Engineering Economy Chapter 3x

Authored by Don Smith, Texas A&M University 2004 1

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

CHAPTER 3

COMBINING FACTORS

Mc

GrawHill

ENGINEERING ECONOMY, Sixth Edition by Blank and

Tarquin



Learning Objectives

Shifted SeriesShifted Series and Single AmountsShifted GradientsDecreasing GradientsSpreadsheets



Section 3.1

Calculations For Uniform Series That

Are Shifted



1. Uniform Series that are SHIFTED

A shifted series is one whose present worth point in time is NOT t = 0.Shifted either to the left of “0” or to the right of t = “0”.Dealing with a uniform series: The PW point is always one period to the

left of the first series value No matter where the series falls on the time

line.



Shifted Series

0 1 2 3 4 5 6 7 8

A = -$500/year

Consider:

P of this series is at t = 2 (P3 or F3)

P3 = $500(P/A,i%,4) or, could refer to as F3

P0 = P3(P/F,i%,2) or, F3(P/F,i%,2)

P3P0



Shifted Series: P and F

A = -$500/year

Consider:

F for this series is at t = 6

F6 = A(F/A,i%,4)

0 1 2 3 4 5 6 7 8

P3P0

F at t = 6



Figure 3.3, Suggested Steps

Draw and correctly label the cash flow diagram that defines the problemLocate the present and future worth points for each seriesWrite the time value of money equivalence relationshipsSubstitute the correct factor values and solve



Section 3.2

Uniform Series and Randomly Placed Single Amounts



3.2 Series with other single cash flows

It is common to find cash flows that are combinations of series and other single cash flows.Solve for the series present worth values then move to t = 0Solve for the PW at t = 0 for the single cash flowsAdd the equivalent PW’s at t = 0



3.2 Series with Other cash flows

Consider:

0 1 2 3 4 5 6 7 8

A = $500

F5 = -$400

F4 = $300

•Find the PW at t = 0 and FW at t = 8 for this cash flow

i = 10%



3.2 The PW Points are:

F5 = -$400

F4 = $300

A = $500

0 1 2 3 4 5 6 7 8

i = 10%

t = 1 is the PW point for the $500 annuity;“n” = 3

1 2 3



3.2 The PW Points are:

F5 = -$400

F4 = $300

A = $500

0 1 2 3 4 5 6 7 8

i = 10%

t = 1 is the PW point for the two other single cash flows

1 2 3

Back 4 periods

Back 5 Periods



3.2 Write the Equivalence Statement

P = $500(P/A,10%,3)(P/F,10%,2)+ $300(P/F,10%,4)-

400(P/F,10%,5)

Substituting the factor values into the equivalence expression and solving….



3.2 Substitute the factors and solve

P = $500( 2.4869 )( 0.8264 )+ $300( 0.6830 )-

400( 0.6209 )=

$831.06

$1,027.58

$204.90

$248.36



Section 3.3

Calculations for Shifted Gradients



3.3 The Linear Gradient - revisited

The Present Worth of an arithmetic gradient (linear gradient) is always located: One period to the left of the first cash

flow in the series ( “0” gradient cash flow) or,

Two periods to the left of the “1G” cash flow



3.3 Shifted Gradient

A Shifted Gradient is one whose present value point is removed from time t = 0.A Conventional Gradient is one whose present worth point is t = 0.



3.3 Example of a Conventional Gradient

Consider:

……..Base Annuity ……..

Gradient Series

0 1 2 … n-1 n

This Represents a Conventional Gradient.

The present worth point is t = 0.



3.3 Example of a Shifted Gradient

Consider:

……..Base Annuity ……..

Gradient Series

0 1 2 … n-1 n

This Represents a Shifted Gradient.

The Present Worth Point for the Base Annuity and the Gradient

would be here!



