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Students are partitioned into groups of equal size $m$ similar to Stable roommates problem except $m$ can be any natural number greater than 1. Each student ranks the other students in strict order of preferences.

For each person $i∈\{{1,…,n}\}$, $i$ can rank all other students from the most preferred to the least preferred. These preferences can be thought of as $p_{ij}$, where $p_{ij}$ is rank (1 highest) of $j$ in $i$'s ordering.

I'm trying to calculate the weighted rate of satisfaction a person would get mathematically from being in a group that contains his preference ranked at n. The weighted rate could be calculated with a function $f(m, p_{ij}) = ?$

The satisfaction rate is a number between 0 and 1 that multiplied by 100 could be converted to percentages.

Each preference should be weighted differently. If the student's group contains one of his preferences ranked at for example $p_{ij}=3$ he wouldn't be satisfied as much as a preference ranked at $p_{ij}=1$.

The ideal group for a person would consist of his preferences ranked at $p_{ij}=1$ to $p_{ij}=m-1$ (amount of students in a group except for $i$ himself) this would give him a satisfaction rate of 1 (meaning fully satisfied), or written mathematically:

$$\sum_{p_{ij}=1}^{m-1} f(m,p_{ij}) =1$$

So what could the definition of function $f(m, p_{ij})$ be?

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    I am not sure what you are trying to do. Let me try and you correct me. You have $n$ students. You want to partition the students into groups, where each group has size $m$ (so that $n$ is divisible by $m$). For each person $i\in\{1,\ldots,n\}$, $i$ can rank all other students from the most preferred to the least preferred. These preferences can be though of as $p_{ij}$, where $p_{ij}$ is rank ($1$ highest) of $j$ in $i$'s ordering. Now satisfaction of $i$ from partition $\mathbf{P}$ depends only on which group $i$ is in? – 2017-02-07
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    The number of students in total is irrelevant in this case. The satisfaction reflects how satisfied the student is with his current group, so only the group the student is placed in is relevant to the satisfaction rate of that person. Each student in a student's group is placed at n in his preference list and has a value that is included in his total satisfaction rate. What you said up to "Now satisfaction of i from partition P depends only on which group i is in?" was correct, but I don't quite understand that line. – 2017-02-07
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    I am trying to understand whether satisfaction of $i$ depends only on the group $i$ is in or on everything else (composition of other groups etc.) Now about the satisfaction. Let $p_{ij}$ have the meaning I explained above and let $i$ be in group $G$. What is wrong with $i$s satisfaction being $\sum_{j\in G\setminus\{i\}}p_{ij}$? Or, when you set $p_{ii}=0$, $\sum_{j\in G}p_{ij}$? – 2017-02-07
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    Partition is exactly what it says. E.g., partition of $\{1,2,3,4\}$ is $\{\{1,2\},\{3,4\}\}$. – 2017-02-07
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    If I understand you correctly then it only depends on the group i is in. – 2017-02-07
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    What do you mean by j∈G Preference j in group G – 2017-02-07
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53184/discussion-between-martin-and-jan). – 2017-02-07
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    I do not have time to chat now. $i$ and $j$ are indexes of people. $p_{ij}$ is rank of $j$ in $i$s preference ordering. That is, $p_{ij}=1$ means that $i$ ranks $j$ as the person $i$ would like to be most with. Your original post calls this $n$ but I want to have notation for who's this $n$ is ($i$) and towards whom ($j$). – 2017-02-07
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    Okay, so you suggest the satisfaction is the sum of preferences (sum of p_ij) in the group? How do I convert from that to rate (normalized value between 0 and 1) or percentages? – 2017-02-07
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    Divide my $p_{ij}$ by the number of people, for example. – 2017-02-07
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    Divide by the group sizes? Also that would give a low number a lower rate, so p_ij should be amount of students on the preference list + 1 - p_ij, right? Say the group sizes are 5 and a student gets his ideal group could you please show me a calculation from your idea that would amount to 1 as I described in my original answer? – 2017-02-07
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    I am not following. I said 'divide by the number of people', not by group size. Problem with higher number means worse can be solve just by multiplying by $-1$. Let me explain why I wrote my first comment: I was hoping you would revise your question to a form anyone can understand. All the comments here are just we going in circles of misunderstanding. My sense is that if you do not understand your own question precisely, then you cannot ask it clearly. Write your question, including all details, in a notation that makes sense. – 2017-02-07
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    Okay, I edited my question to my full capability. Is anything unclear now? – 2017-02-07
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    Good, it is still very messy but better. Now I believe what you are after follows when you transform the $p_{ij}$ to $n+1-p_{ij}$, calculate satisfaction of $i$ being in a group by summing $n+1-p_{ij}$ by going over index $j$ for the players that are in the same group and divide the result by $\sum_{i=1}^{m}n+1-i$, which is the maximum satisfaction a person can have. – 2017-02-07
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    I don't quite understand "summing $n+1−p_{ij}$ by going over index $j$ for the players that are in the same group". Could you try writing that with sigma notation please? – 2017-02-07

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Let me give an example. There are $n=4$ people and group of size $m=3$. The rankings are as follows (row is $i$, column is $j$, that is, those are the $p_{ij}$): $$\begin{array}{cccc}0&1&2&3\\3&0&2&1\\3&2&0&1\\1&2&3&0\end{array}$$ That is, person $1$ wants to be with $2$ most and $4$ least. Person $3$, say, wants to be with $4$ most and $1$ least.

Suppose the group is $G=\{2,3,4\}$. Then the satisfaction for $i=2$ is $\sum_{j\in G}p_{2j}=1+2=3$. The worst satisfaction a player can get from $G$ is $\sum_{j=1}^{m-1}m+1-i=3+2=5$. The best satisfaction a player can get from $G$ is $\sum_{i=1}^{m-1}i=1+2=3$. Denoting these by $S$, $W$ and $B$ respectively, $\frac{W-S}{B-S}$ is a measure that has higher numbers for happier people and ranges on $[0,1]$.

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    I understand what you propose and it's exactly what I want, thank you! Simple and works perfectly. But I don't understand your example and there is a misunderstanding. First of all $n$ in your calculations should be $m-1$ (excluding $i$ himself) and second your final calculation claims you should iterate over the whole group until $m$ but $i$ can't put himself on his own preference list, so it should instead iterate until $n$ – 2017-02-07
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    From what I understand the table says $i=1$ has the preferences $\{0,2,2\}$ but you say person 1 wants to be with 2 most and 3 least. – 2017-02-07
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    The $p_{ij}$ entry in the table is ranking how $i$ wants to be with $j$. $p_{11}=0$ is irrelevant and set to zero so that in the sums below I can sum over $p_{ii}$ as well. Hence $p_{12}=1$, $p_{13}=2$ and $p_{14}=3$ means $1$ wants to be with $2$ most and $4$ least (sorry for the typo, corrected now). – 2017-02-07