Chapter 2

INDR 202

ENGINEERING ECONOMICS

CHAPTER 2

TIME VALUE OF MONEY

SPRING 2015

INSTRUCTOR: BORA ÇEKYAY

Office Hours

•

Arezou Rahimi – Tuesday – 14:00-15:15 – ENG 129F

•

Efe Çötelioğlu – Wednesday – 13:00-14:15 – ENG 218

•

Siamak Khayyati – Thursday – 14:30-15:45 – ENG 129F

•

Gita Tharkhani – Friday – 11:00-12:15 – ENG 216

TIME VALUE OF MONEY

3

Interest: Cost of Money

Economic Equivalence

Interest Formulas

EXAMPLE: POWERBALL

4

David & Erica Harrig and Maureen & Stephen Hickley won the Powerball jackpot in December 11, 2013. They were given two options:

Option 1: Lump sum payment of $34.18M,

EXAMPLE: POWERBALL

5

Year Option 1 Option 2

What Do We Need to Know?



To make such comparisons (the lottery

decision problem), we must be able to

compare

the value of money at different

point in time

.

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 money—a cost to the

borrower and an earning to

the lender _{This a two-edged sword whereby earning}

NOTATION

8

𝑷

: Principal – initial amount of money invested or borrowed

𝒊

: Interest rate per period

𝒏

: Interest period – frequency

𝑨

_𝒏: Cash flow in period 𝑛 – receipt or disbursement

𝑵

: Number of interest periods

EXAMPLE 1: PAYING BACK A LOAN

9

A bank loans $60,000 at 10% annual interest rate, with a loan origination fee of $400 to be paid at the beginning of the loan. It offers two repayment plans:

Plan 1: Equal payments at the end of every year for the

next 5 years,

EXAMPLE 1: PAYING BACK A LOAN

10

End of Year

Receipts

Payments

Plan 1

Plan 2

0

$60,000

$400

$400

1

_𝐴

2

_𝐴

3

_𝐴

4

_𝐴

CASH FLOW DIAGRAM

11

Upward arrow

positive flow, receipt

Downward arrow

negative flow, expenditure

Horizontal axis: Time

CASH FLOW DIAGRAM

12

1 2 3 4 5

59,600

𝑨

𝑨

𝑨

𝑨

𝑨

END-OF-PERIOD CONVENTION

13

Beginning of

Interest Period End Periodof Interest

1 0

METHODS OF CALCULATING INTEREST

14

Simple Interest

interest earned only on

principal

Compound Interest

interest earned on

principal

&

EXAMPLE 2: SIMPLE INTEREST

15

𝑃 = $100

𝑖 = 8%

𝑁 = 3

years

End of

Year Beginning Balance Interest Earned BalanceEnding

0 $100

1 $100 $8 $108

2 $108 $8 $116

3 $116 $8 $124

EXAMPLE 2: COMPOUND INTEREST

16

𝑃 = $100

𝑖 = 8%

𝑁 = 3

years

End of

Year Beginning Balance Interest Earned BalanceEnding

0 $100

1 $100 $8 $108

2 $108 $8.64 $116.64

3 $116.64 $9.33 $125.97

𝐹 = 𝑃 1 + 𝑖

𝑁

COMPOUNDING PROCESS

17

$100

$108

0

1

$108

$116.64

2

$116.64

$125.97

0

$100

$125.97

1 2

3

F

=

$100(1

+

0.08)

3

=

$125.97

COMPOUND INTEREST

19

Fundamental Law of Engineering Economics

End of Year Ending Balance

0 𝑃

1 𝑃(1 + 𝑖)

2 𝑃 1 + 𝑖 2

3 𝑃 1 + 𝑖 3

… …

SIMPLE VS. COMPOUND INTEREST

20

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0

2 4 6 8 10 12

𝜟𝑰

𝑵 = 𝟐 𝑵 = 𝟑 𝑵 = 𝟒

𝒊

ECONOMIC EQUIVALENCE

21

Economic equivalence

exists between cash flows

that have same economic effect & could therefore

be traded for one another.

Cash flows with different size & timing may be

equivalent under an appropriate interest rate.

ECONOMIC EQUIVALENCE

22

Economic equivalence between

𝑃

now &

𝐹

at the

end of period

𝑁

, given interest rate

𝑖

𝐹 = 𝑃 1 + 𝑖

𝑁

Deposit

𝑃

today for

𝑁

periods at interest rate

𝑖

&

have

𝐹

at the end of period

𝑁

.

