Variations of Present Worth
Analysis
Lecture No.17
Chapter 5
Contemporary Engineering Economics
Copyright © 2006
Contemporary Engineering Economics, 4th
edition, © 2007
Future Worth Criterion



Given: Cash flows
and MARR (i)
Find: The net
equivalent worth at a
specified period other
than “present”,
commonly the end of
project life
0
Decision Rule: Accept
the project if the
equivalent worth is
positive.
$24,400
1
$55,760
$27,340
2
3
$75,000
Contemporary Engineering Economics, 4th
edition, © 2007
Project life
Example 5.6 Net Future Worth at the End
of the Project
Contemporary Engineering Economics, 4th
edition, © 2007
Alternate Way of Computing the NFW
FW (15%)inflow  $24,400( F / P,15%,2)  $27,340( F / P,15%,1)
$55,760( F / P,15%,0)
 $119,470
FW (15%)outflow  $75,000( F / P,15%,3)
 $114,066
FW (15%)  $119,470  $114,066
 $5,404  0, Accept
Contemporary Engineering Economics, 4th
edition, © 2007
Excel Solution:
A
B
C
1
Period
Cash Flow
2
0
($75,000)
3
1
$24,400
4
2
$27,340
5
3
$55,760
6
PW(15%)
$3553.46
7
FW(15%)
$5,404.38
=FV(15%,3,0,-B6)
Contemporary Engineering Economics, 4th
edition, © 2007
Solving Example 5.6 with Cash Flow
Analyzer
Payback
period
Project
Cash
Flows
Net
Present
Worth
Net
Future
Worth
Contemporary Engineering Economics, 4th
edition, © 2007
Example 5.7 Future Equivalent at an
Intermediate Time
Contemporary Engineering Economics, 4th
edition, © 2007
Example 5.9 Project’s Service Life is
Extremely Long
• Built a hydroelectric plant using his personal savings of $800,000
• Power generating capacity of 6 million kwhs
• Estimated annual power sales after taxes - $120,000
• Expected service life of 50 years
•
Was Bracewell's $800,000 investment a wise one?
•
How long does he have to wait to recover his initial investment,
and will he ever make a profit?
Contemporary Engineering Economics, 4th
edition, © 2007
Mr. Bracewell’s Hydroelectric Project
V1  V2  $1,101K  $1, 468K
 $367 K  0
V2  120 K ( P / A,8%,50)
 $1, 468K
V1  $50 K ( F / P,8%,9)  $50 K ( F / P,8%,8)
 $100 K ( F / P,8%,1)  60 K
 $1,101K
Contemporary Engineering Economics, 4th
edition, © 2007
How Would You Find P for a Perpetual
Cash Flow Series, A?
Contemporary Engineering Economics, 4th
edition, © 2007
Capitalized Equivalent Worth
 Principle: PW for a project with an annual
receipt of A over infinite service life
Equation:
CE(i) = A(P/A, i,
) = A/i
A
0
P = CE(i)
Contemporary Engineering Economics, 4th
edition, © 2007
Practice Problem
Given: i = 10%, N = ∞
Find: P or CE (10%)
$2,000
$1,000
0
10
P = CE (10%) = ?
Contemporary Engineering Economics, 4th
edition, © 2007
∞
Solution
$2,000
$1,000
0
10
P = CE (10%) = ?
$1,000 $1,000

( P / F,10%,10)
0.10
0.10
 $10,000(1  0.3855)
CE(10%) 
 $13,855
Contemporary Engineering Economics, 4th
edition, © 2007
∞
A Bridge Construction Project
 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, 4th
edition, © 2007
Cash Flow Diagram for the Bridge Construction
Project
Years
15
30
45
60
$500,000
$500,000
$500,000
$500,000
0
$50,000
$2,000,000
Contemporary Engineering Economics, 4th
edition, © 2007
Solution:
 Construction Cost
P1 = $2,000,000
 Maintenance Costs
P2 = $50,000/0.05 = $1,000,000
 Renovation Costs
P3 = $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 = P1 + P2 + P3 = $3,463,423
Contemporary Engineering Economics, 4th
edition, © 2007
Alternate way to calculate P3
 Concept: Find the effective interest rate per payment period
0
15
$500,000
30
$500,000
45
60
$500,000 $500,000
 Effective interest rate for a 15-year cycle
i = (1 + 0.05)15 - 1 = 107.893%
 Capitalized equivalent worth
P3 = $500,000/1.07893
= $463,423
Contemporary Engineering Economics, 4th
edition, © 2007