Compound Interest: 3.11% Annually, 3x/Year

In summary: If ##n## is finite, e.g. if I compound it 12 times per year, then the actual ##A(t)## would be a step function. The formula given in the question would only really hold for integer values of ##t##.The rate...If ##n## is finite, e.g. if I compound it 12 times per year, then the actual ##A(t)## would be a step function. The formula given in the question would only really hold for integer values of ##t##.
  • #1
brake4country
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7
Homework Statement
The formula for compound interest is given by
𝐴(𝑡)=𝐼(1+ (0.01×𝑟/n))^𝑛𝑡,
where 𝐼 is the initial amount invested in dollars, 𝑟 is the annual interest rate (expressed as a percent % rate), 𝑛 is the number of times interest is compounded per year, 𝑡 is the time in years since the initial investment and 𝐴(𝑡) is the amount, or balance, in dollars after 𝑡 years have passed.

Suppose you invest $350 today into an account with an annual rate of 3.11% which is compounded 3 times each year. Answer the following questions about the growth of your investment.
Relevant Equations
1. What is the instantaneous rate at which your investment grows (in dollars per year) as a function of 𝑡 ?

2. What is the rate at which the investment is growing (in dollars per year) after 17 years?

3. What is the true percent increase of the investment at the end of the first year?
1. 350*(1+(0.0311)/3)*ln(1+(0.0311)/3)

2 and 3. Not sure how to answer these questions.
 
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  • #2
Part 3 is easy. How much money do you have after 1 year?
 
  • #3
It is not clear what the instantaneous rate that the investment would grow, since the compounding is incremental. Maybe what they mean is the equivalent rate of growth if the compounding were continuous, but this would not be a function of time.
 
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  • #4
brake4country said:
1. What is the instantaneous rate at which your investment grows (in dollars per year) as a function of 𝑡
brake4country said:
1. 350*(1+(0.0311)/3)*ln(1+(0.0311)/3)
This is not a function of ## t ##.

You are given
brake4country said:
The formula for compound interest is given by 𝐴(𝑡)=𝐼(1+ (0.01×𝑟/n))^𝑛𝑡
which is a function of ## t ##, what do you think you need to do with this? It will be easier if we work in ## \LaTeX ##, I'll give you a head start by rewriting the equation in the question:
$$ A(t) = I \left(1 + \left( 0.01 \frac{r}{n} \right) \right) ^ {nt} $$
Edit: probably best if we define ## a = I \left(1 + \left( 0.01 \frac{r}{n} \right) \right) ## and write
$$ A(t) = a ^ {nt} $$
 
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  • #5
pbuk said:
This is not a function of ## t ##.

You are given

which is a function of ## t ##, what do you think you need to do with this? It will be easier if we work in ## \LaTeX ##, I'll give you a head start by rewriting the equation in the question:
$$ A(t) = I \left(1 + \left( 0.01 \frac{r}{n} \right) \right) ^ {nt} $$
Edit: probably best if we define ## a = I \left(1 + \left( 0.01 \frac{r}{n} \right) \right) ## and write
$$ A(t) = a ^ {nt} $$
That last equation makes no sense to me.
 
  • #6
Chestermiller said:
That last equation makes no sense to me.
Am I missing the point here - surely it's just exponential growth?
 
  • #7
I think @Chestermiller's point is that the model suggested in the question is for discrete values of ##t##, so a derivative is not well defined. For purposes of the question it is probably required to assume that the given equation holds over a continuous interval of ##t##.
 
  • #8
The equivalent continuous compounding interest rate R would satisfy $$A=Ie^{Rt}= I \left(1 + \left( 0.01 \frac{r}{n} \right) \right)^{nt}$$or$$R=n\ln{\left(1 + \left( 0.01 \frac{r}{n} \right)\right)}$$
 
  • #9
etotheipi said:
I think @Chestermiller's point is that the model suggested in the question is for discrete values of ##t##, so a derivative is not well defined. For purposes of the question it is probably required to assume that the given equation holds over a continuous interval of ##t##.
Economics is a social science so you have to accommodate a certain lack of rigour :wink:

However where does it say in the question that ## t ## is discrete?
 
  • #10
pbuk said:
Economics is a social science so you have to accommodate a certain lack of rigour :wink:

However where does it say in the question that ## t ## is discrete?
When they say that interest is compounded 3 times a year, what does that mean to you?
 
  • #11
pbuk said:
Economics is a social science so you have to accommodate a certain lack of rigour :wink:

However where does it say in the question that ## t ## is discrete?

If ##n## is finite, e.g. if I compound it 12 times per year, then the actual ##A(t)## would be a step function. The formula given in the question would only really hold for integer values of ##t## (or at least those that coincide with the time of compounding).

