Financial mathematics

Vukile Xhego
Vukile XhegoTeacher en Azara Secondary School
13 February 2017
 Revision of Grade11 work
Simple and Compound interest (at varying
compounding periods)
Nominal and Effective interest rates
Simple and Compound Decay
Time Lines:
Changing interest rates
Deposits and Withdrawals
 Grade 12 content
Calculating the period of an investment/loan using
logarithms
Future value annuities
Present value annuities
Choosing a better investment or loan option
through critical analysis
Simple Interest
𝐴 = 𝑃(1 + 𝑖. 𝑛)
Compound Interest
𝐴 = 𝑃 1 + 𝑖 𝑛
Simple Decay/ Straight-line Depreciation
𝐴 = 𝑃(1 − 𝑖. 𝑛)
Compound Decay/ Reducing Balance
Depreciation
𝐴 = 𝑃 1 − 𝑖 𝑛
Linda invested R500 at 8% simple interest.
Calculate the balance after 5 years.
Solution
𝑃 = 500, 𝑖 = 8% = 0,08, 𝑛 = 5𝑦𝑒𝑎𝑟𝑠, 𝐴 =?
𝐴 = 𝑃 1 + 𝑖. 𝑛
𝐴 = 500 1 + 0,08 × 5 = 𝑅700
Bofelo invested R500 at 8% p.a compounded
monthly. Calculate his balance after 5 years.
Solution
𝑃 = 500, 𝑖 =
0.08
12
, 𝑛 = 5 × 12 = 60, 𝐴 =?
𝐴 = 𝑃 1 + 𝑖 𝑛
𝐴 = 500 1 +
0.08
12
60
= 𝑅744,92
Vuyo bought a Samsung J2 cellphone for
R2500. If his cellphone depreciates at a rate of
22% p.a on a reducing balance, determine its
value in 3 years time.
Solution
𝑃 = 2500, 𝑖 = 0.22, 𝑛 = 3, 𝐴 =?
𝐴 = 𝑃 1 − 𝑖 𝑛
𝐴 = 2500 1 − 0.22 3 = 𝑅1 186.38
Recall that: nominal rate is the rate that is
quoted (mentioned) in the statement/question
(what the bank offers). The effective rate is the
equivalent annual rate you receive once
compounding has been taken into account.
Formula:
1 + 𝑖 𝑒𝑓𝑓 = 1 +
𝑖 𝑛𝑜𝑚
𝑚
𝑚
Given an interest rate of 12% p.a. compounded
monthly:
a) Determine the effective interest rate
b) Using the effective interest rate, convert an
interest rate of 12% p.a. compounded
monthly to an interest rate of 12% p.a.
compounded quarterly.
Solutions
a) 1 + 𝑖 𝑒𝑓𝑓 = 1 +
𝑖 𝑛𝑜𝑚
𝑚
𝑚
𝑖 𝑒𝑓𝑓 = 1 +
0.12
12
12
− 1 = 12.68%
b) 1 + 𝑖 𝑒𝑓𝑓 = 1 +
𝑖 𝑛𝑜𝑚
𝑚
𝑚
1 + 0.1268 = 1 +
𝑖 𝑛𝑜𝑚
4
4
4
1.1268 = 1 +
𝑖 𝑛𝑜𝑚
4
4
1.1268 − 1 =
𝑖 𝑛𝑜𝑚
4
4
4
1.1268 − 1 = 𝑖 𝑛𝑜𝑚
∴ 𝑖 𝑛𝑜𝑚 = 0.12118 = 12.12%
quarterly
 It is useful to draw a timeline for problems
where there is a change in interest rate as
well as problems where there are additional
deposits and/or withdrawals…
 Ipeleng invested R2 850 at 8% p.a.
compounded quarterly for 15 years. However,
after the first 9 years the bank increased the
rate to 10,5% p.a. compounded monthly for
the remainder of her investment. How much
is the investment worth after 15 years?
 Shelen invests R10 000 into an account that
offers 3,25% p.a. compounded quarterly.
After 2 years he deposits an additional R2
500 into the account and then 3 years later
withdraws R5000. How much will he have in
his account after 10 years?
 𝑥 Rand is invested into an account that offers
an interest rate of 12% p.a. compounded
monthly. 3 years later, 2𝑥 Rand is deposited
into the account. After 7 years there is R27
655,87 in the account. Determine the value of
𝑥.
 An investment of R25 000 grows to an
amount of R55 267,04. If the account offers
an interest rate of 12% p.a compounded
annually, for how long was the money
invested?
 Ntsiki deposits R4000 into a savings account
paying 10% per annum compounded
annually.
a) How long will it take for the savings to
double?
b) Suppose the interest rate for Ntsiki’s
investment was 10% per annum
compounded half-yearly. How long will it
take for the investment to double in this
case?
 A motor car costing R200 000 depreciates at
a rate of 8% p.a. on the reducing balance
method. Calculate how long it took for the
car to depreciate to a value of R90 000.
Financial mathematics
Financial mathematics
 Future Value of an annuity
◦ Savings accounts (equal deposits, regular intervals)
◦ Retirement annuities/ Pension funds
◦ Sinking funds
 An annuity is a series of equal investment
payments or loan repayments at regular
intervals and subject to a rate of interest over
a period of time
 A Future Value Annuity is when money is
invested at regular intervals in order to save
for the future
◦ Compound interest is paid on the money invested,
which means, the more money one has saved the
more interest he/she makes.
 x Rand is paid each month into a savings
account starting in one month’s time. The
interested rate is i p.a. compounded monthly.
