 * under the License.
 * 
 *************************************************************/

function IRR(values, guess) {
  // Credits: algorithm inspired by Apache OpenOffice
  
  // Calculates the resulting amount
  var irrResult = function(values, dates, rate) {
    var r = rate + 1;
    var result = values[0];
    for (var i = 1; i < values.length; i++) {
      result += values[i] / Math.pow(r, (dates[i] - dates[0]) / 365);
    }
    return result;
  }

  // Calculates the first derivation
  var irrResultDeriv = function(values, dates, rate) {
    var r = rate + 1;
    var result = 0;
    for (var i = 1; i < values.length; i++) {
      var frac = (dates[i] - dates[0]) / 365;
      result -= frac * values[i] / Math.pow(r, frac + 1);
    }
    return result;
  }

  // Initialize dates and check that values contains at least one positive value and one negative value
  var dates = [];
  var positive = false;
  var negative = false;
  for (var i = 0; i < values.length; i++) {
    dates[i] = (i === 0) ? 0 : dates[i - 1] + 365;
    if (values[i] > 0) positive = true;
    if (values[i] < 0) negative = true;
  }
  
  // Return error if values does not contain at least one positive value and one negative value
  if (!positive || !negative) return '#NUM!';

  // Initialize guess and resultRate
  var guess = (typeof guess === 'undefined') ? 0.1 : guess;
  var resultRate = guess;
  
  // Set maximum epsilon for end of iteration
  var epsMax = 1e-10;
  
  // Set maximum number of iterations
  var iterMax = 50;

  // Implement Newton's method
  var newRate, epsRate, resultValue;
  var iteration = 0;
  var contLoop = true;
  do {
    resultValue = irrResult(values, dates, resultRate);
    newRate = resultRate - resultValue / irrResultDeriv(values, dates, resultRate);
    epsRate = Math.abs(newRate - resultRate);
    resultRate = newRate;
    contLoop = (epsRate > epsMax) && (Math.abs(resultValue) > epsMax);
  } while(contLoop && (++iteration < iterMax));

  if(contLoop) return '#NUM!';

  // Return internal rate of return
  return resultRate;
}

IRR([-200,5,5,5,5,5]) yields Infinity

Thank you very much!

Thanks you a lot!
This was the most accurate formula I have found

Bro, you're great! 🥇

Thanks, that was perfect

Here's the TypeScript modern version:

// Calculates the resulting amount
const _irrResult = function (values: number[], dates: number[], rate: number) {
  const r = rate + 1
  let result = values[0]
  for (let i = 1; i < values.length; i++) {
    result += values[i] / Math.pow(r, (dates[i] - dates[0]) / 365)
  }
  return result
}

// Calculates the first derivation
const _irrResultDerivation = function (values: number[], dates: number[], rate: number) {
  const r = rate + 1
  let result = 0
  for (let i = 1; i < values.length; i++) {
    const frac = (dates[i] - dates[0]) / 365
    result -= (frac * values[i]) / Math.pow(r, frac + 1)
  }
  return result
}

const IRR = (values: number[], guess: number): number => {
  // Initialize dates and check values contains at least one positive value and one negative value
  const dates = [] as number[]
  let positive = false
  let negative = false
  for (let i = 0; i < values.length; i++) {
    dates[i] = i === 0 ? 0 : dates[i - 1] + 365
    if (values[i] > 0) positive = true
    if (values[i] < 0) negative = true
  }

  // Return error if values does not contain at least one positive value and one negative value
  if (!positive || !negative) {
    return NaN
  }

  // Initialize the guess and the resultRate
  let resultRate = typeof guess === 'undefined' ? 0.1 : guess

  // Set maximum epsilon for the end of the iteration
  const epsMax = 1e-10

  // Set maximum count of iterations
  const iterMax = 50

  // Implement Newton's method
  let newRate, epsRate, resultValue
  let iteration = 0
  let contLoop = true
  do {
    resultValue = _irrResult(values, dates, resultRate)
    newRate = resultRate - resultValue / _irrResultDerivation(values, dates, resultRate)
    epsRate = Math.abs(newRate - resultRate)
    resultRate = newRate
    contLoop = epsRate > epsMax && Math.abs(resultValue) > epsMax
  } while (contLoop && ++iteration < iterMax)

  if (contLoop) {
    return NaN
  }

  // Return internal rate of return
  return resultRate
}