Ekonomi Teknik Kimia

JTK FT UGM 2015

Economic Criteria

• Depending on the situation, the economic

criterion will be one of the following :

Situation Criterion

For fixed input Maximize output

For fixed output Minimize input

Neither input or Maximize

output fixed (output – input)

Applying Present Worth Techniques

Equivalence provides the logic by which we may

adjust the cash flow for a given alternative into

the same equivalent sum or series with

equivalent present consequences, then

compared it to chose the feasible alternatives.

In this case, the consequences of each

alternative must be considered for this period of

time whish is usually called the analysis period

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Applying Present Worth Techniques

The tree criteria for economic efficiency are restated in

terms of present worth analysis:

Situation Criterion

Fixed

input

Amount of money or

other input resources are

fixed

Maximize present worth of

benefits or other outputs

Fixed

output

There is a fixed task,

benefits or other output to

be accomplished

Minimize present worth of

costs or other inputs

Neither

input nor

output is

fixed

Neither amount of money,

or other inputs nor

amount of benefits or

other outputs is fixed

Maximize (present worth of

benefits – present worth of

costs), that is maximize net

present worth

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Three different analysis-period situations in economic

analysis problems

The useful life of each alternative equals the analysis

period

The alternatives have useful lives different from the

analysis period

There is an infinite analysis period, n =

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Assumptions in solving economic analysis

problem(1)

End-of-year convention

The annual expense / revenue occur in end-of-year.

Example :

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Tahun 0 1 Jan

(Akhir) Tahun 1 31 Des 1 Jan

(Akhir) Tahun 2 31 Des

P

A

F

A

Assumptions in solving economic analysis

problem(2)

Middle-of-period convention

The annual expense / revenue occur in middle-of-

year.

Example :

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Tahun 0 1 Jan

Pertengahan Tahun 1 30 Juni

Akhir Tahun 2 31 Des

P

A F

A

Pertengahan Tahun 2 30 Juni

31 Des 1 Jan

Useful lives equal the analysis period(1)

Example 1:

A firm is considering which of two mechanical devices to install to reduce costs in a particular situation. Both devices cost $1000 and have useful lives of five years and no salvage value. Device A can be expected to result in $300 savings annualy. Device B will provide cost savings of $400 the first year but will decline $50 annually, making the second year savings $350, the third year savings $300, and so forth. With interest at 7% which device should the firm purchase ?

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Useful lives equal the analysis period(2)

The analysis period can conveniently be selected

as the useful life of the devices or 5 years.

Devices A Devices B

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A = 300 400

350 300

250 200

PW of benefits PW of benefits

Useful lives equal the analysis period(3)

PW of benefits A = 300(P/A,7%,5) = 300(4,1)

= $1230

PW of benefits B = 400(P/A,7%,5) – 50(P/G,7%,5)

= 400(4,1) – 50(7,647)

= $1257,65

PW of benefits B > PW of benefits A,

therefore devices B preferred aternative

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Useful lives equal the analysis period(4)

Example 2:

Three mutually exclusive investment alternatives for implementing an office automation plan in an engineering design firm are being considered. Each alternative meets the same service (support) requirements, but differences in capital investment amounts and benefits (cost saving) exist among them. The study periods 10 years, and the useful lives af all three alternatives are also 10 years. Market values of all alternatives are assumed to be zero at the end of their useful lives. If the firm’s MARR is 10% per year, which alternaitve should be selected in view of the following etimates ?

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Useful lives equal the analysis period(5) Alternative

A B C

Capital

investment

$390.000 $920.000 $660.000

Annual cost

savings

$69.000 167.000 133.500

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Solution :

PWA = -$390.000 + $69.000 (P/A, 10%, 10) = $33,977

PWB = -$920.000 + $167.000 (P/A, 10%, 10) = $106,148

PWC = -$660.000 + $133.500 (P/A, 10%, 10) = $160,304

Useful lives different from the analysis

period(1)

Example 1 :

A purchasing agent is considering the purchase of some new equipment for the mailroom. Two different manufacturers have provided quotations. An analysis of the quotations indicates the following :

Manufacturer Cost Useful life(year) salvage value

Speedy $ 1500 5 $ 200

Allied $ 1600 10 $ 325

Assume 7% interest and equal maintenance costs.

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Useful lives Different from the analysis

period(2)

If the allied equipment has a useful life of ten

years, and the speedy equipment will last five

years, one solution is to select a ten year analysis

period.

