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1 Annuity

Annuity is defined as a sequence of n periodical payments, where n can be either finite or infinite.

In order to evaluate an annuity or compare annuities, we need to calculate their present value at

a common reference time point. To find the present value of an annuity is to sum up the present

value at the reference time of each payment.

1.1 Discrete Annuity

DotProduct Strategy is a generalized algebraic tool to evaluate present value of annuity. With

this tool, we can coherently generalize all formulas presented in chapter 2.

Outlines of DotProduct Strategy for Discrete Annuity:

1. Construct the dated cashflow diagram, and define the unit of time period based on the

payment frequency.

2. Find the effective interest rate per period for periods under consideration. For example, if

the assumed accumulation function a(t) is of the form (1+j)t, then we can use 1 + i(n)n n =1 + i

(m)

m

m, where n and m are the payment frequency and interestcrediting frequency

respectively.

3. Choose the appropriate reference time point, denoted by t0.

4. Based on the cashflow diagram, define an ncomponent vector, payment schedule, repre-

senting the periodical payment at each payment period.

P =

P0 P1 P2 . . . P n

, where Pk is the payment at time k.

5. Given t0, define another n-component vector, called discount schedule, representing the

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discount factor of corresponding payment at each payment period.

|t0 =

a(t0)a(0)

a(t0)a(1)

a(t0)a(2)

. . .a(t0)a(n)

6.

P Vt0{P} = P |t0 (1)

= P0a(t0)

a(0)+ P1

a(t0)

a(1)+ P2

a(t0)

a(2)+ + Pn

a(t0)

a(n)

Note:

a(t0)a(k) is the accumulation factor for the payment Pk if t0 > k. Otherwise

a(t0)a(k) is present-value

factor if t0 < k.

If a(t) = (1 + j)t, then we can rewrite P Vt0{P} as follow

P Vt0{ P} = P0(1 + j)t0 + P1(1 + j)

t01 + P2(1 + j)t02 + + Pn(1 + j)

t0n

Dot-product strategy separates the cashflow analysis and discounting mechanism.

1.1.1 Examples

Example 1.1.1 (Textbook Exercise 2.1.19) A loan of 10,000 is being repaid by 10 semian-

nual payments, with the first payment made onehalf year after the loan. The first 5 payments

are K each, and the final 5 are K + 200 each. What is K if i(2) = .06?

Soln:

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Let t0 = 0 and j =i(2)

2 = 0.03.

P =

0 K K K K+ 200 K + 200

=

0 K K K

+

0 0 200 200

= P1 + P2

|t0=0 =

1 1(1+j)

1(1+j)10

P V0{ P} = P |t0=0

= ( P1 + P2) |t0=0

= P1 |t0=0 +P2 |t0=0

10000 = Ka10|j + 2005|a5|j

Example 1.1.2 (Textbook Exercise 2.1.3) Smith makes deposits of 1000 on the last day of

each month in an account earning interest at rate i(12) = .12. The first deposit is January 31,

2000 and the final deposit is December 31, 2029. The accumulated account is used ot make

monthly payments ofY

starting January 31, 2030 with the final one on December 31, 2054.

Find Y.

Soln:

Note that there are 30 12 deposits and 25 12 withdrawals.

Denote Pin the payment schedule for deposits (cash-inflow) and Pout the payment schedule

for withdrawals (cash-outflow).

Since the deposits or withdrawals are made monthly, the unit of time period is month.

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Let t0 = 360 and effective rate per period j =i(12)

12 = 0.01.

Pin =

0 1000 1000 1000 0 0

Pout =

0 0 0 Y Y Y

|t0=360 = a(360)a(0)

a(360)

a(1)

a(360)

a(360)

a(360)

a(361)

a(360)

a(660)

=

(1 + j)360 (1 + j)359 1 1(1+j)

1(1+j)300

P Vt0{Pin} = P Vt0{

Pout} Pin |t0=360 =Pout |t0=360

1000s360|j = Y a300|j

Y =1000s360|j

a300|j

Remarks:

1. The annuity-immediate formula an|j is used to evaluate the present value of the uniform pay-

ment schedule at one period before the first non-zero payment. For example, the formula is

the dot-product of the following payment and discount schedule.

P =

0 1 1 . . . 1 1

and =

0 1 2 . . . n

where P and are n + 1component vectors.

Note that there are n non-zero payments in the payment schedule above.

