Comparing alternatives

1. American University of Armenia IE 340 – Engineering Economics Spring Semester, 2016

Ch5- Comparing Alternatives

2. Comparing Alternatives

3. The Objective

• To learn how to properly apply the profitability
measures to select the best alternative out of a set of
mutually exclusive alternatives (MEA)
• The cash-flow analysis methods (previously
described) used in this process:
Present Worth ( PW )
Annual Worth ( AW )
Future Worth ( FW )
Internal Rate of Return ( IRR )
External Rate of Return ( ERR )

4. Three groups of investment alternatives

1. Mutually exclusive
At most one project out of the group can be chosen, i.e. when
one project is chosen all the other ones are excluded
Example: suppose you are shopping for a car. You consider
several cars, but will only buy one from a mutually exclusive
set of choices
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

5. The fundamental purpose of capital investment is to obtain at least the MARR for each dollar invested

• Basic Rule:
Spend the least amount of capital possible unless the
extra capital can be justified by the extra savings or
benefits
In other words, any increment of capital spent (above
the minimum) must be able to pay its own way

6. Investment alternatives and cost alternatives

• An Investment Alternative has an initial investment
producing positive cash flows resulting from
increased revenues, reduced costs, or both
"Do nothing" (DN) is usually an implicit investment alternative
If ∑positive cash flows > ∑ negative cash flows, then IRR exists
If PW(MARR) ≥0, investment is profitable
if PW(MARR)<0, do nothing (DN) is better
• Cost Alternatives have all negative cash flows
except for the salvage value (if applicable)
These alternatives represent “must do” situations, and DN is not
an option
IRR not defined for cost alternatives. Can you explain why?

7. The study period must be appropriate for the decision being made

• Study Period (or planning horizon): The time interval
over which service is needed to fulfill a specified
function or the interval over which the alternatives
are compared
• Useful Life: The period during which an asset is kept
in productive operation
• We will discuss two cases:
Case 1: Study period = Useful life
Case 2: Study period ≠ Useful life
• Fundamental Principle: Compare MEAs over the
same period of time

8. Case 1: Study period = Useful life

9. Equivalent Worth (EW) Methods: PW, AW, FW

Procedure for Selecting the Best MEA using the EW
method:
1. Compute the equivalent worth of each alternative, using the
MARR as the interest rate
2. For Investment Alternatives: Select the alternative having
the greatest equivalent worth
Note: If all equivalent worths are < 0 for investment alternatives, then
"do nothing" is the best alternative
3. For Cost Alternatives: Select the alternative having the
smallest equivalent cost (the one that is least negative)
All three equivalent worth methods (PW, AW, FW) will identify the
same "best" alternative

10. Example: Investment Alternatives Study period = Useful Life

I
II
III
IV
Investment
costs (I)
$ 100,000
$ 152,000
$ 184,000
$ 220,000
Net annual
receipts (A)
$ 15,200
$ 31,900
$35,900
$41,500
0
$ 15,000
$ 20,000
10
10
10
Salvage value $ 10,000
(SV)
Useful life
10
If MARR = 12%, use PW method to select the best alternative

11. Solution

5.6502
0.3220
• PW(12%) = -I + A(P/A, 12%, 10) + SV(P/F, 12%, 10)
PW(Do Nothing) (12%) = 0
PWI (12%) = -$10,897
PWII (12%) = +$28.241
PWIII (12%) = +$23,672
PWIV (12%) = +$20,923
Select Alternative ___ to maximize PW.

12. Example: Cost Alternatives Study period = Useful life

A
B
C
Initial cost (I)
- $ 85,600
- $ 63,200
- $ 71,800
Annual expenses, years
1-7 (AC)
- $ 7,400
- $ 12,100
- $ 10.050
MARR = 12%
Use AW method to select the best alternative

13. Solution

0.2191
AW = I(A/P, 12%, 7) + AC
• AW A =
• AW B =
• AW C =
-$85,600(A|P, 12%, 7) - $7,400 = -$26,155
= -$25,947
= -$25,781
• Assuming one must be chosen (i.e., DN is not an option), select
alternative C to minimize annual equivalent costs