ECONOMIC EQUIVALENCE

23

N

𝑭

𝑷 0

𝑭 = 𝑷 𝟏 + 𝒊

𝑵

Practice Problem

•

Problem Statement

Solution

0 1 2 3 4 5 6 7 8 9 10

$100

$200

F

1 0

8

$ 1 0 0 (1 0 .1 0 ) $ 1 0 0 ( 2 .5 9 ) $ 2 5 9

$ 2 0 0 (1 0 .1 0 ) $ 2 0 0 ( 2 .1 4 ) $ 4 2 9

$ 2 5 9 $ 4 2 9 $ 6 8 8 F

  

  

Practice problem

•

Problem Statement

Consider the following sequence of deposits

and withdrawals over a period of 4 years. If

you earn a 10% interest, what would be the

balance at the end of 4 years?

$1,000 $1,500

$1,210

0 1

2 3

4

?

Solution

End of Period Beginning balance Deposit made Withdraw Ending balance

n = 0 0 $1,000 0 $1,000

n = 1 $1,000(1 + 0.10) =$1,100

$1,000 0 $2,100

n = 2 $2,100(1 + 0.10)

=$2,310

0 $1,210 $1,100

n = 3 $1,100(1 + 0.10)

=$1,210

$1,500 0 $2,710

n = 4 $2,710(1 + 0.10)

=$2,981

EXAMPLE 3

29

Option A:

Receive

$2,100 at the end of 5 years

Option B:

Receive

$1,000 today

Assumption:

No current need for cash

Interest Rate Best Option

0% A

1% A

5% A

10% A

EXAMPLE 3

30

0 1000 2000 3000 4000 5000 6000 7000 8000

0 0,05 0,1 0,15 0,2 0,25 0,3 0,35 0,4 0,45 0,5

$𝟏, 𝟎𝟎𝟎 𝟏 + 𝒊

𝟓

$𝟐, 𝟏𝟎𝟎

EQUIVALENCE OF TWO CASH FLOWS

31

Step 1.

Set

base period

.

Step 2.

Set

interest rate

.

EXAMPLE 3

32

5

$1,000

0

OPTION B

5

$2,100

0

OPTION A

Base period: Year 5 –

FUTURE WORTH

Interest Rate Option A Option B

10% $2,100 $1,000 1 + 0.10 5 = $1,611

16% $2,100 $1,000 1 + 0.16 5 = $2,100

EXAMPLE 3

33

5

$1,000

0

OPTION B

5

$2,100

0

OPTION A

Base period: Year 0 –

PRESENT WORTH

Interest Rate Option A Option B

10% $2,100 1 + 0.10 −5 = $1,304 $1,000

16% $2,100 1 + 0.16 −5 = $1,000 $1,000

EXAMPLE 3

34

5

$1,000

0

OPTION B

5

$2,100

0

OPTION A

Base period: Year 3

Interest Rate Option A Option B

10% $2,100 1 + 0.10 −2 = $1,736 $1,000 1 + 0.10 3 = $1,331

16% $2,100 1 + 0.16 −2 = $1,561 $1,000 1 + 0.16 3 = $1,561

EQUIVALENCE OF TWO CASH FLOWS

35

Interest Rate 16%

Base Period Option A Option B

0 $1,000 $1,000 3 $1,561 $1,561 5 $2,100 $2,100

Equivalent cash flows are equivalent at any

common point in time.

EXAMPLE 4: EQUIVALENT WORTH

36

Interest Rate 10%

$100

$80

$120 $150

$200

$100

0 1 2 3 4 5

Base Period

EXAMPLE 4: EQUIVALENT WORTH

37

Worth of 𝐴₀ = $100 at 𝑛 = 3 : $100 1 + 0.10 3 = $133.10

Worth of 𝐴₁ = $80 at 𝑛 = 3 : $80 1 + 0.10 2 = $96.80

Worth of 𝐴₂ = $120 at 𝑛 = 3 : $120 1 + 0.10 1 = $132

Worth of 𝐴₃ = $150 at 𝑛 = 3 : $150 1 + 0.10 0 = $150

Worth of 𝐴₄ = $200 at 𝑛 = 3 : $200 1 + 0.10 −1 = $181.82

Worth of 𝐴₅ = $100 at 𝑛 = 3 : $100 1 + 0.10 −2 = $82.64

Practice Problem 2

 Find C that makes the two cash flow transactions equivalent at i = 10%

$500

$1,000

0 1 2 3

0 1 2 3 A

B

C C



_Approach

_:

 Step 1: Select a base period to use, say n = 2.

