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Home/ Questions/Q 957717
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Editorial Team
  • 0
Editorial Team
Asked: May 16, 20262026-05-16T00:45:04+00:00 2026-05-16T00:45:04+00:00

I’m going through my code and each time D1 ends up being NaN. The

  • 0

I’m going through my code and each time D1 ends up being NaN. The code looks fine to me, and I’m completely stumped…

double D1;
Data Data = new Data();

PriceSpot = 40;
Data.PriceStrike = 40;
Data.RateInterest = .03;
Data.Volatility = .3;
Data.ExpriationDays = 300;

D1 = 
    (
        Math.Log(PriceSpot/Data.PriceStrike) +
        (
            (Data.RateInterest + (Math.Pow(Data.Volatility,2)/2)) *
            (Data.ExpirationDays/365)
        )
    ) /
    (
        Data.Volatility *
        Math.Pow(Data.ExpirationDays/365,.5)
    );
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T00:45:04+00:00Added an answer on May 16, 2026 at 12:45 am

    Data.Volatility * Math.Pow(Data.ExpirationDays/365,.5) is 0 since 300/365 as int equals to 0

    Assuming ExpriationDays property is of type int indeed, it’ll make the whole expression be 0.

    For example:

    [Test]
    public void Test()
    {
        var val = 300 / 365;
    
        Assert.That(val, Is.EqualTo(0));
    }
    

    Some comment about dividing by 0:

    When dividing two 0 integers an exception will be thrown at runtime:

    [Test]
    public void TestIntDiv()
    {
        int zero = 0;
        int val;
    
        Assert.Throws<DivideByZeroException>(() => val = 0 / zero);
    }
    

    When dividing two 0 doubles the result will be NaN and no exception will be thrown:

    [Test]
    public void TestDoubleDiv()
    {
        double zero = 0;
        double val = 0 / zero;
    
        Assert.That(val, Is.NaN);
    }
    
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