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SEMESTER JANUARY 2010 1 CHAPTER 9 REPLACEMENT ANALYSIS Why you replace? When you replace?

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SEMESTER JANUARY 2010 1

CHAPTER 9

REPLACEMENT ANALYSIS

Why you

replace?

When you

replace?

SEMESTER JANUARY 2010 2

Replacement Concept

Replacement refers to Selection of new assets to replace existing assets. The evaluation of entirely different ways to perform an

asset’s function.

Example:Old trucks can be replaced with new models that operate similarly but have additional features that improve performance or efficiency.

Trucks could be replaced with conveyor system, overhead crane, subcontract for hauling, or manual labour; that serves the needed function.

SEMESTER JANUARY 2010 3

Introduction

The pressure of competition in business such as requiring higher quality goods and services, shorter response times, competitive price often lead to a situation whereby organisations have to decide whether the existing asset should be

Retired from use, Retain the asset for backup, Continued in service, or Replaced with a new assets.

SEMESTER JANUARY 2010 4

INTRODUCTION

In REPLACEMENT ANALYSIS, the reference for comparison is the existing resource --- anything that is used in business such as machine, tools, or equipment.

The question is:

When should we replace the resource?

Or more focused question:

Should we replace the equipment now or sometime later?

Precise question:

Should we budget now to replace the resource during the next financial year?

SEMESTER JANUARY 2010 5

REPLACEMENT ANALYSIS

The evaluation of changes in economics of assets associated with their use in an operating environment.

Considers asset: replacement retirement Augmentation (increase the capability)

SEMESTER JANUARY 2010 6

The Importance of Replacement Decision

Timely replacement decisions are critically important to a company:

A decision to replace a machine because it is temporarily out of order, or untimely replacement for latest technology, can be a serious drain on operating capital of a company.

A decision to postpone replacement until there is no other way to continue production, can place a company in a dangerous position of becoming uncompetitive. Therefore, we need to recognise when an asset is no longer employed efficiently, what replacement should be considered, and when replacement is economically feasible.

The GOAL is to be COST EFFICIENT.

SEMESTER JANUARY 2010 7

DEFENDER – CHALLENGER CONCEPT

Replacement analysis can be conceptualised better by considering the existing resource as a defender as it were trying to defend its continued use.

The one being considered to replace the defender is called challenger.

A complete replacement analysis involves three tasks:

1. Selection of the defender and its analysis

2. Selection of the challenger and its analysis

3. Defender – challenger comparison

SEMESTER JANUARY 2010 8

REASONS FOR REPLACEMENT ANALYSIS

Physical Impairment (Deterioration) Altered Requirements Technology Financing [rental (lease) is more attractive than ownership]

SEMESTER JANUARY 2010 9

PHYSICAL IMPAIRMENT (DETERIORATION)

Efficiency loss resulting from continued use - - - aging Increased routine and corrective maintenance costs Greater energy requirements Increased need for operator intervention Unanticipated problems leading to equipment

deterioration

SEMESTER JANUARY 2010 10

ALTERED REQUIREMENTS

Significant change in demand for related products or services

Significant change in the composition or design of associated products or services

May be considered a form of obsolescence

SEMESTER JANUARY 2010 11

TECHNOLOGY

Impact of technological change varies with associated industry

Technological changes typically reduce cost per unit and improve quality of output

Results in earlier replacement of existing assets with improved assets

May be considered a form of obsolescence

SEMESTER JANUARY 2010 12

FINANCING Considers economic opportunity changes

external to the physical operation or use of the asset(s)

May involve income tax considerations (depreciation and after-tax analysis) EG: rental of assets may become more

attractive than ownership May be considered a form of obsolescence

SEMESTER JANUARY 2010 13

REPLACEMENT ANALYSISTwo types of approach available [when to apply]:

1. EUAC-BASED analysis [zone A and early part of zone B]

2. MARGINAL-COST-BASED analysis [later part of zone B]

Cost

AgeEconomic life

A B

EUAC min

Capital Recovery cost

Operating cost

Total cost

SEMESTER JANUARY 2010 14

MARGINAL-COST

Marginal cost is the cost to keep an asset in service one more year. This concept is applicable to mature, older existing equipment (defender) with increasing operating costs.

