Cells in a culture grow every day

We want to know the number of cells every day: \(\text{ncell(t)}\)

Here \(t\) is the time in days.

We start with an initial number of cells, that we call \(\text{initial}\)

Each day, the number of cells increases by **a factor**
\(\text{rate}\)

\[\text{ncell(t)} = \text{initial} \cdot \text{rate}^{t}\]

**How can we model it?**

We cannot see what happens when values are small

We can see better using a **logarithmic** vertical
scale

We need very little math: arithmetic, algebra, and logarithms

Just remember that if \(x=p^m\) then \[\log_p(x) = m\] For example \[\log_{10}(10000) = 4\]

If we use another base, for example \(q\), then \[\log_q(x) = \log_p(x) /\log_p(q)\] For example \[\begin{aligned} \log_2(10000) &= \log_{10}(10000)/\log_{10}(2)\\ \log_2(10000) &= 4/\log_{10}(2)\\ 13.28771 &= 4 / 0.30103 \end{aligned}\]

So if we use different bases, there is only a scale factor

The “easiest” one is *natural logarithm*

If \[x=\exp(m)\] then \[\log(x)=m\]

They only work with positive numbers. Not with 0

If \(x=p\cdot q\) then \[\log(x)=\log(p)+\log(q)\]

If \(x=a^m\) then \[\log(x)=m\log(a)\]

Basic linear model \[y=A+B\cdot x\] Exponential change (Initial value and growth Rate) \[y=I\cdot R^x\] Power law (Constant and Exponent) \[y=C\cdot x^E\]

Basic linear model \[y=A+B\cdot x\] Exponential: if \(y=I\cdot R^x\) then \[\log(y)=\log(I)+\log(R)\cdot x\] Power of \(x\): if \(y=C\cdot x^E\) then \[\log(y)=\log(C)+E\cdot\log(x)\]

The easiest way to decide is to

- draw several plots, placing
**log()**in different places, - see which one seems more like a straight line

For example, let’s analyze data from Kleiber’s Law

https://www.dry-lab.org/static/kleiber1947.txt

**Exercise:** Make all plots. Which plot seems more
“straight”?

The plot that seems more straight line is the log–log plot

Therefore we need a log–log model.

\[\log(\text{kcal})=β_0 + β_1 \cdot \log(\text{kg})\]

If \(\log(\text{kcal})=4.21 + 0.756\cdot \log(\text{kg})\) then \[\text{kcal}=\exp(4.21) \cdot \text{kg}^{0.756} =67.1 \cdot \text{kg}^{0.756}\]

Therefore:

- For a 1kg animal, the average energy consumption is \(\exp(4.21) = 67.1\) kcal
- The energy consumption increases at a rate of \(0.756\) kcal/kg.

*“An animal’s metabolic rate scales to the ¾ power of the
animal’s mass”.*

Google it

animal | kg | kcal | predicted |
---|---|---|---|

Mouse | 0.021 | 3.6 | 1.285 |

Rat | 0.282 | 28.1 | 3.249 |

Guinea pig | 0.410 | 35.1 | 3.532 |

Rabbit | 2.980 | 167.0 | 5.031 |

Cat | 3.000 | 152.0 | 5.036 |

Macaque | 4.200 | 207.0 | 5.291 |

Dog | 6.600 | 288.0 | 5.632 |

animal | kg | kcal | predicted |
---|---|---|---|

Goat | 36.0 | 800 | 6.915 |

Chimpanzee | 38.0 | 1090 | 6.955 |

Sheep ♂ | 46.4 | 1254 | 7.106 |

Sheep ♀ | 46.8 | 1330 | 7.113 |

Woman | 57.2 | 1368 | 7.265 |

Cow | 300.0 | 4221 | 8.517 |

Young cow | 482.0 | 7754 | 8.876 |

We want to predict the metabolic rate, depending on the weight

The independent variable is \(\text{kg}\), the dependent variable is \(\text{kcal}\)

But our model uses only \(\log(\text{kg})\) and \(\log(\text{kcal})\)

So we have to undo the logarithm, using \(\exp()\)

animal | kg | kcal | predicted |
---|---|---|---|

Mouse | 0.021 | 3.6 | 3.616 |

Rat | 0.282 | 28.1 | 25.762 |

Guinea pig | 0.410 | 35.1 | 34.185 |

Rabbit | 2.980 | 167.0 | 153.113 |

Cat | 3.000 | 152.0 | 153.889 |

Macaque | 4.200 | 207.0 | 198.458 |

Dog | 6.600 | 288.0 | 279.287 |

animal | kg | kcal | predicted |
---|---|---|---|

Goat | 36.0 | 800 | 1007 |

Chimpanzee | 38.0 | 1090 | 1049 |

Sheep ♂ | 46.4 | 1254 | 1220 |

Sheep ♀ | 46.8 | 1330 | 1228 |

Woman | 57.2 | 1368 | 1429 |

Cow | 300.0 | 4221 | 5001 |

Young cow | 482.0 | 7754 | 7157 |

A idea from 1965, by George Moore (Intel)

The simple version of this law states that processor speeds will double every two years

More specifically, “the number of transistors on a CPU would double every two years”

(see paper)

A ‘flops’ is a *floating point operation per second*

In simple words, is the number of multiplications per second that a computer can do

Cost of sequencing human genome

Months since Sept 2000

https://www.dry-lab.org/static/Transistor_count.txt

An article in the “London Review of Books”

He tells this story

- 1992 Russo-American moratorium on nuclear testing
- Just after the Cold War
- 1996 Computer simulations to design new weapons
- Needed more computing power than could be delivered by any existing machine

USA designed

*ASCI Red*, the first supercomputer doing over one teraflopIn 1997, ASCI Red did 1.8 teraflops

It was the most powerful supercomputer in the world until about the end of 2000.

*“I was playing on Red only yesterday – I wasn’t
really, but I did have a go on a machine that can process 1.8
teraflops.*

*“This Red equivalent is called the PlayStation3:
it was launched by Sony in 2005 and went on sale in 2006.*

**Red** characteristics

- the size of a tennis court
- used as much electricity as 800 houses
- cost US$55 million.

The PS3

- fits under the TV,
- runs off a normal power socket,
- costs less than £200.

*“[In 10 years], a computer able to process 1.8 teraflops went
from being something that could only be made by the world’s richest
government […], to something a teenager could expect [as a
gift].*

That was 15 years ago