However, if we make the assumption that we can convert it into an analogous interest model over a continuous interval of ##t##, then we can of course recast it in terms of ##A(t) = Ie^{Rt}##.
 
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  • #12
etotheipi said:
If ##n## is finite, e.g. if I compound it 12 times per year, then the actual ##A(t)## would be a step function. The formula given in the question would only really hold for integer values of ##t## (or at least those that coincide with the time of compounding)..
Yes, at integral multiples of 1/n.
 
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  • #13
etotheipi said:
If ##n## is finite, e.g. if I compound it 12 times per year, then the actual ##A(t)## would be a step function. The formula given in the question would only really hold for integer values of ##t##.
But it doesn't say any of that in the question. You are bringing extraneous information about your knowledge of how a deposit account works in the real world that is not in the question. The question presents a continuous, differentiable model of compound interest.

And as it happens the discrete model is not how it happens anyway - interest is normally accrued on a daily basis (using some conventional calculation) increasing the value of the deposit (almost) continuously. The accrued amount is not included in the amount that is used to calculate interest.
 
  • #14
Chestermiller said:
Maybe what they mean is the equivalent rate of growth if the compounding were continuous, but this would not be a function of time.
The rate of growth in dollars per year (which is what the question asks for) surely would be a function of time.
 
  • #15
Chestermiller said:
When they say that interest is compounded 3 times a year, what does that mean to you?
Sorry, missed this. It means that 3 times a year the balance of accrued interest is added to the principal balance of the account.

It doesn't matter though, the question does not ask you to write an equation for the amount of money in the account based on the information that interest is compounded 3 times a year, it gives you that equation.
 
  • #16
pbuk said:
But it doesn't say any of that in the question. You are bringing extraneous information about your knowledge of how a deposit account works in the real world that is not in the question. The question presents a continuous, differentiable model of compound interest.

And as it happens the discrete model is not how it happens anyway - interest is normally accrued on a daily basis (using some conventional calculation) increasing the value of the deposit (almost) continuously. The accrued amount is not included in the amount that is used to calculate interest.

I think you are right in practice, I cannot imagine financial analysts sitting around wasting their time with discrete models when ##Pe^{rt}## is a very good approximation anyway for finite ##n##.

But in the context of this question, the wording is pretty ambiguous. It doesn't specify the domain of ##t##, and given that we compound only thrice per year, that would seem to imply a very pronounced step function, so we cannot infer immediately that the model is continuous and differentiable...

In any case, the only way to answer the question is to assume it holds for all ##t## and differentiate.
 
  • #17
Poor OP, he must think we are mad. To recap: the question tells you that at time ## t ##, your investment is worth ## A(t) = a ^ {nt} ## dollars. What is the rate of change of your investment at time ## t ##?
 
  • #18
etotheipi said:
I think you are right in practice, I cannot imagine financial analysts sitting around wasting their time with discrete models when ##Pe^{rt}## is a very good approximation anyway for finite ##n##.

But in the context of this question, the wording is pretty ambiguous. It doesn't specify the domain of ##t##, and given that we compound only thrice per year, that would seem to imply a very pronounced step function, so we cannot infer immediately that the model is continuous and differentiable...

In any case, the only way to answer the question is to assume it holds for all ##t## and differentiate.
I think you give financial analysts too much credit. I don't think most financial analysts would know what continuous compounding is if it jumped up and bit them on the butt (or what an exponential function is).
 
  • #19
pbuk said:
Poor OP, he must think we are mad. To recap: the question tells you that at time ## t ##, your investment is worth ## A(t) = a ^ {nt} ## dollars. What is the rate of change of your investment at time ## t ##?
I see what you are saying, I think. The math teacher that dreamt up this problem did not fully understand the implications of discrete compounding and imagined that all the student needed to do was differentiate the equation that was given with respect to t.
 
  • #20
etotheipi said:
I think you are right in practice
I know I am right in practice, I do this for a living. But I am also right in theory.

etotheipi said:
, I cannot imagine financial analysts sitting around wasting their time with discrete models when ##Pe^{rt}## is a very good approximation anyway for finite ##n##.
I think you mean discrete ## n ##. And it's not a 'very good approximation': at every time ## t_i ## that the discrete model is defined it is equal to the value of the continuous model at ## t_i ##.

etotheipi said:
But in the context of this question, the wording is pretty ambiguous.
Not to me, and not to the examiner. The question presents you with an explicit model and asks you about its rate of change. The question would be meaningless if its rate of change were not defined.
 
