1. Variations in Present Worth Analysis
Lecture No. 17
Chapter 5
Contemporary Engineering Economics
Copyright © 2016

2. Future Worth Criterion
Given
Cash flows and MARR (i)
Find
The net equivalent worth at a specified period other than the “present,” commonly at the end of the project life
Decision Rule
Accept the project if the equivalent worth is positive.
$47,309
$37,360
$35,560
$31,850
$34,400 1 2 3
$76,000
Project life

3. Excel Solution A B C 1 Period Cash Flow 2 ($76,000) 3 $35,650 4
($76,000) 3
$35,650 4
$37,360 5
$31,850 6
$34,400 7
PW(12%)
$30,145 8
FW(12%)
$47,434
=FV(12%,4,0,-B7)

4. FW Calculation with the Cash Flow Analyzer
Payback
Period
Project
Cash
Flows
Net
Present
Worth
Net
Future
Worth

5. Example 5.6: Future Equivalent at an Intermediate Time
Figure: 05-09EXM

6. Example 5.8: 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
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?

7. Mr. Bracewell’s Hydroelectric Project
Figure: 05-11EXM

8. Find P for a Perpetual Cash Flow Series, A
Figure: 05-10

9. 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 n
P=CE(i)

10. Practice Problem $2,000 $1,000 ∞ P = CE (10%) = ?
Given: i = 10%, N = ∞
Find: P or CE (10%)
$2,000
$1,000 10 ∞
P = CE (10%) = ?

11. Solution
$2,000
$1,000 10 ∞
P = CE (10%) = ?

12. 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%

13. Cash Flow Diagram for the Bridge Construction Project
Years 15 30 45 60
$50,000
$500,000
$500,000
$500,000
$500,000
$2,000,000

14. Solution Construction Cost P1 = $2,000,000 Maintenance Costs
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

15. Alternate Way to Calculate P3
Concept: Find the effective interest rate per payment period.
Interest rate: Find the effective interest rate for a 15-year cycle.
i = ( )15 − 1
= %
Capitalized equivalent worth
P3 = $500,000/1.0789
= $463,423
Effective interest rate
for a 15-year period 15 30 45 60
$500,000