# Bank savings question

Discussion in 'Physics & Math' started by EoDEo, Dec 9, 2016.

1. ### EoDEoRegistered Senior Member

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I apologies bothering you with a trivial thing, but I’ve been thinking about saving some money in a bank and one of them offers 2.89% nominal benefits on yearly bases, but they award you each and every day a fraction of this yearly benefit. Each subsequent day has the base + previous day’s benefits calculated as a new base for the %.

They cite that If you deposit 10k you get 0.79 every day. What are the actual effective yearly benefits in %? Is it 3%? More? Less?

Thank you

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3. ### EoDEoRegistered Senior Member

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Is it 2.932% ?

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5. ### rpennerFully WiredStaff Member

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$\left( \frac{1000079}{1000000} \right)^{365}$ is between 2.906% and 2.944% APR or between 2.865% and 2.901% compounded continuously. So this figure looks compatible with continuous compounding at 2.89%.

So if the bank quotes 2.89% compounded continuously, then the expected result of investing 10k for a full year is $10000 \times e^{0.0289} \approx 10293.22$ which is a APR of 2.9322% or an average of 0.80 / day over 365 days. But the first day you get $10000 \times \left( e^{\frac{0.0289}{365}} - 1 \right) \approx 0.79$.

In contrast to continuous compounding, simple interest of 0.79 per day on a 1 year term gives you $10000 + 0.79 \times 365 = 10288.50$ which is a bit less.

Last edited: Dec 21, 2016
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7. ### EoDEoRegistered Senior Member

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Thank you kindly!

8. ### rpennerFully WiredStaff Member

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2.89% compounded annually gives 2.890000% APR
2.89% compounded semiannually gives 2.910880% APR
2.89% compounded quarterly gives 2.921472% APR
2.89% compounded monthly gives 2.928589% APR
2.89% compounded daily gives between 2.932046% and 2.932048% APR depending on the technical definition of what constitutes a year in banking terms. This may be what your bank was referring to.

If r is the rate as a percentage, then the APR for compounding n times in a year is $100 \times \left( \left( 1 + \frac{r}{100 \times n} \right)^n - 1 \right)$ (the 100's go away if both the rate and the APR are expressed as a decimal like 0.0289).

Approximation for when r < 3:
APR $\approx r + \frac{n-1}{200 n} r^2 + \frac{(n - 2) (n - 1)}{60000 n^2} r^3$
Which is a truncation of $\sum_{k=1}^{n} \frac{(n-1)!}{k! \, (n-k)! \, 100^{k-1} \, n^{k-1} } r^k$

2.89% compounded continuously gives 2.932166% APR

If r is the rate as a percentage, then the APR for compounding continuously is $100 \times \left( \textrm{exp}\left( \frac{r}{100} \right) - 1 \right)$.

Approximation for when r < 3:
APR $\approx r + \frac{1}{200} r^2 + \frac{1}{60000} r^3$
Which is a truncation of $\sum_{k=1}^{\infty} \frac{1}{k! \, 100^{k-1} } r^k$

9. ### CounterRegistered Senior Member

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That depends. For how long? Forever?!

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There is actually an equation answering your question. Simply enter the day i.e. day seven, and the equation tells you how much money you have. Unfortunately I do not know the equation but should it be found, then you may alter it to also tell you which day it is, depending how much money you have.

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Last edited: Jan 19, 2017