annuities

ANNUITIES

Annuities

Mathematics Learning Centre

Annuities ANNT-A

Objectives ............................ .......................................... ........................... ........................... ............................ ........................... .............

ANNT 1

ANNT-B

Sequences ........................... ......................................... ........................... ........................... ............................ ............................ ................

ANNT 2

ANNT-C

Annuities ............................ ......................................... ........................... ............................ ............................ ........................... ............... ..

ANNT 3

ANNT-D

Present value of an ordinary annuity .......................... ....................................... .......................... .............

ANNT 3

ANNT-E

The Future Value of an ordinary annuity ......................... ...................................... ................... ......

ANNT 6

ANNT-F

Annuity Due .......................... ........................................ ............................ ........................... ........................... ......................... ...........

ANNT 8

ANNT-G

Mixed Questions........................... ......................................... ............................ ............................ ............................ ................. ...

ANNT 10

ANNT-H

Amortisation of Loans ............................ ......................................... ........................... ............................ .................... ......

ANNT 12

ANNT-Y

Index ........................... ......................................... ........................... ........................... ............................ ............................ ...................... ........

ANNT 16

ANNT-Z

Solutions ........................... ......................................... ............................ ............................ ............................ ............................ ................

ANNT 17

ANNT-A

Objectives

1.

To understand the concept of an annuity.

2.

To perform calculations involving annuity lump sums.

3.

To perform calculations involving annuity payments.

4.

To determine if an annuity is an annuity due or an ordinary annuity.

5.

To do calculations on amortised loans.

ANNT 2

Annuities

ANNT-B


Sequences

A sequence is a set of numbers arranged in a set order. For example: 5, 10, 15, 20, 25… is a sequence. The first term is 5, the second is 10, and the the third is 15 and so on. The type of sequence used to obtain Annuity formulas is a Geometric Geometric Sequence. In a geometric sequence any new term is found by multiplying or dividing the previous term by a constant number called the common ratio. The sequence 2, 4, 8, 16, 32, 64… is a geometric sequence because any new value is found by multiplying the previous value by 2. The next term in this sequence is 128 (i.e. 64x2). Let’s consider the Geometric Sequence 1, 3, 9, 27, 81, 243 … The second term was determined by multiplying the first term by 3 (3=1x3). The third term was found by multiplying the second term by 3 (9=3x3). The fourth term was found found by multiplying the third term by 3 (27 = 9x3) and so on. In this example the common ratio ratio is 3. The common ratio is given the pronumeral r , the first term is given the pronumeral a, therefore any sequence can be expressed as: n− a, ar , ar , ar ar , ar , .........ar

2

3

4

1

For the geometric sequence 100, 110, 121, 133.1, 146.41…,, the first term a is 100, the common ratio r is 110 ÷ 100 = 1.1. It is important to note that 121 ÷ 110 & 133.1 ÷ 121 also give 1.1. For the geometric sequence where a =10 and r = = 1.5, the sequence is 10, 15, 22.5, 33.75, 50.625… When the terms of a geometric sequence are added together −1 3 ie: a + ar + ar 2 + ar a r + ......... ar n , this is called a Geometric Series.

An Annuity is simply a special type of Geometric Series.

ANNT 3

Annuities

ANNT C


Annuities

An annuity is a series of regular, equally spaced, payments over a defined period of time (often called the term) at a constant rate of interest. The payments may occur weekly, fortnightly, monthly, quarterly or yearly. Example of annuities includes: regular regular payments into a savings account or superannuation fund, loan payments and periodic payments to a person from a retirement fund. An Ordinary annuity is an annuity where the regular payment is made at the end of the successive successive time periods. periods. Another type of annuity, called an Annuity Due, will be covered later in the module. ANNT D

The Present Value formula of an Ordinary Annuity.

A loan from a bank is an example of an annuity with a present value and repayments for the term of the loan. In other words, the banks gives gives you the lump now (at present) present) and the repayments are made in periodic payments after this.

Present 0

R

R

R

R

R

1

2

3

4

5

R - periodic payment

Each repayment must be changed to its present present value (see Finance module). Doing this and then considering the sum of the series, the sum of o f the series is A = R

1 − (1 + r ) − n r

Where $ A is the present value of an ordinary annuity, $ R is the amount of each payment, r is is the rate per period (payment) and n is the number of periods (payments). An ordinary annuity is an annuity where the payment is made at the end of the payment period. In this module, normal rounding will be used. However, be aware that financial institutions institutions will use a different method.

ANNT 4

Annuities


Examples ANNT-D1 1. Mr. and Mrs. Lyons wish to have an annuity for when their daughter goes to university. They wish to invest into an annuity that will pay their daughter $1000 per month for 4 years. What is the present present value of the annuity given that current interest rates are 8% p.a?

