Список проголосовавших за комментарий "В VBA у меня так считается:<br/> <br/> Public Function d_1(ByVal F As Double, ByVal K As Double, ByVal T As Double, ByVal Sigma As Double) As Double<br/> <br/> d_1 = (Log(F / K) + T * 0.5 * Sigma * Sigma) / (Sigma * Sqr(T))<br/> End Function<br/> <br/> Public Function nd_1(ByVal d1 As Double) As Double<br/> <br/> nd_1 = Math.Exp(-1 * (d1 ^ 2) * 0.5) / Sqr(2 * PI)<br/> End Function"