the measurement of interest

The measurement of interest

Stephen G. Kellison(edited by Gunardi)

March 20, 2012

Stephen G. Kellison (edited by Gunardi) The measurement of interest

Introduction

Companies/people needing money, borrowers

Loanable funds, lenders

The compensation that a borrower of capital pays to lender ofcapital for its use is called interest.

Farmer A may lend a tractor to Farmer B for use in harvestingB’s wheat crop in return for a percentage of the wheatharvested.

The tractor is capital and the portion of wheat that B gives toA is interest.

For almost all applications, both capital and interest areexpressed i terms of money.

Definition of an accumulation function

An accumulation function a(t) is the accumulated value at timet ≥ 0 of an original investment of 1.

The properties of the accumulation function:

a(0) = 1,

a(t) is generally an increasing function,

If interest accrues continuously, as is usually the case, thefunction will be continuous

a(0) = 1,

Definition of an amount function

An amount function A(t) is the accumulated value at time t ≥ 0of an original investment of k.

A(0) = k ,

A(t) = ka(t) is generally an increasing function,

The amount of interest earned during the nth period from the dateof investment: In = A(n) − A(n − 1) for n ≥ 1.

A(0) = k ,

Definition of the effective rate of interest (1)

The effective rate of interest i is the amount of money that oneunit invested at the beginning of a period will earn during theperiod, where interest is paid at the end of the period.

In terms of the accumulation function:

i = a(1) − a(0) (1)

a(1) = 1 + i (2)

The effective rate of interest i is the amount of money that oneunit invested at the beginning of a period will earn during theperiod, where interest is paid at the end of the period.

i = a(1) − a(0) (1)

a(1) = 1 + i (2)

The effective rate of interest i is the ratio of the amount ofinterest earned during the period to the amount of principalinvested at the beginning of the period.

In terms of the amount function:

i =(1 + i) − 1

a(1) − a(0)

A(1) − A(0)

I1A(0)

In the nth period:

in =A(n) − A(n − 1)

A(n − 1)=

InA(n − 1)

for n ≥ 1. (4)

i =(1 + i) − 1

a(1) − a(0)

A(1) − A(0)

I1A(0)

In the nth period:

in =A(n) − A(n − 1)

A(n − 1)=

InA(n − 1)

for n ≥ 1. (4)

i =(1 + i) − 1

a(1) − a(0)

A(1) − A(0)

I1A(0)

In the nth period:

in =A(n) − A(n − 1)

A(n − 1)=

InA(n − 1)

for n ≥ 1. (4)

Definition of simple interest

The accumulated value of 1 at the end of the first period is 1 + i ,at the end of the first period is 1 + 2i , etc. The accruing ofinterest according to this pattern is called simple interest.

a(t) = 1 + it for t ≥ 0. (5)

Let i be the rate of compound interest, the effective rate ofinterest for the nth period:

in =a(n) − a(n − 1)

a(n − 1)=

(1 + in) − (1 + i(n − 1))

1 + i(n − 1)

1 + i(n − 1)for n ≥ 1. (6)

a(t) = 1 + it for t ≥ 0. (5)

in =a(n) − a(n − 1)

a(n − 1)=

(1 + in) − (1 + i(n − 1))

1 + i(n − 1)

1 + i(n − 1)for n ≥ 1. (6)

a(t) = 1 + it for t ≥ 0. (5)

in =a(n) − a(n − 1)

a(n − 1)=

(1 + in) − (1 + i(n − 1))

1 + i(n − 1)

1 + i(n − 1)for n ≥ 1. (6)

The properties of a(t) under simple interest

An initial investment of one unit over t + s periods is equal to theamount of interest earned over t periods plus the amount ofinterest over s periods.

a(t + s) = a(t) + a(s) − 1 for t ≥ 0 and t ≥ 0. (7)

The properties of a(t) under simple interest

An initial investment of one unit over t + s periods is equal to theamount of interest earned over t periods plus the amount ofinterest over s periods.

a(t + s) = a(t) + a(s) − 1 for t ≥ 0 and t ≥ 0. (7)

Example 1.1

Find the accumulated value of 2000 invested for four years, if therate of simple interest is 0.08 annum.

