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Minimum Asset Life to Justify

A Higher Investment

By:Analyn M. Catapang

MSEE major in Computer Engineering

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CONTENT: Definition of asset

Classification of asset

Comparing Alternatives

Compounding-Interest Table

Sample Problem

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What is asset?Are economic resources.

Anything tangible or intangible that is capable of being

owned or controlled to produce value and that is held to

have positive economic value.

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Classification of AssetThe principal distinction normally made for businesspurposes is between:

Fixed assetsex. Debtors, stock, cash and work in progress

Current assetsex. Land, buildings, machinery etc

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How to present a winning justification

for new and replacement equipment

From time to time it becomes clear that an item of

equipment is expensive to maintain and it would becheaper to replace it. To get money for the change it is

necessary to justify the expenditure and prove that

replacement is better decision than keeping the existing

item.

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Comparing Alternatives: Determine the annual cost of an asset.

The general method of computing annual costs can beused for any number of industrial or commercial assetsregardless of whether they are stationary, moving, or

water or air-borne.

The key fact is that accurate cost are required if theannual cost comparison is to have validity.

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Comparing Alternatives: Determine the minimum asset life.

Upon computing the annual cost together with the otherexpenses such as maintenance cost, interest rate, andthe capital-recovery factor results will be used for thecomputation of the minimum asset life.

Used the Compounding-interest table depending uponthe percentage of interest rate used.

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Compound-Interest Table

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Sample problemThe timber floor of a bridge is to be replaced and

consideration is being given to, treating the timber to

prolong its life and reduce maintenance costs. Anuntreated timber floor costs $5000 and has an annual

maintenance cost of $500 and a life of 10 years. A treated

timber floor costs $8500 and has an annual maintenance

cost of $300. How long should the treated timber last tomake it more economical than the untreated timber?

Use an interest rate of 5 percent.

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Compute the annual cost of the untreated timber

floor:Given:

Initial cost (P) = $5000

Maintenance cost (M) = $500

Interest rate (i) = 5%

Useful Life (N) = 10

Capital-recovery factor (CR) = 0.12950

Solve for A:

Formula: A = P(CR) + M

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A = P(CR) + M

A = $5000 (0.12950) + $500

A = $1147.50

The annual cost of the untreated timber floor is:

$1147.50

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Setup an expression for the annualcost of the treated timber floor:

A = $8500 (CR) + $300

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Compute the minimum life required tojustify treating the timber:

Equate the annual cost:

$8500 (CR) + $300 = $1147.50 From the given equation compute the value of CR:

$8500 (CR) = $1147.50 - $300

CR = $847.5 / $8500

CR = 0.099706

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Interpolation Method Interpolating in the compound-interest

table for 5 percent, we find N is within the

range of 14 to 15 years. If we undergo theactual computation of interpolation theactual answer is 14.3 years.

The life of the treated timber floor mustexceed 14.3 years to make it moreeconomical.

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References:Handbook of Civil Engineering Calculations thru Web

Mechanical Engineering Board Reviewer in Engineering

Economics by Ricardo C. Asin

Engineering Economy 2nd Edition

by Hipolito B. Sta Maria

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ThankYou!