TIME VALUE OF MONEY TIME VALUE OF MONEY

&&

INTEREST FORMULASINTEREST FORMULAS7

Time Value of MoneyTime Value of Money

• Time Value of Money

• Money can “make” money if Invested

• Centers around an interest rate

The change in the amount of money over a given time period is called the time value of money

Interest RateInterest Rate

INTERESTINTEREST -- MANIFESTATION OF THE TIME MANIFESTATION OF THE TIME

VALUE OF MONEY. THE AMOUNT PAID TO VALUE OF MONEY. THE AMOUNT PAID TO

USE MONEY. IT REPRESENTS THE GROWTH USE MONEY. IT REPRESENTS THE GROWTH

OF CAPITAL PER UNIT PERIOD.OF CAPITAL PER UNIT PERIOD.

INVESTMENTINVESTMENT

INTEREST = VALUE NOW INTEREST = VALUE NOW -- ORIGINAL AMOUNTORIGINAL AMOUNT

LOANLOAN

INTEREST = TOTAL OWED NOW INTEREST = TOTAL OWED NOW -- ORIGINAL AMOUNT ORIGINAL AMOUNT

RENTAL FEE PAID FOR THE USE OF SOMEONE ELSES MONEY…EXPRESSED AS A %

Economic EquivalenceEconomic Equivalence

•Economic Equivalence

•Two sums of money at two different points in time can be made economically equivalent if:

•We consider an interest rate and,

•No. of Time periods between the two sums

Equality in terms of Economic Value

Equivalence IllustratedEquivalence Illustrated

T=0 t = 1 Yr

RM20,000 is

received here

RM21,800 paid back here

RM20,000 now is economically equivalent to RM21,800 one year from now IF the interest rate is set to equal 9%/year

Simple and Compound InterestSimple and Compound Interest

•Two “types” of interest calculations

•Simple Interest

•Compound Interest

•Compound Interest is more common worldwide and applies to most analysis situations

Simple Interest

• Calculated on the principal amount only

• In a multiperiod situation with simple interest:

•The accrued interest does not earn interest during the succeeding time period

•Normally, the total sum borrowed (lent) is paid back at the end of the agreed time period PLUS the accrued (owed but not paid) interest.

•Simple Interest is:

(principal)(interest rate)(time)(principal)(interest rate)(time)

RMI = (P)(i)(n)RMI = (P)(i)(n)

Simple and Compound InterestSimple and Compound Interest Compound InterestCompound Interest

• To COMPOUND – stop and compute the associated interest and add it to the unpaid balance.

•When interest is compounded, the interest that is accrued at the end of a given time period is added in to form a NEW principal balance.

•That new balance then earns or is charged interest in the succeeding time period

•Interest then “earns interest”

Compound InterestCompound Interest:: ExampleExample

• Assume:

•P = RM1,000

• i = 5% per year compounded annually (C.A.)

•N = 3 years

II22=RM52.50=RM52.50

II11=RM50.00=RM50.00

1 2 3

P=RM1,0P=RM1,0

0000

II33=RM55.13=RM55.13

Owe at t = 3 years:

RM1,000 + 50.00 + 52.50 + 55.13

= RM1157.63

Interest formulasInterest formulas

1. F/P and P/F Factors

2. P/A and A/P Factors

3. F/A and A/F Factors

4. P/G and A/G Factors

5. Effective Interest Rate

P=Present value, F=Future Value, A=Annual equivalent amount, P=Present value, F=Future Value, A=Annual equivalent amount,

i=Interest Rate, n=No. of interest periods,i=Interest Rate, n=No. of interest periods,

1.Single1.Single--payment Compound Amountpayment Compound Amount

F/P FactorF/P Factor To find F given PTo find F given P

P0

Fn

n

………….

Compound forward in time

Fn = P(F/P,i%,n) = P(1+i)n

ExampleExample-- F/P AnalysisF/P Analysis

Example: P= RM1,000;n=3;i=10%Example: P= RM1,000;n=3;i=10%

What is the future value, F? What is the future value, F?

0 1 2 3

P=RM1,00P=RM1,00

00

F = ??F = ??

i=10%/yeari=10%/year

F3 = RM1,000[F/P,10%,3] = RM1,000[1.10]3

= RM1,000[1.3310] = RM1,331.00

2.Single2.Single--Payment Present Worth AmountPayment Present Worth Amount

P/F FactorP/F Factor To find P given FTo find P given F

P

Fn

n

………….

P/F factor brings a single P/F factor brings a single

future sum back to a specific future sum back to a specific

point in time.point in time.

P = F(P/F,i%,n) = F(1+i)-n

ExampleExample –– P/F AnalysisP/F Analysis

Assume F = RM100,000, 9 years from now. What is Assume F = RM100,000, 9 years from now. What is

the present worth of this amount now if i =15%?the present worth of this amount now if i =15%?

