Topic 9: Equipment Replacement Analysis

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Topic 9 Equipment Replacement Analysis

• Introduction
• Defender and Challenger Concepts in Replacement
  Analysis
• Replacement Analysis Using a Specified Planning
  Horizon
• Conventional and Cash Flow Approaches to
  Replacement Analysis
• Replacement Analysis for 1-Additional Year
  Retention
• Minimum- Cost Life Analysis

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Introduction
After selecting and using an alternative, as time
passes the new question is Should the in-place
alternative be replaced, if so, with what and
when. i.e. Has the economic life of the asset
been reached? Which alternative should be
selected as its replacement?
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Replacement requirement reasons are

• Reduced performance
• due to
• - deterioration of parts
• - reliability productivity decrease
• a increased costs of operation
• higher scrap, lost sales,
• higher maintenance costs
• a need to replace.

4
Cont.

• Changing requirements
• - New requirements on
• . Accuracy
• . Speed
• . Other specifications (e.g. Pollution
  reduction)
• may not be met by existing asset or system.
• (e.g. EC regulations for eastern European
  countries.)

5
Cont.

• Obsolescence
• New technological innovations make existing
  assets obsolete.
• The decrease in the time of the development
  cycle of new products causes replacement analysis
  to be performed before the completion of expected
  economic life of assets.

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Defender and Challenger Concepts in Replacement
Analysis
As done until the objective in replacement
analysis is again comparison of two or more
alternatives. Difference a Defender is the
asset already owned. Challengers are the
alternatives with which the already owned asset
is compared with. In this comparison, the
consultants view is taken a We assume (for
comparison purposes) that we do not own either
asset (i.e. including the already owned asset)
7
Cont

• This means that when estimating the first cost of
  the defender, the going market value of the asset
  is used, not the book value.
• This going market value may differ considerably
  from the original data.
• Past data is of no significance, it is just used
  to calculate
• Sunk cost present book value-present realizable
  value.
• Sunk costs are results of bad decisions made in
  the past, and past economic decisions must not
  influence future economic decisions.

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Example
A truck was purchased 3 yrs. Ago for 12,000
with an estimated life of 8 years, a salvage
value of 1,600 at the end of its estimated life,
and annual operating costs of 3,500. The current
book value of the truck is 8,100. A new truck is
now offered for 11,000 with a trade-in value of
7,500 for the old truck. The company estimates
the new trucks economic life as 10 yrs., with a
salvage value of 2,000 at the end of its
economic life. Annual operating costs are
estimated to be 1,800/yr. The existing truck has
a remaining life of 3 yrs., with a realizable
salvage value of 2,000 then, and same operating
costs.
9
Cont.
The comparison should be based on the following
cost estimates
old truck new truck Initial cost(P)
7,500 11,000 Annual
operating 3,500/yr.
1,800/yr. costs(AOC) Salvage value(SV)
2,000 2,000 Economic life (n)
3 yrs. 10 yrs.
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Replacement Analysis Using a Specified Planning
Horizon

• Planning horizon of years used in economic
  analysis.
• 2 situations possible in replacement analysis
• The anticipated remaining life of defender equals
  the life of the challenger
• The life of challenger is longer than that o
  defender.

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Cont.
In the first case, there is no problem. The
alternatives are compared over a common planning
horizon.
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Example
The 2 vans were purchased 2 years ago for 60,000
each. The company plans to keep the vans for 10
more years. Fair market value for a
2-year-old-van is 42,000 and for a
12-year-old-van 8,000. Annual fuel, maintenance,
tax, etc. costs are 12,000/yr. Leasing cost is
9,000/yr. With annual operating costs of
14,000. Should the company lease its vans if a
12/yr. Rate of return is required?consider a 10
yr. Life for the currently owned vans.
13
Cont.
A transport company owns 2 vans deteriorating
faster than expected. The available options
are-going on using the existing vans.-leasing
on a yearly basis Old
vans Leasing P
42,000 Least cost 9,000/yr. AOC
12,000/yr. 14,000/yr. SV
8,000 ---- n
10yrs.
10yrs.
14
Cont.
EUAWo(42,000-8,000)(A/P,12,10)8000(0,12)
12,000 18,977/yr. EUAWl9,000
14,00023,000/yr. Since EUAWlgt EUAWo, the
firm should go on using its own vans and not
lease it.
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In the second case,
i.e. When the lives of the defender and
challengers are different, the length of the
planning horizon should be selected. Selection
of a planning horizon is a difficult decision- it
should be based on sound judgment and data.
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Cont.

