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Chapter 15 Replacement Decisions. Replacement Analysis Fundamentals Economic Service Life Replacement Analysis When Required Service is Long Replacement Analysis with Tax Consideration. Sunk cost : any past cost unaffected by any future decisions

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Chapter 15 replacement decisions
Chapter 15Replacement Decisions

  • Replacement Analysis Fundamentals

  • Economic Service Life

  • Replacement Analysis When Required Service is Long

  • Replacement Analysis with Tax Consideration

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

Sunk cost: any past cost unaffected by any future decisions

Trade-in allowance: value offered by the vendor to reduce the price of a new equipment

Operating Cost

Defender: an old machine

Challenger: new machine

Current market value: selling price of the defender in the market place

Replacement Terminology

(c) 2001 Contemporary Engineering Economics


Sunk cost associated with an asset s disposal
Sunk Cost associated with an Asset’s Disposal

Original investment

$20,000

Lost investment

(economic depreciation)

Market value

Repair cost

$10,000

$5000

$10,000

Sunk costs= $15,000

$0 $5000 $10,000 $15,000 $20,000 $25,000 $30,000

(c) 2001 Contemporary Engineering Economics


Replacement decisions

Cash Flow Approach

Treat the proceeds from sale of the old machine as down payment toward purchasing the new machine.

Can be used in the analysis period is same for all alternatives.

Use NPW or AE analysis to decide

Opportunity Cost Approach

Treat the proceeds from sale of the old machine as the investment required to keep the old machine.

Replacement Decisions

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

Replacement Analysis – Cash Flow Approach

Sales proceeds

from defender

$10,000

$5500

$2500

0 1 2 3

0 1 2 3

$6000

$8000

(a) Defender

(b) Challenger

$15,000

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

Annual Equivalent Cost - Cash Flow Approach

 Defender:

PW(12%)D = $2,500 (P/F, 12%, 3) - $8,000 (P/A, 12%, 3)

= - $17,434.90

AE(12%)D = PW(12%)D(A/P, 12%, 3)

= -$7,259.10

 Challenger:

PW(12%)C = $5,500 (P/F, 12%, 3) - $5,000

- $6,000 (P/A, 12%, 3)

= -$15,495.90

AE(12%)C = PW(12%)C(A/P, 12%, 3)

= -$6,451.79

Replace the defender now!

(c) 2001 Contemporary Engineering Economics


Opportunity cost approach
Opportunity Cost Approach

Challenger

Defender

$5500

$2500

0 1 2 3

0 1 2 3

$6000

$8000

$10,000

Proceeds from sale viewed as

an opportunity cost of keeping

the asset

$15,000

(c) 2001 Contemporary Engineering Economics


Opportunity cost approach1
Opportunity Cost Approach

 Defender:

PW(12%)D = -$10,000 - $8,000(P/A, 12%, 3) + $2,500(P/F, 12%, 3)

= -$27,434.90

AE(12%)D = PW(12%)D(A/P, 12%, 3)

= -$11,422.64

 Challenger:

PW(12%)C = -$15,000 - $6,000(P/A, 12%, 3) + $5,500(P/F, 12%, 3)

= -$25,495.90

AE(12%)C = PW(12%)C(A/P, 12%, 3)

= -$10,615.33

Replace the

defender now!

(c) 2001 Contemporary Engineering Economics


Economic service life
Economic Service Life

  • Def:Economic service life is the useful life of a defender, or a challenger, that results in the minimum equivalent annual cost

  • Why do we need it?: We should use the respective economic service lives of the defender and the challenger when conducting a replacement analysis.

Minimize

Ownership (Capital)

Cost (init.+salvg.)

+

Operating

cost

(c) 2001 Contemporary Engineering Economics


Mathematical relationship
Mathematical Relationship

  • Capital Recov. Cost.

