3. 1. Present Worth Comparison
Cash flows are known.
Cash flows do not include effect of inflation.
The interest rate (discounting rate) is known.
Comparisons are made with before tax cash flows.
.
ASSUMPTIONS
5. Comparing projects with equal lives
The competing alternatives have equal lives.
The alternative with the maximum present worth is
the most economical alternative.
For cost dominated cash flow diagrams the alternative
with the lowest present cost is chosen.
In case of cash flow diagrams involving both costs and
revenues the net or difference of present worth of
revenues and costs are found. This is referred to as net
present worth or net present value (NPV).
6. Alternatives with equal lives
The method of comparison of NPV is quite popular for
evaluation of alternatives.
NPV is also used to calculate profitability for an
investment alternative.
7. Example
Compare an alternative
Project A, costing 500,000 Birr and
requiring cash outlays of 250,000 Birr a
year for 10 years,
Project B, costing 750,000 and requiring
prospective cash outlays of 200,000 Birr a
year for 10 years, if interest is 10%.
8.
9. Calculation of present worth
Present worth of Alternate A
= 500000+250000(P/A, 10%, 10yrs) =
500000+250000(6.1446) =
2,036,150.00
Present worth of Alternate B
= 750000+200000(P/A, 10%, 10yrs) =
750000+200000(6.1446)
=1,978,920.00
Decision: In this case, since the present worth of cost for
alternative B is less than that of alternative B, It is
preferable to choose alternative B.
10. Alternatives with unequal lives
the alternatives do not have an equal life period of service
(they are not co-terminal).
decision to choose between two batching plants, which
may have different service lives – say 5 years and 10 years.
The common multiple method and the study period
method are two approaches to solve this class of problems.
11. Common multiple method
In this method a coterminous life period is chosen
for the alternatives using the least common
multiples of the different life periods.
For example if the alternatives have life period of 2,
3, 4 and 6 years, they will be put in use for a period
equal to the least common multiple of their life
periods.
In the case above it is 12 years.
12. Common multiple method
This means that the alternative with two year life
period shall be replaced 6 times; the one with
three years shall be replaced four times and so on
for achieving coterminous life period for all the
alternatives.
It is assumed here that the alternatives shall be
replaced after their service life with same cost
characteristics.
This assumption stands valid if the common
multiple of alternative life periods is small, and the
possibility of a technically different option
emerging during this time is ignored.
13. Study period method
In this method a study period is chosen on the
basis of the length of the project or the service
lives of the alternatives.
An appropriate study period reflects the
replacement circumstances. Thus study period
may be chosen as the shortest life period of the
alternatives as a protection against technological
obsolescence.
In this method we assume that all the assets will
be disposed off at the end of the analysis period.
14. Example
Assets A and B have the capability of performing a
required function.
Asset B has an initial cost of 160,000 and expected salvage
value of 20,000 Birr at the end of its 4 year service life.
Asset A costs 45,000 less initially, with an economic life 1
year shorter than that of B; but A has no salvage value, and
its annual operating costs exceed those of B by 12,500.
When the required rate of return is 15%; state which
alternative is preferred when comparison is by
The common multiple method
study period (assuming the assets are needed for only 3
years.
15. Common multiple method
Net present worth = -160,000 –(160,000/1.154)-
(160,000/1.158 )+(20,000/1.154) + (20,000/1.158 +
( 20,000/1.1512)= -282,074.00
Asset-B
16. Solution
Net present worth= -115,000 –(115,000/1.153)-
(115,000/1.156 )-(115,000/1.159)-12,500*[(1.1512-
1)/(0.15*1.1512)]
= -340,779.50
A
17. Home work
Compare the following Equipments
0
Initial cost
1
Operation and
maintenance
cost
2 3
Equipment A 450,000 225,000 450,000 -
Equipment B 7,25,000 3,00,000 3,00,000 4,00,000
Year
18. Alternatives with infinite lives
when the alternatives involved have long lives, for
example the appraisal of different alternatives
involving construction of civil engineering
structures – dams, power projects, tunnel projects
etc., which have a reasonably long life.
A very popular approach used in such problems is
the application of capitalized equivalent method,
which is described in the following section.
19. Capitalized equivalent method
Capitalized equivalent (CE) is the present (at time zero)
worth of cash inflows and outflows.
In other words, CE is a single amount determined at time
zero, which at a given rate of interest, will be equivalent to the
net difference of receipts and disbursements if the given cash
flow pattern is repeated in infinity.
If n is not infinitive, we use this formula
Mathematically, CE = A x (P/A, i, n=∞)
= A x [((1+i)∞-1)/ i(1+i)∞]= A/i
CE analysis is very useful to compare long-term projects.
20. Example
Dam A initially costs 160,000 Birr to construct and cost
750,000 Birr a year to operate and maintain.
Another dam design B costs 7,500,000 Birr to build and
cost 500,000 a year to operate and maintain.
Both installations are felt to be permanent.
The minimum required rate of return is 5%.
Which alternative should be preferred?
21. Dam A
NPV = -160,000-750,000/0.05
= -15,160,000
22. Dam B
NPV = -7,500,000 - 500,000/0.05
= -17,500,000
The cost of dam A is less, so Construct Dam A.
23. 2.Future Worth Comparison
The future worth of each component of cash flow is
evaluated
There is no discounting of each component of cash
flow to the present
There is no special advantage over the present worth
method,
frequently used in cases when the owner wants an
estimate of net worth at some future date such as
planning for retirement.
comparison and evaluation seems to be more
meaningful as it provides some insight into future
receipts.
24. Example
Compare an alternative
A, costing 500,000 Birr and requiring cash
outlays of 250,000 Birr a year for 10 years,
B, costing 750,000 and requiring
prospective cash outlays of 200,000 Birr a
year for 10 years, if interest is 10%.
25.
26. 3. Annual Cost and Worth Comparison
Most widely used method for comparing alternatives because of
its simplicity, and ease of computations, and understanding
All payments and disbursements are converted into an annual
cost series at a given interest rate
The alternative that yields the least cost is chosen.
The method could be used to compare alternatives with equal
and unequal lives (common multiple method and study period
method).
27. Example
Compare an alternative
Project ‘A’ initially costs 500,000 Birr
and requiring cash outlays of 250,000
Birr a year for 10 years,
B, costs 750,000 and requiring
prospective cash outlays of 200,000 Birr
a year for 10 years, if interest is 10%.
Compare the annual cost
