theory of interest, 2nd ed - kellison, stephen

THE

THEORY

OF

INTEREST

Second Edition

STEPHEN G. KELLISON

Study Notes Prepared by

Kevin Shand, FSA, FCIA

Assistant Professor

Warren Centre for Actuarial

Studies and Research

Contents

1 The Measurement of Interest 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 The Accumulation Function and the Amount Function . . . . . . . . . . . . . . . . 21.3 The Eective Rate of Interest: i . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.6 Present Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.7 The Eective Rate of Discount: d . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.8 Nominal Rate of Interest and Discount Convertible mthly: i(m), d(m) . . . . . . . . 71.9 Forces of Interest and Discount: in, dn . . . . . . . . . . . . . . . . . . . . . . . . . 91.10 Varying Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.11 Summary of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Solution of Problems in Interest 142.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Obtaining Numerical Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Determining Time Periods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 The Basic Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Equations of Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.6 Unknown Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.7 Unknown Rate of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.8 Practical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3 Basic Annuities 243.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.2 Annuity-Immediate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.3 AnnuityDue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4 Annuity Values On Any Date . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.5 Perpetuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.6 Nonstandard Terms and Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . 423.7 Unknown Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.8 Unknown Rate of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.9 Varying Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.10 Annuities Not Involving Compound Interest . . . . . . . . . . . . . . . . . . . . . . 50

4 More General Annuities 514.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.2 Annuities Payable At A Dierent Frequency Than Interest Is Convertible . . . . . 514.3 Further Analysis of Annuities Payable Less Frequency Than Interest Is Convertible 524.4 Further Analysis of Annuities Payable More Frequency Than Interest Is Convertible 624.5 Continuous Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.6 Basic Varying Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.7 More General Varying Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 934.8 Continuous Varying Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994.9 Summary Of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

1

5 Yield Rates 1015.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.2 Discounted Cash Flow Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.3 Uniqueness Of The Yield Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1035.4 Reinvestment Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1045.5 Interest Measurement Of A Fund . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1065.6 Time-Weighted Rates Of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1095.7 Portfolio Methods and Investment Year Methods . . . . . . . . . . . . . . . . . . . 1125.8 Capital Budgeting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1125.9 More General Borrowing/Lending Models . . . . . . . . . . . . . . . . . . . . . . . 112

6 Amortization Schedules and Sinking Funds 1136.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.2 Finding The Outstanding Loan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.3 Amortization Schedules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1146.4 Sinking Funds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1166.5 Diering Payment Periods and Interest Conversion Periods . . . . . . . . . . . . . 1196.6 Varying Series of Payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1256.7 Amortization With Continuous Payments . . . . . . . . . . . . . . . . . . . . . . . 1306.8 Step-Rate Amounts Of Principal . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

7 Bonds and Other Securities 1317.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1317.2 Types Of Securities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1317.3 Price of A Bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1327.4 Premium And Discount Pricing Of A Bond . . . . . . . . . . . . . . . . . . . . . . 1357.5 Valuation Between Coupon Payment Dates . . . . . . . . . . . . . . . . . . . . . . 1407.6 Determination Of Yield Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1417.7 Callable Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1447.8 Serial Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1457.9 Some Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1457.10 Other Securities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1457.11 Valuation Of Securities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

8 Practical Applications 1468.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.2 Truth In Lending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.3 Real Estate Mortgages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.4 Approximate Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.5 Depreciation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.6 Capitalized Cost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1518.7 Short Sales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1528.8 Modern Financial Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

9 More Advanced Financial Analysis 1569.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1569.2 An Economic Rationale for Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . 1569.3 Determinants of the Level of Interest Rates . . . . . . . . . . . . . . . . . . . . . . 1569.4 Recognition of Ination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1569.5 Reecting Risk and Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

2

9.6 Yield Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1579.7 Interest Rate Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1589.8 Duration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1589.9 Immunization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1619.10 Matching Assets and Liabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

3

1 The Measurement of Interest

1.1 Introduction

Interest

compensation a borrower of capital pays to a lender of capital

lender has to be compensated since they have temporarily lost use of their capital

interest and capital are almost always expressed in terms of money

1.2 The Accumulation Function and the Amount Function

The Financial Transaction

an amount of money or capital (Principal) is invested for a period of time

at the end of the investment period, a total amount (Accumulated Value) is returned

dierence between the Accumulated Value and the Principal is the Interest Earned

Accumulation Function: a(t)

let t be the number of investment years (t 0), where a(0) = 1 assume that a(t) is continuously increasing

a(t) denes the pattern of accumulation for an investment of amount 1

Amount Function: A(t) = k a(t) let k be the initial principal invested (k > 0) where A(0) = k

A(t) is continuously increasing

A(t) denes the Accumulated Value that amount k grows to in t years

Interest Earned during the nth period: In = A(n) A(n 1) interest earned is the dierence between the Accumulated Value at the end of a period and

the Accumulated Value at the beginning of the period

1.3 The Eective Rate of Interest: i

Denition

i is the amount of interest earned over a one-year period when 1 is invested

let in be the eective rate of interest earned during the nth period of the investment whereinterest is paid at the end of the period

i is also dened as the ratio of the amount of Interest Earned during the period to theAccumulated Value at the beginning of the period

in =A(n) A(n 1)

A(n 1) =In

A(n 1) , for integral n 1

2

1.4 Simple Interest

assume that interest accrues for t years and is then reinvested for another s years where s < 1

let interest earned each year on an investment of 1 be constant at i and

a(0) = 1, a(1) = 1 + i

simple interest is a linear accumulation function, a(t) = 1 + it, for integral t 0 simple interest has the property that interest is NOT reinvested to earn additional interest

a constant rate of simple interest implies a decreasing eective rate of interest:

in =A(n) A(n 1)

A(n 1) =k a(n) k a(n 1)

k a(n 1)=

a(n) a(n 1)a(n 1) =

1 + i n [1 + i (n 1)]1 + i (n 1)

=i

1 + i (n 1)

Nonintegral Value of t


if no interest is credited for fractional periods, then a(t) becomes a step function with dis-continuities

assume that interest accrues proportionately over fractional periods

a(t+ s) = a(t) + a(s) 1 = 1 + it+ 1 + is 1 = 1 + i(t + s)

amount of Interest Earned to time t is

I = A(0) it

1.5 Compound Interest

let interest earned each year on an investment of 1 be constant at i and

a(0) = 1, a(1) = 1 + i

compound interest is an exponential accumulation function a(t) = (1 + i)t, for integral t 0 compound interest has the property that interest is reinvested to earn additional interest

compound interest produces larger accumulations than simple interest when t > 1

3

a constant rate of compound interest implies a constant eective rate of interest

in =A(n) A(n 1)

A(n 1) =k a(n) k a(n 1)

k a(n 1)=

a(n) a(n 1)a(n 1) =

(1 + i)n (1 + i)n1(1 + i)n1

= i

Nonintegral Value of t


a(t+ s) = a(t) a(s) = (1 + i)t (1 + i)s = (1 + i)t+s

1.6 Present Value

Discounting

Accumulated Value is a future value pertaining to payment(s) made in the past

Discounted Value is a present value pertaining to payment(s) to be made in the future

discounting determines how much must be invested initially (X) so that 1 will be accumulatedafter t years

X (1 + i)t = 1 X = 1(1 + i)t

X represents the present value of 1 to be paid in t years

let v =1

1 + i, v is called a discount factor or present value factor

X = 1 vt

Discount Function: a1(t)

let a1(t) =1

a(t)

simple interest: a1(t) =1

1 + it

compound interest: a1(t) =1

(1 + i)t= vt

compound interest produces smaller Discount Values than simple interest when t > 1

4

1.7 The Eective Rate of Discount: d

Denition

an eective rate of interest is taken as a percentage of the balance at the beginning of theyear, while an eective rate of discount is at the end of the year.

eg. if 1 is invested and 6% interest is paid at the end of the year, then the AccumulatedValue is 1.06

eg. if 0.94 is invested after a 6% discount is paid at the beginning of the year, then theAccumulated Value at the end of the year is 1.00

let dn be the eective rate of discount earned during the nth period of the investment wherediscount is paid at the beginning of the period

d is also dened as the ratio of the amount of interest (amount of discount) earned duringthe period to the amount invested at the end of the period

dn =A(n) A(n 1)

A(n)=

InA(n)

, for integral n 1

if interest is constant, im = i, then discount is constant, dm = d

Relationship Between i and d

if 1 is borrowed and interest is paid at the beginning of the year then 1 d remains the accumulated value of 1 d at the end of the year is 1:

(1 d)(1 + i) = 1

interest rate is the ratio of the discount paid to the amount at the beginning of the period:

i =I1

A(0)=

d

1 d

discount rate is the ratio of the interest paid to the amount at the end of the period:

d =I1

A(1)=

i

1 + i

the present value of interest paid at the end of the year is the discount paid at the beginningof the year

iv = d

the present value of 1 to be paid at the end of the year is the same as borrowing 1 d andrepaying 1 at the end of the year (if both have the same value at the end of the year, thenthey have to have the same value at the beginning of the year)

1 v = 1 d

5

the dierence between interest paid at the end and at the beginning of the year depends onthe dierence that is borrowed at the beginning of the year and the interest earned on thatdierence

i d = i[1 (1 d)] = i d 0

Discount Function: a1(t)

let dn = d

under the simple discount model, the discount function is

a1(t) = 1 dt for 0 t < 1/d

under the compound discount model, the discount function is

a1(t) = (1 d)t = vt for t 0

a constant rate of simple discount implies an increasing eective rate of discount

dn =A(n) A(n 1)