Calculate the total accumulated at the end of
n months.
Let F be the total accumulated at the end of n
months, immediately after the nth payment
0 1 2 3 nn-1
x x x x x
i is the interest rate p.a. compounded monthly
𝐹 = 𝑥 + 𝑥 1 + 𝑖 1
+ 𝑥 1 + 𝑖 2
+ 𝑥 1 + 𝑖 3
+. . . 𝑡𝑜 𝑛 𝑡𝑒𝑟𝑚𝑠
We can use formula for a geometric series to
calculate the sum
𝑆 𝑛 =
𝑎 𝑟 𝑛−1
𝑟−1
𝑎 = 𝑥 𝑎𝑛𝑑 𝑟 = (1 + 𝑖)
∴ 𝐹 =
𝑥 1+𝑖 𝑛−1
(1+𝑖)−1
=
𝑥 1+𝑖 𝑛−1
𝑖
 Formula for Future Value of an annuity
𝐹 =
𝑥 1 + 𝑖 𝑛 − 1
𝑖
𝑤ℎ𝑒𝑟𝑒; 𝐹 → future value
𝑥 → amount deposited at regular intervals
𝑖 → interest rate as a decimal =
r
100
𝑛 →
number of payments or number of periods of
the investment
 R1000 is deposited into a bank account. One
month later, another R1000 is deposited and
then a further R1000 one month after this. If
the interest rate is 6% per annum
compounded monthly, how much will have
been saved after two months?
 Use at least two methods…
 Suppose R1500 is invested every month,
starting immediately, for a period of 12
months. Interest is 18% per annum
compounded monthly. Calculate the
accumulated amount at the end of the
12month period
 Avela starts saving money into a Unit Trust
Fund. He immediately deposits R800 into the
fund. Thereafter, he deposits R800 at the end
of each month and continues to do this for
10 years. Interest is 8% per annum
compounded monthly. Calculate the future
value of his investment at the end of the 10-
year period.
 On the 31st of December 2009 Lerato decided
to start saving money for a period of 8 years.
At the end of January 2010 (in one month’s
time), she deposited R2 300 into the savings
plan. Thereafter, she continued making
deposits of R2 300 at the end of each month
for the duration of the 8 year period. The
interest rate remained fixed at 10% per
annum compounded monthly. How much will
she have saved at the end of the plan?
 General has just turned 20 years old and has
a dream of saving R8 000 000 by the time he
reaches 50. He starts to pay equal monthly
installments into a retirement annuity which
pays 18% per annum compounded monthly.
His first payment starts on his 20th birthday
and his last payment is made on his 50th
birthday. How much will he have to deposit
each month?
 It is often the case that in annuity
investments, the last payment is made one
month before the end of the investment.
◦ In such cases, the balance after the last deposit is
made then accrues (gains) interest for the
remainder of the investment period
 Formula for payments in advance:
𝐹 =
𝑥 1 + 𝑖 𝑛 − 1
𝑖
(1 + 𝑖)
 A father decides to start saving for his baby
daughter’s future education. On opening the
account, he immediately deposits R2000 and
continues to make monthly payments at the end
of each month thereafter for a period of 16 years.
The interest rate remains fixed at 15% per annum
compounded monthly.
a) How much money will he have accumulated at
the end of the 16th year?
b) At the end of the sixteen-year period, he leaves
the money in the account for a further year.
How much will he then have accumulated?
 In order to supplement his state pension after
retirement, a school teacher aged 30 takes
out a retirement annuity. He makes monthly
payments of R1000 into the fund and the
payments start immediately. The payments
are made in advance, which means that the
last payment of R1000 is made one month
before annuity is out. The interest rate for the
annuity is 12% per annum compounded
monthly. Calculate the future value of the
annuity in twenty-five years’ time.
 A sinking fund is a savings account set up to
save money in order to replace an item in
future
A company purchases a new vehicle for R200 000. the
vehicle is expected to depreciate at a rate of 24% p.a.
on a reducing balance. It is expected that the vehicle
will have to be replaced after 5 years. A sinking fund is
set up for this purpose. If the replacement cost of the
vehicle is expected to increase by 18% p.a compounded
annually, calculate:
a) The value of the current vehicle after 5 years
b) The cost of a new vehicle in 5 years time
c) The total required value of the sinking fund, if the old
vehicle is sold and the proceeds contribute towards the
new vehicle
d) The monthly installments paid into the sinking fund if the
interest rate is 15% p.a. compounded monthly and
payments start one month after the vehicle is initially
purchased
a) Reducing balance: book value
𝑃 = 200 000, 𝑖 = 0,24, 𝑛 = 5𝑦𝑒𝑎𝑟𝑠, 𝐴 =?
𝐴 = 𝑃 1 − 𝑖 𝑛
= 200 000 1 − 0,24 5
= 𝑅50 710,51
b) Compound interest: cost of a new vehicle
𝐴 = 𝑃 1 + 𝑖 𝑛
= 200 000 1 + 0,18 5
= 𝑅457 551,55
c) Sinking fund value = cost of new – scrap value
= 457 551,55 − 50 710,51
= 𝑅406 841, 04
d) Monthly installments:
𝐹 =
𝑥 1+𝑖 𝑛−1
𝑖
∴ 406 841,04 =
𝑥 1+
0,15
12
60
−1
0,015
12
406 841,04×
0,15
12
1+
0,15
12
60
−1
= 𝑥
∴ 𝑥 = 𝑅4 593,21
 It is the 31st of December 2016. Shuaib
decides to start saving money and wants to
save R300 000 by paying monthly amounts of
R4000, starting in one month’s time (31
January 2017), into a savings account paying
15% per annum compounded monthly. How
many payments of R4000 will be made? The
duration starts on the 31st December 2016,
even though the first payment is made on the
31st January 2017.
