MATH 1053 – Quantitative Methods for Business

1MATH 1053 –
Quantitative Methods for Business
Week 2 –
Time Value of Money
Annuities and their
Applications
1
2Course outline
Time value
of money
Annuities &
Net Present Value
Linear
programming
Making Good
Business Decisions
Mathematics for Business
Simple &
compound
interest
Percentages
and
proportions
Break-even
analysis
Sampling,
data displays
& elementary
probability
Correlation
& Linear
Regression
Hypothesis
TestingStatistics for BusinessCLT &
confidence
intervalsDescriptive Statistics
of a sample
Statistical Inference
from a sampleNormal
distributionSummary
measures
School of Info. Tech. & Mathematical Sciences 2Week 2
2
3Topics to be covered
 Working with multiple
cash flows:
 Annuities
 Present value
 Future value
 Payment size
 Sinking funds
 Amortisation
 Net Present Value
 Case Studies
School of Info. Tech & Mathematical Sciences 33
4School of Info. Tech & Mathematical Sciences
Case Study 1
Set for Life!
“Imagine waking up one day finding
that you were Set for Life, and that you’d
won $20,000 each and every month for
20 years. Imagine the possibilities …”
4Would you be better off with $20,000 every
month? How much would you really win?
 You’ve just won Set for Life!
 You can take holidays, drive luxury cars,
start your own business or invest the
income and grow your wealth!
4
5Case study 2
What’s the best option?
School of Info. Tech & Mathematical Sciences
Samsung 65” Full HD LED LCD
3D Capable SMART TV
NO DEPOSIT
12 MONTHS INTEREST FREE*
with monthly repayments *Offer available on advertised or ticketed price. Minimum financed amount $825.
Establishment fee ($35 for new accounts), account service fee (currently $2.95
per month) and other fees and charges are payable.
55
6Annuity
 Series of equal payments
 often made under contract
 paid at equal intervals
(e.g. quarterly or monthly)
 from an agreed date
 for a specified period of time.
 Also the sum of money that
makes a periodic payment.
 Examples:
 Mortgage or home loan
repayments
 Hire purchase repayments
 Insurance premiums
 Body corporate sinking
fund
 Lease payments on cars.
School of Info. Tech & Mathematical Sciences
 In this course, all annuities are simple and ordinary :
 Interest is compounded at the same times as the annuity
payments.
 The first payment is made at the end of the first period.
66
7Lecture example 1:
Simple ordinary annuity
 Suppose you set aside $100 at the end of each year for
three years. What will the accumulated amount be at the
end of three years?
 Assume compound interest at 10% per annum.  Using what you learned last week:
0 1 2 3
$100 $100 $100
Use with each payment:
n 1
n  2 100 1 1 $121 2  . 
100 1 1 $110 1  . 
100 1 1 $100 0  . 
S = $331
 
n M  P  1 i
School of Info. Tech & Mathematical Sciences 77
8School of Info. Tech & Mathematical Sciences
Future value of an ordinary annuity
 where:
S = future or accumulated value
R = annuity payment per period
i = interest rate per period
n = number of payments
0 1 2 … n – 1 n
R R … R R
S  ?
88
9School of Info. Tech & Mathematical Sciences
Lecture example 1 revisited
    100 3.31 $331
0.10
1.10 1
100 1 1
3
10%
$100
3
   




 
  




  
 



i
i
S R
n
i
R
n
9Same answer!
 Suppose you set aside $100 at the end of each year for
three years. What will the accumulated amount be at the
end of three years?
 Assume compound interest at 10% per annum.  Using the future value formula:
𝑆 = 𝑅 ×
1 + 𝑖 ௡ − 1
𝑖 = 100 ×
1.10 ଷ − 1
0.10 = 100 × 3.31 = $3319
10Summary: Future Value
 
n M P  1 i
Compound Interest Annuities
M = future value
P = present value
i = interest rate per period
n = number of compounding periods
S = future value
R = annuity payment per periodi = interest rate per period
n = number of payments
0 1 2 … n
P M
0 1 2 … nS
10A
R R … R
௡
10
11School of Info. Tech & Mathematical Sciences
 Suppose you set aside $100 at the end of the year for
three years. What is the value today of those three
deposits?
 Assume compound interest at 10% per annum.
0 1 2 3
$100 $100 $100
Use with each payment: P  M  1i  n
n 1
n  2
100 1 1 $75.13 3
 

.
100 1 1 $82.64 2
 

.
100 1 1 $90.91 1
 

. A = $248.69
n  3
11Lecture example 2:
Simple ordinary annuity
11
12School of Info. Tech & Mathematical Sciences
Present (discounted) value of an
ordinary annuity
 where:
A = present value
R = annuity payment per period
i = interest rate per period
n = number of payments
0 1 2 … n – 1 n
R R … R R
12A  ?
12
13School of Info. Tech & Mathematical Sciences
Lecture example 2 revisited
 Using the present value formula:
    100 2.48 $248.69
0.10
1 1.10 100 1 1
3
10%
$100
3
   