2. The perpetuity-immediate formula a|j = limn an|j

3. The annuity-due formula an|j is used to evaluate the present value of the uniform payment

schedule at the period of the first non-zero payment. For example, the formula is the dot-

product of the following payment and discount schedule.

P =

1 1 1 . . . 1 0

and =

0 1 2 . . . n

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Note that there are n non-zero payments in the payment schedule above.

4. The perpetuity-due formula a|j = limn an|j

5. The k -deferred annuity-immediate k|an|j is used to evaluate the present value of the uniform

payment schedule at k +1 periods before the first non-zero payment. In other words, the first

non-zero payment is deferred k periods.

6. The k -deferred annuity-due k|an|j is used to evaluate the present value of the uniform payment

schedule at k periods before the first non-zero payment. In other words, the first non-zero

payment is deferred k periods.

7. The annuity-immediate formula sn|j is used to evaluate the present value of the uniform pay-

ment schedule at the period of the last non-zero payment. For example, the formula is the

dot-product of the following payment and discount schedule.

P =

0 1 1 . . . 1 1

and =

n 1n 2n . . . nn


Note that there are n non-zero payments.

8. The annuity-due formula sn|j is used to evaluate the present value of the uniform payment

schedule at one period after the last non-zero payment. For example, the formula is the

dot-product of the following payment and discount schedule.

P =

0 1 1 . . . 1 1

and =

n 1n 2n . . . nn


Note that there are n non-zero payments.

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1.1.2 More Examples

Example 1.1.3 (Textbook Example 2.9) A perpetuityimmediate pay X per year. Brain re-

ceives the first n payments, Colleen receives the next n payments, and Jeff receives the re-

maining payments. Brians share of the present value of the original perpetuity is 40%, and

Jeffs share is K. Calculate K.

Let P denote the payment schedule for the perpetuity-immediate, PB denote the payment

schedule for the portion that Brian receives, and PJ denote the portion that Jeff receives. Let

t0 = 0. Then

P = 0 X X X |t0=0 =

1 2 3

PV0{ P} = P |t0=0 = Xa|j

As for Brian, PB =

0 X X X

PV0{ PB} = PB |t0=0 = Xan|j

Xan|j = 0.4Xa|j

X1 n

j= 0.4

X

j

n = 0.6

Similarly, PJ =

0 0 0 X X

PV0{ PJ} = PJ |t0=0 = X2n|an|j = 2n X

j

= (n)2X

j= 0.36 PV0{ P}

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Example 1.1.4 (Textbook Exercise 2.3.6) Smith has 100,000 with which he buys a perpetu-

ity on January 1, 2005. Suppose that i = 0.045 and the perpetuity has annual payments

beginning January 1, 2006. The first three payments are 2000 each, the next three payments

are 2000(1+r) each, . . . , increasing forever by a factor of 1+r every three years. What is r?

Let t0 = 0. Then

P =

0 2000 2000 2000 2000(1 + r) 2000(1 + r) 2000(1 + r) 2000(1 + r)2 2000(1 + r)2

=

1 2 3

, where = 11.045 .

PV0{ P} = P = 100000

100000 = 2000 + 20002 + 20003 + 2000(1 + r)4 + 2000(1 + r)5 + 2000(1 + r)6 +

=

2000 + 20002 + 20003

+

2000(1 + r)4 + 2000(1 + r)5 + 2000(1 + r)6

+

= 2000

1 + + 2

+ 2000(1 + r)4

1 + + 2

+ 2000(1 + r)27

1 + + 2

+

=

1 + + 2

2000 + 2000(1 + r)4 + 2000(1 + r)27 +

= 1 + + 2

2000(1 + (1 + r)3 + (1 + r)26 + )

=

1 3

1

2000

k=0

k

, where = (1 + r)3.

100000 =

1 3

1

2000

1

1

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Example 1.1.5 (Textbook Exercise 2.2.15) A loan of 10,000 is made on January 1, 2005 at

an interest rate i(12) = .12. The loan calls for payments of 500 on the first day of April, July,

and October each year, with an additional fractional payment on the next scheduled payment

date after the final regular payment of 500. Find the date and amount of the additional

fractional payment.

Since the payments are made on the first day of April, July, and October each year, the unit

of year should be a quarter of year.

j =i(4)

4= 1.013 1 = 0.030301

P =

0 500 500 500 0 500 500

=

1 2 n

, where = 11+j .

