TIME VALUE OF MONEY

CHAPTER 2

1
 Time Value of Money

 The Cost of Money is established and measured by an interest rate, a

percentage that is periodically applied and added to an amount of money over
a specified length of time.

 Economic Equivalence

 Interest Formulas – Single Cash Flows

 Equal-Payment Series

 Dealing with Gradient Series

 Composite Cash Flows.

2
 Time Value of Money

 Money has a time value because it can earn more money over time
(earning power).

 Money has a time value because its purchasing power changes over time
(inflation).

 Time value of money is measured in terms of interest rate.

 Interest is the cost of having money available for use - a cost to the
borrower and an earning to the lender

3
4
 Elements of Transactions involve Interest

1. Initial amount of money in transactions involving debt or investments is

called the principal (P).

2. The interest rate ( i ) measures the cost or price of money and is expressed
as a percentage per period of time.

3. A period of time, called the interest period (n), determines how frequently
interest is calculated.

4. A specified length of time marks the duration of the transactions and

thereby establishes a certain number of interest periods (N).

5. A plan for receipts or disbursements (An) that yields a particular cash flow
pattern over a specified length of time. [monthly equal payment]

6. A future amount of money (F) results from the cumulative effects of the
interest rate over a number of interest periods.

5
 EXAMPLE OF INTEREST TRANSACTION

 Suppose that you apply for an education loan of

$30,000 from a bank at a 9% annual interest rate. In
addition you pay a $300 loan origination fee when the
loan begins.

 The bank offers two repayment plans, one with equal

payments made at the end of every year for the next
five years (installment plan) and the other with a single
payment made after the loan period of five years
(deferment plan).

6
 Which Repayment Plan?
Table 2.1 Repayment plan offered by the lender

End of Year Receipts Payments

Plan 1 Plan 2
Year 0 $30,000.00 $300.00 $300.00

Year 1 $7,712.77 0

Year 2 $7,712.77 0

Year 3 $7,712.77 0

Year 4 $7,712.77 0

Year 5 $7,712.77 46,158.72

The amount of loan = $30,000 & origination fee = $300 & interest rate = 9% APR
F = P (1 + i )N = P (1+0.09)5 = $30,000 x (1.09)5 = $46,158.72

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  i (1 + i ) N   0.09(1 + 0.09) 5 
( A / P, i , N ) = A = P   A = P (30,000)   = $7,712.77
 (1 + i ) − 1  (1 + 0.09) − 1 
N 5

 i 
( A / F , i, N ) = A = F    0.09 
 (1 + i ) − 1
N A = $46,158.72   = $7,712.77
 (1 + 0.09) − 1
5

Figure 2-2 A cash flow diagram for plan 1 of the loan repayment example
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Methods of Calculating Interest

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12
 Practice Problem

Problem Statement

If you deposit $100 now (n = 0) and $200 two years from now (n = 2) in a
savings account that pays 10% interest, how much would you have at the end of
year 10?

13
Solution
F

0 1 2 3 4 5 6 7 8 9 10

$100(1 + 0.10)
=10
=
$100(2.59) $259
$200(1 + 0.10)
= 8
=
$200(2.14) $429
$100
$200 F = $259 + $429 = $688

14
 Economic Equivalence

 What do we mean by “economic equivalence?”

 Why do we need to establish an economic equivalence?

 How do we establish an economic equivalence?

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 Economic Equivalence
How do we know, whether we should prefer to have $20,000 today and
$50,000 ten years from now, or $8,000 each year for the next ten years?

Which option would you prefer?

16
 Equivalence Calculation: A Simple example

Figure 2-4 Using compound interest to establish economic equivalence

17
Equivalence relation between P and F.

18
$1,500

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 EXAMPLE 2.3 Equivalence Calculation
FIND: V3 (equivalent worth at n = 3) and i = 10%.

Step 1: $100(1+0.1)3+ $80(1+0.1)2+$120(1+0.1)1+$150 = $511.90

Step 2: $200(1+0.1)-1+ $100(1+0.1)-2 = $264.46
Step 3: V3= $511.90 + $264.46 = $776.36

Figure 2-6 Equivalent worth calculation at n = 3

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 Interest Formulas for Single Cash Flows
Compound Amount Factor

Figure 2-7 Compounding process: Find F, given P, i, and N

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Interest Rate Factors (10 %)

23
Example 2.4 If you had $1,000 now and invested it at 7% interest compounded annually,
how much would it be worth in 8 years?

