Create an Installment Plan for Mr. Budi's Annuity

[Monoxdifly]

Mr. Budi borrows 2,000,000IDR which will be amortized with 10 annuities. The first annuity will be paid in 1 year with 10% per year interest. Make the installment plan!
For the annuity I got A = \(\displaystyle M\times\frac1{\sum_{n=1}^p(1+i)^{-n}}\) = \(\displaystyle 2,000,000\times\frac1{\sum_{n=1}^{10}(1+0,1)^{-n}}\) = \(\displaystyle 2,000,000\times\frac1{\sum_{n=1}^{10}(1,1)^{-n}}\) = \(\displaystyle 2,000,000\times\frac1{\sum_{n=1}^{10}(1,1)^{-n}}\) = \(\displaystyle 2,000,000\times\frac1{(1,1)^{-1}+(1,1)^{-2}+(1,1)^{-3}+(1,1)^{-4}+(1,1)^{-5}+(1,1)^{-6}+(1,1)^{-7}+(1,1)^{-8}+(1,1)^{-9}+(1,1)^{-10}}\)
(\(\displaystyle (1,1)^{-1} + (1,1)^{-2} + … + (1,1)^{-10}\) is a geometric sequence with the first term \(\displaystyle a = (1,1)^{-1}\) and ratio \(\displaystyle r = (1,1)^{-1}\))
= \(\displaystyle 2,000,000\times\frac1{\frac{1-1.1^{-10}}{0.1}}\) = \(\displaystyle 2,000,000\times\frac{0.1}{1-1.1^{-10}}\) = 2,000,000 × 0.61445671 = 1,228,913.42
Thus, the annuity is 1,228,913.42.
On the end of first year:
Annuity = 1,228,913.42IDR
Interest : 10% × 2,000,000IDR = 200,000IDR
Installment : 1,228,913.42IDR – 200.000IDR = 1,028,913.42IDR
Remaining loan : 2,000,000IDR – 1,028,913.42IDR = 971,086.58IDR
On the end of second year:
Annuity = 1,228,913.42IDR
Interest : 10% × 971,086.58IDR = 97,108.66IDR
Installment : 1,228,913.42IDR – 97,108.66IDR = 1,131,804.76IDR
Remaining loan : 971,086.58IDR – 1,131,804.76IDR = -159.718,58IDR
Why am I seeing negatives already?

 

[Olinguito]

Thus, the annuity is 1,228,913.42.

I’m no expert on these things but this figure looks way too high to me. It means that at the end of the first of ten years you would have to pay back more than half the amount of your loan already.

 

[tkhunny]

Mr. Budi borrows 2,000,000IDR which will be amortized with 10 annuities. The first annuity will be paid in 1 year with 10% per year interest. Make the installment plan!

Try that again...

$2,000,000\times\frac{0.1}{1-1.1^{-10}} = 2,000,000 \times \dfrac{0.1}{0.61445671}$

 

[Monoxdifly]

It's weird. My calculator app gives the same result for both \(\displaystyle 1-1.1^{-10}\) and \(\displaystyle \frac{0.1}{1-1.1^{-10}}\).

 

[tkhunny]

It's weird. My calculator app gives the same result for both \(\displaystyle 1-1.1^{-10}\) and \(\displaystyle \frac{0.1}{1-1.1^{-10}}\).

1) Delete that app, or
2) Add parentheses.

 

[Monoxdifly]

2) Add parentheses.

Both 1-1.1^(-10) and 0.1/(1-1.1^(-10)) yield the same results on the calculator. And it seems like it's Windows calculator.

 

[Olinguito]

$$1-1.1^{-10}\ =\ 0.61445671057046825263559635552114$$

$$\frac{0.1}{1-1.1^{-10}}\ =\ 0.1627453948825116076230715715748$$

on my Windows calculator.

 

[tkhunny]

Both 1-1.1^(-10) and 0.1/(1-1.1^(-10)) yield the same results on the calculator. And it seems like it's Windows calculator.

Okay, that leaves the other option.

This is why I do not recommend carrying such expressions as $1.1^{-10}$. Too much confusion.

I MUCH prefer this version:

$v = \dfrac{1}{1.1}$

$v^{10}$

So much cleaner.

Your ten payments are $v + v^{2} + ... + v^{10}$

Your sum is $\dfrac{v-v^{11}}{1-v}$, which is just as easily converted to $\dfrac{v\cdot\left(1-v^{10}\right)}{v\cdot i} = \dfrac{1-v^{10}}{i}$ if you like.

So much less to go wrong.

 

[Wilmer]

STANDARD formula (google would have given it to you Mr.Fly!):
P = Ai / (1-v) where v = 1 / (1+i)^n (same as TK's)

P = ?
A = 2,000,000
i = .10
n = 10