# I have one question about mathematical modeling

My school does Orienteering in PE, and basically the grades are from 0 to 20, the is the best,0 is the worst and 10 is 50%, well my teacher calculates the grades this way, if you do the task in 5 minutes you get a 20, if you do in 10 you get a 10, and calculates the rest, with a linear equations(I did it myself and it is: $$y= -2x + 30$$.

(Not so important to read, my opinion):Well is this really the best model? I don't think so, the effort someone makes to get a 15(knowing he had 10) is the same someone with a 15 makes to get a 20, which is 2 minutes and 30 seconds, obviously 2 minutes and 30 isn't the same for someone everyone, but the effort for someone who has a 18 to get a 19 or even twenty, at least to me, should be mucher higher(less time), than the effort someone makes to get a 13, knowing he has a 10

What kind of model should I choose, so in general between the grades 15 and 20, the difference of time is less than between 15 and 10(and how can I change it, like, if i wanted it to be from 17-20 and 0-17. But still:

f(10)=10

f(5)=20

f(x)=20 <=> x<=5 I can asure this by making the function recursion

is it a logistic model?

There are many functions that would do what you want for $$x\ge 5$$, but if you want a logistic $$f$$ it would be $$f(x)=\frac{40}{1+\exp\frac{x-5}{5/\ln 3}}.$$This implies $$f(10)=10,\,f(5)=20,\,f(7.5)=\frac{40}{1+\sqrt{3}}\approx 14.64,\,f(x)=15\implies x=5+\frac{5\ln\frac{5}{3}}{\ln 3}\approx 7.32.$$That achieves what you want, albeit only just in that example. You might want to look for a quadratic solution instead; if nothing else, it's easier to explain.
• @NunoMateus I solved for $a/(1+\exp (x-5)/b)$ to get the right values of $f(5),\,f(10)$. – J.G. Mar 14 at 16:13