This document discusses methods for evaluating capital investment alternatives that have different cash flows over time. It provides two rules: 1) for investment alternatives, select the one with the greatest positive equivalent worth using the MARR, and 2) for cost alternatives, select the one that minimizes costs. It also discusses comparing alternatives using net present value, internal rate of return, and the incremental investment analysis procedure. Key points covered include ensuring a comparable basis, dealing with alternatives that have unequal lives, and common errors to avoid when using rates of return for analysis.

1. Comparison and Selection
among Alternatives
Created By Eng.Maysa Gharaybeh

2. Quiz
 1, 2, 7, 15,19, 20, 22, 26, 36, 40.

3. The objective of chapter 6 is
to evaluate correctly capital
investment alternatives when
the time value of money is a
key influence.

4. Making decisions means
comparing alternatives.
 In this chapter we examine feasible design
alternatives.
 The decisions considered are those selecting
from among a set of mutually exclusive
alternatives (when selecting one excludes the
choice of any of the others).

5. Principle 2 from Chapter 1.
The alternative that requires the minimum investment of
capital and produces satisfactory functional results will be
chosen(the base alternative)
unless the incremental capital associated with an alternative
having a larger investment can be justified with respect to its
incremental benefits.
The alternative requiring the least investment is the base
alternative.

6. For alternatives that have a larger
investment than the base…
If the extra benefits obtained by investing additional
capital are better than those that could be obtained from
investment of the same capital elsewhere in the
company at the MARR, the investment should be made.
(Please note that there are some cautions when
considering more than two alternatives, which will be
examined later.)

7. Types of Alternatives
 Investment alternatives (capital & Revenue)
 Cost alternatives (Negative & salvage value) same
expected revenue

8. Ensuring Comparable Basis
Rule 1. When revenues and other economic benefits
are present(Investment alternatives ), select
alternative that has greatest positive equivalent
worth at i %= MARR% and satisfies project
requirements.
Rule 2. When revenues and economic benefits are
not present(Cost alternatives ), select alternative
that minimizes cost.

9. In Other Words:
Select the alternative that gives you
the most money!
 For investment alternatives the PW of all cash flows
must be positive, at the MARR, to be attractive.
Select the alternative with the largest PW.
 For cost alternatives the PW of all cash flows will be
negative. Select the alternative with the largest
(smallest in absolute value) PW.

10. Example: Investment Alternative
Use a MARR of 10% and useful life of 5 years to select
between the investment alternatives below.
Alternative
A B
Capital investment -$100,000 -$125,000
Annual revenues less expenses $34,000 $41,000
Both alternatives are attractive, but Alternative B provides
a greater present worth, so is better economically.

11. Cost alternative example
Use a MARR of 12% and useful life of 4 years to select
between the cost alternatives below.
Alternative
C D
Capital investment -$80,000 -$60,000
Annual expenses -$25,000 -$30,000
Alternative D costs less than Alternative C, it has a greater
PW, so is better economically.

12. Investment
Alternative
PW(10%)A= 9,738$
PW(10%)B= 10,131$
PW(10%)B-A = 393$

13. Cost
Alternative
with selvage
value
PW(10%)C= -477,077$
PW(10%)D= -463,607$
PW(10%)D-C = 13,470$

14. Determining the study period.
 A study period (or planning horizon) is the time
period over which mutually exclusive alternatives
are compared, and it must be appropriate for the
decision situation.
 mutually exclusive alternatives can have equal
lives (in which case the study period used is these
equal lives), or they can have unequal lives, and at
least one does not match the study period.
 The equal life case is straightforward.

15. Useful Lives of all Alternatives
is Equal to the Study Period
1. Equivalent worth methods
2. Rate of return methods

16. 1.Equivalent Worth Methods
 If : PWA (i) < PWB (i)
Then
 AWA (i) < AWB (i)
and
 FWA (i) < FWB (i)
Select Alternative B

22. Example:
When lives are equal adjustments to cash flows are not
required. The MEAs can be compared by directly comparing
their equivalent worth (PW, FW, or AW) calculated using the
MARR. The decision will be the same regardless of the
equivalent worth method you use. For a MARR of 12%, select
from among the MEAs below.
Alternatives
A B C D
Capital investment -$150,000 -$85,000 -$75,000 -$120,000
Annual revenues $28,000 $16,000 $15,000 $22,000
Annual expenses -$1,000 -$550 -$500 -$700
Market Value $20,000 $10,000 $6,000 $11,000
(EOL)
Life (years) 10 10 10 10

23. Selecting the best alternative.
Present worth analysis  select Alternative A (but C is close).
Annual worth analysis—the decision is the same.

24. Using rates of return is
another way to compare
alternatives.

25.  The return on investment (rate of return) is a popular
measure of investment performance.
 Selecting the alternative with the largest rate of return
can lead to incorrect decisions do not compare the
IRR of one alternative to the IRR of another
alternative.
 Remember, the base alternative must be attractive
(rate of return greater than the MARR), and the
additional investment in other alternatives must itself
make a satisfactory rate of return on that increment.

