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I have \$100000. I want to divide the amount proportionately to 300 employees. I want the money divided inversely from the top down. So, the top earner get \$0 and the lowest paid employee(s) get the most. If there are 10 employees at the lowest pay rate, they all get the same amount. Is there a formula or program I can create to create this?

Xoque55
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Roberto
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  • Two questions: 1) How is the distribution of ranked enployees? (v.g. Do you have 290 employees at each own ranking and the bottom 10 employees sharing a 291th rank?) 2) By inversely, do you have some pattern? (v.g. lineal, harmonic, geometrical?) – Carlos Eugenio Thompson Pinzón Dec 06 '13 at 15:34
  • Distribution is not linear. There may be 1 that makes 100000 so he gets 0 but there may be 5 that make 50000 and 3 that make 100. By inversely I mean the highest paid gets no increase and the lowest paid gets the most, going from 0 at the highest salary to the max. amount for the lowest paid. – Roberto Dec 06 '13 at 15:38
  • Okay, let's say we have tree employees, each one earns respectively $100,000 , $50,000 and $10,000. Now, you have a $9.000 bonus. In this simpler problem do you have an idea of how to distribute? $0; $3,000; $6.000? $0; $1.500; $7.500? $0; $4,000; $5,000? (each one of these sample distribution has it's own logic) – Carlos Eugenio Thompson Pinzón Dec 06 '13 at 15:50
  • Yes I understand what you're saying. So it cannot be solved linearly? I would want it linearly if possible. – Roberto Dec 06 '13 at 15:54
  • Actually the $0, $4,000, $5,000 example is an inverse linear distribution, taking as variable how less each employee has to get the top salary (respectively $0, $40,000, $50,000) and have a direct linear relationship between this variable and how much each one gets. – Carlos Eugenio Thompson Pinzón Dec 06 '13 at 15:58
  • You'll get the most sensible results if you open excel, fill column A with the number of people in each income group, fill column C with Ai*Bi and fiddle with column B until SUM(C:C) is exactly $1000000. – Karolis Juodelė Dec 06 '13 at 15:59
  • What is Ai and Bi? – Roberto Dec 06 '13 at 16:00
  • What if there are 10 making 40,000? – Roberto Dec 06 '13 at 16:01

1 Answers1

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According to conversation in the comment session, employees are ranked by the salary group and the distribution of the \$100,000 bonus should be distributed inversely. (If it is proportionally inversely, then top earner will get slightly more than \$0.)

So, let $a_i$ for $i=1,2,\dots,300$ be the salary of each employee, and $A=\max a_i$ be the salary of the top earner. One posibility is to have an inverse linear distribution using the variable $x_i=A-a_i$, and have each bonus $b_i$ proportional to $x_i$.

In this case we know that $b_i=\alpha x_i$, and that $$\sum_{i=1}^{300}b_i=100000.$$ Replacing $b_i$: \begin{align} \sum_{i=1}^{300}\alpha x_i&=100000 \\ \alpha\sum_{i=1}^{300} x_i&=100000 \\ \alpha\sum_{i=1}^{300} A-a_i&=100000 \\ \alpha\left(\sum_{i=1}^{300} A-\sum_{i=1}^{300}a_i\right)&=100000 \\ \alpha\left(300A-\sum_{i=1}^{300}a_i\right)&=100000 \end{align} Where $\sum a_i$ is sum of all salaries: the total wage package $W$. So, from hear we have: \begin{align} \alpha(300A-W)&=100000 \\ \alpha&=\frac{100000}{300A-W} \end{align} So the bonification of each employee will be: \begin{align} b_i &= \alpha x_i\\ b_i &= \frac{100000}{300A-W} x_i\\ b_i &= \frac{100000}{300A-W}(A-a_i)\\ \end{align}


If you prefer an inverse proportional distribution, the distribution variable would be $x_i=\frac1{a_i}$. However in this distribution the top earner will have a part of the bonus (although a smaller part than the other employees).

A fix to this distribution is to make $x_i=\frac1{a_i}-\frac1A$.


I have treated each salary independently, however if you have only a few possible salaries you can adopt a different strategy:

Let $A_1,A_2,\dots,A_n$ with $n\ll 300$ be the possible salaries and let's assume that $A_1>A_2>\dots>A_n$ (salaries are ordered).

Let $k_i$ be the number of employees that earn $A_i$.

So the total number of employees are $\sum_{i=1}^{n}k_i=300$ and the total wage package is $\sum_{i=1}^{n}k_iA_i=W$.

We define the distribution variable $X_i$ with some inverse criteria, v.g. $X_i = A_1-A_i$ (for the lineal distribution), $X_i=\frac1{A_i}$ (for the inverse proportional distribution), $X_i=\frac1{A_i}-\frac1{A_1}$ (for the shifted inverse proportional distribution), $X_i=i-1$ for a ranked distribution, etc.

Let $B_i$ be the bonus for each salary rank, and have it proportional to the distribution variable $B_i=\alpha X_i$

Now, you have: $$\sum_{i=1}^{n}k_iB_i=\alpha\sum_{i=1}^{n}k_iX_i=100000,$$ so $$B_i=\alpha X_i=X_i\frac{10000}{\sum_{i=1}^{n}k_iX_i}.$$

The result of $\sum_{i=1}^{n}k_iX_i$ (or $\sum_{i=1}^{200}x_i$) depends on your criteria for choosing $X_i$.

  • Instead of solving this mathematically you can use Excel, as suggested by @KarolisJuodelė. So write in cell A1 the salary $A_1$, in A2 write $A_2$ etc. In B1 write how many employees earn $A_1$, same for B2, etc. In C1 write $X_1$ according to the formula, v.g. =A1-A$1, if you copy this to C2 you will have =A2-A$1, etc. In D1 write =B1*C1. Copy down the column. Then at D301 have =SUM(D1:D300) and have C1 set to =C1*100000/D$301 (copy in the next cells) – Carlos Eugenio Thompson Pinzón Dec 06 '13 at 16:40
  • you mean E1 set to C1*10000/d$301 in las sentenece? – Roberto Dec 06 '13 at 17:10
  • Right. E1 set to =C1*100000/D$300. – Carlos Eugenio Thompson Pinzón Dec 06 '13 at 17:12
  • To play with other distributions of $X_i$ you can write other formulas in C1:C300 such as =1/A1, =1/A1-1/A$1, 0 (when dragged C2 will be set to 1, C3 to 2, etc.), or any arbitrary values. – Carlos Eugenio Thompson Pinzón Dec 06 '13 at 17:15
  • when i do this in excel the sum of bonuses (I made a column F) does not equal 100000 – Roberto Dec 06 '13 at 17:17
  • You should create F1 with =B1*E1, then the sum of F1:F300 should be 100000. That is because E1... contains how much you pay to each employee in that salary group, while F1... will be the sum of the payments to that salary group. – Carlos Eugenio Thompson Pinzón Dec 06 '13 at 17:18
  • I've rechecked it and it doesn't add up. It's close but not exact. – Roberto Dec 06 '13 at 17:24
  • Finally! I had typos in my Excel sheet. It worked great! – Roberto Dec 06 '13 at 20:02