Time value of money

Contents
• Interest: Cost of Money
• Simple Interest & Compound Interest
• Discounting & Compounding
• Comparison of Different Money series by
Economic Equivalence

1. Interest: Cost of Money
The Interest Rate reflects the market rate, which takes into account the
earning power as well as the effect of inflation perceived in the market place.
The rate at which money earn interest should be higher than the inflation
rate to make any economic sense of the delayed purchase
The economic value of money depends on when it is received; because
money has both earning and purchasing power

Elements of Transactions
Initial
Amount of
Money
(Principal)
Time
Period
(Interest
Period)
Cost of
Money
(Interest
Rate)
Number
of Interest
Periods
Plan (Cash
Flow
Pattern)
Future
Amount of
Money

Elements of Transactions
An = A discrete amount occurring at the end of some interest period
i = The interest rate for a period
N = Total number of interest periods
P = Initial amount at time zero, referred as Present Value (Present Worth)
F = Future some of money at the end of period
A = An end-of-period payment in a uniform series that continues for N
periods; where A1 = A2 = ----- = AN
Vn = An equivalent amount of money at the end of a specific period ‘n’ that considers
the effect of time value of money; that is, V0 = P and VN = F

Cash Flow Diagram
• Cash Flow Diagrams give a convenient summary of all
the important elements of a problem in a graphical
form to determine whether the statement of the
problem has been converted into its appropriate
parameters.
• Cash flow diagram represents time by a horizontal line
marked off with the number of interest periods.
• Upward arrows denote positive flows (receipts) and
downward arrows negative flows (payments). Arrows
represents net cash flows.
• End of period convention is the practice of placing all
cash flow transactions at the end of an interest period.

Cash Flow Diagram
Period-1 Period-2 Period-3 Period-4 Period-5
Outflows
Inflow
Period-0

2. Methods of Calculating Interest
1. Simple Interest
Total amount available in N periods
F = P+I = P(1+iN)
For a deposit of P at a simple interest rate i for N periods;
Total Interest
I = (iP)N
Interest earned only on the principal amount during each
interest period
2. Compound Interest
Total amount at the end of N periods
For a deposit of P at interest rate i, total amount at end of
period-1
P+iP = P(1+i)
Interest earned in each period is calculated based on total
amount at end of previous period.
Total amount = principal amount + accumulated interest

Illustration Question
If you deposit SR 1000 in a bank savings account
that pays interest rate at a rate of 10%
compounded annually. Assume that you don’t
withdraw the interest earned at the end of
each period (one year), but let it accumulate.
How much money would you have at the end
of year 3? Compare your answer with the
simple interest.

Answer
P = 1000 N = 3 i = 10% = 0.10
Simple Interest
F = P(1+iN) = 1000 [1+(0.10)3]
= 1000 x 1.3= 1300
Compound Interest
 N
iPF )1(
3
)10.01(1000  3
)1.1(1000 )331.1(1000 1331
Comparison of Interest = 1331-1300 = SR31 more under Compound Interest

3. Economic Equivalence
 N
iPF )1(
5
)12.01(1000  34.1762SR
At 12% interest, SR1000 received now is equivalent to SR1762.34 received in 5 years

Principles of Equivalence Calculation
1. Equivalence calculations made to compare
alternatives require a Common Time basis
(Present worth of future worth)
2. Equivalence depends on Interest Rate.
3. Equivalence calculations may require the
conversion of multiple payment cash flows to
a Single Cash flow
4. Equivalence is maintained regardless of
point of view (borrower or lender)

Five Types of Cash Flows
Years 0 1 2 3 4 5
Years 0 1 2 3 4 5
Years 0 1 2 3 4 5
Years 0 1 2 3 4 5
Years 0 1 2 3 4 5
Single Cash
Flow Series
Equal (Uniform)
Payment Series
Linear-gradient
Series
Geometric-
gradient Series
Irregular
Payment Series

Types of Cash Flows
Cash flow Series Payment Characteristics
1. Single Cash Flow Series Deal with only Single amount: Present
amount P & its future worth F
2. Equal (Uniform) Payment Series Equal cash flows at regular intervals
3. Linear-gradient Series Cash flows increases (decreases) at
uniform fixed amount
4. Geometric-gradient Series Cash flow increases (decreases) at
uniform fixed rate expressed as
percentages
5. Irregular Payment Series Cash flows are irregular

