There's probably a very simple answer to this, but I can't put my finger on it.

I have two numbers. I want one to be large, and the other to be small. I'd like to identify these with a single value.

For instance (hypothetical example follows), students should have high test scores and low absence rates. Therefore, a student with a test score of 100 and an absence rate of 0 is the ideal student.

For the purpose of this example, the absence rate is the more important number. We'd rather have a student with a test score of 50 and has an absence rate of 1 than a student with a test score of 100 and an absence rate of 2. In other words, it's most important that the small number is actually small.

So taking test score / absence rate won't work. Because in the aforementioned example, it scores both students equally. Additionally, a student with perfect attendance throws a divide-by-zero error.

Any suggestions? Is there actually a simple answer? I apologize if the question is too elementary for this site.


P.S. I couldn't identify an appropriate tag, so please re-tag if you can think of a better one.

EDIT: Reading through my question, I see that it could depend on how much more important the small number is than the big number. Really, I'm not concerned with the particulars. A general answer is all I need to craft a solution to my specific problem.

  • $\begingroup$ Maybe you are looking for a function like $f(x,y)=xe^{-y}$, which is linearly increasing in $x$ for fixed $y$, and exponentially decreasing in $y$ for fixed $x$. $\endgroup$
    – Shai Covo
    May 29, 2011 at 15:35

1 Answer 1


This is somewhat of an "ad-hoc" answer, drawn from experience with weighted grading, etc.

Attendance: You may want to subtract absence rate from 100, e.g., so high scores uniformly express desirability, with 100, e.g., "perfect". In doing so, you could simply subtract "attendance rate" from 100 (so a student attending 24 of 25 classes would have a score of 96), or 2. you could assign a "point penalty" for every class missed, which gets subtracted from 100 possible attendance "points". In that sense, say there are 50 classes in all, and a student attends only 25 of 50 classes. By the initial attendance rate (1), the student would receive an attendance score of "50"; whereas, if you penalize students 4 points for each missed class, that student would receive no attendance credit. I.e. you can establish a minimum threshold where students do not receive "any" attendance credit unless they attend more than 50% of the classes.

Test Score:

  • Average the test scores to calculate average test score (for maximum 100 score). Four tests at $100$ points each would thereby each contribute $25$%, or $25$ possible points, to the total test score (of course, if you have 3 tests, it would probably be "nicer" arithmetically speaking to add all of a students test scores, then divide by number of tests.

    • You can also weight particular exams, like a final: E.g. You could give more weight to the final exam: $2$ midterms ($100$ points possible, for each) and $1$ final exam (200 points possible, or $100$ doubled), add, and then divide by 4, to arrive at a test score, with $100$ the maximum score.

    • You could also simply break up the $100$ maximum test score by assigning, for example, $40$ possible points for final, $20$ possible points for each of two midterms, and $5$ points for each of $4$ quizzes.

    • Total score: If you now have decided on a means to rate attendance, with maximum score $100$, and the same for test score, you then need only assign a "weight" factor of greater than 1 to multiply by $100$ to arrive at the weighted attendance score: sum test score* $:= T$ with the weighted attendance score $:= A$, and divide by "$F$",( 1 + "the weight factor you chose for attendance"), and you'll have a single score, with highest possible score @ 100.

E.g.: Suppose you consider a students attendance rate to be twice as important as his/her test score.

Then in this case $F = 2$. $T$ is the test score, and $A = 2 \times\;\;$ attendance rate. Then compute: $\displaystyle \text{Final Score} = \frac{(T + A)}{F} = \frac{T + A}{1 + 2} = \frac {T+A}{3}$

  • $\begingroup$ The use of students and tests was entirely hypothetical, but your answer is very informative notwithstanding. Thank you very much for helping so thoroughly. $\endgroup$
    – Peter
    May 29, 2011 at 14:14
  • $\begingroup$ In my real-world problem, what I exemplified by "absence rate" is actually the standard deviation for a sequence of values (I simplified things for the purpose of easy explanation). I like how you determined a "score" for absence... but how would you suggest doing the same for standard deviation? I suppose the minimum standard deviation is 0, and the maximum is equal to half of the range? $\endgroup$
    – Peter
    May 29, 2011 at 14:25

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