PV0{ P} = P = 10000

10000 = 500 + 5002 + 5003 + 5005 + 5006 + 5007 + + 500n1 + 500n

= (500

+ 5002

+ 5003

) + (5005

+ 5006

+ 5007

) +

= 500( + 2 + 3) + 5004( + 2 + 3) + + 5004n(+ 2 + 3)

, where n is the number of years required to repay the loan

= 500a3|j + 5004a3|j + + 500

4na3|j

= 500a3|j[1 + 4 + + (4)n

]

10000 = 500a3|j

1 (4)n

1 4

n = 13.32388587

Assignment: Complete the example

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1.2 Continuous Annuity

When the annuity is continuous, few modifications are required to evaluate the present value of a

continuous annuity. First, as the counterpart of payment schedule, we define a payment function

h(t). The payment function h(t) represents the instantaneous rate of payment at time t. Think

of h(t) as a density function. The amount of payment made between [t, t + t) is approximately

equal to h(t)t. Of course, h(t) is not a density function since it does not integrate to 1. Next,

we define a discount function as (t; t0) =a(t0)a(t) , where t0 is a chosen reference time point. The

discount function (t; t0) is the counter part of the discount schedule. Furthermore, given the

force of interest (t), we know that the accumulation function a(t) = et

0 (s)ds is differentiable, and

thus continuous. The continuity of a(t) implies that (t; t0) is also continuous, except at some

time t such that a(t) = 0. The dot-product of two continuous functions on [0, ) is defined as

f g =0 f(x)g(x) dx . Therefore, P Vt0{h} = h =

0 h(t)(t; t0) dt.

Next, I would like to derive an expression for the present value evaluated at the reference time t0,

where 0 t0 < , given the payment function h(t) and the force of interest (t). Assuming that

the payment function h(t) is defined on [0, ), I can calculate the present value at time t0 of the

payment function h(t), i.e. PVt0{h} =0 h(t)(t; t0) dt, where

(t; t0) =a(t0)

a(t)=

expt0

0 (s) ds

expt

0 (s) ds

= exp

t00

(s) ds

t0

(s) ds

= exp

t0t

(s) ds

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1.2.1 Examples

Example 1.2.1 Define the payment function h(t) and the force of interest (t) as follow:

h(t) =

1000 for t 0

0 otherwise

(t) =

0.02t where 0 t < 3

0.025 where t 3

0 otherwise

Calculate the present value at time 0.

(t; 0) = exp

0t

(s) ds

= exp

t0

(s) ds

=

exp

0.01t2

where 0 t < 3

exp(0.09 0.025(t 3)) where t 3

1 otherwise

Then

PV0{h} =

0

h(t)(t; 0) dt

=

30

1000 exp

0.01t2

dt +

3

1000exp(0.09 0.025(t 3)) dt

Example 1.2.2 (SOA Exam FM Sample #21) Payments are made to an account at a con-

tinuous rate of(8k+tk), where0 t 10. Interest is credited at a force of interest (t) = 18+t .

After 10 years, the account is worth 20000. Calculate k.

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h(t) = 8k + tk = k(8 + t)

(t) =1

8 + t

t0 = 10

(t; t0 = 10) =a(10)

a(t)=

exp{100 (s)ds}

exp{t0 (s)ds}

= exp

10t

(s)ds

= exp

10t

1

8 + sds

= exp

ln(8 + s)|10t

= exp

ln

18

8 + t

=18

8 + t

P V10{h} = h =100

h(t)(t; t0 = 10)dt

20000 =

100

k(8 + t)18

8 + tdt = 180k

k =20000

180

Remarks:

1. When the payment function is of the following functional form, the annuity is said to be an

unit n-year continuous annuity.

h1(t) =

1 if 0 t n

0 otherwise

The present value at time 0 is denoted by an|; the present value at time n is denoted by sn|.


unit n-year continuous perpetuity.

h2(t) =

1 if t 0

0 otherwise

The present value at time 0 is denoted by a|.

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unit k-deferred n-year continuous annuity.

h3(t) =

1 if k t n + k

0 otherwise

The present value at time 0 is denoted by k|an|; the present value at time n + k is denoted by

k|sn|.


unit n-year increasing continuous annuity.

h4(t) =

t if 0 t n

0 otherwise

The present value at time 0 is denoted by Ian|; the present value at time n + k is denoted by

Isn|.


unit n-year increasing continuous perpetuity.

h5(t) =

t if t 0

0 otherwise

The present value at time 0 is denoted by Ia|.


unit n-year decreasing continuous annuity.

h6(t) =

n t if 0 t n

0 otherwise

The present value at time 0 is denoted by Dan|; the present value at time n + k is denoted

by Dsn|.

