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Compound Interest Systems of Equations Problem

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Romeo was given a gift of $10,000 when he turned 16. He invested it at 3% per annum. Three years later, Juliet was given $10,000, which she invested at 5% per annum. When will the two amounts be equal in value?

    2. Relevant equations
    Compound Interest Formula
    Total = Capital(1+interest)^Years
    t = c(1+i)^n
    3. The attempt at a solution
    Deciding how to create my formula is where things get fuzzy. The way I initially attempted this was to add a 3 year head start onto Romeo's interest formula. This is what results (Used wolfram to save time rewriting):
    CYAiJmG.png
    This eventually results in "4.61103." However, this is not the answer at the back of my textbook. They decided to opt for a different formula where the time is subtracted from Juliet's interest function. This is shown here:
    cSZJimk.png
    This results in the correct answer of 7.61103. What I'm wondering, why do I have to subtract 3 years from Juliet's compound interest formula, and why does adding 3 years to Romeo's compound interest formula produce the wrong answer?
     
  2. jcsd
  3. Apr 16, 2015 #2

    BvU

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    So clearly the book starts counting at the beginning of the story: when Romeo brings his money to the bank.
    That means Julia's interest clock is at x-3 when Romeo's is at x.
     
  4. Apr 16, 2015 #3

    SammyS

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    What is Romeo's age when then two accounts have equal value?

    Both answers agree.
     
  5. Apr 16, 2015 #4
    Is this always the convention when dealing with problems like this? From my experience with physics courses, I assumed that time shares a relation with the total amount of money generated by compound interest. The interest and initial capital is thus fixed, and you are only left with a relation between time and the total sum of money. If you were to treat time as a vector which you could progress backwards and forwards on, each snapshot through time depending on which way you were progressing would net either a larger or smaller amount of the total. I don't understand why the book is forcing me to start at the beginning of the story. The story is like pages of a book which can be turned towards or away from the end of the novel. The content of the novel does not change and its contents are already written.

    I don't understand what Romeo's age has to do with this? Is this some kind of statement that alludes to time only progressing forward or something?
     
  6. Apr 16, 2015 #5

    BvU

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    Nothing so fancy. The book is sloppy by not stating when it wants the clock (calendar in this case) to show t = 0. From the book answer it appears that it assumes t = 0 when Romeo turns 16. The exercise could equally well have assumed t = 0 when Julia receives the ten grand. If I were grading this, I'd have to allow both answers, certainly when the calculation steps are shown.
     
  7. Apr 16, 2015 #6
    I actually just figured it out while discussing this question with a friend just a few moments before you posted. I took the delta of the two times from both of the ways of solving this question and realized that this is nothing more than a semantics issue of where you measure time from. This question is poorly setup, and I am going to voice my irritation to the person marking.
     
  8. Apr 16, 2015 #7

    Ray Vickson

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    A more serious problem is that when compounding annually, the answer is "never". At Romeo's 23rd birthday, Romeo's account contains more money than Juliet's, while at his 24th birthday, Juliet's account contains more than Romeo's. Unless there is something like continuous compounding, the two amounts never match exactly; that is, unless you can withdraw your money part-way through a year and earn a part-year's interest, you could never find a time where the two withdrawals are equal. (However, I would not raise this issue with your teacher if I were you; just "go with the flow".)

    Also, instead of expressing irritation, it would be wiser for you to say that you will give two answers (depending on where to start measuring time) and point out (nicely) that the question is a bit ambiguous.
     
  9. Apr 16, 2015 #8

    SammyS

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    RGV makes an excellent point. This is a short-coming of many problems of this type.

    It's often the case that for any fraction of a single compounding period that's "left over", simple interest is computed for that extra amount of time. With this method, find the value of the account at x = 4, then see if the values can be equal at any time in the 4th year using simple interest.

    (Edited slightly.)
     
    Last edited: Apr 16, 2015
  10. Apr 16, 2015 #9
    I don't wish to sound like an imbecile, but I have no clue what that means. How would I set up a systems of equation if I subbed in 4 for both x's? I believe I would be left with only one unknown for either equation, or am I misunderstanding you? Please clarify what you mean by using x=4.
     