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

 Step 3: Equate both equivalent values and solve for unknown C.

$500

$1,000

0 1 2 3

0 1 2 3 A B C C 2 1 2 2

For A: $500(1 0.10) $1,000(1 0.10) $1,514.09

For B: (1 0.10) 2.1

2.1 $1,514.09 $721

V

V C C

EXAMPLE 7: EQUIVALENT WORTH

39

𝑁 =?

𝑭 = 𝟒𝑷

𝑷 0

𝐹 = 4𝑃 = 𝑃 1 + 0.15

𝑁

log 4 = 𝑁 log 1.15

𝑁 ≅ 10

years

EXAMPLE 8: EQUIVALENCE

40

At what interest rate the following two options

are economically equivalent?

$500

$1,000

0 1 2 3

$502 $502 $502

0 1 2 3

EXAMPLE 8: EQUIVALENCE

41

Step 1.

Set base period, say period 3.

Step 2.

Find equivalent worth of each at interest rate

𝑖

.

Step 3.

Equate the equivalent worths & solve for

𝑖

.

OPTION 1:

𝐹 = $500 1 + 𝑖

3

+ $1,000

OPTION 2:

𝐹 = $502 1 + 𝑖

2

+ 1 + 𝑖

1

+ 1 + 𝑖

0

$𝟓𝟎𝟎 𝟏 + 𝒊

𝟑

+ $𝟏, 𝟎𝟎𝟎 = $𝟓𝟎𝟐 𝟏 + 𝒊

𝟐

+ 𝟏 + 𝒊 + 𝟏

EXAMPLE 8: EQUIVALENCE

42

$𝟓𝟎𝟎 𝟏 + 𝒊

𝟑

+ $𝟏, 𝟎𝟎𝟎 = $𝟓𝟎𝟐 𝟏 + 𝒊

𝟐

+ 𝟏 + 𝒊 + 𝟏

Interest Rate LHS RHS

0% $1,500 $1,506

5% $1,578.8 $1,582.6

10% $1,665.5 $1,661.6

7.5% $1,629.9 $1,629.7

By

TYPES OF CASH FLOWS

43

SINGLE PAYMENT

UNEVEN PAYMENT SERIES

EQUAL (UNIFORM) PAYMENT SERIES

LINEAR GRADIENT SERIES

SINGLE PAYMENT

45

𝑁

𝑭

𝑷

0

𝐹 = 𝑃 1 + 𝑖

𝑁

= 𝑃 𝐹|𝑃, 𝑖, 𝑁

Compound-amount (growth) factor:

𝐹|𝑃, 𝑖, 𝑁 = 1 + 𝑖

𝑁

Example: 𝑃 = $2,000, 𝑖 = 10% , 𝑁 = 8

𝐹 = $2,000 1 + 0.10 8 = $2,000 𝐹|𝑃, 10%, 8 = $4,287.18

SINGLE PAYMENT

46

𝑁

𝑭

𝑷

0

𝑃 = 𝐹 1 + 𝑖

−𝑁

= 𝐹 𝑃|𝐹, 𝑖, 𝑁

Present-worth (discount) factor:

𝑃|𝐹, 𝑖, 𝑁 = 1 + 𝑖

−𝑁

Example: 𝐹 = $1,000, 𝑖 = 12% , 𝑁 = 5

𝑃 = $1,000 1 + 0.12 −5 = $1,000 𝑃|𝐹, 12%, 5 = $567.4

Example 3.7 Single Amounts: Find F, Given i, N, and P

Given: P = $2,000, i = 10%, N = 8 years

Find: F

 Excel Solution:

 Single Cash Flow Formula –

Compound Amount Factor

P F N 0 (1 ) ( / , , ) N

F P i

F P F P i N

 



8

A Typical

Compound Interest Table – say 12%

To find the compound interest factor when the interest rate is 12% and the number

Example 3.8 Single Amounts: Find P, Given i, N, and F

Given: F = $1,000, i = 12%, N = 5 years

Find: P

 Excel Solution:

 Single Cash Flow Formula – Present Worth Amount Factor

P F N 0 (1 ) ( / , , ) N

P F i

P F P F i N



 



5

Example 3.9 Single Amounts: Find i, Given P, F, and N

Given: F = $40, P = $20, N = 5 years

Find: i

 Excel Solution:

Example 3.10 Single Amounts: Find N, Given P, F, and i

Given: P = $6,000, F = $12,000, and i= 20%

Find: N

 Excel Solution:

 Solving for N

2 (1 0.20)

2 1.2

lo g 2 lo g 1.2

lo g 2

lo g 1.2

3.80 ye a rs

N N

F P P

EXAMPLE 9: UNEVEN PAYMENT SERIES

52

A company predicts the following expenses for the next four years:

Year 1: $50,000 for computer hardware & software, Year 2: $10,000 for additional hardware,

Year 3: none,

Year 4: $15,000 for software upgrades.

EXAMPLE 9: UNEVEN PAYMENT SERIES

53

Interest Rate 8%

0

3 2

1 4

𝑃

$50,000

$10,000

EXAMPLE 9: UNEVEN PAYMENT SERIES

55

𝑃

₁

= $50,000 𝑃|𝐹, 8%, 1 = $46,296

𝑃

₂

= $10,000 𝑃|𝐹, 8%, 2 = $8,573

𝑃

₄

= $15,000 𝑃|𝐹, 8%, 4 = $11,026

EQUAL PAYMENT SERIES

56

𝑨

0 1 2 3 4 5 𝑁 − 1 𝑁

…

𝑨 𝑨

𝑨 𝑨

𝑨 𝑨

EQUAL PAYMENT SERIES

57

𝐹 = 𝐴

𝑛=1 𝑁

1 + 𝑖

𝑁−𝑛

= 𝐴

1 + 𝑖

𝑁

− 1

𝑖

= 𝐴 𝐹|𝐴, 𝑖, 𝑁

Compound-amount factor:

𝐹|𝐴, 𝑖, 𝑁 =

1 + 𝑖

𝑁

− 1

𝑖

EQUAL PAYMENT SERIES

58

𝑨

0 1 2 3 4 5 𝑁 − 1 𝑁

…

𝑨 𝑨

𝑨 𝑨

𝑨 𝑨

0 1 2 3 4 5 𝑁 − 1 𝑁

EQUAL PAYMENT SERIES

59

𝐴 = 𝐹

𝑖

1 + 𝑖

𝑁

_{− 1}

= 𝐹 𝐴|𝐹, 𝑖, 𝑁

Sinking-fund factor:

𝐴|𝐹, 𝑖, 𝑁 =

𝑖

1 + 𝑖

𝑁

_{− 1}

Sinking-Fund Factor: Find A, Given i, N, and F

 Given: F = $5,000, N= 5 years, and i= 7% per year

 Find: A

 Excel Solution:

 Formula – Sinking Fund Factor

$5, 000( / , 7% , 5) $869.50

A  A F



$5, 000( / , 7% , 5) $869.50

A  A F



$5,000

A

0 1 ₅

TIME SHIFTS IN UNIFORM SERIES

61

0 1 2 3 4 5 𝑁 − 1 𝑁 𝑭 𝑨

0 1 2 3 4 5 𝑁 − 1 𝑁 𝑨

𝑨

…

𝑨 𝑨

𝑨 𝑨

EXAMPLE 10: TIME SHIFTS

62

$4K

0 1 2 3

$4K $4K

$4K

0 1 2 3

$4K $4K

𝐹 = $4,000 𝐹|𝐴, 10%, 3 = $13,240

Interest Rate 10%

EQUAL PAYMENT SERIES

63

𝑨

0 1 2 3 4 5 𝑁 − 1 𝑁

…

𝑨 𝑨

𝑨 𝑨

𝑨 𝑨

EQUAL PAYMENT SERIES

64

𝑃 = 𝐴

1 + 𝑖

𝑁

_{− 1}

𝑖 1 + 𝑖

𝑁

= 𝐴 𝑃|𝐴, 𝑖, 𝑁

Present-worth factor:

𝑃|𝐴, 𝑖, 𝑁 =

1 + 𝑖

𝑁

− 1

𝑖 1 + 𝑖

𝑁

EQUAL PAYMENT SERIES

65

𝑨

0 1 2 3 4 5 𝑁 − 1 𝑁

…

𝑨 𝑨

𝑨 𝑨

𝑨 𝑨

0 1 2 3 4 5 𝑁 − 1 𝑁

EQUAL PAYMENT SERIES

66

𝐴 = 𝑃

𝑖 1 + 𝑖

𝑁

1 + 𝑖

𝑁

_{− 1}

= 𝑃 𝐴|𝑃, 𝑖, 𝑁

Capital-recovery (annuity) factor:

𝐴|𝑃, 𝑖, 𝑁 =

𝑖 1 + 𝑖

𝑁

1 + 𝑖

𝑁

_{− 1}

EXAMPLE 10: DEFERRED LOAN REPAYMENTS

67

You borrow 40K TL from a bank to finance your one-year educational expenses. You wish to repay the loan in five equal (annual) installments starting at the end of the 3rd year after borrowing. The interest rate is 8%.

How much should you pay in each installment? How much will you pay in total?

EXAMPLE 10: DEFERRED LOAN REPAYMENTS

68

Grace period

0 1 2 3 4 5 6 7

𝑨

𝑨 𝑨

𝑨 𝑨

EXAMPLE 10: DEFERRED LOAN REPAYMENTS

69

0 1 2 3 4 5 6 7

𝑨

𝑨 𝑨

𝑨 𝑨

𝟒𝟎, 𝟎𝟎𝟎 𝑭|𝑷, 𝟖%, 𝟐 = 𝟒𝟔, 𝟔𝟓𝟔

Each installment: 𝐴 = 46,656 𝐴|𝑃, 8%, 5 = 11,687 Total payment: 5 × 11,687 = 58,435

EXAMPLE 11: SAVINGS PLAN

70

Early Savings Plan: Annual deposits of $2,000 for 10 years starting at the end of the first year

Deferred Savings Plan: Annual deposits of $2,000 until the end of year 44 starting at the end of year 11

EXAMPLE 11: SAVINGS PLAN

71

0 1 2 3 4 5 6 7 8 9 10

44

Early Savings Plan

𝐴 = $2,000

0 1 2 3 4 5 6 7 8 9 10 11 12

Deferred Savings Plan

EXAMPLE 11: SAVINGS PLAN

72

0 1 2 3 4 5 6 7 8 9 10

44

Early Savings Plan

𝐴 = $2,000

EXAMPLE 11: SAVINGS PLAN

73

0 1 2 3 4 5 6 7 8 9 10 11 12

Deferred Savings Plan

𝐴 = $2,000

PERPETUITIES

74

Perpetuity:

equal payment series that continue forever

𝑃 = lim

𝑁→∞

𝐴

1 + 𝑖

𝑁

− 1

𝑖 1 + 𝑖

𝑁

= 𝐴 lim

_𝑁→∞

𝑃|𝐴, 𝑖, 𝑁 =

𝐴

𝑖

Example: 𝐴 = $1,000, 𝑖 = 10%

𝑃 = $1,000

LINEAR GRADIENT SERIES

75

0 1 2 3 4 5 𝑁 − 1 𝑁

…

𝟒𝑮

𝑵 − 𝟐 𝑮

𝟑𝑮 𝟐𝑮

𝑮

LINEAR GRADIENT SERIES

76

0 1 2 3 𝑁

…

𝑵 − 𝟏 𝑮

𝟐𝑮 𝑮

0 1 2 3 𝑁

LINEAR GRADIENT SERIES

77

𝑃 = 𝐺

1 + 𝑖

𝑁

− 𝑖𝑁 − 1

𝑖

2

_{1 + 𝑖}

𝑁

= 𝐺 𝑃|𝐺, 𝑖, 𝑁

Linear-gradient-series present-worth factor:

LINEAR GRADIENT SERIES

78

𝑃 =

_1+𝑖0

+

_1+𝑖𝐺 ₂

+

_1+𝑖2𝐺 ₃

+ ⋯ +

𝑁−1 𝐺_{1+𝑖 𝑁}

𝑃 = 𝐺

_𝑛=1𝑁

𝑛 − 1 1 + 𝑖

−𝑛

= 𝐺𝑥

_𝑛=1𝑁

𝑛 − 1 𝑥

𝑛−1

Arithmetic-Geometric Series

:

𝑆 = 𝑥 + 2𝑥

2

+ 3𝑥

3

+ ⋯ + 𝑁 − 1 𝑥

𝑁−1

Geometric Series

:

LINEAR GRADIENT SERIES

79

Composite Series

0 1 2 3 𝑁

…

𝑨 + 𝑵 − 𝟏 𝑮 𝑨 + 𝟐𝑮

𝑨 + 𝑮 𝑨

0 1 2 3 𝑁

…𝑨 − 𝑵 − 𝟏 𝑮

𝑨 − 𝑮

EXAMPLE 12: LINEAR GRADIENT SERIES

80

$1,000 $1,250

$1,500 $1,750

$2,000

1 2 3 4 5 0

𝑷 =?