 
  




  
 



 
i
i
A R
n
i
R
n
13Same answer!
 Suppose you set aside $100 at the end of the year for three
years. What is the value today of those three deposits?
 Assume compound interest at 10% per annum.
( ) ( )
𝐴 = 𝑅 ×
1 − 1 + 𝑖 ି௡
𝑖
= 𝑅 ×
1 − 1.10 ିଷ
0.10 = 100 × 2.48 = $248.6913
14Summary: Present Value
P = present value
M = future value
i = interest rate per period
n = number of compounding periods
A = present value
R = annuity payment per periodi = interest rate per period
n = number of payments
Compound Interest Annuities
0 1 2 … n
P
R R … RS
0 1 2 … nA
P  M  1i  n
M
14ି௡14
15Lecture exercise 1
 An investment pays $50 at the end of every 6 months for 15 years.  What is the value of all the payments today if money can earn
8.5% per year compounded semi-annually?
School of Mathematics and Statistics 15The value today of all the payments is
15
16Have a go!
 Josh deposits $300 every three months into an investment account
that pays interest at 6% per year compounded quarterly.
 How much will Josh have in his account at the end of four years?School of Mathematics and Statistics 16Josh will have in his account at the end of four years.
16
17Case study 1
Set for Life!
 Channel 7’s Sunrise asked exactly how much a winner will be paid,
over the 20 year horizon.
 Kochie said “I’ve done the maths” and her face changed as below.
What was discussed that had her so worried?
School of Info. Tech & Mathematical Sciences 1717
18Case study 1
Set for Life!
 If $20,000 in 20 years’ time is worth $11,000 in today’s terms, this
suggests an interest rate of about 3% p.a. compounded monthly.
 How much do you win in total over 20 years? Is it $4.8 million
(20,000 x 12 x 20)?
R  $20,000
i  3% = 0.03/12  0.0025
n 12  20  240
Interpretation:
 If the winner received
a single equivalent
payoff today, it would
be $3.6 million.
 The Lottery
Commission needs to
set aside only $3.6
million today to pay
the prize money!
School of Info. Tech & Mathematical Sciences 18 
 
20,000 180.3109 $3,606,218
0.0025
1 1.0025 20,000
(1 1 )
240
  

 





  
 


i
i
A R
n
( )
18
19Saving up or borrowing – how much
is enough?
School of Info. Tech & Mathematical Sciences
 You want to have $12,000 one year from now.
 Should you set aside $1,000 per month?
 You borrow $12,000 today for a year.
 Will $1,000 per month be enough to pay it off?
1919
20Sinking Funds vs Amortisation
Sinking Funds Amortisation
Want to accumulate a
nominated amount of money
Want to discharge a debt
School of Info. Tech & Mathematical Sciences 2020
21School of Info. Tech & Mathematical Sciences
Sinking fund
 Periodic payments made so as to accumulate a
nominated amount of money in a specified period.
 where:
S = future value (target amount)
R = annuity payment per period
i = interest rate per period
n = number of payments
0 1 2 … n – 1 n
R R … R R ?21S
21
22School of Info. Tech & Mathematical Sciences
Amortisation
 A steady stream of even payments (constant
dollar amount) over the life of a loan.
 Each period you pay interest and repay some of
the principal.
 E.g. car loans, home loans.
2222
23School of Info. Tech & Mathematical Sciences
Amortisation
 Periodic payments that will discharge a debt :
 where:
A = present value (borrowed amount)
R = annuity payment per period
i = interest rate per period
n = number of payments
0 1 2 … n – 1 n
R R … R R ?23A
23
24School of Info. Tech & Mathematical Sciences
Lecture example 5
 If you take out a $55,000 car
loan for 10 years, what will
the monthly repayments be?
 The interest rate on this loan
is set at 8.5% per annum
compounded monthly.
 How much will you still owe at the end of two
years?
2424
25School of Info. Tech & Mathematical Sciences
Lecture example 5 solution
The monthly repayment will be $681.92.
2525
26School of Info. Tech & Mathematical Sciences
Lecture example 5 continued
Outstanding Balance = PV(remaining payments)
8 12 96
0 007083
12
0 085
$681 91
  