4.Equivalence in Single Cash Flow
0 N
0 N
P
P
F
F
Compounding Process
Discounting Process

4.1 Compounding Process
The process of finding F is called Compounding Process
N
i)1(  is called Compounding Factor (Table Values)Where
Functional Notation for Single payment Compound-amount Factor
),,/( NiPF which is read as ‘Find F, given P, i, and N’

4.2 Discounting Process
Finding the Present Worth of a future sum is
simply the reverse of compounding and is known
as the Discounting Process
N
N
iFP
i
FP










)1(
)1(
1
N
iWhere 
 )1(, is known as Present worth factor
Or Discounting factor (Table Value)
Functional Notation for Single payment Present Worth Factor (Discounting Factor)
),,/( NiFP which is read as ‘Find P, given F, i, and N’

Illustration Question: 1
If you had SR 2000 now and invest it at 10%,
how much would it be worth in eight years?

Answer to Question: 1
P = SR 2000
i = 10% = 0.10
N = 8 years
18.4287
)10.01(2000
)1(
8
SRF
F
iPF N




Suppose that SR 1000 is to be received in five
years. At an annual interest rate of 12%, what
is the Present Worth of this amount?

F = SR 1000
i = 12% = 0.12
N = 5 years
43.567
)56743.0(1000)12.01(1000
)1(
5
SRP
P
iFP N






• You have just purchased 100 shares of SABIC
at SR 60 per share. You will sell the stock when
its market price has doubled. If you expect the
stock price to increase 20% per year, how long
do you anticipate waiting before selling the
stock?

Share price = SR60
Number of shares = 100
P = 60 x 100 = SR 6000
F = SR 12000 (Doubled)
i = 20% = 0.20
yearsN
N
N
iPF
N
N
N
N
480.3
20.1log
2log
20.1log2log
20.12
20.1
6000
12000
)20.01(600012000
)1(








5. Equivalence in Uneven payment
(Irregular) Series
We can find the Present Worth of any uneven
stream of payments by calculating the present
value of each individual payments and summing
the results
Once the Present Worth if found, we can
calculate other Equivalence Calculations

If your business wishes to set aside money now to invest over the next
4 years to use to automate its customer service department. The
business can earn 10% on a lump sum deposited now, and it wishes
to withdraw the money in the following increments.
Year-1: SR 25,000 to purchase a computer and database software
designed for customer service use.
Year-2: SR 3,000 to purchase additional hardware to accommodate
anticipated growth in use of the system.
Year-3: No expenses
Year-4: SR 5,000 to purchase software upgrades
How much money must be deposited now to cover the anticipated
payments over the next 4 years? (Assume each transaction occur at
the end of each year)

(Cash Flow Diagram)
0 1 2 3 4
SR 25,000
SR 3,000
SR 5,000
P
P1
P2
P4
P = P1 + P2 + P3 + P4

Answer (Continues)
F1 = SR 25,000
F2 = SR 3,000
F3 = SR 5,000
P = P1 + P2 + P3 + P4
622,28415,30479,2727,224321
415,3)6830.0(5000)10.1(5000)10.01(50004
03
479,2)8264.0(3000)10.1(3000)10.01(30002
727,22)9091.0(25000)10.1(25000)10.01(250001
)1(
44
22
11
SRPPPPP
SRP
P
SRP
SRP
iFP N











Suppose that you have a savings account in a Bank. By
looking at the history of the account, you learned the
interest rate in each period during the last five years as
shown in the cash flow diagram below. Show how the
bank calculated your balance in the fifth year?
0 1 2 3 4 5
SR 500
SR 300
SR 400
F = ?
5% 6% 6% 4% 4%

P1 = SR 300; i = 5% (N=1)+ 6% (N=2)+4% (N= 2)
P2 = SR 500; i = 6%(N=1) + 4% (N=2)
P3 = SR 400; i = 4% (N=1)
82.382)0816.1(934.353)04.1(934.353)04.01(934.353
)934.353&2%(4
934.353)1236.1(315)06.1(315)06.01(315
)315&2%(6
315)05.01(300
)1%(5
300
)1(
22
22
1








