Laurence Booth
Sean Cleary
5
Time Value of Money
LEARNING OBJECTIVES
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
Explain the importance of the time value of money and how it is
related to an investor’s opportunity costs.
Define simple interest and explain how it works.
Define compound interest and explain how it works.
Differentiate between an ordinary annuity and an annuity due, and
explain how special constant payment problems can be valued as
annuities and, in special cases, as perpetuities.
Differentiate between quoted rates and effective rates, and explain
how quoted rates can be converted to effective rates.
Apply annuity formulas to value loans and mortgages and set up an
amortization schedule.
Solve a basic retirement problem.
Estimate the present value of growing perpetuities and annuities.
5.1 OPPORTUNITY COST
• Money is a medium of exchange.
• Money has a time value because it can be invested
today and be worth more tomorrow.
• The opportunity cost of money is the interest rate
that would be earned by investing it.
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 3
5.1 OPPORTUNITY COST
• Required rate of return (k) is also known as a
discount rate.
• To make time value of money decisions, you will
need to identify the relevant discount rate you
should use.
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 4
5.2 SIMPLE INTEREST
• Simple interest is interest paid or received only on
the initial investment (principal).
• The same amount of interest is earned in each year.
• Equation 5-1:
Value (time n)  P  (n  P  k )
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 5
5.2 SIMPLE INTEREST
EXAMPLE: Simple Interest
The same amount of interest is earned in each year.
Example
Simple Interest
You invest $500 today for five years and receive 10 percent annual simple interest.
Annual interest = $500 × 0.1 = $50 per year
Year
Beginning Amount
Ending Amount
1
$500
$550
2
$550
$600
3
$600
$650
4
$650
$700
5
$700
$750
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 6
5.3 COMPOUND INTEREST
• Compound interest is interest that is earned
on the principal amount and on the future
interest payments.
• The future value of a single cash flow at any
time ‘n’ is calculated using Equation 5.2.
FVn  PV0 (1  k ) n
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 7
USING EQUATION 5.2
• Given three known values, you can solve for the
one unknown in Equation 5.2
FVn  PV0 (1  k ) n
•
•
•
•
•
[5.2]
Solve for:
FV given PV, k, n (finding a future value)
PV given FV, k, n (finding a present value)
k given PV, FV, n (finding a compound rate)
n given PV, FV, k (find holding periods)
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 8
COMPOUND VERSUS SIMPLE INTEREST
• Simple interest grows principal in a linear manner.
• Compound interest grows exponentially over time.
$120,000
Compound interest
Future Value
$100,000
Simple interest
$80,000
$60,000
$40,000
$20,000
$0
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50
Year
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 9
5.3 COMPOUND INTEREST
EXAMPLE: Compounding (Computing Future Values)
FVn  PV0 (1  k ) n
Example
[5.2]
Compound Interest
You invest $500 today for five years and receive 10 percent annual compound interest.
Year
1
2
3
4
5
Booth • Cleary – 3rd Edition
Beginning Amount
$500
$550
$605
$666
$732
Interest
$500 × 0.1 = $50
$550 × 0.1 = $55
$605 × 0.1 = $60.50
$666 × 0.1 = $66.66
$732 × 0.1 = $73.20
© John Wiley & Sons Canada, Ltd.
Ending Amount
$550
$605
$666
$732
$805
Page 10
5.3 COMPOUND INTEREST
Compounding (Computing Future Values)
• Compound value interest factor (CVIF) represents the
future value of an investment at a given rate of interest
and for a stated number of periods.
CVIFn?,k ?  (1  k ) n
• The CVIF for 10 years at 8% would be:
CVIFn10,k 0.08  (1  0.08)10  2.1589
• $100 invested for 10 years at 8% would equal:
FV10  $100  (1  0.08)10  $100  2.1589  $215.89
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 11
5.3 COMPOUND INTEREST
EXAMPLE: Using the CVIF
Find the FV20 of $3,500 invested at 3.25%.
FV20  P0  CVIFn  20,k 3.5% 
 $3,500  (1  0.035) 20
 $3,500 1.99
 $6,964.26
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 12
5.3 COMPOUND INTEREST
Discounting (Computing Present Values)
• The inverse of compounding is known as
discounting.
• You can find the present value of any future single
cash flow using Equation 5.3.
PV0 
Booth • Cleary – 3rd Edition
FV0
(1  k ) n
[5.3]
© John Wiley & Sons Canada, Ltd.
Page 13
5.3 COMPOUND INTEREST
Discounting (Computing Present Values)
Present value interest factor (PVIF) is the inverse of the
CVIF.
PVIFn ?,k ?
Booth • Cleary – 3rd Edition
1