  • #21
Articulating and defending my answers:

1. Instantaneous rate as a function of t: FIND THE DERIVATIVE (d/dx a^x=a^x*ln(a). Derivative A'(t)=1050*(3.0311/3)^(3t)*ln(3.0311/3)
2. Total amount invested after 17 years: A(t)=350*(3.0311/3)^3t. A(17)=$592.24.
3. Rate at which the investment IS growing after 17 years: USE DERIVATIVE. Thus, A'(t)=1050*(3.0311/3)^(3t)*ln(3.0311/3). A'(17)=18.32%.
4. True percent increase at the end of the first year: (361-350)/350=3.14%.

I think I got it after some thought and DESMOS. Thank you!
 
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  • #22
brake4country said:
Articulating and defending my answers:

1. Instantaneous rate as a function of t: FIND THE DERIVATIVE (d/dx a^x=a^x*ln(a). Derivative A'(t)=1050*(3.0311/3)^(3t)*ln(3.0311/3)
2. Total amount invested after 17 years: A(t)=350*(3.0311/3)^3t. A(17)=$592.24.
3. Rate at which the investment IS growing after 17 years: USE DERIVATIVE. Thus, A'(t)=1050*(3.0311/3)^(3t)*ln(3.0311/3). A'(17)=18.32%.
4. True percent increase at the end of the first year: (361-350)/350=3.14%.

I think I got it after some thought and DESMOS. Thank you!
I haven't checked your arithmetic but I'm sure its right - as you have discovered, this is actually a pretty easy question when you realize that you have to throw away some extraneous information.
 
  • #23
pbuk said:
I haven't checked your arithmetic but I'm sure its right - as you have discovered, this is actually a pretty easy question when you realize that you have to throw away some extraneous information.
Edit: as pointed out in #24 and #30 I am not very good at adding fractions.

Oops - almost right. You had the term 1+(0.0311)/3 in your original answer which was correct but now you have 3.0311/3 which is not.
 
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  • #24
pbuk said:
Oops - almost right. You had the term 1+(0.0311)/3 in your original answer which was correct but now you have 3.0311/3 which is not.
Check his algebra again. It is right.
 
  • #25
Chestermiller said:
I think you give financial analysts too much credit. I don't think most financial analysts would know what continuous compounding is if it jumped up and bit them on the butt (or what an exponential function is).
Having revealed my profession in #20 (it's on my profile anyway) I am going to ignore that remark.
 
  • #26
Chestermiller said:
Check his algebra again. It is right.
Oh nuts, really? See, when it comes to numbers I'm lost without a spreadsheet :biggrin:
 
  • #27
Sad but true:
The top two jobs for [http://www.careers.cam.ac.uk/sectors/maths/index.asp] mathematics graduates are finance and investment analyst and adviser, and chartered or certified accountant. Other roles in the top five include programmer, software developer and actuary.

A fifth of mathematics graduates are in further study. Of these, 40% continue their education in mathematics and a further 30% are trainee teachers.
 
  • #28
The only instantaneous interest rate that makes sense to me is $$100\frac{A'}{A}=100n\ln{\left(1 + 0.01 \frac{r}{n} \right)}$$which is a constant, independent of t.
 
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  • #29
Chestermiller said:
The only instantaneous interest rate that makes sense to me is $$100\frac{A'}{A}=100n\ln{\left(1 + 0.01 \frac{r}{n} \right)}$$which is a constant, independent of t.
Yes, a constant with dimensions ## T^{-1} ## . If you substitute the original equation 𝐴(𝑡)=𝐼(1+ (0.01×𝑟/n))^𝑛𝑡 into that then you recover the answer to Q1, the rate of growth in $ per year as a function of t.
 
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  • #30
pbuk said:
Oops - almost right. You had the term 1+(0.0311)/3 in your original answer which was correct but now you have 3.0311/3 which is not.
I just used common denominator =)
 
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  • #31
pbuk said:
I know I am right in practice, I do this for a living.
So much for my theory on peanut butter ( pb) ;)
 
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  • #32
brake4country said:
I just used common denominator =)
Yes of course o:)
 
  • #33
pbuk said:
I think you mean discrete ## n ##. And it's not a 'very good approximation': at every time ## t_i ## that the discrete model is defined it is equal to the value of the continuous model at ## t_i ##.

