help with finance (1 Viewer)

stressedadfff · Nov 13, 2021

i dont really get wha tit means by 'charged monthly' and then '5 annual isntallments?

Screen Shot 2021-11-13 at 11.14.44 pm.png

Greninja340 · Nov 13, 2021

the interest is applied monthly and the principal amount is paid off in 5 yearly instalments

stressedadfff · Nov 13, 2021

Greninja340 said:
the interest is applied monthly and the principal amount is paid off in 5 yearly instalments

could you pleasee explain that i dont get it

stressedadfff · Nov 13, 2021

does anyone know which formula to use for c also?

Screen Shot 2021-11-13 at 11.48.20 pm.png

cossine · Nov 13, 2021

stressedadfff said:
i dont really get wha tit means by 'charged monthly' and then '5 annual isntallments?

View attachment 33692

The "charged monthly" is the rate at which interest in being charged. So since the rate is 15% p.a. 15% is the yearly rate. You need to divide by 12 to get the monthly rate you are being charged.

Annual meaning yearly. Note p.a is per annum meaning per a year.

Installment is referring to the payment of the loan.

So '5 annual installments' is basically saying make 5 payments to pay off the loan and the payment is made each year.

cossine · Nov 13, 2021

stressedadfff said:
does anyone know which formula to use for c also?
View attachment 33694

Do you have any answers?

cossine · Nov 13, 2021

stressedadfff said:
does anyone know which formula to use for c also?
View attachment 33694

nevermind flat interest rate is just simple interest. So interest is being applied on 62500

stressedadfff · Nov 14, 2021

cossine said:
The "charged monthly" is the rate at which interest in being charged. So since the rate is 15% p.a. 15% is the yearly rate. You need to divide by 12 to get the monthly rate you are being charged.

Annual meaning yearly. Note p.a is per annum meaning per a year.

Installment is referring to the payment of the loan.

So '5 annual installments' is basically saying make 5 payments to pay off the loan and the payment is made each year.

oh i see thanks! i just dont get how to solve it would it monthly or yearly.... tkrjlerjbd

stressedadfff · Nov 14, 2021

cossine said:
Do you have any answers?

yep 18.8%?

cossine · Nov 14, 2021

stressedadfff said:
yep 18.8%?

1st question https://boredofstudies.org/threads/annoying-loan-reyament-questions.109719/

2nd question

a) 5 *2022 * 12= 121320
b) I assume you got this
c) 62500(1+p*5)= 121320

=> p = [(121320/62500) - 1 ]/5 = 18.8%

CM_Tutor · Nov 14, 2021

stressedadfff said:
oh i see thanks! i just dont get how to solve it would it monthly or yearly.... tkrjlerjbd

The important thing to recognise is:

if interest is charged at 12% pa, and compounded yearly, then interest will be charged once at 12%, so if you owe $A at the start of the year, the amount owing at the end of the year (before any payment is made) is

$A \times \left(1 + \cfrac{12}{100}\right)^1 = A(1.12)$

if interest is charged at 12% pa, and compounded monthly, then interest will be charged at 1% but at the end of each month, and compounding, so if you owe $A at the start of the year, the amount owing at the end of the year (before any payment is made) is

$A \times \left(1 + \cfrac{1}{100}\right)^{12}=A(1.01)^{12}$

stressedadfff · Nov 14, 2021

CM_Tutor said:
The important thing to recognise is:

if interest is charged at 12% pa, and compounded yearly, then interest will be charged once at 12%, so if you owe $A at the start of the year, the amount owing at the end of the year (before any payment is made) is

$A \times \left(1 + \cfrac{12}{100}\right)^1 = A(1.12)$

if interest is charged at 12% pa, and compounded monthly, then interest will be charged at 1% but at the end of each month, and compounding, so if you owe $A at the start of the year, the amount owing at the end of the year (before any payment is made) is

$A \times \left(1 + \cfrac{1}{100}\right)^{12}=A(1.01)^{12}$

i think what im confused about is some of the annuities go like
A1= P(r)^12
A2 = P(r)^11
how come the n is decreasing as the year is going on, im not sure when to use this type?

CM_Tutor · Nov 14, 2021

stressedadfff said:
i think what im confused about is some of the annuities go like
A1= P(r)^12
A2 = P(r)^11
how come the n is decreasing as the year is going on, im not sure when to use this type?

Suppose I invest $1000 on 1 Jan each year, and it earns interest at 12% pa, compounded monthly. I start on 1 Jan 2022. How much will I have on 31 Dec 2030?