 Exercise 3 on page 78 (Mind Action Series)
 Exercise 8 on page 97 (Mind Action Series)
Financial mathematics
 Revision of Hire Purchase loans
 Lender’s schedules
 Present Value of an Annuity
◦ Definition of PV annuity
◦ Deriving a formula for calculating PV of an annuity
◦ Calculating payments in arrear and payments in
advance
◦ Calculating the outstanding balance of a loan
◦ Critically analyse loan options and make informed
financial decisions
 In grade 10 and 11 you studied hire purchase
loans in which the interest payable is calculated
at the start of the loan using the formula for
simple interest
 The total interest on the loan must be paid in full
 Hence, even if you were to pay-off the loan
earlier, you would still pay the interest owed in
full…no advantage
 The monthly repayments are calculated by
dividing the loan amount by the number of
months
 You are required to pay all these amounts and
there is no advantage in paying off the loan early
 Sally buys a tumble dryer for R4000. she
takes out a hire purchase loan involving
monthly payments over three years. The
interest rate is calculate at 14% per annum
simple interest. Calculate:
a) The actual amount paid for the tumble dryer
b) The interest paid
c) How much must be paid each month
a) 𝐴 = 𝑃 1 + 𝑖𝑛
∴ 𝐴 = 4000 1 + 0,14 × 3 = 𝑅5 680
b) Interest paid: R5 680 – R4000= R1 680
c) The monthly loan repayments:
5680
36
= 𝑅157,78
 You borrow money now (to buy a house/car),
and pay it back with interest, in equal
repayments until the loan and all the interest
has been paid.
 This is called a Present Value Annuity
 The furniture store requires a deposit of 10%
of the cost of the item furniture at the time of
the purchase, and then 24 equal monthly
payments for the remaining 90% of the cost.
 The car dealer requires a deposit of 12.5% of
the cost of the car and then 60 equal monthly
payments for the remaining 87.5% of the
purchase price.
 The micro-lender lends cash amounts to
customers at a rate of over 20% per month.
 Gladys borrowed R750 from a micro-lender
to pay the school fees for her children and
she chose the option of making six monthly
installments of R240 starting one month after
she received the loan.
◦ 6 monthly installments of R240 = R1440
◦ Why the extra R690?
◦ Interest rate calculated at approximately 22.6%
◦ Is the amount of R240 per month for 6 months the
correct installment?
Month Owing at
the
beginning
of month
Interest
added at
the end of
month
Total
owing at
the end of
the month
Installment
subtracted
New
amount
owing at
the end of
the month
1 R750 R169.17 R919.17 R240 R679.17
2 R679.17 R153.20 R832.37 R240 R592.37
3 R592.37 R133.62 R725.99 R240 R485.99
4 R485.99 R109.62 R595.61 R240 R355.61
5 R355.61 R80.21 R435.82 R240 R195.82
6 R195.82 R44.17 R239.99 R240 -0.01
 But what if Gladys cant afford to pay R240
and she only pays R160 per month? Maybe
she will take a little longer but surely she will
still pay-off her debt?
Month Owing at
the
beginning
of month
Interest
added at
the end of
month
Total
owing at
the end of
month
Installment
subtracted
New
amount
owing at
the end of
month
1 750 169.17 919.17 160 759.19
2 759.19 171.25 930.44 160 770.44
3 770.44 173.78 944.22 160 784.22
4 784.22 176.89 961.11 160 801.11
5 801.11 180.70 981.81 160 821.81
6 821.81 185.37 1007.18 160 847.18
 Notice that the interest at the end of the first
month (R169.17) is more than her repayment of
R160.
 Her installments pay only part of the interest and
so Gladys will never finish paying off her debt.
 The size of installments cannot be found by
guessing, but has to be calculated using a
mathematical formula.
 A PRESENT VALUE ANNUITY is the whole process
of paying off a debt by a sequence of payments
made at regular intervals in time.
 The Present Value (P) of the loan is the sum
of the present values of the six payment (P):
 𝑃 = 240 1 + 𝑖 −1 + 240 1 + 𝑖 −2 + 240 1 + 𝑖 −3 +
240 1 + 𝑖 −4
+ 240 1 + 𝑖 −5
+ 240 1 + 𝑖 −6
 This is a geometric series with:
◦ 𝑎 = 240 1 + 𝑖 −1, 𝑟 = 1 + 𝑖 −1
𝑆 𝑛 =
𝑎 𝑟 𝑛
− 1
𝑟 − 1
=
240 1 + 𝑖 −1
1 + 𝑖 −1
− 1
1 + 𝑖 −1 − 1
 if we replace 𝑆 𝑛 by 𝑃 in the previous and assume that there were
up to n payments of amount 𝑥, then we would have:
◦ 𝑎 = 𝑥 1 + 𝑖 −1
, 𝑟 = 1 + 𝑖 −1
, number of terms = 𝑛
∴ 𝑃 =
𝑥 1+𝑖 −1[ 1+𝑖 −𝑛−1]
1+𝑖 −1−1
=
𝑥[ 1+𝑖 −𝑛−1]
1+𝑖 [ 1+𝑖 −𝑛−1)
=
𝑥 1+𝑖 −𝑛−1
1+𝑖 0−(1+𝑖)
=
𝑥 1+𝑖 −𝑛−1
−𝑖
=
−𝑥 1+𝑖 −𝑛−1
𝑖
∴ 𝑃 =
𝑥 1− 1+𝑖 −𝑛
𝑖
𝑷 =
𝒙[𝟏 − 𝟏 + 𝒊 −𝒏
]
𝒊
 Where;
 𝑃 is the present value/ Outstanding balance
 𝑥 is the regular installments
 𝑖 is the interest rate (as a decimal)
 𝑛 is the number of payments (deposits) that
still have to be made
 P represents the outstanding balance on a
loan with n payments still to go.
 n is always the number of payments left on
the loan
 Usually payments start one period after a loan
is granted
 Suppose that a loan is repaid by means of of
a payment of R1000 one month after the loan
was granted and a further payment of R1000
one month after the first payment of R1000.
calculate the amount borrowed (present value
of the loan). The interest rate is 6% per
annum compounded monthly.
