Topic 4: Methods of Evaluating Alternatives

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Topic 4 Methods of Evaluating Alternatives

• MARR
• Present Worth Method
• Future Worth Method
• The Annual Worth Method
• Rate of Return Method
• The Payback Period Method
• Capitalized Cost Comparison
• Benefit/ Cost Ratio (B/C) Method

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Minimum Attractive rate of Return (MARR)
Profit or net income
Current amount-original investment

• iROR 100

original investment
-IS COMMONLY USED WHEN ESTIMATING THE
PROFITABILITY OF A PROPOSED ALTERNATIVE OR WHEN
EVALUATING THE RESULTS OF A COMPLETED PROJECT OR
INVESTMENT.
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-INVESTMENT ALTERNATIVES ARE EVALUATED UPON A
REASONABLE ROR EXPECTED. THE REASONABLE RATE IS
CALLED THE MINIMUM ATTRACTIVE RATE OF RETURN
(MARR). - MARR IS ALSO REFERRED TO AS THE
HURDLE RATE FOR PROJECTS THAT IS, TO BE
CONSIDERED FINANCIALLY VIABLE THE EXPECTED ROR
MUST MEET OR EXCEED THE MARR.
4
Cont.

• MARR OR HURDLE RATE MUST BE STATED OR ESTABLISHED
  TO EVALUATE A SINGLE PROPOSAL OR COMPARE
  ALTERNATIVES,
• THAT ESTIMATED PROJECT RORs LESS THAN THE MARR
  SHOULD BE REGARDED AS ECONOMICALLY UNACCEPTABLE.

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THE PRESENT WORTH METHOD

• ONE ALTERNATIVE
• IF PW gt 0, THE REQUESTED RATE OF RETURN IS MET
  OR EXCEED AND THE ALTERNATIVE IS FINANCIALLY
  VIABLE.

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• TWO OR MORE ALTERNATIVES
• WHEN ONLY ONE CAN BE SELECTED (I.E., ALTERNATIVES
  ARE MUTUALLY EXCLUSIVE)
• FOR EACH ALTERNATIVE
• FIND THE PW OF ALL COSTS BY DISCOUNTING. THEN
  SELECT THE ALTERNATIVE WITH THE LOWEST PW.
• IF ALTERNATIVES INVOLVE PROFITS, FIND PW OF EACH
  ALTERNATIVE. THEN SELECT THE ALTERNATIVE WITH THE
  HIGHEST PW.

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THE FUTURE WORTH METHOD

• ONE ALTERNATIVE
• IF FW gt 0, THE REQUESTED RATE OF RETURN IS MET
  OR EXCEED AND THE ALTERNATIVE IS FINANCIALLY
  VIABLE.

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• TWO OR MORE ALTERNATIVES
• WHEN ONLY ONE CAN BE SELECTED (I.E., ALTERNATIVES
  ARE MUTUALLY EXCLUSIVE)
• FOR EACH ALTERNATIVE
• FIND THE FW OF ALL COSTS BY DISCOUNTING. THEN
  SELECT THE ALTERNATIVE WITH THE LOWEST FW.
• IF ALTERNATIVES INVOLVE PROFITS, FIND FW OF EACH
  ALTERNATIVE. THEN SELECT THE ALTERNATIVE WITH THE
  HIGHEST FW.

9
Cont.

• PW AND FW COMPARISONS CAN ONLY BE DONE OVER SAME
  YEARS OF FUTURE SERVICE FOR ALL ALTERNATIVES.

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• UNLESS A PLANNING HORIZON OVER WHICH THE
  COMPARISON IS TO BE DONE IS SPECIFIED, THE
  SHORTEST PERIOD TO USE IS THE LEAST-COMMON-MULTIPL
  E (LCM) OF YEARS OF SERVICE OF DIFFERENT LIVED
  ALTERNATIVES.

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• THE INTEREST RATE TO BE USED FOR DISCOUNTING
  CANNOT BE LOWER THAN THE MARR THE COMPANY CAN
  INVEST ITS FUNDS IN PROJECTS OF COMPARABLE COST.