PNiWhen
PNiWhen
NiWhen
SRofonContributi
iPF N

Answer Continues
25.573)0816.1(530)04.1(530)04.01(530
)530&2%(4,
530)06.1(500)06.01(500
)500&1%(6
500
)1(
22
1
SR
PNiWhen
PNiWhen
SRofonContributi
iPF N






416)04.1(400)04.01(400
)400&1%(4,
400
)1(
11
SR
PNiWhen
SRofonContributi
iPF N




Total Balance after 5 years = 382.82+573.25+416 =SR1,372.06

6. Equivalence in Equal Payment Series
If an amount A is invested at the end of each period for N periods, the
total amount F that can be withdrawn at the end of the N periods
will be the sum of the compound amounts of the individual
deposits. This means there exists a series of the following form.
:&
)4()1()1(
:lim32
)3()1(..............)1()1()1(
:)1(1
)2()1(.............)1()1((
exp1
)1()1(...........)1()1(
2
12
21
givewillAFforSolving
iAAFiF
gettotermscommoninateetoEquationfromEquationgSubtractin
iAiAiAFi
ibyEquationgMultiplyin
iAiAiAAF
followsaselyalternativressedisEquation
AiAiAiAF
N
N
N
NN











Compound-amount Factor & Sinking-
fund Factor in Equal Payment Series
)5(
1)1(





 

i
i
AF
N
The term in the bracket is known as Compound-amount Factor in Equal Payment
Series
)6(
1)1(







 N
i
i
FA
The term in the bracket is known as Sinking –fund Factor in Equal Payment Series

Capital-recovery Factor (Annuity
factor) & Present-worth factor
The term in the bracket is knows as Capital-recovery Factor & A is Annuity Factor
)8(
1)1(
)1(
)7(
1)1(
)1(
:,)1(6Re


















N
N
N
N
N
i
ii
PA
OR
i
i
iPA
getweiPbyEquationinFforplacing
)9(
)1(
1)1(
:8










N
N
ii
i
AP
givesPforsolvingEquationIn
The term in the bracket is knows as Present-worth Factor

Suppose you make an annual contribution of SR 3000 to
your savings account at the end of each year for 10
years. If the account earns 7% interest annually, how
much can be withdrawn at the end of 10 years?
0 1 2 3 4 5 6 7 8 9 10
SR 3000
i=7%
Cash Flow Diagram
F=?

A = SR 3000
N = 10 years
i = 7% = 0.07
20.449,41)8164.13(3000
07.0
1)07.1(
3000
07.0
1)07.01(
3000
1)1(
10
10
SRF
F
i
i
AF
N






 






 






 


A Biotechnology Company has borrowed SR
250,000 to purchase laboratory equipments. The
loan carries an interest rate of 8% per year and is
to be repaid in equal installments over the next 6
years. Compute the amount of annual installment
0 1 2 3 4 5 6
SR 250,000
i= 8%
A= ?

P = SR 250,000
i = 8% = 0.08
N = 6 years
 
54075
2163.0000,250
1)08.1(
)08.1(08.0
000,250
1)08.01(
)08.01(08.0
000,250
1)1(
)1(
6
6
6
6
SRA
i
ii
PA N
N




























Suppose you are joining in an insurance scheme, where you
have to deposit an amount now, so that you will receive an
annual amount of SR 10,576,923 for next 26 years at an
interest rate of 5%. Estimate the initial deposit amount?
0 1 2 ----------------- 25 26
A = SR 10,576,923
i= 5%
P= ?

i = 5% = 0.05
A = SR 10,576,923
N = 26 years
383,045,152
)3752.14(923,576,10
)05.1(05.0
1)05.1(
923,576,10
)05.01(05.0
1)05.01(
923,576,10
)1(
1)1(
26
26
26
26
SRP
P
P
ii
i
AP N
N







 



















