Embed Size (px)
Citation preview
Contemporary Engineering
Economics, 4th edition, © 2007
Variations of Present Worth Analysis
Lecture No.17Chapter 5Contemporary Engineering EconomicsCopyright © 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
Decision Rule: Accept the project if the equivalent worth is positive.
$75,000
$24,400 $27,340$55,760
01 2 3
Project life
Contemporary Engineering
Economics, 4th edition, © 2007
Example 5.6 Net Future Worth at the End of the Project
Contemporary Engineering
Economics, 4th edition, © 2007
FW F P F P
F P
FW F P
FW
( $24, ( / , ) $27, ( / , )
$55, ( / , )
$119,
( $75, ( / , )
$114,
( $119, $114,
$5, ,
15%) 400 15%,2 340 15%,1
760 15%,0
470
15%) 000 15%,3
066
15%) 470 066
404 0
inflow
outflow
Accept
Alternate Way of Computing the NFW
Contemporary Engineering
Economics, 4th edition, © 2007
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)
Excel Solution:
Contemporary Engineering
Economics, 4th edition, © 2007
Solving Example 5.6 with Cash Flow Analyzer
NetPresentWorth
Net FutureWorth
Paybackperiod
ProjectCash Flows
Contemporary Engineering
Economics, 4th edition, © 2007
Example 5.7 Future Equivalent at an Intermediate Time
Contemporary Engineering
Economics, 4th edition, © 2007
• 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?
Example 5.9 Project’s Service Life is Extremely Long
Contemporary Engineering
Economics, 4th edition, © 2007
Mr. Bracewell’s Hydroelectric Project
1 $50 ( / ,8%,9) $50 ( / ,8%,8)
$100 ( / ,8%,1) 60
$1,101
V K F P K F P
K F P K
K
2 120 ( / ,8%,50)
$1,468
V K P A
K
1 2 $1,101 $1,468
$367 0
V V K K
K
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
10
$1,000
$2,000
P = CE (10%) = ?
0
Given: i = 10%, N = ∞Find: P or CE (10%)
∞
Contemporary Engineering
Economics, 4th edition, © 2007
Solution
CE P F($1,
.
$1,
.( / , )
$10, ( . )
$13,
10%)000
0 10
000
0 1010%,10
000 1 0 3855
855
10
$1,000
$2,000
P = CE (10%) = ?
0∞
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
$500,000 $500,000 $500,000 $500,000
$2,000,000
$50,000
0 15 30 45 60
Years
Cash Flow Diagram for the Bridge Construction Project
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
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
15 30 45 600
$500,000 $500,000 $500,000 $500,000