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Example
There are 2 possible methods of performing a
metal-forming operation. One method requires the
purchase of m/c A, and the other method requires
the purchase of m/c B.
Using a 6 rate of return, which m/c should you
prefer based on the PW method?
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CAPITALIZED COST COMPARISON
WHEN THE SERVICE OF AN ASSET WILL BE REQUIRED
INDFINITELY INTO THE FUTURE i.e. CALCULATING PW
OF PERPETUAL SERVICE (CAPITILAZED COST) IS MORE
APPROPRIATE.
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• USING THE PREVIOUS EXAMPLE
• m/c A
• Purchase cost.6000
• PW of infinite
• of renewals of the m/c..3580
• 5000(A/F,6,15)(1/0.06)
• (PA/i)
• PW of perpetual
• operating costs83,333
• PA/i5000/0.06
• ------------------------------------------
• Total PW of m/c A.92913

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• m/c B
• Purchase cost.10,000
• PW of infinite
• of renewals of the m/c..12,645
• 10,000(A/F,6,10)(1/0.06)
• (PA/i)
• PW of perpetual
• operating costs50,000
• PA/i3000/0.06
• ------------------------------------------
• Total PW of m/c B.72,645
• then select m/c b.

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GENERAL PROCEDURE IN CALCULATING CAPITALIZED COST

• DRAW THE CASH FLOW DIAGRAM SHOWING ALL
  NON-RECURRING EXPENDITURE/RECEIPTS, AND AT LEAST
  2 CYCLES OF ALL RECURRING EXPENDITURES.
• FIND THE PW OF ALL NON-RECURRING EXPENDITURES.
• FIND THE ANNUAL EQUIVALENT (A) OF ALL RECURRING
  EXPENDITURES THROUGH ONE LIFE CYCLE.
• FIND THE CAPITALIZED COST OF a FOUND IN (3).
• ADD (2) AND (4).

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EXAMPLE
2 SITES UNDER CONSIDERATION FOR A BRIDGE TO CROSS
A RIVER NORTH SUSPENSION BRIDGE. SOUTH-TRUSS
BRIDGE WITH A NEW ROAD CONSTRUCTION.
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Suspension bridge First cost..30m
illion Annual maintenance cost15,000/yr. Res
urfacing of concrete deck every 10 years
..50,000 Cost of purchasing right-of-way(k
amulastirma).800,000
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Truss bridge First cost of bridge
and Approached roads12million Annual
maintenance cost.8,000/yr. Painting cost
of bridge every 3 years..10,000 Sandbla
sting and major Painting every 10
years45,000 Cost of purchasing
right-of-way10.3million Which bridge should
be selected if interest rate is 6/yr.?
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THE EQUIVALENT UNIFORM ANNUAL WORTH METHOD(EUAW)

• FOR EACH ALTERNATIVE, ALL INCOMES AND
  DISBURSEMENTS (IRREGULAR AND UNIFORM) ARE
  CONVERTED INTO EQUIVALENT ANNUAL UNIFORM AMOUNT
  (A) AND COMPARED.
• THE ALTERNATIVE WITH MINIMUM EUAW (WHEN ONLY
  COSTS ARE INVOLVED), THE ALTERNATIVE WITH MAXIMUM
  EUAW (WHEN PROFITS ARE INVOLVED) WILL BE SELECTED.

21
Cont.

• NO ADJUSTMENTS IS REQUIRED WHEN ALTERNATIVES HAVE
  UNEQUAL LIVES. SINCE EUAW IS THE SAME NO MATTER
  HOW MANY LIFE CYCLES ARE CONCERNED AS LONG AS
  CASH FLOWS ARE THE SAME FOR EACH CYCLE.
• HOWEVER SINCE THIS IS ANOTHER FORM OF PW IT AGAIN
  ASSUMES COMPARISON OVER EQUAL LIVES.
• WHEN MAKING AW COMPARISON ONLY ONE LIFE CYCLE OF
  EACH ALTERNATIVE SHOULD BE CONSIDERED.

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Example
The following costs are estimated for 2 equal
service tomato-peeling machines in a
food-canning plant?
m/c A m/c B First cost
26,000 36,000 Annual
800/yr.
300/yr. maintenance cost Annual labor cost
11,000/yr. 7,000/yr. Extra income
taxes - 2,600 Salvage
value 2,000
3,000 Estimated life 6 yrs.
10 yrs. If the minimum rate of return
is 15/yr. Which m/c should be selected?
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Example
If the company in the previous example is
planning to get out of the tomato canning
business in 4 yrs., and the company expects to be
able to sell m/c A for 12,000, and m/c B for
15,000 at that time with all other expected to
remain the same, which m/c should be purchased?
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AW of A Permanent Investment
Comparison of alternatives which have long lives
that they may be considered infinite in economic
analysis terms. For example public works
investments have very long lives (means
infinity), then calculate APi.
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Example
2 proposals for increasing the capacity of the
main canal of irrigation system. Proposal A-
dredging the canal, dredging equipment
accessories have a first cost of 65,000,
expected life of 10 yrs. With 7,000 salvage
value. Annual labor and operating costs are
22,000/yr. Yearly cost of weed control program
including labor cost is 12,000. Proposal
B-Lining the canal with concrete initial cost-
650,000. The lining is permanent but minor
maintenance cost- 1,000/yr. Lining repairs every
5 yrs. At a cost of 10,000. Which alternative
is better if MARR 5 per year?
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Rate of Return Method