Require knowledge of the future MV of the asset.

The marginal cost is calculated for each year of the asset’s life.

The marginal cost is used to compare against EUAC of the proposed replacement.

SEMESTER JANUARY 2010 15

ECONOMIC LIFE

ECONOMIC LIFE is the period of time (years) that results in the minimum Equivalent Uniform Annual Cost (EUAC) of owning and operating an asset

Assuming good asset management, economic life should coincide with time from date of acquisition to date of abandonment, demotion in use, or replacement from primary intended service

Replacement studies are usually made as EUAC (annual

equivalent) calculations to take advantage of data commonly collected as annual charges {maintenance costs, operating expenses, salaries, inflation, depreciation, taxes, etc}.

SEMESTER JANUARY 2010 16

ECONOMIC LIFE

Economic life is sometimes called minimum-cost life or optimum replacement interval (for cyclic replacement of assets)

For a new asset, economic life can be computed if capital investment, annual expenses, and year-by-year market values are known or can be estimated

SEMESTER JANUARY 2010 17

OWNERSHIP LIFE

Period between date of acquisition and date of disposal by a specific owner

A given asset may have different categories of use during this period

SEMESTER JANUARY 2010 18

PHYSICAL LIFE

Period of time between original acquisition and final disposal of an asset over its succession of owners

SEMESTER JANUARY 2010 19

USEFUL LIFE

The time period in years that an asset is kept in productive service either in primary or backup mode

An estimate of how long an asset is expected to be used in a trade or business to produce income

SEMESTER JANUARY 2010 20

ECONOMIC LIFE FOR CYCLIC REPLACEMENT

Many mechanical items used are replaced by essentially the same machine when the original one wears out.

Informal rules may be used to establish cyclic replacement times. A company replaces a car whenever it exceeds 5 years of service. A rental car company replaces their cars whenever they exceed 100,000km.

Such replacement rules recognise that the automobiles or similar machines become less efficient and accumulate higher and higher repair bills as they age.

The total lifetime cost continues to increase with age, but average annual cost passes through a minimum.

Refer to Figure 9.1

SEMESTER JANUARY 2010 21

$

Age at replacement, years

Cumulative total cost, CTC

Cumulative operating cost, COC

Cumulative capital cost =

(initial cost – resale cost)

Min average annual cost

1 2 3 5 6 7 84

20,000

60,000

40,000

Figure 9.1

Average annual cost=

CTC / age of replacement

SEMESTER JANUARY 2010 22

1 2 3 4 5Operating cost 2,000 3,000 4,620 8,000 12,000 Market value (MV) 15,000 11,250 8,500 6,500 4,700

Example 1: Cyclic Replacement Analysis:

A delivery service company owns a fleet of small delivery trucks for store-to-home deliveries. The purchase price per truck is $20,000, and the anticipated schedule of future operating costs and salvage value is shown below.

We are required to calculate the least-cost replacement interval.

Table 1

SEMESTER JANUARY 2010 23

The above tabular solution method reveals that trading-in the trucks after 3 years for new unit is the minimum-cost replacement cycle.

This is the economic life of the challenger, where EUAC is at the lowest.

Table 2

Least cost

Cycle time iterative results

Loss in Cost of capital operating Total marginal EUAC

EOY MV @ MV during 10% of begin'g cost cost for year through

k EOY k year k of year MV (Ek) (TCk) year k

(A) (B) (C)=(A)*0.1 (D) (B)+(C)+(D)

0 20,000

1 15,000 5,000 2,000 2,000 9,000 9,000 2 11,250 3,750 1,500 3,000 8,250 8,643 3 8,500 2,750 1,125 4,620 8,495 8,598 4 6,500 2,000 850 8,000 10,850 9,084 5 4,750 1,750 650 12,000 14,400 9,954

From Table 1 Given

SEMESTER JANUARY 2010 24

Table 3

Sample calculation of EUAC for N at year 2.