  11. Apr 16, 2015 #10

    SammyS

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    When x = 4, (7years for Romeo, 4 years for Juliet), what is the value in each in each account?

    Start from that point for each & using simple interest, and see how long it takes for the values to be equal.
     
  12. Apr 16, 2015 #11
    I am not familiar with the simple interest formula, so I googled it. I am operating under the assumption that it is different than the compound interest formula.

    I am to assume this is correct?
    Simple interest Formula
    I = "the grown money"
    p = the initial capital
    r = interest rate per year
    t = t
    (I) = (p)(r)(t)

    I think I know what you're getting at, so let me make an attempt.

    Romeo Compound Interest Calculation Total
    Ar = 10,000(1.03)^4+3
    Ar = 10,000(1.03)^7
    Ar = 12,298.73865

    Julie Compound Interest Calculation Total
    Aj = 10,000(1.05)^4
    Aj = 12,155.0625

    Romeo Simple Interest Calculation
    (Ir) = (12,298.73865)(0.03)(t)
    (Ir) = (368.94)(t)

    Juliet Simple Interest Calculation
    (Ir) = (12,155.0625)(0.05)(t)
    (Ir) = 607.753125(t)

    Equate the two to equal each other.

    (368.94)(t) = 607.753125(t)
    (368.94)(t) -607.753125(t) = 0
    t(368.94-607.753125)=0

    This doesnt work though...
     
  13. Apr 16, 2015 #12

    Ray Vickson

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    It's not that complicated. Let ##R_t, J_t## be the amounts in Romeo's and Juliet's accounts at time ##t## (in years, measured from Romeo's 16th birthday). We have:
    [tex] \begin{array}{ccc}
    & t=7 & t=8 \\
    R_t & 12298.74 & 12667.70 \\
    J_t & 12155.06 & 12762.82
    \end{array} [/tex]
    You can draw two straight-line plots, one going from x = 0 at (7,R7) to x = 1 at (8,R8) and another from x = 0 at (7,J7) to x = 1 at (8,J8). Where the two lines cross is the point where simple interest calculations (for the partial year) would give you equality of monetary values. This will be close to (but not exactly equal to) the value x = 0.611---that is, for time 7.611 ---that your previous calculation would have produced. (The difference is due to the slight curvature in the money vs.time graph over the year, due to continuous compounding over the year, as compared with the straight line obtained from simple interest accumulation over the year.)
     
  14. Apr 16, 2015 #13
    Could you show me an example of this because I am simply not understanding what you're saying, or can you show me how to equate the two together using algebra?
     
  15. Apr 17, 2015 #14
    This is getting out of hand, so I am going to try to be as clear and as concise as possible. I know how to calculate the correct answer for this question, and I realize where I first made my mistake. I am now wondering what everyone is talking about when they mention using simple interest. I have literally 0 experience doing any simple interest calculations, and I have no idea what sammyS is talking about here.

    If anyone is holding information back because they believe an attempt has not been made and it would violate forum rules, please stop. I understand how to do the book's question using the books method, and I know and where the discrepancy between the 4.61103 and 7.61103 came from. I have made a wholehearted attempt so please for sake of my sanity just give me an example or step by step breakdown. I learn through examples, I look at the process involved, and I create a model in my head how everything interacts and relates with each other.

    What I am seeking is an explanation and a simple example of what this simple interest method is for calculating the 0.61103 year after the exact 7 years has passed.

    What I am understanding so far.
    - First use the compound interest formula to create outputs for both Romeo's and Juliet's accounts after being influenced by exactly 7 years of interest.
    - This is done by assigning "4" to the variable x
    - Get said products, in this case being:
    Romeo's Total After Exactly 7 Years = 12,298.73865
    Juliet's Total After Exactly 7 Years = 12,155.0625
    - Since you know that the two amounts are still not equal, you must calculate the remaining time between the two using the Simple interest formula
    - This is what I am having problems with and what I would appreciate having an example for. Preferably shown algebraically, so I am not plotting nearly straight lines in Desmos.

    Not to sound contentious, but when you try to find the point of intersect of two nearly straight lines with a slope that is in hundredths of a unit, it is just plainly byzantine. I am probably doing this wrong, but graphically representing each simple interest formula and trying to find where they meet is insane.
     
