Transcript Variations of Present Worth Analysis

Lecture No.17
Chapter 5
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th
edition, © 2010
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
 Decision Rule: Accept
the project if the
equivalent worth is
positive.
$47,309
$35,560 $37,360 $31,850 $34,400
0
1
$76,000
Contemporary Engineering Economics, 5th
edition, © 2010
2
3
Project life
Example 5.6 Future Equivalent at an
Intermediate Time
Contemporary Engineering Economics, 5th
edition, © 2010
Example 5.8 Project’s Service Life is Extremely
Long
•
Q1: 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, 5th
edition, © 2010
Ex. 5.8: Mr. Bracewell’s Hydroelectric Project
V1  V 2   $1,101 K  $1, 468 K
 $367 K  0
V 2  120 K ( P / A , 8% , 50)
 $1, 468 K
V1   $50 K ( F / P , 8% , 9)  $50 K ( F / P , 8% , 8)
 $100 K ( F / P , 8% ,1)  60 K
  $1,101 K
Contemporary Engineering Economics, 5th
edition, © 2010
How Would You Find P for a Perpetual Cash
Flow Series, A?
Contemporary Engineering Economics, 5th
edition, © 2010
Capitalized Equivalent Worth
 Principle: PW for a
project with an
annual receipt of A
over infinite service
life
A

0
Equation:
CE(i) = A(P/A, i,  )
= A/i
P =CE(i)
Contemporary Engineering Economics, 5th
edition, © 2010
Practice Problem
Given: i = 10%, N = ∞
Find: P or CE (10%)
$2,000
$1,000
0
10
P = CE (10%) = ?
Contemporary Engineering Economics, 5th
edition, © 2010
∞
Solution:
$2,000
$1,000
0
10
∞
CE(10% ) 
P = CE (10%) = ?
$1, 000

$1, 000
(P / F ,10% ,10)
0.10
0.10
 $10 , 000(1  0.3855)
 $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, 5th
edition, © 2010
Cash Flow Diagram for the Bridge
Construction Project
15
Years
30
45
60
$500,000
$500,000
$500,000
$500,000
0
$50,000
$2,000,000
Contemporary Engineering Economics, 5th
edition, © 2010
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, 5th
edition, © 2010
Alternate way to calculate P3
 Concept: Find the
effective interest
rate per payment
period
Effective interest rate
for a 15-year period
Effective interest
0
15
30
45
60
rate for a 15-year
cycle
i = (1 + 0.05)15 - 1
= 107.893%
$500,000 $500,000 $500,000 $500,000
Capitalized equivalent worth
P3 = $500,000/1.0789
= $463,423
Contemporary Engineering Economics, 5th
edition, © 2010