(1  k ) n
© John Wiley & Sons Canada, Ltd.
Page 14
5.3 COMPOUND INTEREST
Discounting (Computing Present Values)
EXAMPLE: Using the PVIF
Find the PV0 of receiving $100,000 in 10 years time if
the opportunity cost is 5%.
PV0  FV10  PVIFn 10,k 5% 
1
 $100,000 
(1  0.05)10
1
 $100,000 
1.629
 $100,000  0.6139
 $61,391.33
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 15
5.3 COMPOUND INTEREST
Solving for Time or “Holding Periods”
Equation 5.3 is reorganized to solve for n:
PV0 
n
Booth • Cleary – 3rd Edition
FV0
(1  k ) n
[5.3]
ln FVn / PV0 
ln 1  k 
© John Wiley & Sons Canada, Ltd.
Page 16
5.3 COMPOUND INTEREST
EXAMPLE: Solving for ‘n’
How many years will it take $8,500 to grow to $10,000
at a 7% rate of interest?
ln FVn / PV0 
n
ln 1  k 
ln $10,000 / $8,500 ln[1.17647 ]
n

ln 1  .07
ln[1.07]
0.1625
n
 2.4 years
0.06766
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 17
5.3 COMPOUND INTEREST
Solving for Compound Rate of Return
Equation 5.3 is reorganized to solve for k:
PV0 
FV0
(1  k ) n
 FV 
k  n
 PV0 
Booth • Cleary – 3rd Edition
[5.3]
1/ n
1
© John Wiley & Sons Canada, Ltd.
Page 18
5.3 COMPOUND INTEREST
EXAMPLE: Solving for ‘k’
Your investment of $10,000 grew to $12,500 after 12
years. What compound rate of return (k) did you earn
on your money?
 FVn 
k

PV
 0
1/ n
1
1
12
 $12,500 
0.083
k

1

1
.
25
1

 $10,000 
k  0.01877  1.88%
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 19
5.4 ANNUITIES AND PERPETUITIES
• An annuity is a finite series of equal and periodic
cash flows.
• A perpetuity is an infinite series of equal and
periodic cash flows.
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 20
5.4 ANNUITIES AND PERPETUITIES
• An ordinary annuity offers payments at the end of
each period.
• An annuity due offers payments at the beginning of
each period.
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 21
5.4 ANNUITIES AND PERPETUITIES
The formula for the compound sum of an ordinary
annuity is:
 (1  k ) n  1
FVn  PMT 

k


Booth • Cleary – 3rd Edition
[5.4]
© John Wiley & Sons Canada, Ltd.
Page 22
5.4 ANNUITIES AND PERPETUITIES
EXAMPLE: Find the Future Value of an Ordinary Annuity
You plan to save $1,000 each year for 10 years. At 11%
how much will you have saved if you make your first
deposit one year from today?
FVA10  PMT  FVIFAn ,k 
 1  k n  1
FVA10  $1,000  

k


 1.1110  1
FVA10  $1,000  

0
.
11


FVA10  $1,000  16.722  $16,722.01
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 23
5.4 ANNUITIES AND PERPETUITIES
The formula for the compound sum of an annuity due
is:
 (1  k ) n  1
FVn  PMT 
(1  k)
k


Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
[5.6]
Page 24
5.4 ANNUITIES AND PERPETUITIES
EXAMPLE: Find the Future Value of an Annuity Due
You plan to save $1,000 each year for 10 years. At 11%
how much will you have saved if you make your first
deposit today?
FVA10  PMT  FVIFAn ,k  1  k 
 1  k n  1
FVA10  $1,000  
 (1  k )
k


 1.1110  1
FVA10  $1,000  
 (1.11)
 0.11 
FVA10  $1,000  16.722  1.11  $18,561.43
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 25
5.4 ANNUITIES AND PERPETUITIES
The formula for the present value of an annuity is:
1

1

 (1  k ) n
PV0  PMT 
k


Booth • Cleary – 3rd Edition





© John Wiley & Sons Canada, Ltd.
[5.5]
Page 26
5.4 ANNUITIES AND PERPETUITIES
EXAMPLE: Find the Present Value of an Ordinary Annuity
What is the present value of an investment that offers
to pay you $12,000 each year for 20 years if the
payments start one year from day? Your opportunity
cost is 6%.
PVA0  PMT  PVIFAn  20,k 0.06
1 

1

 (1.06) 20 
PVA0  $12,000  

0
.
06




PVA0  $12,000 11.47  $137,639.06
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 27
5.4 ANNUITIES AND PERPETUITIES
The formula for the present value of an annuity is:
1

1

 (1  k ) n
PV0  PMT 
k


Booth • Cleary – 3rd Edition


 (1  k)