Figure 2-8 Cash flow diagram

Given: P = $1,000, i = 7 %, and N = 8 years; Find: F

F = $1,000 (1+0.07)8 = $1,718.19 or using this
F = P (F/P, i, N) factor notation together with table value
F = $1,000 (1.7182) = $1,718.19

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Present -Worth Factor

25
 Example 2.5
A zero coupon (installment) bond is a popular variation on the bond theme
for some investors. What should be the price of an eight year (maturity)
zero-coupon with a face value (future value) of $1,000 if similar, nonzero
coupon bonds are yielding 6% annual interest?

Figure 2-10 Cash flow diagram

Given: F = $1000, i = 6%, and N = 8 years

Find: P P = $1,000 (1+0.06)-8 = $1,000 (0.6274) = $627.40

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 Example 2.6
Suppose you buy a share of stock for $10 and sell it for $20; your profit is thus $10. It takes five
years to gain this profit, what would be the rate of return on your investment?

Given: P = $10, F = $20 and N = 5 years, Find: i

F = $20 = $10 (1+ i )5 $2 = (1+ i )5 = (F / P, i, 5)

1+ i = = 1.14869 i = 1.14869 – 1 = 0.14869 or 0.1487 = 14.87%

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 Example 2.7
You have just purchased 200 shares of GE stock at $15 per share. You will
sell the stock when its market price doubles. If you expect the stock price
to increase 12% per year, how long do you expect to wait before selling the
stock?

Given: P = $3,000, F = $6,000 and i = 12%, Find: N

F = P (1+ i )N = P (F/P, i, N) …… 6,000 = $3,000 (1+ 0.12)N

2 = (1.12) N ………… Log 2 = N log 1.12 solve for N gives

N = log 2 / log 1.12 = 0.301 / 0.049 = 6.116 ≈ 6.12 years

Figure 2-12 Cash flow diagram

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 Practice Problem

$1,000

$500

A
Given: i = 10%,
0 1 2 3
Find: C that makes the two cash
flow streams to be indifferent
C C

0 1 2 3

29
 Approach

Step 1: $1,000
Select the base period to use, say n = 2
$500

Step 2:
Find the equivalent lump sum value at n = 2 for A 0 1 2 3
both A and B.

Step 3: C C
Equate both equivalent values and solve for
unknown C.
B
0 1 2 3

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 Solution

For A:
$1,000
V2 = $500(1 + 0.10) 2 + $1, 000(1 + 0.10) −1
= $1, 514.09 $500
A
For B:
V2 =C (1 + 0.10) + C 0 1 2 3
= 2.1C

To Find C: C C

2.1C = $1, 514.09 B

C = $721
0 1 2 3

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 Practice Problem
$1,000

$500

A
0 1 2 3
At what interest rate would
you be indifferent between the $502 $502 $502
two cash flows?

B
0 1 2 3

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 Approach
Step 1:
$1,000
Select the base period to compute
the equivalent value (say, n = 3) $500

A
0 1 2 3
Step 2:
Find the net worth of each at n = 3.
$502 $502 $502

B
0 1 2 3

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 Establish Equivalence at n = 3

Option A : =
F3 $500(1 + i ) 3 + $1, 000
Option B :=
F3 $502(1 + i ) 2 + $502(1 + i ) + $502

Find the solution by trial and error, say i = 8%

=
Option A : F3 $500(1.08) 3 + $1, 000
= $1, 630
Option B : F3 = $502(1.08) 2 + $502(1.08) + $502
= $1, 630

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 Practice Problem

 You want to set aside a lump sum amount today in a

savings account that earns 7% annual interest to meet a
future investment in the amount of $10,000 to be
incurred in 6 years.

 How much do you need to deposit today?

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Solution
$10,000

F = $10,000; N = 6 years; i = 7 %; Find P

=P $10, 000(1 + 0.07) −6

= $10, 000( P / F , 7%, 6)
P = $6, 663

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Figure 2.13 Decomposition of uneven cash flow series

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Equal - Payment Series
• There is an equal payment (a uniform series) over a number of periods
• Example; rental payment, tuition saving plans,….
• The concern is to find P or F of such a series of A’s.

