Javascript IRR (Internal rate of return) Formula A

问题:

I'm using a IRR function in javascript to create calculation a that is done in excel using its own IRR function. The problem is mine is little off and I have no idea why. Here's the code below.

var IRRval = [];

IRRval.push(-financed);
for (i = 0; i < period; i++) {
    IRRval.push(rental);
}

var IRR = IRRCalc(IRRval, 0.001) * 0.01;

function IRRCalc(CArray, guest) {
    inc = 0.000001;
    do {
        guest += inc;
        NPV = 0;
        for (var j=0; j < CArray.length; j++) {
            NPV += CArray[j] / Math.pow((1 + guest), j);
        }
    } while (NPV > 0);
    return guest * 100;
}

Now if you use these figures:

---

Period 24

Financed 22000

Rental 1017.5000

My Result is: 0.008523000000000175

Excel Result is: 0.008522918

---

OR

---

Period 42

Financed 218000

Rental 5917.1429

My Result is: 0.006247000000000489

Excel Result is: 0.00624616

---

The Excel function is called: =IRR(T12:T73,0.01) T12-T73 is the same figures I'm using.

Any help would be much appreciated, Thanks Jason

UPDATE

I've solved it by changing the values below. But now the performance is too slow. Any ideas on how to improve this?

IRRCalc(IRRval, 0.001)

//to

IRRCalc(IRRval, 0.0001)

inc = 0.000001;

//to

inc = 0.00000001;

回答1:

After a quick skim read through your code, the error seems to be related to floating point precision error. More information can be found here: http://ajaxian.com/archives/crock-on-floating-points-in-javascript

In older javascript engines, if you did 0.3 + 0.3 you get something like 0.600000000001

Though most javascript engines today return 0.6, under the hood, the problem remains. Adding floats together causes unexpected results. So in your case the

inc = 0.000001;
guest += inc;

seems to me, to be the problem.

A way to solve this would be to use whole numbers. So instead of 0.000001 you would use 1 and instead of 0.001 you would use 1000. Then divide your return result by 100000

回答2:

We modified the code to achieve performance and accuracy. Try this:

function IRRCalc(CArray) {

  min = 0.0;
  max = 1.0;
  do {
    guest = (min + max) / 2;
    NPV = 0;
    for (var j=0; j<CArray.length; j++) {
          NPV += CArray[j]/Math.pow((1+guest),j);
    }
    if (NPV > 0) {
      min = guest;
    }
    else {
      max = guest;
    }
  } while(Math.abs(NPV) > 0.000001);
  return guest * 100;
}

回答3:

It's a floating point issue to be sure. You're going to want to use a library like BigDecimal.js to handle your values. It's really the only way to avoid this issue.

回答4:

Try this.

function NPV(discountRate, cashFlow){
    var npv = 0;
    for(var t = 0; t < cashFlow.length; t++) {
        npv += cashFlow[t] / Math.pow((1+ discountRate),t);
    }
    return npv;
}


function IRR(cashFlow,guess){
    guess = guess ? guess : 0.1;
    var npv;
    var cnt = 0;
    do
    {
        npv = NPV(guess,cashFlow);
        guess+= 0.001;

        cnt++;
    }
    while(npv > 0)

    return guess;
}

回答5:

I would like to build on from @Zohaib Answer, but what I would like to do is show undefined where appropriate. The best I can do is get it to equal to zero. I am using this simple dataset irr_arr=[-100, 100, 100, 100, 100,100]. I would appreciate some advice.

//IRRCALC funtion that handles irr going to infinity 

function IRRCalc_test(CArray) {

  min = 0.0;
  max = 1.0;
  c=0;
  do {
    guest = (min + max) / 2;
    NPV = 0;
    for (var j=0; j<CArray.length; j++) {
          NPV += CArray[j]/Math.pow((1+guest),j);
    }
    if (NPV > 0) {
      min = guest;
      c++; 
    }
    else {
      max = guest;
      c++;
    }

    if(c>=15){ return guest*100; }
  } while(Math.abs(NPV) > 0.000001);
  return guest*100;
}


// some testing 

irr_arr=[-100, 100, 100, 100, 100,100]
irr_res_arr_expected=[0,0,61.8,83.93,92.76,96.6]
for(i=1;i<=irr_arr.length;i++){
console.log("irr_arr - ",irr_arr.slice(0,i));
console.log("IRRCalc - ",IRRCalc(irr_arr.slice(0,i)))
//console.log("irr_expected - ", irr_res_arr_expected[i-1])
//if(IRRCalc(irr_arr.slice(0,i))===parseFloat(irr_res_arr_expected[i-1]).toFixed(2)){console.log(i,"- TRUE")} else {console.log(i,"- FALSE")}
}

this is the output

irr_arr -  [-100]
IRRCalc -  0.00      <<<<<<<<<<<<<<<<<<<------- this should be 'undefined' and not 'zero'
irr_arr -  [-100, 100]
IRRCalc -  0.00
irr_arr -  [-100, 100, 100]
IRRCalc -  61.80
irr_arr -  [-100, 100, 100, 100]
IRRCalc -  83.93
irr_arr -  [-100, 100, 100, 100, 100]
IRRCalc -  92.76
irr_arr -  [-100, 100, 100, 100, 100, 100]
IRRCalc -  96.60

here is the excel of what I am trying to produce

回答6:

Don't set your min IRR to 0. That is what bounds your answer. Try allowing it to go negative. Also don't try to calc an IRR on one number.