Lecture Engineering economics - Chapter 5: Annual amount and gradient functions

In this chapter, students will be able to understand: Calculating future values from annual amounts, calculating present values from annual amounts, calculating future and present values from gradient amounts, calculating present value of a future perpetual amounts, calculating deferred annuities.

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Chapter 5 – Unit 1

Annual Amount and Gradient

Functions

IET 350 Engineering Economics

Learning Objectives – Chapter 5

Upon completion of this chapter you should understand:

Calculating future values from annual amounts

Calculating present values from annual amounts

Calculating future and present values from gradient

amounts

Calculating present value of a future perpetual amounts

Calculating deferred annuities

2

Learning Objectives – Unit 1

Upon completion of this unit you should understand:

amounts

3

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The prior chapter covered single‐payment functions where a

cash inflow occurred at one point in time and a cash outflow

occurred at a second point in time

Many financial transactions have elements that occur at y

multiple points in time. These can include:

Equal annual cash flow

Linear gradient cash flow

Non‐linear gradient cash flow

Mixed annual cash flow

4

Introduction

This chapter covers three types of multiple‐payment

situations:

Equal annual amounts (A) – equal dollar amounts flow

into or out of an investment or project each year.p j y

Linear gradient amounts (G) – dollar amounts flowing

into or out of an investment or project increase/decrease

each year by a constant amount (linear)

Mixed annual amounts – differing dollar amount flow into

and/or out of an investment or project each year

5

Equal Annual Amounts

Assumptions for equal annual amount analysis include:

Cash flow occurs at the end of each year

All cash flows are equal and occur each year

Note that most interest table such as those in appendix B of

the Bowman text are based on end of year transactions.

Interest table are available that use the beginning or middle

of time periods.

If using a time value of money function on your calculator,

check the manual to determine if the time basis is end of

period (year) or some other basis

6

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Notations used for time value of money calculation

Future Value (one‐time occurrence) → F

Present Value (one‐time occurrence) → P

Equal Annual Amount → A

Cash flow diagrams represent annual amounts as equal

length lines as illustrated in Figure 5‐1:

7

Future Value Calculations

Future value for an equal annual amount is determined by the

following equation:

( )

⎥

⎤

⎢

⎡ + 1 in‐ 1

8

( )

⎥

⎦

⎤

⎢

⎣

⎡ +

×

=

i 1

i 1 A F

Where: F = Future Value ($)

A = Annual Amount ($)

n = Time (years)

i = Interest (% per year)

Solution methods for finding future values:

Use the F/A column on a Interest Factors table (Bowman

text appendix B, page 580).

Notation F/A is interpreted as → Find F given A

Notation (F/A, n, i) is interpreted as → Find F given A

for n years at i interest rate.

Use the Excel function1→ FV(rate, nper, pmt, pv, type)

Use the formula and calculator

9

1 Note that the cash outflows are entered as a negative number.

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(/ )

Future Value/Annual – Example

Your plan is to save $100 at the end of each year at 8% interest

What will be the size of the account in 10 years?

( )

$1,448.70 F 14.487

$100 8%

10, F/A, A F

=

×

=

×

=

10

FValue/Annual – Example (continued)

Solution using Excel®:

Note that the

annual amount

11

was entered as

a negative

number which

indicates a cash

outflow.

( )

1

‐ 0.08) (1

$

i 1 ‐

i 1 A F

10 n

⎤

⎡ +

⎥

⎤

⎢

⎡ +

×

=

FValue/Annual – Example (continued)

Solution using formula:

F = Future Value = ?

A = Annual Amount = $100

$1,448.66 F 14.48656

$100

0.08 1 ‐ 2.158925

$100

0.08 1 0.08) (1

$100

=

×

=

⎥

⎤

⎢

⎡ +

×

=

12

n = Time = 10 years

i = Interest = 8% per year

The slight difference

between this amount and

the amount determined by

the factor from the tables is

due to rounding.

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Equal annual amounts for a future value is determined by the

following equation:

⎤

Note that this

formula is the

13

( ) ⎥ ⎦ ⎤

⎢

⎣

⎡ +

×

=

1 ‐

i 1

i F

Where: F = Future Value ($)

n = Time (years)

inverse of the

formula to find

F given A.

Solution methods for finding annual amounts:

Use the A/F column on a Interest Factors table (Bowman

Notation A/F is interpreted as → Find A given F

Notation (A/F, n, i) is interpreted as → Find F given A

Use the Excel function1→ PMT(rate, nper, pv, fv, type)

14

Annual/Future Value – Example

Your goal is to save $7,500 for a car down payment in 4 years

by investing part of your end‐of‐year bonus.

How much to you need to save annually at 4% interest?