The answer:Given k = 2000, n = 4, i = 0.08A(4) =?

A(n) = ka(n) = k(1 + in) (8)

A(4) = 2000 ∗ (1 + 0.08 ∗ 4) = 2640 (9)

Example 1.1

Find the accumulated value of 2000 invested for four years, if therate of simple interest is 0.08 annum.

A(n) = ka(n) = k(1 + in) (8)

A(4) = 2000 ∗ (1 + 0.08 ∗ 4) = 2640 (9)

Definition of compound interest

The accumulated value of 1 at the end of the first period is 1 + i ,this balance of 1 + i can be as principal at the beginning of thesecond period. and will earn interest of i(1 + i) during the secondperiod. The balance at the end of the second period is(1 + i) + i(1 + i) = (1 + i)2, etc. The accruing of interestaccording to this pattern is called compound interest.

a(t) = (1 + i)t for t ≥ 0. (10)

in =a(n) − a(n − 1)

a(n − 1)=

(1 + i)n − (1 + i)(n−1)

(1 + i)(n−1)

=(1 + i) − 1

1= i for n ≥ 1. (11)

a(t) = (1 + i)t for t ≥ 0. (10)

in =a(n) − a(n − 1)

a(n − 1)=

(1 + i)n − (1 + i)(n−1)

(1 + i)(n−1)

=(1 + i) − 1

1= i for n ≥ 1. (11)

a(t) = (1 + i)t for t ≥ 0. (10)

in =a(n) − a(n − 1)

a(n − 1)=

(1 + i)n − (1 + i)(n−1)

(1 + i)(n−1)

=(1 + i) − 1

1= i for n ≥ 1. (11)

The properties of a(t) under compound interest

An initial investment of one unit over t + s periods is equal to theamount of interest earned if the investment is terminated at theend of t periods and the accumulated value at that point isimmediately reinvested for an additional s periods.

a(t + s) = a(t)a(s) for t ≥ 0 and t ≥ 0. (12)

The properties of a(t) under compound interest

An initial investment of one unit over t + s periods is equal to theamount of interest earned if the investment is terminated at theend of t periods and the accumulated value at that point isimmediately reinvested for an additional s periods.

a(t + s) = a(t)a(s) for t ≥ 0 and t ≥ 0. (12)

Example 1.2

Find the accumulated value of 2000 invested for four years, if therate of compound interest is 0.08 annum.

A(n) = ka(n) = k(1 + i)n (13)

A(4) = 2000 ∗ (1 + 0.08)4 = 2720.98 (14)

Example 1.2

Find the accumulated value of 2000 invested for four years, if therate of compound interest is 0.08 annum.

A(n) = ka(n) = k(1 + i)n (13)

A(4) = 2000 ∗ (1 + 0.08)4 = 2720.98 (14)

Definition of present value

It is often necessary to determine how much a person must investinitially so that the balance will be 1 at the end of one period. Theanswer is (1 + i)−1 = ν. The term ν is often called a discountfactor. The term νt is said to be the present value of 1 to be paidat the end of t periods.

In terms of the discount function:

a−1(t) =1

1 + itsimple interest (15)

a−1(t) =1

(1 + i)t= νt compound interest. (16)

Definition of present value

It is often necessary to determine how much a person must investinitially so that the balance will be 1 at the end of one period. Theanswer is (1 + i)−1 = ν. The term ν is often called a discountfactor. The term νt is said to be the present value of 1 to be paidat the end of t periods.

a−1(t) =1

1 + itsimple interest (15)

a−1(t) =1

(1 + i)t= νt compound interest. (16)

Example 1.3

Find the amount must be invested at a rate of simple interest of0.09 per annum in order to accumulate 1000 at the end of threeyears.

