# Exponential vs. Linear Formula

1. ## Exponential vs. Linear Formula

Experts:

I need some assistance with potentially *tweaking* an exponential function/formula.

Attached spreadsheet includes background and specific requirements that I'm trying to solve. As a "picture is worth a 1000 words" the explanations should make more sense once you see the data (vs. me listing all of the background in this message).

Also, I included a 2nd tab for a "Linear" calculation... focus should be on the exponential scenario though.

I'd welcome any feedback/recommendations for modifying the formulae.

EEH

2. ## Re: Exponential vs. Linear Formula

Formula:
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In Linear sheet. Rest also same only range will change.

3. ## Re: Exponential vs. Linear Formula

avk - I am well familiar with that formula.

My question was focused on the "Exponential" scenario. As far as the linear model is concerned, the MEDIAN does NOT address relative scoring.

4. ## Re: Exponential vs. Linear Formula

I see a math problem. Solve the math problem, then it will be easier to solve the programming problem. Mathophobes beware!

Now, instead, I would prefer an inverse scenario. That is, I'd like to have smaller delta values at the top (intervals 1, 2, 3...) and have larger delta valus as I near the 10th, 20th, 30th, or 100th requirement.
If the solution gives me the larger delta at the bottom, I still must end up with zero (0) in "Y" for the last program requirments.
Looking at the sheet, Delta is essentially y(i+1)-y(i). Because the x(i)'s are evenly spaced integers (1,2,3,...), Delta is essentially an expression of "slope". Which, to me, distills the problem down to a differential equation, initial value problem. https://calculushowto.com/initial-value-problem/

At this point, you've only given a vague definition of dy/dx. You've said that dy/dx should be "small" when x is small (near 0) and that dy/dx should get "larger" as x increases towards x(max). At x(max), y should be 0. At this point, I would figure out more precisely what I want dy/dx to look like, then solve the initial value problem.

Example using polynomials (because, why not?).
Assume dy/dx is a straight line that is 0 at x=0 and decreases by 0.12 (so I can use the 0.06 that is currently in R1) for each x dy/dx=-0.12*x
This gives me a quadratic/parabola for y y=-0.06*x^2+C
y must be 0 at xmax -- 0=-0.06*xmax^2+C --> C=0.06*xmax^2
final expression for y =-0.06*x^2+0.06*xmax^2=0.06*(xmax^2-x^2)

At this point, the problem boils down to defining exactly what you want Delta to look like (with an adjustable parameter if you want). Then solve the initial value problem for that delta curve.

5. ## Re: Exponential vs. Linear Formula

Mr. Shorty:

I think you provided the perfect solution... I'll do some more reading on this topic but I'm very happy w/ the outcome.

Please attached attached snapshot which I labeled "Scenario #2" (for now -- with 0.01 growth rate).

The "Delta" are growing in DESC order... that's what I wanted.

Again, thank you for the help!
EEH

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