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unit n-year step-increasing continuous annuity.

h7(t) =

t if 0 t n

0 otherwise

The present value at time 0 is denoted by Ian|; the present value at time n is denoted by

Isn|.


unit n-year step-decreasing continuous annuity

h8(t) =

n t if 0 t n

0 otherwise

The present value at time 0 is denoted by Dan|; the present value at time n is denoted by

Dsn|.

2 Applications

We are going two important applications regarding annuity. One is reinvestment rate; the other is

loan repayment schemes. In the application of reinvestment rate, we are interested in the accumu-

lated value of $1 invested at time 0 if the interest is reinvested at another rate. As for the loan

repayment schemes, the aims are to understand different ways to pay off a loan and to calculate

the amount of payment as well as the yield rate with respect to each schemes In Chapter 3, we are

going to further study the loan/mortgage repayment system by decomposing each payment into

two parts principal reduction and interest.

2.1 Reinvestment Rate

Consider $1 deposited at time 0 with the compound-interest accumulation function a(t) = (1 +j)t,

where j is the effective rate per period. One of the implicit assumptions associated with such an

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accumulation function is that the interest is automatically reinvested at the same rate from which

it was generated.

Let relax this particular assumption. Suppose that the interest is reinvested at different rate j,

instead of j. Image that at the time you open an bank account, which I will call the principal

account, and so is another invisible account, called interest account. The interest account collects

the amount of interest generated by the principal account. While the per-period interest rate of

the principal account is j, the per-period of interest rate of the interest account is j. In other

words, the interest, which generated by the rate j, is reinvested at the rate j. Note that j is call

the investment rate, whereas j is called reinvestment rate.

We are interested in the accumulated value at time n of $1 deposited at time 0. Remember the

interest account exists but being invisible. Therefore, the accumulated value is the sum of the

balances of the principal and interest account.

Let A(t) denote the accumulated value at time t of $1 deposited at time 0, AP(t) denote the balance

of the principal account at time t and AI(t) denote the balance of the interest account at time t.

At time 0, $1 is deposited in the principal account. AP(0) = 1 and AI(0) = 0 .

At time 1, the amount of interest generated by $1 in the principal account is j, but $j is de-

posited (reinvested) into the interest account instead of being credited in the principal account.

AP(1) = 1 and AI(1) = j .

At time 2, the amount of interest generated by $1 in the principal account is also j (why?). Again

$j is deposited (reinvested) into the interest account.

AP(2) = 1 and AI(2) = j(1 + j) + j .

At time 3, the amount interest generated by A(2) is again $j, which is deposited (reinvested) into

the interest account.

AP(3) = 1 and AI(3) = j(1 + j)2 + j(1 + j) + j .

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...

At time n, AP(n) = 1 and AI(n) = j(1 + j)n1 + j(1 + j)n2 + + j(1 + j) + j .

Therefore, at time n, the accumulated value is A(n) = AP(n) + AI(n) = 1 + jsn|j .

The average yield rate or rate of return per period j is then defined as A(n) = A(0)(1 + j)n For

example, if j = j, then we know that A(n) = (1 +j)n and j = j. On the other hand, when j = j,

A(n) = AP(n) + AI(n) = A(0)(1 + j)n 1 + jsn|j = (1 + j)

n.

Remarks:

1. The account balance of principal account is always $1.

2. the cash-flow diagram for the interest account can be represented by PI =

0 j j . . . j

.

3. j is always between j and j. (Q # 2.4.10)

2.2 Loan Repayment Scheme and the Yield Rate

Consider the following three schemes which a loan of L at the interest (investment) rate of j can

be repaid:

Present Value Scheme A sequence of n periodical payments is used to pay off the loan, which

is the most common repayment scheme in the financial industry. It is understood implicitly

that payments and interest are calculated based on the same investment rate. For instance,

the periodical level mortgage payment P is calculated L = P an|j, where n is called the

amortization period, and j is called the mortgage/loan rate. Note that each payment usually

contains a portion to reduce the principal and a portion to pay off the interest. In Chapter 2,

we study how to calculate the size of each payment under some specifications. In Chapter 3,

we study how to decompose of these payments into principalreduction and interest portion.