© John Wiley & Sons Canada, Ltd.
[5.7]
Page 28
5.4 ANNUITIES AND PERPETUITIES
EXAMPLE: Find the Present Value of an Annuity Due
What is the present value of an investment that offers
to pay you $12,000 each year for 20 years if the
payments start one today? Your opportunity cost is 6%.
PVA0  PMT  PVIFAn ,k  1  k 
1 

1

 (1.06) 20 
PVA0  $12,000  
 (1  .06)
 0.06 


PVA0  $12,000 11.47 1.06  $145,897.40
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 29
5.4 ANNUITIES AND PERPETUITIES
A perpetuity is an infinite series of equal and periodic
cash flows.
PV0 
Booth • Cleary – 3rd Edition
PMT
k
[5.8]
© John Wiley & Sons Canada, Ltd.
Page 30
5.4 ANNUITIES AND PERPETUITIES
EXAMPLE: Find the Present Value of a Perpetuity
What is the present value of a business that promises
to offer you an after-tax profit of $100,000 for the
foreseeable future if your opportunity cost is 10%?
PV0 
Booth • Cleary – 3rd Edition
P1 $100,000

 $1,000,000
k
0.1
© John Wiley & Sons Canada, Ltd.
Page 31
5.5 QUOTED VERSUS EFFECTIVE RATES
• A nominal rate of interest is a ‘stated rate’ or quoted
rate (QR).
• An effective annual rate (EAR) rate takes into
account the frequency of compounding (m).
m
 QR 
EAR  k  1 
 1
m 

Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
[5.9]
Page 32
5.5 QUOTED VERSUS EFFECTIVE RATES
EXAMPLE: Find an Effective Annual Rate
Your personal banker has offered you a mortgage rate
of 5.5 percent compounded semi-annually. What is the
effective annual rate charged (EAR)on this loan?
QR m
0.055 2
) - 1  (1 
) -1
m
2
EAR  1.02752 - 1  5.58%
EAR  (1 
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 33
5.5 QUOTED VERSUS EFFECTIVE RATES
EXAMPLE: Effective Annual Rates
EARs increase as the frequency of compounding
increase.
Example
Effective Annual Rates
QR = 8%
Frequency of
Compounding
Annual
Semi-annual
Quarterly
Monthly
Daily
Continuous
Booth • Cleary – 3rd Edition
Effective Annual
Rate
8.0%
8.16%
8.24322%
8.29995%
8.32776%
8.32781%
© John Wiley & Sons Canada, Ltd.
Page 34
5.6 LOAN OR MORTGAGE
ARRANGEMENTS
• A mortgage loan is a borrowing arrangement where
the principal amount of the loan borrowed is
typically repaid (amortized) over a given period of
time making equal and periodic payments.
• A blended payment is one where both interest and
principal are retired in each payment.
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 35
5.6 LOAN OR MORTGAGE
ARRANGEMENTS
EXAMPLE: Loan Amortization Table
Determine the annual blended payment on a five –year
$10,000 loan at 8% compounded semi-annually.
1

1

 (1  k ) n
PV0  PMT 
k







1


1

 (1  0.0816)5 
$10,000  PMT  

0
.
0816




$10,000
PMT 
 $2,515.14
3.9759
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
[5.5]
Page 36
5.6 LOAN OR MORTGAGE
ARRANGEMENTS
EXAMPLE: Loan Amortization Table
The loan is amortized over five years with annual
payments beginning at the end of year 1.
Example 5 -4
QR = 8%
EAR = 8.16%
Period
1
2
3
4
5
Booth • Cleary – 3rd Edition
Loan Amortization Table
Principal Borrowed = $10,000
Amortization Period = 5 years
(1)
(2)
(3)
Beginning
Principal
PMT
Interest
$10,000
$2,515
$816
$8,301
$2,515
$677
$6,463
$2,515
$527
$4,475
$2,515
$365
$2,325
$2,515
$190
© John Wiley & Sons Canada, Ltd.
(4)
Principal
Repayment
$1,699
$1,838
$1,988
$2,150
$2,325
(5)
Ending
Principal
$8,301
$6,463
$4,475
$2,325
$0
Page 37
5.6 LOAN OR MORTGAGE
ARRANGEMENTS
EXAMPLE: Mortgage
• Determine the monthly blended payment on a $200,000 mortgage
amortized over 25 years at a QR = 4.5% compounded semi-annually.
Number of monthly payments = 25 × 12 = 300
0.045 2
)  1  4.550625%
2
• Find EAR:
 (1 
• Find EMR:
4.550625%  (1  EMR)12  1
1.04550625 12  (1  EMR)
1
EMR  0.3715318%
• Determine monthly payment:
Booth • Cleary – 3rd Edition
PMT 
$200,000
 $1,106.85
1


