Question 83310
3)

Using this formula {{{A=P(1+r/n)^(nt)}}} calculate the following:


a)Calculate the return (A) if the bank compounds annually (n = 1)

{{{A=10000(1+0.10/1)^(1*2)}}}       Start with the given expression
{{{A=10000(1+0.1)^(2)}}}     Divide 0.10 by 1 and multiply the exponents 1 and 2
{{{A=10000(1.1)^(2)}}}       Add
{{{A=10000(1.21)}}}      Raise 1.1 to 2
{{{A=12100}}}       Multiply


So the return is $12,100



b)Calculate the return (A) if the bank compounds quarterly (n = 4)

{{{A=10000(1+0.10/4)^(4*2)}}}       Start with the given expression
{{{A=10000(1+0.025)^(8)}}}     Divide 0.10 by 4 and multiply the exponents 4 and 2
{{{A=10000(1.025)^(8)}}}       Add
{{{A=10000(1.21840289750992)}}}      Raise 1.025 to 8
{{{A=12184.0289750992}}}       Multiply


So the return is $12184.03



c)Calculate the return (A) if the bank compounds monthly(n = 12)

{{{A=10000(1+0.10/12)^(12*2)}}}       Start with the given expression
{{{A=10000(1+0.00833333333333333)^(24)}}}     Divide 0.10 by 12 and multiply the exponents 12 and 2
{{{A=10000(1.00833333333333)^(24)}}}       Add
{{{A=10000(1.22039096137556)}}}      Raise 1.00833333333333 to 24
{{{A=12203.9096137556}}}       Multiply


So the return is $12203.91


d)Calculate the return (A) if the bank compounds daily (n = 365)

{{{A=10000(1+0.10/365)^(365*2)}}}       Start with the given expression
{{{A=10000(1+0.000273972602739726)^(730)}}}     Divide 0.10 by 365 and multiply the exponents 365 and 2
{{{A=10000(1.00027397260274)^(730)}}}       Add
{{{A=10000(1.22136930163979)}}}      Raise 1.00027397260274 to 730
{{{A=12213.6930163979}}}       Multiply


So the 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.