1000,, r = The information given is: R = 1000

0.08

0.0066 6666 6667 67 per per month onth,, n = 12 × 4 = 48 month onthss . = 0.00 12 This annuity is considered to be an ordinary annuity, that is, each payment will occur at the end of the payment period. A = R

1 − (1 + r )

−n

r

1 − (1 + 0.00666667)−48 A = 1000 0.00666667 0.273079396 A = 1000 × 0.00666667 A = 40961.91 This means $40 961.91 invested now at 8% will provide $1000 per month for 4 years.

2.

Roger Little borrows $20 000 to buy a car. He wishes to make monthly payments for 4 years. The interest rate he is charged is 10.5% p.a. What is the size size of each monthly payment?

The lump sum of $20 000 0 00 is given at the beginning beg inning of the loan, this indicates it is the present 0.105 = 0.00 $20000 00,, r = 0.0087 875 5 per per month onth,, n = 12 × 4 = 48 month onthss . This value of an annuity: A = $200 12 annuity is considered to be an ordinary annuity because each payment occurs at the end of the payment period. A = R R = R =

1 − (1 + r )

−n

r Ar

1 − (1 + r ) − n 20000 000 × 0.008 00875 1 − (1 + 0.00875)−48

R = $512.07

This means Roger will pay $512.07 per month for 4 years to repay the loan. ANNT 5

Annuities


Exercise ANNT-D1 For each of the following situations, find the present value of the ordinary an nuity described. 1. $2000 per month for 5 years at a rate of 6% compounded monthly.

2. $5500 per quarter for 6

1

years at 5.6% compounded quarterly. 2 3. $150 per month for 3 years at 8% compounded semi-annually. 4. Joe wants to invest a lump sum of money now to cover a monthly commitment of $100 over 5 years. If the lump sum is invested at 7.5% compounded monthly, what amount will Joe invest? 5. Jane’s car insurance company charges a monthly premium of $41.50. What is the present value of two year’s premiums if the rate of inflation is 0.3% per month? Exercise ANNT-D2 For each of the following situations, find the periodic payment necessary for the annuity given the present value. 1. $25000 to fund 4 years of monthly payments at a rate of 5.25% 2. A public servant retires with $325000 in superannuation. He is offered an annuity for the next 15 years where he will receive monthly payments. If the interest rate is 8% compounded monthly, determine the size of the monthly payment. 3. The executor of a will has to distribute an inheritance of $28 000 to a sole beneficiary in equal monthly payments over 4 years while the beneficiary undertakes university study. study. How much will each monthly payment be? (assume (assume an interest rate of 7.5%) 4. If an annuity is purchased for $50 000, how much will each quarterly payment be over 10 years if the rate of interest is 5% compounded quarterly.

ANNT 6

Annuities


ANNT E

The Future Value formula of an Ordinary Annuity.

Where regular payments are made with a lump sum at the end, the lump sum at the end is called the Future Value of an annuity. A good example of this is is a saving scheme where regular payments are made to build to a lump sum at the end of a period of time. In business, this is called a sinking fund. It is used to save for the future replacement replacement of major capital items. items. 0 1 2 3 8

Future Value ($) 0 R

1

2

R

R

3

8

R

R

9 R

10 R

The future value of an annuity is the value of all payments at the end of the term. It is given by S=R

(1 + r ) n − 1

r the equation: , where S is the future value of the ordinary annuity, R is the periodic payment, r is the interest rate per period and n is the number of payments. Examples ANNT E1:

1. George deposits $150 into a bank account at the end of the month for 5 years at a rate of 7% compounded monthly. S=R

(1 + r ) n − 1 r

R = 150, r = 0.07 ÷ 12 = 0.00583&3 n = 5 × 12 = 60

(1 + 0.005833& )60 − 1 S = 150 0.005833& 150 × 71. 71.592 592900 90091 S = 150 S = 10738.94

The future value of the annuity is $10 738.94

ANNT 7

Annuities


2. Wally is planning for his retirement 20 years away. When he retires he wants a lump sum of $300 000. His financial advisor suggested that that 5% p.a. was a suitable suitable interest rate to consider. How much will he have to pay per month into his retirement fund (assume ordinary annuity). Given: S = 300000, r = 0. 0.05 ÷ 12 = 0.0041&6, n = 12 12× 20 = 240 S=R Sr

(1 + r ) n − 1 30000 0000 × 0.0 0.0041 0416& (1.00416& )240 − 1

(1 + r ) n − 1 r

= R = R

R = 729.87

Wally will have to pay $729.87 per month until he retires in 20 years.

Exercise ANNT-E1 1. Find the future value of an annuity where $1200 is paid at the end of each year for 15 years at a rate of 7% compounded annually. 2. The ABC concreting company set up a sinking fund to assist in buying a new truck in 5 years time. They can only afford $3000 a quarter which is paid into a savings account with an interest rate of 8% p.a. Assuming quarterly compounding, what will will be the size of the sinking fund after 5 years? ye ars? (Assume ordinary annuity) 3. Colin invests $500 per month, paid into a savings account for 10 years. What is the balance of the account at the end of the period assuming an interest rate of 7% compounded monthly? 4. In 5 years, a printing machine machine is to be replaced. A new machine is expected to cost $33000. Assuming an annual interest rate rate of 8% compounded monthly, what will be the size of each monthly payment?