• The Internal Rate of Return Method (IRR)
• The External Rate of Return Method (ERR)
• Incremental rate of Return Analysis (For
  Multiple Projects)

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The Internal Rate of Return Method (IRR)
Rate of return of a project such that
PW of disbursement PW of receipts or
EUAWDEUAWR
IRR is not a rate of return on initial
investment, but is the rate of return on
unrecovered investment balances such that it
causes exact recovery of investment over the life
of the project plus a return on the unrecovered
balances throughout the life of the project.
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Cont.
Economically meaningful interest rates IRRs
must lie in the range 0ltilt , since negative
interest rates imply either partial or complete
nonrecovery of capital. Selection criteria for
independent projects Select the project if
i(IRR) MARR or firms marginal investment
rate Reject, otherwise.
29
Cont.
Unrecovered investment balance at time t1 aFt1
can be found from the recursive relationship
Ft1 Ft (1i) Ct1 where Ft
unrecovered investment at time t Ft1
unrecovered investment at time t Ct
cash flow at the end of period t
i interest rate
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Example
Suppose a single project has the following
estimated cash flows IRRi is the i that
satisfies -100002000(P/F,i,1)4000(P/F,i,2)
7000(P/F,i,3)5000(P/F,i,4)3000(P/F,i,5)0
Solving by trial and error
i28.35
31
Cont.

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Multiple and Indeterminate Rates of Return

• When solving for IRR for projects with
  certain forms of cash flow stream, it is possible
  to
• Find that a unique soln. does not exist means
  more than one interest rate will satisfy the
  relation. PWD PWR a Multiple rates of return
• Find that no real value of i will satisfy the
  relation PWD PWR a no soln. exists a
  indeterminate rate of return

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Difficulties Associated with the IRR Method

• Recovered funds are reinvested at i(IRR) rather
  than at MARR. This assumption may not mirror
  reality in some problems.
• Not easy to compute.

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The External Rate of Return Method

• Directly takes into account the interest rate
  external to a project at which net cash flows
  generated (or required) by the project over its
  life can be reinvested.(borrowed)
• If this external reinvestment rate, which is
  usually the firms MARR, equal IRR of project,
  then the ERR method results the IRR method
  results
• Solve PWD(i)FW(i) FWR(i)
• Find i ERR

35
Cont.

• Selection criteria
• If i(ERR)gt MARR then accept project
• Two advantages over the IRR
• Solved for directly rather than by trial and
  error.
• Not subject to the possibility of multiple rates
  of return.

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Soln.
25,000(F/P,i,5)8,000(F/A,20,5)5,000
(F/P,i,5) 2.5813
i20.88 Because igt MARR, the
alternative is barely justified.
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Example
A piece of new equipment has been proposed be
engineers to increase the productivity of a
certain manual welding operation.The investment
cost is 25,000, and the equipment will have a
salvage value of 5,000 at the end of its
expected life of five years. Increased
productivity attributable to the equipment will
amount to 8,000 per year after extra operating
costs have been subtracted from the value of the
additional production. Evaluate the ERR of the
proposed equipment. Is the investment a good one?
MARR is 20/yr..
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Incremental Rate of Return Analysis
When comparing mutually exclusive alternatives
with different cash flow streams, one can not
choose the alternative with the higher ROR, since
ROR is the return on the unrecovered investment
balances. So one must use an incremental
approach. That is the alternative with the higher
initial investment is chosen if the incremental
rate of return MARR
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Example
A corporation is considering the purchase of an
additional machine, the mutually exclusive
alternatives are
Extra investment
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Example
2 mutually exclusive alternatives are as follows
at MARR15
A B First cost ()
-8,000 -13,000 Annual
disbursements () -3,500
-1,600 Salvage value () 0
2,000 Life (yrs.)
10 5
Since the alternatives
have unequal lives to calculate the incremental
rate of return the net cash flow should be
prepared for the LCM of years.(if PW method is
to be used).
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Selection from Mutually Exclusive Multiple
Alternatives

• An alternative should never be compared with one
  for which the incremental investment has not been
  justified.
• The procedure
• Order the alternatives in ascending order of
  initial investments.
• For alternatives which have positive cash flows,
  do-nothing alternative is the alternative with
  which the lowest initial investment alternative
  is compared.