Figure 2-14 Equal payment series: Find equivalent P or F

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Subsections of equal-payment series

1. Find F, given A, i, and N (Compound amount factor)

2. Find A, given F, i, and N (Sinking fund factor)

3. Find A, given P, i, and N (Capital recovery factor)

4. Find P, given A, i, and N (Present-worth factor)

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Equal Payment Series Compound Amount Factor: Find F, Given A, i, and N
(Future Value of an annuity)

Figure 2-15 Cash flow diagram of the relationship between A and F

The future value of the amount A paid at the end of the 1st period = A(1+i)N-1

The future value of the amount A paid at the end of the 2nd period = A(1+i)N-2

F = A(1+i)N-1 + A(1+i)N-2+ A(1+i)N-3 …………+ A(1+i) + A Geometric sequence

(1 + i ) N − 1
F = A
i
= A( F / A, i , N )

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Example 2.9: Suppose you make an annual contribution of $5,000 to your
savings account at the end of each year for five years. If your savings
account earns 6% interest annually, how much can be withdrawn at the
end of five years?

Given: A = $5,000, N = 5 years, and i = 6%

Find: F
Solution: F = $5,000(F/A, 6%, 5) =
$28,185.46
use interest factor table and get the value, F =
$5,000 (5.6371) = $28,185.46

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Validation

$5, 000(1 + 0.06) 4 =

$6, 312.38
$5, 000(1 + 0.06)3 =
$5, 955.08
$5, 000(1 + 0.06) 2 =
$5, 618.00
$5, 000(1 + 0.06)1 =
$5, 300.00
$5, 000(1 + 0.06) 0 =
$5, 000.00
$28.185.46

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Example 2.10 Handling Time Shifts in a Uniform Series
First deposit of the five deposit series was made at the end of period one and the remaining four
deposits were made at the end of each following period. Suppose that all deposits were made at the
beginning of each period instead. Compute the balance at the end of period five.

F=?
F5 = $5, 000( F / A, 6%, 5)(1.06)
= $29, 876.59

First deposit occurs at n = 0 i = 6%

0 1 2 3 4 5

$5,000 $5,000 $5,000 $5,000 $5,000

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Sinking fund (Find A, given F, i, and N )

Definition
 A sinking fund is a fund established by an economic entity by setting
aside revenue over a period of time to fund a future capital expense,
or repayment of a long-term debt.
 It is created by making periodic deposits (usually equal) at compound
interest in order to accumulate a given sum at a given future time for
some specific purpose.
 It is an interest-bearing account into which a fixed sum is deposited
each interest period; The term within the colored area is called
sinking-fund factor. F

i
0 1 2 3 A = F
N (1 + i ) N − 1
= F ( A / F ,i, N )
A
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Example 2.11 – College Savings Plan:
You want to set up a college savings plan for your daughter. She is currently
10 years old and will go to college at age 18. You assume that when she
starts college, she will need at least $100,000 in the bank. How much do you
need to save each year in order to have the necessary funds if the current
rate of interest is 7%? Assume that end-of-year deposits are made.

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Solution
Given: F = $100,000 $100,000
i = 7%
N = 8 years
Current age: 10 years old

0
1 2 3 4 5 6 7 8

A= ?

i = 7%
Find: A from table (0.0975)

Solution: A = $100,000 (A/F, 7%, 8) = $9,746.78

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Capital Recovery Factor (Annuity Factor) Find A, given P, i, and N
Annuity:
 An amount of money payable to a recipient at regular intervals for a prescribed period of
time out of a fund reserved for that purpose.
 A series of equal payments occurring at equal periods of time.

Annuity factor:
 The function of interest rate and time that determines the amount of periodic annuity that may
be paid out of a given fund.
 Is the colored area which is designated (A/P, i, N). In finance, this A/P factor is referred to as
the annuity factor which indicates a series of payments of a fixed amount for a specified number
of periods.
P i (1 + i ) N
A = P
(1 + i ) N − 1
= P( A / P, i , N )
1 2 3
0 N

A=?

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Example 2.12 Paying Off Educational Loan

You borrowed $21,061.82 to finance the educational expenses for your senior year of college.
The loan will be paid off over five years. The loan carries an interest rate of 6% per year and is
to be repaid in equal annual installments over the next five years. Assume that the money was
borrowed at the beginning of your senior year and that the first installment will be due a year
later. Compute the amount of the annual installments.