A(n)=

k a(n) k a(n 1)k a(n)

= 1 a(n 1)a(n)

= 1 a1(n)

a1(n 1)= 1 (1 d n)

1 d(n 1) =1 d n+ d 1 + d n

1 d(n 1)=

d

1 d(n 1)

a constant rate of compound discount implies a constant eective rate of discount

dn =A(n) A(n 1)

A(n)=

k a(n) k a(n 1)k a(n)

= 1 a(n 1)a(n)

= 1 a1(n)

a1(n 1)= 1 (1 d)

n

(1 d)n1= 1 (1 d)= d

6

1.8 Nominal Rate of Interest and Discount Convertible mthly: i(m), d(m)

Denition

an eective rate of interest (discount) is paid once per year at the end(beginning) of the year

a nominal rate of interest (discount) is paid more frequently during the year (m times) andat the end (beginning) of the sub-period (nominal rates are also quoted as annual rates)

nominal rates are adjusted to reect the rate to be paid during the subperiod

i(2) = 10% i(2)

2=

10%2

= 5% paid every 6 months

Equivalency to Eective Rates of Interest: i, i(m)

with eective interest, you have interest, i, paid at the end of the year

with nominal interest, you have interesti(m)

m, paid at the end of each sub-period and this is

done m times over the year (m sub-periods per year)

(1 + i) =(1 +

i(m)

m

)m

if given an eective rate of interest, a nominal rate of interest can be determined

i(m) = m[(1 + i)1/m 1]

the interest rate per sub-period can be determined, if given the eective interest rate

i(m)

m= (1 + i)1/m 1

Equivalency to Eective Rates of Discount: d, d(m)

with eective discount, you have discount, d, paid at the beginning of the year

with nominal discount, you have discountd(m)

m, paid at the beginning of each sub-period

and this is done m times over the year (m sub-periods per year)

(1 d) =(1 d

(m)

m

)m

if given an eective rate of discount, a nominal rate of discount can be determined

d(m) = m[1 (1 d)1/m]

the discount rate per sub-period can be determined, if given the eective discount rate

d(m)

m= 1 (1 d)1/m

7

Relationship Betweeni(m)

mand

d(m)

m

when using eective rates, you must have (1 + i) or (1 d)1 by the end of the year

(1 + i) =1v

=1

(1 d) = (1 d)1

when replacing the eective rate formulas with their nominal rate counterparts, you have(1 +

i(m)

m

)m=(1 d

(p)

p

)p

when p = m (1 +

i(m)

m

)m=(1 d

(m)

m

)m

1 +i(m)

m=(1 d

(m)

m

)1

1 +i(m)

m=

m

m d(m)i(m)

m=

m

m d(m) 1 =mm + d(m)

m d(m)i(m)

m=

d(m)

m d(m)

i(m)

m=

d(m)

m

1 d(m)

m

the interest rate over the sub-period is the ratio of the discount paid to the amount at thebeginning of the sub-period (principle of the interest rate still holds)

d(m)

m=

i(m)

m

1 +i(m)

m

the discount rate over the subperiod is the ratio of interest paid to the amount at the endof the sub-period (principle of the discount rate still holds)

the dierence between interest paid at the end and at the beginning of the sub-period dependson the dierence that is borrowed at the beginning of the sub-period and on the interestearned on that dierence (principle of the interest and discount rates still holds)

i(m)

m d

(m)

m=

i(m)

m

[1

(1 d

(m)

m

)]

=i(m)

m d

(m)

m 0

8

1.9 Forces of Interest and Discount: in, dn

Denitions

annual eective rate of interest and discount are applied over a one-year period

annual nominal rate of interest and discount are applied over a sub-period once the rateshave been converted

annual force of interest and discount are applied over the smallest sub-period imaginable (ata moment in time) i.e. m

Annual Force of Interest At Time n : in

recall that the interest rate over a sub-period is the ratio of the Interest Earned during thatperiod to the Accumulated Value at the beginning of the period

i(m)

m=

A(n + 1m

)A(n)A(n)

if m = 12,i(12)

12=

A(n+ 112

) A(n)A(n)

= monthly rate; monthly rate x 12 = annual rate

if m = 365,i(365)

365=

A(n + 1365

)A(n)A(n)

= daily rate; daily rate x 365 = annual rate

if m = 8760,i(8760)

8760=

A(n+ 18760

)A(n)A(n)

= hourly rate; hourly rate x 8760 = annual rate

if m, limm

i(m)

m= lim

mA(n+ 1

m

) A(n)A(n)

= instantaneous rate

let in = limm i

(m) = limm

[A(n+ 1

m

) A(n)1m

]A(n)

= Force of Interest At Time n

in =

d

dnA(n)

A(n)=

d

dnk a(n)

k a(n) =d

dna(n)

a(n)

in =d

dnln[A(n)] =

d

dnln[a(n)]

9

Accumulation Function Using the Force of Interest

recall that the Force of Interest is dened as

in =d

dnln[a(n)] in dn = d(ln[a(n)])

integrating both sides from time 0 to t results in t0

in dn = t0

d(ln[a(n)])

= ln[a(t)] ln[a(0)]

= ln[a(t)a(0)

] t0

in dn = ln[a(t)])

taking the exponential function of both sides results in

e

t0

in dn= a(t)

the Accumulation Function can therefore be dened as an exponential function where theannual force of interest is converted into an innitesimally small rate [in dn]; this small rateis then applied over every existing moment from time 0 to time t

Interest Earned Over t Years Using the Force of Interest

recall that the Force of Interest is also dened as

in =

d

dnA(n)

A(n) A(n) in dn = d(A(n))

integrating both sides from time 0 to t results in t0

A(n) in dn = t0

d(A(n)) t0

A(n) in dn = A(t) A(0)

the Interest Earned over a t year period can be found by applying the interest rate thatexists at a certain moment, in dn, to the balance at that moment, A(n), and evaluating itfor every moment from time 0 to t

10

Annual Force of Discount At Time n = Annual Force of Interest At Time n : dn = in

recall that the interest rate over a sub-period is the ratio of the Interest Earned during thatperiod to the Accumulated Value at the end of the period

d(m)

m=

A(n+ 1

m

) A(n)A(n+ 1

m)

if m = 12,d(12)

12=

A(n + 1

12

) A(n)A(n + 112)

= monthly rate; monthly rate x 12 = annual rate

if m = 365,d(365)

365=

A(n+ 1365

) A(n)A(n+ 1365)

= daily rate; daily rate x 365 = annual rate

if m = 8760,d(8760)

8760=

A(n+ 18760

) A(n)A(n+ 18760)

= hourly rate; hourly rate x 8760 = annual rate

if m, limm

d(m)

m= lim

mA(n+ 1m

)A(n)A(n + 1m )

= instantaneous rate

let dn = limm d(m) = lim

m

[A(n+ 1m

) A(n)1m

]A(n+ 1m)

= Force of Discount At Time n

dn = limm

d

dnA(n)

A(n+ 1m) A(n)A(n)

dn =

d

dnA(n)

A(n) limm

A(n)A(n+ 1m))

=

d

dnA(n)

A(n) 1

dn = in

an alternative approach to determine the Force of Discount is to take the derivative of thediscount functions (remember that in took the derivative of the accumulation functions)

dn = d

dna1(n)

a1(n)= d

dn

1a(n)1

a(n)

=(1) 1

a(n)2d

dna(n)

1a(n)

=

d

dna(n)

a(n)

dn = in

from now on, we will use n instead of in or dn

11

Force of Interest When Interest Rate Is Constant

n can vary at each instantaneous moment

let the Force of Interest be constant each year: n = in = i, then

a(t) = e

t0

in dn= e

t0

dn

= et = (1 + i)t e = 1 + i =ln[1+i]

e = v

note that nominal rates can now be introduced

1 + i =(1 +

i(m)

m

)m=(1 d

(p)

p

)p= e

Force of Interest Under Simple Interest

a constant rate of simple interest implies a decreasing force of interest

n =

d

dna(n)

a(n)=

d

dn(1 + i n)1 + i n

=i

1 + i n

Force of Interest Under Simple Discount

a constant rate of simple discount implies an increasing force of interest

n = d

dna1(n)

a1(n)= d

dn(1 d n)

1 d n=

d

1 d n for 0 t < 1/d

12

1.10 Varying Interest

Varying Force of Interest

recall the basic formula

a(t) = e

t0

ndn

if n is readily integrable, then a(t) can be derived easily

if n is not readily integrable, then approximate methods of integration are required

Varying Eective Rate of Interest

the more common application

a(t) =t

k=1

(1 + ik)

and

a1(t) =t

k=1

1(1 + ik)

1.11 Summary of Results

Rate of interest or discount a(t) a1(t)

Compound interest

i (1 + i)t vt = (1 + i)t

i(m)[1 +

i(m)

m

]mt [1 +

i(m)

m

]mtd (1 d)t (1 d)t

d(m)[1 d

(m)

m

]mt [1 d

(m)

m

]mt et et

Simple interest

i 1 + it (1 + it)1

Simple discount

d (1 dt)1 1 dt

13

2 Solution of Problems in Interest

2.1 Introduction

How to Solve an Interest Problem

use basic principles

develop a systematic approach

2.2 Obtaining Numerical Results

using a calculator with exponential functions is the obvious rst choice

in absence of such a calculator, using the Table of Compound Interest Functions: AppendixI (page 376 392) would be the next option

series expansions could be used as a last resort

e.g. (1 + i)k = 1 + ki+k(k 1)