The answer:Given A(3) = 1000, n = 3, i = 0.09k =?

A(n) = ka(n) = k(1 + in) (17)

k =A(n)

1 + in(18)

1 + 0.09 ∗ 3= 787.40 (19)

Example 1.3

Find the amount must be invested at a rate of simple interest of0.09 per annum in order to accumulate 1000 at the end of threeyears.

A(n) = ka(n) = k(1 + in) (17)

k =A(n)

1 + in(18)

1 + 0.09 ∗ 3= 787.40 (19)

Example 1.4

Find the amount must be invested at a rate of compound interestof 0.09 per annum in order to accumulate 1000 at the end of threeyears.

A(n) = ka(n) = k(1 + i)n (20)

k =A(n)

(1 + i)n= kνn (21)

= 1000 ∗ (1 + 0.09)−3 = 772.18 (22)

Example 1.4

Find the amount must be invested at a rate of compound interestof 0.09 per annum in order to accumulate 1000 at the end of threeyears.

A(n) = ka(n) = k(1 + i)n (20)

k =A(n)

(1 + i)n= kνn (21)

= 1000 ∗ (1 + 0.09)−3 = 772.18 (22)

Definition of the effective rate of discount

The effective rate of discount d is a measure of interest paid at thebeginning of the period.

d =A(1) − A(0)

I1A(1)

In the nth period:

dn =A(n) − A(n − 1)

InA(n)

for n ≥ 1. (24)

d =A(1) − A(0)

I1A(1)

In the nth period:

dn =A(n) − A(n − 1)

InA(n)

for n ≥ 1. (24)

d =A(1) − A(0)

I1A(1)

In the nth period:

dn =A(n) − A(n − 1)

InA(n)

for n ≥ 1. (24)

Definition of present value under d

From the basic definition of i as the ratio of the amount ofinterest(discount) to the principal,

1 − d(25)

1 + i= iν = 1 − ν (26)

a−1(t) = 1 − dt for 0 ≤ t < 1/d simple interest (27)

a−1(t) = νt = (1 − d)t for t ≥ 0 compound interest. (28)

Definition of present value under d

From the basic definition of i as the ratio of the amount ofinterest(discount) to the principal,

1 − d(25)

1 + i= iν = 1 − ν (26)

a−1(t) = 1 − dt for 0 ≤ t < 1/d simple interest (27)

a−1(t) = νt = (1 − d)t for t ≥ 0 compound interest. (28)

Example 1.5

Find the amount must be invested at a rate of simple discount of0.09 per annum in order to accumulate 1000 at the end of threeyears.

The answer:Given A(3) = 1000, n = 3, d = 0.09k =?

A(n) = ka(n) (29)

k =A(n)

a(n)= A(n) ∗ a−1(n) = A(n) ∗ (1 − dn)

= 1000 ∗ (1 − 0.09 ∗ 3) = 730 (30)

Example 1.5

Find the amount must be invested at a rate of simple discount of0.09 per annum in order to accumulate 1000 at the end of threeyears.

A(n) = ka(n) (29)

k =A(n)

a(n)= A(n) ∗ a−1(n) = A(n) ∗ (1 − dn)

= 1000 ∗ (1 − 0.09 ∗ 3) = 730 (30)

Example 1.6

Find the amount must be invested at a rate of compound discountof 0.09 per annum in order to accumulate 1000 at the end of threeyears.

A(n) = ka(n) (31)

k =A(n)

a(n)= A(n) ∗ a−1(n)

= A(n) ∗ νn = A(n) ∗ (1 − d)n

= 1000 ∗ (1 − 0.09)3 = 1000 ∗ (0.91)3 = 753.57 (32)

Example 1.6

Find the amount must be invested at a rate of compound discountof 0.09 per annum in order to accumulate 1000 at the end of threeyears.