No, I mean finite ##n##. If we take the function to be continuous on the interval containing all real ##t##, then ##A(t) = I(1+\frac{r}{n})^{nt}## approaches ##A_{2}(t) = Ie^{rt}## in the limit of large ##n##. However since the former is very close to the latter even for finite ##n##, the latter is a very useful time-saving approximation. The former is still exactly an exponential function, ##A(t) = Ie^{Rt}##, but with a slightly smaller rate constant ##R =n\ln{\left(1 + \frac{r}{n} \right)}##. You can show that ##\lim_{n \rightarrow \infty} n\ln{\left(1 + \frac{r}{n} \right)} = r##. My point is that an analyst would probably approximate discrete compounding with an exponential of unchanged rate constant :wink:
 
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  • #34
etotheipi said:
No, I mean finite ##n##. If we take the function to be continuous on the interval containing all real ##t##, then ##A(t) = I(1+\frac{r}{n})^{nt}## approaches ##A_{2}(t) = Ie^{rt}## in the limit of large ##n##. However since the former is very close to the latter even for finite ##n##, the latter is a very useful time-saving approximation. The former is still exactly an exponential function, ##A(t) = Ie^{Rt}##, but with a slightly smaller rate constant ##R =n\ln{\left(1 + \frac{r}{n} \right)}##. You can show that ##\lim_{n \rightarrow \infty} n\ln{\left(1 + \frac{r}{n} \right)} = r##. My point is that an analyst would probably approximate discrete compounding with an exponential of unchanged rate constant :wink:
I see. However we are not interested in large ## n ##, we are only interested in two values, ## n = 3 ## and ## n = 1 ##. If we use ## n = 3 ## then the value of the investment after 1 year using the headline rate ## r ## is ## I (1 + \frac{r}{3} ) ^ 3 ## and if we use ## n = 0 ## then the value of the investment after 1 year using the equivalent annual rate ## r_a ## is ## I ( 1+ r_a ) ^ 1 ##. From the definition of the equivalent annual rate ## r_a = (1 + \frac{r}{n})^n - 1 ## you can immediately see that these two calculations are exactly equal.

Because we only define the annual rate of return at integral numbers of years, the continuous model is not an approximation, it is exactly equal to the discrete model at every point that the discrete model is defined.

A model where a headline rate is compounded continuously to give an effective annual rate which you have correctly calculated as ## e^r - 1 ## is not useful either in investment theory or practice.
 
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  • #35
That is all fine, but when I wrote that I was referring to this :wink:
pbuk said:
And as it happens the discrete model is not how it happens anyway - interest is normally accrued on a daily basis (using some conventional calculation) increasing the value of the deposit (almost) continuously. The accrued amount is not included in the amount that is used to calculate interest.
 
<h2>What is compound interest?</h2><p>Compound interest is the interest earned on both the initial principal amount and the accumulated interest from previous periods. This means that the interest earned in each period is added to the principal amount, and the interest for the next period is calculated based on the new total.</p><h2>What is the annual interest rate of 3.11%?</h2><p>The annual interest rate of 3.11% refers to the percentage of interest that will be earned on the initial principal amount in one year. This means that for every $100 invested, $3.11 will be earned in interest each year.</p><h2>What does 3x/year mean in compound interest?</h2><p>3x/year means that the interest is compounded three times per year. This means that the interest earned in each period will be added to the principal amount, and the interest for the next period will be calculated based on the new total. In this case, the interest will be compounded every 4 months.</p><h2>How does compound interest differ from simple interest?</h2><p>Compound interest differs from simple interest in that simple interest is only calculated on the initial principal amount, while compound interest takes into account the accumulated interest from previous periods. This means that compound interest will result in higher earnings over time compared to simple interest.</p><h2>How does the frequency of compounding affect the total amount earned?</h2><p>The more frequent the compounding, the higher the total amount earned will be. This is because with more frequent compounding, the interest earned in each period is added to the principal amount more often, resulting in a larger total amount. In this case, compounding 3 times per year will result in a higher total amount compared to compounding once per year.</p>

What is compound interest?

Compound interest is the interest earned on both the initial principal amount and the accumulated interest from previous periods. This means that the interest earned in each period is added to the principal amount, and the interest for the next period is calculated based on the new total.

What is the annual interest rate of 3.11%?

The annual interest rate of 3.11% refers to the percentage of interest that will be earned on the initial principal amount in one year. This means that for every $100 invested, $3.11 will be earned in interest each year.

What does 3x/year mean in compound interest?

3x/year means that the interest is compounded three times per year. This means that the interest earned in each period will be added to the principal amount, and the interest for the next period will be calculated based on the new total. In this case, the interest will be compounded every 4 months.

How does compound interest differ from simple interest?

Compound interest differs from simple interest in that simple interest is only calculated on the initial principal amount, while compound interest takes into account the accumulated interest from previous periods. This means that compound interest will result in higher earnings over time compared to simple interest.

How does the frequency of compounding affect the total amount earned?

The more frequent the compounding, the higher the total amount earned will be. This is because with more frequent compounding, the interest earned in each period is added to the principal amount more often, resulting in a larger total amount. In this case, compounding 3 times per year will result in a higher total amount compared to compounding once per year.

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