Assume the replacement speedy equipment will

also cost $1500 five years hence

Speedy Allied

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1500 1500 1600

200 200 325

Useful lives Different from the analysis period(3)

PW of cost Speedy

= 1500 + (1500-200) (P/F,7%,5) – 200(P/F,7%,10)

= 1500 + 1300(0,713) – 200(0,5083)

= $ 2325,24

PW of cost Allied

= 1600 – 325(P/F,7%,10) = 1600 – 325(0,5083)

= $ 1434,803

PW of cost Allied < PW of cost Speedy ,

therefore Allied preferred aternative

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Useful lives Different from the analysis period(4)

Example 2 :

The following data have been estimated for two mutually

exclusive investment alternatives, A and B associated with a

small engineering project for which revenues as well as

expenses are involved. They have useful lives of four and six

years, respectively. If the MARR = 10% per year, show which

alternative is more desirable by using equivalent worth methods.

Use the repeatability assumption.

A B

Capital Investment $3.500 $5.000

Annual Cash Flow 1.255 1.480

Useful Life (years) 4 6

Market value at end of useful

life

0 0

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Useful lives Different from the analysis period(5)

Solution :

The least common multiple of the useful lives of

alternatives A and B is 12 years.

PWA = -$3.500 - $3.500 [(P/F,10%,4) + (P/F,10%,8)]

+ $1.255 (P/A,10%,12)

= $1.028

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Infinite Analysis Period (1)

In governmental analysis, at times there are

circumstances where a service or condition is to be

maintained for an infinite period. We call this particular

analysis Capitalized Cost

Capitalized Cost, is the present sum of money that

would need to be set aside now, at some interest rate,

to yield the funds required to provide the service

indefinitely. Assume cost of maintenance is equal every

year.

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Infinite Analysis Period (2)

For one year period :

If iP is cost of maintenance every period,F = P +

iP, if iP adalah biaya perawatan untuk setiap

periode, maka di setiap periode terjadi

pengeluaran sebesar iP, sehingga untuk n =

dapat dinyatakan sebagai A = iP dan

Capitalized cost (P) = A / i.

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Infinite Analysis Period (3)

example :

A city plans a pipeline to transport water from a distance

watershed area to the city. The pipeline will cost $8 million and

have an expected live of seventy years. The city anticipates it will

need to keep the water line in service indefinitely.

Compute the capitalized cost assuming 7% interest?

Solution :

Here we have renewals of the pipeline every seventy years

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P

$ 8 juta $ 8 juta $ 8 juta

70 tahun 140 tahun

$ 8 juta

n =

Infinite Analysis Period (4)

To compute the capitalized cost, it is necessary to first compute an end-of-period disbursement A that is equivalent to $8 million every seventy years

A = F(A/F,i,n) = $ 8 juta(A/F,7%,70)

= $ 8 juta( 0,00062) = $ 4960

Capitalized cost P = $ 8juta + A/i = $ 8juta +4960/0,07

= $ 8.071.000

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$ 8 juta

n = 70

A

Present Worth analysis for multiple alternatif(1)

Example :

An investor paid $8000 to a consulting firm to analyze what he might do with a small parcel of land on the edge of town that can be bought for $30.000. In their report, the consultants suggested four alternatives :

Assuming MARR is 10%, what should the investor do?

Alternative Total

investment

Annual

benefit

Terminal value at

end of 20 year

A: Do nothing $ - $ - $ -

B: Vegetable Market 50000 5100 30000

C: Gas Station 95000 10500 30000

D: Small Motel 350000 36000 150000

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Present Worth analysis for multiple alternatif(2)

This Problem is one of neither fixed input nor fixed output, so our criterion will be to maximize the present worth of benefit minus the present worth of cost, or, simply stated, maximize net present worth.

Alternative A : NPW = 0 (do nothing)

Alternative B (vegetable market) :

NPW = -50000 + 5100(P/A,10%,20) +30000(P/F,10%,20)

= -50000 + 5100(8,514) + 30000(0,1486)

= -2120,6

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Present Worth analysis for multiple alternatif(3)

Alternative C (gas station):

NPW = -95000 + 10500(P/A,10%,20) + 30000(P/F,10%,20)

= -1145

Alternative D (small motel) :

NPW = -350000 + 36000(P/A,10%,20) + 150000(P/F,10%,20)

= -21206

In this situation alternative A has NPW equal to zero, and others alternative have negative NPW. We will select the best alternatives, namely, the do nothing (alt. A).

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