A(n) = ka(n) (31)

k =A(n)

a(n)= A(n) ∗ a−1(n)

= A(n) ∗ νn = A(n) ∗ (1 − d)n

= 1000 ∗ (1 − 0.09)3 = 1000 ∗ (0.91)3 = 753.57 (32)

Definition of nominal rate of interest

The symbol for a nominal rate of interest payable m times perperiod is i (m),m > 1. It means a rate payable mthly, i.e. the rateof interest is i (m)/m for each mth of a period.

From the definition of equivalency:

1 + i = (1 +i (m)

m)m (33)

i = (1 +i (m)

m)m − 1 (34)

i (m) = m[(1 + i)1/m − 1]. (35)

Definition of nominal rate of interest

The symbol for a nominal rate of interest payable m times perperiod is i (m),m > 1. It means a rate payable mthly, i.e. the rateof interest is i (m)/m for each mth of a period.

1 + i = (1 +i (m)

m)m (33)

i = (1 +i (m)

m)m − 1 (34)

i (m) = m[(1 + i)1/m − 1]. (35)

Definition of nominal rate of discount

The symbol for a nominal rate of discount payable m times perperiod is d (m),m > 1. It means a rate payable mthly, i.e. the rateof discount is d (m)/m for each mth of a period.

1 − d = (1 − d (m)

m)m (36)

d = 1 − (1 − d (m)

m)m (37)

d (m) = m[1 − (1 − d)1/m] = m[1 − ν1/m]. (38)

Definition of nominal rate of discount

The symbol for a nominal rate of discount payable m times perperiod is d (m),m > 1. It means a rate payable mthly, i.e. the rateof discount is d (m)/m for each mth of a period.

1 − d = (1 − d (m)

m)m (36)

d = 1 − (1 − d (m)

m)m (37)

d (m) = m[1 − (1 − d)1/m] = m[1 − ν1/m]. (38)

The relationship between i (m) and d (m)

The following relationship holds, since both sides of the equationare equal to 1 + i ,

(1 +i (m)

m)m = (1 − d (p)

p)−p (39)

(1 +i (m)

m)m = (1 − d (m)

m)−m if m = p (40)

Proof:

(1 +i (m)

m)m = 1 + i = 1 +

1 − d(41)

= (1 − d)−1 = (1 − d (m)

m)−m. (42)

The following relationship holds, since both sides of the equationare equal to 1 + i ,

(1 +i (m)

m)m = (1 − d (p)

p)−p (39)

(1 +i (m)

m)m = (1 − d (m)

m)−m if m = p (40)

Proof:

(1 +i (m)

m)m = 1 + i = 1 +

1 − d(41)

= (1 − d)−1 = (1 − d (m)

m)−m. (42)

Analogous to formula,

m− d (p)

Proof:

Analogous to formula,

m− d (p)

Proof:

Example 1.7

Find the accumulated value of 5000 invested for five years at 0, 08per annum convertible quarterly.

The answer:Given k = 5000, n = 5,m = 4, i (m) = 0.08A(5) =?

A(n) = ka(n) = k(1 +i (m)

A(5) = 5000(1 +i (4)

4)4∗5

= 5000 ∗ (1 + 0.08/4)20

= 5000 ∗ (1.02)20 = 742.9727 (44)

Example 1.7

Find the accumulated value of 5000 invested for five years at 0, 08per annum convertible quarterly.

The answer:Given k = 5000, n = 5,m = 4, i (m) = 0.08A(5) =?

A(n) = ka(n) = k(1 +i (m)

A(5) = 5000(1 +i (4)

4)4∗5

= 5000 ∗ (1 + 0.08/4)20

= 5000 ∗ (1.02)20 = 742.9727 (44)

Example 1.8

Find the present value of 1000 to be paid at the end of six years at0, 06 per annum payable in advance and convertible semiannually.

The answer:Given A(6) = 1000, n = 6,m = 2, d (m) = 0.06k =?

A(n) = ka(n)

k =A(n)

a(n)= A(n) ∗ a−1(n)

= A(n) ∗ νn = A(n) ∗ (1 − d)n = (1 − d (m)

m)m∗n

= 1000 ∗ (1 − 0.06/2)2∗6

= 1000 ∗ (0.97)12 = 693.8424 (45)

Example 1.8

Find the present value of 1000 to be paid at the end of six years at0, 06 per annum payable in advance and convertible semiannually.

