A blog partially about the MA faction and partially about random things I think up. Enjoy! (the poll will never close, muahahaha)
11/04/2009
The Industrial Exponent
In this post I will explain a quality of industrial systems that I stumbled across on October 30, 2009. This quality is difficult to explain, so I will use examples with lots of equations. Keep a graphing calculator handy so you won't get confused. My first example is rather simple, but it is complicated enough to illustrate my quality.
Suppose you have a constant number of workers represented by w = 1
Each unit of workers can create a factory in one unit of time: f = wt
Each factory can create one unit of product in one unit of time: p = ft
Thus p = t^2
Those who are logical inclined will notice an error. Each factory must be completed before it can create product, therefore my equation for p should be something more like p= .5t^2 - .5t.
My second example is more complicated, so prepare to be slightly more confused.
Suppose you have one factory type a: a = 1
Each factory type a produces one factory type b per unit of time: b = at
Each factory type b produces one factory type c per unit of time: c = bt
Each factory type c produces one unit of product per unit of time: p = ct
After my reality check, this comes out to about p= (1/6)x^3-.5x^2+(1/3)x-.00000000002 (I don't know where the -.00000000002 came from, but my calculator spat it out) The significance of this is that the more steps you add to an industrial equation, the larger the exponent.
For the first example, let's look at how many units of product after the first several time units: t 1 2 3 4 5 6 7 f 1 2 3 4 5 6 7 p 0 1 3 6 10 15 21
The first two numbers in p are 0 and 1 and the rest of the numbers are the sum of the value of t-1 and the previous value of p. Or, you can look at it as the sum all previous values of t. The formula for that is the classic formula p = t * (t - 1) / 2 which I have told you a story about before: when adding up an even number of positive numbers, the sum of the first and last number that are being added up is a constant n+1, and there are n/2 of them. For example, 1+6=2+5=3+4=7, and 7*(6/2)=21. This example's formula includes "t-1" instead of "t+1" because we don't include the number t itself.
If you rearrange the terms of the formula we get p = (t^2)/2 - t/2 which is exactly what you came up with. Good work!
After googling for that sequence of numbers, I find that the formula is a simple example of the general concept in math called Triangular Numbers: http://en.wikipedia.org/wiki/Triangular_number
Please show more work with the second example to verify if it is correct. I wouldn't be surprised if it were another kind of Triangular Number formula. I am suspicious of that tiny number your calculator spat out, it could just be a rounding error.
Ah, and here I thought you had deduced the formula directly. Plugging it into the calculator like that is cheating a little, I think. That's how you get those weird tiny constants and don't really have to understand what's going on. I'm not sure that you gave enough input to the calculator to be able to figure out a good formula. Did you only plug in those two points?
Scott here! So the second set would be "Tetrahedral Numbers," where we have progressively larger right tetrahedra evenly filled with dots? Thus the non-simplified formula would be t(t-1)(t-2)/6 would it not?
For the first example, let's look at how many units of product after the first several time units:
ReplyDeletet 1 2 3 4 5 6 7
f 1 2 3 4 5 6 7
p 0 1 3 6 10 15 21
The first two numbers in p are 0 and 1 and the rest of the numbers are the sum of the value of t-1 and the previous value of p. Or, you can look at it as the sum all previous values of t. The formula for that is the classic formula
p = t * (t - 1) / 2
which I have told you a story about before: when adding up an even number of positive numbers, the sum of the first and last number that are being added up is a constant n+1, and there are n/2 of them. For example, 1+6=2+5=3+4=7, and 7*(6/2)=21. This example's formula includes "t-1" instead of "t+1" because we don't include the number t itself.
If you rearrange the terms of the formula we get
p = (t^2)/2 - t/2
which is exactly what you came up with. Good work!
After googling for that sequence of numbers, I find that the formula is a simple example of the general concept in math called Triangular Numbers:
http://en.wikipedia.org/wiki/Triangular_number
Please show more work with the second example to verify if it is correct. I wouldn't be surprised if it were another kind of Triangular Number formula. I am suspicious of that tiny number your calculator spat out, it could just be a rounding error.
Dad
My second example is basically an extension of the first.
ReplyDeletea=1 1 1 1 1 1 1
b=1 2 3 4 5 6 7
c=0 1 3 6 10 15 21
p=? 0 1 4 10 20 35
I plugged the points, (2,0)-(7,35) into the calculator and it spat out the formula I gave you.
Ah, and here I thought you had deduced the formula directly. Plugging it into the calculator like that is cheating a little, I think. That's how you get those weird tiny constants and don't really have to understand what's going on. I'm not sure that you gave enough input to the calculator to be able to figure out a good formula. Did you only plug in those two points?
ReplyDeleteDad
No, of course not. I plugged in about 6 points, and the model exactly intersected all of them (r=1)
ReplyDeleteScott here!
ReplyDeleteSo the second set would be "Tetrahedral Numbers," where we have progressively larger right tetrahedra evenly filled with dots? Thus the non-simplified formula would be
t(t-1)(t-2)/6
would it not?
Right. Good job thinking that up.
ReplyDelete