 Use three methods…
 The present value at T0 of the payment at T1:
𝑃 = 𝐴 1 + 𝑖 −𝑛
∴ 𝑃 = 1000 1 +
0,06
12
−1
= 𝑅995,02
 The present value of T0 of the payment at T2:
∴ 𝑃 = 1000 1 +
0,06
12
−2
= 𝑅990,07
The total amount borrowed (the loan) is the sum
of the two present values:
𝑃 = 1000 1 +
0,06
12
−1
+1000 1 +
0,06
12
−2
= 𝑅1 985,10
 You will probably notice that the payments generate a
geometric series:
1000 1 +
0,06
12
−1
+1000 1 +
0,06
12
−2
 𝑎 = 1000 1 +
0,06
12
−1
𝑎𝑛𝑑 𝑟 = 1 + 𝑖 −1
 𝑆 𝑛 =
𝑎 𝑟 𝑛−1
𝑟−1
=
1000 1+
0,06
12
−1
1+
0,06
12
−1
−1
1+
0,06
12
−1
−1
= 𝑅1 985,10
 𝑃 =
𝑥[1− 1+𝑖 −𝑛]
𝑖
 ∴ 𝑃 =
1000 1− 1+0,
06
12
−2
0,
06
12
= 𝑅1 985,10
 The formula for P can only be used if there is
a gap between the loan and the first payment
 If payments are monthly, then there must be
a one month gap between the loan and the
first payment
 Sindisiwe takes out a bank loan to pay for her
new car. She repays the loan by means of
monthly repayments of R5000 for a period of
five years starting one month after the loan
has been granted. The interest rate is 24% per
annum compounded monthly. Calculate the
purchase price of her new car.
 If a loan is taken and a payment is made at
the same time (immediately), this must be
subtracted from the original loan…
 it is a deposit and must be deducted
 Prudence take a bank loan for her new car.
She pays an initial amount (deposit) of R10
000. She then makes monthly payments for a
period of 5 years, starting one month after
the loan is granted. The interest rate is 24%
per annum compounded monthly. Calculate
the monthly payments if the car originally
cost her R173 804,43.
 How much will you have to win from a Lottery
in order to receive equal monthly payments
of R10 000 from a bank for a period of
twenty years starting one month after
winning the money? The bank grants you an
interest rate of 12% per annum compounded
monthly.
Financial mathematics
 Melaine takes out a twenty-year loan of
R100000. She repays the loan by means of
monthly payments starting three months
after the granting of the loan. The interest
rate is 18% per annum compounded monthly.
Calculate the monthly repayments.
 Rufayda borrows R500 000 from a bank and
repays the loan by means of monthly
payments of R8000, starting one month after
the granting of the loan. The interest rate is
fixed at 18% per annum compounded
monthly.
a) How many payments of R8000 will be made
and what will the final lesser payment be?
b) How long is the savings period?
 Page 90 Mind Action Series
 You decide to take a bank loan of R800 000. repayments
are made on a monthly basis starting one month after the
granting of the loan. You are offered three loan options:
Option A
Pay of the loan at 18% p.a. compounded monthly
Option B
Pay of the loan over thirty years at 18% p.a. compounded
monthly
Option C
Pay off the loan over thirty years at 15% p.a. compounded
monthly
a) Calculate the monthly amount payable for each option
b) Which option would be the best financially for you?
Motivate your answer by using appropriate calcualtions
 Page 92 Mind Action Series
 It is sometimes useful to calculate the
balance still owed on a loan at a given time
during the course of the loan.
 James takes out a one year bank loan of R18
000 to pay for an expensive laptop. The
interest rate is 18% per annum compounded
monthly and monthly repayments of R1
650,24 are made starting one month after the
loan was granted.
a) Calculate his balance outstanding after he
has paid the sixth installment
b) Calculate his balance outstanding after he
has paid his ninth installment
 Page 94 Mind Action Series
 Payments which start immediately
◦ Payment at To becomes an additional/extra
payment
 Payments in advance
◦ Payments that end one or more periods before the
end of the investment term
 Sinking funds
◦ Savings set up in order to replace an item in future
 Fixed payments
◦ Payments which start one period (e.g. one month) after a
loan is granted
 Skipped payments
◦ When repayments start more than one period (e.g. in
three months’ time) after the loan is granted
◦ In such cases, the original loan amount accrues interest
and payments are made to pay off a new Present Value
 Balance outstanding on a loan
◦ This is the present value of a loan with n payments still
to go
◦ Position of the last installment becomes position of your
new present value(balance outstanding)
 Page 98-102 Mind Action Series
 Mind Action Series textbook
 Classroom Mathematics Grade 12 learner’s
book
 Maths Hand Book and Study Guide (Gr 12) –
Kevin Smith
 Maths and Stats for Bussiness – X-Kit
Undergraduate – J Bassons et al.
