1 CHAPTER 3 LEARNING OBJECTIVES Sct 3.1 Calculations for ...

ENGINEERING ECONOMY, Sixth Edition 

M c 

Graw 

Hill 

Slide Sets to accompany Blank & Tarquin, Engineering 

Economy, 6 th Edition, 2005 



by Blank and Tarquin 

CHAPTER 3 

COMBINING FACTORS 

3-1 

3-3 

© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 

Sct 3.1 Calculations for Uniform 

Series that are Shifted 

For a shifted series the present worth point 

in time is NOT t = 0. 

It is shifted either to the left of “0” or to the 

right of “0”. 

Remember, when 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. 


Example of Shifted Series P and F 

0 1 2 3 4 5 6 7 8 

P0 

P 2 



A = $-500/year 

• F for this series is at t = 6; F 6 = A(F/A,i%,4) 

• P 0 for this series at t = 0 is 

P 0 = -500(P/A,i%,4)(P/F,i%,2) 

3-5 

F 6 


LEARNING OBJECTIVES 



Consider: 



1. Dealing with 

shifted series 

2. Shifted series 

and single 

amounts 

3. Shifted 

gradients 

4. Decreasing 

gradients 

5. Spreadsheet 

applications 

3-2 

Shifted Uniform Series 

3-4 


0 1 2 3 4 5 6 7 8 

P 0 

P 2 

A = $-500/year 

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

P 2 = -500(P/A,i%,4) or could refer to as F 2 

P 0 = P 2 (P/F,i%,2) or could be F 2 (P/F,i%,2) 



3-6 


Sct 3.2 Calculations Involving 

Uniform-Series Uniform Series and Randomly-Placed 

Randomly Placed 

Single Amounts 

Draw and correctly label the cash flow diagram that 

defines the problem 

Locate the present and future worth points for each 

series 

Write the time value of money equivalence 

relationships 

Substitute the correct factor values and solve 


1



Series with additional 

single cash flows 

It is common to find cash flows that are combinations of 

series and single, randomly-placed cash flows 

For present worth, P 

Solve for the series present worth values then move to t = 0 

Then solve for the P at t = 0 for the single cash flows using the 

P/F factor for each cash flow 

Add the equivalent P values at t = 0 

For future worth, F 

Convert all cash flows to an equivalent F using the F/A and F/P 

factors in year t = n 

Add the equivalent F values at t = n 

3-7 

3.2 The PW Points are: 



A = $500 

1 2 3 

F 4 = $300 


0 1 2 3 4 5 6 7 8 

i = 10% 

F 5 = -$400 

t = 1 is the PW point for the $500 annuity; 

“n” = 3 

3-9 


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…. 



3-11 


3.2 Series with Other cash flows 

♦ Consider: 

0 1 2 3 4 5 6 7 8 



A = $500 

i = 10% 

F 4 = $300 

F 5 = -$400 

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



3-8 

3-10 


3.2 The PW Points are: 

Back 4 periods 

A = $500 

1 2 3 

F 4 = $300 

0 1 2 3 4 5 6 7 8 

i = 10% 

Back 5 Periods 



F 5 = -$400 

t = 1 is the PW point for the two other 

single cash flows 

3-12 


3.2 Substitute the factors and 

solve 

P = $500( 2.4869 )( 0.8264 ) 

+ 

$300( 0.6830 ) 

- 

400( 0.6209 ) 

= 

$984.12 

$1,027.58 

$204.90 

$248.36 


2

Sct 3.3 Calculation for Shifted 

Gradients 

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 



3-13 


Example of a Conventional Gradient 

♦ Consider: 



Gradient Series 

……..Base Series …….. 

0 1 2 … n-1 n 

This represents a conventional gradient 

The present worth point is t = 0 

3-15 


Shifted Gradient: Numerical Example 

0 



3-17 

G = $+100 

Base Series = $500 

1 2 3 4 ……….. ……….. 9 10 

Cash flows start at t = 3 

$500/year increasing by $100/year through 

year 10; i = 10%; Find P at t = 0 


A Shifted Gradient 

has its present value 

point removed from 

time t = 0 

A Conventional 

Gradient has its 

present worth point at 

t = 0 





Shifted Gradient 

3-14 

3-16 


Example of a Shifted Gradient 

Gradient Series 

……..Base Series …….. 

0 1 2 … n-1 n 

The present worth point for the 

base series and the gradient is 

here! 