The answer:Given A(6) = 1000, n = 6,m = 2, d (m) = 0.06k =?

A(n) = ka(n)

k =A(n)

a(n)= A(n) ∗ a−1(n)

= A(n) ∗ νn = A(n) ∗ (1 − d)n = (1 − d (m)

m)m∗n

= 1000 ∗ (1 − 0.06/2)2∗6

= 1000 ∗ (0.97)12 = 693.8424 (45)

Example 1.9

Find the nominal rate of interest convertible quarterly which isequivalent to a nominal rate of discount of 0, 06 per annumconvertible monthly.

The answer:Given m = 4, p = 12, d (p) = 0.06i (m) =?

(1 +i (m)

m)m = (1 − d (p)

p)−p

(1 +i (4)

4)4 = (1 − d (12)

12)−12

(1 +i (4)

4)4 = (1 − 0.06

12)−12

i (4) = 4[(0.995)−3 − 1] = 0.0606 (46)

Example 1.9

Find the nominal rate of interest convertible quarterly which isequivalent to a nominal rate of discount of 0, 06 per annumconvertible monthly.

The answer:Given m = 4, p = 12, d (p) = 0.06i (m) =?

(1 +i (m)

m)m = (1 − d (p)

p)−p

(1 +i (4)

4)4 = (1 − d (12)

12)−12

(1 +i (4)

4)4 = (1 − 0.06

12)−12

i (4) = 4[(0.995)−3 − 1] = 0.0606 (46)

Definition of the force of interest

The force of interest is the measure of the intensity with whichinterest is operating at each moment of time, i.e. overinfinitesimally small interval of time.

The force of interest at time t:

δt =A′(t)

dtlogeA(t) (47)

=a′(t)

dtlogea(t) (48)

Definition of the force of interest

The force of interest is the measure of the intensity with whichinterest is operating at each moment of time, i.e. overinfinitesimally small interval of time.

The force of interest at time t:

δt =A′(t)

dtlogeA(t) (47)

=a′(t)

dtlogea(t) (48)

The relationship between δt and a(t)

The expression for the value of a(t) in terms of the function δt ,

a(t) = e∫ t0 δrdr (49)

Proof: ∫ t

0δrdr =

dtlogea(r)dr (50)

= logea(t) (51)

The relationship between δt and a(t)

The expression for the value of a(t) in terms of the function δt ,

a(t) = e∫ t0 δrdr (49)

Proof: ∫ t

0δrdr =

dtlogea(r)dr (50)

= logea(t) (51)

The relationship between δt and A(t)

The expression for the value of A(t) in terms of the function δt ,

A(t) = A(0)e∫ t0 δrdr (52)

A(t) − A(0) =

0A(r)δrdr (53)

Proof: ∫ t

0δrdr =

dtlogeA(r)dr = loge

A(0)(54)

0A(r)δrdr =

0A′(r)dr = A(t) − A(0) (55)

The relationship between δt and A(t)

The expression for the value of A(t) in terms of the function δt ,

A(t) = A(0)e∫ t0 δrdr (52)

A(t) − A(0) =

0A(r)δrdr (53)

Proof: ∫ t

0δrdr =

dtlogeA(r)dr = loge

A(0)(54)

0A(r)δrdr =

0A′(r)dr = A(t) − A(0) (55)

The relationship between δt = δ and a(t)

If the force of interest is constant over an interval of time, then theeffective rate of interest will also be constant over that interval,

e∫ n0 δrdr = enδ if δt = δ for 0 ≤ t ≤ n

= a(n) = (1 + i)n (56)

so that

eδ = 1 + i (57)

i = eδ − 1 (58)

δ = loge(1 + i) (59)

The relationship between δt = δ and a(t)

If the force of interest is constant over an interval of time, then theeffective rate of interest will also be constant over that interval,

e∫ n0 δrdr = enδ if δt = δ for 0 ≤ t ≤ n

= a(n) = (1 + i)n (56)

so that

eδ = 1 + i (57)

i = eδ − 1 (58)

δ = loge(1 + i) (59)

Definition of the force of discount

The force of discount δ′t bears a relationship to nominal andeffective rates of discount similar to the relationship that the forceof interest bears to nominal and effective rates of interest, δ′t = δt .

