Question 82927
{{{A=p(1+r/n)^(n*t)}}} Start with the given equation


{{{A=10000(1+0.1/n)^(n*2)}}}  Plug in p=10000, r=0.1


a) 

Lets calculate the return if the bank compounds annually

Let n=1 and plug it into  {{{A=10000(1+0.1/n)^(n*2)}}}

{{{A=10000(1+0.1/1)^(1*2)}}}       Start with the given expression
{{{A=10000(1+0.1)^(1*2)}}}     Divide 0.1 by 1 to get 0.1
{{{A=10000(1+0.1)^(2)}}}     Multiply the exponents 1 and 2 to get 2
{{{A=10000(1.1)^(2)}}}       Add 1 and 0.1 to get 1.1
{{{A=10000(1.21)}}}      Raise 1.1 to 2 to get 1.21
{{{A=12100}}}       Multiply 10000 and 1.21 to get 12100

So our return is $12100
b) 

Lets calculate the return if the bank compounds quarterly

Let n=4 and plug it into  {{{A=10000(1+0.1/n)^(n*2)}}}

{{{A=10000(1+0.1/4)^(4*2)}}}       Start with the given expression
{{{A=10000(1+0.025)^(4*2)}}}     Divide 0.1 by 4 to get 0.025
{{{A=10000(1+0.025)^(8)}}}     Multiply the exponents 4 and 2 to get 8
{{{A=10000(1.025)^(8)}}}       Add 1 and 0.025 to get 1.025
{{{A=10000(1.21840289750992)}}}      Raise 1.025 to 8 to get 1.21840289750992
{{{A=12184.0289750992}}}       Multiply 10000 and 1.21840289750992 to get 12184.0289750992

So our return is $12184.03
c) 

Lets calculate the return if the bank compounds monthly

Let n=12 and plug it into  {{{A=10000(1+0.1/n)^(n*2)}}}

{{{A=10000(1+0.1/12)^(12*2)}}}       Start with the given expression
{{{A=10000(1+0.00833333333333333)^(12*2)}}}     Divide 0.1 by 12 to get 0.00833333333333333
{{{A=10000(1+0.00833333333333333)^(24)}}}     Multiply the exponents 12 and 2 to get 24
{{{A=10000(1.00833333333333)^(24)}}}       Add 1 and 0.00833333333333333 to get 1.00833333333333
{{{A=10000(1.22039096137556)}}}      Raise 1.00833333333333 to 24 to get 1.22039096137556
{{{A=12203.9096137556}}}       Multiply 10000 and 1.22039096137556 to get 12203.9096137556

So our return is $12203.91
d) 

Lets calculate the return if the bank compounds daily

Let n=365 and plug it into  {{{A=10000(1+0.1/n)^(n*2)}}}

{{{A=10000(1+0.1/365)^(365*2)}}}       Start with the given expression
{{{A=10000(1+0.000273972602739726)^(365*2)}}}     Divide 0.1 by 365 to get 0.000273972602739726
{{{A=10000(1+0.000273972602739726)^(730)}}}     Multiply the exponents 365 and 2 to get 730
{{{A=10000(1.00027397260274)^(730)}}}       Add 1 and 0.000273972602739726 to get 1.00027397260274
{{{A=10000(1.22136930163979)}}}      Raise 1.00027397260274 to 730 to get 1.22136930163979
{{{A=12213.6930163979}}}       Multiply 10000 and 1.22136930163979 to get 12213.6930163979

So our return is $12213.69


e) What observation can you make about the size of the increase in your return as your compounding increases more frequently?


As the compounding frequency increases, the return slowly approaches some finite number (which in this case appears to be about $12213.69). Think about it, banks wouldn't be too fond of shelling out an infinite amount of cash.



f)Calculate A with continuous compounding

Using the contiuous compounding formula {{{A=Pe^(rt)}}} where e is the constant 2.7183 and letting r=0.1, P=10,000, and t=2 we get


{{{A=10000(2.7183)^(0.1*2)}}} Start with the given equation


{{{A=10000(2.7183)^(0.2)}}} Multiply 0.1 and 2


{{{A=10000(1.22140439115573)}}} Raise 2.7183 to 0.2


{{{A=12214.0439115572}}} Multiply


So using continuous compounding interest we get a return of $12,214.04 (which is real close to what we got from a daily compounding frequency)



g)Now suppose, instead of knowing t, we know that the bank returned to us $15,000 with the bank compounding continuously. Using natural logarithms, find how long we left the money in the bank (find t)


{{{15000=10000e^(0.1t)}}} 


{{{15000/10000=e^(0.1t)}}} Divide both sides by 10,000


{{{1.5=e^(0.1t)}}} 

{{{ln(1.5)=0.1t}}} Take the natural log of both sides. This eliminates "e".The natural log (pronounced "el" "n") is denoted "ln" on calculators.


{{{ln(1.5)/0.1=t}}} Divide both sides by 0.1


So we get


{{{t=0.4054/0.1=4.054}}}

{{{t=4.054}}}


So it will take about 4 years to generate $15,000



h) A commonly asked question is, “How long will it take to double my money?” At 10% interest rate and continuous compounding, what is the answer?


Since we want to double our money, let A=2*10,000. So A=20,000. Now solve for t:

{{{20000=10000e^(0.1t)}}} 


{{{20000/10000=e^(0.1t)}}} Divide both sides by 10,000


{{{2=e^(0.1t)}}} 

{{{ln(2)=0.1t}}} Take the natural log of both sides. This eliminates "e".The natural log (pronounced "el" "n") is denoted "ln" on calculators.


{{{ln(2)/0.1=t}}} Divide both sides by 0.1


So we get


{{{t=0.69314/0.1=6.9314}}}

{{{t=6.9314}}}


So it will take about 7 years to double your money.