# Using A Finite Resource

We will develop the algebra of what happens during unconstrained growth in resource utilization – growth not limited by outside influences, only by internal circumstances.

click for our discussions about our natural resources

Given a fixed-sized resource – if we spend it, how long will it take to completely use it up?  We will look at a particular situation.  The result is that if usage of the reserve increases at a given rate, we will get exponential growth.  Exponential growth can be demonstrated by the looking at the Chess Board Fable,   or the Rabbit growth problem. (In the menu bar below the Blog’s Banner picture, click on PDF Resources and select the one you would like to read.)   We use general algebra here and it is is a bit messy, but using calculus ideas, the solutions come quickly.

The basic results: Q is quantity in reservoir, Y0 is lifetime of reservoir (years), r is annual rate of growth in withdrawal, m is number of withdrawals/yr (withdraw in steps), t is time (yr) to deplete reservoir to Q = 1 – f.

 [Eqn 1] Q(t) = Q0 — $quad big( frac{Q_0}{r Y_0} big)$ $big( (1+i)^{mt+1} -1big)$ Steps [Eqn 2] t = $frac{1}{m} big( frac{ln (1+r Y_0 f)}{ln (1+r/m)} - 1 big)$ Steps [Eqn 3] t = $frac{1}{r}$ $ln(1 + r Y_0 f)$ Continuous

### 1. Example: Spending a hoard

A 20 year old Arizona prospector finds a cache of a 30,000 one-oz gold coins. He uses straight line planning so that at the rate he will spend his treasure, it will last more than his life span.

•  Spend 300 coins per year …  = 30,000/100
•  Monthly, he gets 25 gold coins …  = 300/12

In the first month, he spends the 25 coins with ease, and the second month he wants just a little tiny bit more. No worry! The amount gone from his huge reservoir is negligible …

By the end of the first year, he overspent only 5% more than expected (1 and quarter coins out of a reservoir that will to last longer than his entire life).  Over the succeeding years, his spending continues to grow, each year 5% more than the preceding one.

Summary of his plan:

 Model Reservoir quantity: Q0 = 30,000  coins Reserve to last: Y0 = 100 years Withdrawal cycle: m =   12/yr … monthly cycle Usage allotment: U0 = 25 coins/month = $quadfrac{Q_0}{m cdot Y_0}$ Growth rate: rate = 5% /yr  … annual rate Cycle rate: i = 0.05/12  …  rate/cycle

### 2. Question: How do the withdrawals work?

Let’s look at the first several months of his usage of the coin hoard.  The key factor is that the Amount A withdrawn at any cycle is a bit bigger than the previous one. The difference depends on the ever-growing size of the last withdrawal.  The cycle rate i controls the proportional growth in the withdrawn amount.

 Month Coins Withdrawn Algebra form First month k = 1 25 coins, initial amount taken A1 = U0 =$frac{Q_0} {m cdot Y_0}$ as planned Second month k = 2 last Amount + rate· last Amount 25 + (0.05/12)×25 ≈ 25.1 Use calculator exact amount A2 = A1 + i A1 A2 = A1(1+ i) Third month k = 3 last Amount + rate· last Amount 25.1 + (0.05/12)×25.1 ≈ 25.21 Use calculator exact amount A3 = A2 + i A2 A3 = A2(1+ i) A3 = A1(1+ i)2 Fourth month k = 4 last Amount + rate· last Amount 25.21 + (0.05/12)×25.21 ≈ 25.31 Use calculator exact amount A4 = A3 + i A3 A4 = A3(1+ i) A4 = A1(1+ i)3 Month N k = N Amount withdrawn during Nth withdrawal follows the same pattern AN = AN-1+ i AN-1 AN = AN-1(1+ i) AN = A1(1+ i)N-1

Here is his utilization of the coin hoard: On the average, he removes each month a bit more than the previous withdrawal because what he takes out depends on the size of his last withdrawal.  The form of the Nth withdrawal is: AN = A1 (1+ i)N-1.  This is very similar to the compound interest investment formula.

### 3. Amount Remaining in reservoir during depletion

Now is the time for algebra.  The content of a fixed reserve decreases as the withdrawals mount up. The total quantity that has been removed by the time of the Nth withdrawal is q(N) which is the sum of each separate amount removed: A1 + A2 + … + AN

 Let X = (1+ i). Use the relations of the last section q(N) = A1 (1 + X + X2 + … + XN-1) The part in parentheses is a geometric series so we apply the series formula. Total quantity removed q(N) = A1 (1 + X + X2 + … + XN-1) = $: big( frac{Q_0}{m cdot Y_0} big) frac {X^N -1}{X-1}$ Expand the X q(N) = $: big( frac{Q_0}{m cdot Y_0} big) frac {(1+i)^N -1}{1+i-1}$ Use i = $: frac{r}{m}$ in denominator q(N) = $: big( frac{Q_0}{r Y_0} big)$ $big( (1+i)^N -1big)$ Shift from number of withdrawals N to the time t needed for the withdrawals to happen … the number of cycles is N-1 = m·t = (the cycles/yr) times (the number of years). N = m t + 1. The total withdrawn becomes q(t). q(t) = $: big( frac{Q_0}{r Y_0} big)$ $big( (1+i)^{mt+1} -1big)$ The total quantity left in the reservoir Q(t) is the difference between the amount at the start and the total withdrawn … Q0 — q(t) [Eqn 1] Q(t) = Q0 — $quad big( frac{Q_0}{r Y_0} big)$ $big( (1+i)^{mt+1} -1big)$