ANNT 8

Annuities


ANNT F Annuity Due So far, all calculations have been performed with ordinary annuities, where periodic payments are made at the end of the period. An annuity due differs from an ordinary annuity since periodic payments are made at the beginning of the period. Some payments such as house rents and insurance premiums are paid at the beginning of the period before the the service is provided. If the payment is made at the beginning of the period, each payment will be subject to an extra month’s interest when compared to ordinary annuities. A simple modification to the ordinary annuity formulae give: PRESENT VALUE OF AN ANNUITY DUE A D = (1 + r ) AO A D = R (1 + r )

1 − (1 + r )

−n

r

FUTURE VALUE OF AN ANNUITY DUE S D = (1 + r ) S O S D = R (1 + r )

n (1 + r ) − 1

r

Examples ANNT F1 1. A company wishes to deposit an amount of money into an account at the beginning of each year for the next 5 years to to purchase a new machine costing $50000. How much will each yearly payment be if the current interest rate is 7.2%p.a? This is a sinking fund fund scenario. It is an annuity due. 0.072, S = 50000 50000 Given: n = 5, r = 0.072, S D = R (1 + r )

n (1 + r ) − 1

r

S D r

= R ⎡(1 + r )n − 1⎤ (1 + r ) ⎣ ⎦ 5 0 0 0 0 × 0 .0 7 2 = R ⎡(1.072) 5 − 1⎤ (1.072) ⎣ ⎦ 3600 = R 0.445639816 R = 8078.27 $8 078.27 is required at the beginning of the year for 5 years at 7.2% to have $50 000 in the sinking fund at the end of 5 years.

ANNT 9

Annuities


2. Joe pays $250 rent per week at the beginning of each week. He is considering paying a whole year’s rent in advance; given the the interest rate is 5.2% p.a. How much is this amount? This is a present value calculation. This is an annuity due situation. situation. 0.052 ÷ 52 = 0.001, n = 52 Given: R = 250, r = 0. A D = R (1 + r )

1 − (1 + r )

250 (1.00 1.001 1) A D = 250

−n

r

1 − (1.001)

−52

0.001

A D = 12674.28

Joe would have to pay $12 674.28 now to cover his rent for the year. Exercise ANNT – F1 1. Find the future value of an annuity due if $800 is paid into an account at the beginning of each month for 5 years at a rate of interest of 5% p.a. compounded monthly.

2. Find the present value of an annuity if the periodic amount is $450 per quarter for 20 years at the rate of 4.5% p.a. compounded quarterly. 3. A company leases office space for a period of 12 months. The monthly rent of $2500 is paid at the beginning of each month. If the company is to cover all rents with a single lump sum at the beginning of the year and invests this at 6.3% p.a. how much will the lump sum be?

ANNT 10

Annuities

ANNT G


Mixed Questions

For the following questions, follow these steps. 1. Read the question thoroughly 2. Decide on the type of annuity: either ordinary annuity, where periodic payments are made at the end of the period OR annuity due, where periodic payments are made at the beginning of the period. 3. Decide if the annuity contains a lump sum in the present, the lump sum is at the beginning of the term of the annuity, OR future, the lump sum is at the end of the annuity. 4. Select the appropriate formula Ordinary Annuity t A n e e u s l e r a P V S e r e u u l t a u F V

A = R

Annuity Due

1 − (1 + r ) − n

A D = R (1 + r )

r

1 − (1 + r )

−n

r n

(1 + r )n − 1 S=R r

S D = R (1 + r )

(1 + r ) − 1 r

Example ANNT G1 Kevin Smith won $820 000 in a lotto game. With this lump lump sum he purchases an annuity to give him a monthly income for the next twenty years. The first payment occurs a month month after the start of the annuity. If the interest rate rate is 6.8% p.a. compounded monthly, calculate the amount of each payment. This is an ordinary annuity (The periodic payments occur at the end of the period.) This annuity involves a present value (The lump sum sum is at the beginning of the term of the annuity.) The formula required is A = R

1 − (1 + r ) − n r

ANNT 11

Annuities


Information given is A = 820000 r =

0.068

= 0.005666& n = 20 ×12 = 240

12 Rearranging the formula to make R the subject: A = R

1 − (1 + r )− n

Ar

1 − (1 + r ) − n R =

r

= R

& 820000 0000× 0.0 0.00566 05666 6 1 − (1.005666& )−240

R = $6259.38

Kevin will receive $6259.38 per month for 20 years.