 

n
i
R
.
.
.
So there will be $47,381.44 still owing at the end of two years.
26ି௡ ିଽ଺26
27Amortisation schedule
 Table detailing the effect of each periodic payment
on the loan balance.
Calculations:
 Interest Amount = Interest rate per period x Opening Balance Principal Amount = Payment Amount – Interest Amount
 Closing Balance = Opening Balance – Principal Amount
 New Opening Balance = Previous Closing Balance
School of Info. Tech & Mathematical Sciences 2727
28Amortisation schedule:
Our own ‘loan calculator’
Total = $55,000
28
29Our own ‘loan calculator’
School of Info. Tech & Mathematical Sciences
Outstanding balance decreases slowly over time.
29$-
$10,000.00
$20,000.00
$30,000.00
$40,000.00
$50,000.00
$60,000.00
0 10 20 30 40 50 60 70 80 90 100 110 120 Dollar amount
Month
Outstanding Balance
29
30Our own ‘loan calculator’
School of Info. Tech & Mathematical Sciences
Initially, repayments go almost entirely towards interest.
3030
31Case study 2
How much is that TV?
 No deposit
 12 months interest free
 Monthly repayments
 $35 establishment fee
 $2.95 monthly account keeping feeSchool of Info. Tech & Mathematical Sciences
 What cash amount should the store be willing to accept
instead of the interest-free plan on this TV?
 The store can use surplus cash to pay down the balance
on its operating loan on which interest accrues at 11.5%
per year compounded monthly.
3131
32Case study 2
How much is that TV?
 Monthly repayment = $4,999/12 = $416.58
School of Info. Tech & Mathematical Sciences
0 1 2 … 11 12 (months)
$35 $416.58 $416.58 … $416.58 $416.58$2.95 $2.95 … $2.95 $2.95Now
No-interest plan
This is an annuity.
How much is it worth?
3232
33Case study 2
How much is that TV?
 The cash amount the store should be willing to accept is
the present value of all the no-interest plan cash flows:
School of Info. Tech & Mathematical Sciences
12 1 12
0 009583
12
0 115
416 58 2 95 $419 53
  
 
  
n
.
.
i
R . . .
PV = Establishment fee + A = $4,734.31 + $35
= $4,769.31 Equivalent cash price
33ି௡ ିଵଶ
33
34School of Info. Tech & Mathematical Sciences 3434
35Net Present Value (NPV)
NPV Investment Criterion:
 Accept the investment if NPV >0
 Reject the investment if NPV < 0
School of Info. Tech & Mathematical Sciences
NPV =  Present value of
cash inflows
Present value of
cash outflows
3535
36Lecture example 6
 A firm is contemplating the purchase of a
$10,000 machine that would reduce labour costs
by $3,000 each year for the next 4 years.
 The firm expects to sell the machine for $1,000
at the end of four years.
 Should the machine be purchased if the firm’s
cost of capital is 15% compounded annually?
School of Info. Tech & Mathematical Sciences 3636
37Lecture example 6 solution
School of Info. Tech & Mathematical Sciences
0 1 2 3 4
($10,000)
$1,000
$3,000 $3,000 $3,000 $3,000
37$1000
4
0 15
$4 000
S ,
n
i .
R ,




M
$3,000
 
 n
n
i
S
i
i NPV R

 




  
   

1
1 1
Initial investment ( ) MNPV  10,000 3, 000 
1  1.15 4
0.15






1, 000
  1.15 4
 10, 000 8,564.9350  571.7532
 $863.31
( )
37
38Lecture example 6 solution
 Decision:
 Since the NPV < 0, the machine should not be
purchased.
 Interpretation:
 Investing in the machine will cost the firm $863.31 in
the long run, and so the purchase will not pay for itself.
School of Info. Tech & Mathematical Sciences 3838
39NPV function in Excel – It’s wrong!!!
School of Info. Tech & Mathematical Sciences 3939
40Case study 3
Should MotionMedia invest in an App? MotionMedia want to invest in developing a
new iPhone app that streams live TV and is
designed to ‘Take TV With You’.
 The initial outlay for design and filming is $80,000
now and a second payment of $50,000, in one year’s time.  Annual operating profits of $28,000 per year will be generated for
the next 10 years, through sponsorship.
 Technology-related updates costing $12,000 will be required after
5 years. In 10 years time, the product resale value would be $70,000. School of Info. Tech & Mathematical Sciences 40Should MotionMedia invest in developing the iPhone app if its
cost of capital is 15% compounded annually?
40
41Case study 3
Should MotionMedia invest in an App?School of Info. Tech & Mathematical Sciences
0 1 2 3 4 5 6 7 8 9 10 (yrs)
(80) (50)
28 28 28 28 28 28 28 28 28 28 ($’000)
70 ($’000)
(12) ($’000)
 
   
 
$28,384.07
80,000 43,478.2609 5,966.1208 140,525.5000 17,302.9294
1.15
70,000
0.15
1 1.15 28,000
1.15
12,000
1.15
50,000 80,000 10
10
5

     
 




 
     

NPV
4141
42 Decision:
 Since the NPV > 0, MotionMedia should invest in the
iPhone app.
 Interpretation:
 Additional profits from producing the iPhone app will
be more than enough to repay the initial outlay and
subsequent technology update.
 Given MotionMedia’s cost of capital, the investment
produces an operating surplus of $28,384.
School of Info. Tech & Mathematical Sciences 42Case study 3
Should MotionMedia invest in an App?42
43School of Info. Tech & Mathematical Sciences
Next week
 Algebra in business
 Making good business decisions (linear equations)
 Break-even analysis
4343

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