# histogram of data

may be this very easy. because of lack of maths skills I'm posting this.

ex: a has 710444 computers and 23461 cables b has 1047577 computers and 39211 cables...........

a   710444  23461
b   1047577 39211
c   247186  8792
d   518264  13886
e   116696  2045
f   1044267 42573
g   470177  18562
h   499851  16525
i   444441  10608
j   981386  40677
k   732669  27183
l   1143795 37862
m   305696  12324
n   106570  4422
o   381875  12230
p   1478656 31646
q   327575  9624


How can I calculate who holds highest number of cables. generally "f". But as you can see bias in number of computers and thus difference in the cables. Is there any way to calculate remove this bias calculate who has highest cables ?

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"710444 computers"? Is that supposed to be remotely realistic? –  lhf Aug 19 '11 at 13:45
You don't calculate who has the highest number of cables; you just find it by comparing the number of cables each one has. The number of computers does not enter into this, does it? –  lhf Aug 19 '11 at 13:47
Maybe the OP is thinking about the number of cables per computer ... ? –  AlexPof Aug 19 '11 at 14:10
@lhf- It is estimated that Google has 900,000 machines alone. See "GROWTH IN DATA CENTER ELECTRICITY USE 2005 TO 2010" by Koomey. So, certainly reasonable. –  lewellen Aug 19 '11 at 14:45

You can divide the number of cables (second column) by the number of computers (first column). That gives you the ratio $$\frac{\text{cables}}{\text{computers}}.$$

Doing the division tells you the number of cables per computer.
Here are couple of results, so that you will know you are doing things right.

(a) $0.033023$

(b) $0.0374302$

(c) $0.0355684$

(d) $0.0267933$

So far, (b) has the highest ratio of cables to computers. The division compensates for the disparities in the numbers of computers. The numbers we get are kind of ugly looking. So you might want to multiply my numbers by $1000$, round them suitably, and label the new column "Number of cables per $1000$ computers." It looks as if rounding to $1$ decimal place would be plenty good enough.

Another approach is to divide the number of computers by the number of cables. If "big" ratios are good with the first approach, then "small" ratios are good with the second approach. For example, in (a) we would have roughly $30.3$, for $b$ roughly $26.7$, and so on. This approach gives you a sorting opposite to the order given by the first approach.

Plot all the points $(x,y)$ where $x$ is the number of computers someone has and $y$ is the number of cables that person has, that is, plot the points $(710444, 23461)$ and $(1047577, 39211)$ and so on. Do they look like they lie more-or-less on a straight line? If they do, there's a method called "least squares" that will find you the line that comes closest to going through all the points. Some points will be above that line, some will be below. The point the farthest above the line (in techspeak: the point with the greatest residual) corresponds to the guy who has the most cables in comparison to the number of computers.
If the points don't look like they mostly lie on a line, then maybe they look like they lie on an $x^2$ curve, or a $\sqrt x$ curve, or a logarithmic curve, or an exponential curve, or.... Whatever, there's a variation of least squares that will handle it, give you a best fit, and leave you looking at the residuals to find the guy with the most cables, relatively speaking.