2!i2 +

k(k 1)(k 2)3!

i3 +

e.g. ek = 1 + k +(k)2

2!+

(k)3

3!+

A Common Problem

using compound interest for integral periods of time and using simple interest for fractionalperiods is an exercises in linear interpolation

e.g. (1 + i)n+k (1 k)(1 + i)n + k(1 + i)n+1= (1 + i)n[(1 k) + k(1 + i)]= (1 + i)n(1 + ki)

e.g. (1 d)n+k (1 k)(1 d)n + k(1 d)n+1= (1 d)n(1 kd)

2.3 Determining Time Periods

when using simple interest, there are 3 dierent methods for counting the days in an invest-ment period

(i) exact simple interest approach: count actual number of days where one year equals 365days

(ii) ordinary simple interest approach: one month equals 30 days; total number of daysbetween D2,M2, Y2 and D1,M1, Y1 is

360(Y2 Y1) + 30(M2 M1) + (D2 D1)

(iii) Bankers Rule: count actual number of days where one year equals 360 days

14

2.4 The Basic Problem

there are 4 variables required in order to solve an interest problem

(a) original amount(s) invested

(b) length of investment period(s)

(c) interest rate

(d) accumulated value(s) at the end of the investment period

if you have 3 of the above variables, then you can solve for the unknown 4th variable

2.5 Equations of Value

the value at any given point in time, t, will be either a present value or a future value(sometimes referred to as the time value of money)

the time value of money depends on the calculation date from which payment(s) are eitheraccumulated or discounted to

Time Line Diagrams

it helps to draw out a time line and plot the payments and withdrawals accordingly

0 1 2 ... t ... n-1 n

P1 P2 ... Pt ... Pn-1 Pn

W1 W2 Wt Wn-1 Wn

15

Example

a $600 payment is due in 8 years; the alternative is to receive $100 now, $200 in 5 years and$X in 10 years. If i = 8%, nd $X, such that the value of both options is equal.

0

100

600

X

5 8 10

200

compare the values at t = 0

0

100

600

X

5 8 10

200

600v88% = 100 + 200v58% +Xv

108%

X =600v88% 100 200v58%

v108%= 190.08

16


0

100

600

X

5 8 10

200

600v3 = 100(1 + i)5 + 200 + Xv5

X =600v3 100(1 + i)5 200

v5= 190.08


0

100

600

X

5 8 10

200

600(1 + i)2 = 100(1 + i)10 + 200(1 + i)5 +X

X = 600(1 + i)2 100(1 + i)10 200(1 + i)5 = 190.08

all 3 equations gave the same answer because all 3 equations treated the value of the paymentsconsistently at a given point of time.

17

2.6 Unknown Time

Single Payment

the easiest approach is to use logarithms

Example

How long does it take money to double at i = 6%?

(1.06)n = 2n ln[1.06] = ln[2]

n =ln[2]

ln[1.06]= 11.89566 years

if logarithms are not available, then use an interest table from Appendix I (page 376 392)and perform a linear interpolation

(1.06)n = 2go to page 378, and nd:

(1.06)11 = 1.89830 and

(1.06)12 = 2.01220

n = 11 + 2 1.898302.01220 1.89830 = 11.89 years

Rule of 72 for doubling a single payment

n =ln[2]

ln[1 + i]=

0.6931i

iln[1 + i]

=0.6931

i(1.0395), when i = 8%

n 0.72i

Rule of 114 for tripling a single payment

n =ln[3]

ln[1 + i]=

1.0986i

iln[1 + i]

=1.0986

i(1.0395), wheni = 8%

n 1.14i

18

An Approximate Approach For Multiple Payments

let St represent a payment made at time t such that

0 t1 t2 ... tn-1 tn

S1 S2 ... Sn-1 Sn

we wish to replace the multiple payments with a single payment equal ton

k=1

Sk such that

the present value of this single payment at a single moment in time (call it t) is equal to thepresent value of the multiple payments.

to nd the true value of t:

(S1 + S2 + + Sn) vt = S1vt1 + S2vt2 + + Snvtn(n

k=1

Sk

) vt =

nk=1

Skvtk

vt =

nk=1

Skvtk

nk=1

Sk

t ln[v] = ln

nk=1

Skvtk

nk=1

Sk

t =

ln

nk=1

Skvtk

nk=1

Sk

ln[v]

t =

ln

[n

k=1

Skvtk

] ln

[n

k=1

Sk

]ln[v]

19

to nd an approximate value of t:

let t equal the weighted average of time (weighted by the payments)

t =S1 t1 + S2 t2 + + Sn1 tn1 + Sn tn

S1 + S2 + + Sn1 + Sn

t =

nk=1

Sktk

nk=1

Sk

method of equated time

if t > t, then the present value using the method of equated time will be less than the presentvalue using exact t

Algebraic Proof: t > t

let vtk be the present value of a future payment of 1 at timetk and let Sk be the number ofpayments made at time k

(a) arithmetic weighted mean of present values

S1vt1 + S2vt2 + + SnvtnS1 + S2 + + Sn =

nk=1

Skvtk

nk=1

Sk

(b) geometric weighted mean of present value

[(vt1)S1 (vt2)S2 (vtn)Sn] 1S1+S2++Sn

=[vS1t1+S2 t2++Sntn

] 1S1+S2++Sn

= vS1 t1+S2 t2++Sntn

S1+S2++Sn

= vt

Since geometric means are less than arithmetic means,

vt k.

an i1 = k1an i = kan i2 = k2

}i i1 + k1 k

k1 k2 (i2 i1)

3. Successive Approximation (Iteration)

considered the best way to go if no calculator and precision is really important

there are two techniques that can be used

i. Solve for i

an i = k1 vni

i= k

is+1 =1 (1 + is)n

k

What is a good starting value for i0 (s = 0)?

one could use linear interpolation to nd i0

one could use the rst two terms of 1a n and solve for i

1k

=1n

+(n+ 12n

)i

i0 =2(n k)k(n + 1)

another approach to derive a starting value is to use

i0 =1 ( k

n

)2k

ii. Newton-Raphson Method

This method will see convergence very rapidly

is+1 = is f(is)

f (is)

44

let

an i = kan i k = 01 (1 + is)n

i k = 0

f(is) = 1 (1 + is)n isk = 0f (is) = (n)(1 + is)n1 k

is+1 = is 1 (1 + is)n isk

n(1 + is)n1 k= is

[1 +

1 (1 + is)n k is1 (1 + is)n1{1 + is(n+ 1)}

]

What about sn i = k?

1. Algebraic Techniques

sn i = k = n +n(n+ 1)

2!i+

n(n 1)(n 2)3!

i2 1

sn i=

1k

=1n (n 1)

2ni+

(n2 1)12n

i2 +

2. Linear Interpolationsn i1 = k1sn i = ksn i2 = k2

}i i1 + k1 k

k1 k2 (i2 i1)

3. Successive Approximation (Iteration)

there are two techniques that can be used

i. Solve for i

sn i = k(1 + i)n 1

i= k

is+1 =(1 + is)n 1

k

What is a good starting value for i0 (s = 0)?

one could use linear interpolation to nd i0

45

one could use the rst two terms of 1sn

and solve for i

1k

=1n(n 12n

)i

i0 =2(n k)k(n 1)

another approach to derive a starting value is to use

i0 =

(kn

)2 1k

ii. Newton-Raphson Method

is+1 = is f(is)

f (is)

let

sn i = ksn i k = 0(1 + is)n 1

i k = 0

f(is) = (1 + is)n 1 isk = 0f (is) = n(1 + is)n1 k

is+1 = is (1 + is)n 1 isk

n(1 + is)n1 k= is

[1 +

(1 + is)n 1 k is(1 + is)n1{1 is(n 1)} 1

]= is

[1 +

(1 + is)n 1 k isk is n is(1 + is)n1

]

46

3.9 Varying Interest

if the annual eective rate of interest varies from one year to the next, i = ik, then thepresent value and accumulated value of annuity payments needs to be calculated directly

the varying rate of interest can be dened in one of two ways:

(i) eective rate of interest for period k, ik, is only used for period k

(ii) eective rate of interest for period k, ik, is used for all periods

If ik Is Only Used For Period k

A payment made during year k will need to be discounted or accumulated over each past orfuture period at the interest rate that was in eect during that period.

The time line diagram below illustrates the varying interest rates for an annuity-immediate.

0 1 2 ... n - 1 n

1 1 ... 1 1 i1 i2 in-1 in

The present value of an annuityimmediate is determined as follows:

47

an =1

1 + i1+

11 + i1

11 + i2

+ + 11 + i1

11 + i2

11 + in

=n

k=1

kj=1

11 + ij

The accumulated value of an annuityimmediate is determined as follows:

sn = 1 + (1 + in) + (1 + in) (1 + in1) + + (1 + in) (1 + in1) (1 + i2)

= 1 +n1k=1

kj=1

(1 + inj+1)

The time line diagram below illustrates the varying interest rates for an annuity-due.