14. IRR Method: Why not select the investment opportunity that maximizes IRR?

A
B
B-A
Investment (I)
- $ 100
- $ 10,000
- $ 9,900
Lump-sum receipt next
year
$ 1,000
$ 15,000
$ 14,000
IRR
900%
50%
41.4%
MARR = 20%
If MARR = 20%, would you rather have A or B if comparable risk is
involved?
If MARR = 20%, PW(A) = $733 and PW(B )= $2,500
Never simply maximize the IRR!
Never simply select the MEA that maximizes the IRR! Do not
maximize the rate of return. Look at the increment
Never compare the IRR to anything except the MARR!
IRR(A→B): PW(B-A) = 0 = -9,900 + 14,000(P/F, i'%, 1)
9,900/14,000 = (P/F, i'%, 1)
i' = 41.4% > MARR . Select B

15. IRR Method: Another Example

1
2
3
Investment (FC)
- $ 28,000
- $ 16,000
- $ 23,500
Net cash flow/year
$ 5,500
$ 3,300
$ 4,800
Salvage value
$ 1,500
0
$ 500
Useful life
10 yrs
10 yrs
10 yrs
Study period
10 yrs
10 yrs
10 yrs
Given three MEAs and MARR = 15% per year
Use the Incremental IRR procedure to choose the best
alternative.

16. Solution steps

Step 1 DN
2
3
1
Rank in order alternatives
from low capital investment to
high capital investment
• Step 2: compare DN with alternative 2
∆ cash flow
• ∆ Investment
= -$16,000 – 0
= $3,300 – 0
= 0–0
∆ Annual receipts
∆ Salvage value
Compute ∆ IRR(2-DN)
PW(∆i') = 0 = -$16,000 + $3,300(P/A, ∆i'%, 10)
i‘(2-DN) ≈ 15.9%
=-$16,000
= $3,300
= 0
• Step 3. Since ∆i' > MARR, keep alt. 2 (higher FC) as
current best alternative. Drop DN from further
consideration

17. Solution steps

• Step 4: compare alternative 2 with alternative 3
∆ cash flow
• ∆ Investment
• ∆ Annual receipts
• ∆ Salvage value
• Compute ∆ IRR(3-2)
= -$23,500 – (-16,000)
= $ 4,800 - 3,300
= $500 – 0
=-$7,500
= $1,500
= $500
• PW(∆ i') = 0 = -$7,500 + $1,500(P/A, ∆i'%, 10) + $500(P/F, ∆i'%, 10)
• i‘(3-2) ≈ 15.5%
• Since ∆i' > MARR, keep alt. 3 (higher FC) as current best
alternative
• Drop alt. 2 from further consideration

18. Solution steps

• Step 4: next comparison: alternative 3 with
alternative 1
∆ cash flow
• ∆ Investment
= -$28,000 – (-23,500)
= $ 5,500 - 4,800
= $1,500 -500
=-$4,500
= $700
= $1,000
∆ Annual receipts
∆ Salvage value
Compute ∆ IRR(1-3)
PW(∆ i') = 0 = -$4,500 + $700(P|A, ∆i'%, 10) + $1,000(P/F, ∆i'%, 10)
∆ i‘(1-3) ≈ 10.9%
Since ∆i' < MARR, keep alt. 3 (lower FC) as current best
alternative. Drop alt. 1 from further consideration
• Step 5. All alternatives have been considered
• Recommend alternative 3 for investment

19. Case 2: Study period ≠ Useful life

20. Study period ≠ Useful life

• Up until now, study periods and useful lives have
been the same length. The study period is frequently
taken to be a common multiple of the alternatives’
lives when study period ≠ useful life
• Repeatibility Assumption
Conditions:
1. Study period is either indefinitely long or equal to a common
multiple of the lives of the alternative
2. The cash flows associated with an alternative's initial life span
are representative of what will happen in succeeding life spans
Actual situations in engineering practice seldom meet these two
conditions, which limits the use of the repeatability assumption

