Difficult Interest Scenario

[Timcago]

You invest $1500 at a local bank offering 4.5% interest that compounds its interest quarterly. I invest $1250 at another bank offering 5.5% interest that compounds its interest monthly.

a) who has more money after 9 years, and by how much.
b) Is there ever a point in time when the accounts will have the same amount? If so, when? How much will be in each account?

I can easily do part a

1500(1+(.045/4))^(4(9))=2243.87420154
1250(1+(.055/12))^(12(9))=2048.30501785

2243.87420154-2048.30501785= 195.5691837


b) here is my progress so far

1516.875^(4x)=1255.72916667^(12x)

4x*ln(1516.875)=12x*ln(1255.72916667)

I do not know where to go from here. I am not ever sure if that is the best way to set it up. Anyone know how to figure part b out?

 

[tkhunny]

1516.875^(4x)=1255.72916667^(12x)

??? Whence came those values?

Shouldn't it be

1500*(1 + 0.045/4)^(4*x) and 1250*(1 + 0.055/12)^(12*x)

That exponentiation preceeds the multiplication. Remember you rorder of operations.

Next, what does "Is there ever a point in time?" actually mean? If interest is credited monthly or quarterly, only whole quarters would constitute a positive response. Technically, they are not continuous functions.

 

[Denis]

You need to convert the 4.5% compounding quarterly to its
compounding monthly equivalent, which is ~4.4832%

Then you have:
1500(1 + .044832/12)^n = 1250(1 + .055/12)^n

Solve for n.

I assume you know how to get the 4.4832...; if not:
(1 + .045/4)^4 = (1 + i/12)^12
Solve for i.

 

[Timcago]

??? Whence came those values?

Shouldn't it be

1500*(1 + 0.045/4)^(4*x) and 1250*(1 + 0.055/12)^(12*x)

That exponentiation preceeds the multiplication. Remember you rorder of operations.

Next, what does "Is there ever a point in time?" actually mean? If interest is credited monthly or quarterly, only whole quarters would constitute a positive response. Technically, they are not continuous functions.

So your saying that, most likely, there is never a point in time since they are not continuous functions? The only chance they have at being equal is when they are both widthdrawn at the same time(the end of the year)?

The problem i see there is that the question does not say "is there ever a point in time *when you can widthdraw them both* and have an equal value. It only wants to know if there ever is a point in time when both are equal reguardless of whether you can widthdraw them or not.

You need to convert the 4.5% compounding quarterly to its
compounding monthly equivalent, which is ~4.4832%

Then you have:
1500(1 + .044832/12)^n = 1250(1 + .055/12)^n

Solve for n.

I assume you know how to get the 4.4832...; if not:
(1 + .045/4)^4 = (1 + i/12)^12
Solve for i.

Ok, i still dont knw how to solve for n or I

This is as far as i know how to take it

n*ln(1500(1 + .044832/12)=n*ln(1250(1 + .055/12)

Now what?

 

[soroban]

Hello, Timcago!

Your set-up for part (b) is incorrect . . .

You invest $1500 at a local bank offering 4.5% interest that compounds its interest quarterly.
I invest $1250 at another bank offering 5.5% interest that compounds its interest monthly.

b) Is there ever a point in time when the accounts will have the same amount?
If so, when? How much will be in each account?

Your balance after n years is: \(\displaystyle \,1250\left(1\,+\,\frac{0.055}{12}\right)^{12n} \;\approx\;1250(1.00458)^{12n}\)

My balance after n years is: \(\displaystyle \,1500\left(1\,+\,\frac{0.045}{4}\right)^{4n} \;= \;1500(1.01125)^{4n}\)

If the balance are equal: \(\displaystyle \,1250(1.00458)^{12n}\;=\;1500(1.01125)^{4n}\)

Divide by 250: \(\displaystyle \;5(1.00458)^{12n}\;=\;6(1.01125)^{4n}\)

\(\displaystyle \;\;\)On the left side, we have: \(\displaystyle \,(1.00458)^{12n}\;=\;\left[(1.00458)^3\right]^{4n}\;\approx\;(1.01380)^{4n}\)

The equation becomes: \(\displaystyle \:5(1.01380)^{4n}\;=\;6(1.01125)^{4n}\)

Then: \(\displaystyle \L\,\frac{(1.01380)^{4n}}{(1.01125)^{4n}} \;=\;\frac{6}{5}\;\;\Rightarrow\;\;\left(\frac{1.01380}{1.01125}\right)^{4n}\;=\;1.2\)

\(\displaystyle \;\;\)which is approximately: \(\displaystyle \1.00252)^{4n} \;= \;1.2\)


Take logs: \(\displaystyle \:\log(1.00252)^{4n}\:=\:\log(1.2)\;\;\Rightarrow\;\;4n\cdot\log(1.00252)\:=\:\log(1.2)\)

Therefore: \(\displaystyle \L\:n\:=\:\frac{\log(1.2)}{4\log(1.00252)} \:= \:18.11023667\)

The balances will be (almost) equal in about 18 years.


For \(\displaystyle n\,=\,18\), your balance is: \(\displaystyle \,1250(1.00458)^{216}\:\approx\:\$3354.04\)

My balance will be: \(\displaystyle \,1500(1.01125)^{72} \:\approx\:\$3356.65\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For more accuracy, don't use rounded-off decimals like I did.

 

[Timcago]

Awesome, thank you so much. That problem took hours away from my life, but now i understand it.

 

[tkhunny]

It is not a matter of "withdraw", rather crediting. If one credits onthly and the other quarterly, it could be a trick business to track down an equivalence. There amy be one, but it is non-trivial task without the assumption of continuous functions.

18.11023667 is 18 years and change
0.11023667*12 = 1.32284 months.

18 years
Monthly Crediting: 3356.443
Quarterly Crediting: 3356.648 <== Ahead

18 years + 1 month
Monthly Crediting: 3371.826 <== Ahead
Quarterly Crediting: 3356.648

18 years + 2 month
Monthly Crediting: 3387.281 <== Ahead
Quarterly Crediting: 3356.648

18 years + 3 month
Monthly Crediting: 3402.806
Quarterly Crediting: 3394.410

Were they EVEr "the same"? Now, I am almost certain you are not expected to make this sort of examination. However, it really should not be overlooked tacitly. It emphasizes the importance of reading your contract VERY CAREFULLY and making sure you understand the terms. Meddling in interest rates really is not for the casual observer.