Exercise ANNT G1 1. Sally borrows $10 000 to buy a car. car. If the interest rate charged is 10.5% p.a. calculate the monthly repayment over the term of the loan, 5 years. The first payment is made made a month after the loan lump sum is advanced. 2. Brett wishes to set aside all his rent for one year. This money will be put into into an account paying 6.6% p.a. compounded monthly and his rent is $1040 per month. Rent is always paid in advance. Calculate the amount Brett must deposit. 3. From the time Jane and John’s daughter was born, they decided to save for her university education. Jane and John assume their daughter will require $1000 per month for for her four years of study, payments being made at the beginning of each month. If Jane and John save for 18 years, calculate the amount they must save at the beginning of each month. Assume 6% p.a. interest is compounded monthly. Hint: there are two annuities here, calculate the lump sum required to fund the $1000 p.m. allowance first, then the monthly amount the parents must save. 4. The PS Transport Company decide that they must start saving for a new vehicle in 5 years time. In an account that pays 5.4% p.a. compounded monthly monthly they deposit a one off payment of $20 000 and $500 at the end of each month. How much much will they have at the end of 5 years? (Hint: treat the two two amounts separately, one is an annuity and the other compound growth)

ANNT 12

Annuities

ANNT H


Amortization of Loans

The term amortization refers to loans where each payment is made up of principal and a nd interest components. An example of where amortization is commonly used is housing mortgage loans loans which usually have a term of many years. Early in the term of the loan, the amount of principal outstanding is large, hence the interest component of the loan is high and the amount paid off the loan is small. As the loan progresses, the amount of principal paid each repayment increases increases and the amount of interest decreases. The operation of the loan can be shown shown using an amortization schedule. Example ANNT H1 For this example, a loan of $100 000 is being considered over a term of 10 years at an interest rate of 9%p.a. with monthly repayments. Repayments on loans are made at the the end of the month so this is an ordinary annuity. The lump sum is at the beginning of the annuity so the present value formula is used. The first step is to calculate the repayment amount of the loan. 0.09 = 0.0075, n = 10 ×12 = 120 A = $100000, r = 12 A = R R = R =

1 − (1 + r )

−n

r Ar

1 − (1 + r )− n 10000 0000 × 0.007 0075 1 − (1 + 0.0075)−120

R = $1266.76

The monthly repayment for this loan (annuity) is $1266.76 per month. The amount of interest for the first month is $100000 × 0.0075 = $750 , so the amount of principal paid off the loan during the first month is $1266.76 - $750 = $516.76. So the balance of the loan for the second month is $100000 - $516.76 = $99483.24, and this continues until the loan is paid off after 10 years. The amortization schedule for the first six months of the loan is shown below.

ANNT 13

Annuities

Month

1 2 3 4 5 6


Principal Outstanding at the beginning of the month $100 000 $99 483.24 $98 962.60 $98 438.06 $97 909.59 $97377.15

Interest per month

Payment at the end of the month

$750 $746.12 $742.22 $738.29 $734.32 $730.33

$1266.76 $1266.76 $1266.76 $1266.76 $1266.76 $1266.76

Principal paid off outstanding principal $516.76 $520.64 $524.54 $528.47 $532.44 $536.43

Principal Outstanding at the end of the month $99 483.24 $98 962.60 $98 438.06 $97 909.59 $97 377.15 $96 840.72

The total amount of interest paid over the 10 years of the loan is calculated by calculating the total paid in repayments 120 × 1266.76 = $152011.20 subtract the amount borrowed $100 000 to give $52 011.20 The amount of principal outstanding at the beginning of the 25th month: At the beginning of the 25th month, 24 payments have been made, resulting in 96 months left on the loan. Using the present value of an annuity, the amount of principal outstanding is given by: A = R

1 − (1 + r )

−n

r

A = 1266.76

1 − (1 + .0075)

−96

0.0075

A = 86467.06

The amount of principal outstanding at the beginning of the 25th month (which is also the balance at the end of the 24th month) is $86 467.06 Because this is the amount of principal outstanding at the beginning of the 25th month, the th amount of interest for the 25 month is 0.0075 × $86467.06 = $648.50 Consequentially, the amount of principal paid in the 25th month will be $1266.76-$648.50 = $618.26 Example ANNT H2 A person amortizes a loan of $30000 for a new car by obtaining a 10 year loan at a rate of 12% with monthly payments. Find (a) the monthly payment (b) the total interest charged (c) the principal remaining after 4 years (d) the interest paid and the principal paid during the 49th month ANNT 14

Annuities


(a) The monthly repayment: A = 30000 r = 0.12 ÷ 12 = 0.01 n = 10 × 12 = 120 A = R ×

1 − (1 + r )

−n

(b) The total interest charged:

I = n × R − A

I = 120 × 430.41 − 30000 I = $21649.20

r

rearranging R = R =

The total interest charged is $ 21 649.20

Ar

1 − (1 + r )− n 30000 × 0.01 1 − (1.01)−120

R = $430.41

The monthly payment is $430.41 per month. (c) The principal remaining after 4 years yea rs or 48 months. After 48 months, there are 120 − 48 = 72 months remaining. A = R ×