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Cont.

• If the IRR(i)lt MARR for the lowest initial
  investment alternative, remove that alternative
  from further consideration. Compute the overall
  IRR for the next-higher initial investment
  alternative. Repeat this until igt MARR for one
  of the alternatives. When igt MARR , that
  alternative is tentatively accepted and the next
  higher initial investment alternative is compared
  with this tentatively accepted alternative.
• Calculate the incremental rate of return between
  the tentatively accepted alternative and next
  higher initial investment alternative.

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Cont.

• If the incremental ROR calculated in (4) gt MARR,
  the higher initial investment alternative becomes
  the new tentatively accepted and the previous
  alternative is removed from further
  consideration.
• Repeat steps 4 and 5 until no other alternative
  is left.

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Example4 different building locations, one will
be selected. MARR10/yr.
A B
C D Building cost ()
-200,000 -275,000
-190,000 -350,000 Cash flow(/yr.)
22,000 35,000 19,500
42,000 Life(yrs.) 30
30 30
30 In ascending order of initial
investment C
A B
D Building cost () -190,000
-200,000 -275,000 -350,000 Cash
flow(/yr.) 19,500 22,000
35,000 42,000 Projects compared
C to nothing A to nothing B to
A D to B Incremental cost -190,000
-200,000 -75,000
-75,000 Incremental cash 19,500
22,000 13,000
7,000 flow Incremental i 9.63
10.49 17.28
8.55 Increment justified No
Yes Yes
No Projects selected None
A B
B
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When alternatives consist of disbursements only,
the income is the difference between costs. In
this case, no need to compare any alternative
with do-nothing alternative. The lowest cost
alternative is the one with the others are to be
compared with.
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Example 4 mutually exclusive m/c alternatives.
The costs are as follows
1 2
3 4 First cost () 5000
6500 10000 15000 Annual operating
3500 3200 3000
1400 Cost(/yr.) Salvage value () 500
900 700 1000 Life (yrs.)
8 8 8
8 Which m/c to select if
MARR13.5/yr.?
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When alternatives have different lives then make
the comparison over the LCM of years between
the alternatives when using the PW method.
Example 3 mutually exclusive alternatives. MARR
15/yr. Which alternative should be selected?
A B
C Initial investment ()
-6000 -7000
-9000 Salvage value () 0
200 300 Cash flow (/yr.)
2000 3000
3000 Life (yrs.) 3
4 6
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When the lives of alternatives are very long then
can be considered as infinite, use capitalized
cost comparison.
Example 6 different sites for the construction
of a dam. If the life of a dam can be considered
to be infinite, and MARR 6/yr., which site
should be selected? C
E A B D
F P(million ) -3 -5
-6 -8 -10
-11 A(1000 ) 125 350
350 420 400
700 Comparison C to not. E to not. A to
E B to E D to E F to E
P(million ) 3 5 1
3 5 6
A(1000 ) 125 350
0 70 50
350 Incremental i 4.17 7 0
2.33 1 5.83 Increment
No Yes No
No No No justified
Alternative None E
E E E
E selected Select alternative E.
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The Payback Period Method
Payback period of years required for recovering
the investment made in a project-without taking
the time value of money into consideration. i.e.
Let q payback period yt cash flow
of project in period t, t0,1,..,n n
life of the project yt lt 0 when cash outlay yt gt
0 when cash receipt
i.e. It measures the speed with which invested
funds are returned to the business.
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Cont.

• -The earnings or savings figure, i.e. When yt gt 0
  , should be cash earnings or net savings after
  taxes.
• Payback period is used as a limit rather than the
  criterion itself.
• i.e. A firm will have some maximum acceptable
  payback period for a given class of projects, and
  any or all projects having payback periods
  greater than this maximum will be rejected.
• - Payback period is used as a ranking device a
  projects with the shortest payback period are
  given the highest rankings.

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Example
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From this example

• Payback period as an isolated criterion by itself
  is not a very reliable one.
• Fails to distinguish between projects AB, i.e.
  It ignores the cash flows after the payback
  period.
• It does not consider the time value of money.
• It may incorrectly rank projects. A B are
  ranked as preferable to C D. But C D might be
  preferred due to higher cash flows (total).

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Cont.

• So, although managers like to use it because it
  is
• simple to calculate,
• Managers would like to recover their money as
  soon as possible in an uncertain environment,
• It should not be used on its own, but be used
  together with the other criteria in economic
  decision making.