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Wizard
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April 18th, 2020 at 3:34:01 AM permalink
A microbe, let’s call it Covid-20 can spawn a new microbe at any time. The probability of a particular microbe spawning at any given time is always the same, regardless of the time since the last spawning. The average time between spawnings from the same microbe is one day.

Once a microbe enters your lungs, what is the expected number of microbes will you have after seven days?
Last edited by: Wizard on Apr 18, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Wizard
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April 18th, 2020 at 11:51:38 AM permalink
Does anyone not understand what's being asked?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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April 18th, 2020 at 12:04:31 PM permalink
Quote: Wizard

Does anyone not understand what's being asked?


The only thing I don't understand is, what is meant by "an average rate of one per day." Average in terms of what?

Even if you say, "If there are an infinite number of them created simultaneously, the mean time before they reproduce is 24 hours," what is the curve of time to probability of reproducing (or, alternatively, not reproducing) by that time?
IndyJeffrey
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DJTeddyBear
April 18th, 2020 at 12:07:55 PM permalink
The easy guess is 2^6=64 microbes -- but that sounds too obvious.

Phrases that cause some ambiguity for me: "average rate", "same rate", "memory-less property"

I should have taken more mathematics classes in college.
charliepatrick
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April 18th, 2020 at 1:25:21 PM permalink
Quote: Wizard

Does anyone not understand what's being asked?

Yes!

If a microbe always creates another one 24 hours after it was born, then clearly the number of microbes just multiplies by 2 every day. After 7 days there are 2^7=128 (assuming we're talking Monday 0h01 to Sunday 25h59).

However suppose a microbe either creates another one 12 hours or 36 hours, then the ones born earlier have a chance of multiplying earlier, which has a much bigger effect on the growth rate than the ones born later (average about 1154). As a silly idea suppose a microbe can either replicate after 1 minute (P=1/9) or about 8 days (Pr=8/9), then the average is 1 day but on average a new one will be born every 9 minutes.

btw I don't know how to work it out (I'm too lazy to do integrals etc.) if, as I suspect, it's an exponential distribution.
DJTeddyBear
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April 18th, 2020 at 2:41:48 PM permalink
Quote: Wizard

Does anyone not understand what's being asked?

Me.

I’m sure the answer isn’t 2^6=64.....
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Wizard
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April 18th, 2020 at 4:12:36 PM permalink
I said the AVERAGE time between spawninngs is one day. That implies is not is always exactly one day. Otherwise, the answer would indeed be 64.

I just changed the wording as follows:

Quote:

A microbe, let’s call it Covid-20 can spawn a new microbe at any time. The probability of a particular microbe spawning at any given time is always the same, regardless of the time since the last spawning. The average time between spawnings from the same microbe is one day.

Once a microbe enters your lungs, what is the expected number of microbes will you have after seven days?



I'm trying to ask this without using mathematical jargon.

The time between spawnings is similar to the time between royal flushes in video poker if you averaged getting one per day. On any given day you might get zero or you could get 10, or anything. The difference is that with royal flushes there are only so many opportunities to get them in a day, about 40,000, but with the microbe there is no limit and they can happen anytime, even a nano-second after the last spawn.
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gordonm888
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April 18th, 2020 at 4:27:48 PM permalink
If N is the number of microbes at any time t, the instantaneous growth rate is given by:

dN/dt = + N/tm

where tm is the mean time to double/spawn

Integrating between any arbitrary zero time, No and a later time t gives us:


N = No eλt

where λ = 1/tm

So, starting with 1 microbe, after 7 days there should be e7 = 1096.6 microbes approximately.
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Wizard
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April 18th, 2020 at 5:17:32 PM permalink
Quote: gordonm888

If N is the number of microbes at any time t, the instantaneous growth rate is given by:

dN/dt = + N/tm

where tm is the mean time to double/spawn

Integrating between any arbitrary zero time, No and a later time t gives us:


N = No eλt

where λ = 1/tm

So, starting with 1 microbe, after 7 days there should be e7 = 1096.6 microbes approximately.



I agree!

Here is my solution (PDF)

Gordon, I would be interested in your thoughts on how to word the problem in plain simple English. I am having a hard time getting across the point how the microbes grow at an exponential rate.
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gordonm888
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April 18th, 2020 at 5:49:47 PM permalink
Quote: Wizard

Quote: gordonm888

If N is the number of microbes at any time t, the instantaneous growth rate is given by:

dN/dt = + N/tm

where tm is the mean time to double/spawn

Integrating between any arbitrary zero time, No and a later time t gives us:


N = No eλt

where λ = 1/tm

So, starting with 1 microbe, after 7 days there should be e7 = 1096.6 microbes approximately.



I agree!

Here is my solution (PDF)

Gordon, I would be interested in your thoughts on how to word the problem in plain simple English. I am having a hard time getting across the point how the microbes grow at an exponential rate.



I thought you worded it almost perfectly. I immediately recognized from your write-up that this problem was identical to radioactive decay of an unstable atom, except that it involves growth -hence exponential growth.
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Wizard
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April 18th, 2020 at 7:40:28 PM permalink
Quote: gordonm888

I thought you worded it almost perfectly. I immediately recognized from your write-up that this problem was identical to radioactive decay of an unstable atom, except that it involves growth -hence exponential growth.



Thank you! At least one person understood what was being asked. I have changed the wording several times so am not sure which version you approve of. Here is how I have it now:

Quote:

A microbe, let’s call it Covid-20 can spawn a new microbe at any time. The probability of a particular microbe spawning at any given time is always the same, regardless of the time since the last spawning. The average time between spawnings from the same microbe is one day.

Once a microbe enters your lungs, what is the expected number of microbes will you have after seven days?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ssho88
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April 18th, 2020 at 7:42:21 PM permalink
Quote: Wizard

Quote: gordonm888

If N is the number of microbes at any time t, the instantaneous growth rate is given by:

dN/dt = + N/tm

where tm is the mean time to double/spawn

Integrating between any arbitrary zero time, No and a later time t gives us:


N = No eλt

where λ = 1/tm

So, starting with 1 microbe, after 7 days there should be e7 = 1096.6 microbes approximately.



I agree!

Here is my solution (PDF)

Gordon, I would be interested in your thoughts on how to word the problem in plain simple English. I am having a hard time getting across the point how the microbes grow at an exponential rate.




I agree with your solution.

The question clearly stated that a microbe spawns at rate rate of ONE per day, meaning it will be DOUBLE IN A DAY, and it should be 1, 2, 4, 8....... The question now is how to explain it to general people(in layman's term) that it should be 1, 2.718, 2.718^2, 2.718^3....and NOT 1, 2, 4, 8....... I find it is very HARD to explain it.

General people will argue that DOUBLE means 2 and NOT 2.718 ! LOL
Ace2
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April 18th, 2020 at 8:14:31 PM permalink
Quote: ssho88

The question clearly stated that a microbe spawns at rate rate of ONE per day, meaning it will be DOUBLE IN A DAY, and it should be 1, 2, 4, 8....... The question now is how to explain it to general people(in layman's term) that it should be 1, 2.718, 2.718^2, 2.718^3....and NOT 1, 2, 4, 8....... I find it is very HARD to explain it.

General people will argue that DOUBLE means 2 and NOT 2.718 ! LOL

Many people would understand in terms of compound interest. One dollar that earns 100% over 10 years without compounding becomes $2.00. Compounded once per year it becomes $1.00 * (1 + 1/10)^10 = $2.59. Compounded continuously it becomes $1.00 * (1 + 1/inf)^inf = e = $2.72
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ssho88
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April 18th, 2020 at 8:21:16 PM permalink
Quote: Ace2

Many people would understand in terms of compound interest. One dollar that earns 100% over 10 years without compounding becomes $2.00. Compounded once per year it becomes $1.00 * (1 + 1/10)^10 = $2.59. Compounded continuously it becomes $1.00 * (1 + 1/inf)^inf = e = $2.72



Not agree. 1, 2, 4, 8 . . .and 1, 2.718, 2.718^2, 2.718^3......BOTH are compound...
IndyJeffrey
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April 18th, 2020 at 8:26:28 PM permalink
Quote: ssho88

General people will argue that DOUBLE means 2 and NOT 2.718 ! LOL



"General people"?

I understand compound interest. I don't understand it in the context of the original question.

And, yes, I should have taken more mathematics in college.
Ace2
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April 18th, 2020 at 8:33:18 PM permalink
Quote: ssho88

Not agree. 1, 2, 4, 8 . . .and 1, 2.718, 2.718^2, 2.718^3......BOTH are compound...

Yes, both are compound. 1, 2, 4, 8 is 100% growth compounded ONCE per period. 1, e, e^2, e^3 is 100% growth compounded INFINITELY per period...call it a trillion times per period if that’s easier to grasp.

Continuous growth is the same as continuously compounded interest
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ssho88
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April 18th, 2020 at 8:40:47 PM permalink
Quote: Ace2

Yes, both are compound. 1, 2, 4, 8 is 100% growth per period compounded ONCE per period. 1, e, e^2, e^3 is 100% growth per period compounded INFINITELY per period...call it a trillion times per period if that’s easier to grasp.

Continuous growth is the same as continuously compounded interest




Agree
Ace2
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April 18th, 2020 at 8:52:22 PM permalink
Incidentally, e was “discovered” in 1683 by Jacob Bernoulli while finding a solution for continuously compounded interest.

I find it interesting that the most important constant in math was discovered quite recently in human history.
Last edited by: Ace2 on Apr 18, 2020
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Wizard
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April 18th, 2020 at 9:04:33 PM permalink
Quote: ssho88

The question clearly stated that a microbe spawns at rate rate of ONE per day, meaning it will be DOUBLE IN A DAY,



It said the average time to double was a day.
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ssho88
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April 18th, 2020 at 9:16:35 PM permalink
Quote: Wizard

It said the average time to double was a day.



Meaning it can be doubled in nano sec, in 3 seconds, in 1.2 days......I am not good at English, I may not fully understand your question .
Wizard
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April 18th, 2020 at 9:40:14 PM permalink
Quote: ssho88

Meaning it can be doubled in nano sec, in 3 seconds, in 1.2 days......I am not good at English, I may not fully understand your question .



Yep.
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IndyJeffrey
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April 19th, 2020 at 6:59:33 AM permalink
OK. So my problem is e is irrational and I am a rational kind of guy.

Would it be fair to say that failing to factor in the "average per day" is what tripped up this general person? Using e in an expected-value problem allows one to factor in the possibility that sometimes multiple will be spawned in a day and sometimes zero will be spawned in a day; am I close?
Wizard
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April 19th, 2020 at 3:26:22 PM permalink
Quote: IndyJeffrey

Would it be fair to say that failing to factor in the "average per day" is what tripped up this general person?



That seems to have been the stumbling block, but I don't see why so many people interpreted an average of one per day to be exactly one per day.

Quote:

Using e in an expected-value problem allows one to factor in the possibility that sometimes multiple will be spawned in a day and sometimes zero will be spawned in a day; am I close?



I'm not sure I see your point. e pops up because to get the answer you have to integrate 1/x.
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