( )

year

$1,766.25/

A

0.2355

$7,500 4%

4, A/F, F A

=

×

=

×

=

15

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Annual/FValue – Example (continued)

Note that the

function returns

ti

16

a negative

number which

indicates a cash

outflow.

( )1 i ‐1

i F

A= ×⎢⎡ + n ⎥⎤

Annual/FValue – Example (continued)

Solution using the formula:

F = Future Value = $7,500

A = Annual Amount = ? ( )

$1,766.17 A

0.235490

$7,500

1 ‐ 1.169859

0.04

$7,500

1 ‐ 0.04) (1

0.04

$7,500

1

i 1

4

=

×

=

⎥⎦

⎤

⎢⎣

⎡

×

=

⎥

⎤

⎢

⎡ +

×

=

⎦

⎣ +

17

n = Time = 4 years

The slight difference

between this amount and

the amount determined by

the factor from the tables is

due to rounding.

End Unit 1 Material

Additional Reading Ö Financial Functions:

http://www.functionx.com/excel/Lesson12.htm

Go to Unit 2 Present Value Amounts

18

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Present Value Amounts

amounts

20

Present Value Calculations

Present value for an equal annual amount is determined by the

following equation:

( )

⎥

⎤

⎢

1

‐ i 1

21

( ) ( ) ⎥ ⎦ ⎤

⎢

⎣

⎡ +

+

×

i 1 i 1

i 1 A P

Where: P = Present Value ($)

n = Time (years)

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Solution methods for finding present values:

Use the P/A column on a Interest Factors table (Bowman

Notation P/A is interpreted as → Find P given A

Notation (P/A, n, i) is interpreted as → Find P given A

Use the Excel function1→ PV(rate, nper ,pmt, fv, type)

22

Present Value/Annual – Example

You scheduled to receive $15,000 at the end of the next 7 years

If the current interest rate is 6%, what is the equivalent amount

today?

( )

$83,736 P

5.5824

$15,000 6%

7, P/A, A P

=

×

=

×

=

23

PValue/Annual – Example (continued)

Note that the

24

function returns a

negative number

which indicates a

cash outflow.

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( ) ( )1 i

i 1

i 1 A

n

⎥

⎤

⎢

⎡ +

− +

×

=

PValue/Annual – Example (continued)

Solution using the formula:

P = Present Value = ?

A = Annual Amount = $15 000

$83,735.72 P

5.582381

$15,000

(1.50363) (0.06)

1 1.50363

$15,000

0.06) (1 0.06

1 0.06) (1

$15,000 7

7

=

×

=

⎥

⎤

⎢

⎡

×

−

×

=

⎥

⎤

⎢

⎡ +

×

− +

×

=

25

A = Annual Amount = $15,000

n = Time = 7 years

Equal annual amounts for a present value is determined by the

following equation:

( )

⎥

⎤

⎢

⎡ i 1 + in

Note that this

formula is the

26

( ) ( ) ⎥ ⎦

⎤

⎢

⎣

⎡ +

+

×

=

1 ‐

i 1

i 1 i P

Where: P = Present Value ($)

n = Time (years)

inverse of the

formula to find

P given A.

Use the A/P column on a Interest Factors table (Bowman

Notation A/P is interpreted as → Find A given P

Notation (A/P, n, i) is interpreted as → Find A given P

Use the Excel function1→ PMT(rate, nper, pv, fv, type)

27

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(A/P 8 3%)

P

Annual/Present Value – Example

You $5,000 invest in an account that returns 6% annual interest

How much can you withdraw each semester (twice/year) over

the next 4 years for books and supplies?

( )

mester

$712.50/se A

0.1425

$5,000 3%

8, A/P, P A

=

×

=

×

=

28

Time periods other than a year can be used, however, the tabulated

interest rate is a yearly rate so it must be adjusted to match the number

of periods/ year → 6% per year/2 periods per year = 3% per period . Also

the total number of periods is used → 4 yrs x 2 periods/yr = 8 periods.

Annual/PValue – Example (continued)

Remember that i

and n must be

29

and n must be

adjusted for time

periods other

than yearly.

1 ‐ m r 1 m r 1 m r P m A

nm nm

⎥

⎦

⎤

⎢

⎣

⎡

⎟

⎜

⎛ +

⎟

⎜

⎛ +

×

=

Annual/PValue – Example (continued)

Solution using the formula

(see page 187 Bowman text):

P = Present Value = $5 000

mester

$712.28/se m A 0.142456

$5,000

1 ‐ 1.26677 1.26677 0.03

$5,000

1 ‐ ) 2 0.06 (1

) 2 0.06 (1 2 0.06

$5,000

2 2

=

×

=

⎥⎦

⎤

⎢⎣

×

=

⎥

⎦

⎤

⎢

⎣

⎡ +

+

×

=

×

30

P = Present Value = $5,000

A = Annual Amount = ?

n = Time = 4 years

M = #Periods/year = 2

r = Annual Interest = 6%

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Occasionally an engineering economic analysis will occur

when the number of years (n) or the interest rate (i) is

unknown

Like single‐payment calculations, if three of the four factors g p y ,

are known, we can solve for the unknown

Future value factors → F, A, i, n

Present value factors → P, A, i, n

31

Unknown i and n Calculations

Solution methods for finding interest or time period values:

Interpolate using the appropriate column on a Interest

Factors table (Bowman text appendix B, page 580).