0 1 2 ... n - 1 n

1 1 1 ... 1 i1 i2 in-1 in

48

The present value of an annuitydue is determined as follows:

an = 1 +1

1 + i1+

11 + i1

11 + i2

+ + 11 + i1

11 + i2

11 + in1

= 1 +n1k=1

kj=1

11 + ij

The accumulated value of an annuitydue is determined as follows:

sn = (1 + in) + (1 + in) (1 + in1) + + (1 + in) (1 + in1) (1 + i1)

=n

k=1

kj=1

(1 + inj+1)

Note that the accumulated value of an annuityimmediate can also be solved by using theabove annuitydue formula and applying Basic Relationship 6 from Section 3.3: sn =1 + sn 1 .

Note that the present value of an annuitydue can also be solved by using the above annuityimmediate formula and applying Basic Relationship 5 from Section 3.3: an = 1 + an 1 .

If ik Is Used For All PeriodsA payment made during year k will be discounted or accumulated at the interest rate thatwas in eect at the time of the payment. For example, if the interest rate during year 10 was6%, then the payment made during year 10 will discounted back or accumulated forward at6% for each year.

The present value of an annuityimmediate is an =n

k=1

1

(1 + ik)k.

The accumulated value of an annuityimmediate is sn = 1 +n1k=1

(1 + ink)k.

The present value of an annuitydue is an = 1 +n1k=1

1

(1 + ik+1)k.

The accumulated value of an annuitydue is sn =n

k=1

(1 + ink+1)k.

49

Note that the accumulated value of an annuityimmediate can also be solved by using theabove annuitydue formula and applying Basic Relationship 6 from Section 3.3: sn =1 + sn 1 .

Note that the present value of an annuitydue can also be solved by using the above annuityimmediate formula and applying Basic Relationship 5 from Section 3.3: an = 1 + an 1 .

3.10 Annuities Not Involving Compound Interest

material not tested in SoA Exam FM

50

4 More General Annuities

4.1 Introduction

in Chapter 3, annuities were described as having level payments payable at the same frequencyas what the interest rate was being converted at

in this chapter, non-level payments are examined as well as the case where the interestconversion period and the payment frequency no longer coincide

4.2 Annuities Payable At A Dierent Frequency Than Interest Is Con-vertible

let the payments remain level for the time being

when the interest conversion period does not coincide with the payment frequency, one cantake the given rate of interest and convert to an interest rate that does coincide

ExampleFind the accumulated value in 10 years if semi-annual due payments of 100 are being madeinto a fund that credits a nominal rate of interest at 10%, convertible semiannually.

FV10 = 100s 10 12 j

Interest rate j will need to be a monthly rate and is calculated based on the semiannual ratethat was given:

j =i(12)

12=

1 + i(2)2

6 month rate

1/6

1 =(1 +

10%2

)1/6 1 = 0.8165%

Therefore, the accumulated value at time t = 10 is

FV10 = 100s 120 0.8165% = 20, 414.52

51

4.3 Further Analysis of Annuities Payable Less Frequency Than Inter-est Is Convertible

material not tested in SoA Exam FM

AnnuityImmediate

let i(k) be a nominal rate of interest convertible k times a year and let there be level end-of-year payments of 1

after the rst payment has been made, interest has been converted k times

after the second payment has been made, interest has been converted 2k times

after the last payment has been made, interest has been converted n times

therefore, the term of the annuity (and obviously, the number of payments) will ben

kyears

(i.e. if n = 144 and i(12) is used, then the term of the annuity is14412

= 12 years).

The time line diagram will detail the above scenario:

years 0 1 2 ... ... (n/k) - 1 n/k

1 1 ... ... 1 1

conversion

periods 0 k 2k ... ... n - k n

the present value (at t = 0) of an annual annuityimmediate where payments are made every

52

k conversion periods and where the rate of interest is j =i(k)

kshall be calculated as follows:

PV0 = (1)v1i + (1)v2i + + (1)vn/k1i + (1)vn/ki

= (1)vkj + (1)v2kj + + (1)vnkj + (1)vnj

= vkj(1 + vkj + + vn2kj + vnkj

)= vkj

(1 (vkj )n/k

1 vkj

)

= vkj

(1 vn1 vkj

)

=1

(1 + j)k

(1 vn1 vkj

)

=(

1 vn(1 + j)k 1

) jj

=an js k j

There is an alternative approach in determining the present value of this annuityimmediate:

The payment of 1 made at the end of each year can represent the accumulated value ofsmaller level end-of-conversion-period payments that are made k times during the year.

P s k j = 1

These smaller level payments are therefore equal to P =1

s k j. If these smaller payments

were to be made at the end of every conversion period, during the term of the annuity, andthere are n conversion periods in total, then the present value (at t = 0) of these n smallerpayments is determined to be

PV0 =1

s k j an j .

the accumulated value (at t = n/k years or t = n conversion periods) of an annual annuityimmediate where payments are made every k conversion periods and where the rate of interest

53

is j =i(k)


FVnk= 1 + (1)(1 + i)1 + + (1)(1 + i)nk2 + (1)(1 + i)nk1= 1 + (1)(1 + j)k + + (1)(1 + j)n2k + (1)(1 + j)nk

=1 [(1 + j)k]n/k

1 (1 + j)k

=1 (1 + j)n1 (1 + j)k

=[(1 + j)n 1(1 + j)k 1

] j

j

=sn js k j

Another approach in determining the accumulated value would be to go back to a basicrelationship where a future value is equal to its present value carried forward with interest:

FVnk= FVn= PV0 (1 + j)n

=an js k j

(1 + j)n

=sn js k j

54

The accumulated value can also be derived by again considering that each end-of-year pay-ment of 1 represents the accumulated value of smaller level end-of-conversion-period paymentsthat are made k times during the year:

P s k j = 1


s k j. If these smaller payments are

made at the end of every conversion period, over the term of the annuity, and there are nconversion periods in total, then the accumulated value (at t = n/k years or t = n conversionperiods) of these smaller payments is determined to be

FVnk=

(1

s k j

) sn j .

55

AnnuityDue

let i(k) be a nominal rate of interest convertible k times a year and let there be level beginning-of-year payments of 1

after the second payment has been made, interest has been converted k times

after the third payment has been made, interest has been converted 2k times

after the last payment has been made, interest has been converted n k times

therefore, the term of the annuity (and obviously, the number of payments) will be 1+n k

k=

n

kyears (i.e. if n = 144 and i(12) is used, then the term of the annuity is

14412

= 12 years).

The time line diagram will detail the above scenario:

years 0 1 2 ... ... (n/k) - 1 n/k

1 1 1 ... ... 1

conversion

periods 0 k 2k ... ... n - k n

the present value (at t = 0) of an annual annuitydue where payments are made every k

conversion periods and where the rate of interest is j =i(k)


PV0 = 1 + (1)v1i + (1)v2i + + (1)vn/k1i

= 1 + (1)vkj + (1)v2kj + + (1)vnkj

= 1 + vkj + + vn2kj + vnkj

=1 (vkj )n/k

1 vkj=

1 vn1 vkj

=

(1 vn1 vkj

) jj

=an ja k j

56

There is an alternative approach in determining the present value of this annuitydue:

The payment of 1 made at the beginning of each year can represent the present value ofsmaller level end-of-conversion-period payments that are made k times during the year.

P a k j = 1


a k j. If these smaller payments

were to be made at the end of every conversion period, during the term of the annuity, andthere are n conversion periods in total, then the present value (at t = 0) of these n smallerpayments is determined to be

PV0 =1

a k j an j .

the accumulated value (at t = n/k years or t = n conversion periods) of an annual annuitydue where payments are made every k conversion periods and where the rate of interest is

j =i(k)


FVnk= (1)(1 + i)1 + (1)(1 + i)2 + + (1)(1 + i)nk1 + (1)(1 + i)nk= (1)(1 + j)k + (1)(1 + j)2k + + (1)(1 + j)nk + (1)(1 + j)n

= (1 + j)k(1 [(1 + j)k]n/k

1 (1 + j)k)

= (1 + j)k(

1 (1 + i)n1 (1 + j)k

)

=1vkj

((1 + j)n 1(1 + j)k 1

)

=

[(1 + j)n 1

1 vkj

] j

j

=sn ja k j

Another approach in determining the accumulated value would be to go back to a basicrelationship where a future value is equal to its present value carried forward with interest:

FVnk= FVn= PV0 (1 + j)n

=an ja k j

(1 + j)n

=sn ja k j

57

The accumulated value can also be derived by again considering that each beginning-of-yearpayment of 1 represents the present value of smaller level end-of-conversion-period paymentsthat are made k times during the year:

P a k j = 1


a k j. If these smaller payments are

made at the end of every conversion period, over the term of the annuity, and there are nconversion periods in total, then the accumulated value (at t = n/k years or t = n conversionperiods) of these smaller payments is determined to be

FVnk=

(1

a k j

) sn j .

58

Other Considerations

PerpetuityImmediate

a perpetuityimmediate with annual end-of-year payments of 1 and where the nominal cred-ited interest rate is convertible more frequently than annually, can be illustrated as:

years 0 1 2 ... ... n ...