This represents a shifted gradient 



PW of the base series 

P 0 

0 

P 2 



n series = 8 time periods 

Base Series = $500 

1 2 3 4 ……….. ……….. 9 10 

P 2 = 500(P/A,10%,8) = 500(5.3349) = $2667.45 

P 0 = 2667.45(P/F,10%,2) = 2667.45(0.8264) 

= $2204.38 

3-18 


3


P 0 

0 

P 2 



PW for the gradient component 

1 2 3 4 ……….. ……….. 9 10 

3-19 

G = +$100 

P 2 = $100(P/G,10%,8) = $100(16.0287) = $1,602.87 

P 0 = $1,602.87(P/F,10%,2) = $1,602.87(0.8264) 

= $1,324.61 


To Find A for a Shifted Gradient 

1) Find the present worth of the gradient 

at actual time 0 

2) Then apply the (A/P,i,n) factor to 

convert the present worth to an 

equivalent annuity (series) 



3-21 


Shifted Geometric Gradient 




A 1 

0 1 2 3 … … … n 

Present worth point is at t = 2 for this example 

3-23 


Example: Total Present Worth Value 

For the base series 

P 0 = $2204.38 

For the arithmetic gradient 

P 0 = $1,324.61 

Total present worth 

P = $2204.38 + $1,324.61 = $3528.99 





3-20 

3-22 



A 1 

Conventional Geometric Gradient 

0 1 2 3 … … … n 

Present worth point is at t = 0 for a conventional 

geometric gradient 


Shifted Geometric Gradient Example 

0 1 2 3 4 5 6 7 8 

A = $700/year 

A 1 = $400 in year t = 5 



i = 10%/year 

3-24 

12% increase/year 


4

Geometric Gradient Example 

0 1 2 3 4 5 6 7 8 

A = $700/year 

PW point for the A 

series is t = 0 

PW point for the 

gradient is t = 4 



i = 10%/year 

3-25 

12% increase/year 


3.3 The Annuity Present Worth 

♦ PW of the Annuity 

0 1 2 3 4 5 6 7 8 

A = $700/yr 

P 0 = $700(P/A,10%,4) 



i = 10%/year 

= $700( 3.1699 ) = $2,218.94 

3-27 


Sct 3.4 Shifted Decreasing Arithmetic 

Gradients 

Given the following shifted, decreasing gradient 

i = 10%/year 



F 3 = $1,000; G = $-100 

0 1 2 3 4 5 6 7 8 

Find the present worth at t = 0 

3-29 


3.3 The Gradient Amounts 



t Base Amt 

51 

$400.00 

62 

$448.00 

73 

$501.76 

8 

4 $561.97 

Present Worth of the Gradient at t = 4 

3.73674 

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

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

P 0 = $1,020,88 



3-26 

3-28 


3.3 Total Present Worth 

♦ Geometric Gradient @ t = 

♦ P 0 = $1,020,88 

♦ Annuity 

♦ P 0 = $2,218.94 

♦ Total Present Worth” 

♦ $1,020.88 + $2,218.94 

♦ = $3,239.82 


PW for Shifted Decreasing Gradient 



First, find PW at t = 2 

F 3 = $1,000; G = -$100 

0 1 2 3 4 5 6 7 8 

PW point at t = 2 

3-30 

i = 10%/year 


5

Shifted Decreasing Gradient Example 

i = 10%/year 

0 1 2 3 4 5 6 7 8 

P 0 here 

P 2 



Second, find the PW at t = 0 

F 3 = $1,000; G = $-100 

The gradient amount is subtracted from the base 

amount 

3-31 


3.4 Time Periods Involved 

0 1 2 3 4 5 6 7 8 

P 0 here 




Economy, 6th Edition, 2005 

F 3 = $1,000; G=-$100 

1 2 3 4 5 

P 2 or, F 2 : Take back to t = 0 

Dealing with n = 5. 

3-33 

Summary 

i = 10%/year 


Chapter summarizes cash flow 

patterns that are shifted away from time 

t = 0 

Illustrations of using multiple factors 

to perform PW or FW analysis for 

shifted cash flows 

Illustrations of shifted arithmetic and 

geometric gradients 

Illustrations of the power of Excel 


financial functions3-35 

Shifted Decreasing Gradient Example 

i = 10%/year 

0 1 2 3 4 5 6 7 8 

P 0 here 

P 2 





F 3 = $1,000; G = -$100 

Base amount = $1,000 

3-32 

3-34 


3.4 Time Periods Involved 

F 3 = $1,000; G=-$100 

$1,000 

G = -$100/yr 

1 2 3 4 5 

i = 10%/year 

0 1 2 3 4 5 6 7 8 

P 2 = $1,000( P/A,10%,5 ) – 100( P/G,10%.5 ) 

P 2 = $1,000( 3.7908 ) - $100( 6.8618 ) = $3,104.62 

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





End of Slide Set 

3-36 



6

1 CHAPTER 3 LEARNING OBJECTIVES Sct 3.1 Calculations for ...

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