Engineering Economy 8th, Summaries of Engineering Economy

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Lecture slides to accompany Engineering Economy 8 th^ edition

Leland Blank Anthony Tarquin

Chapter 2

Factors: How

Time and

Interest Affect

Money

LEARNING OUTCOMES

1. F/P and P/F Factors
2. P/A and A/P Factors
3. F/A and A/F Factors
4. Factor Values
5. Arithmetic Gradient
6. Geometric Gradient
7. Find i or n

Equation

. . .

Single Payment Factors (F/P and

P/F)

Standard Factor Notation

As if they are numerator and denominat

F/P and P/F for

Spreadsheets

Future value F is calculated using FV

function:

= FV( )

Present value P is calculated using PV

function:

= PV( )

Note the use of double commas in each function

8

Introduction to Spreadsheet

Functions

Excel financial functions

Present Value, P: = (i%,n,A,F) Future Value, F: = (i%,n,A,P) Equal, periodic value, A: = (i%,n,P,F) Number of periods, n: = (i%,A,P,F) Compound interest rate, i: = (n,A,P,F) Compound interest rate, i: = (first_cell:last_cell) Present value, any series, P: = (i%,second_cell:last_cell) +

Example: Estimates are P = $5000 n = 5 years i = 5% per year Find A in $ per year Function and display: = (5%, 5, -5000) displays A = $1154.

Assume

Transaction s

Example: Finding Present Value

A small company wants to make a single deposit now so it will have enough money to purchase a backhoe costing $50,000 five years from now. If the account will earn interest of 10% per year, the amount that must be deposited now is nearest to:

(A) $10,000 (B) $ 31,050 (C) $ 33,250 (D) $319,

The cash flow diagram is: Solution: P = F(P/F,i,n ) = 50,000( )

)

Answer is ( )

Uniform Series Involving P/A

and A/P

The cash flow diagrams are:

Standard Factor Notation

P = A(P/A,i,n) A = P(A/P,i,n)

Note: P is of first A value

(1) Cash flow occurs in interest periods

The uniform series factors that involve P and A are derived as fo

(2) Cash flow amount is in each interest period

Uniform Series Involving F/A

and A/F

(1) Cash flow occurs in consecutive interest periods

The uniform series factors that involve F and A are derived as fo

(2) Last cash flow occurs in same period as F

Note: F takes place in the ______ period as last A

Cash flow diagrams are:

Standard Factor Notation

F = A(F/A,i,n) A = F(A/F,i,n)

Example: Uniform Series

Involving F/A An industrial engineer made a modification to a chip manufacturing process that will save her company $10,000 per year. At an interest rate of 8% per year, how much will the savings amount to in 7 years?

(A) $45,300 (B) $68,500 (C) $89,228 (D) $151, The cash flow diagram is:

A = $10,

F

i = 8% 0 1 2 3 4 5 6 7

Solution:

F = 10,000( )

Answer is ( )

Example: Untabulated i

Determine the value for (F/P, 8.3%,10)

Formula: F = (1 + 0.083)^10 = 2. Spreadsheet: = FV(8.3%,10,,1) = 2. Interpolation: 8% ------ 2. 8.3% ------ x 9% ------ 2.

x = 2.1589 + [(8.3 - 8.0)/(9.0 - 8.0)][2.3674 – 2.1589] = 2.

Absolute Error = 2.2215 – 2.2197 = 0.

OK OK

(Too high)

Arithmetic Gradients

Arithmetic gradients change by the ______ amount

e ach period

The cash flow diagram for

the PG

of an arithmetic gradient is:

G starts between periods 1 and 2 (not between 0 and 1)

This is because cash flow in year 1 is usually not equal to G and is handled separately as a base amount (shown on next slide)

Note that PG is located Ahead of the first change that is equal to G Standard factor notation is PG = G(P/G,i,n)

Converting Arithmetic Gradient

to A

i = 10% 0 1 2 3 4 5

G 2G (^) 3G 4G

i = 10% 0 1 2 3 4 5

A =?

Arithmetic gradient can be converted into equivalent A value using G

General equation when base amount is involved is A = base amount + G(A/G,i,n) 0 1 2 3 4 5

4G

3G

2G G

For decreasing gradients, change plus sign to minus

A = base amount - G(A/G,i,n)

Example: Arithmetic Gradient

= 400(3.6048) + 30(6.3970) = $1,633. Answer is ( )

PT = 400( ,12%,5) + 30( ,12%,5)

The cash flow could also be converted A = ____ + 30(into an^ A^ value as follows: ,12%,5) = 400 + 30(1.7746) = $453.

Solution: