Lecture No.17
Chapter 5
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5 th edition, © 2010

 Given : Cash flows and
MARR (i)
 Find : The net equivalent worth at a specified period other than “present”, commonly the end of project life
 Decision Rule : Accept the project if the equivalent worth is positive.
0
$35,560
$76,000
1
$37,360
2
$47,309
$31,850 $34,400
3
Project life
Contemporary Engineering Economics, 5 th edition, © 2010

Example 5.6 Future Equivalent at an
Intermediate Time
Contemporary Engineering Economics, 5 th edition, © 2010

Example 5.8 Project’s Service Life is Extremely
Long
 : Was Bracewell's
$800,000 investment a wise one?
 Q2 : How long does he have to wait to recover his initial investment, and will he ever make a profit?
Contemporary Engineering Economics, 5 th edition, © 2010

Ex. 5.8: Mr. Bracewell’s Hydroelectric Project
V
1

V
2
 
$1,101 K

$1, 468 K

$367 K

0
V
2


$1, 468 K
V
1
 
$50 (

K F P

$100 (

 
$1,101 K
K
Contemporary Engineering Economics, 5 th edition, © 2010

How Would You Find P for a Perpetual Cash
Flow Series, A?
Contemporary Engineering Economics, 5 th edition, © 2010

 Principle : PW for a project with an annual receipt of A over infinite service life 0
A
 Equation :
P =CE(i)
= A/i
Contemporary Engineering Economics, 5 th edition, © 2010


Given : i = 10%, N = ∞
Find : P or CE (10%)
$2,000
$1,000
0
10
P = CE (10%) = ?
Contemporary Engineering Economics, 5 th edition, © 2010
∞

$2,000
0
P = CE (10%) = ?
$1,000
10 ∞
CE(10%)


0.10
0.10
$10,000(1 0.3855)

$13,855
Contemporary Engineering Economics, 4 th edition, © 2007

 Construction cost = $2,000,000
 Annual Maintenance cost = $50,000
 Renovation cost = $500,000 every 15 years
 Planning horizon = infinite period
 Interest rate = 5%
Contemporary Engineering Economics, 5 th edition, © 2010

Cash Flow Diagram for the Bridge
Construction Project
0
15
Years
30 45
$50,000
60
$2,000,000
$500,000 $500,000 $500,000
Contemporary Engineering Economics, 5 th edition, © 2010
$500,000

 Construction Cost
P
1
= $2,000,000
 Maintenance Costs
P
2
= $50,000/0.05 = $1,000,000
 Renovation Costs
P
3
= $500,000(P/F, 5%, 15)
+ $500,000(P/F, 5%, 30)
+ $500,000(P/F, 5%, 45)
+ $500,000(P/F, 5%, 60)
.
= {$500,000(A/F, 5%,15)}/0.05
= $463,423
 Total Present Worth
P = P
1
+ P
2
+ P
3
= $3,463,423
Contemporary Engineering Economics, 5 th edition, © 2010

Alternate way to calculate P
3
 Concept : Find the effective interest rate per payment period
Effective interest rate for a 15-year period
 Effective interest rate for a 15-year cycle
0 15 30 45 60
i = (1 + 0.05) 15 - 1
= 107.893%
$500,000 $500,000 $500,000 $500,000
 Capitalized equivalent worth
P
3
= $500,000/1.0789
= $463,423
Contemporary Engineering Economics, 5 th edition, © 2010