Let $A_n be the amount the nth deposit grows to. So, $A₁ is invested for all of 2022, 2023, 2024, ... and 2030, so for 9 years:

$A_1 = 1000\left(1 + \cfrac{1}{100}\right)^{12 \times 9} = 1000(1.01)^{108}$

$A₂ is invested on 1 Jan 2023 and so grows for all of 2023, 2024, 2025, ... and 2030, so for 8 years:

$A_2 = 1000\left(1 + \cfrac{1}{100}\right)^{12 \times 8} = 1000(1.01)^{96}$

$A₃ is invested on 1 Jan 2024 and so grows for all of 2024, 2025, 2026, ... and 2030, so for 7 years:

$A_3 = 1000\left(1 + \cfrac{1}{100}\right)^{12 \times 7} = 1000(1.01)^{84}$

Continuing on until the investment on 1 Jan 2030, which grows for only 1 year. It is the 9th investment, and so is $A₉:

$A_9 = 1000\left(1 + \cfrac{1}{100}\right)^{12 \times 1} = 1000(1.01)^{12}$

So, on 31 Dec 2030, the amount that I will have is

$\begin{align*} A_1 + A_2 + A_3 + ... + A_9 &= 1000(1.01)^{108} + 1000(1.01)^{96} + 1000(1.01)^{84} + ... + 1000(1.01)^{12} \\ &= 1000(1.01)^{12} + 1000(1.01)^{24} + 1000(1.01)^{36} + ... + 1000(1.01)^{108} \qquad \text{as the GP and pattern is easier to see with increasing terms} \\ &= 1000(1.01)^{12}\big[1 + (1.01)^{12} + (1.01)^{24} + ... + (1.01)^{96}\big] \qquad \text{where we can see a GP of $n = 9$ terms with $a=1$ and $r=1.01^{12}$} \\ &= 1000(1.01)^{12} \times \cfrac{1 \times \bigg[\left[(1.01)^{12}\right]^9 - 1\bigg]}{(1.01)^{12} - 1} \\ &= \cfrac{1000(1.01)^{12}\left[(1.01)^{108} - 1\right]}{(1.01)^{12} - 1} \\ &\approx \$17\ 138.27 \qquad \text{(nearest cent)} \end{align*}$

Do you see that the powers decrease as each subsequent investment is compounding for a shorter period of time?

Many would approach this by defining $A_n as the total amount at the time the nth deposit is made, but this makes for a more complicated approach (in my opinion) as you can equally treat each investment as a single account that grows without interruption and a new account is set up for the next deposit. You can then simply add up the accounts to get the total investment.

stressedadfff · Nov 14, 2021

CM_Tutor said:
Suppose I invest $1000 on 1 Jan each year, and it earns interest at 12% pa, compounded monthly. I start on 1 Jan 2022. How much will I have on 31 Dec 2030?

Let $A_n be the amount the nth deposit grows to. So, $A₁ is invested for all of 2022, 2023, 2024, ... and 2030, so for 9 years:

$A_1 = 1000\left(1 + \cfrac{1}{100}\right)^{12 \times 9} = 1000(1.01)^{108}$

$A₂ is invested on 1 Jan 2023 and so grows for all of 2023, 2024, 2025, ... and 2030, so for 8 years:

$A_2 = 1000\left(1 + \cfrac{1}{100}\right)^{12 \times 8} = 1000(1.01)^{96}$

$A₃ is invested on 1 Jan 2024 and so grows for all of 2024, 2025, 2026, ... and 2030, so for 7 years:

$A_3 = 1000\left(1 + \cfrac{1}{100}\right)^{12 \times 7} = 1000(1.01)^{84}$

Continuing on until the investment on 1 Jan 2030, which grows for only 1 year. It is the 9th investment, and so is $A₉:

$A_9 = 1000\left(1 + \cfrac{1}{100}\right)^{12 \times 1} = 1000(1.01)^{12}$

So, on 31 Dec 2030, the amount that I will have is

$\begin{align*} A_1 + A_2 + A_3 + ... + A_9 &= 1000(1.01)^{108} + 1000(1.01)^{96} + 1000(1.01)^{84} + ... + 1000(1.01)^{12} \\ &= 1000(1.01)^{12} + 1000(1.01)^{24} + 1000(1.01)^{36} + ... + 1000(1.01)^{108} \qquad \text{as the GP and pattern is easier to see with increasing terms} \\ &= 1000(1.01)^{12}\big[1 + (1.01)^{12} + (1.01)^{24} + ... + (1.01)^{96}\big] \qquad \text{where we can see a GP of $n = 9$ terms with $a=1$ and $r=1.01^{12}$} \\ &= 1000(1.01)^{12} \times \cfrac{1 \times \bigg[\left[(1.01)^{12}\right]^9 - 1\bigg]}{(1.01)^{12} - 1} \\ &= \cfrac{1000(1.01)^{12}\left[(1.01)^{108} - 1\right]}{(1.01)^{12} - 1} \\ &\approx \$17\ 138.27 \qquad \text{(nearest cent)} \end{align*}$

Do you see that the powers decrease as each subsequent investment is compounding for a shorter period of time?

Many would approach this by defining $A_n as the total amount at the time the nth deposit is made, but this makes for a more complicated approach (in my opinion) as you can equally treat each investment as a single account that grows without interruption and a new account is set up for the next deposit. You can then simply add up the accounts to get the total investment.

thank you thank you!!

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