The proof:

δ′t = −ddt a−1(t)

a−1(t)(60)

=a−2(t) d

dt a(t)

a−1(t)(61)

=a−2(t)a(t)δt

a−1(t)(62)

= δt (63)

Definition of the force of discount

The force of discount δ′t bears a relationship to nominal andeffective rates of discount similar to the relationship that the forceof interest bears to nominal and effective rates of interest, δ′t = δt .

The proof:

δ′t = −ddt a−1(t)

a−1(t)(60)

=a−2(t) d

dt a(t)

a−1(t)(61)

=a−2(t)a(t)δt

a−1(t)(62)

= δt (63)

The force of interest under simple interest

The force of interest under simple interest is

δt =i

1 + it(64)

The proof:

δt =a′(t)

=ddt (1 + it)

1 + it=

1 + it(65)

The force of interest under simple interest

The force of interest under simple interest is

δt =i

1 + it(64)

The proof:

δt =a′(t)

=ddt (1 + it)

1 + it=

1 + it(65)

The force of interest under simple discount

The force of interest under simple discount is

δt = δ′t =d

1 − dtfor 0 ≤ t < 1/d (66)

The proof:

δt = δ′t = −ddt a−1(t)

a−1(t)

= −ddt (1 − dt)

1 − dt=

1 − dt(67)

The force of interest under simple discount

The force of interest under simple discount is

δt = δ′t =d

1 − dtfor 0 ≤ t < 1/d (66)

The proof:

δt = δ′t = −ddt a−1(t)

a−1(t)

= −ddt (1 − dt)

1 − dt=

1 − dt(67)

The relationship between δ and i (m) (1)

The nominal rate of interest convertible mthly:

i (m) = m[eδm − 1] (68)

proof:

[1 +i (m)

m]m = 1 + i = eδ

1 +i (m)

i (m) = m[eδm − 1] (69)

The nominal rate of interest convertible mthly:

i (m) = m[eδm − 1] (68)

proof:

[1 +i (m)

m]m = 1 + i = eδ

1 +i (m)

i (m) = m[eδm − 1] (69)

δ can be interpreted as the nominal rate of interest convertiblecontinuously:

limm→∞

i (m) = δ (70)

proof:

limm→∞

i (m) = limm→∞

m[eδm − 1]

= limm→∞

m]3 + ...]

= limm→∞

δ +δ2

3!m2+ ...] = δ (71)

δ can be interpreted as the nominal rate of interest convertiblecontinuously:

limm→∞

i (m) = δ (70)

proof:

limm→∞

i (m) = limm→∞

m[eδm − 1]

= limm→∞

m]3 + ...]

= limm→∞

δ +δ2

3!m2+ ...] = δ (71)

The relationship between δ and d (m) (1)

The nominal rate of discount convertible mthly:

d (m) = m[1 − e−δm ] (72)

proof:

[1 − d (m)

m]−m = (1 − d)−1 = ν−1 = 1 + i = eδ

1 − d (m)

m= e−

d (m) = m[1 − e−δm ] (73)

The nominal rate of discount convertible mthly:

d (m) = m[1 − e−δm ] (72)

proof:

[1 − d (m)

m]−m = (1 − d)−1 = ν−1 = 1 + i = eδ

1 − d (m)

m= e−

d (m) = m[1 − e−δm ] (73)

δ can be interpreted as the nominal rate of discount convertiblecontinuously:

limm→∞

d (m) = δ (74)

proof:

limm→∞

d (m) = limm→∞

m[1 − e−δm ]

= limm→∞

m− 1

m]3 − ...]

