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I have two points (x2,y2) and (x3,y3) that represent points in an exponential decay curve of discounted cash flows (x2 is less than x3):

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My question is: What is the decay curve equation for the decay function that passes through two points and the y-axis at the maximum value of y, where x1=0 and y1=Maximum Possible?

Mike
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1 Answers1

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First you have to evaluate the lower bound of your curve. Let denote it as $L$.

The function for exponential decay is $y=A\cdot e^{-bx}+L$

You have the two points $(x_2/y_2)$ and $(x_3/y_3)$. They can be inserted into the function:

$y_2=A\cdot e^{-bx_2}+L \quad (1)$

$y_3=A\cdot e^{-bx_3}+L \quad (2)$

Dividing (1) by (2).

$\frac{y_2-L}{y_3-L}=e^{b(x_3-x_2)}$

Taking logs

$log\left(\frac{y_2-L}{y_3-L} \right)=b(x_3-x_2)$

$b=\large{\frac{log\left(\frac{y_2-L}{y_3-L} \right)}{x_3-x_2}}$

Now you can use the value of $b$ to calculate $A$. Just insert the value of $b$ in one of the equation (1) or (2) and solve for $A$. $A$ is the y-intercept. This is the y-value of the point $(0/y_1)$.

callculus42
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  • What if L is approaching y3 as close as possible? What I mean by this is, what is the largest decay function that can be applied through these two points while still crossing the y axis? Apologies if your solution shows this, but as I see it the curve you describe solves for the least decay function through the y-axis and the two points. I am looking for the steepest curve possible. – Mike Nov 28 '15 at 21:03
  • In general you are restricted. Assuming exponential decay you have only 2 parameters for which you can choose values. My answer refer to your question. Now you have some additional ideas/ assumptions. What if L is approaching $y_3$ as close as possible? I don´t understand this question, because $y_3$ is given and $L$ is given. Here you have no choice. What do you mean by the largest decay function ? I have a general question: Must the function cross the two points exactly ? – callculus42 Nov 28 '15 at 21:33
  • Thank you for your attention. I will make some example curves to illustrate my question. I expect I am not understanding something obvious. I will post at a later time. – Mike Nov 28 '15 at 21:37
  • Ok. But what about my last question ? : Must the function cross the two points exactly ? – callculus42 Nov 28 '15 at 21:39
  • Yes it must go through the two points exactly. As I see it, for the steepest curve, then the lower bound of the curve would be as close to the y3 as possible. While A should be the maximum possible with any continuously decaying curve. – Mike Nov 28 '15 at 21:52
  • A have made an edit and came back to my origin answer. Since you have two data points the curve is determinated because you have two parameters ($A,b$). You can calculate the values of A and b like it is written in the answer. And the value of A is the value of the NVP, if the interest rate is zero. Therefore A is the sum of all (undiscounted) cashflows. This is the maximum possible value of the NVP since the interest rate cannot be negative. – callculus42 Nov 28 '15 at 23:36
  • I made a rollback to my second answer. L is the value of the acquisition costs. If the interest rate goes to infinity the profits have a present value of zero. And the largest value of A you can get if L is zero. And we are back to my first answer. – callculus42 Nov 29 '15 at 00:03
  • Studying your answer, I am sure you have answered my original question. My issue is determining the correct question. Rather than continue this thread I asked hopefully a simpler question here. http://math.stackexchange.com/questions/1551643/solving-the-curve-equation-for-logarithmic-decay-using-two-anchor-points Thank you for your help! – Mike Nov 29 '15 at 16:30