Use the Excel functions1:

Use the Excel functions :

RATE(nper, pmt, pv, fv, type, guess) → returns the

interest rate per period for a cash flow

NPER(rate, pmt, pv, fv, type) → returns the number of

periods for a cash flow with a constant interest rate

Rearrange the appropriate formula and solve with your

calculator

32

Unknown Interest – Example

You have $5,000 to invest in an account and would like to

withdraw $1,550 per year for the next four years

What interest rate will be required to meet the needs?

33

You will need to

invest the $5,000

at 9.2% annual

interest rate.

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Unknown Time – Example

You plan to invest $1,250 per year in a security with a 4.75%

annual return rate.

How many years before the account grows to $12,500?

34

Time Required

8 years

4 months

15 days

End Unit 2 Material

Go to Unit 3 Gradient Amounts

35

Gradient Amounts

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amounts

37

Gradient Amounts

Unlike equal annual amounts, gradient amounts increase or

decrease each time period. Types:

Linear – change in cash flow is by an equal amount for

each time period. Gradient factors tabulated in the p

interest tables or determined by formula

Non‐linear – change is cash flow varies between time

periods. Non‐linear gradient functions must be calculated

with a series of P/A or A/P for each time period

38

Gradient Amounts

Assumptions for linear gradient amount analysis include:

Cash flow occurs at the end of each year

Change in cash flow year to year is at a constant rate. The

amount of change is designated → Gg g

Initial cash flow is designated → A’

39

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Use the A/G column on a Interest Factors table (Bowman

text appendix B, page 580). Future or present values can

then be determined using the annual amount (A):

Notation A/G is interpreted as → Find A given G

Notation (A/G, n, i) is interpreted as → Find A given P

Use the Excel function1→ XNPV(rate, values, dates)

40

Gradient Calculations

Equal annual amounts for a linear gradient values is determined

by the following equation:

⎥

⎤

⎢

⎡

−

×

±

′

A

41

( ) ⎥ ⎦

⎢

×

±

=

1 ‐

i 1 i G A

Where: A = Annual Amount ($)

A′ = Initial Cash Flow($)

G = Gradient Amount ($)

n = Time (years)

When the change

is increasing

between time

periods, the

gradient is added

(+) to the initial

value and

subtracted (‐)

when decreasing.

Gradient – Example

Your 1styear’s salary is $45,000. Your contract states that

your raise will be $5,000/year in years 2 through 6

What is the present value of the contract at 5% interest?

Cash Flow Diagram:

42

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(/ )

′

Gradient – Example (continued)

Solution method:

Find annual value (A) of the gradient (G)

Convert the annual amount (A) into the present value (P)

0

$288,246.5 P

(5.0757)

$56,789.50 5%) 6, A(P/A, P

/year

$56,789.50 A

2.3579

$5,000

$45,000

5%

6, A/G, G A A

=

×

=

× +

= +

′

=

43

Solution using Excel®. You must create a schedule of amounts

with a date. The schedule must start at time = 0 (today)

Non‐linear gradients can be solved with this method

44

( ) ( ) 6 1 ‐ 0.05 1 6 0.05 1

$5,000

$45,000

1 ‐

i 1 n i 1 G A A

6 n

⎤

⎡

⎥

⎢ − +

× +

=

⎥

⎢ − +

×

±

′

=

Solution using the formulas:

P = Present value = ?

A = Annual Amount = ?

[ ] ( )

( )

[ ] 0

$288,246.9 P

5.075697

$56,789.61

0.05) (1 0.05 1 0.05) (1

$56,789.61

i 1

i 1 A P /year

$56,789.61 A

2.357922

$5,000

$45,000

1 ‐ 1.340096 6 ‐ 20

$5,000

$45,000

6 6 n n

=

×

=

⎥

⎢ +× + −

×

=

⎥

⎢ ++−

×

=

× +

=

⎥⎦

⎤

⎢⎣

⎡

× +

=

45

A′ = Initial Amount = $45,000

G = Gradient = $5,000

n = Time = 6 years

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Go to Unit 4 Perpetual Amounts and

Deferred Annuities

46

Perpetual Amounts and Deferred

Annuities

amounts

48

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