= limm→∞

δ − δ2

3!m2− ...] = δ (75)

δ can be interpreted as the nominal rate of discount convertiblecontinuously:

limm→∞

d (m) = δ (74)

proof:

limm→∞

d (m) = limm→∞

m[1 − e−δm ]

= limm→∞

m− 1

m]3 − ...]

= limm→∞

δ − δ2

3!m2− ...] = δ (75)

Example 1.10

Find the accumulated value of 1000 invested for ten years if theforce of interest is 0.05.

The answer:Given k = 1000, n = 10, δ = 0.05A(n) =?

A(n) = ka(n) = k ∗ eδn

A(10) = 1000 ∗ e0.05∗10

= 1000 ∗ e0.5 = 1648.721 (76)

Example 1.10

Find the accumulated value of 1000 invested for ten years if theforce of interest is 0.05.

The answer:Given k = 1000, n = 10, δ = 0.05A(n) =?

A(n) = ka(n) = k ∗ eδn

A(10) = 1000 ∗ e0.05∗10

= 1000 ∗ e0.5 = 1648.721 (76)

Varying interest

The first type of variation considered is a continuously varyingforce of interest,

a(t) = e∫ t0 δrdr (77)

The second type of variation considered involves changes in theeffective rate of interest over a period of time. Let in denote theeffective rate of interest during the nth period from the date ofinvestment,

a(t) = (1 + i1)(1 + i2)...(1 + it)

(1 + ik) (78)

Varying interest

The first type of variation considered is a continuously varyingforce of interest,

a(t) = e∫ t0 δrdr (77)

The second type of variation considered involves changes in theeffective rate of interest over a period of time. Let in denote theeffective rate of interest during the nth period from the date ofinvestment,

a(t) = (1 + i1)(1 + i2)...(1 + it)

(1 + ik) (78)

Example 1.11

Find the accumulated value of 1 at the end of n years if

δt =1

1 + t(79)

The answer:

a(n) = e∫ n0 δtdt

= e∫ n0

= e loge(1+n) = 1 + n (80)

Example 1.11

Find the accumulated value of 1 at the end of n years if

δt =1

1 + t(79)

The answer:

a(n) = e∫ n0 δtdt

= e∫ n0

= e loge(1+n) = 1 + n (80)

Example 1.12

Find the accumulated value of 1000 at the end of 15 years if theeffective rate of interest is 0.05 for the first 5 years, 0.045 for thesecond 5 years, and 0.04 for the third 5 years,.

The answer:Given k = 1000, n = 15,ik = 0.05 for k = 1, 2, ..., 5, ik = 0.045 for k = 5, 6, ..., 10,ik = 0.04 for k = 11, 12, ..., 15 A(15) =?

A(k) = ka(n) = k15∏k=1

(1 + ik)

A(10) = 1000 ∗5∏

(1 + 0.05) ∗10∏k=6

(1 + 0.045) ∗15∏

(1 + 0.04)

= 1000 ∗ (1.05)5 ∗ (1.045)5 ∗ (1.04)5 = 1648.721 (79)

Example 1.12

Find the accumulated value of 1000 at the end of 15 years if theeffective rate of interest is 0.05 for the first 5 years, 0.045 for thesecond 5 years, and 0.04 for the third 5 years,.

The answer:Given k = 1000, n = 15,ik = 0.05 for k = 1, 2, ..., 5, ik = 0.045 for k = 5, 6, ..., 10,ik = 0.04 for k = 11, 12, ..., 15 A(15) =?

A(k) = ka(n) = k15∏k=1

(1 + ik)

A(10) = 1000 ∗5∏

(1 + 0.05) ∗10∏k=6

(1 + 0.045) ∗15∏

(1 + 0.04)

= 1000 ∗ (1.05)5 ∗ (1.045)5 ∗ (1.04)5 = 1648.721 (79)

the measurement of interest

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