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How to calculate annually compounded interest?

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0

$begingroup$

I am having some troubles understanding this topic of interest rates.

For example, if i invest 1 dollar at a continuous compounding rate of 11% , then my end of year value is equal to $e^{0.11}=1.116$ dollars. From here it says that investing at 11% a year continuously compounded is the same as investing 11.6 a year annually compounded.

Now where is this 11.6 coming from ? Is it coming from the 1.(116) dollars value? So i just take the decimal 0.116 and transform it into a percentage?

Another question I have regards this problem:

Suppose the annually compounded rate is 18.5%. The present value of a $100$ perpetuity, with each cash flow received at the end of the year, is $100/.185 =$540.54.$ If the cash flow is received continuously, we must divide $100 by 17%, because 17% continuously compounded is equivalent to 18.5% annually compounded (with the explanation that $e^{0.17}=1.185).$

What is the relationship between the annually compounded rate and the continuously compounded rate ? How do I use one to calculate the other ? I am very confused.

logarithms exponential-function

share|cite|improve this question

edited Mar 15 at 17:46

saulspatz

17.1k31435

asked Mar 15 at 17:42

BM97BM97

778

$endgroup$

add a comment | 

0

$begingroup$

I am having some troubles understanding this topic of interest rates.

For example, if i invest 1 dollar at a continuous compounding rate of 11% , then my end of year value is equal to $e^{0.11}=1.116$ dollars. From here it says that investing at 11% a year continuously compounded is the same as investing 11.6 a year annually compounded.

Now where is this 11.6 coming from ? Is it coming from the 1.(116) dollars value? So i just take the decimal 0.116 and transform it into a percentage?

Another question I have regards this problem:

Suppose the annually compounded rate is 18.5%. The present value of a $100$ perpetuity, with each cash flow received at the end of the year, is $100/.185 =$540.54.$ If the cash flow is received continuously, we must divide $100 by 17%, because 17% continuously compounded is equivalent to 18.5% annually compounded (with the explanation that $e^{0.17}=1.185).$

What is the relationship between the annually compounded rate and the continuously compounded rate ? How do I use one to calculate the other ? I am very confused.

logarithms exponential-function

share|cite|improve this question

edited Mar 15 at 17:46

saulspatz

17.1k31435

asked Mar 15 at 17:42

BM97BM97

778

$endgroup$

add a comment | 

$begingroup$

I am having some troubles understanding this topic of interest rates.

For example, if i invest 1 dollar at a continuous compounding rate of 11% , then my end of year value is equal to $e^{0.11}=1.116$ dollars. From here it says that investing at 11% a year continuously compounded is the same as investing 11.6 a year annually compounded.

Now where is this 11.6 coming from ? Is it coming from the 1.(116) dollars value? So i just take the decimal 0.116 and transform it into a percentage?

Another question I have regards this problem:

Suppose the annually compounded rate is 18.5%. The present value of a $100$ perpetuity, with each cash flow received at the end of the year, is $100/.185 =$540.54.$ If the cash flow is received continuously, we must divide $100 by 17%, because 17% continuously compounded is equivalent to 18.5% annually compounded (with the explanation that $e^{0.17}=1.185).$

What is the relationship between the annually compounded rate and the continuously compounded rate ? How do I use one to calculate the other ? I am very confused.

logarithms exponential-function

share|cite|improve this question

edited Mar 15 at 17:46

saulspatz

17.1k31435

asked Mar 15 at 17:42

BM97BM97

778

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:46

saulspatz

17.1k31435

asked Mar 15 at 17:42

BM97BM97

778

share|cite|improve this question

edited Mar 15 at 17:46

saulspatz

17.1k31435

asked Mar 15 at 17:42

BM97BM97

778

edited Mar 15 at 17:46

saulspatz

17.1k31435

edited Mar 15 at 17:46

saulspatz

17.1k31435

asked Mar 15 at 17:42

BM97BM97

778

asked Mar 15 at 17:42

BM97BM97

778

1 Answer
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oldest

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0

$begingroup$

$11%$ per year, compounded continuously is (approximately) equivalent to $11.6%$ per year, compounded annually. If $i$ is the annual interest rate, the equivalent continuous rate is $ln(1+i)$. The reason is simply that $$e^{ln(1+i)}=1+i,$$
so that if you invest a dollar at a continuous rate of $ln(1+i),$ at the end of a year, you have exactly what you would have had you invested the dollar at a rate of $i,$ compounded annually.

Of course, if you know the continuous rate $delta$ and you want the equivalent annual rate $i,$ it's just $$i=e^delta-1$$

share|cite|improve this answer

answered Mar 15 at 17:58

saulspatzsaulspatz

17.1k31435

$endgroup$

add a comment | 

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0

$begingroup$

$11%$ per year, compounded continuously is (approximately) equivalent to $11.6%$ per year, compounded annually. If $i$ is the annual interest rate, the equivalent continuous rate is $ln(1+i)$. The reason is simply that $$e^{ln(1+i)}=1+i,$$
so that if you invest a dollar at a continuous rate of $ln(1+i),$ at the end of a year, you have exactly what you would have had you invested the dollar at a rate of $i,$ compounded annually.

Of course, if you know the continuous rate $delta$ and you want the equivalent annual rate $i,$ it's just $$i=e^delta-1$$

share|cite|improve this answer

answered Mar 15 at 17:58

saulspatzsaulspatz

17.1k31435

$endgroup$

add a comment | 

0

$begingroup$

$11%$ per year, compounded continuously is (approximately) equivalent to $11.6%$ per year, compounded annually. If $i$ is the annual interest rate, the equivalent continuous rate is $ln(1+i)$. The reason is simply that $$e^{ln(1+i)}=1+i,$$
so that if you invest a dollar at a continuous rate of $ln(1+i),$ at the end of a year, you have exactly what you would have had you invested the dollar at a rate of $i,$ compounded annually.

Of course, if you know the continuous rate $delta$ and you want the equivalent annual rate $i,$ it's just $$i=e^delta-1$$

share|cite|improve this answer

answered Mar 15 at 17:58

saulspatzsaulspatz

17.1k31435

$endgroup$

add a comment | 

$begingroup$

$11%$ per year, compounded continuously is (approximately) equivalent to $11.6%$ per year, compounded annually. If $i$ is the annual interest rate, the equivalent continuous rate is $ln(1+i)$. The reason is simply that $$e^{ln(1+i)}=1+i,$$
so that if you invest a dollar at a continuous rate of $ln(1+i),$ at the end of a year, you have exactly what you would have had you invested the dollar at a rate of $i,$ compounded annually.

Of course, if you know the continuous rate $delta$ and you want the equivalent annual rate $i,$ it's just $$i=e^delta-1$$

share|cite|improve this answer

answered Mar 15 at 17:58

saulspatzsaulspatz

17.1k31435

$endgroup$

share|cite|improve this answer

answered Mar 15 at 17:58

saulspatzsaulspatz

17.1k31435

answered Mar 15 at 17:58

saulspatzsaulspatz

17.1k31435

answered Mar 15 at 17:58

saulspatzsaulspatz

17.1k31435