1 1 ... ... 1 ...

conversion

periods 0 k 2k ... ... n x k ...

the present value (at t = 0) of an annual perpetuityimmediate where payments are made

every k conversion periods and where the rate of interest is j =i(k)

kshall be calculated as

follows:

PV0 = (1)v1i + (1)v2i + (1)v

3i +

= (1)vkj + (1)v2kj + (1)v

3kj +

= vkj (1 + vkj + v

2kj + v

3kj + )

= vkj

(1 (vk)1 vkj

)

= vkj

(1 01 vkj

)

=1

(1 + j)k

(1

1 vkj

)

=(

1(1 + j)k 1

) jj

=1

j s k j

one could also derive the above formula by simply substituting n = into the originalpresent value formula

PV0 =a js k j

=1j 1

s k j

59

PerpetuityDue

a perpetuitydue with annual beginning-of-year payments of 1 and where the nominal cred-ited interest rate is convertible more frequently than annually, can be illustrated as:

years 0 1 2 ... ... n ...

1 1 1 ... ... 1 ...

conversion

periods 0 k 2k ... ... n x k ...

the present value (at t = 0) of an annual perpetuitydue where payments are made every k

conversion periods and where the rate of interest is j =i(k)


PV0 = 1 + (1)v1i + (1)v2i + (1)v

3i +

= 1 + (1)vkj + (1)v2kj + (1)v

3kj +

= 1 + vkj + v2kj + v

3kj +

=

(1 (vk)1 vkj

)

=

(1 01 vkj

)

=

(1

1 vkj

) jj

=1

j a k j

one could also derive the above formula by simply substituting n = into the originalpresent value formula

PV0 =a ja k j

=1j 1

a k j

60

Interest Is Convertible Continuously: i() =

the problem under this situation is that k is innite (so is the total number of conversionperiods over the term). Therefore, the prior formulas will not work.

for example, the present value (at t = 0) of an annuityimmediate where payments of 112 aremade every month for n years (or 12n periods) and where the annual force of interest is can be calculated as follows:

PV0 =(112

)v112i + (

112

)v212i + + (

112

)v1112i + (

112

)v1212i (1st year)

+(112

)v1312i + (

112

)v1412i + + (

112

)v2312i + (

112

)v2412i (2nd year)

...

+(112

)v12(n1)+1

12i + (

112

)v12(n1)+2

12i + + (

112

)v12n112

i + (112

)v12n12

i (last year)

=(112

)v112i

[1 + v

112i + + v

1012i + v

1112i

](1st year)

+(112

)v1312i

[1 + v

112i + + v

1012i + v

1112i

](2nd year)

...

+(112

)v12(n1)+1

12i

[1 + v

112i + + v

1012i + v

1112i

](last year)

=((112

)v112i + (

112

)v1312i + + (

112

)v12(n1)+1

12 )i

)[1 + v

112i + + v

1012i + v

1112i

]

=(112

)v112i

(1 + v

1212i + + v(n1)i

)[1 (v 112i )121 v 112i

]

=(112

)1

(1 + i) 112

(1 vni1 v1i

)[

1 v1i1 v 112i

]

=(112

)1

(1 + i) 112

(1 vni1 v 112i

)

=(112

)(

1 vni(1 + i)

112 1

) i

i

=(112

)(

i

(1 + i)112 1

) an i = (

112

)

(1 + i

(12)

12

)12 1

i(12)

12

an i

=[(112

) s 12i(12)12

] an i =

[(112

) s 12i(12)12 =e

12 1

] an i=e1

in this case, the monthly payments for each year are converted to end-of-year lump sums thatare discounted back to t = 0 at an annual eective rate of interest, i, which was convertedfrom .

61

4.4 Further Analysis of Annuities Payable More Frequency Than In-terest Is Convertible

AnnuityImmediate

payments of 1m are made at the end of every1mth of year for the next n years

the present value (at t = 0) of an mthly annuityimmediate, where the annual eective rateof interest is i, shall be denoted as a(m)n i and is calculated as follows:

a(m)n i

=(1m

)v1m

i + (1m

)v2m

i + + (1m

)vm1

m

i + (1m

)vmm

i (1st year)

+(1m

)vm+1

m

i + (1m

)vm+2

m

i + + (1m

)v2m1

m

i + (1m

)v2mm

i (2nd year)

...

+(1m

)v(n1)m+1

m

i + (1m

)v(n1)m+2

m

i + + (1m

)vnm1

m

i + (1m

)vnmm

i (last year)

=(1m

)v1m

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](1st year)

+(1m

)vm+1

m

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](2nd year)

...

+(1m

)v(n1)m+1

m

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](last year)

=((1m

)v1m

i + (1m

)vm+1

m

i + + (1m

)v(n1)m+1

m )i

)[1 + v

1m

i + + vm2

m

i + vm1

m

i

]

=(1m

)v1m

i

(1 + v

mm

i + + v(n1)i)[1 (v 1mi )m

1 v 1mi

]

=(1m

)1

(1 + i)1m

(1 vni1 v1i

)[

1 v1i1 v 1mi

]

=(1m

)1

(1 + i)1m

(1 vni1 v 1mi

)

=(1m

)(

1 vni(1 + i) 1m 1

)

=1 vni

m[(1 + i) 1m 1

]=

1 vnii(m)

=(

1mm

) 1 v

ni

i(m)

62

the accumulated value (at t = n) of an mthly annuityimmediate, where the annual eectiverate of interest is i, shall be denoted as s(m)n i and is calculated as follows:

s(m)n i

=(1m

) + (1m

)(1 + i)1m + + ( 1

m)(1 + i)

m2m + (

1m

)(1 + i)m1

m (last year)

+(1m

)(1 + i)mm + (

1m

)(1 + i)m+1

m + + ( 1m

)(1 + i)2m2

m + (1m

)(1 + i)2m1

m (2nd last year)

...

+(1m

)(1 + i)(n1)m

m + (1m

)(1 + i)(n1)m+1

m + + ( 1m

)(1 + i)nm2

m + (1m

)(1 + i)nm1

m (rst year)

=(1m

)[1 + (1 + i)

1m + + (1 + i)m2m + (1 + i)m1m

](last year)

+(1m

)(1 + i)mm

[1 + (1 + i)

1m + + (1 + i)m2m + (1 + i)m1m

](2nd last year)

...

+(1m

)(1 + i)(n1)m

m

[1 + (1 + i)

1m + + (1 + i)m2m + (1 + i)m1m

](rst year)

=((1m

) + (1m

)(1 + i)mm + + ( 1

m)(1 + i)

(n1)mm )

)[1 + (1 + i)

1m + + (1 + i)m2m + (1 + i)m1m

]

=(1m

)(1 + (1 + i)

mm + + (1 + i)(n1)

)[1 ((1 + i) 1m )m1 (1 + i) 1m

]

=(1m

)(1 (1 + i)n1 (1 + i)1

)[1 (1 + i)11 (1 + i) 1m

]

=(1m

)(

1 (1 + i)n1 (1 + i) 1m

)

=(1 + i)n 1

m[(1 + i)

1m 1

]=

(1 + i)n 1i(m)

=(

1mm

) (1 + i)

n 1i(m)

63

Basic Relationship 1 : 1 = i(m) a(m)n + vn

Basic Relationship 2 : PV (1 + i)n = FV and PV = FV vn

if the future value at time n, s(m)n , is discounted back to time 0, then you will have itspresent value, a(m)n

s(m)n vn =

[(1 + i)n 1

i(m)

] vn

=(1 + i)n vn vn

i(m)

=1 vni(m)

= a(m)n

if the present value at time 0, a(m)n , is accumulated forward to time n, then you willhave its future value, s(m)n

a(m)n (1 + i)n =

[1 vni(m)

](1 + i)n

=(1 + i)n vn(1 + i)n

i(m)

=(1 + i)n 1

i(m)

= s(m)n

Basic Relationship 3 :1

m a(m)n=

1

m s(m)n+

i(m)

m

Consider a loan of 1, to be paid back over n years with equal mthly payments of P madeat the end of each mth of a year. An annual eective rate of interest, i, and nominalrate of interest, i(m), is used. The present value of this single payment loan must beequal to the present value of the multiple payment income stream.

(P m) a(m)n i = 1

P =1

m a(m)n i Alternatively, consider a loan of 1, where the mthly interest due on the loan, (1) i(m)m ,

is paid at the end of each mth of a year for n years and the loan amount is paid backat time n.

In order to produce the loan amount at time n, payments of D at the end of each mth

of a year, for n years, will be made into an account that credits interest at an mthlyrate of interest i

(m)

m .

64

The future value of the multiple deposit income stream must equal the future value ofthe single payment, which is the loan of 1.

(D m) s(m)n i = 1

D =1

m s(m)n i The total mthly payment will be the interest payment and account payment:

i(m)

m+

1

m s(m)n i Note that

1

a(m)n i

=i(m)

1 vn (1 + i)n

(1 + i)n=

i(m)(1 + i)n

(1 + i)n 1

=i(m)(1 + i)n + i(m) i(m)

(1 + i)n 1 =i(m)[(1 + i)n 1] + i(m)

(1 + i)n 1

= i(m) +i(m)

(1 + i)n 1 = i(m) +

1

s(m)n

Therefore, a level mthly annuity payment on a loan is the same as making an mthlyinterest payment each mth of a year plus making mthly deposits in order to save for theloan repayment.