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 P =$21,061.82

i = 6%

0 1 2 3 4 5

A A A A A

Solution
i (1 + i ) N
A = P
(1 + i ) N − 1
 Given: P = $21,061.82, N = 5 years, and i = 6% Find: A
= P( A / P, i , N )
(0.2374) from table
 Solution: A = $21,061.82(A/P,6%,5) = $5,000

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Example 2.13 Deferred (delayed) Loan Repayment Plan
Suppose that you want to negotiate with the bank to defer the
first loan installment until the end of year 2. (But you still
desire to make five equal payments at 6% interest.) If the bank
wishes to earn the same profit, what should be the new annual
installment?

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 P =$21,061.82

i = 6%
Old policy

0 1 2 3 4 5 6

Grace period
A A A A A

P ’ = $21,061.82(F / P, 6%, 1) = $22,325.53

New Policy 0 1 2 3 4 5 6

A’ A’ A’ A’ A’

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 Two - Step Procedure

P' = $21, 061.82( F / P , 6%,1)

= $22, 325.53
A = $22, 325.53( A / P , 6%, 5)
= $5, 300

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Present Worth Factor (Annuity Factor) Find P, given A, i, and N

P= ?

1 2 3
0 N

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Example:
A truck driver won Powerball, a multistate lottery game. The winner could
choose between a single lump sum of $116.5 million or a total of $195
million paid out over 20 annual installments ($9.75 million per year and
the first installment being paid out immediately). The truck driver opted for
the lump sum. Assume that the annual interest rate is 8%.
a) From a strictly economic standpoint, did he make the lucrative
choice?
b) What is the minimum rate of return at which accepting the $116.5
million lump sum would make sense?

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 Linear Gradient Series
Engineers frequently meet situations involving periodic payments that increase or
decrease by a constant amount (G) from period to period.

Figure 2-25 Cash flow diagram of a strict gradient series

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Present-Worth Factor: Linear Gradient (P/G, i, N)

Figure 2-27 Cash flow diagram of a strict gradient series

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Gradient Series as a Composite Series of a Uniform Series of N Payments of A1
and the Gradient Series of Increments of Constant Amount G.

63
Example 2.16 Creating a Graduated (divided into regular stages)
Loan Repayment with a Linear Gradient Series; Given P, A1, N and i
Find: G
You borrowed $10,000 from a local bank, with the agreement that you will pay back the loan
according to a graduated payment plan. If your first payment is set at $1,500, what would the
remaining payment look like at a borrowing rate of 10% over five years?

Figure 2-28 Cash flow diagram representing a graduated payment plan

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Example 2.16

SOLUTION:
Since the loan payment series consist of two parts – (1) a $1,500 equal payment series
and (2) A strict gradient series (unknown, yet to be determined) – we can calculate
the present value of each series and equate them with $10,000.
$10,000 = $1,500 (P/A, 10%, 5) + G (P/G, 10%, 5) get the table values from
appendix B …… (3.7908) (6.8618)

$10,000 = $5,686.20 + 6.8618 G $10,000 - $5,686.20 = 6.8618 G

$4,313.80 = 6.8618 G G = $628.67 65

Present value calculation
$2,000
for a gradient series $1,750
$1,500
$1,250
$1,000

0
1 2 3 4 5

How much do you have to deposit now in a savings

account that earns a 12% annual interest, if you want
to withdraw the annual series as shown in the figure?

P =?

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Method 1: Using the (P/F, i, N) Factor
$2,000
$1,750
$1,500
$1,250
$1,000

0
1 2 3 4 5

$1,000(P/F, 12%, 1) = $892.86

$1,250(P/F, 12%, 2) = $996.49
$1,500(P/F, 12%, 3) = $1,067.67
$1,750(P/F, 12%, 4) = $1,112.16
P =? $2,000(P/F, 12%, 5) = $1,134.85

$5, 204.03

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 Method 2: Using the Gradient Factor

P1 = $1,000( P / A,12%,5)
= $3,604.80

P2 = $250( P / G,12%,5)
= $1,599.20

P = $3,604.08 + $1,599.20
= $5,204

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 Geometric Gradient Series
Many engineering economic problems, particularly those relating to construction costs, involve
cash flows that increase over time, not by a constant amount, but rather by a constant percentage
(geometric), called compound growth.

69
An alternative way

P=

70
 Example
A company is considering purchasing a new machine tool. In addition to the initial
purchase and installation costs, management is concerned about the machine’s
maintenance costs, which are expected to be $1,000 at the end of the first year of
the machine’s life and increase 8 percent/year there- after. The machine tool’s
expected life is 15 years. Company management would like to know the present
worth equivalent for expected costs. If the firm’s time value of money is 10
percent/year compounded annually, what is the present worth equivalent?

