Subject:
|
On a scale of 0 to 10?
|
Newsgroups:
|
lugnet.off-topic.geek
|
Date:
|
Wed, 26 Mar 2003 17:24:50 GMT
|
Viewed:
|
310 times
|
| |
|
|
I've got a function that represents a quality rating, based on 2 parameters,
which are on a scale of 0 to 1:
(d-k)*k => quality (q)
Notice that this yields a sort of wacky result, insofar as the quality can
vary between -1 and 0.25, not -1 and 1 or -1 and 0 or 0 and 1. Icky. Anyway,
what I'd *like* to do is translate the resulting quality rating (q) onto a
scale (s) of 0 to 10. In other words, for q = -1, you get a 0, and for a q
of .25, you get a 10.
Problem. Assuming a linear distribution for 0.25 => 10, and -1 => 0, we get:
s = 8q + 8
Well, that's all well and good, except for when q = 0, we get s = 8, and I'd
rather that a quality of 0 translated to 5 rather than 8, since an 8 implies
that such a quality is really good, and in reality, a quality of 0
represents 'middle of the road'.
So. I figured why not assume a quadratic solution? IE: s = Aq^2 + Bq + C,
with known points (-1,0), (0,5), and (0.25,10). However. This seems to yield:
s = 12q^2 + 17q + 5
So, what's wrong with that? Well, for some values of q (like q = -0.72), s
results in a *negative* value (in this case -1.0192). ACK! That means that
*some* things that have better quality ratings actually have *LOWER* 0-to-10
rankings!
I assume I want a logarithmic scale function instead... How do I set that up
& solve for it? Or do I want something totally different?
DaveE
|
|
Message has 1 Reply:
 | | Re: On a scale of 0 to 10?
|
| (...) How about a double-linear solution? from -1 to 0, translate linearly from 0 to 5. From 0 to .25, translate linearly from 5 to 10. You won't have a single equation, but that seems to accomplish what you want, right? Adrian (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
8 Messages in This Thread:
 
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