21. Example: 5-24 Cost alternative: Study period>Useful life

Example: 5-24
Cost alternative: Study period>Useful life
A
B
Investment costs
- $ 14,000
- $ 65,000
Annual costs
- $ 14,000
- $ 9,000
5
20
$ 8,000
$ 13,000
20 yrs
20 yrs
Useful life
Market value at the end of
useful life
Study period
Given two MEAs and MARR = 15% per year, which alternative is
preferred based on the repeatability assumption?
Hint: first draw a cash flow diagram

22. Example: 5-24 - Solution

Consider the AW over the useful life of each Alternative
AW(A )= -14,000(A/P, 15, 5)- 14,000 + 8,000(A/F, 15, 5) = -16,990
Life 1: AW (1-5) = -16,990
Life 2: AW (6-10) = -14,000(A/P, 15, 5)- 14,000 + 8,000(A/F, 15, 5) = 16,990
Life 3: AW (11-15 )= -16,990
Life 4: AW (16-20) = -16,990
AW(B)= -65,000(A/P, 15, 20)- 9,000 + 13,000(A/F, 15, 20) = -19.262
AW(A )> AW(B)
• Shortcut: If the study period equals a common multiple of the
alternatives' lives, simply compare AW computed over the
respective useful lives (assuming repeatability is valid)

23. Example: 5-24 – PW Solution based on the total study period

• PW(A) = -14,000 - 14,000(P/A,15,20) + 8,000(P/F, 15,
20) - 6,000(P/F, 15, 5) -6,000(P/F, 15, 10) -6,000(P/F, 15,
15) = -$106,345
• PW (B) = -65,000 - 9,000(P/A, 15, 20) + 13,000(P/F, 15,
20) = -$120,539
Select A to minimize costs
Also,
AW(A) = -$106,345(A/P, 15, 20) = -$16,990
AW(B)= -$120,539(A/P, 15, 20) = -$19,262

24.

• What if the study period is not a
common multiple of the alternatives‘
lives or repeatability is not
applicable?
• An appropriate study period need to
be selected (coterminated
assumption)

25. Study period > Useful life

Study period > Useful life
Use the Cotermination Assumption
Procedure: The cash flows of the alternatives need to be adjusted
so that all the alternatives are compared over the same study period
• Case 1: Suppose study period > useful life
• Cost alternatives: Since each cost alternative has to provide the
same level of service over the study period, contracting for the
service or leasing the needed equipment for the remaining years
may be appropriate
• Investment alternatives: Assume all cash flows will be reinvested
at the MARR to the end of the study period (i.e., calculate FW at end
of useful life and move this to the end of the study period using the
MARR). PW can also be used.
See Examples 5-7 and 5-8, page 210

26. Case 2: Study period < useful life

Case 2: Study period < useful life
When the study period is explicitly stated to be
shorter than the useful life, use the Cotermination
assumption
Procedure: The cash flows of the alternatives need to
be adjusted to terminate at the end of the study period
Truncate the alternative at the end of the study period
using an estimated Market Value

27. Example: Cotermination

A
B
Investment costs
$ 50,000
$ 120,000
Annual costs
- $ 9,000
- $ 6,000
Useful life
20 yrs
40 yrs
Market value at the end of
useful life
$ 10,000
$ 20,000
Study period
20 yrs
20 yrs
Useful life of A = 20 years = study period
Useful life of B = 40 years > study period
Assume Market Value (B) @ EOY20 = $50,000
The MARR is 10% per year.
Which boiler do you recommend?

28. Example: Solution

AW(A) (10%) = -50,000(A/P,10, 20) -9,000
+10,000(A/F,10,20) = -14,700
AW(B) (10%) = - 6,000 –[120,000(A/P, 10, 20) 50,000(A/F, 10, 20)] = -19,225
Still select boiler A to minimize costs
What would the market value of Boiler B @ EOY 20
have to be in order to select Boiler B instead of A?

29. Example: Solution

Market Value of Boiler B to reverse the decision = ?
Set AW(A) = AW(B) , let X be the unknown market
value
-14,700 = -6,000 - {120,000(A/P, 10, 20) - X(A/F,10, 20)}
-8,700 = -{14,100 - 0.0175X}
5,400 = 0.0175X
X = $308,571 therefore, MV(B) > $308,571 to favor B
Such a value is very unlikely because X is more than
the initial cost of Boiler B