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When does return on interest, equal your annual principle savings

  1. Sep 9, 2011 #1
    This is a question I *might* have already got the answer to, I'd just like for someone very good with calculus and algebra to verify/answer the question themselves.

    To be more exact with the problem, we are putting 1 dollar (a variable principle) into a continuously compounded interest account at rate r. At what moment does one year's interest (e^r-1), equal our principle of 1 dollar per year. Or at what time can we stop putting the dollar in completely.

    I solved it as the most complicated substitution problem I, personally, have ever done. Feel free to do it any way you please, I'm looking for both answers and/or possible mistakes in my math.

    Let F = Final, P = principle, r = interest rate, y = years.
    Given that:
    F = Pe^(ry)

    And

    P1 = ($)1 x y
    (P1 because I'm going to have to use another P later)

    Then replacing 1y for P we get

    F = ye^(ry)

    This gives us F for any time y, and rate r.

    But we want a specific F, to get that I first defined what P2 is required for the next year's interest to be equal to the 1 dollar we are putting in every year.

    P2(e^r-1) = 1

    P2=1/(e^r-1)

    In order to substitute this in we must substitute P2 into a separate F = Pe^(rt2)

    We know that t2 = 1 because in the question we asked when does (e^r-1) of the *last* year equal $1.

    We get

    F = e^r/(e^r-1)

    Then substitute this final into our F = ye^ry

    we get

    e^r/(e^r-1) = ye^ry

    0 = ye^ry - e^r/(e^r-1)

    at 15% interest I get y = 3.96 years. Graphing on my calculator because as far as I know that is unsolvable algebraically

    at 8% interest I get y = 7.27 years.

    Sounds too good to be true. Put $20k in a savings account for 4-7 years and you will raise your wage $20k/year for the rest of your life? Start investing! It will be free soon. lol.

    This is only my most recent approach to this problem. Other answers which I have debunked are (for 8% interest) 13.8 years, 20 years, and 7.5 years. I'd like confirmation, or a correct answer please.
     
    Last edited: Sep 9, 2011
  2. jcsd
  3. Sep 10, 2011 #2

    HallsofIvy

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    If I understand this correctly, you want r such that [itex]P= P(e^{r}- 1[/itex]. That is the same as [itex]1= e^r- 1[/itex] or [itex]e^r= 2[/itex] so [itex]r= ln(2) which is about 69%.
     
  4. Sep 10, 2011 #3
    Hmm, that's an avenue I took once I believe, but that only tells us what interest rate is required for a non-variable principle to double in 1 year.

    If you change your equation from P = P(e^r - 1) to P1 = P2(e^r - 1) that would give the answer I have above.

    Let me clarify what I'm looking for with some real numbers.

    You have an investing account which is gaining 8% interest continously compounded, it currently has 0 dollars in it. Starting now you will continously invest 10% of your $40k yearly income, or $4k dollars. At what time in years, could you stop investing your $4k/year and gain $4k in *interest alone* over the next year?

    Givens:
    Gain in interest over 1 year= e^(.08) - 1
    Final = $4k*years*e^(.08years)

    Math:
    $4k = P2(e^.08 - 1) (adaptation of your equation)
    P2 = $4k/(e^.08 - 1)
    P2 = $48,026.664
    P2 is the principle required to gain $4k in interest over the next year, at this point we just need to solve for years...and to solve for years we can substitute this into the F= equation

    $48,026.664 = $4k*y*e^(.08y)
    0 = $4k*y*e^(.08y) - $48,026.664

    Graphing this, and looking for the x intercept should give me my answer of 7.27 years, if my math is correct in the original post, and here.
     
  5. Sep 10, 2011 #4
    If I understand, the ammount of money in your account at the end of year n (after putting a dollar in it at the beginning of year n) is:

    [TEX]P_n = (P_{n-1} + 1)e^r[/TEX]

    This recurrance can be solved (I'd enlist the help of wolfram alpha) to get a closed form expression for [TEX]P_n[/TEX]. Then I think you can solve this equation for n:

    [TEX]P_n-P_{n-1}-1 = 1[/TEX]

    to get what you need. This is just a quick thought while I'm at work; that is, it might be way off.
     