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Financial mathematics

  • 2.  Revision of Grade11 work Simple and Compound interest (at varying compounding periods) Nominal and Effective interest rates Simple and Compound Decay Time Lines: Changing interest rates Deposits and Withdrawals
  • 3.  Grade 12 content Calculating the period of an investment/loan using logarithms Future value annuities Present value annuities Choosing a better investment or loan option through critical analysis
  • 4. Simple Interest 𝐴 = 𝑃(1 + 𝑖. 𝑛) Compound Interest 𝐴 = 𝑃 1 + 𝑖 𝑛 Simple Decay/ Straight-line Depreciation 𝐴 = 𝑃(1 − 𝑖. 𝑛) Compound Decay/ Reducing Balance Depreciation 𝐴 = 𝑃 1 − 𝑖 𝑛
  • 5. Linda invested R500 at 8% simple interest. Calculate the balance after 5 years. Solution 𝑃 = 500, 𝑖 = 8% = 0,08, 𝑛 = 5𝑦𝑒𝑎𝑟𝑠, 𝐴 =? 𝐴 = 𝑃 1 + 𝑖. 𝑛 𝐴 = 500 1 + 0,08 × 5 = 𝑅700
  • 6. Bofelo invested R500 at 8% p.a compounded monthly. Calculate his balance after 5 years. Solution 𝑃 = 500, 𝑖 = 0.08 12 , 𝑛 = 5 × 12 = 60, 𝐴 =? 𝐴 = 𝑃 1 + 𝑖 𝑛 𝐴 = 500 1 + 0.08 12 60 = 𝑅744,92
  • 7. Vuyo bought a Samsung J2 cellphone for R2500. If his cellphone depreciates at a rate of 22% p.a on a reducing balance, determine its value in 3 years time. Solution 𝑃 = 2500, 𝑖 = 0.22, 𝑛 = 3, 𝐴 =? 𝐴 = 𝑃 1 − 𝑖 𝑛 𝐴 = 2500 1 − 0.22 3 = 𝑅1 186.38
  • 8. Recall that: nominal rate is the rate that is quoted (mentioned) in the statement/question (what the bank offers). The effective rate is the equivalent annual rate you receive once compounding has been taken into account. Formula: 1 + 𝑖 𝑒𝑓𝑓 = 1 + 𝑖 𝑛𝑜𝑚 𝑚 𝑚
  • 9. Given an interest rate of 12% p.a. compounded monthly: a) Determine the effective interest rate b) Using the effective interest rate, convert an interest rate of 12% p.a. compounded monthly to an interest rate of 12% p.a. compounded quarterly.
  • 10. Solutions a) 1 + 𝑖 𝑒𝑓𝑓 = 1 + 𝑖 𝑛𝑜𝑚 𝑚 𝑚 𝑖 𝑒𝑓𝑓 = 1 + 0.12 12 12 − 1 = 12.68% b) 1 + 𝑖 𝑒𝑓𝑓 = 1 + 𝑖 𝑛𝑜𝑚 𝑚 𝑚 1 + 0.1268 = 1 + 𝑖 𝑛𝑜𝑚 4 4 4 1.1268 = 1 + 𝑖 𝑛𝑜𝑚 4 4 1.1268 − 1 = 𝑖 𝑛𝑜𝑚 4 4 4 1.1268 − 1 = 𝑖 𝑛𝑜𝑚 ∴ 𝑖 𝑛𝑜𝑚 = 0.12118 = 12.12% quarterly
  • 11.  It is useful to draw a timeline for problems where there is a change in interest rate as well as problems where there are additional deposits and/or withdrawals…
  • 12.  Ipeleng invested R2 850 at 8% p.a. compounded quarterly for 15 years. However, after the first 9 years the bank increased the rate to 10,5% p.a. compounded monthly for the remainder of her investment. How much is the investment worth after 15 years?
  • 13.  Shelen invests R10 000 into an account that offers 3,25% p.a. compounded quarterly. After 2 years he deposits an additional R2 500 into the account and then 3 years later withdraws R5000. How much will he have in his account after 10 years?
  • 14.  𝑥 Rand is invested into an account that offers an interest rate of 12% p.a. compounded monthly. 3 years later, 2𝑥 Rand is deposited into the account. After 7 years there is R27 655,87 in the account. Determine the value of 𝑥.
  • 15.  An investment of R25 000 grows to an amount of R55 267,04. If the account offers an interest rate of 12% p.a compounded annually, for how long was the money invested?
  • 16.  Ntsiki deposits R4000 into a savings account paying 10% per annum compounded annually. a) How long will it take for the savings to double? b) Suppose the interest rate for Ntsiki’s investment was 10% per annum compounded half-yearly. How long will it take for the investment to double in this case?
  • 17.  A motor car costing R200 000 depreciates at a rate of 8% p.a. on the reducing balance method. Calculate how long it took for the car to depreciate to a value of R90 000.
  • 20.  Future Value of an annuity ◦ Savings accounts (equal deposits, regular intervals) ◦ Retirement annuities/ Pension funds ◦ Sinking funds
  • 21.  An annuity is a series of equal investment payments or loan repayments at regular intervals and subject to a rate of interest over a period of time  A Future Value Annuity is when money is invested at regular intervals in order to save for the future ◦ Compound interest is paid on the money invested, which means, the more money one has saved the more interest he/she makes.
  • 22.  x Rand is paid each month into a savings account starting in one month’s time. The interested rate is i p.a. compounded monthly. Calculate the total accumulated at the end of n months.