SinkFund Scheme A borrower pays creditors the interest, $Lj, on the loan at the end of each

period, and a lumpsum equal to the original loan amount L at time n. In addition, the

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borrow also makes a sequence of n periodical payments into another account, called sinking

fund, so that the account balance of sinkingfund equals exactly the original loan amount at

time n. For example, L = Pssn|j , where Ps is the deposit (payment) to the sinkingfund,

and j is the interest rate of sinkingfund. A borrower chooses the sinkingfund scheme if

there is a more favorable investment option, i.e. j > j.

LumpSum Scheme A borrower pays off the loan with a lump sum of L(1 + j)n at time n.

2.3 Borrowers Payments Dedicated to Loan

Someday, you are going to be a borrower. It is then useful to choose the loan repayment scheme

based on your ability to pay. In this section, I would like to illustrate the total amount of payment(s)

that is dedicated to the loan at the end of each period under each repayment scheme.

Set L = 10000, n = 10, and j = 5% = j.

Present Value Scheme

L = P an|j 10000 = P a10|.05

P =10000

a10|.05= 1295.04575

Therefore, $1,295.05 is dedicated toward the loan at end of each period.

SinkingFund Scheme

P = Lj P = 10000 .05 = 500

L = Pssn|j 10000 = Pss10|.05

Ps =10000

s10|.05= 795.04575

Therefore, the amount of payments dedicated toward the loan is $1,295.04575 (= P + Ps),

which is the same amount under the PV scheme.

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Now suppose the interest rate of sinking fund is slightly lower than the loan rate. Lets set L =

10000, n = 10, j = 5%, and j = 3%.


L = P an|j 10000 = P a10|.05

P =10000

a10|.05= 1295.04575

Therefore, $1,295.05 is dedicated toward the loan at end of each period.

SinkingFund Scheme

P = Lj P = 10000 .05 = 500

L = Pssn|j 10000 = Pss10|.03

Ps =10000

s10|.03= 872.30507

Therefore, the amount of payments dedicated toward the loan is $1,372.30507, which is higher

than the payment under the PV scheme.

2.4 Lenders Yield Rate

Keep in mind that the repayment scheme determine how the loan is repaid, and what amount

the per-period payment is. How well the lend reinvests those payments plays a role on the yield

rate as well. Lets consider the payment specified in each scheme, and the goal is to calculate the

lenders average yield rate per period j. Remember that the initial investment for the lender is

A(0) = L. In order to calculate the average yield rate j, we need the accumulated value at time n

of the lenders initial investment. All calculations essentially involves the following 3 steps.

1. Calculate the size of payments which the borrower is supposed to pay under the specifications,

i.e. 10 level payments per period for 10 periods at a loan rate of 5% .

2. Calculate the account balance of lenders account at the end of loan periods, i.e. Alender(n)

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3. Calculate the average yield rate per period over nperiod based on Alender(n) = Alender(0)(1+

j)n.

Denote the loan rate j, and the reinvestment rate j. Consider the case of j = j. For illustration

purpose, lets set L = 10000, n = 10, and j = .05 = j.


L = P an|j 10000 = P a10|.05

P =10000

a10|.05= 1295.04575

Alender(n) = P sn|j Alender(10) = 1295.04575s10|.05

= 16288.94627

Alender(n) = Alender(0)(1 + j)n 16288.94627 = 10000(1 + j)10

j = .05

SinkingFund Scheme

P = L j P = 10000 .05

= 500

Alender(n) = L + P sn|j Alender(10) = 10000 + 500s10|.05

= 16288.94627

Alender(n) = Alender(0)(1 + j)10 j = .05

LumpSum Scheme

Alender(n) = Alender(0)(1 + j)n Alender(10) = 10000(1.05)

10

Alender(n) = Alender(0)(1 + j)n j = .05 = j

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Now suppose that L = 10000, n = 10, j = .05 and j = 0.3.


L = P an|j 10000 = P a10|.05

P =

10000

a10|.05 = 1295.04575

Alender(n) = P sn|j Alender(10) = 1295.04575s10|.03

= 14846.24818

Alender(n) = Alender(0)(1 + j)n 14846.24818 = 10000(1 + j)10

j = .040307361

SinkingFund Scheme

P = L j P = 10000 .05

= 500

Alender(n) = L + P sn|j Alender(10) = 10000 + 500s10|.03

= 15731.93966

Alender(n) = Alender(0)(1 + j)10 15731.93966 = 10000(1 + j)10

j = 0.046353008

LumpSum Scheme

Alender(n) = Alender(0)(1 + j)n Alender(10) = 10000(1.05)

10

Alender(n) = Alender(0)(1 + j)n j = .05 = j

Remark:

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1. i is always between j and j.

