# calculate value/price with multiple uncomparable values

how does one calculate value/price with several values but only one price and the values are incomparable between them and have different "weight" on the total value.

example:

product A:

price=100 value of characteristics: x=10; y=100; z=1000;

product B:

price=105 value of characteristics: x=11; y=100; z=1000;

product C:

price=110 value of characteristics: x=12; y=100; z=1000;

And characteristic x is (I.E.) 2 times as important as the other characteristics so an increase of 10% in x doesn't make the total value go up by one third of 10% it would go up by 2 thirds if I'm not mistaken.

how do I calculate the correct value/price for each?

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[Due to your phrasing, I'm not certain of what you're asking for.]

$$V = \left(\frac {a}{30} + \frac {b}{300} + \frac {c}{3000}\right) \times 100 ?$$

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no, definitely not x100. 100 is the price so it should be the value divided by price. but I don't know how to calculate the value. sorry for my phrasing. althuogh now I realise the price thing is unimportant I'll edit the question to make it simpler. –  Jo Rijo Jan 28 '13 at 5:40
From your now-deleted comment (probably my fault for editing mine — sorry!), it seems that what you're looking for is a function such that increasing any of $x$, $y$, or $z$ by a fixed percentage always increases the value/price $v$ by a corresponding percentage, no matter what the relative magnitudes of $x$, $y$, and $z$. In that case, you can use a function of the form $$v=(x\cdot y\cdot z)^k$$ for some exponent $k$. For example, when $k=1$, increasing any of $x$, $y$, or $z$ by $10\%$ increases the value by the same $10\%$. When $k=\frac13$, if you increase all three of $x$, $y$, and $z$ by $10\%$, the value will increase by $10\%$. It's up to you to choose which $k$ makes sense for your purposes. The examples in your question, where the increase in $v$ is roughly half the increase in $x$, correspond to $k$ of about $\frac12$.
If you want different weights, you can do $v=x^i\cdot y^j\cdot z^k$ for different exponents $i$, $j$, and $k$. –  Rahul Jan 29 '13 at 0:00