# Pow 15

Pages: 5 (1429 words) Published: April 2, 2013
POW 15 Growth of rat populations Problem Statement This problem is composed of the growth of a rat population over the course of one year from 2 rats. Four assumptions for this problem were made: Each new liter is composed of 6 rats; 3 males, 3 females. The original pair give birth to 6 rats on the first day and then ever 40 days after. There is a 120 day "gestation" period before a newborn rat can reproduce. After this gestation period the rats will give birth every 40 days. Also over the course of a year no rats die. The key question is how many rats are there after one year. Process In order to solve this problem I first focused on getting an answer for the number of rats. Originally I had planned to use a large sheet of poster board paper to draw out the family tree. But quickly realized that this would be quite impractical. I then settled on a similar but more efficient method. In this method I focused on the 10 liters that the original parents produced. The 10 liters comes from the fact that there is 365.25 days in a year and the parent liter produce every 40 days from day zero. Meaning that they will have a liter on day 0, 40, 80, 120, 160, 200, 240, 280, 320, and 360. I started on the day 360 liter and worked backwards. After finishing this method I compiled the data in a spreadsheet. From which I was able to sum each interval of rat births. The data collected in family tree method is shown in the following data table: Couple: Original D0 D40 D80 D120 D160 D200 D240 D280 D320 D360 Sum 0 2 6 40 80 120 160 200 240 280 320 360 Total Sum after birth 2 2 6 8 6 14 6 20 24 44 42 86 60 146 132 278 258 536 438 974 6 834 1808 6 1808

6 6 18 6 18 18 18 18 18 72 18 18 18 126 72 18 18 180 126 72 18 396 180 126 72 836 438 258 132

6 6 6 18 18 18 54 6 18 18 36 6 18 18 6 6

I then found the pattern that emerged in the upper data table interesting. So I decided to look into it further. At first I looked at the graph of the sum after birth to see what kind of relationship it was.

Sum vs Day
2000 1800 1600 1400 1200 1000 800 600 400 200 0 0

50

100

150

200 Series 1

250

300

350

400

As clearly shown the growth is exponential. I then tried to get a model for this data but could not get anything that fit it perfectly, which is what I was aiming for. So I moved on to looking for relationships in the data. I looked at the ratios and differences between the days and the sums, and rats born. But was unable to come up with anything. I then tried to look at other relationships between sterile rats, reproducible rats, and the sum before and after birth. Sum after birth # rats reproducable sum before birth # sterile 2 8 2 2 0 14 2 8 6 20 2 14 12 44 8 20 12 86 14 44 30 146 20 86 66 278 44 146 102 536 86 278 192 974 146 536 390 1808 278 974 696

Solution My solution for this problem was 1808 rats at the end of 1 year. I believe that this solution is correct because of the pattern noticed when looking at the compiled data table. Also because after creating the table shown above (last page) it was logical that that was how it worked out. Through the course of working on the problem I looked at various ways of solving and trying to find patterns and became familiar enough with the growth of the rats that my answer appears to be completely logical with no questions or discrepancies left open. Generalizations In order to generalize this problem I first looked into jumping straight to an explicit formula but quickly found that doing so would be highly improbable. So I then looked into a recursive function that would model the population. Looking at the growth logically I noticed that the first 3 dates 6 rats were due. Then on the 4th date (120) the number of rats born were going to be 6+ the number born 120 days earlier, or 3 dates back. Then if you continue this a recursive function emerges; such that:

2

U0 = 6 U1 = 6 U2 = 6 U n = (3(U n − 3 ) + 3(U n − 4 ) + 3(U n − 5 ) + ... + 3(U n − n ) )+ 6

What the...

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