  • Operating Cost:

  • Total Cost:

  • Objective: Find n* that minimizes AEC

AEC

OC(i)

CR(i)

n*

(c) 2001 Contemporary Engineering Economics


Economic service life for a lift truck
Economic Service Life for a Lift Truck

(c) 2001 Contemporary Engineering Economics


Economic service life calculation example 15 4
Economic Service Life Calculation (Example 15.4)

  • N = 1

    AEC1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000

    = $11,700

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

  • N = 2

    AEC2 = [$18,000 + $1,000(P/A,15%, 15%, 2)](A/P, 15%, 2)

    - $7,500 (A/F, 15%, 2)

    = $8,653

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

N = 3, AEC3 = $7,406

N = 4, AEC4 = $6,678

N = 5, AEC5 = $6,642

N = 6, AEC6 = $6,258

N = 7, AEC7 = $6,394

Minimum cost

Economic Service Life

(c) 2001 Contemporary Engineering Economics


Required assumptions and decision frameworks
Required Assumptions and Decision Frameworks

  • Planning horizon (study period)

  • Technology

  • Relevant cash flow information

  • Decision Frameworks

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

Replacement Strategies under the Infinite Planning Horizon

  • Replace the defender now: The cash flows of the challenger will be used from today and will be repeated because an identical challenger will be used if replacement becomes necessary again in the future. This stream of cash flows is equivalent to a cash flow of AEC* each year for an infinite number of years.

  • Replace the defender, say, x years later: The cash flows of the defender will be used in the first x years. Starting in year x+1,the cash flows of the challenger will be used indefinitely.

(c) 2001 Contemporary Engineering Economics


Example 15 5
Example 15.5

  • Defender: Find the remaining useful (economic) service life.

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

N = 1 year: AE(15%) = $7,500

N = 2 years: AE(15%) = $6,151

N = 3 years: AE(15%) = $5,847

N = 4 years: AE(15%) = $5,826

N = 5 years: AE(15%) = $5,897

NC*=4 years

AEC*=$5,826

(c) 2001 Contemporary Engineering Economics


Replacement decisions1
Replacement Decisions

NC*=4 years

AEC*=$5,826

  • Should replace the defender now?No, because AED < AEC

  • If not, when is the best time to replace the defender? Need to conduct marginal analysis.

(c) 2001 Contemporary Engineering Economics


Marginal analysis
Marginal Analysis

  • Question: What is the additional (incremental) cost for keeping the defender one more year from the end of its economic service life, from Year 2 to Year 3?

  • Financial Data:

  • Opportunity cost at the end of year 2: Equal to the market

  • value of $3,000

  • Operating cost for the 3rd year: $5,000

  • Salvage value of the defender at the end of year 3: $2,000

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

$2000

2

3

  • Step 1: Calculate the equivalent cost of retaining the defender one more from the end of its economic service life, say 2 to 3.

    $3,000(F/P,15%,1) + $5,000

    - $2,000 = $6,450

  • Step 2: Compare this cost with AEC = $5,826 of the challenger.

  • Conclusion: Since keeping the defender for the 3rd year is more expensive than replacing it with the challenger, DO NOT keep the defender beyond its economic service life.

$3000

$5000

2

3

$6,450

(c) 2001 Contemporary Engineering Economics


Replacement analysis under the finite planning horizon
Replacement Analysis under the Finite Planning Horizon

Some likely replacement patterns

under a finite planning horizon of

8 years

(c) 2001 Contemporary Engineering Economics


Example 15 6 replacement analysis under the finite planning horizon pw approach
Example 15.6 Replacement Analysis under the Finite Planning Horizon (PW Approach)

  • Option 1: (j0, 0), (j, 4), (j, 4)

    PW(15%)1=$5,826(P/A, 15%, 8)

    =$26,143

  • Option 2: (j0, 1), (j, 4), (j, 3)

    PW(15%)2=$5,130(P/F, 15%, 1)

    +$5,826(P/A, 15%, 4)(P/F, 15%, 1)

    +$5,857(P/A, 15%, 3)(P/F, 15%, 5)

    =$25,573

(c) 2001 Contemporary Engineering Economics


Example 15 6 continued
Example 15.6 continued Horizon (PW Approach)