Basic Relationship 4 : a(m)n =i

i(m) an , s(m)n =

i

i(m) sn

Consider payments of 1m

made at the end of every 1mth of year for the next n years. Over

a one-year period, payments of 1m

made at the end of each mth period will accumulateat the end of the year to a lump sum of

(1m m

) s(m)1

. If this end-of-year lump sumexists for each year of the n-year annuity-immediate, then the present value (at t = 0)of these end-of-year lump sums is the same as

(1m m

) a(m)n :(1mm

) a(m)n =

(1mm

) s(m)

1 an

a(m)n =

i

i(m) an

Therefore, the accumulated value (at t = n) of these end-of-year lump sums is the sameas(1m m

) s(m)n : (1mm

) s(m)n =

(1mm

) s(m)

1 sn

s(m)n =

i

i(m) sn

65

AnnuityDue

payments of 1m

are made at the beginning of every 1mth of year for the next n years

the present value (at t = 0) of an mthly annuitydue, where the annual eective rate ofinterest is i, shall be denoted as a(m)n i and is calculated as follows:

a(m)n i

=(1m

) + (1m

)v1mi + + (

1m

)vm2

mi + (

1m

)vm1

mi (1st year)

+(1m

)vmmi + (

1m

)vm+1

mi + + (

1m

)v2m2

mi + (

1m

)v2m1

mi (2nd year)

...

+(1m

)v(n1)m

m

i + (1m

)v(n1)m+1

m

i + + (1m

)vnm2

m

i + (1m

)vnm1

m

i (last year)

=(1m

)[1 + v

1m

i + + vm2

m

i + vm1

m

i

](1st year)

+(1m

)vmm

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](2nd year)

...

+(1m

)v(n1)m

m

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](last year)

=((1m

) + (1m

)vmm

i + + (1m

)v(n1)m

m )i

)[1 + v

1m

i + + vm2

m

i + vm1

m

i

]

=(1m

)(1 + v

mm

i + + v(n1)i)[1 (v 1mi )m

1 v 1mi

]

=(1m

)(1 vni1 v1i

)[

1 v1i1 v 1mi

]

=(1m

)

(1 vni1 v 1mi

)

=(1m

)(

1 vni1 (1 d) 1m

)

=1 vni

m[1 (1 d) 1m

]=

1 vnid(m)

=(

1mm

) 1 v

ni

d(m)

66

the accumulated value (at t = n) of an mthly annuitydue, where the annual eective rateof interest is i, shall be denoted as s(m)n i and is calculated as follows:

s(m)n i

=(1m

)(1 + i)1m + (

1m

)(1 + i)2m + + ( 1

m)(1 + i)

m1m + (

1m

)(1 + i)mm (last year)

+(1m

)(1 + i)m+1

m + (1m

)(1 + i)m+2

m + + ( 1m

)(1 + i)2m1

m + (1m

)(1 + i)2mm (2nd last year)

...

+(1m

)(1 + i)(n1)m+1

m + (1m

)(1 + i)(n1)m+2

m + + ( 1m

)(1 + i)nm1

m + (1m

)(1 + i)nmm (rst year)

=(1m

)(1 + i)1m

[1 + (1 + i)

1m + + (1 + i)m2m + (1 + i)m1m

](last year)

+(1m

)(1 + i)m+1

m

[1 + (1 + i)

1m + + (1 + i)m2m + (1 + i)m1m

](2nd last year)

...

+(1m

)(1 + i)(n1)m+1

m

[1 + (1 + i)

1m + + (1 + i)m2m + (1 + i)m1m

](rst year)

=((1m

)(1 + i)1m + (

1m

)(1 + i)m+1

m + + ( 1m

)(1 + i)(n1)m+1

m )

)[1 ((1 + i) 1m )m1 (1 + i) 1m

]

=(1m

)(1 + i)1m

(1 + (1 + i)

mm + + (1 + i)(n1)

)[1 ((1 + i) 1m )m1 (1 + i) 1m

]

=(1m

)(1 + i)1m

(1 (1 + i)n1 (1 + i)1

)[

1 (1 + i)11 (1 + i) 1m

]

=(1m

)(1 + i)1m

(1 (1 + i)n1 (1 + i) 1m

)

=(1 + i)n 1

m v 1mi[(1 + i)

1m 1

]=

(1 + i)n 1m[1 v 1mi

]=

(1 + i)n 1m[1 (1 d) 1m

]=

(1 + i)n 1d(m)

=(

1mm

) (1 + i)

n 1d(m)

67

Basic Relationship 1 : 1 = d(m) a(m)n + vn


if the future value at time n, s(m)n , is discounted back to time 0, then you will have itspresent value, a(m)n

s(m)n vn =

[(1 + i)n 1

d(m)

] vn

=(1 + i)n vn vn

d(m)

=1 vnd(m)

= a(m)n

if the present value at time 0, a(m)n , is accumulated forward to time n, then you willhave its future value, s(m)n

a(m)n (1 + i)n =

[1 vnd(m)

](1 + i)n

=(1 + i)n vn(1 + i)n

d(m)

=(1 + i)n 1

d(m)

= s(m)n

Basic Relationship 3 :1

m a(m)n=

1

m s(m)n+

d(m)

m

Consider a loan of 1, to be paid back over n years with equal mthly payments of Pmade at the beginning of each mth of a year. An annual eective rate of interest, i, andnominal rate of discount, d(m), is used. The present value of this single payment loanmust be equal to the present value of the multiple payment income stream.

(P m) a(m)n i = 1

P =1

m a(m)n i Alternatively, consider a loan of 1, where the mthly discount due on the loan, (1) d(m)m ,

is paid at the beginning of each mth of a year for n years and the loan amount is paidback at time n.

In order to produce the loan amount at time n, payments of D at the beginning of eachmth of a year, for n years, will be made into an account that credits interest at an mthlyrate of discount d

(m)

m .

68


(D m) s(m)n i = 1

D =1

m s(m)n i The total mthly payment will be the discount payment and account payment:

d(m)

m+

1

m s(m)n i Note that

1

a(m)n i

=d(m)

1 vn (1 + i)n

(1 + i)n=

d(m)(1 + i)n

(1 + i)n 1

=d(m)(1 + i)n + d(m) d(m)

(1 + i)n 1 =d(m)[(1 + i)n 1] + d(m)

(1 + i)n 1

= i(m) +d(m)

(1 + i)n 1 = d(m) +

1

s(m)n

Therefore, a level mthly annuity payment on a loan is the same as making an mthlydiscount payment each mth of a year plus making mthly deposits in order to save forthe loan repayment.

Basic Relationship 4 : a(m)n =d

d(m) an , s(m)n =

d

d(m) sn

Consider payments of 1m

made at the beginning of every 1mth of year for the next n

years. Over a one-year period, payments of 1m made at the beginning of each mth period

will accumulate at the end of the year to a lump sum of(1m m

) s(m)1

. If this end-of-year lump sum exists for each year of the n-year annuity-immediate, then the presentvalue (at t = 0) of these end-of-year lump sums is the same as

(1mm) a(m)n :(

1mm

) a(m)n =

(1mm

) s(m)

1 an

a(m)n =

i

d(m) an

Therefore, the accumulated value (at t = n) of these end-of-year lump sums is the sameas(1mm) s(m)n : (

1mm

) s(m)n =

(1mm

) s(m)

1 sn

s(m)n =

i

i(m) sn

69

Basic Relationship 5: Due= Immediate (1 + i) 1m

a(m)n =

1 vnd(m)

=1 vn(

i(m)

1+ i(m)m

) = a(m)n (1 +

i(m)

m

)= a(m)n (1 + i)

1m

s(m)n =

(1 + i)n 1d(m)

=(1 + i)n 1(

i(m)

1+ i(m)m

) = s(m)n (1 +

i(m)

m

)= s(m)n (1 + i)

1m

An mthly annuitydue starts one mth of a year earlier than an mthly annuity-immediate andas a result, earns one mth of a year more interest, hence it will be larger.

Basic Relationship 6 : a(m)n =1m

+ a(m)n 1m

a(m)n = (

1m

) +[(1m

)v1m + + ( 1

m)v

nm2m + (

1m

)vnm1

m

]= (

1m

) + (1m

)v1m

[1 + v

1m + + v nm3m + v nm2m

]

= (1m

) + (1m

)v1m

1

(v

1m

)nm11 v 1m

= (1m

) +1

m(1 + i)1m

[1 vn 1m1 v 1m

]

= (1m

) +

1 vn 1mm[(1 + i) 1m 1

]

= (1m

) +

[1 vn 1mm i(m)

m

]

=1m

+1 vn 1m

i(m)

=1m

+ a(m)n 1m

An additional payment of 1m

at time 0 results in a(m)n 1m

becoming nm (= nm 1 + 1)payments that now commence at the beginning of each mth of a year which is a(m)n .

70

Basic Relationship 7 : s(m)n =1m

+ s(m)n 1m

s(m)n = (

1m

) +[(1m

)(1 + i)1m + + ( 1

m)(1 + i)

nm2m + (

1m

)(1 + i)nm1

m

]= (

1m

) + (1m

)(1 + i)1m

[1 + (1 + i)

1m + + (1 + i)nm3m + (1 + i)nm2m

]

= (1m

) + (1m

)(1 + i)1m

1

((1 + i)

1m

)nm11 (1 + i) 1m

= (1m

) +1

m v 1m

[1 (1 + i)n 1m1 (1 + i) 1m

]

= (1m

) +

1 (1 + i)n 1m

m[v

1m 1

]

= (1m

) +

(1 + i)n 1m 1

m[1 v 1m

]

= (1m

) +

(1 + i)n 1m 1m[1 (1 d) 1m

]

= (1m

) +

[(1 + i)n

1m 1

m d(m)m

]

=1m

+(1 + i)n

1m 1

d(m)

=1m

+ s(m)n 1m

An additional payment of 1m

at time n results in s(m)n 1m

becoming nm (= nm 1 + 1)payments that now commerce at the end of each mth of a year which is s(m)n .