7. Linear-Gradient Series
A linear gradient includes an initial amount A during period-1, which
increases/ decreases by G during the interest periods. Each Cash
Flow consists of a Uniform series of N payments of amount A1 and
Gradient increment/ decrement of G amount
0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5
0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5
+
--
=
=
Increasing Gradient
Decreasing Gradient
A1
A1
G
2G 3G
4G
G 2G
3G 4G
A1
A1 A1 -- G
A1 + G
A1 + 4G
A1 -- 4G

Present worth Factor (Discounting)
Derivation
To find P, the single payment Present worth Factor is applied to each term
:,&Re
)3(
)1(
)1(1
2
)1(
)1(1
)1(20
})1(,.......,2,,0{
)2()1(20[
)1(20
)1/(1
)1()1()1(
)1(
)1(
)1(
2
)1(
0
2
1
2
2
1
12
12
12
32
1
32
obtainwexAforvaluesoriginalplacing
x
xNNx
axP
asrewrittenisEquation
x
xNNx
xxNXx
sumfinitethehasxNxxseriesgeometricarithmetictheSince
xNxxax
axNaxaxP
yieldsxiandaGLetting
iGnP
OR
i
GN
i
G
i
G
P
NN
NN
N
N
N
N
N
n
n
N









































Present Worth Factor
The factor in the bracket is called as Present
Worth factor in the gradient series.








 N
N
ii
iNi
GP
)1(
1)1(
2

Annuity Factor (A)
The figure in the bracket is known as Gradient
uniform series factor












]1)1[(
1)1(
:
N
N
ii
iNi
GA
obtaintoseriespayment
equalinfactorAnnuity
invaluesPngsubstitutiBy

A textile mill has just purchased a lift truck that has
a useful life of 5 years. The engineer estimates
that maintenance cost of for the truck during the
first year will be SR 1000. As the truck ages,
maintenance costs are expected to increase at a
rate of SR 250 per year over the remaining life.
Assuming that the maintenance costs occur at
the end of each year. The firm wants to set up a
maintenance account that earns 12% annual
interest. All future maintenance expenses will be
paid out of this account. How much does the firm
have to deposit in the account now?

Answer
0 1 2 3 4 5
1500 1750 2000
1250
1000
P=?
i=12%
1000 1000 1000 1000
250 500
750
1000

Answer Continues
A1 = 1000
G = 250
i = 12% = 0.12
N = 5
P = P1 (Equal payment series) + P2 (Gradient Series)
8.36041
)6048.3(10001
)12.1(12.0
1)12.1(
1000
)12.01(12.0
1)12.01(
10001
)1(
1)1(
1
5
5
5
5
SRP
P
P
ii
i
AP N
N





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
 


















520425.15998.360421
25.1599)397.6(2502
)12.01(12.0
15*12.0)12.01(
250
)1(
1)1(
2 52
5
2



















PPP
P
ii
iNi
GP N
N

Mr. X and Mr. Y have just opened two savings accounts in a Bank. The
accounts earn 10% annual interest. Mr. X wants to deposit SR 1000 in his
account at the end of the first year and increase this amount by SR 300 for
each of the next 5 years. Mr. Y wants to deposit an equal amount each
year for the next 6 years. What would be the size of Mr. Y’s annual deposit
so that the two accounts will have equal balances at the end of 6 years.
0 1 2 3 4 5 6
1000 1000 1000 1000 1000 1000
300 600 900 1200 1500
1300
1600
1900
2200
2500
Mr. X’s
Deposit
Plan
1000

Answer
A1 = 1000
G = 300
i = 10% = 0.10
N = 6
A = A1 (Equal payment ) + A2 (Gradient payment)
A1 = 1000
08.166708.667100021
08.667)2236.2(3002
]1)10.1[(10.0
16*10.0)10.1(
300
]1)10.01[(10.0
16*10.0)10.01(
3002
]1)1[(
1)1(
6
6
6
6




























AAA
A
A
ii
iNi
GA N
N

8. Geometric Gradient Series
If ‘g’ is the percentage change in cash flow, the magnitude of nth
payment ‘An’ is related to the first payment A1 by the following formula
Cash flows that increase or decrease over time by a constant
percentage (Compound Growth)
decreasewillseriesthegincreasewillseriesthegIf
NngAA n
n

