This document provides an overview of present worth analysis and cash flow analysis methods for evaluating investment alternatives. It discusses measures of profitability like present worth, annual worth, and future worth. It covers evaluating single alternatives as well as comparing alternatives of equal lives and different lives. It also introduces the concepts of planning horizon, coterminated assumptions, and capitalized worth for comparing long-term alternatives. Examples are provided to illustrate how to apply these concepts and calculate present worth, annual worth, capitalized worth, and to compare investment alternatives.

1. CH 5
Present Worth Analysis –
Part1

2. 08/07/142
Chapter 5 Subjects
Measures of Profitability
 PW method
The planning horizon (study period);
Equal lives using PW & AW;
Different lives using PW & AW;
 FW method
 AW method
 IRR method & MARR (Next Chapter)
 ERR method & MARR (is not covered in text)
Measures of Liquidity
 Payback period
 Investment balance diagram
PW of Bonds

3. Measure of Profitability
1)PW, FW, & AW methods
2)IRR method & MARR
3)ERR method & MARR

4. 08/07/144
PW, AW, & FW
For a single alternative, compute either
PW, AW, FW. If PW, AW, or FW is equal
or greater than zero then the project is
viable, otherwise it should be rejected.

5. 08/07/14Dr. Ayman Abu Hammad5
SELECT THE EQUIVALENT WORTH ALTERNATIVE WITH
THE GREATER WORTH
If : PWA (i) < PWB (i)
then
PWA (i) ( A / P,i,N ) < PWB (i) ( A / P,i,N )
and
AWA (i) < AWB (i)
similarly
PWA (i) ( F / P, i, N ) < PWB (i) ( F / P, i, N )
and
FWA (i) < FWB (i)
Select alternative B

6. 08/07/14 6
COMPARING COST ALTERNATIVES
• For cost alternatives that are compared using the PW
method, the alternative that has the least negative
PW is most economically desirable.
• For cost alternatives that are compared using the AW
method, the alternative that has the least negative
AW is most economically desirable.
• For cost alternatives that are compared using the FW
method, the alternative that has the least negative
FW is most economically desirable.

7. 08/07/147
THREE GROUPS OF MAJOR INVESTMENT ALTERNATIVES
1. Mutually exclusive :
At most one project out of the group can be chosen
2. Independent :
The choice of a project is independent of the choice of any
other project in the group, so that all or none of the projects
may be selected or some number in between
3. Contingent :
The choice of the project is conditional on the choice of one or
more other projects

8. 08/07/148
CASH FLOW ANALYSIS METHODS
The cash-flow analysis methods (previously
described in Chapter 4) for a single project are:
Present Worth ( PW )
Annual Worth ( AW )
Future Worth ( FW )
Internal Rate of Return ( IRR )
External Rate of Return ( ERR )

9. 08/07/149
COTERMINATED ASSUMPTION
You cannot compare alternatives using PW method if
life is not equal.
 A finite and identical study period is used for all
alternatives. Cash flows are repeated per the least
common multiple of lives.
A planning horizon is proposed when AW method is
used. Truncate the remaining useful life by
substituting a salvage value. This planning horizon,
combined with appropriate adjustments to the
estimated cash flows, puts the alternatives on a
common comparable basis

10. 08/07/1410
PLANNING HORIZON
The selected time period over which mutually
exclusive alternatives are compared -- study
period
May be influenced by factors including:
 service period required
 useful life of the shorter-lived alternative
 useful life of the longer-lived alternative
 company policy
It is key that the study period be appropriate
for the decision situation under investigation

11. 08/07/1411
COTERMINATED ASSUMPTION
Guidelines when useful life(s) different in length
than study period
Useful life < study period
a. Cost alternatives -- each cost alternative must
provide same level of service as study period : 1)
contract for service or lease equipment for remaining
time; 2) repeat part of useful life of original alternative
until study period ends
b. investment alternatives -- assume all cash flows
reinvested in other opportunities at MARR to end of
study period