  6. Sep 10, 2011 #5
    The future value of a continuous annuity is
    [tex] F = \frac{R}{r} (e^{rt} - 1) [/tex]
    Where R is the continuous rate of deposit (dollars per year(t) )

    r is the annual non-compounded interest per dollar. (Also known as the APR)

    When will the interest on the Future value (F) equal R (one dollar per year) ?

    Leave out R since it equals 1
    [tex] Fr = 1 = (e^{rt} - 1) [/tex]
    add one to both sides
    [tex] 2 = e^{rt} [/tex]
    take logs both sides
    [tex] \ln 2 = rt [/tex]
    [tex] t = \frac{\ln 2}{r} [/tex]
    with your .08 percent interest rate, you must make continuous deposits (at the rate of $4k per year) for 8.6643 years. At that time the future value will be $50,000 and the interest alone will be $4K per year.
     
    Last edited: Sep 10, 2011
  7. Sep 10, 2011 #6
    Good, that's closer to what I'm looking for. Where did you get the $50,000 number? And the F = R/r(e^rt - 1) equation? What's the difference exactly between my F=te^rt and your F=(t/r)(e^rt - 1)?

    Why does yours have an extra r, and why the - 1?

    The problem I have with that answer though, is that when I plug in $50,000 into the commonly accepted F = Ae^rt, I don't get $54,000. However when I plug in my $48,026.664 I get $52,026.664, exactly $4,000 more.

    I also don't see how you got Fr = 1, and R = 1. Shouldn't R = t?
     
  8. Sep 12, 2011 #7
    The following equation calculates the future value of a series of deposits where the deposit is made at the start of the time period. It took me about 2 hours of work because I had to solve for a geometric series.

    Example you start by depositing an amount (a) into a bank account with zero balance. At the start of the next year you make another deposit of the same amount (a). You can make these deposits for as long as you want. At the end of (t) years the future value will = (F)

    This is a more practical method as it is easier to implement.

    One thing to note is that you can't use fractions of a year. (t) must be an integer value
    The nice thing about this equation is it allows continuous compounding with periodic deposits - something that is not too commonly seen.


    [tex] F = a \left ( \left ( \frac {e^{r(t+1)}-1}{e^r-1} \right ) -1 \right) [/tex]
    [tex] a= \frac { F} { \left ( \left ( \frac {e^{r(t+1)}-1}{e^r-1} \right ) -1 \right) }[/tex]

    Let F = 48026.66
    let t=8
    let r = .08
    calc a = 4118.84

    you must make a $4118.84 deposit at the start of each year for 8 years. At the end of the 8th year the balance will be 48026.66 which is what you want.
     
    Last edited: Sep 12, 2011
  9. Sep 13, 2011 #8
    A little bit more on that last equation. I find it interesting that the actual bank balance does not exactly match the curve produced by the equation as t varies through all values. Although it matches exactly at the time the deposit is made. (when (t) is at integer values)

    I think it is called a jump discontinuity. When I derived the equation I doubted that it would work and I was kind of astounded by the results.

    By the way, that is a plus sign in the exponent

    Venik have you tried the equation?
     
  10. Sep 13, 2011 #9
    Yea thanks, I think it's a slightly different equation than the one in my first post. Which helps verify it. You got the same exact P2 as me, 48026.66.

    I'm glad you did it, because it's helpful when you arrive at similar answers 2 different ways it usually means you're doing it right. But, this personal curiousity I have for this problem and it's outcome in the real world includes compounding principle multiple times a year, every two weeks or more. I'm not very good with geometric series, is it a hassle to change the principle integer to compound 30 times a year? That should give an answer somewhere between my 7.27 and your 8 years, and give a real-world value for anyone interested in a maverick savings plan.

    Another good thing about your series is that it tells us it cannot take *more* than 8 years to get the annual principle back in one year's interest.

    8% is actually a very conservative estimate for stocks, as they are hedged against inflation. A stock with 6% yield, if inflation is 4% that year, can actually gain 10% interest a year. Selling options on that stock can yield an extra 15% a year, but you're bound to got a stock called (however you may sell the contract as soon as you want out). An aggressive investment strategy, using options, which is actually much less riskier than trading, can yield you 15-25% interest minus (or including) any stocks which could get called from your options.

    I'm convinced, now, the difference between the rich and the poor, is a conservative but aggressive investment plan, and a very disciplined savings plan. With these numbers a $40k/year worker starting at 20y/o can retire by the time he's 30, if he saves a good portion of his money.
     
    Last edited: Sep 14, 2011
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