in Present Worth Analysis Lecture No 17 Chapter 5 Contemporary Engineering Economics Copyright 2016 Future Worth Criterion Given Cash flows and MARR i Find The net equivalent worth at a specified period other ID: 623631
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Slide1
Variations in Present Worth Analysis
Lecture No
. 17
Chapter 5
Contemporary Engineering Economics
Copyright ©
2016Slide2
Future Worth Criterion
Given
Cash
flows and MARR (
i
)FindThe net equivalent worth at a specified period other than the “present,” commonly at the end of the project life Decision RuleAccept the project if the equivalent worth is positive.
$
76,000
$35,560
$37,360
$34,400
0
1
2
3
Project life
$31,850
$47,309Slide3
Excel Solution
A
BC1PeriodCash Flow
20($76,000)31$35,65042$37,36053
$31,850
6
4
$34,400
7
PW(12%)
$30,145
8
FW(12%)
$47,434
=
FV(12%,4,0,-B7)Slide4
FW Calculation with the Cash Flow Analyzer
Net
PresentWorthNet FutureWorthPaybackPeriodProjectCash FlowsSlide5
Example 5.6:
Future Equivalent at an Intermediate TimeSlide6
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?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 yearsSlide7
Mr. Bracewell’s
Hydroelectric ProjectSlide8
Find P for a Perpetual Cash Flow Series,
ASlide9
Capitalized Equivalent Worth
Principle
: PW for a project with an annual receipt of A over infinite service lifeEquation CE(i) = A(P/A, i, ) = A/iA
0P=CE(i)nSlide10
Practice Problem
10
$1,000$2,000
P = CE (10%) = ?0Given: i = 10%, N = ∞Find: P or CE (10%)∞Slide11
Solution
10
$1,000
$2,000P = CE (10%) = ?0∞Slide12
A Bridge Construction Project
Construction cost = $
2,000,000Annual maintenance cost = $50,000Renovation cost = $500,000 every 15 yearsPlanning horizon = infinite periodInterest rate = 5%Slide13
Cash Flow Diagram for the Bridge Construction Project
$500,000
$500,000$500,000$500,000$2,000,000
$50,0000 15304560YearsSlide14
Solution
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
=
$3,463,423Slide15
Alternate
Way
to Calculate P3Concept: Find the effective interest rate per payment period.Interest rate: Find the effective interest rate for a 15-year cycle.i = (1 + 0.05)15 − 1 = 107.893%Capitalized equivalent worth P3 = $500,000/1.0789 = $463,42315304560
0$500,000$500,000$500,000$500,000Effective interest rate for a 15-year period