eechap4 replacement analysis

7/28/2019 EEChap4 Replacement Analysis

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After completing chapter 4, students should be ableto :

Identify principles and approach used in

replacement analysis. Determine the economic life of an asset. Conduct a replacement analysis to select between

existing asset (Defender) and its challenger. Conduct one year retention analysis.

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Why need replacement? Performance of machine/equipment has

deteriorated.

Requirements has changed.

Development of new technology.

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Approach used in analysis: Latest information Sunk cost

New facility (Challenger) and existingfacility (Defender)

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Annual worth of all costs

Total capital recovery is the annual worth of the firstinvestment and salvage value.

Operation annual cost. Total annual worth = - (Total capital recovery) (annual

worth of capital cost)

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cost

Economic life Years

Total annualworth

Annualoperation cost

Total capitalrecovery


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Example 4.1

i = 12%

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Year (j) Market value atyear j (RM)

Operation cost atyear j (RM)

0 100,000

1 80,000 5,000

2 60,000 6,0003 45,000 7,000

4 25,000 8,000

5 0 9,000


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Solution:

Cash flow diagram if asset is used for :

a. 1 year only b. 2 years only c. 3 yearsonly

d. 4 years only e. 5 years

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Economic life is 4 years

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Year(j)

Capital recovery(RM)

Annual operation cost(RM)

Total cost(RM)

1 -100,000(A/P,0.12,1) +80,000(A/F,0.12,1) = 32,000

5,000 37,000

2 -100,000(A/P,0.12,2) +60,000(A/F,0.12,2) = 30868

5,000 + 1,000(A/G,0.12,2) =5,471.86

38,339.90

3 -100,000(A/P,0.12,3) +45,000(A/F,0.12,3) = 28,299.25

5,000 + 1,000(A/G,0.12,3) =5,924.78

34,224

4 -100,000(A/P,0.12,4) +25,000(A/F,0.12,4) = 28,299.25

5,000 + 1,000(A/G,0.12,4) =6,358.85

*34,051.10

5 -100,000(A/P,0.12,5) + 0 = 27,741 5,000 + 1,000(A/G,0.12,5) =6,774.60

34,515.60


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Consultant approach Example 4.2

- Decision between selling existing vehicle and renting.- Existing vehicle was bought 5 years ago at the price of

RM60,000.- Useful life is 10 more years only.- Salvage value is RM5,000- Operation cost is RM15,000 per year.- If vehicle is sold now, market value is RM40,000- Renting a vehicle will cost RM15,000 a year and the cost

of petrol is RM9,000 a year

Determine if the company should keep the existing vehicleor sell it and rent another vehicle if growth rate is 15% andthe vehicle is required for 10 more years.

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Solution:

a. Cash flow diagram if vehicle is retained.

Aretain = -40,000(A/P,0.15,10) 15,000 +

5,000(A/F,0.15,10)

= -40,000(0.19925) 15,000 + 5,000(0.04925)

= -22,723

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b. Cash flow diagram if rent.

Arent = -24,000Aretain < Arent

Notes: Consultant approach latest data

Market value is RM40,000. Machine value (price) RM60,000 is considered as sunk cost.

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Cash flow approach Example 4.3

Refer to example 4.2

Solution:

a. Cash flow diagram if vehicle is retained

Aretain = -15,000 + 5,000(A/F,0.15,10)= -15,000 + 5,000(0.04925)

= -14,753

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b. Cash flow diagram if rent

Arent = 40,000 (A/P,0.15,10) -24,000

= 40,000(0.19925) 24,000

= -16,030

Aretain < Arent

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Example 4.4

* Challenger X trade-in value for defender is RM4,000 andChallenger Y trade-in value for defender is RM3,000

Determine the decision that should be made.

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Defender Challenger X Challenger Y

Machine cost(RM)

* 10,000 20,000

Operation cost(RM)

3,000 1,500 1,000

Salvage value(RM)

1,000 2,000 1,000

Life (Year) 5 5 5


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Solution : Cash flow diagram for:

a. Defender S

b. Challenger X c. Challenger Y

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Incremental Analysis:

Cash flow diagram for X-S

Cash flow diagram for Y-S

It is more economical to keep existing machine S (defender)

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Reject X, accept S

Reject Y, accept S


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OR Cash flow diagram for Defender

AS = -3,000 + 1,000(A/F,0.25,5)

= -3,000 + 1,000(0.12185)

= -2878.15

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Cash flow diagram of Challenger X

AX = -6,000(A/P,0.25,5) 1,500 +

2,000(A/F,0.25,5)= -6,000(0.37185) 1,500 + 2,000(0.12185)

= -3,487

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Cash flow diagram of Challenger Y

AY = -17,000(A/P,0.25,5) 1,000 +1,000(A/F,0.25,5)

= -17,000(0.37185) 1,000 +1,000(0.12185)= -7,199

AY > AX > AS,Select Defender

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When an asset near the end of its useful life.

Example 4.5a. Machine L

- Bought 5 years ago and it is expected that it may be used for 3 moreyears.

- Current market value is RM40,000

- If it is retained for 1 more year, operation cost is RM45,000- Salvage value after 1 year is RM30,000

b. Machine P- Capable of replacing machine L- Useful life of machine P is 5 years with a first cost of RM90,000 and

annual operation cost of RM40,000- Salvage value is RM20,000

Should machine L be used for 1 more year if MARR is 25%.

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Solutiona. If retain machine L for 1 more year

AL = -40,000(A/P,0.25,1) 15,000(A/F,0.25,1)

= -40,000(1.25) 15,000(1.00)

= -65,000

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b. If buy machine P

AP = -90,000(A/P, 0.25,5) 40,000 + 20,000(A/F,0.25,5)= -90,000(0.37185) 40,000 + 20,000(0.12185)= -71,029

AP > ALRetain machine L for 1 more year

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Analysis

Economic life-To determineestimated lifewhere costs are atminimum.

-Calculate A if use1 year, 2 years and

so on until (m+1)years whereAm+1>AmAchallenger,

retain Defender.

-IfADefender AChallenger, replace

Defender with Challengernow.

eechap4 replacement analysis

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