    Last edited: Apr 17, 2015
  16. Apr 17, 2015 #15

    Ray Vickson

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    How could I have made my explanation any simpler? There was no hidden information or holding back of anything. I said: just draw two straight-line graphs for Romeo and Juliet, where the lines connect their (time,money) points at the start and end of the final year.

    The true graphs of time vs. money will be curved, because of compound interest, but over a short period such as one year the "curvature" will be small, and the graph will look almost like a straight line; replacing the curve by a straight line ---only for that single year---would give you the "simple" interest schedule for that year. It will be almost the same as the compound interest schedule, because the curvature is small over short times such as a year.

    If you don't believe me, just draw the graphs of money vs. time for Romeo and for Juliet. You will see that they curve up, but are almost straight over short periods. The only way to really understand it is to sit down and do it.
     
  17. Apr 17, 2015 #16
    I am going to assume the two lines would be plotted as
    y=(12155.0625)(0.05)(x)
    and
    y=(12298.73865)(0.03)(x)

    If that's the case, I don't understand how any human being is supposed to find the point of intersect graphically like this.

    I did sit down and do it. I have been sitting down and ripping my hair out doing this for sometime.
    5YBnlHU.png
    Edit - In the event that anyone else would like to question how committed I am to "sitting down and doing it." I've had enough of that nonsense from teachers with superiority issues and over bloated opinions that like to question how hard I am making an attempt.
    3gWKXKc.jpg
     
    Last edited: Apr 17, 2015
  18. Apr 17, 2015 #17

    Ray Vickson

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    As long as you persist in trying to plot the wrong thing you will need to keep pulling out your hair. Go back and read what I wrote in #12. Do the plot exactly as I indicated there.

    Better still plot the two curves y = 10000 * (1.03)^x and y = 10000 * (1.05)^(x-3) for x = from 3 to 8. They are curved, aren't they? Don't they look almost straight over the shorter interval 7 \leq x \leq 8? in each case if you were to plot a straight line from (7,y(7)) to (8,y(8)) it would look very close to the actual curve (x,y(x)) for x between 7 and 8. The "simple interest" effect between 7 and 8 would be the straight line, while the "compound interest" effect would be the curve. They both pass through the same points at 7 and 8, but they differ in between.
     
  19. Apr 17, 2015 #18
    As I have stated before.
    I was completely uncertain of what to plot. I am being told to use the simple interest formula, but the two formulas you have provided are compound interest formulas:
    y = 10000 * (1.03)^x
    y = 10000 * (1.05)^(x-3)
    I plotted these long before I even got to this whole simple interest business when trying to figure out the original discrepancy between 7.61103 and 4.61103. I am fully aware that the compound interest lines are nearly straight. I don't see how the average rate of change between x =7 and x =8 is going to help me discover the remaining time of 0.61103 years. I am still unclear of what to plot? Do I plot the compound interest based functions for Romeo and Juliet, or do I plot the Simple interest functions? I have done both and I am not seeing anything that is giving me any kind of better understanding. Is this operation you describe too difficult represent algebraically?

    I am getting far too irritated by this and far too frustrated and am quickly losing the desire to continue.
     
  20. Apr 17, 2015 #19
    I was on the right track here. There is no need for this graphing nonsense. I forgot a single number which was my "+ 3" in my simple interest formula. I was a fool not to recognize that the lack of a constant would render my simple interest system of equation unsolvable. The lack of anyone speaking up about this is worrisome. In the future, when referencing the slope of a secant line between the range of two x values of a curve, please write in the notation of 7 < x < 8. I had absolutely no clue what you were talking about.

    So let's try that last bit again with a fixed formula.

    Romeo Simple Interest Calculation
    (Ir) = (12,298.73865)(0.03)(t+3)
    (Ir) = (368.94)(t+3)

    Juliet Simple Interest Calculation
    (Ir) = (12,155.0625)(0.05)(t)
    (Ir) = 607.753125(t)

    Now the Systems of Equation
    (607.753125)(t) = (368.94)(t+3)
    (607.753125)(t) = (368.94)(t)+ 1106.82
    t(607.753125-368.94) = 1106.82
    t(238.813125) = (1106.82)
    t = (1106.82)/(238.813125)
    t = 4.634669891
    Not too far away from 4.61103. I wonder why...