1 − (1 + r )

−n

r

1 − (1.01) −72 A = 430.41× 0.01 A = $22015.64 At the end of the 48th month (or the beginning of the 49th month) the amount of principal still outstanding in $22 015.64

(d) The interest paid and the principal paid during the 49th month. The start amount for the 49th month is $22 015.64 (see (c)). Interest, I = r × 22015.64 I = 0.01 × 22015 I = $220.16

The interest for the 49 th month is $220.16 The principal paid during the 49th month is R − I 49 th month = 430.41 − 220.16 = $210.15

Exercise ANN-H1 1. Construct an amortization schedule for the following scenario: A loan of $1200 is repaid by 6 quarterly payments, interest is 10% compounded quarterly. 2. John and Joanne take out a mortgage housing loan for $250 000 over 20 years at an interest rate of 8% with monthly payments. Calculate the size of each repayment. 3. John and Joanne’s interest rate (from question 2) has just been increased to 9%, by how much will their repayments increase by so they can still repay the loan after 20 years? 4. A $45 000 mortgage loan for 25 years for home additions is obtained at a rate of 7.75% repaid in monthly repayments. repayments. Find a. The monthly repayment ANNT 15

Annuities


b. c. d. e.

The principal outstanding at the beginning of the 36th month. The interest in the 36th payment. The principal in the 36th payment. The total interest paid.

5. Marilyn and Murray take out a home mortgage loan for $150 000 over 20 years at an interest rate of 6% with monthly payments. a. Calculate the amount of each monthly repayment. After 10 years Marilyn receives a lump sum payout of $30 000 after being made redundant from her university position. position. She decides to pay this off the the principal of the loan. She also decides to shorten the term term of the loan to just just 5 more years (15 in total). b. What will the size of the new repayment be?

ANNT 16

Annuities

TRIG-Y


Index

Topic

Page

Amortization of Loans Amortization Schedule Annuities Annuity Due

ANNT 12 ANNT 12 ANNT 3 ANNT 3, ANNT 8

Future Value of an Ordinary Annuity Future Value of an Annuity Due

ANNT 6 ANNT 8

Geometric Sequence Geometric Series

ANNT 2 ANNT 2

Interest Paid

ANNT 13

Ordinary Annuity

ANNT 2

Periodic Payment Present Value of an Ordinary Annuity Present Value of an Annuity Due Principal Outstanding

ANNT 3 ANNT 3 ANNT 8 ANNT 13

Sequence

ANNT 2

Term

ANNT 3

ANNT 17

Exercise ANNT-D1 Solutions 1. $2000 per month for 5 years at a rate of 6% compounded monthly. Both n & i must be expressed in months. R = 2000, n = 5 ×12 A = R

= 60,

i

4. Joe wants to invest a lump sum of money now to cover a monthly commitment of $100 over 5 years. If the lump-sum is invested at 7. 5% compounded monthly, what amount will Joe invest?

= 0.06 ÷12 = 0.005

1 − (1 + r )− n

0.075 ÷12 = 0.00625 R = 100, n = 5×12 = 60, i = 0. A = R

1− (1+ r )−n r

r

1 − (1 + 0.005) −60 A = 2000 0.005 2000× 51.7 51.725 2556 5607 075 5 A = 2000

A = 100

0.00625 100×49.90 49.9053 5308 0818 18 A = 100 A = 4990.53

A = 103451.12

If Joe invests $4 990.53 now he will meet his monthly commitment of $100 per month for 5 years.

The present value of this ordinary annuity is $103 451.12

5. Jane’s car insurance company charges a monthly premium of $41.50. What is the present value of two year’s premiums if the rate of inflation is 0.3% per month?

1 6 years at 5.6% 2 compounded quarterly. Both n & i must be

2. $5500 per quarter for

0.03 R = 41.50, n = 2×12 = 24, i = 0.

expressed in quarters. R

= 5500,

n

1

= 6 × 4 = 26,

A = R

2 i = 0.056 ÷ 4 = 0.014 A = R

1 − (1 + r )

1− (1+ 0.003)−24 A = 41.50 0.003 1.50× 23.1 23.12 22934 2934 A = 41.50

−n

1 − (1 + 0.014) −26

A = 959.60

0.014 5500 × 21.6 21.668 6802 0257 573 3 A = 5500

The present value of Jane’s car i nsurance premiums is $959.60

A = 119174.14

The present value of this ordinary annuity is $119 174.14 3. $150 per month for 3 years at 8% compounded semi-annually. Semi-annually means means half yearly. yearly. R, n & i must be expressed in half years.

