# how to normalise these values

First of all, i don't know if the correct word is normalise or not, but I'll try to explain my issue.

I have a relationship between an object A and an object B equals 0.5 (max is 1)

you can consider this relationship as a vector that its length is 0.5

I have another vector between A and C and its value is 0.3

I have another vector between A and D and its value is 0.2

so these are the values that I have:

A -> B = 0.5
A -> C = 0.3
A -> D = 0.2


Now I want to give more weight to the relationships. In other words, I really care about A -> B a lot, but I don't care so much about A -> C

so I want to give weight to those numbers, the weights are as this:

the value of A -> B should be multiply by 4
the value of A -> C should be multiply by 1 (so no change)
the value of A -> D should be multiply by 2


my problem is that if i multiply A -> B by 4, then the result is 2, which is bigger than 1. all the values must be between 0 and 1.

my question is how (and what is the abstract equation) to normalise all these values in a correct way?

Regards

### update 1

In the example I gave to you, it was co incidence that the sum of the values is 1, that's not necessary at all

You normalise back to a sum of $1.0$ by dividing each new weight by the sum of all the new weights:

$\begin{array}{rrrrrrr}A\rightarrow B&=0.5&~~~&0.5 \times 4 =& 2.0&&2.0/2.7=0.74074\\A\rightarrow C&=0.3&&0.3 \times 1 =&0.3&&0.3/2.7=0.11111\\A\rightarrow D&=0.2&&0.2\times2=&\underline{0.4}&&0.4/2.7=\underline{0.14815}\\&&&&\underline{2.7}&&\underline{1.00000}\end{array}$

Your question is still kind of vague but it seems to me that you are looking for one of these:

• If it is important (and not just coincidence in your example) that all values sum up to $1$, you should take a look at Frentos answer.

• If the values are independent of each other (i.e. several can be equal to $1$), then you should divide all factors by the biggest factor. In your case, instead of multiplying by $4,1,2$, you should multiply by $1, \frac{1}{4}, \frac{1}{2}$.

Both methods will yield values which between $0$ and $1$.

• I didn't get this  then you should divide all factors by the biggest factor. In your case that would mean to multiply by ... could you make the example please ? Feb 8, 2016 at 15:02
• @MarcoDinatsoli Sorry, that sentence was kind of confusing. I edited my answer and hope it is clearer now. Feb 9, 2016 at 8:36
• Your first method won't give a sum of 1, except in case the original weights are equal. Feb 9, 2016 at 8:38
• @orion You are right, thanks. Obviously, Frentos had the right answer for that. Feb 9, 2016 at 8:48