• The use of a short horizon may bias the economic
  decision in favor of the defender. (by not
  allowing proper capital recovery of the
  challenger over its anticipated life)
• Use of a longer horizon may be undesirable
  because
• future is uncertain
• international competition, and rapid obsolescence
  may necessitate shorter planning horizons.

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What to Do?

• In general, in replacement analysis the planning
  horizon is taken to be equal to the life of the
  longer-lived alternative. The assumption made in
  this case is
• - EUAW of the shorter-lived asset is the same
  throughout the planning horizon. I.e. The service
  performed by the shorter-lived asset can be
  acquired at the same EUAW as presently computed
  for its expected life.

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Example
Using the truck example given in the previous
slay Planning horizon 10 yrs. If MARR
10/yr. EUAWo (7,500-2,000)(A/P,10,3)2,000(0,1
)3,500 5,911.605/yr. EUAWN
(11,000-2,000)(A/P,10,10)2,000(0,1)1,800
3,464.75/yr. So the old truck should be
replaced with the new trucks since EUAWo gt
EUAWN.
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• International competition, uncertainty about the
  future impose the use of shorter horizons a
  recovery of invested capital the required
  return over a shorter period of time than the
  expected lives of the alternatives.

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Example
The same example as previous, except the planning
horizon should be 5 years due to technological
progress. Assuming that the salvage value of
challenger will remain at 2,000 at the end of
the 5th year, EUAWD 5,911.605 EUAWN(11,000-2,0
00)(A/P,10,5) 2,000(0.1)1,800
4,374.2 Since EUAWNlt EUAWD replace the
old trucks.
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• A common practice is to use augmentation a
  defender is augmented with a newly purchased
  asset to make it comparable in ability (speed,
  volume, etc.) with the challenger.

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Example

• 3 years ago the fire department purchased a new
  fire truck but due to population growth new
  fire-fighting capacity is needed.
• Plan A- An additional identical tuck can be
  bought to augment the existing one
• OR
• Plan B- A new double capacity truck can replace
  the presently owned asset.
• MARR12/yr.
• Compare the alternatives,
• Using a 12 yr. planning horizon,
• 9 yr planning horizon

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Cont.
Plan A
Presently owned
Augmentation P 18,000
P58,000
AOC1,500/yr. AOC
1,500/yr. SV 5,100
SV 6,960 n 9 yrs.
n 12 yrs.
Plan B
Double capacity
P 72,000 AOC 2,500/yr.
SV 7,200 n 12
yrs.
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Cont.

• Over a 12 years planning horizon
• EUAWA (18,000-5,100)(A/P,12,9)
• 5,100(0.12)1,500(58,000-6,960)
• (A/P,12,12)6,960(0.12)1,500
• EUAWA 4,53310,575 15,108/yr.
• EUAWB (72,000-7,200)(A/P,12,12)
• 7,200(0.12)2,500
• 13,825/yr.
• Purchase double capacity truck (B) with an annual
  cost advantage of 1,283/yr.

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Cont.
b)Over a 9 years planning horizon Assuming that
the SV will remain the same. EUAWA
(18,000-5,100)(A/P,12,9)
5,100(0.12)1,500(58,000-6,960)
(A/P,12,9)6,960(0.12)1,500
16,447/yr. EUAWB (72,000-7,200)(A/P,12,9)
7,200(0.12)2,500
15,526/yr. Again plan B should be selected.
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Conventional and Cash Flow Approaches to
Replacement Analysis
. Equally correct . Equivalent way of handling
replacement analysis

• Conventional Approach
• Cash Flow Approach

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Conventional Approach

• What we used up till now.
• The defenders P is the highest value attainable
  through disposal (sale, scrap, trade-in, etc.)
• Cumbersome when more than one challenger with
  each having different trade-in values for the
  defender.