= 150 × 6 = 900, i = 0.08 ÷ 2 = 0.04

R

A = R

n

= 3 × 3 = 6,

1 − (1 + r ) − n r

A = 900

1 − (1 + 0.04) −6

0.04 900 × 5.24 5.2421 2136 3685 857 7 A = 900 A =

1− (1+ r )−n r

r A = 5500

1− (1+ 0.00625)−60

4717.92

The present value of this ordinary annuity is $4 717.92

ANNT 17

Exercise ANNT-D2 Solutions 1. $25000 to fund 4 years of monthly payments payments at a rate of 5.25% A =

= 4 × 12 = 48, r = 0.0525 ÷ 12 = 0.004375 25000,

A = R

n

1 − (1 + r ) − n

A =

= 4 ×12 = 48, r = 0.075 ÷ 12 = 0.00625

r

⇒ R = R

R

= =

Ar

1 − (1 + r ) − n

250 25000 × 0.00 0.0043 4375 75 1 − (1 + 0.004375) −48

A = 325000, n

= 15 ×12 = 180, & r = 0.08 ÷ 12 = 0.006666

R

= =

R

=

R

=

2166.666

0.697603911 R = 3105.87

1 − (1 + r ) − n 280 28000 × 0.00 .00625 625 1 − (1 + 0.00625) −48 175

A = 50000, n r = 0.05 ÷ 4 R

=

R

=

R

=

32500 25000 0 × 0.00 .0066& 1 − (1 + 0.0066& ) −180

Ar

4. If an annuity is purchased for $50 000, how much will each quarterly payment be over 10 years if the rate of interest is 5% compounded quarterly.

Ar

1 − (1 + r ) − n

n

The student will receive a monthly payment of $677.01

2. A public servant servant retires with $325000 $325000 in superannuation. He is offered an annuity for the next 15 years where he will r eceive monthly payments. If the interest rate is 8% compounded monthly, determine the size of the monthly payment.

R

=

28000,

0.258489819 R = 677.01

The monthly payment will be $578.57

=

R

109.375

0.189044389 R = 578.57

R

3. The executor of a will has to distribute an inheritance of $28000 to a sole beneficiary in equal monthly payments over 4 years while the beneficiary undertakes university study. How much will each monthly payment be?

= 10 10 × 4 = 40,

= 0.0125

Ar

1 − (1 + r ) − n 50000 × 0.0125 1 − (1 + 0.0125) −40 625

0.391586664 R = 1596.07 The quarterly payment will be $1596.07.

The monthly payment will be $3105.87

ANNT 18

Exercise ANNT-E1 Solutions 1. Find the future value of an annuity where $1200 $1200 is paid at the end of each year for 15 years at a rate of 7% compounded annually. R S

= 120 1200, 0,

=R

r

= 0.07, 0.07,

n

= 15.

= 500, r = 0.07 ÷ 12 = 0.00583&, n = 10 10 ×12 = 120.

R

(1 + r ) n − 1 r

S = 1200

(1.07)15 − 1

S =R

S = 500

2. The ABC concreting company set up a sinking fund to assist in buying a new truck in 5 years time. They can only afford $3000 a quarter which is paid into a savings account with an interest rate of 8%p.a. Assuming quarterly compounding, what will be the size of the sinking fund after 5 years?

S

=R

r

= 0.08 ÷ 4 = 0.02,

n

(1.00583& )120 − 1

0.00583& S = $86542.40

The future of this ordinary annuity is $30154.83

= 3000,

(1 + r ) n − 1 r

0.07 S = $30154.83

R

3. Colin invests $500 per month, paid paid into a savings account for 10 years. What is the balance of the account at the end of the period assuming an interest rate of 7% compounded monthly?

Colin will have a savings account balance of $86542.40 4. In 5 years, a printing machine is to be replaced. A new machine is expected to cost $33000. Assuming an annual interest rate of 8% compounded monthly, what will be the size of each monthly payment?

= 33000, r = 0.08 ÷ 12 = 0.006&, n = 5 × 12 = 60.

S

= 5 × 4 = 20.

(1 + r ) n − 1

S

=R

r

r S = 3000

(1 + r ) n − 1

Sr

(1.02) 20 − 1

(1 + r ) n − 1

0.02 S = $72892.11

220 0.489845649

The company will have saved $72892.11 towards the new truck.

R

= R = R = $449.12

A payment of $449.12 per month will be required to raise $33000 for new machinery in 5 years ti me.

