Poll : Hire all for 2 possibilities

[paulwoman]

For the time being, "engage all" engages people from the highest performers to the lowest performers in one go.

In this case, it would start by engaging the person at 85, 77 and so on.

I propose to propose the reverse. So you start at the bottom.
Like: you hire the one at 5, at 8, 20, etc.

There would then be the possibilities:

Hire the most competent
Hire the least competent
Hire an average

The average could be : If there is 11 people, hire 3

1. 100
2. 95
3. 80
4. 79
5. 78
6. 70
7. 60
8. 64
9. 50
10. 45
11. 40

Most competent :

• Number 1, 2, 3

Leas competent

• Number 11,10,9

The average v1 :

• (100+95+80+79+78+70+60+64+50+45+40)/11) = 69
  • 1rs : The closer that 69 = 70
  • 2nd : the less competent : 60
  • 3e : the more compétent : 78

The average v1 :

• There is 11 people. 11/2 = 5,5
  
  • 1rs and 2nd : The closer that 5,5= --->5 and 6 --> people number 5 ( 78) and 6 (70)
  • 3th : 5-1 = 4 (79)
  Note : if there is more to hire, it could be
  
  6+1 = 7

Note :