12. 08/07/1412
Comparing Projects of Equal Lives

13. 08/07/14
13
PW of equal live alternatives
Perform a PW analysis of three machines. MARR=
10%. Revenues are common for the three options
Example 5.1 Electric
Powered
Gas-
Powered
Solar-
Powered
First cost -2500 -3500 -6000
Annual Operating Cost (AOC), $ -900 -700 -50
Salvage Value SV, $ 200 350 100
Life (n), years 5 5 5
• PWE=-2500-900(P/A,10%,5)+200(P/F,10%,5)=$-5788
• PWP=-3500-700(P/A,10%,5)+350(P/F,10%,5)=$-5936
• PWS=-6000-50(P/A,10%,5)+100(P/F,10%,5)=$-6127

14. 08/07/1414
Comparing Projects of Different Lives

15. 08/07/1415
PW of different live alternatives
 Perform a PW analysis of following two lease options. MARR= 15%. If a
study period of 5 years and returns would not change, which option
should be selected
Location A Location B
First cost, $ -15,000 -18,000
Annual Operating Cost (AOC), $ -3,500 -3,100
Salvage Value SV, $ 1,000 2,000
Life (n), years 6 9
• Compare using LCM of lives= 18 years
• For life cycles after the first year, the first cost is repeated in year 0 of each new
cycle, which is the last year of the previous cycle. These are years 6 and 12 for A,
and 9 for B.
• Draw cash flow of repeated cycles for the two options over 18 years.
• Calculate the PW at MARR=15%

16. 08/07/14Dr. Ayman Abu Hammad16
 PWA= -15000-15000(P/F,15%,6)+1000(P/F,15%,6)-
15,000(P/F,15%,12)+1000(P/F,15%,12)+1000(P/F,15%,18)-
3500(P/A,15%,18)=$-45,036
 PWB= -18000-18000(P/F,15%,9)+2000(P/F,15%,9)+2000(P/F,15%,18)-
3100(P/A,15%,18)=$-41,384
PW of different live alternatives
0 1 2 3 4
A=$3,100
5 6 7 8 9 10 11 12 13 14 15 16 17 18
PWB=?
A=$3,500
7 8 9 10 11 12 13 14 15 16 17 180 1 2 3 4 5 6
PWA=?
$15,000
$1,000 $1,000
$15,000
$1,000
$15,000
$1,000
$2,000
$18,000 $18,000
$2,000

17. 08/07/1417
PW of alternatives using a Planning Horizon
For a 5 year planning horizon:
PWA= -15000 -3500(P/A,15%,5) +1000(P/F,15%,5)
= $-26,236
PWB = -18000 -3100(P/A,15%,5) +2000(P/F,15%,5)
= $-41,384

18. 08/07/1418
COMPARING ALTERNATIVES USING THE
CAPITALIZED WORTH METHOD
Capitalized Worth (CW) method -- Determining the
present worth of all revenues and / or expenses over
an infinite length of time
Capitalized cost -- Determining the present worth of
expenses only over an infinite length of time
Capitalized worth or capitalized cost is a convenient
basis for comparing mutually exclusive alternatives
when a period of needed services is indefinitely long
and the repeatability assumption is applicable

19. 08/07/1419
CAPITALIZED WORTH METHOD
Capitalized worth of a perpetual series of end-
of-period uniform payments, A, with interest i%
per period:
A ( P /A, i%, )
CW = PWN --> = A ( P / A, i%, )
( 1+i )N
- 1
= A lim ------------- = A ( 1 / i )
N -->
i ( 1 + i )N 8
8
8
8

20. 08/07/1420
Capitalized Cost Calculation
PW Computations when N → ∞
 These cases are also called "perpetuities" - e.g. scholarships,
endowments, etc.
 There are many engineering projects where the life of the
project is so long as to be considered ‘forever’. Usually ≥50
years as a rule of thumb, but assume cash flows continue
indefinitely.
 There are other projects where a sum of money is provided at
the beginning of the project and it is then invested to yield the
amount used for the project (annuities for ≥50 years)
 For both, the PW of the infinitely long uniform series of cash
flows becomes the Capitalized Value

21. 08/07/1421
PW when N → ∞
Until now, we assumed
that the horizon N is a
fixed number of years.
The present worth of a
very long and uniform
series of cash flows is
calculated as
→∞
→∞
→∞
=
 + −
=  + 
 
− +
=  
 
  
=
lim ( / , , )
(1 ) 1
lim
(1 )
1
1
(1 )
lim
N
N
NN
N
N
P A P A i N
i
P A
i i
i
P A
i
A
P
i

22. Example 5.4
The system has an installed cost of $150,000 and
an additional cost of $50,000 after 10 years. The
annual software maintenance contract cost is
$5000 for the first 4 years and $8000 thereafter. In
addition, there is expected to be a recurring major
upgrade cost of $15,000 every 13 years. Assume
that i = 5% per year for county funds.

