-1

I need to develop a grading system for a course. The first assignment is evaluated based on only one criteria, a. The second assignment based on the first criteria, a, plus one new criteria, b, the third assignment based on three criteria, a, b, and c, etc. The 16th assignment is graded based on 16th criteria, a through p.

Assignment 1: a
Assignment 2: a+b
Assignment 3: a+b+c
…

I initially devised this distribution:

Assignment 1: 1
Assignment 2: 1/2+1/2
Assignment 3: 1/3+1/3+1/3
…
Assignment 16: 1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16+1/16, 1/16+1/16+1/16

This does not suit the problem well though, because the first criteria introduced are the most important, so the earlier criteria should always take more weight than later criteria. I need a system where a is always greater than b and b is always greater than c, meanwhile, making sure that p never gets to be too much smaller than ~1/50 or so.

 Assignment 16: a > b > c > d > e > f > g > h > i > j > k > l > m > n > o > p

How can I distribute the value of each criteria for the assignments?

Village
  • 203
  • I don't quite understand what you're looking for. The first display says that assignment 1 has distribution $a$, and assignment 2 has distribution $a+b$. But then when I match this up with the second display, it seems to say that for assignment 1, $a=1$ and for assignment 2, $a=\frac12$. I don't understand what $a, b, c\ldots$ are supposed to represent here. Could you please edit your question to clarify this? – MJD Sep 05 '13 at 03:50
  • As an example, for assignment 1, "Correct spelling", a, is worth 1/1, or 100%. For assignment 2, "Correct spelling" is worth 1/2 or 50% and the new criteria, "Correct punctuation", b, is worth 1/2, or 50%. a is how important the first criteria is to the grade, but this value gets smaller in later assignments, but should always be larger than b. – Village Sep 05 '13 at 04:01
  • There are obvious many choices of assignment. I think a decision you need to make is how much weight you want to assign to $a$ in $n^{th}$ assignment as $n$ changes. Do you want it goes like $\frac{1}{n}$ i.e. every choices roughly equal at the end or the weight of $a$ remains finite as $n \to \infty$. – achille hui Sep 05 '13 at 04:22

1 Answers1

1

Do you really need fractions? You can make them an arithmetic progression. For assignment $k$, the average weight is $\frac 1k$. Let the difference between terms be $d$. The weights range from $\frac 1k+\frac{(k-1)d}2$ to $\frac 1k-\frac {(k-1)d}2$. So if $k=15, d=0.005, \frac 1k \approx 0.066667$ and the weights range from $0.1016667$ down to $0.0316667$. Now you just have to pick a $d$ for each $k$ that makes you happy. Then you can convert these back to fractions if you want.

Community
  • 1
Ross Millikan
  • 374,822