(last updated: 1:57:38 PM, October 08, 2020)
Interest: \(I=Prt\)
Future value: \(A=P+I\)
Credit card interest (also called finance charges) can be calculated using simple interest (or at least approximated well).
I have seen the average daily balance method used. There are other methods, but I will only show you this one.
\(I=Prt\)
\(I=\) interest = finance charge for a billing cycle.
\(P=\) average daily balance during billing cycle
\(r=\) annual interest rate
\(t=\) portion of the year that is covered by the billing cycle, you can do \(t=1/12\) for a month, or \(t=(\text{number of days})/365\) or \(t=(\text{number of days})/360\) as well. Typically if the billing cycle is a month, \(t=1/12\) is the easiest choice.
Average daily balance
To calculate the average daily balance, we need to know the closing balance on each day of the billing cycle: \(B_1,B_2,B_3,\ldots,B_{29},B_{30}\) where \(B_n\) is the balance on day \(n\). Of course, the billing cycle could have any number of days, but is usually a month of 28 to 31 days.
The average daily balance is just the average of all the \(B_n\)’s: \(P=\frac1n(B_1+B_2+\cdots+B_{30})\).
Example:
Suppose we start with a balance of $0 for a billing cycle with 31 days, and a purchase for $500 is made that posts on the 13\(^{th}\) day. Then the balance is $0 for 12 days and $500 for 19 days. The average daily balance is \(\frac{0\times12+500\times19}{31}=306.45\).
Suppose the interest rate is 17%. Now we can calculate the finance charge: \(I=Prt=306.45 \times 0.17 \times 1/12=4.34\). So we would pay a \(\$4.34\) interest charge on that credit card account for this billing cycle.
Example: Suppose a credit card has a billing cycle of 28 days and an interest rate of 24.35%. Also suppose the balance is initially $500, then a $400 refund posts on day 3, and a $150 charge posts on day 12, then a $300 charge posts on day 25. Calculate the finance charge using the average daily balance method.
\(P=\frac{500\cdot 2+100\cdot9+250\cdot 13+550\cdot4}{28}=262.50\)
\(I=262.50\cdot 0.2425\cdot \frac{28}{365}= \ \boxed{\$ 4.88}.\)
\(A=P(1+\frac rn)^{nt}\)
\(A=e^{rt}\)
\((\text{future dollar buying power})=\frac{(\text{current dollar buying power})}{(1+r)^{t}}\)
\((\text{future price})=(\text{current price})\times(1+r)^{t}\)
For the US, the target inflation rate is \(r=2\%\).
Buying power of the dollar
We can think of inflation as the devaluation of currency. $1 today will be “worth” \(\frac{\$1}{(1+r)^t}\) in \(t\) years
Prices over time
Inflation is best understood as the general increase in prices over time.
An item that costs $1 today will cost \(\$1(1+r)^t\) in \(t\) years.
Example: What will the buying power of the US dollar be in 15 years?
Solution: \(\frac{1}{1.02^{15}}=0.74\) so a dollar will only be “worth” 74 cents in 15 years. To understand this, you should realize that the dollar will still be “worth” a dollar, but its “buying power” in 15 years will be equivalent to the buying power of 74 cents today. In other words, something that costs 74 cents today, will cost 1 dollar in 15 years.
Example: If inflation is 2% per year and an item costs $800, how much do you expect it to cost in 25 years?
Solution: \(800(1.02)^{25}=1312.49\) so we expect it to cost \(\$1,312.49\) in 25 years.
\(A=m\left[\frac{(1+\frac rn)^{nt}-1}{\frac rn}\right]\)
A savings/investment example with initial deposit and monthly deposits:
Let’s say we deposit $5,000 in an account and thereafter deposit $100 monthly for 10 years. Suppose the account earns 11% interest compounded monthly. Calculate the future value at the end to the decade.
Solution: Use the compound interest formula on the initial deposit, then the annuity formula on the monthly deposits, and add the results.
Initial deposit: \(A=5000\left(1+\frac{0.11}{12}\right)^{120}=\$14,945.75\)
Monthly deposits: \(A=100\cdot\left[\frac{(1+\frac {0.11}{12})^{120}-1}{\frac {0.11}{12}}\right]=\$21,699.81\)
So the total future value is \(\$14,945.75+\$21,699.81=\$36,645.56\)
This is generally how monthly payments are calculated on most modern loans such as those for cars and homes.
\(P=\) loan amount (home price minus down payment plus any added fees that will be financed)
\(r=\) interest rate
\(n=12\) for monthly payments
\(t=\) term of loan in years, usualy 15, 20, 25, or 30. \(t=30\) is the most common
Monthly payment: \(m=\frac{P \frac rn}{1-(1+\frac rn)^{-nt}}\)
To calculate the loan you can afford, solve the above for \(P\):
Maximum loan amount: \(P=m\frac{1-(1+\frac rn)^{-nt}}{\frac rn}\)
To break down your first monthly payment into principal and interest:
Amount of interest in first monthly payment: \(P\cdot\frac rn\)
Amount of principle in first monthly payment: \(m-P\cdot\frac rn\)
Example:
A home loan of $250,000 at 4.5% interest over 30 years gives a monthly payment of \(m=\frac{P \frac rn}{1-(1+\frac rn)^{-nt}} =\frac{250000\cdot \frac{0.045}{12}}{1-(1+\frac{0.045}{12})^{-360}} =\$1,266.71\)
Calculate the principal and interest portions of the first monthly payment.
Interest: \(P\cdot\frac rn=250000\cdot \frac{0.045}{12}=\$937.50\)
Principal: \(m-P\cdot\frac rn=250000\cdot \frac{0.045}{12}=1266.71-937.50=\$329.21\)
Example:
If we can afford $1,200 per month and can secure of home loan for 30 years at 4.5%, then the maximum loan amount we can take is
\(P=m\frac{1-(1+\frac rn)^{-nt}}{\frac rn} =1200\cdot \frac{1-(1+\frac{0.045}{12})^{-360}}{\frac{0.045}{12}} =\$236,833.40\)
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