ANNT 19

Exercise ANNT-F1 Solutions 1. Find the future value of an annuity due if $800 is paid into an account at the beginning of each month for 5 years at a rate of i nterest of 5% p.a. compounded monthly R

= 800,

r

=

0.05 12

= 0.0041666&,

n

= 5 × 12 = 60

S

= R(1 + r )

(1 + r ) n − 1 r

(1.0041666& ) 60 − 1 & = × × 800 1.00 1.004 41666 1666 S 800 0.0041666& S = 54631.55

The future value of this annuity due is $54 631.55 2. Find the present value of an annuity if the periodic amount is $450 per quarter for 20 years at a rate of 4.5% p.a. compounded quarterly R

= 450,

r

=

0.045 4

= 0.01125,

n

= 20 × 4 = 80

A = R (1 + r )

1 − (1 + r ) − n r

1 − (1.01 1.0112 125 5) −80 A = 450 450 ×1.01 1.0112 125 5× 0.01125 A = 23921.41 The present value of this annuity due is $23 921.41

3. A company leases office space for a period of 12 months. The monthly rent of $2500 is paid at the beginning of each month. If the company is to cover all rents with a single lump sum at the beginning of the year and invests this at 6.3% p.a. how much will the lump sum be? This is the present v alue of an annuity due. R

= 2500,

r

=

0.063 12

= 0.00525,

n

= 1×12 = 12

A = R (1 + r )

1 − (1 + r ) − n r

A =

2500 500 ×1.0 1.00525 ×

A =

29153.10

1 − (1.00525) −12 0.00525

The company must invest $29 153.10 to cover rent for the entire year.

ANNT 20

Exercise ANNT G1 Solutions 1. Sally borrows $10 000 to buy a car. If the interest rate charged is 10.5%p.a. calculate the monthly repayment over the term of the loan, 5 years. The first payment is made one month after the loan lump sum is advanced. This involves the present value of an ordinary annuity. R = 10000, i = 0.105 ÷ 12 = 0.00875, n = 5 ×12 = 60 A = R

1 − (1 + r ) − n

Amount required to give $1000 per month for 48 months is: present value of an annuity due.

r R

R

= =

R = 1000,

Ar

1 − (1 + r )

3. From the time Jane and John’s daughter was born, they decided to save for her university education. Jane and John assume their daughter will require $1000 per month for her four years of study, payments being made at the beginning of each month. If Jane and John save for 18 years, calculate the amount they must save at the beginning of each each month. Assume 6% p.a. interest is compounded monthly.

−n

100 10000× 0.00 .00875 875 1 − (1.008 .0087 75)

= 0.06 ÷12 = 0.005,

r

1 − (1 + r ) − n

A D

= R(1+ r )

A D

= 1000 1000×1.00 .005 ×

A D

= 42793.22

r

−60

R = 214.94 Sally will pay $214.94 per month.

n = 4 ×12 = 48

1 − (1.005) −48 0.005

The monthly deposit for 18 years is: future value of an annuity due.

= 42793.22,

S

Sr

R = 1040, r

A D

= 0. 0.066 ÷12 = 0.0055,

= R(1 + r )

n = 1×12 = 12

1 − (1 + r ) − n

A D

= 1040 ×1.0055 ×

A D

= 12111.31

0.0055

r

4279 42793. 3.22 22× 0.00 0.005 5 216 1.005 × [(1.005) − 1]

R

=

R

= 109.93

= 20000, i = 0. 0.0045, n

n

= 5 × 12 = 60

= 20000(1.0045) 60 = 26183.43

$500 per month is used to calculate the future value of an ordinary annuity (sinking fund). R S

Brett must deposit $12 111.31

(1 + r ) n − 1

Using the compound interest (growth) formula, the $20000 will grow to: A = P (1 + i )

1 − (1.0055) −12

= 18 ×12 = 216

Jane and John should deposit $109.93 at the beginning of each month for 18 years. 4. The PS Transport Company decide that they must start saving for a new vehicle in 5 years time. In an account that pays 5.4% p.a. compounded monthly, they deposit a one off payment of $20 000 and $500 at the end of each month. How much will they have at the end of 5 years?

P

r

n

= R

(1 + r ) ⎡⎣ (1 + r )n − 1⎤⎦

This involves the present value of an annuity due.

= 0.0 0.005,

= R(1+ r )

S

2. Brett wishes to set aside all his rent for one year. This money will be put into an account paying 6.6% p.a. compounded monthly and his rent is $1040 per month. Rent is always paid in advance. Calculate the amount Brett must deposit.

r

= 500, r = 0.0045, 0.0045, n = 60

=R

(1 + r )n − 1 r

(1.0045) 60 − 1 0.0045 S = 34352.36 Altogether, there will be $60 535.79 available to purchase a new vehicle. S = 500 ×

ANNT 21

Exercise ANN H1 1. Construct an amortization schedule for the following scenario: a loan of $1200 is repaid by 6 quarterly payments, interest is 10% compounded quarterly. To do this a repayment figure is required. This scenario involves the present value of an ordinary annuity. R = 1200, i = 0.10 ÷ 4 = 0.025, n = 6 A = R

1 − (1 + r ) − n r

Quarter 1 2 3 4 5 6

Beginning Principal 1200 1012.14 819.58 622.21 419.91 212.55

Ar

R

=

R

=

R

= 217.86

1 − (1 + r ) − n 1200 × 0.025 1 − (1.025) −6

Interest 0.025x 1200=30 0.025x1012.14=25..30 0.025x819.58=20.49 0.025x622.21=15.56 0.025x419.91=10.50 0.025x212.55=5.31