  • 23. Let F be the total accumulated at the end of n months, immediately after the nth payment 0 1 2 3 nn-1 x x x x x i is the interest rate p.a. compounded monthly
  • 24. 𝐹 = 𝑥 + 𝑥 1 + 𝑖 1 + 𝑥 1 + 𝑖 2 + 𝑥 1 + 𝑖 3 +. . . 𝑡𝑜 𝑛 𝑡𝑒𝑟𝑚𝑠 We can use formula for a geometric series to calculate the sum 𝑆 𝑛 = 𝑎 𝑟 𝑛−1 𝑟−1 𝑎 = 𝑥 𝑎𝑛𝑑 𝑟 = (1 + 𝑖) ∴ 𝐹 = 𝑥 1+𝑖 𝑛−1 (1+𝑖)−1 = 𝑥 1+𝑖 𝑛−1 𝑖
  • 25.  Formula for Future Value of an annuity 𝐹 = 𝑥 1 + 𝑖 𝑛 − 1 𝑖 𝑤ℎ𝑒𝑟𝑒; 𝐹 → future value 𝑥 → amount deposited at regular intervals 𝑖 → interest rate as a decimal = r 100 𝑛 → number of payments or number of periods of the investment
  • 26.  R1000 is deposited into a bank account. One month later, another R1000 is deposited and then a further R1000 one month after this. If the interest rate is 6% per annum compounded monthly, how much will have been saved after two months?  Use at least two methods…
  • 27.  Suppose R1500 is invested every month, starting immediately, for a period of 12 months. Interest is 18% per annum compounded monthly. Calculate the accumulated amount at the end of the 12month period
  • 28.  Avela starts saving money into a Unit Trust Fund. He immediately deposits R800 into the fund. Thereafter, he deposits R800 at the end of each month and continues to do this for 10 years. Interest is 8% per annum compounded monthly. Calculate the future value of his investment at the end of the 10- year period.
  • 29.  On the 31st of December 2009 Lerato decided to start saving money for a period of 8 years. At the end of January 2010 (in one month’s time), she deposited R2 300 into the savings plan. Thereafter, she continued making deposits of R2 300 at the end of each month for the duration of the 8 year period. The interest rate remained fixed at 10% per annum compounded monthly. How much will she have saved at the end of the plan?
  • 30.  General has just turned 20 years old and has a dream of saving R8 000 000 by the time he reaches 50. He starts to pay equal monthly installments into a retirement annuity which pays 18% per annum compounded monthly. His first payment starts on his 20th birthday and his last payment is made on his 50th birthday. How much will he have to deposit each month?
  • 31.  It is often the case that in annuity investments, the last payment is made one month before the end of the investment. ◦ In such cases, the balance after the last deposit is made then accrues (gains) interest for the remainder of the investment period  Formula for payments in advance: 𝐹 = 𝑥 1 + 𝑖 𝑛 − 1 𝑖 (1 + 𝑖)
  • 32.  A father decides to start saving for his baby daughter’s future education. On opening the account, he immediately deposits R2000 and continues to make monthly payments at the end of each month thereafter for a period of 16 years. The interest rate remains fixed at 15% per annum compounded monthly. a) How much money will he have accumulated at the end of the 16th year? b) At the end of the sixteen-year period, he leaves the money in the account for a further year. How much will he then have accumulated?
  • 33.  In order to supplement his state pension after retirement, a school teacher aged 30 takes out a retirement annuity. He makes monthly payments of R1000 into the fund and the payments start immediately. The payments are made in advance, which means that the last payment of R1000 is made one month before annuity is out. The interest rate for the annuity is 12% per annum compounded monthly. Calculate the future value of the annuity in twenty-five years’ time.
  • 34.  A sinking fund is a savings account set up to save money in order to replace an item in future
  • 35. A company purchases a new vehicle for R200 000. the vehicle is expected to depreciate at a rate of 24% p.a. on a reducing balance. It is expected that the vehicle will have to be replaced after 5 years. A sinking fund is set up for this purpose. If the replacement cost of the vehicle is expected to increase by 18% p.a compounded annually, calculate: a) The value of the current vehicle after 5 years b) The cost of a new vehicle in 5 years time c) The total required value of the sinking fund, if the old vehicle is sold and the proceeds contribute towards the new vehicle d) The monthly installments paid into the sinking fund if the interest rate is 15% p.a. compounded monthly and payments start one month after the vehicle is initially purchased
  • 36. a) Reducing balance: book value 𝑃 = 200 000, 𝑖 = 0,24, 𝑛 = 5𝑦𝑒𝑎𝑟𝑠, 𝐴 =? 𝐴 = 𝑃 1 − 𝑖 𝑛 = 200 000 1 − 0,24 5 = 𝑅50 710,51 b) Compound interest: cost of a new vehicle 𝐴 = 𝑃 1 + 𝑖 𝑛 = 200 000 1 + 0,18 5 = 𝑅457 551,55
  • 37. c) Sinking fund value = cost of new – scrap value = 457 551,55 − 50 710,51 = 𝑅406 841, 04 d) Monthly installments: 𝐹 = 𝑥 1+𝑖 𝑛−1 𝑖 ∴ 406 841,04 = 𝑥 1+ 0,15 12 60 −1 0,015 12 406 841,04× 0,15 12 1+ 0,15 12 60 −1 = 𝑥 ∴ 𝑥 = 𝑅4 593,21
  • 38.  It is the 31st of December 2016. Shuaib decides to start saving money and wants to save R300 000 by paying monthly amounts of R4000, starting in one month’s time (31 January 2017), into a savings account paying 15% per annum compounded monthly. How many payments of R4000 will be made? The duration starts on the 31st December 2016, even though the first payment is made on the 31st January 2017.