2. When j = j, the average yield rate per period i is always equal to j, independent of repayment

schemes. However, it is also shown that when j = j, the average yield rate per period depends

on the type of loan repayment scheme.

2.4.1 Examples

Example 2.4.1 (Textbook Exercise 2.1.16) $10,000 can be invested under two options:

Option 1 Deposit the 10,000 into a fund earning an effective annual rate of i; or

Option 2 Purchase an annuityimmediate with 24 level annual payments at an effective

annual rate of 10%. These payments are then deposited into a fund earning an effective

annual rate of 5%.

Both options produce the same accumulated value at the end of 24 years. Calculate i.

Let PV(2)24 denote the present value at the end of 24 years under option 1, and PV

(2)24 denote the

present value at the end of 24 years under option 2.

Under Option 1, PV(2)24 {10000 @ t = 0} = 10000(1 + i)

24.

Under Option 2, 10000 = P a24|10% P = 1112.997764. Therefore, at the end of each year,

$1,113.00 is then being deposited into an account, earning at the rate of 5%. Immediately after the

deposit of the 24th payment, the account balance is

PV(2)24 {10000 @ t = 0} = P s24|5% = 495306.2522

Hence i = 6.8939376%.

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2.5 Borrowers Yield Rate

Lets focus on the sinkingfund scheme from the borrowers point of view. A borrower receives $L

from the lender at the interest rate of j at time 0 and agrees to make a sequence of n periodical

interest payment of Lj as well as a lump sum ofL at the end of loan period (time n). Of course, a

person does not borrow money so that he can just simply put the money under his pillow. Suppose

that the borrower uses this $L to purchase an nyear annuity, which makes $P1 at time 1, $P2 at

time 2, . . . , Pn at time n. At the end of period k, the borrows receives $Pk and pays $Lj. The net

amount is Pk Lj. If the borrow deposits the net amount into a sinkingfund, earning interest

at the rate of j at the end of each periods, then account balance of sinkingfund at time n will

be just enough for the borrower to complete pay off the loan. In other words, ASF(n) = L. The

sinkingfund method is usually used to determine how much to borrow. In other words, determine

L.

The payment and the discount schedule of sinkingfund are

PSF =

0 P1 Lj P2 Lj . . . P n Lj

= 0 P1 P2 . . . P n 0 Lj Lj . . . Lj |t0=n =

(1 + j)n (1 + j)n1 (1 + j)1 1

If we assume that all payments are the same i.e. Pk = P for k = 1, 2, . . . , n, then

PSF =

0 P P . . . P

0 Lj Lj . . . Lj

P Vn{ PSF} = P sn|j Ljsn|j = L

L =P sn|j

1 + jsn|j(**)

The above equation involves 5 quantities, namely, L, P, n, j, and j. If any 4 of these are given,

then the last unknown can be solved easily. Cost-conscious practitioners are often interested in

determining the maximum amount of loan (the cost) relative to the potential returns that the

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investment can bring. Therefore, n can usually be predetermined, j and j can be projected

within a reasonable time frame in a politically stable economy, P or, more generally, {Pk}nk=1 can

be projected with experience and knowledge.

2.5.1 Examples

Example 2.5.1 (Textbook Example 2.30) A manufacturer is considering the purchase of some

equipment to increase the production that will generate an extra income of $15,000 per year,

payable at the end of each year for 8 years. After 8 years, the residual value or salvage value

of the equipment is$0. What the price of equipment should be if the manufacturer wants to re-

alize an average annual rate of return of 10% while recovering the principal in a sinkingfund

earning at 7%?

Think of the manufacturer is both the lender and borrower. If the manufacturer wants to

realize an average annual rate of return of 10%, then he must borrow $L from himself to

purchase the equipment. Let L denote the price of equipment.

j = 0.1, j = 0.07 n = 8, Pk = 15000 for k = 1, 2, , 8

L =15000s8|.07

1 + 0.1s8|.07

Example 2.5.2 (Textbook Exercise 2.4.11) A purchaser pays 245,000 for a mine which will

be exhausted at the end of 18 years. What level annual revenue is required in order for the

purchaser to receive a 5% annual return on his investment if he can recover his principal in

a sinking fund earning 3.5% per year?

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j = 0.05, j = 0.035 n = 18, L = 245000

245000 =P s18|.035

1 + 0.05s18|.035

P =

245000(1 + 0.05s18|.035)

s18|.035

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