8.3
TLW find compound interest and depreciation.
Compound Interest
• You earn interest on capital AND previously
earned interest.
• Interest on interest
Simple Interest
• A gradual increase
• Initial value used to find interest
• Unfair to the investor since there is actually more
value than the capital
• Total Value = 𝐶 1
𝑟 𝑛
+
𝑛
• Total Interest = 𝐶 1 +
𝑟 𝑛
𝑛
−𝐶
Why?
Example
• Lars invests 10 000 Swedish krona (SEK) at a rate of 5.1%
compounding yearly. Calculate the amount in Lars’s account
after 4 years and find how much of that amount is interest,
Give answers correct to 2 d.p.
Another Example
• Giovanni’s bank manager told him that if he invests
3000 EUR now, compounding yearly, it will be worth
4600 EUR in 5 years’ time.
Another, another example
• Marina is saving to buy a small boat that costs 35 000 USD.
She has 28 000 USD in an account that pays 5.34% interest
compounding yearly. How must Marina wait before she can
buy the boat?
• Setup:
• Let’s use Solver to find the answer.
Compound Interest and the GDC
• In APPS you will find Finance
• Use TVM Solver
• N = # of time periods ( # of years)
• I = interest rate
• Can be negative if we are computing depreciation
• PV = principal (C)
• This is relative to the investor. Money invested is outgoing and therefore entered with
a negative sign.
• If the money is borrowed, then it is positive.
• PMT = extra payments (none)
• Negative if money is borrowed
• FV = final value
• This is positive because it is paid to the investor at the end of investment
•
•
•
•
P/Y = payments per year, leave as 1
C/Y = compounding events per year
PMT END BEGIN = when to apply interest in the period
Place the cursor on what you’re solving for and press ALPHA ENTER to
solve.
Again but with GDC
• Marina is saving to buy a small boat that costs 35 000 USD.
She has 28 000 USD in an account that pays 5.34% interest
compounding yearly. How must Marina wait before she can
buy the boat?
Subdivided Time Periods
• Formula: 𝑇 = 𝐶 1
𝑘𝑛
𝑟
+
100𝑘
• k = how many times compounded in one year
• 𝐼 =𝐶 1+
𝑘𝑛
𝑟
100𝑘
− 𝐶 finds just the interest
• Possible “k” in a year
•
•
•
•
•
Quarterly – 4 times
Weekly – 52
Daily – 365
Monthly – 12
Half yearly – 2
• Actual Interest: Divide interest rate (nominal rate) by the
number of times compounded in a year.
• The bank in Grabiton is advertising a nominal yearly rate of 5%
with compounding applied quarterly. State the number of
compounding periods for a 3-year investment and find the
actual interest rate applied after each time period.
• Fleur invests 500 GBP in this bank for three years. Calculate
the total capital in her account after this time.
• Suppose Fleur invests this money at a rate of 5% p.a.
compounding only once a year. How much less interest would
she receive?
A Real-life Example
• A bank in Australia is offering a “term-deposit” account with a choice of
interest rates. As long as you leave your money with them for 2 years, you
can get a nominal rate of
• 5.05% compounded monthly OR
• 5.10% compounded quarterly OR
5.15% compounded half-yearly OR
• An actual yearly rate of 5.2%
Abigail has 7000 AUD to invest. Which is her best option?
Remember you
must show your
method on
Papers 1 and 2
or lose all the
marks.
• Annoushka has 2736.74 EUR in her bank account. She has left her money
there for exactly 3 years at a nominal rate of 4.1% p.a. compounding daily.
Calculate, correct to the nearest EUR, how much Annoushka put in the
account when she opened it. (Assume there were no leap years in that
time.)
• Check with GDC.
How does this connect to geometric sequence?
• Simple interest– we add equal amounts
• Compounded interest– we multiply equal
amounts
• This makes it a geometric sequence
• A geometric sequence has third term 212.24 and fifth
term 220.82.
• If these are capital amounts in a an account offering r%
interest at time periods, after 3 and 5 time periods, then find
r and the initial amount, C, in the account to the nearest
whole number.
• Amanda has paid school fees for her son for seven years. In
the first year, the fees were $2000. They have increased by
5% p.a. every year. Find how much Amanda has paid in total.
• Sum a geometric sequence: 𝑆𝑛 =
𝑢1 𝑟 𝑛 −1
𝑟−1
• Joe is saving for retirement. He pays 10 000 GBP into a bank
account at the start of each year for n years. The account pays
7.1% interest compounding annually. Show that, after n years,
the amount, A, that Joe has in the account is
• 𝐴 = 10 710
1.071𝑛 −1
0.071
• If n is 12, calculate A.
𝐺𝐵𝑃
Depreciation
• Loss of value
• Rate r is negative
• Vijay has paid 300 000 Indian rupees (INR) for a car. The car
depreciates at a rate of 9% p.a. Calculate the value, V, of the
car in 4 years’ time, giving your answer correct to the nearest
INR.
• Find the percentage loss over the 4-year period.
1. Find the new value.
2. Percentage loss =
𝑜𝑙𝑑−𝑛𝑒𝑤
𝑥100%
𝑜𝑙𝑑
• Anthony bought a house for 380 000 USD 7 years ago. Since
then, in his area, houses have increased in value by an
average of 10 % p.a. for the first 5 years but then lost value at a
rate of 4% p.a. for the last 2 years.
• What is the value, V, of the house now? Give the answer to the
nearest 1000 USD.