  • 39.  Exercise 3 on page 78 (Mind Action Series)  Exercise 8 on page 97 (Mind Action Series)
  • 41.  Revision of Hire Purchase loans  Lender’s schedules  Present Value of an Annuity ◦ Definition of PV annuity ◦ Deriving a formula for calculating PV of an annuity ◦ Calculating payments in arrear and payments in advance ◦ Calculating the outstanding balance of a loan ◦ Critically analyse loan options and make informed financial decisions
  • 42.  In grade 10 and 11 you studied hire purchase loans in which the interest payable is calculated at the start of the loan using the formula for simple interest  The total interest on the loan must be paid in full  Hence, even if you were to pay-off the loan earlier, you would still pay the interest owed in full…no advantage  The monthly repayments are calculated by dividing the loan amount by the number of months  You are required to pay all these amounts and there is no advantage in paying off the loan early
  • 43.  Sally buys a tumble dryer for R4000. she takes out a hire purchase loan involving monthly payments over three years. The interest rate is calculate at 14% per annum simple interest. Calculate: a) The actual amount paid for the tumble dryer b) The interest paid c) How much must be paid each month
  • 44. a) 𝐴 = 𝑃 1 + 𝑖𝑛 ∴ 𝐴 = 4000 1 + 0,14 × 3 = 𝑅5 680 b) Interest paid: R5 680 – R4000= R1 680 c) The monthly loan repayments: 5680 36 = 𝑅157,78
  • 45.  You borrow money now (to buy a house/car), and pay it back with interest, in equal repayments until the loan and all the interest has been paid.  This is called a Present Value Annuity
  • 46.  The furniture store requires a deposit of 10% of the cost of the item furniture at the time of the purchase, and then 24 equal monthly payments for the remaining 90% of the cost.
  • 47.  The car dealer requires a deposit of 12.5% of the cost of the car and then 60 equal monthly payments for the remaining 87.5% of the purchase price.
  • 48.  The micro-lender lends cash amounts to customers at a rate of over 20% per month.
  • 49.  Gladys borrowed R750 from a micro-lender to pay the school fees for her children and she chose the option of making six monthly installments of R240 starting one month after she received the loan. ◦ 6 monthly installments of R240 = R1440 ◦ Why the extra R690? ◦ Interest rate calculated at approximately 22.6% ◦ Is the amount of R240 per month for 6 months the correct installment?
  • 50. Month Owing at the beginning of month Interest added at the end of month Total owing at the end of the month Installment subtracted New amount owing at the end of the month 1 R750 R169.17 R919.17 R240 R679.17 2 R679.17 R153.20 R832.37 R240 R592.37 3 R592.37 R133.62 R725.99 R240 R485.99 4 R485.99 R109.62 R595.61 R240 R355.61 5 R355.61 R80.21 R435.82 R240 R195.82 6 R195.82 R44.17 R239.99 R240 -0.01
  • 51.  But what if Gladys cant afford to pay R240 and she only pays R160 per month? Maybe she will take a little longer but surely she will still pay-off her debt?
  • 52. Month Owing at the beginning of month Interest added at the end of month Total owing at the end of month Installment subtracted New amount owing at the end of month 1 750 169.17 919.17 160 759.19 2 759.19 171.25 930.44 160 770.44 3 770.44 173.78 944.22 160 784.22 4 784.22 176.89 961.11 160 801.11 5 801.11 180.70 981.81 160 821.81 6 821.81 185.37 1007.18 160 847.18
  • 53.  Notice that the interest at the end of the first month (R169.17) is more than her repayment of R160.  Her installments pay only part of the interest and so Gladys will never finish paying off her debt.  The size of installments cannot be found by guessing, but has to be calculated using a mathematical formula.  A PRESENT VALUE ANNUITY is the whole process of paying off a debt by a sequence of payments made at regular intervals in time.