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Solution

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 Example 2.19 Required Cost-of-living Adjustment Calculation

Suppose that your retirement benefits during your first year of retirement
are $50,000. Assume that this amount is just enough to meet your cost of
living during the first year. However, your cost of living is expected to
increase at an annual rate of 5%, due to inflation. Suppose you do not expect
to receive any cost of living adjustment in your retirement pension. Then,
some of your future cost of living has to come from your savings other than
retirement pension. If your savings account earns 7% interest a year, how
much should you set aside in order to meet this future increase in the cost of

living over 25 years?

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 Example 2.18 Required Cost- of - living Adjustment Calculation
Find P, Given A1, g, i, N

Figure 2-31 Cash flow diagram

Given: g = 5%, i = 7%, N = 25 years, A1 = $50,000

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77
Alternatively;

So, the required additional savings = $940,167 – $582,679 = $ 357,488

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 COMPOSITE CASH FLOWS
Compute the equivalent present worth for mixed payment series at 15%.

METHOD 1

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METHOD 2: GROUP THE CASH FLOW COMPONENTS

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PRACTICE PROBLEMS

CHAPTER 2

81
82
1)
Suppose you have the alternative of receiving either $15,000 at the end of seven
years or P dollars today. Currently, you have no need for the money, so you could
deposit the P dollars into a bank account that pays 6% interest compounded
annually. What value of P would make you indifferent in your choice between P
dollars today and the promise of $15,000 at the end of seven years?

2)
Suppose that, to purchase a car, you are obtaining a personal loan from your uncle
in the amount of $75,000 (now) to be repaid in three years. If your uncle could
earn 9% interest (compounded annually) on his money invested in various sources,
what minimum lump-sum payment three years from now would make your uncle
happy economically?

3)
State the amount accumulated each of the following present investments:
$5,000 in 6 years at 7% compounded annually.
$13,000 in 8 years at 6% compounded annually.
$16,000 in 25 years at 10% compounded annually.
$10,000 in 10 years at 5% compounded annually.

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4) State the present worth of the following future payments:
$8,000 five years from now at 8% compounded annually.
$10,000 six years from now at 10% compounded annually.
$12,000 eight years from now at 7% compounded annually.
$18,000 ten years from now at 9% compounded annually.

5) How many years will it take to triple your investment of $8,000 if it has an interest
rate of 7% compounded annually?

6) A project is expected to generate a cash flow of $8,000 in year 1, $2,000 in year 2,

and $5,000 in year 3. At the interest rate of 8%, what the maximum amount that you
could in the project at year zero?

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7) If you desire to withdraw the following amounts over the next five years from a
savings account that earns 7% interest compounded annually, how much do you
need to deposit now?
Year Amount
2 $5,000
3 $6,000
4 $8,200
5 $4,500

8) If $4,000 is invested now, $7,000 two years from now, and $5,000 four years
from now at an interest rate of 9% compounded annually, what will be the total
amount in 8 years?

9) How much invested now at an interest rate of 10% compounded annually would
be just sufficient to provide three payments as follows: the first payment in the
amount of $5,000 occurring two years from now, the second payment in the amount
of $7,000 four years thereafter, and the third payment in the amount of $9,000 six
years thereafter.

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10)
Suppose that an oil well is expected to produce 1,200,000 barrels of oil during its first
year in production. However, its subsequent production (yield) is expected to
increase by 9% over the previous year's production. The oil well has a proven reserve
of 10,500,000 barrels.

(a)
Suppose that the price of oil is expected to be $120 per barrel for the next six years.
What would be the present worth of the anticipated revenue stream at an interest rate
of 10% compounded annually over the next six years?
(b)
Suppose that the price of oil is expected to start at $120 per barrel during the first
year, but to increase at the rate of 3% over the previous year's price. What would be
the present worth of the anticipated revenue stream at an interest rate of 10%
compounded annually over the next six years?
(c)
Consider part (b) again. After three years' production, you decide to sell the oil well.
What would be a fair price?

86
11)
What is the amount of 10 equal annual deposits that can provide five annual withdrawals
when a first withdrawal of $3,000 is made at the end of year 11 and subsequent
withdrawals increase at the rate of 6% per year over the previous year's withdrawal if
a) The interest rate is 8% compounded annually?

b) The interest rate is 6% compounded annually?

87