    Let's try the simplified interest formula that measures time from Julias deposit date.

    Romeo Simple Interest Calculation
    (Ir) = (12,298.73865)(0.03)(t)
    (Ir) = (368.94)(t)

    Juliet Simple Interest Calculation
    (Ir) = (12,155.0625)(0.05)(t-3)
    (Ir) = 607.753125(t-3)

    607.753125(t-3)= (368.94)(t)
    (607.753125)(t) - 1823.259375 = (368.94)(t)
    (607.753125)(t) - (368.94)(t) =1823.259375
    t(607.753125 - 368.94) = 1823.259375
    t(238.813125) = 1823.259375
    t = 7.634.634669891
    Not too far away from 7.61103. Surprise surprise....

    Look Ma! No graphs.
    7.634.634669891 - 4.634669891 = 3 just like 7.61103 - 4.61103 =3. I wonder if they're related?
    I don't think I am wrong by stating that the time after the 7 year mark would actually be 0.634669891 of a year instead of 0.61103 if fractional compound interest is indeed calculated with simplified interest.

    We can take things a step further and look at the secant line's slope of the curve that models interest from Romeo's deposit date during 7 < x < 8. Let's look at the average rate of change for both Romeo's and Juliet's account.

    Romeo Compound Interest AROC: 7 < x < 8

    While x = 4
    Ar = 10,000(1.03)^4+3
    Ar = 10,000(1.03)^7
    Ar = 12,298.73865
    x = 4
    y =12,298.73865

    While x =5
    Ar = 10,000(1.03)^5+3
    Ar = 10,000(1.03)^8
    Ar = 12,667.70081
    x = 5
    y = 12,667.70081

    Romeo's AROC
    = (12,667.70081 - 12,298.73865) / (5 -4)
    = 368.96216

    Juliet's Compound Interest AROC 7 < x < 8

    While x = 4
    Aj = 10,000(1.05)^4
    Aj = 12,155.0625
    x = 4
    y =12,155.0625

    While x = 5
    Aj = 10,000(1.05)^5
    Aj = 12,762.81563
    x = 5
    y = 12,762.81563

    Juliet's AROC
    = (12,762.81563 - 12,155.0625) / (5-4)
    =607.7531

    After that huge bunch of calculations we now have slopes that we can create linear relations with.

    Romeo's Linear Relation
    y = (368.96216)(x + 3)

    Juliet's Linear Relation
    y = 607.7531(x)

    Let's see what the Solution is for these in equations when put into a system.

    (368.96216)(x + 3) = 607.7531(x)
    (368.96216)(x) + 1106.88648 = 607.7531(x)
    607.7531(x) -(368.96216)(x) = 1106.88648
    x(607.7531-368.96216) = 1106.88648
    x(238.783884) = 1106.88648
    x = 4.635515854

    Notice a pattern?

    Time Measured from Romeo's Deposit Compound Calculation Product
    x = 4.61103
    Time Measured from Romeo's Deposit Simplified Interest Product
    x = 4.634669891
    Time Measured from Romeo's Deposit AROC: 7 < x< 8 Product
    x = 4.635515854

    I think its safe to say that they matched sometime in the range of 0.635515854 - 0.634669891- 0.61103 of the start of the 8th year. I think I made my point. I don't need any graphs to know how to rock, and I certainly showed that I sat down and did the work.
     
  21. Apr 17, 2015 #20

    Ray Vickson

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    Of course it can be done without graphs. Graphs can be of help in setting up the algebraic equations that must, eventually, be solved without graphs. However, you seemed to not understand how simple interest works (your claim, not mine) and to see that a graphical representation can sometimes be helpful---not always, just sometimes. If it was not helpful to you, fine. And, of course I suggested plotting the compound-interest graphs, but I guess you missed out the part where I said that during a short period, such as 1 year, the graph looks almost straight, and that a straight line replacement for the graph (but ONLY in that single year) gives you the simple-interest effect within the year.

    Your writeup is almost incomprehensible to me, but your final answer seems OK. You say you still do not know why it is different from the original answer, and that is precisely what I was speaking of when I mentioned graphs: you could see right away why there is a difference.
     
    Last edited: Apr 17, 2015
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