$1,000 𝑃|𝐹, 12%, 1 = $892.9 $1,250 𝑃|𝐹, 12%, 2 = $996.5 $1,500 𝑃|𝐹, 12%, 3 = $1,067. 7 $1,750 𝑃|𝐹, 12%, 4 = $1,112.13 $2,000 𝑃|𝐹, 12%, 5 = $1,134.8

𝑃 = $5,204.03

EXAMPLE 12: LINEAR GRADIENT SERIES

81

$1,000 $1,000 $1,000 $1,000 $1,000

1 2 3 4 5 0

𝑷_𝟏

$250 $500

$750 $1,000

1 2 3 4 5 0

𝑷_𝟐

𝑃₂ = $250 𝑃|𝐺, 12%, 5 = $1,599.25

𝑃₁ = $1,000 𝑃|𝐴, 12%, 5 = $3,604.8

Gradient-to-Equal-Payment Series Conversion Factor, (A/G,

i, N)

 Given: G = $1,000, N = 10 years, i= 12%

 Find: A

 Solution:

$1, 000( / ,12% ,10) $1, 000(3.5847)

$3, 584.70

A  A G

 



_{Cash Flow Series}

Example 3.22 – Linear Gradient: Find A, Given A₁, G, i, and N

 Given: A₁= $1,000, G = $300, N= 6 years, and i= 10% per year

Example 3.23 Declining Linear Gradient Series

 Given: A₁ = $1,200, G = -$200, N = 5 years, and i = 10% per year

 Find: F

GEOMETRIC GRADIENT SERIES

85

0 1 2 3 𝑁 − 1 𝑁

…

𝑨_𝟏 𝟏 + 𝒈 𝟐 𝑨_𝟏 𝟏 + 𝒈

𝑨_𝟏

𝑨_𝟏 𝟏 + 𝒈 𝑵−𝟐

GEOMETRIC GRADIENT SERIES

86

0 1 2 3 𝑁 − 1 𝑁

…

𝑨_𝟏 𝟏 − 𝒈 𝟐 𝑨_𝟏 𝟏 − 𝒈

𝑨_𝟏

𝑨_𝟏 𝟏 − 𝒈 𝑵−𝟐

GEOMETRIC GRADIENT SERIES

87

𝑃 =

𝐴

₁

1 − 1 + 𝑔

𝑁

1 + 𝑖

−𝑁

𝑖 − 𝑔

if 𝑖 ≠ 𝑔,

𝑁𝐴

₁

GEOMETRIC GRADIENT SERIES

88

𝑃 =

𝑛=1 𝑁

𝐴

_𝑛

1 + 𝑖

−𝑛

=

𝐴

1

1 + 𝑔

𝑛=1 𝑁

1 + 𝑔

1 + 𝑖

𝑛

= 𝑎

𝑛=1 𝑁

𝑥

𝑛

𝑥 ≠ 1

(

𝑖 ≠ 𝑔

):

𝑃 = 𝑎

𝑥−𝑥_1−𝑥𝑁+1

GEOMETRIC GRADIENT SERIES

89

Example: 𝐴₁ = $1,000, 𝑔 = 8%, 𝑖 = 12%, 𝑁 = 5

𝑃 = $1,000 1 − 1 + 0.08

5 _{1 + 0.12} −5

0.12 − 0.08 = $4,157

0 1 2 3 4 5

$𝟏, 𝟏𝟔𝟔 $𝟏, 𝟎𝟖𝟎

$𝟏, 𝟎𝟎𝟎

$𝟏, 𝟐𝟔𝟎

EXAMPLE 13: REACHING A GOAL

90

An investor wishes to collect 3M TL in 10 years by investing annually in a bank account that earns interest at rate 10% per year. She expects her annual income to grow at rate 5% every year and her annual deposits will increase at the same rate. How much should she deposit at the end of the first year in order to reach her goal?