26. The Incremental Investment
Analysis Procedure.
 Arrange (rank order) the feasible alternatives based on
increasing capital investment.
 Establish a base alternative.
 Cost alternatives the first alternative is the base.
 Investment alternatives the first acceptable alternative
(IRR>MARR) is the base.

27.  Iteratively evaluate differences (incremental cash flows)
between alternatives until all have been considered.
a. If incremental cash flow between next alternative
and current alternative is acceptable, choose the next.
b. Repeat, and select as the preferred alternative the last
one for which the incremental cash flow was
acceptable

28. Example 6-4
A B C D E F
Capital 900 1,500 2,500 4,000 5,000 7,000
Annual R – 150 276 400 925 1,125 1,425
Annual E
IRR 10.6% 13% 9.6% 19.1% 18.3% 15.6%
 The alternative is arranged based on
increasing capital investment.
 IRR for alternative (C) 9.6<MARR(10%)
 Then C is rejected

29. A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)
ΔCapital 900
Δ(Annual R 150
– Annual E)
IRRΔ 10.6%
Is Yes
increment
justified
A is the base alternative then we will compare A
with B

30. A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)
ΔCapital 900 600
Δ(Annual R 150 126
– Annual E)
IRRΔ 10.6% 16.4%
Is Yes Yes
increment
justified
B is better then A because the additional capital
investment gives IRR >MARR(16.4%>10%)
Then we will compare B with D since C is rejected from
the beginning

31. A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)
ΔCapital 900 600 2,500
Δ(Annual R 150 126 649
– Annual E)
IRRΔ 10.6% 16.4% 22.6%
Is Yes Yes Yes
increment
justified
D is better then B because the additional capital
investment gives IRR >MARR(22.6%>10%)
Then we will compare D with E

32. A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)
ΔCapital 900 600 2,500 1,000
Δ(Annual R 150 126 649 200
– Annual E)
IRRΔ 10.6% 16.4% 22.6% 15.1%
Is Yes Yes Yes Yes
increment
justified
E is better then D because the additional capital
investment gives IRR >MARR(15.1%>10%)
Then we will compare E with F

33. A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)
ΔCapital 900 600 2,500 1,000 2,000
Δ(Annual R 150 126 649 200 300
– Annual E)
IRRΔ 10.6% 16.4% 22.6% 15.1% 8.1%
Is Yes Yes Yes Yes No
increment
justified
F is NOT better then E because the additional
capital investment gives IRR < MARR(8.1%>10%)
Then E is the best alternative

34. Three Errors Common To Incremental Investment
Analysis Procedure Applied To IRR
Choosing the feasible Alternative with:
1. the highest overall IRR on total cash flow (choose D)
A B C D E F
Capital 900 1,500 2,500 4,000 5,000 7,000
Annual R – 150 276 400 925 1,125 1,425
Annual E
IRR 10.6% 13% 9.6% 19.1% 18.3% 15.6%

35. Three Errors Common To Incremental Investment
Analysis Procedure Applied To IRR
Choosing the feasible Alternative with:
2.the highest IRR on an incremental capital investment(choose
D-B instead of E-D)
A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)
ΔCapital 900 600 2,500 1,000 2,000
Δ(Annual R 150 126 649 200 300
– Annual E)
IRRΔ 10.6% 16.4% 22.6% 15.1% 8.1%
Is Yes Yes Yes Yes No
increment
justified

36. Three Errors Common To Incremental Investment
Analysis Procedure Applied To IRR
3.the largest capital investment that has an IRR greater
than or equal to the MARR (choose F)
A B C D E F
Capital 900 1,500 2,500 4,000 5,000 7,000
Annual R – 150 276 400 925 1,125 1,425
Annual E
IRR 10.6% 13% 9.6% 19.1% 18.3% 15.6%

40. Incremental analysis must be
used with rate of return
methods to ensure the best
alternative is selected

43. Comparing MEAs with Unequal lives.
• Repeatability assumption
• Coterminated assumption

48. Rate of Return with unequal lifes
A B
Capital investment $3,500 $5,00
Annual cash flow 1,255 1,480
Useful life 4 6
When the repeatability applies
1.we need to find the AW of each alternative over it’s own
useful life
2.Find the interest rate that make them equal
AW A(i*%) = AW B(i*%)
-3,500(A/i*%,4) +1,255 = -5,000(A/i*%,6) +1,480
Trail and error i= 26% >MARR (10%) the increment is justified
and B is preferred

49. Useful life > Study period
 The imputed Market Value Technique
 i. e find the estimated market value at any year that is
before the end of the useful life
 MVT = PW at EOY T of remaining CR amounts + PW at
EOY T of original market value at end of useful life

50.  Example
 If the capital investment is 47,600 useful life = 9 years
market value at the end of 9 years = $5,000 find the
market value et the end of 5 years if MARR = 20%
Step 1 : CR = 47,000 (A/P,20%,9) – 5,000(A/F,20%,9)
Step 2 : PW at EOY 5 of the remaining CR
PW(20%) CR = CR (P/A, 20%,4) = $29,949
Step 3 : PW at EOY 5 of MV 9
PW(20%) MV = 5,000 (P/F, 20%,4) = $2,412
The estimated market value at EOY 5 =
MV 5 = PW(20%) CR + PW(20%) MV = 29,949 + 2,412 =
32,361