Usually, mt is so much greater than 1 that (mt+1) is written (mt). The interesting questions are all about how long it takes to use up some fraction of a given reservoir. We will solve [Eqn 1] for t = so that we can respond to the questions.

 First define Depletion Fraction f = $quadfrac{Q_0-Q(t)}{Q_0}$ = $frac{Q_0-Q}{Q_0}$ f = 0 … Reservoir full  [Q = Q0] 1/2 … half gone [Q = Q0/2] 1 … Reservoir empty  [Q = 0] Apply algebra and, in 2 steps, get the desired result $1+r Y_0 f$ = $(1+i)^{mt+1}$ [Eqn 2] t = $frac{1}{m} big( frac{ln (1+r Y_0 f)}{ln (1+r/m)} - 1 big)$

As with [Eqn 1] the 1 is usually ignored because the ratio of logarithms is so much bigger (see comment)

### 4. Apply To Gold Prospector

Apply [Eqn 2] to the lucky prospector. Suppose he lets his spending continue to increase by a mere 5% each year, without actually looking at what was left in the treasure chest. How long would his gold hoard last? What is the time T when the quantity used up = original amount (f = 1)?

 T = $frac{1}{12} big( frac{ln (1+0.05*100*1)}{ln (1+0.05/12)} - 1 big)$ T ≈ 36 yrs He found treasure at 20, he will run out of gold when he is 56 years old. Suppose, instead, that he noticed that the coin level in the chest was dropping too fast when it was about 1/2 full. When would that have occurred (f=0.5)? T = $frac{1}{12} big( frac{ln (1+0.05*100*0.5)}{ln (1+0.05/12)} - 1 big)$ T ≈ 25 years. He redoes his budget, sticks to it this time, and lives happily ever after.

If reservoir is drained in 36 years, why does the half-filled point happen after 36/2 = 18 years? The answer is that the usage grows exponentially, and the the fastest use occurs near the end. This is a topic for a different posting. Look here for additional situations where a reservoir is drained in discrete steps.

### 5. Continuous depletion of a reservoir

Something is continuous if it has no interruptions.  Extraction from the reservoir with 1 withdrawal cycle per minute, m=1, means: wait 1 minute until the next withdrawal.   Increase the number of cycles in a minute and there is less time between withdrawals. m=10 means 10 cycles per minute, m=1,000 is a thousand cycles in a second. As m grows without ending the time between cycles squeezes ever closer to zero and the process becomes “continuous.”  Extraction in distinct cycle steps may be called “withdrawing.”  Continuous extraction could be called “draining.”

Recall [Eqn 1]: Q(t) = Q0$quad big( frac{Q_0}{r Y_0} big)$ $big[(1+i)^{mt} -1big]$. For simplicity, the (mt + 1) was replace by (mt). Time from start of extraction is  t, Q0 is the original reservoir size, i = r/m where m is the cycle is a time step between successive withdrawals and r is the growth rate of the withdrawal amounts. Y0 is the life-time of the reservoir given r = 0.

Again, define the “depletion fraction” f as was done just before [Eqn 2]. Rewrite

 f = $frac{1}{r Y_0}$ $big[ (1+r/m)^{mt}-1big]$ 1 + rY0f = (1+r/m)mt Use standard techniques to convert this to continuous draining 1 + rY0f = ert Solve this for t: [Eqn 3] t = $frac{1}{r}$ $ln(1 + r Y_0 f)$

Equation 3 is the form to use to estimate the continuous draining of natural resources. We would get exactly the same form as [Eqn 3] if we had left the (mt+1) form and converted to continuous processes. The proof is left to the student. Although, in the dotEDU game, we call this proof by intimidation, a blog site really is not the right place to do the most complicated things possible.

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### Note A: Treasure Usage vs. Compound Interest Generation

 Reservoir Depletion Compound Interest Growth rate: Amount of any withdrawal exceeds the previous by a fixed fraction of the previous amount. This is set by the growth rate per cycle i = r/m. Growth rate: Principal after any compounding exceeds the previous by a fixed fraction of that previous principal. This is set by the interest rate per cycle i = r/m. Amount of last withdrawal is AN = A1 (1+ i)N-1, where N is the number of withdrawals Amount of final principal is PN = P0 (1+ i)N, where N is the number of interest cycles.