23. Example – cont.
 Find the present worth of the nonrecurring costs of $150,000 now
and $50,000 in year 10 at i = 5%. Label this CCI‘
 CCI = - 150,000 - 50,000(P/F,5%,10) = $-180,695
 Convert the recurring cost of $15,000 every 13 years into an annual
worth AI for the first 13 years.
 AI = - 15 ,000(A/ F,5 %, 13) = $-847
 The same value, AI = $- 847, applies to all the other 13-year periods
as well.
 The capitalized cost for the two annual maintenance cost series
may be determined in either of two ways:
 (I) consider a series of $- 5000 from now to infinity and find the
present worth of -$8000 - ($-5000) = $-3000 from year 5 on; or
 (2) find the CC of $-5000 for 4 years and the present worth of $-
8000 from year 5 to infinity. Using the first method, the annual
cost (A2) is $- 5000 forever. The capitalized cost CC2 of $- 3000
from year 5 to infinity is found using Equation [5.1] times the P /
F factor.
 CC2=-3000/0.05(P/F,5%,4)=$-49,362

24. Example –cont.
The two annual cost series are converted into a
capitalized cost CC3.
 CC3=(A1+A2)/I = (-847+ -5000) / 0.05 = -116,940
The total capitalized cost CCT is obtained by adding the
three CC values.
 CCT = - 180,695 - 49,362 - 116,940 = $- 346,997
determines the A value forever.
A = Pi = CCT(i) = $346,997(0.05) = $17,350
Correctly interpreted, this means Marin County officials
have committed the equivalent of $17,350 forever to
operate and maintain the property appraisal software.

25. Problem 2
Determine the capitalized cost of an expenditure
of $200,000 at time 0, $25,000 in years 2
through 5, and $40,000 per year from year 6 on.
Use an interest rate of 12% per year.
Soln:
CC = -200,000 – 25,000(P/A,12%,4)(P/F,12%,1)
– [40,000/0.12])P/F,12%,5)
= -200,000 – 25,000(3.0373)(0.8929) –
[40,000/0.12])(0.5674)
= $-456,933

26. Problem 3
What is the capitalized cost (absolute value) of
the difference between the following two plans at
an interest rate of 10% per year? Plan A will
require an expenditure of $50,000 every 5 years
forever (beginning in year 5). Plan B will require
an expenditure of $100,000 every 10 years
forever (beginning in year 10).
Soln:

27. Problem 3- Cont.
Find AW of each plan, then take difference, and
divide by i.
AWA = -50,000(A/F,10%,5)
= -50,000(0.16380)
= $-8190
AWB = -100,000(A/F,10%,10)
= -100,000(0.06275)
= $-6275
CC of difference = (8190 - 6275)/0.10
= $19,150

28. Problem 4
Two processes can be used for producing a
polymer that reduces friction loss in engines.
Process K will have a first cost of $160,000, an
operating cost of $7000 per quarter, and a
salvage value of $40,000 after its 2-year life.
Process L will have a first cost of $210,000, an
operating cost of $5000 per quarter, and a
$26,000 salvage value after its 4-year life. Which
process should be selected on the basis of a
present worth an alysis at an interest rate of 8%
per year, compounded quarterly?
Soln:

29. Problem 4 – Cont.
PWK = -160,000 – 7000(P/A,2%,16) –
120,000(P/F,2%,8) + 40,000(P/F,2%,16)
= -160,000 – 7000(13.5777) –120,000(0.8535) +
40,000(0.7284)
= $-328,328
PWL = -210,000 – 5000(P/A,2%,16) +
26,000(P/F,2%,16)
= -210,000 – 5000(13.5777) + 26,000(0.7284)
= $-258,950
Select process L