𝑖 = 10% ≠ 𝑔 = 5%

3𝑀 𝑃|𝐹, 10%, 10 = 𝐴₁ 1 − 1 + 0.05 10 1 + 0.10 −10

0.10 − 0.05

EXAMPLE 14

91

You deposit 5K TL in your bank account every year for six years. The annual interest rate is 8%. You will withdraw the money accumulated in your account in equal annual amounts over the subsequent 6 years. How much money will you withdraw each year?

METHOD 1. Compare present worth.

5,000 𝑃|𝐴, 8%, 6 = 𝐴 𝑃|𝐴, 8%, 6 𝑃|𝐹, 8%, 6

METHOD 2. Compare worth at the end of year 6.

5,000 𝐹|𝐴, 8%, 6 = 𝐴 𝑃|𝐴, 8%, 6

METHOD 3. Match annual amounts.

EXAMPLE 15: COMPOSITE CASH FLOWS

92

$𝟓𝟎 𝑷|𝑭, 𝟏𝟓%, 𝟏 = $𝟒𝟑. 𝟒𝟖

$𝟏𝟎𝟎 𝑷|𝑨, 𝟏𝟓%, 𝟑 𝑷|𝑭, 𝟏𝟓%, 𝟏 = $𝟏𝟗𝟖. 𝟓𝟒

$𝟏𝟓𝟎 𝑷|𝑨, 𝟏𝟓%, 𝟒 𝑷|𝑭, 𝟏𝟓%, 𝟒 = $𝟐𝟒𝟒. 𝟖𝟓

$𝟐𝟎𝟎 𝑷|𝑭, 𝟏𝟓%, 𝟗 = $𝟓𝟔. 𝟖𝟓

𝑷 = $𝟒𝟑. 𝟒𝟖 + $𝟏𝟗𝟖. 𝟓𝟒 + $𝟐𝟒𝟒. 𝟖𝟓 + $𝟓𝟔. 𝟖𝟓 = $𝟓𝟒𝟑. 𝟕𝟐

$100 $100 $100

$150 $150 $150 $150

$200

0 1 2 3 4 5 6 7 8 9 $50

EXAMPLE 16: VARYING INTEREST RATES

93

$300

$500 $400

5% 6% 6% 4% 4%

0 1 2 3 4 5

𝒏 = 𝟏: $300 𝐹|𝑃, 5%, 1 = $315

𝒏 = 𝟐: $315 𝐹|𝑃, 6%, 1 + $500 = $833.9

𝒏 = 𝟑: $833.9 𝐹|𝑃, 6%, 1 = $883.93

𝒏 = 𝟒: $883.93 𝐹|𝑃, 4%, 1 + $400 = $1,319.29

EXAMPLE 17: MISSING PAYMENTS

94

$100

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

missing payment

𝑖 = 10%

$100

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Add & subtract 10th _payment.

EXAMPLE 18: UNCONVENTIONAL REGULARITY

95

$10,000

0

1 _{2 3 4 5 6 7 8 9 10 11 12 13 14}

𝑪 𝑪 𝑪 𝑪 𝑪 𝑪 𝑪

EXAMPLE 18: UNCONVENTIONAL REGULARITY

96

METHOD 1.

Modify cash flows.

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨

EXAMPLE 18: UNCONVENTIONAL REGULARITY

97

0

1 _{2 3 4 5 6 7 8 9 10 11 12 13 14}

𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨 𝑨

$10,000

1 _{2 3 4 5 6 7 8 9 10 11 12 13 14}

EXAMPLE 18: UNCONVENTIONAL REGULARITY

98

𝑪

𝑨 𝑨

≡

EXAMPLE 18: UNCONVENTIONAL REGULARITY

99

METHOD 2.

Modify interest rate.

Two-year interest rate:

𝑖

′

= 1 + 0.1

2

− 1 = 21%

𝐶 = $10,000 𝐴|𝑃, 21%, 7 = $2,850.67

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

𝑪 𝑪 𝑪 𝑪 𝑪 𝑪 𝑪

SUMMARY

100

Time value

of money originates from its earning &

purchasing powers.

Economic equivalence

exists between cash flows that

have same economic effect.

Equivalent worth

of a cash flow series at a time point

depends on

interest rate

as well as amount & timing

of payments.