71

Other Considerations

PerpetuityImmediate

payments of 1m are made at the end of every1mth of year forever.

years 0 1 2 ... ... n ...

conversion

periods 0 k 2k ... ... n x k ...

1

m m mm m m m m m m m m m m

1 1 1 1 1 1 1 1 1 1 1 1 1

the present value (at t = 0) of an mthly perpetuityimmediate, where the annual eective

72

rate of interest is i, shall be denoted as a(m) i and is calculated as follows:

a(m) i =(

1m

)v1mi + (

1m

)v2mi + + (

1m

)vm1

mi + (

1m

)vmmi (1st year)

+(1m

)vm+1

m

i + (1m

)vm+2

m

i + + (1m

)v2m1

m

i + (1m

)v2mm

i (2nd year)

...

+(1m

)v(n1)m+1

m

i + (1m

)v(n1)m+2

m

i + + (1m

)vnm1

m

i + (1m

)vnmm

i (nth year)

...

=(1m

)v1m

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](1st year)

+(1m

)vm+1

m

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](2nd year)

...

+(1m

)v(n1)m+1

m

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](nth year)

...

=((1m

)v1mi + (

1m

)vm+1

mi + + (

1m

)v(n1)m+1

m )i +

)[1 + v

1mi + + v

m2m

i + vm1

mi

]

=(1m

)v1m

i

(1 + v

mm

i + + v(n1)i + )[1 (v 1mi )m

1 v 1mi

]

=(1m

)1

(1 + i) 1m

(1 vi1 v1i

)[

1 v1i1 v 1mi

]

=(1m

)1

(1 + i) 1m

(1 0

1 v 1mi

)

=(1m

)(

1(1 + i)

1m 1

)

=1

m[(1 + i)

1m 1

]=

1i(m)

=(

1mm

) 1i(m)

one could also derive the above formula by simply substituting n = into the originalpresent value formula:

a(m) i =

1 vi(m)

=1 0i(m)

=1

i(m)

73

PerpetuityDue

payments of 1m

are made at the beginning of every 1mth of year forever.

years 0 1 2 ... ... n ...

conversion

periods 0 k 2k ... ... n x k ...

1

m m mm m m m m m m m m m

1 1 1 1 1 1 1 1 1 1 1 1

m

1

the present value (at t = 0) of an mthly perpetuitydue, where the annual eective rate of

74

interest is i, shall be denoted as a(m) i and is calculated as follows:

a(m) i =(

1m

) + (1m

)v1mi + + (

1m

)vm2

mi + (

1m

)vm1

mi (1st year)

+(1m

)vmm

i + (1m

)vm+1

m

i + + (1m

)v2m2

m

i + (1m

)v2m1

m

i (2nd year)

...

+(1m

)v(n1)m

m

i + (1m

)v(n1)m+1

m

i + + (1m

)vnm2

m

i + (1m

)vnm1

m

i (nth year)

...

=(1m

)[1 + v

1mi + + v

m2m

i + vm1

mi

](1st year)

+(1m

)vmmi

[1 + v

1mi + + v

m2m

i + vm1

mi

](2nd year)

...

+(1m

)v(n1)m

m

i

[1 + v

1m

i + + vm2

m

i + vm1

m

i

](nth year)

...

=((1m

) + (1m

)vmm

i + + (1m

)v(n1)m

m )i +

)[1 + v

1m

i + + vm2

m

i + vm1

m

i

]

=(1m

)(1 + v

mm

i + + v(n1)i + )[1 (v 1mi )m

1 v 1mi

]

=(1m

)(1 vi1 v1i

)[

1 v1i1 v 1mi

]

=(1m

)

(1 0

1 v 1mi

)

=(1m

)(

11 (1 d) 1m

)

=1

m[1 (1 d) 1m

]=

1d(m)

=(

1mm

) 1d(m)

one could also derive the above formula by simply substituting n = into the originalpresent value formula:

a(m) i =

1 vd(m)

=1 0d(m)

=1

d(m)

75

4.5 Continuous Annuities

payments are made continuously every year for the next n years (i.e. m =) the present value (at t = 0) of a continuous annuity, where the annual eective rate of interest

is i, shall be denoted as an i and is calculated as follows:

an i = n0

vtdt

= n0

etdt

= 1et

]n0

= 1

[en e0]

=1

[1 en]

=1 vni

one could also derive the above formula by simply substituting m = into one of the originalmthly present value formulas:

an i =a()n i

=1 vnii()

=1 vni

=a()n i =1 vnid()

=1 vni

the accumulated value (at t = n) of a continuous annuity, where the annual eective rate ofinterest is i, shall be denoted as sn i and is calculated as follows:

sn i = n0

(1 + i)ntdt

= n0

(1 + i)tdt

= n0

etdt

=1etdt

]n0

=1

[en e0]

=(1 + i)n 1

76

Basic Relationship 1 : 1 = an + vn


if the future value at time n, sn , is discounted back to time 0, then you will have itspresent value, an

sn vn =[(1 + i)n 1

] vn

=(1 + i)n vn vn

=1 vn

=an

if the present value at time 0, an , is accumulated forward to time n, then you will haveits future value, sn

an (1 + i)n =[1 vn

](1 + i)n

=(1 + i)n vn(1 + i)n

=(1 + i)n 1

=sn

Basic Relationship 3 :1an

=1sn

+

Consider a loan of 1, to be paid back over n years with annual payments of P that arepaid continuously each year, for the next n years. An annual eective rate of interest,i, and annual force of interest, , is used. The present value of this single payment loanmust be equal to the present value of the multiple payment income stream.

P an i = 1

P =1

an i

Alternatively, consider a loan of 1, where the annual interest due on the loan, (1) , ispaid continuously during the year for n years and the loan amount is paid back at timen.

In order to produce the loan amount at time n, annual payments of D are paid continu-ously each year, for the next n years, into an account that credits interest at an annualforce of interest, .

77


D sn i = 1

D =1

sn i

The total annual payment will be the interest payment and account payment:

+1

sn i

Note that

1an i

=

1 vn (1 + i)n

(1 + i)n=

(1 + i)n

(1 + i)n 1

=(1 + i)n +

(1 + i)n 1 =[(1 + i)n 1] +

(1 + i)n 1

= +

(1 + i)n 1 = +1sn

Therefore, a level continuous annual annuity payment on a loan is the same as makingan annual continuous interest payment each year plus making level annual continuousdeposits in order to save for the loan repayment.

Basic Relationship 4 : an =i

an , sn = i

sn

Consider annual payments of 1 made continuously each year for the next n years. Overa one-year period, the continuous payments will accumulate at the end of the year toa lump sum of s 1 . If this end-of-year lump sum exists for each year of the n-yearannuity-immediate, then the present value (at t = 0) of these end-of-year lump sums isthe same as an :

an =s 1 an=

i

an

Therefore, the accumulated value (at t = n) of these end-of-year lump sums is the sameas sn :

sn =s 1 sn=

i

sn

78

Basic Relationship 5 :d

dts t = 1 + s t ,

d

dta t = 1 a t

First o, consider how the accumulated value (as at time t) of an annuity-immediatechanges from one payment period to the next:

s t + 1 =1 + s t (1 + i)=1 + s t + i s t

s t + 1 s t =1 + i s ts t =1 + i s t

Therefore, the annual change in the accumulated value at time t will simply be theinterest earned over the year, plus the end-of-year payment that was made.

For an annuity-due, the annual change in the accumulated value will be:

s t + 1 =1 (1 + i) + s t (1 + i)=1 (1 + i) + s t + i s t

s t + 1 s t =1 (1 + i) + i s ts t =1 (1 + i) + i s t

Note that, in general, a continuous annuity can be expressed as:

s t + h =1 h0

(1 + i)htdt+ s t (1 + i)h

The change in the accumulated value over a period of h is then:

s t + h s t =1 h0

(1 + i)htdt+ s t [(1 + i)h 1]

The derivative with respect to time of the accumulated value can be dened as:d

dts t = lim

h 0s t + h s t

h

= limh 0

1 h0 (1 + i)htdth

+ limh 0

s t [(1 + i)h 1]h

= limh 0

1 ddh h0(1 + i)htdtddh h

+ limh 0

[(1 + i)t+h (1 + i)t]h (1 + i)t s t

=1 + s t The derivative with respect to time of the present value can be dened as:

d

dta t =

d

dt

(s t vt

)=(

d

dts t

) vt + s t

(d

dtvt)

=(1 + s t ) vt + s t (vt ln[v])

=(vt + a t

)+ a t ()

=1 a t

79

4.6 Basic Varying Annuities

in this section, payments will now vary; but the interest conversion period will continue tocoincide with the payment frequency

3 types of varying annuities are detailed in this section:

(i) payments varying in arithmetic progression

(ii) payments varying in geometric progression

(iii) other payment patterns

Payments Varying In Arithmetic Progression

Annuity-Immediate

An annuity-immediate is payable over n years with the rst payment equal to P and eachsubsequent payment increasing by Q. The time line diagram below illustrates the abovescenario:

0 1 2 ... n - 1 n

P P + (1)Q ... P + (n-2)Q P + (n-1)Q

80

The present value (at t = 0) of this annual annuityimmediate, where the annual eectiverate of interest is i, shall be calculated as follows:

PV0 = [P ]v + [P + Q]v2 + + [P + (n 2)Q]vn1 + [P + (n 1)Q]vn= P [v + v2 + + vn1 + vn] + Q[v2 + 2v3 + (n 2)vn1 + (n 1)vn]= P [v + v2 + + vn1 + vn] + Qv2[1 + 2v + (n 2)vn3 + (n 1)vn2]= P [v + v2 + + vn1 + vn] + Qv2 d

dv[1 + v + v2 + + vn2 + vn1]

= P an i + Qv2d

dv[an i ]

= P an i + Qv2d

dv

[1 vn1 v

]

= P an i + Qv2[(1 v) (nvn1) (1 vn) (1)

(1 v)2]

= P an i +Q

(1 + i)2

[nvn(v1 1) + (1 vn)(i/1 + i)2

]

= P an i + Q[(1 vn) nvn1 nvn

i2

]

= P an i + Q[(1 vn) nvn(v1 1)

i2

]

= P an i + Q[(1 vn) nvn(1 + i 1)

i2

]

= P an i + Q

(1 vn)i

nvn(i)i

i

= P an i + Q[an i nvn

i

]

The accumulated value (at t = n) of an annuityimmediate, where the annual eective rateof interest is i, can be calculated using the same approach as above or calculated by usingthe basic principle where an accumulated value is equal to its present value carried forwardwith interest:

FVn = PV0 (1 + i)n

=(P an i + Q

[an i nvn

i

])(1 + i)n

= P an i (1 + i)n + Q[an i (1 + i)n nvn (1 + i)n

i

]

= P sn i + Q[sn i n

i

]

81

Let P = 1 and Q = 1. In this case, the payments start at 1 and increase by 1 every yearuntil the nal payment of n is made at time n.

0 1 2 ... n - 1 n

1 2 ... n - 1 n

The present value (at t = 0) of this annual increasing annuityimmediate, where the annualeective rate of interest is i, shall be denoted as (Ia)n i and is calculated as follows:

(Ia)n i = (1) an i + (1) [an i nvn

i

]=

1 vni

+an i nvn

i

=1 vn + an i nvn

i

=an i nvn

i

The accumulated value (at t = n) of this annual increasing annuityimmediate, where theannual eective rate of interest is i, shall be denoted as (Is)n i and can be calculated usingthe same general approach as above, or alternatively, by simply using the basic principlewhere an accumulated value is equal to its present value carried forward with interest:

(Is)n i = (Ia)n i (1 + i)n

=(an i nvn

i

) (1 + i)n

=an i (1 + i)n nvn (1 + i)n

i

=sn i n

i

82

Let P = n and Q = 1. In this case, the payments start at n and decrease by 1 every yearuntil the nal payment of 1 is made at time n.

0 1 2 ... n - 1 n

n n - 1 ... 2 1

The present value (at t = 0) of this annual decreasing annuityimmediate, where the annualeective rate of interest is i, shall be denoted as (Da)n i and is calculated as follows:

(Da)n i = (n) an i + (1) [an i nvn

i

]= n 1 v

n

i an i nv

n

i

=n nvn an i + nvn

i

=n an i

i

The accumulated value (at t = n) of this annual decreasing annuityimmediate, where theannual eective rate of interest is i, shall be denoted as (Ds)n i and can be calculated byusing the same general approach as above, or alternatively, by simply using the basic principlewhere an accumulated value is equal to its present value carried forward with interest:

(Ds)n i = (Da)n i (1 + i)n

=(n an i

i

) (1 + i)n

=n (1 + i)n sn i

i

83

Annuity-DueAn annuity-due is payable over n years with the rst payment equal to P and each subsequentpayment increasing by Q. The time line diagram below illustrates the above scenario:

0 1 2 ... n - 1 n

P P + (1)Q P + (2)Q ... P + (n-1)Q

84

The present value (at t = 0) of this annual annuitydue, where the annual eective rate ofinterest is i, shall be calculated as follows:

PV0 = [P ] + [P +Q]v + + [P + (n 2)Q]vn2 + [P + (n 1)Q]vn1= P [1 + v + + vn2 + vn1] +Q[v + 2v2 + (n 2)vn2 + (n 1)vn1]= P [1 + v + + vn2 + vn1] +Qv[1 + 2v + (n 2)vn3 + (n 1)vn2]= P [1 + v + + vn2 + vn1] +Qv d

dv[1 + v + v2 + + vn2 + vn1]

= P an i + Qvd

dv[an i ]

= P an i + Qvd

dv

[1 vn1 v

]

= P an i + Qv[(1 v) (nvn1) (1 vn) (1)

(1 v)2]

= P an i +Q

(1 + i)

[nvn(v1 1) + (1 vn)(i/1 + i)2

]

= P an i + Q[(1 vn) nvn1 nvn

i2

] (1 + i)

= P an i + Q[(1 vn) nvn(v1 1)

i2

] (1 + i)

= P an i + Q[(1 vn) nvn(1 + i 1)

i2

] (1 + i)

= P an i + Q

(1 vn)i

nvn(i)i

i

(1 + i)


i

] (1 + i)


d

]

This present value of the annual annuity-due could also have been calculated using the basicprinciple that since payments under an annuity-due start one year earlier than under anannuity-immediate, the annuitydue will earn one more year of interest and thus, will begreater than an annuity-immediate by (1 + i):

PV due0 = PVimmediate0 (1 + i)

=(P an i +Q

[an i nvn

i

]) (1 + i)

= (P an i) (1 + i) + Q[an i nvn

i

] (1 + i)


d

]

85

The accumulated value (at t = n) of an annuitydue, where the annual eective rate ofinterest is i, can be calculated using the same general approach as above, or alternatively,calculated by using the basic principle where an accumulated value is equal to its presentvalue carried forward with interest:

FVn = PV0 (1 + i)n

=(P an i + Q

[an i nvn

d

])(1 + i)n

= P an i (1 + i)n + Q[an i (1 + i)n nvn (1 + i)n

d

]

= P sn i + Q[sn i n

d

]

86

Let P = 1 and Q = 1. In this case, the payments start at 1 and increase by 1 every yearuntil the nal payment of n is made at time n 1.

0 1 2 ... n - 1 n

1 2 3 ... n

The present value (at t = 0) of this annual increasing annuitydue, where the annual eectiverate of interest is i, shall be denoted as (Ia)n i and is calculated as follows:

(Ia)n i = (1) an i + (1) [an i nvn

d

]=

1 vnd

+an i nvn

d

=1 vn + an i nvn

d

=an i nvn

d

The accumulated value (at t = n) of this annual increasing annuityimmediate, where theannual eective rate of interest is i, shall be denoted as (Is)n i and can be calculated usingthe same approach as above or by simply using the basic principle where an accumulatedvalue is equal to its present value carried forward with interest:

(Is)n i = (Ia)n i (1 + i)n

=(an i nvn

d

) (1 + i)n

=an i (1 + i)n nvn (1 + i)n

d

=sn i n

d

87

Let P = n and Q = 1. In this case, the payments start at n and decrease by 1 every yearuntil the nal payment of 1 is made at time n.

0 1 2 ... n - 1 n

n n - 1 n - 2 ... 1

The present value (at t = 0) of this annual decreasing annuitydue, where the annual eectiverate of interest is i, shall be denoted as (Da)n i and is calculated as follows:

(Da)n i = (n) an i + (1) [an i nvn

d

]= n 1 v

n

d an i nv

n

d

=n nvn an i + nvn

d

=n an i

d

The accumulated value (at t = n) of this annual decreasing annuitydue, where the annualeective rate of interest is i, shall be denoted as (Ds)n i and can be calculated using thesame approach as above or by simply using the basic principle where an accumulated valueis equal to its present value carried forward with interest:

(Ds)n i = (Da)n i (1 + i)n

=(n an i

d

) (1 + i)n

=n (1 + i)n sn i

d

Basic Relationship 1 : an = i (Ia)n + nvnConsider an nyear investment where 1 is invested at the beginning of each year. The presentvalue of this multiple payment income stream at t = 0 is an .

Alternatively, consider a nyear investment where 1 is invested at the beginning of eachyear and produces increasing annual interest payments progressing to n i by the end of thelast year with the total payments (n 1) refunded at t = n.

The present value of this multiple payment income stream at t = 0 is i (Ia)n + nvn.

Note that (Ia)n =an n vn

i an = i (Ia)n + nvn. Therefore, the present value

of both investment opportunities are equal.

88

Payments Varying In Geometric Progression

Annuity-Immediate

An annuity-immediate is payable over n years with the rst payment equal to 1 and eachsubsequent payment increasing by (1+k). The time line diagram below illustrates the abovescenario:

0 1 2 ... n - 1 n

1 1(1+k) ... 1(1+k) 1(1+k) n-2 n-1

The present value (at t = 0) of this annual geometrically increasing annuityimmediate,where the annual

theory of interest, 2nd ed - kellison, stephen

Documents

unknown rate

basic varying annuities

general varying annuities

basic annuities

general annuities

analysis of annuities

eective rate of discount

nominal rate