  • Option 3 (j0, 2), (j, 4), (j, 2)

    PW(15%)3=$5,116(P/A, 15%, 4)(P/F, 15%, 2)

    +$5,826(P/A, 15%, 4)(P/F, 15%, 2)

    +$6,151(P/A, 15%, 2)(P/F, 15%, 6)

    = $25,217 minimum cost

  • Option 4 (j0, 3), (j, 5)

    PW(15%)4= $5,500(P/A, 15%, 3)

    +$5,897(P/A, 15%, 5)(P/F, 15%, 3)

    =$25,555

(c) 2001 Contemporary Engineering Economics


Example 15 6 continued1
Example 15.6 continued Horizon (PW Approach)

  • Option 5: (j0, 3), (j, 4), (j, 1)

    PW(15%)5= $5,500(P/A, 15%, 3)

    + $5,826(P/A, 15%, 4)(P/F, 15%, 3)

    + $7,500(P/F, 15%, 8)

    = $25,946

  • Option 6: (j0, 4), (j, 4)

    PW(15%)6= $5,826(P/A, 15%, 4)(P/F, 15%, 4)

    + $5,826(P/A, 15%, 4)(P/F, 15%, 4)

    = $26,529

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

Planning horizon = 8 years Horizon (PW Approach)

(j0, 0), (j, 4), (j, 4),

(j0, 1), (j, 4), (j, 3),

(j0, 2), (j, 4), (j, 2),

(j0, 3), (j, 5),

(j0, 3), (j, 4), (j, 1),

(j0, 4), (j, 4),

Option 1

Option2

Option 3

Option 4

Option 5

Option 6

0 1 2 3 4 5 6 7 8

Years in service

(c) 2001 Contemporary Engineering Economics


Replacement analysis with tax consideration
Replacement Analysis with Tax Consideration Horizon (PW Approach)

  • Whenever possible, replacement decisions should be based on the cash flows after taxes. (Example 15.8)

  • When computing the net proceeds from sale of the old asset, any gains or losses must be identified to determine the correct amount of the opportunity cost. (Example 15.7)

  • All basic replacement decision rules including the way of computing economic service life remain unchanged. (Example 15.10)

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

Depreciation basis Horizon (PW Approach)

$20,000

$20,000

Total

depreciation

Book value

$14,693

$5307

Book loss

Market value

$10,000

$4693

Loss

tax credit

Market value

$10,000

Net proceeds from disposal ($11,877)

$0 $4000 $8000 $12,000 $16,000 $20,000

(c) 2001 Contemporary Engineering Economics


Summary
Summary Horizon (PW Approach)

  • In replacement analysis, the defenderis an existing asset; the challenger is the best available replacement candidate.

  • The current market value is the value to use in preparing a defender’s economic analysis. Sunk costs—past costs that cannot be changed by any future investment decision—should not be considered in a defender’s economic analysis.

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

  • Two basic approaches to analyzing replacement problems are the cash flow approach and the opportunity cost approach.

    • The cash flow approach explicitly considers the actual cash flow consequences for each replacement alternative as they occur.

    • The opportunity cost approach views the net proceeds from sale of the defender as an opportunity cost of keeping the defender.

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

  • Economic service life the is the remaining useful life of a defender, or a challenger, that results in the minimum equivalent annual cost or maximum annual equivalent revenue. We should use the respective economic service lives of the defender and the challenger when conducting a replacement analysis.

  • Ultimately, in replacement analysis, the question is not whether to replace the defender, but when to do so.

  • The AE method provides a marginal basis on which to make a year-by-year decision about the best time to replace the defender.

  • As a general decision criterion, the PW method provides a more direct solution to a variety of replacement problems, with either an infinite or a finite planning horizon, or a technological change in a future challenger.

(c) 2001 Contemporary Engineering Economics


Chapter 15 replacement decisions

  • The role of the technological change in asset improvement should be weighed in making long-term replacement plans

  • Whenever possible, all replacement decisions should be based on the cash flows after taxes.

(c) 2001 Contemporary Engineering Economics