Interest compounded monthly and quartely

**Kristina** · December 2nd 2010, 05:21 AM

HI

A person wants to take out a loan of DKK 300,000. The person gets two offers.
Offer 1: Yearly nominal interest of 4.2 % compounded monthly.
Offer 2: Yearly nominal interest of 4.3 % compounded quarterly.
Which offer should the person choose?

I wanted to ask, I'm doing every thing all right?
Offer 1 $APR=100 (1+{\frac{\frac{4,2}{12}}{100}}^{12}-1)=4,281800720$
Offer 2 $APR=100 (1+{\frac{\frac{4,3}{4}}{100}}^{4}-1)=4.369835754$

**e^(i*pi)** · December 2nd 2010, 07:01 AM

Originally Posted by Kristina

HI

A person wants to take out a loan of DKK 300,000. The person gets two offers.
Offer 1: Yearly nominal interest of 4.2 % compounded monthly.
Offer 2: Yearly nominal interest of 4.3 % compounded quarterly.
Which offer should the person choose?

I wanted to ask, I'm doing every thing all right?
Offer 1 $APR=100 (1+{\frac{\frac{4,2}{12}}{100}}^{12}-1)=4,281800720$
Offer 2 $APR=100 (1+{\frac{\frac{4,3}{4}}{100}}^{4}-1)=4.369835754$

Why are you subtracting 1 in the brackets? Your APR is given in the question so I don't get why you want to work that out. I'd just take a time period of a year and use the compound interest formula below:

The compound interest formula is $P(n) = P_0\left(1+\dfrac{r}{n}\right)^n$

Where:

$P(n)$ = amount owed after n compounding periods
$P_0$ = the initial amount borrowed/owed
$r$ = the nominal annual rate expressed as a decimal
$n$ = number of compounding periods

For offer 1: $n_1 = 12$ and $r = 0.042$

For offer 2: $n_2 = 4$ and $r = 0.043$

In both cases $P_0 = 3 \cdot 10^5$

Plug and chug to see which is lower

**Kristina** · December 2nd 2010, 07:13 AM

why n2 is 3 ? not 4?

Frequent Compounding of Interest:

What if interest is paid more frequently?
Here are a few examples of the formula:
Annually = P × (1 + r) = (annual compounding)
Quarterly = P (1 + r/4)4 = (quarterly compounding)
Monthly = P (1 + r/12)12 = (monthly compounding)

**e^(i*pi)** · December 2nd 2010, 07:24 AM

Originally Posted by Kristina

why n2 is 3 ? not 4?

Frequent Compounding of Interest:

What if interest is paid more frequently?
Here are a few examples of the formula:
Annually = P × (1 + r) = (annual compounding)
Quarterly = P (1 + r/4)4 = (quarterly compounding)
Monthly = P (1 + r/12)12 = (monthly compounding)

Because I made a mistake, n should be 4. I got confused with quarterly.

If interest is paid more frequently then you'll find that the amount of interest paid converges to $e$

If you compound more and more frequently you'll eventually have an infinitesimal compounding period. Mathematically it will be the limit to infinity:

$\displaystyle \lim_{n \to \infty} \left(1+\dfrac{r}{n}\right)^n = e$

Spoiler:

**Kristina** · December 2nd 2010, 07:54 AM

okay then, but I have one more question
b) The person chooses offer 1. The loan must be repaid over 6 years with 72 equal monthly payments. The first repayment is to be paid one month after the raising of the loan.
Calculate the amount of the monthly payment.

Which formula I should jus?
$P(n) = P_0\left(1+\dfrac{r}{n}\right)^6$ this or APR?

**e^(i*pi)** · December 2nd 2010, 09:05 AM

Originally Posted by Kristina

okay then, but I have one more question
b) The person chooses offer 1. The loan must be repaid over 6 years with 72 equal monthly payments. The first repayment is to be paid one month after the raising of the loan.
Calculate the amount of the monthly payment.

Which formula I should jus?
$P(n) = P_0\left(1+\dfrac{r}{n}\right)^6$ this or APR?

For t years use the formula $P(n) = P_0\left(1+\dfrac{r}{n}\right)^{nt}$ which is the normal formula raised to the power $t$ .

Alternatively you can say that there are $12 \cdot 6 = 72$ compounding periods and set $n=72$

$P(n) = 3 \times 10^5 \left(1+\dfrac{0.042}{72}\right)^{72}$

**Kristina** · December 2nd 2010, 09:24 AM

so if i get 312864,5124, I need 312864,5124-300000=12864,51240 Yes?

**Kristina** · December 2nd 2010, 09:41 AM

And Last PART OF EXERCISE
Just before payment no 52 the borrower wins the Lotto and therefore wishes to pay off the entire loan at the same time as payment no 52 has to be paid.
How much must the borrower pay in addition to payment no 52 to pay off the loan
$P(n) = 3 \times 10^5 \left(1+\dfrac{0.042}{52}\right)^{52}=312863,0397$
312863,0397-300000=12863,0397
The borrower for 52 need to pay 12863,0397
So if total is 72 payments. 72-52=20 Is left 20 payments
So logical I need to do like this 12863,0397*20=257260,7940
Somenting is WRONG It's to huge numbers

**Wilmer** · December 2nd 2010, 06:20 PM

Originally Posted by Kristina

okay then, but I have one more question
b) The person chooses offer 1. The loan must be repaid over 6 years with 72 equal monthly payments. The first repayment is to be paid one month after the raising of the loan.
Calculate the amount of the monthly payment.
Which formula I should jus?
$P(n) = P_0\left(1+\dfrac{r}{n}\right)^6$ this or APR?

i = .042 / 12
Payment = 300000(i) / [1 - 1/(1+i)^72] = 4720.9424...
Formula:
P = A(i) / [1 - 1/(1+i)^n]

Where are you getting your formulas? They sure look strange...

**Wilmer** · December 2nd 2010, 06:41 PM

Originally Posted by Kristina

And Last PART OF EXERCISE
Just before payment no 52 the borrower wins the Lotto and therefore wishes to pay off the entire loan at the same time as payment no 52 has to be paid.
How much must the borrower pay in addition to payment no 52 to pay off the loan
$P(n) = 3 \times 10^5 \left(1+\dfrac{0.042}{52}\right)^{52}=312863,0397$
312863,0397-300000=12863,0397
The borrower for 52 need to pay 12863,0397
So if total is 72 payments. 72-52=20 Is left 20 payments
So logical I need to do like this 12863,0397*20=257260,7940
Somenting is WRONG It's to huge numbers

No; sorry, but that's VERY wrong!
Has to be done in 2 steps:
1: Calculate FV of 300,000 after 52 months
2: Calculate FV of monthly payments of 4,720.9424... after 52 months
The difference is what will be owing.

i = .042/12
1: 300000(1+i)^52
2: 4720.9424[(1+i)^52 - 1] / i
Difference will be ~91,036.25

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