Least cost

k for year (TCk)01 9,000 9,000 2* 8,250 8,643 3 8,495 8,598 4 10,850 9,084 5 14,400 9,954

* sample calculation for 2 years ownership:EAC (N=2) = ($9,000)(P/F, 10%, 1)(A/P, 10%, 2)

+ $8,250(A/F, 10%, 2)= $4,714.40 + $3,928.65= $8,643

SEMESTER JANUARY 2010 25

REPLACEMENT STUDY CONSIDERATIONS

1. Recognition and acceptance of past errors

2. Sunk costs

3. Existing asset value and the outsider viewpoint

4. Income tax considerations

5. Economic life of the proposed replacement (Challenger)

6. Remaining (economic) life of the old asset (defender)

SEMESTER JANUARY 2010 26

PAST ESTIMATION ERRORS

Past estimation errors are irrelevant unless there are income tax implications

Example:

1. when BV > current MV, frequently attributed to estimation error,

2. inadequate capacity,

3. maintenance costs higher than anticipated.

The above are mainly due to the inability to foresee future conditions better at the time of original estimates.

Must focus on valid estimation of future replacement, without consideration of loss which may have occurred in the past.

SEMESTER JANUARY 2010 27

THE SUNK COST TRAP

Only present and future cash flows should be considered in replacement studies

Unamortized values of existing asset considered for replacement are the result of past decisions -- (Sunk costs = BV - MV)

Sunk costs are irrelevant to replacement decisions, except to extent they affect income taxes

When tax considerations are involved, sunk costs must be included in study

SEMESTER JANUARY 2010 28

EXISTING ASSET INVESTMENT VALUE --- AN OUSIDER VIEWPOINT

Perspective of impartial third party in establishing fair market value (MV) of your used asset

Present realizable MV defines correct investment amount for asset in replacement studies

Consider the opportunity cost of retaining the asset -- the defender

SEMESTER JANUARY 2010 29

THE OUTSIDER VIEWPOINT

The total investment in the defender is the opportunity cost of not selling the existing asset for its current MV, plus the cost of upgrading to be competitive with best available challenger

SEMESTER JANUARY 2010 30

Example 9.1 The purchase price of a new automobile (challenger) is

RM 21,000. The present automobile (defender) can be sold for RM 10,000. The defender was purchased three years ago and its current BV is RM 12,000. To make defender comparable in continued service to the challenger, your firm would need to make some repairs at an estimated cost of RM 1,500.

1. Total capital investment in the defender (if kept):

RM 10,000 + RM 1,500 = RM 11,500

2. Unamortised value of the defender is:

RM 12,000 – RM 10,000 = RM 2,000 (loss)

This is the difference between the current market value and the current book value of the defender.

This represents sunk cost and has no relevance to the replacement decision.

SEMESTER JANUARY 2010 31

Example 9.2

A old pressure vessel has annual O&M expenses of $60,000 per year and it can be kept for 5 years more at which time it will have $0 MV. The present MV is $30,000 if it were sold now.A new pressure vessel cost $120,000. It will have a MV of $50,000 in 5 years and will have O&M expenses of $30,000 per year.Using before tax MARR of 20% per year, determine whether or not the old pressure vessel should be replaced.

Solution:The 1 st step is to determine the investment value of the defender (old vessel). Using outsider viewpoint, the investment value of the defender is $30,000, its present MV.

SEMESTER JANUARY 2010 32

Example 9.2 cont’d

The problem can be solved using PW, FW or AW method.

Defender:

AW(20%) = -$30,000(A/P,20%,5) - $60,000

= -$70,032

Challenger:

AW(20%) = -120,000(A/P,20%,5) - $30,000

+ $50,000(A/F,20%,5)

= -$63,408

EUAC challenger < EUAC defender.

Thus, the old pressure vessel should be replaced.

SEMESTER JANUARY 2010 33

THREE CASES OF USEFUL LIFE USED IN REPLACEMENT ANALYSIS

Useful lives of the defender and challenger are known and the same and also equal study period.

Useful lives of the defender and challenger may or may not be known but economic life can be determined.

Useful lives of the defender and challenger are known but not the same.