 
,0&,0
)1(.,,.........3,2,1,)1( 1
1


A1
P
P
A1 (1+g)
A1 A1 (1+g)
Geometric Gradient Series
0 1 2 3 4 5
0 1 2 3 4 5
2
1 )1( gA 
1
1 )1( 
 N
gAincreasewillseriesthegWhen  ,0
decreasewillseriesthegIf  ,0
2
1 )1( gA  1
1 )1( 
 N
gA

Estimation of P
:
)6()1(
1
)(
)()1(
)(
:45
)5()(
:4
)4()(
:3.
1
1
1
)3(
1
1
)1(
:)1(
)2()1()1(
1
1
1
1432
32
1
1
1
1
1
1
1
1
xandaforvaluesoriginalreplacingBy
x
x
xxa
P
xxaxP
xxaxPP
equationfromequationgSubstratin
xxxxaxP
gettoxbyequationMultiply
xxxxaP
asequationrewriteThen
i
g
xand
g
A
aLet
i
g
g
A
P
yieldssummationtheoutsidegABringing
igAP
N
N
N
N
N
nN
n
N
n
nn








































Formulae of P




















N
i
g
gi
A
P
giIf
1
1
11
i
NA
P
giIf



1
1
The factor in bracket is known as Geometric-gradient series present-worth factor

Illustration Question
Ansell, Inc., a medical device manufacturer uses compressed air in solenoids
and pressure switches in its machines to control various mechanical
movements. Over the years, the manufacturing floor has changed layouts
numerous times. With each new layout, more piping was added to the
compressed-air delivery system to accommodate new locations of
manufacturing machines. None of the extra, unused old pipe was capped
or removed; thus the current compressed-air delivery system is inefficient
and fraught with leaks. Because of the leaks, the compressor is expected
to run 70% of the time that the plant will be in operation during the
upcoming year. This will require 260kWh of electricity at a rate of SR
0.05/kWh for 6000 hours. The Plant runs 250 days a year, 24 hours per
day. If Ansell continues to operate the current air delivery system, the
compressor run time will increase by 7% per year for the next 4 years
because of ever-worsening leaks (After 5 years, the current system will not
be able to meet the plant’s compressed air requirements, so it will have to
be replaced). If Ansell decides to replace all of the old piping now, it will
cost SR 28,570. The compressor will still run the same number of days;
however, it will run 23% (or will have 70% (1-0.23) = 53.9% usage during
the day) less because of the reduced air pressure loss. If Ansell’s interest
rate is 12%, is the machine worth fixing now? (Calculate A1 and P)

Answer
Current Power Consumption (g) = 7%
i = 12% = 0.12
N = 5
Step: 1
Calculate the cost of power consumption of
current piping for 1st year
Power Cost = (% of operating days) (Days operating per year) (Hours
per day) (kWh) ($/KWh)
= (70%) (250) (24) (260) (0.05) = SR 54,600

54,600
Step: 2
Each year, the annual power cost increase @ 7%. The anticipated
power cost over 5 years is as shown in the diagram below. The
equivalent present lump-sum cost @ 12% for this geometric
gradient series is:
0 1 2 3 4 5
58,422
62,512
66,887
71,569
g=7%
222937
12.1
07.1
1
05.0
54600
)(
12.01
07.01
1
07.012.0
54600
1
1
1)(
5
5
1
SRoldP
i
g
gi
A
oldP
N























































Step: 3
If Ansell replaces the current compressed air system with the
new one; the annual power cost will be 23% less during the
1st year and will remain at that level over the next 5 years.
The equivalent Present lump-sum cost at 12% is:
A = 54600 (1- 23%) = 54600 (1 - 0.23) = 54600 (0.77)
= SR42042
552,151
)6048.3(42042
)12.01(12.0
1)12.01(
42042
)1(
1)1(
5
5
SR
ii
i
AP
SeriesPaymentEqualIn
N
N





















Step: 4
The net cost for not replacing the old system =
= 222,937 – 151,552 = SR 71,385
Since the new system costs only SR28,570, the
replacement should be made now

Time value of money

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