  • 54.  The Present Value (P) of the loan is the sum of the present values of the six payment (P):  𝑃 = 240 1 + 𝑖 −1 + 240 1 + 𝑖 −2 + 240 1 + 𝑖 −3 + 240 1 + 𝑖 −4 + 240 1 + 𝑖 −5 + 240 1 + 𝑖 −6  This is a geometric series with: ◦ 𝑎 = 240 1 + 𝑖 −1, 𝑟 = 1 + 𝑖 −1 𝑆 𝑛 = 𝑎 𝑟 𝑛 − 1 𝑟 − 1 = 240 1 + 𝑖 −1 1 + 𝑖 −1 − 1 1 + 𝑖 −1 − 1
  • 55.  if we replace 𝑆 𝑛 by 𝑃 in the previous and assume that there were up to n payments of amount 𝑥, then we would have: ◦ 𝑎 = 𝑥 1 + 𝑖 −1 , 𝑟 = 1 + 𝑖 −1 , number of terms = 𝑛 ∴ 𝑃 = 𝑥 1+𝑖 −1[ 1+𝑖 −𝑛−1] 1+𝑖 −1−1 = 𝑥[ 1+𝑖 −𝑛−1] 1+𝑖 [ 1+𝑖 −𝑛−1) = 𝑥 1+𝑖 −𝑛−1 1+𝑖 0−(1+𝑖) = 𝑥 1+𝑖 −𝑛−1 −𝑖 = −𝑥 1+𝑖 −𝑛−1 𝑖 ∴ 𝑃 = 𝑥 1− 1+𝑖 −𝑛 𝑖
  • 56. 𝑷 = 𝒙[𝟏 − 𝟏 + 𝒊 −𝒏 ] 𝒊  Where;  𝑃 is the present value/ Outstanding balance  𝑥 is the regular installments  𝑖 is the interest rate (as a decimal)  𝑛 is the number of payments (deposits) that still have to be made
  • 57.  P represents the outstanding balance on a loan with n payments still to go.  n is always the number of payments left on the loan  Usually payments start one period after a loan is granted
  • 58.  Suppose that a loan is repaid by means of of a payment of R1000 one month after the loan was granted and a further payment of R1000 one month after the first payment of R1000. calculate the amount borrowed (present value of the loan). The interest rate is 6% per annum compounded monthly.  Use three methods…
  • 59.  The present value at T0 of the payment at T1: 𝑃 = 𝐴 1 + 𝑖 −𝑛 ∴ 𝑃 = 1000 1 + 0,06 12 −1 = 𝑅995,02  The present value of T0 of the payment at T2: ∴ 𝑃 = 1000 1 + 0,06 12 −2 = 𝑅990,07 The total amount borrowed (the loan) is the sum of the two present values: 𝑃 = 1000 1 + 0,06 12 −1 +1000 1 + 0,06 12 −2 = 𝑅1 985,10
  • 60.  You will probably notice that the payments generate a geometric series: 1000 1 + 0,06 12 −1 +1000 1 + 0,06 12 −2  𝑎 = 1000 1 + 0,06 12 −1 𝑎𝑛𝑑 𝑟 = 1 + 𝑖 −1  𝑆 𝑛 = 𝑎 𝑟 𝑛−1 𝑟−1 = 1000 1+ 0,06 12 −1 1+ 0,06 12 −1 −1 1+ 0,06 12 −1 −1 = 𝑅1 985,10
  • 61.  𝑃 = 𝑥[1− 1+𝑖 −𝑛] 𝑖  ∴ 𝑃 = 1000 1− 1+0, 06 12 −2 0, 06 12 = 𝑅1 985,10
  • 62.  The formula for P can only be used if there is a gap between the loan and the first payment  If payments are monthly, then there must be a one month gap between the loan and the first payment
  • 63.  Sindisiwe takes out a bank loan to pay for her new car. She repays the loan by means of monthly repayments of R5000 for a period of five years starting one month after the loan has been granted. The interest rate is 24% per annum compounded monthly. Calculate the purchase price of her new car.
  • 64.  If a loan is taken and a payment is made at the same time (immediately), this must be subtracted from the original loan…  it is a deposit and must be deducted
  • 65.  Prudence take a bank loan for her new car. She pays an initial amount (deposit) of R10 000. She then makes monthly payments for a period of 5 years, starting one month after the loan is granted. The interest rate is 24% per annum compounded monthly. Calculate the monthly payments if the car originally cost her R173 804,43.
  • 66.  How much will you have to win from a Lottery in order to receive equal monthly payments of R10 000 from a bank for a period of twenty years starting one month after winning the money? The bank grants you an interest rate of 12% per annum compounded monthly.
  • 68.  Melaine takes out a twenty-year loan of R100000. She repays the loan by means of monthly payments starting three months after the granting of the loan. The interest rate is 18% per annum compounded monthly. Calculate the monthly repayments.
  • 69.  Rufayda borrows R500 000 from a bank and repays the loan by means of monthly payments of R8000, starting one month after the granting of the loan. The interest rate is fixed at 18% per annum compounded monthly. a) How many payments of R8000 will be made and what will the final lesser payment be? b) How long is the savings period?
  • 70.  Page 90 Mind Action Series
  • 71.  You decide to take a bank loan of R800 000. repayments are made on a monthly basis starting one month after the granting of the loan. You are offered three loan options: Option A Pay of the loan at 18% p.a. compounded monthly Option B Pay of the loan over thirty years at 18% p.a. compounded monthly Option C Pay off the loan over thirty years at 15% p.a. compounded monthly a) Calculate the monthly amount payable for each option b) Which option would be the best financially for you? Motivate your answer by using appropriate calcualtions
  • 72.  Page 92 Mind Action Series
  • 73.  It is sometimes useful to calculate the balance still owed on a loan at a given time during the course of the loan.
  • 74.  James takes out a one year bank loan of R18 000 to pay for an expensive laptop. The interest rate is 18% per annum compounded monthly and monthly repayments of R1 650,24 are made starting one month after the loan was granted. a) Calculate his balance outstanding after he has paid the sixth installment b) Calculate his balance outstanding after he has paid his ninth installment
  • 75.  Page 94 Mind Action Series
  • 76.  Payments which start immediately ◦ Payment at To becomes an additional/extra payment  Payments in advance ◦ Payments that end one or more periods before the end of the investment term  Sinking funds ◦ Savings set up in order to replace an item in future
  • 77.  Fixed payments ◦ Payments which start one period (e.g. one month) after a loan is granted  Skipped payments ◦ When repayments start more than one period (e.g. in three months’ time) after the loan is granted ◦ In such cases, the original loan amount accrues interest and payments are made to pay off a new Present Value  Balance outstanding on a loan ◦ This is the present value of a loan with n payments still to go ◦ Position of the last installment becomes position of your new present value(balance outstanding)
  • 78.  Page 98-102 Mind Action Series
  • 79.  Mind Action Series textbook  Classroom Mathematics Grade 12 learner’s book  Maths Hand Book and Study Guide (Gr 12) – Kevin Smith  Maths and Stats for Bussiness – X-Kit Undergraduate – J Bassons et al.