Payment 217.86 217.86 217.86 217.86 217.86 217.86

Principal paid off 217.86-30 = 187.86 217.86-25.30=192.56 217.86-20.49=197.37 217.86-15.56=202.30 217.86-10.50=207.36 217.86-5.31-212.55

End Principal 1200-187.86=1012.14 1012.14-192.56=819.58 819.58-197.37=622.21 622.21-202.30=419.91 419.91-207.36=212.55 212.55-212.55=0

2. John and Joanne take out a mortgage housing loan for $250 000 over 20 years at an interest rat e of 8% p.a. with monthly payments. Calculate the size of each payment. payment. A loan like this is an example of an ordinary annuity with a present value (the amount borrowed). A = 250000, r = A = R

1 − (1 + r )

0.08 12

& , n = 20×12 = 240 = 0.00 .00666

−n

r Ar

R

=

R

=

R

= 2091.10

A monthly repayment of $2091.10 is required to service the loan.

1 − (1 + r ) − n 2500 250000 00 × 0.00 0.0066 666 6&

& ) −240 1 − (1.0 1.00666 0666)

3. John and Joanne’s interest rate (from question 2) has just been increased to 9%, by how much will their payments increase by so they can still repay the loan after 20 years? A = 250000, r = A = R

0.09 12

= 0.00 .0075, n = 20×12 = 240

1 − (1 + r ) − n r Ar

R

=

R

=

R

= 2249.31

1 − (1 + r ) − n 250 250000× 0.00 0.007 75 1 − (1.0075) −240

The new monthly repayment is $2249.31, an increase of $158.21

ANNT 22

4. A $45 000 mortgage loan for 25 years for home additions is obtained at a rate of 7.75% repaid in monthly repayments. Find (a) The monthly repayment: A loan li ke this is an example of an ordinary annuity with a present value (the amount borrowed). A = 45000, r A = R

=

0.0775 12

& , n = 25×12 = 300 = 0.00 .0064583

1 − (1 + r ) − n r Ar

R

=

R

=

R

= 290.62

The monthly repayment is $290.62

1 − (1 + r ) − n 4500 45000 0 × 0.00 0.0064 6458 583 3& 1 − (1.00 1.0064 6458 583 3& ) −300

(b) The principal outstanding at the beginning of the 36 th month. At the beginning of the 36 th month,35 payments have been made, so there are 300-35=265 months remaining. A = R

1 − (1 + r ) − n r

1 − (1 + 0.0064583& ) −265 The principal outstanding at the beginning beginning of the 36th month is $36 827.42 & 0.0064583 A = 36827.42 A =

290.62 ×

(c) The interest in the 36th payment. Interest for the 36th payment is

0.0064583& × 36827.42 = 237.84

(d) The principal in the 36th payment. Principal = $290.62-$237.84 = $52.78 In the 36th month, the principal outstanding will reduce by $52.78 (e) The total interest paid. Total paid off the loan i s $290.62 x 300=$87 186, total interest paid = $87 186 - $45 000 = $42 186 5. Marilyn and Murray take out a home mortgage loan for $150 000 over 20 years at an interest rate of 6% with monthly payments. (a) Calculate the amount of each monthly payment. A = 150000, r A = R

=

0.06 12

= 0.005, n = 20×12 = 240

1 − (1 + r ) − n r Ar

R

=

R

=

R

= 1074.65

1 − (1 + r ) − n 150 150000 000 × 0.0 0.005 1 − (1.005) −240

Each repayment will be $1074.65

ANNT 23

(b) After 10 years, Marilyn receives a lump sum payout of $30 000 after being made redundant from her university position. She decides to pay this off the principal of the loan. She also decides to shorten the term of the loan to just 5 more years (15 in total). What will the size of the new repayments be? After 10 years (120 months) the $30 000 is paid off the principal of the loan, therefore the lump sum is paid off during the 121 st payment. The outstanding principal at the end end of the 121st month, which is the same as the outstanding principal at the beginning of the 122nd month. After the 121st month, there are 240-121 = 119 months remaining A = R

1 − (1 + r ) − n r

A = 1074.65 ×

1 − (1 + 0.005) −119 0.005

A = 96206.77

The balance of the loan at t he end of the 121st month (beginning of the 122 nd month) is $96 206.77 . Because $30 000 is paid off the l oan during the 121st month, the new principal outstanding will be $96206.77 $30 000 = $66 206.77 . If the loan is now shortened to 4 years 11 months at the beginning of 122nd month, a new repayment figure can be calculated. A = 662 66206.7 06.77, 7, r = A = R

0.06 12

= 0.005, 0.005, n = 59

1 − (1 + r ) − n r Ar

R

=

R

=

R

= 1298.57

1 − (1 + r ) − n 6620 66206. 6.77 77 × 0.00 0.005 5 1 − (1.005) −59

The new repayment figure is $1298.57

ANNT 24

annuities

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