The math of resource depletion and compound interest differ the way that cloths pins differ from dish towels. The way you count them hanging on a drying line differs by one count.

 Reservoir Depletion Compound Interest Depletion begins with a withdrawal and ends with one. We count the transactions, not the time between as important. 4 withdrawal cycles would have N=5 actual transactions, one between each cycle and one at each end.Count the cloths pins holding the towels (cycle times) on the line. Investment cycles begin with the deposit of the initial principal. Interest is not awarded until the first cycle is complete. The cycle ends with an interest payment. 4 investment cycles would have N=4 interest payments at the end of each cycle. The cycle time is the important activity.Count the towels(cycle periods) not the pins.

Back to Section 2:How does it work?

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### Note B: Discussion of math details

The N–1 in [Eqn 1] and mt–1 in [Eqn 2] are due to the fact that the reservoir exists has its full content until the first withdrawal. Since the number 1 draw happens at N=1, or t=0, it must be that Q(t=0) ≠ Q0. This is why f ≠ 0 at the start. If you set f = 0 and find the time, you will get a negative value, one cycle prior to t=0. That is, if f = 0 then t = –1/m years (one month before the first withdrawal, if we our cycle is months).

Back to Start of Section 3: Amount Remaining

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### Note C: Budgeting during scarcity

1. Jill will work the summer for a university that pays after the work is completed.  There is no income between May 15 through Sept 15, 120 days.  She has \$5,000 in savings, must budget 5000/120 = \$41.67 /day total.  On day 1 she survives on this, with difficulty.  Day 2 she spent a mere 21 cents too much, about 0.5% overspending.  Each day she overspent on average about 0.50% greater than the previous day.  m=1 cycle/day, Y0=120 days.  Show each point:

(A)  Jill’s  savings would be \$0.00 if the 0.5% growth went on for 94 days.
(B)  Her savings drops to 1/2 its value  (\$2,500) in 53 days.
(C)  When (B) occurred, she was spending \$54.28 per day. (use compound growth start is \$41.67/day. 53 days, 0.5%/day, m = 1)
(D)  Jill acted at the 1/2 savings point, and her new budget became = \$37.88

2. Contractors are frequently paid upon completion of the job.  Jack took a job lasting 6 months.  He had carefully budgeted \$1200/month, \$300/wk to survive and had \$7,500 in the bank, enough for 25 weekly cycles, or 6.25 months. He was in budget the first month but it was really uncomfortable. At the end of the second month, he had spent \$30 too much, 10% more than budgeted. This had little effect on his account, so he formed the habit of spending 10% more every month than the previous month.  Assume m=4 cycles/month, Y0=6.25 months, r = 10% per month growth rate.  Show each of these points:

(A)  \$0.00 would remain in Jack’s bank after 4.7 months of this spending pattern.
(B)  Jack would see his savings had dropped to \$3,750 after 2.5 months (10 weeks)
(C)  What was Jack’s weekly  spending when he was half way through his savings?  Use compound growth: 300(1+0.1/4)4·2.5 = \$384.03

Back to start of Section 4: Applications

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### Note D: Making a continuous function

Here is a method that is used in Calculus-I.

 Start work with an expression that comes up in most growth-rate problems $big(1 + frac{r}{m}big)^{mt}$ Math is all about finding and using patterns, so we use x = m/r to rewrite the expression $big(1 + frac{1}{x}big)^{xrt}$ $big[ big(1+frac{1}{x}big)^x big]^{rt}$ Here is the pattern: a simple base expression inside the […] is raised to the power (r·t). The base expression converts from step changes at each cycle into smoothly continuous changes as m grows without bound (and x grows proportionally). Look at the […] expression as x grows very large $x = 1phantom{0^3} quadbig(1+frac{1}{x}big)^x = 2$ $x = 10^3 quadbig(1+frac{1}{x}big)^x = 2.7169ldots$ $x = 10^6 quadbig(1+frac{1}{x}big)^x = 2.718280ldots$ $x = 10^9 quadbig(1+frac{1}{x}big)^x = 2.718281ldots$ $x rightarrow infty quadbig(1+frac{1}{x}big)^x rightarrow$ e e = 2.7 1828 1828 45 904… is a “universal” irrational number similar to π

e is the base value for standard exponential growth function ert, r is the rate of increase in the function value and t is the time. Although e is the base value that arises from calculus derivations, it is not the only base that can be used. The tall tale of stacking coins on a chess board uses base 2, for a function that doubles with every step: N = 2t. When the true story of the overpowering presence of gray rabbits on the continent of Australia is modeled, the growth function is N = 2 · 19t. This is a population growth that is overwhelmingly difficult to comprehend.

Back to start of Section 5: Continuous Depletion

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### Final Note: Calculus makes life simpler

We have a use-rate that grows in proportion to its current size. This is the setup for a standard e-base function dU = rU dt. Given an exponential growth, the total use is the integral of the ert plus a constant. Put in the end values, and you get the numbered equations.