3.3 Shifted Gradient: ExampleGiven:

Base Annuity = $100

G = +$100

0 1 2 3 4 ……….. ……….. 9 10

Let C.F start at t = 3:

$500/ yr increasing by $100/year through year 10; i = 10%; Find the PW at t = 0



3.3 Shifted Gradient: ExamplePW of the Base Annuity

Base Annuity = $100

0 1 2 3 4 ……….. ……….. 9 10

P2 = $100( P/A,10%,8 ) = $100( 5.3349 ) = $533.49

Nannuity = 8 time periods

P0 = $533.49( P/F,10%,2 ) = $533.49( 0.8264 )

= $440.88



3.3 Gradient Present Worth

For the gradient component

0 1 2 3 4 ……….. ……….. 9 10

G = +$100

• PW of gradient is at t = 2:

•P2 = $100( P/G,10%,8 ) = $100( 16.0287 ) = $1,602.87

•P0 = $1,602.87( P/F,10%,2 ) = $1,602.87( 0.8264 )

• = $1,324.61



3.3 Example: Final Solution

For the Base Annuity P0 = $440.88

For the Linear Gradient P0 = $1,324.61

Total Present Worth: $440.88 + $1,324.61 = $1,765.49



3.3 Shifted Geometric Gradient

Conventional Geometric Gradient

0 1 2 3 … … … n

A1

Present worth point is at t = 0 for a conventional geometric gradient!



3.3 Shifted Geometric Gradient

Conventional Geometric Gradient

0 1 2 3 … … … n

A1

Present worth point is at t = 2 for this example



3.3 Geometric Gradient Example

0 1 2 3 4 5 6 7 8

A = $700/yr

12% Increase/yr

i = 10%/year

A1 = $400 @ t = 5



3.3 Geometric Gradient Example

0 1 2 3 4 5 6 7 8

A = $700/yr

12% Increase/yr

i = 10%/year

PW point for the gradient

PW point for the annuity



3.3 The Gradient Amounts

t Base Amt

1 $400.00

2 $448.00

3 $501.76

4 $561.97

5

6

7

8

Present Worth of the Gradient at t = 4

P4 = $400{ P/A1,12%,10%,4 } = 1,494.70$

3.73674

P0 = $1,494.70( P/F,10%,4) = $1,494.70( 0. 6830 )

P0 = $1,020,88



3.3 The Annuity Present Worth

PW of the Annuity

0 1 2 3 4 5 6 7 8

i = 10%/year

P0 = $700(P/A,10%,4)

= $700( 3.1699 ) = $2,218.94

A = $700/yr



3.3 Total Present Worth

Geometric Gradient @ t = P0 = $1,020,88

Annuity P0 = $2,218.94

Total Present Worth”

$1,020.88 + $2,218.94

= $3,239.82



Section 3.4

Shifted Decreasing Arithmetic Gradients



3.4 Shifted Decreasing Linear Gradients

Given the following shifted, decreasing gradient:

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

Find the Present Worth @ t = 0




Given the following shifted, decreasing gradient:

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100 i = 10%/year

PW point @ t = 2




0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100 i = 10%/year

P2 or, F2: Take back to t = 0

P0 here




0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year


P0 here

Base Annuity = $1,000



3.4 Time Periods Involved

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year


P0 here

1 2 3 4 5

Dealing with n = 5.



3.4 Time Periods Involved

0 1 2 3 4 5 6 7 8

F3 = $1,000; G=-$100

i = 10%/year

1 2 3 4 5

P2 = $1,000( P/A,10%,5 ) – 100( P/G,10%.5 )

$1,000 G = -$100/yr

P2= $1,000( 3.7908 ) - $100( 6.8618 ) = $3,104.62

P0 = $3,104.62( P/F,10%,2 ) = $3104.62( 0 .8264 ) = $2,565.65



Section 3.5

Spreadsheet Applications



3.5 Spreadsheet Applications

Assume Excel is the spreadsheet of choice Instructors may vary on the degree of emphasis placed on spreadsheet useStudent’s Goal: Learn the Excel Financial Functions Create your own spreadsheets to solve

a variety of problems



3.5 NPV Function in Excel

NPV function is basicRequires that all cell in the range so defined have an entry.The entry can be $0…but not blank!Incorrect results can be generated if one or more cells in the defined range is left blank . A “0” value must be entered.



3.5 Spreadsheets

It is assumed if an instructor desires to apply spreadsheets, he or she will provide examples and go over each example and the associated cell formulas.See Appendix A for further details on Excel applications



End of Slide Set

Mc

GrawHill

ENGINEERING ECONOMY, Sixth Edition

Blank and Tarquin

Engineering Economy Chapter 3x

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