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Suppose that Canada has a total of \$10 billion in \$20 bills in circulation, and each day \$40 million of these \$20 bills passes through one bank or another. A new harder-to-forge version of the \$20 bill is developed, and the banks replace the old bills with new ones whenever they can. How long does it take for the new bills to reach 90% of the total number of $20 bills in circulation?

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Since the tag is differential equations, we use a continuous model. Ten billion in $20$ dollar bills is $500$ million bills. Of these, $2$ million pass daily through a bank.

Let $x=x(t)$ be the number of new bills in circulation after time $t$. We will find an expression for $\dfrac{dx}{dt}$.

Every day, the fraction $\dfrac{2}{500}$ of bills in circulation passes though a bank. If there are $x$ new bills in circulation, then the number of old bills that pass through the bank is $$\frac{2}{500}(500000000 -x).$$ These are replaced. We conclude that $$\frac{dx}{dt}=\frac{2}{500}(500000000-x).$$ To avoid typing such large numbers, let $x=500000000y$. Then $\dfrac{dy}{dt}=500000000\dfrac{dx}{dt}$, and quickly we arrive at $$\frac{dy}{dt}=\frac{1}{250}(1-y).$$ Initially, $y=0$. We want to find out when $y$ reaches $0.9$. Not that the above differential equation looks very much like the one we get in Newton's Law of Cooling.

Make the change of variable $z=1-y$. Then $z$ is the proportion of old bills after time $t$. We get $$\frac{dz}{dt}=-\frac{1}{250}z.$$ This is the familiar differential equation of exponential decay. It has the solution $$z=z(0)e^{-t/250}.$$ We have $z(0)=1$, and want to find the time $t$ such that $z(t)=0.10$. So we solve the equation $$e^{-t/250}=0.1.$$ Take logarithms to the base $e$. We get $-\dfrac{t}{250}=\ln(0.1)$. I prefer to note that $\ln(1/10)=-\ln(10)$, which tells us that $$t=250\ln(10).$$ Finally, calculate. We get something like $t\approx 575$.

Remark: It is important to note that we were working with a model of reality. That model cannot be fully accurate. The round numbers supplied are obviously only approximate, and the total money supply may grow. And old $20$ dollar bills may be hidden in socks and circulate only every few years. And there are light days and heavy days at banks. And a new bill put into circulation today unlikely to return to the bank tomorrow. And we are using continuous modelling for a discrete situation. So it would be unreasonable to treat the answer of $575$ as anything more than a ballpark figure.

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Let's say that on say $n$ the proportion of new bills in the population is $0\leq P(n)\leq 1$. Then each day a fraction $f$ of all the bills go through the banks. $1-P(n)$ of those will be old bills, which will immediately be replaced with new. Therefore

$$P(n+1) = P(n) + f(1-P(n))$$

which can be rearranged to

$$P(n+1) = (1-f) P(n) + f$$

If we solve the homogeneous part, we find that $P(n)=A(1-f)^n$ for some constant $A$. Now if we guess that $P(n)=c$ solves the whole equation, we get

$$c = (1-f)c + f$$

from which you can deduce $c=1$. We know that initially there are no new bills in circulation (i.e. $P(0)=0$) so we have

$0 = 1 + A(1-f)^0 = 1 + A$

and hence $A = -1$, so the general solution is

$$P(n) = 1 - (1-f)^n$$

The question wants to know when this is equal to 90%. We have $f=4\cdot 10^7 / 10^{10} = 0.004$ and hence $1-f=0.996$, so

$$0.9 = 1 - 0.996^n$$

and hence $n=\log(0.1)/\log(0.996) = 574.5$ days.

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