1

 (1  0.003715) 300 


0
.
003715




© John Wiley & Sons Canada, Ltd.
Page 38
5.6 LOAN OR MORTGAGE
ARRANGEMENTS
EXAMPLE: Mortgage Amortization
The mortgage is amortized over 25 years with annual
payments beginning at the end of the first month.
Example 5 -5
Mortgage Amortization Table
QR = 4.5%
Principal Borrowed = $200,000
EAR = 4.55% Amortization Period = 25 years
EMR =0.372%
(1)
(2)
(3)
Beginning
Month
Principal
PMT
Interest
1
$200,000
$1,107
$743
2
$199,636
$1,107
$742
3
$199,271
$1,107
$740
4
$198,905
$1,107
$739
5
$198,537
$1,107
$738
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
(4)
Principal
Repayment
$364
$365
$366
$368
$369
(5)
Ending
Principal
$199,636
$199,271
$198,905
$198,537
$198,168
Page 39
5.7 COMPREHENSIVE EXAMPLES
• Time value of money (TMV) is a tool that can be
applied whenever you analyze a cash flow series over
time.
• Because of the long time horizon, TMV is ideally
suited to solve retirement problems.
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 40
COMPREHENSIVE EXAMPLE:
Retirement Problem
• Kelly, age 40 wants to retire at age 65 and currently has no
savings.
• At age 65 Kelly wants enough money to purchase a 30 year
annuity that will pay $5,000 per month.
• Monthly payments should start one month after she reaches
age 65.
• Today Kelly has accumulated retirement savings of $230,000.
• Assume a 4% annual rate of return on both the fixed term
annuity and on her savings.
• How much will she have to save each month starting one
month from now to age 65 in order for her to reach her
retirement goal?
*NOTE – these are ordinary annuities
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 41
COMPREHENSIVE EXAMPLE:
Retirement Problem
How much will the fixed term annuity cost at age 65?
Steps in Solving the Comprehensive Retirement Problem
1. Calculate the present value of the retirement annuity as
at Kelly’s age 65.
2. Estimate the value at age 65 of her current accumulated
savings.
3. Calculate gap between accumulated savings and
required funds at age 65.
4. Calculate the monthly payment required to fill the gap.
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 42
COMPREHENSIVE EXAMPLE:
Retirement Problem
Example Solution – Preliminary Calculations
Preliminary calculations Required
• Monthly rate of return when annual APR is 4%
4%  (1  k m )12  1
1  km  1  .04  (1.04).083  1.00326
1
12
km  0.326%
• Number of months during savings period
n  25 12  300
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
Page 43
COMPREHENSIVE EXAMPLE: Retirement Problem
Time Line & Analysis Required to Identify Savings Gap
GAP  $1,533,728  $613,142
 $920,586
30 year fixed-term retirement
annuity = 30 ×12 =360 months
$1,533,728
Additional
monthly
savings
(1  k ) n  1
FVA25  PMT
k
FV25  P0 (1  k annual ) 25  $230,000(1.04) 25
Existing
Savings
 $613,142
Age 40
65
25 year asset
accumulation phase
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
95
30 year asset
depletion phase
(retirement)
Page 44
COMPREHENSIVE EXAMPLE: Retirement Problem
Monthly Savings Required to fill Gap
GAP  $1,533,728  $613,142
 $920,586
Monthly
savings to fill
gap?
FVA25
$920,586

n
(1  k )  1 (1.00326)300  1
k
0.00326
$920,586

 $1,813.46
507.64
PMT 
Additional
monthly
savings
Your Answer
FV25  P0 (1  k annual ) 25  $230,000(1.04) 25
 $613,142
Existing
Savings
Age 40
65
25 year asset
accumulation phase
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
95
30 year asset
depletion phase
(retirement)
Page 45
Appendix 5A
GROWING ANNUITIES & PERPETUITIES
Growing Perpetuity
• A growing perpetuity is an infinite series of periodic
cash flows where each cash flow grows larger at a
constant rate.
• The present value of a growing perpetuity is
calculated using the following formula:
PV0 
Booth • Cleary – 3rd Edition
PMT0 (1  g ) PMT1

kg
kg
© John Wiley & Sons Canada, Ltd.
[5A-2]
Page 46
Appendix 5A
GROWING ANNUITIES & PERPETUITIES
Growing Annuity
• An annuity is a finite series of periodic cash flows
where each subsequent cash flow is greater than the
previous by a constant growth rate.
• The formula for a growing annuity is:
n
PMT1   1  g  
PV0 
 1  
 
k  g   1  k  
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
[5A-4]
Page 47
WEB LINKS
Wiley Weekly Finance Updates site (weekly news updates):
http://wileyfinanceupdates.ca/
Textbook Companion Website (resources for students and
instructors): www.wiley.com/go/boothcanada
Booth • Cleary – 3rd Edition
© John Wiley & Sons Canada, Ltd.
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