Understanding Coefficients V0 & K - out of one's depth

[dtex]

Hi,

I first want to say, I'm kind of out of my depth here.

My apologies in advance if this is the wrong section, or I say something that is complete nonsense.

After much research experimenting with exponential decay, I found a website that used different models to fit curves to a specific dataset you give it. I found a specific formula that seems to fit my needs.



My problem is trying to understand the coefficients, V0 & K. How they're calculated.

Dataset:

X	Y
0	6
500	3.6
1000	3.5

K = 0.0063561076607244
V0 = 0.0159179044024703

So what I've found is the formula y(t) = a × ekt, which you can use to workout K. Which involves dividing by a, and taking the natrual log from both sides etc... but i just can't seem to get to K.

I also can't seem to find what V0 is. How it is calculated, I'm kind of going around in circles. I want to replicate this in Excel.

Is there anyone who can point me to some resources I need to learn to understand this, or maybe explain in a more intuitive way as to what these two coefficients are?

Thanks!

 

[Dr.Peterson]

The basic technique is to plug in your three pairs of values to make three equations in the three unknowns [MATH]Y_0[/MATH], [MATH]V_0[/MATH], and [MATH]K[/MATH].

Your equation is [MATH]y = Y_0 + \frac{V_0}{K}\left(1-e^{-Kx}\right)[/MATH]. The three pairs yield

[MATH]Y_0 + \frac{V_0}{K}\left(1-e^{-K(0)}\right) = 6[/MATH]​

[MATH]Y_0 + \frac{V_0}{K}\left(1-e^{-K(500)}\right) = 3.6[/MATH]​

[MATH]Y_0 + \frac{V_0}{K}\left(1-e^{-K(1000)}\right) = 3.5[/MATH]​

The first, simplified, becomes [MATH]Y_0 = 6[/MATH], so the other two equations become

[MATH]\frac{V_0}{K}\left(1-e^{-500K}\right) = -2.4[/MATH]​

[MATH]\frac{V_0}{K}\left(1-e^{-1000K}\right) = -2.5[/MATH]​

My first thought for the next step is to divide one equation by the other, which will reduce it to a single variable. Then you'll have an equation with two exponentials, which typically requires some tricks; but in this special case, you can make it "easy" by substituting [MATH]u = e^{-500K}[/MATH].

Backing up a bit, though, we need to be sure that the form you found is really appropriate for your unstated problem. It clearly is intended for some specific application, in which [MATH]V_0[/MATH] and [MATH]K[/MATH] mean something particular; if your application is different, you could just as well have used the form [MATH]y = A - Be^{-Kx}[/MATH]. What is your problem, and where did you get the formula you propose?

 

[dtex]

Wow, thanks for that.

The problem:

I have 2 known values: (0, 6) & (1000, 3.5) - Which are the start and finish points of the curve.

I'm trying to find a exponential equation that starts and 'finishes' (doesn't go below 3.5) at these two points. The know point (500, 3.6) is made up. I'm trying to see how it effects the curve.

I want to try plug in different values in the middle or at X = 250. The outcome is an equation i can plug any X into that gives me the value of Y on that curve, dictated by the points I've defined.

I got the formula from - https://mycurvefit.com/ - I was trying different points and formula's to see how it effected the curve.

I hope that makes sense, this is a world I'm not familiar with. There is a massive gap in my knowledge.

Edit: The 'made-up' known point, isn't just 500 it could be N number of times between 0 - 1000. If i have 5-6 known points, can i generate an equation that gives me to rest of the points on that curve.

Thanks!

 

[Dr.Peterson]

Your choice of x=500 makes it easy to find the parameters, because you end up with a quadratic equation; another number would probably make it a good bit harder (at least requiring a different method).

More than 3 points would overspecify the problem, and it would no longer be able to fit an exponential function of the form you gave.

But since as far as I can tell, you don't specifically need an exponential equation, just something that smoothly passes through some points that generally resemble an exponential, something else might work. It would just be harder to say what you want. If we knew what the points mean, and why you expect it to be exponential, there might be more to be done.

 

[dtex]

But since as far as I can tell, you don't specifically need an exponential equation, just something that smoothly passes through some points that generally resemble an exponential, something else might work. It would just be harder to say what you want. If we knew what the points mean, and why you expect it to be exponential, there might be more to be done.

You're correct, i just need something that resembles an exponential curve.

So X points, are cost of a product and the Y points are markup of that cost. As the cost increases the markup decreases, but I want it to decrease in such a way is resembles exponential decay.

Like mentioned previously i only know the start and end point, but i want some way of changing the 'steepness' of that curve. Maybe i want the markup to decrease rapidly early and slow down as the cost increases. (Which was why i was theorising points in the middle or 1/4th of the way down to see how it effected that curve).

I guess what i was looking at wasn't the correct solution to this problem.

Thanks for your time.

 

[dtex]

I also want to add that anything past the known end point doesn't decay further.

So last known X is (1000, 3.5), if i put 2000 as X it remains (2000, 3.5).

 

[Dr.Peterson]

You're correct, i just need something that resembles an exponential curve.

So X points, are cost of a product and the Y points are markup of that cost. As the cost increases the markup decreases, but I want it to decrease in such a way is resembles exponential decay.

Like mentioned previously i only know the start and end point, but i want some way of changing the 'steepness' of that curve. Maybe i want the markup to decrease rapidly early and slow down as the cost increases. (Which was why i was theorising points in the middle or 1/4th of the way down to see how it effected that curve).

I guess what i was looking at wasn't the correct solution to this problem.

My answer in #2 should meet your needs (once finished); if you don't like the look of the result, you could play with different values at 500. Or use the site you found, and try varying things until it looks right to you. There is no one "right" solution.

I also want to add that anything past the known end point doesn't decay further.

So last known X is (1000, 3.5), if i put 2000 as X it remains (2000, 3.5).

This would be handled by a "piecewise" function, where you just use the formula up to 1000, and a constant beyond that. No one formula could do this.

 

[HallsofIvy]

There are, of course, infinitely many functions whose graphs pass through two give points, even if you require that they be exponential. You say "I'm trying to find a exponential equation that starts and 'finishes' (doesn't go below 3.5) at these two points." It sounds like you are saying, not that the function value must be 3.5 past some point, but that you want the graph to approach 3.5 from above as x goes to 3.5.

To do that use something of the form \(\displaystyle y= 3.5+ ae^{-x}\). With y(0)=6 that gives \(\displaystyle 6= 3.5+ a\) so a= 2.5 and \(\displaystyle y= 3.5+ 2.5e^x\).

 

[Deleted member 4993]

There are, of course, infinitely many functions whose graphs pass through two give points, even if you require that they be exponential. You say "I'm trying to find a exponential equation that starts and 'finishes' (doesn't go below 3.5) at these two points." It sounds like you are saying, not that the function value must be 3.5 past some point, but that you want the graph to approach 3.5 from above as x goes to 3.5.

To do that use something of the form \(\displaystyle y= 3.5+ ae^{-x}\). With y(0)=6 that gives \(\displaystyle 6= 3.5+ a\) so a= 2.5 and \(\displaystyle y= 3.5+ 2.5e^x\).

You say:

"..... and \(\displaystyle y=3.5+2.5e^x \ \)"

shouldn't that be:

\(\displaystyle y=3.5+2.5e^{-x} \)

 

[Dr.Peterson]

Of course, if the goal is to actually reach 3.5, this won't work. The extra parameter in the suggested form allows it to do so, while approaching at a tunable rate.