0 1 2 3 8 9…………

F9 =

RM100,000

P= ??

i = 15%/yr

P0 = RM100,000(P/F, 15%,9) = RM100,000(1/(1.15)9)

= RM100,000(0.2843) = RM28,430 at time t = 0

3.Equal3.Equal--Payment Series Compound AmountPayment Series Compound Amount

F/A FactorF/A Factor To find F given ATo find F given A

0

…………..N

RM A per periodRM A per period

RMF

Find RM F given the Find RM F given the

RM A amountsRM A amounts

F = A (F = A (F/A, i, nF/A, i, n)= A )= A i

in 1)1(

ExampleExample

Formosa Plastics has major fabrication plants in Formosa Plastics has major fabrication plants in

Texas and Hong Kong. It is desired to know the Texas and Hong Kong. It is desired to know the

future worth of RM1,000,000 invested at the end of future worth of RM1,000,000 invested at the end of

each year for 8 years, starting one year from now. each year for 8 years, starting one year from now.

The interest rate is assumed to be 14% per year.The interest rate is assumed to be 14% per year.

• A = RM1,000,000/yr; n = 8 yrs, i = 14%/yr

Solution:Cash flows are indicated in RM1000 units. The F value in 8 years is

F = l000(F/A,14%,8) = 1000( 13.23218) = RM13,232.80 = 13.232 million 8 years from now.

4.Equal4.Equal--Payment Series Sinking FundPayment Series Sinking Fund

A/F FactorA/F Factor

RM A per periodRM A per period

0

…………..

N

RMF

Find RM A given the Find RM A given the

Future amt. Future amt. –– RM FRM F

(1 ) 1n

i

iFA = F (A = F (A/F, i, nA/F, i, n)=)=

ExampleExample

How much money must Carol deposit every How much money must Carol deposit every year startingyear starting,, l year from now at 5.5% per l year from now at 5.5% per year in order to accumulate RM6000 seven year in order to accumulate RM6000 seven years from now?years from now?

SolutionSolution

The cash How diagram from Carol's perspective fits the The cash How diagram from Carol's perspective fits the

A/F factor.A/F factor.

A= RM6000 (A= RM6000 (A/F,5.5%,7A/F,5.5%,7) = 6000() = 6000(0.120960.12096) = ) =

RM725.76RM725.76 per yearper year

The A/F factor Value 0f 0.12096 was computed using The A/F factor Value 0f 0.12096 was computed using

the A/F factor formulathe A/F factor formula

5.Equal5.Equal--Payment Series Present Worth AmountPayment Series Present Worth Amount

P/A FactorP/A Factor Desire an expression for the present worth Desire an expression for the present worth –– P of a P of a

stream of equal, end of period cash flows stream of equal, end of period cash flows –– AA

A = givenA = given

0 1 2 3 n-1 n

P = ??

n

n

i)i(1

1i)(1An)i,(P/A,AP

Find RM P given Find RM P given

the RM A amountthe RM A amount

6.Equal6.Equal--Payment Series Capital Recovery AmountPayment Series Capital Recovery Amount

A/P FactorA/P Factor

0 1 2 3 n-1 n

A = ??

P

1)1(

)1(),,/(

n

n

i

iiPniPAPA

Find RMA given Find RMA given

the the RMRM P amountP amount

7.Arithmetic Gradient7.Arithmetic Gradient

Increasing uniform seriesIncreasing uniform series

0 1 2 3 n-1 N

A1+G

A1+2G

A1+(n-2)G

A1+(n-1)G

This represents a positive, increasing arithmetic gradientThis represents a positive, increasing arithmetic gradient

7.Arithmetic Gradient7.Arithmetic Gradient

n

n

n

n

ii

iniG

ii

iA

niGPGniAPAP

)1(

1)1(

)1(

1)1(

),,/(),,/(

2

iii

iniGA

niGAGAA

n

n

)1(

1)1(

),,/(

1

1

• P/G Factor To find P given G

• A/G Factor To find A given G

Gradient ExampleGradient Example

• Consider the following cash flow

0 1 2 3 4 5

RM100

RM200

RM300

RM400

RM500

Find the present worth if i = 10%/yr; n = 5 yrs

Present Worth Point is here!And the G amt. = RM100/period

Gradient ExampleGradient Example

•PW(10%) of the base annuity = RM100(P/A,10%,5)

•PBase = RM100(3.7908)= RM379.08

•Not Finished: We need the PW of the gradient component and then add that value to the RM379.08 amount

•Calculating or looking up the P/G,10%,5 factor yields the following:

•Pt=0 = RM100(6.8618) = RM686.18 for the gradient PW

•Total PW(10%) = RM379.08 + RM686.18

•Equals RM1065.26

8.Nominal and Effective 8.Nominal and Effective

InterestInterestNominal Interest Rate, rNominal Interest Rate, r::

An interest rate that does not include any consideration of An interest rate that does not include any consideration of compoundingcompounding

Format: Format: ““rr% per time period, t% per time period, t””

Ex: 5% per yearEx: 5% per year””

1.5% per month for 12 months1.5% per month for 12 months

Same as: (1.5%)(12) = 18% per 12 monthsSame as: (1.5%)(12) = 18% per 12 months

Effective Interest Rates, iEffective Interest Rates, iaa::

An interest rate taking into account the effect of any An interest rate taking into account the effect of any

compounding during the yearcompounding during the year

Format: Format: ““rr% per time period, compounded % per time period, compounded ‘‘mm’’ times a times a

year.year.

‘‘mm’’ denotes or infers the number of times per year that denotes or infers the number of times per year that

interest is compounded. interest is compounded.

Ex: 18% per year, compounded monthlyEx: 18% per year, compounded monthly

EIR NotationEIR Notation

rr = the nominal interest rate per year.= the nominal interest rate per year.

mm = the number of compounding periods within the = the number of compounding periods within the

year.year.

ii = the effective interest rate per compounding = the effective interest rate per compounding

period (period (rr//mm))

iiaa = effective interest rate per year = effective interest rate per year

iiaa = (1 + r/m)= (1 + r/m)mm -- 1 = (1 + 1 = (1 + i)i)mm -- 1. 1.

•Remember: Always apply the Effective Interest Rate in solving problems.

ExampleExample

If a savings bank pays 1.5% interest per 3 months, If a savings bank pays 1.5% interest per 3 months,

what are the nominal and effective interest rates per what are the nominal and effective interest rates per

year?year?

SolutionSolution

Nominal interest rate r = 4 x 1.5% = 6%Nominal interest rate r = 4 x 1.5% = 6%

Effective interest rate per year iEffective interest rate per year iaa = (1+r/m)= (1+r/m)mm--11

== (1+.06/4)(1+.06/4)44--11

= 6.1%= 6.1%

ExampleExample

A person deposits RM100 into a bank that A person deposits RM100 into a bank that pays 5% interest, compounded semiannually. pays 5% interest, compounded semiannually. How much would be in the saving account at How much would be in the saving account at the end of one year?the end of one year?

SolutionSolution

5% interest, compounded semiannually, means that the 5% interest, compounded semiannually, means that the bank pays 2.5% every 6 months. Thus, the initial amount bank pays 2.5% every 6 months. Thus, the initial amount P=RM100 would be credited with .025(100)=RM2.50 P=RM100 would be credited with .025(100)=RM2.50 interest at the end of 6 months, or interest at the end of 6 months, or

PP P + PP + Pii = 100+100(.025) = RM102.50= 100+100(.025) = RM102.50

The RM102.50 is left in the saving account; at the end of The RM102.50 is left in the saving account; at the end of the 2the 2ndnd 66--month period, the interest earned is month period, the interest earned is .025(102.50)=RM2.56, for a total in the account at the end.025(102.50)=RM2.56, for a total in the account at the end

The RM102.50 is left in the saving account; at the end The RM102.50 is left in the saving account; at the end

of the 2nd 6of the 2nd 6--month period, the interest earned is month period, the interest earned is

.025(102.50)=RM2.56, for a total in the account at the .025(102.50)=RM2.56, for a total in the account at the

endend

(P + P(P + Pii)) (P + P(P + Pii)+ )+ ii (P + P(P + Pii) = P(1+) = P(1+ii)2)2

=100(1+.025)2=100(1+.025)2

=RM105.06=RM105.06

We saw that RM100 left in the saving account for one year increased to

RM105.06, so the interest paid was RM5.06. the effective interest rate per

year , ia , is RM5.06/RM100 = .0506 =5.06%

ExampleExample (cont’) ExampleExample

A person invests a sum of RM5000 in a bank at a A person invests a sum of RM5000 in a bank at a

nominal interest rate of 12% for 10 years. The nominal interest rate of 12% for 10 years. The

compounding is quarterly. Find the maturity compounding is quarterly. Find the maturity

amount of the deposit after 10 years.amount of the deposit after 10 years.

Solution:Solution:

M=4, r = 12%M=4, r = 12%

iiaa = (1+.12/4)= (1+.12/4)44 -- 1 = 12.55%1 = 12.55%

F = P(1+iF = P(1+iaa))nn = 5000(1+.1255)= 5000(1+.1255)1010==

RM16,308.91RM16,308.91

THE ENDTHE END