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Cash Flow Approach

• Can only be used if the challenger and defender
  have equal lives, or when the comparison is done
  over a pre-specified planning horizon.
• The first cost, P, of the defender is set equal
  to zero, and the first cost of the challenger is
  set equal to
• challenger cost- defender trade-in value.
• So, when a challenger is selected, the defender's
  market value is a cash inflow to the challenger
  alternative. If defender is selected no actual
  cash outlay.

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Example
A 7 yr. old asset may be replaced with either of
2 new assets. If MARR18/yr. What should be the
decision? Defender
Possible Replacements
(Current Asset) Challenger 1 Challenger 2
First cost()
10,000 18,000 Defender
3,500
2,500 trade-in () AOC(/yr.) 3,000
1,500 1,200 SV()
500 1,000
500 Life (yrs.)
5 5 5
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Cont.
Cash Flow Approach EUAWD 3,000-500(A/F,18,5)
2,930.11/yr. EUAWc1 (10,000-3,500-1,000)(A/P,18
,5) 1,000(0.18)1,500
3,438.79/yr. EUAWc2 (18,000-2,500-500)(A/P,18,
5) 500(0.18)1,200
6,086.7/yr. Since the defender has the least
EUAW retain defender.
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Cont.
Conventional Approach EUAWD when compared with
challenger 1 EUAWD(3,500-500)(A/P,18,5)500(0.
18) 3,000 4,049.34/yr. EUAWc1
(10,000-1,000)(A/P,18,5)
1,000(0.18)1,500 4,558.02/yr. So choose
defender to challenger 1.
32
Cont.
EUAWD when compared with challenger
2 EUAWD(2,500-500)(A/P,18,5)500(0.18)
3,000 3,729.56/yr. EUAWc2
(18,000-500)(A/P,18,5)
500(0.18)1,200 6,886.15/yr. Again retain
defender, since EUAWD is the least.
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Conventional Approach can be used to determine
Replacement Value (RV) of the defender that would
make its retention or replacement equally
attractive by solving EUAWD EUAWC for the
unknown RV.
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Replacement Analysis for 1-Additional Year
Retention

• When a currently owned asset is close to the end
  of its useful life, should it be replaced by a
  challenger or retained in service
• one more year
• More than one more year up until its remaining
  life
• So, there are 3 alternatives
• 1- Select the challenger
• 2- Retain the defender for one more year
• 3- Retain the defender for its remaining life.

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Cont.

• In this case, the comparison should be based on
  EUAWC CD(1)a EUAWD for one more year
• If CD(1)lt EUAWC, retain the defender one more
  year.
• If CD(1)gt EUAWC, calculate EUAWD over its
  remaining life.
• a) If EUAWDlt EUAWC, the defender is
  retained until next year the analysis is
  repeated.
• b) If EUAWDgt EUAWC, replace the defender.

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Minimum- Cost Life Analysis
Often one wants to determine the amount of time
an asset should be kept in service so as to
minimize its total cost. This time in years is an
n value referred to as - minimum cost life
- economic life - retirement life -
replacement life The n value is the of years
which give or yield a minimum annual cost.
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Cont.

• To find n-minimum cost life
• increase the life index k from 1 to the largest
  expected life N.
• For each k determine the value of EUAWk
• where
• EUAWkP(A/P,i,k)-SVk(A/F, i,k)
• 
  (A/P,i,k), k 1,2,....,N.
• where
• SVk- salvage value if the asset is
  retained k years.
• AOCj-annual operating cost in year j, j
  1,2,...,k.

38
Cont.
The minimum- cost life is the k value for which
EUAWk is the smallest.
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Example
An asset has a first cost of 3,000. Given the
following data and MARR12/yr. find the assets
economic life.
End of year SV Operating
costs (k) ()
during year k ()
1 1,500
1,000 2 1,000
1,700 3 500
2,400 4 0
3,100