SEMESTER JANUARY 2010 34

ECONOMIC LIFE OF THE CHALLENGER

Economic life of an asset minimizes equivalent uniform annual cost of owning and operating an asset

Economic life is often shorter than useful or physical life Economic data regarding challengers are periodically

(often annually) updated Replacement studies then repeated to ensure an on-going

evaluation of improvement opportunities.

SEMESTER JANUARY 2010 35

ECONOMIC LIFE OF DEFENDER

Often one year Because different lives of the challenger and defender are

involved, care should be taken when comparing defender with challenger

Defender should be kept longer than apparent economic life as long as its marginal cost < minimum equivalent uniform annual cost of challenger over its economic life

SEMESTER JANUARY 2010 36

Example 2:

It is desired to determine how much longer a fork lift should remain in service before it is replaced by new unit (challenger) discussed in Example 1.

The defender in this case is 2 years old, originally cost $13,000, and has a present MV of $5,000. If kept, its market values and annual expenses are as tabulated below:

Loss in Cost of capital operating Total marginalEOY MV @ MV during 10% of begin'g expenses cost for year

k EOY k year k of year MV (Ek) (TCk)0 5,000 1 4,000 1,000 500 5,500 7,000 2 3,000 1,000 400 6,600 8,000 3 2,000 1,000 300 7,800 9,100 4 1,000 1,000 200 8,800 10,000

Since the defender’s marginal cost increases during the four-year analysis period, we compare its marginal cost, year by year, with the corresponding EUAC of the challenger to decide when to replace it.

SEMESTER JANUARY 2010 37

EOY Defender's Challenger's Decisionk marginal cost EUAC to replace0 (from Table 2)1 7,000 9,000 NO2 8,000 8,643 NO3 9,100 8,598 YES4 10,000 9,084

Since the defender’s marginal cost at year 3 ($9,100) exceeds the EUAC of the challenger ($8,598), the defender is replaced at year 3.

SEMESTER JANUARY 2010 38

RETENTION OF THE DEFENDER

The defender should be kept longer than the apparent economic life of the defender as long as its marginal cost (total cost for an additional year of service) is less than the minimum EUAC for the best alternative challenger

SEMESTER JANUARY 2010 39

PROJECT RETIREMENT WITHOUT REPLACEMENT -- ABANDONMENTTwo assumptions apply: Once capital investment made, firm desires to postpone project

abandonment as long as its present equivalent value (PW) is not decreasing

The project will be terminated at the best abandonment time and will not be replaced by the firm

Note:

In abandonment problems, annual benefits are present.

In economic life analysis, costs are dominant.

For both cases the objective is to increase the wealth of the firm by: a) finding the life that maximises profits,

b) finding the life that minimises the costs.

See Example 9-7

SEMESTER JANUARY 2010 40

Example 9.7 A $50,000 baling machine for recycled paper is being considered

by the XYZ company. Annual revenues less expenses and end-of-year abandonment values (MV) for the machine have been estimated for the project. The MARR is 12% per year. What is the best time to abandon the project if the firm decided to acquire the baling machine and use it for no longer than 7 years?

[finding the life that maximises profits]

1 2 3 4 5 6 7Annual Rev - Exp 10,000 15,000 18,000 12,000 7,000 4,000 3,500 MV of machine 40,000 32,000 25,000 19,000 15,000 12,000 10,000

End of Year, $

SEMESTER JANUARY 2010 41

Keep for one year$40,000

$10,000

0 1$50,000

PW (12%) = -$50,000 + ($10,000 + $40,000) ( P/F, 12%, 1)= ($5,355)

Keep for two years $32,000

$10,000 $15,000

0 1 2$50,000

PW (12%) = -$50,000 + $10,000 (P/F, 12%, 1) + ($15,000 + $32,000) (P/F, 12%, 2)= ($3,603)

Sample calculation:

SEMESTER JANUARY 2010 42

In the same manner, the PW (12%) for years three through seven can be computed. The results are as follows:

Keep for three years PW (12%) = $1,494

Keep for four years PW (12%) = $3,400

Keep for five years PW (12%) = $3,802

Keep for six years PW (12%) = $3,403

Keep for seven years PW (12%) = $3,430

Conclusion:

It is obvious that PW is maximised ($3,802) by retaining the machine for a total of 5 years.