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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

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.

-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.

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.

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.

- 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.

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.

- 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.

Cont.

- PW AND FW COMPARISONS CAN ONLY BE DONE OVER SAME

YEARS OF FUTURE SERVICE FOR ALL ALTERNATIVES.

- 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.

- 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.

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?

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.

- 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

- 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.

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).

EXAMPLE

2 SITES UNDER CONSIDERATION FOR A BRIDGE TO CROSS

A RIVER NORTH SUSPENSION BRIDGE. SOUTH-TRUSS

BRIDGE WITH A NEW ROAD CONSTRUCTION.

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

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.?

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.

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.

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?

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?

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.

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?

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)

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.

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.

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

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

Cont.

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

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.

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

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.

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.

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..

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

Example

A corporation is considering the purchase of an

additional machine, the mutually exclusive

alternatives are

Extra investment

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).

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.

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.

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.

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

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.

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.?

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

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.

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.

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.

Example

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).

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.