Lecture no46 replacement fundamentals

Replacement Analysis Fundamentals

Lecture No. 46
Chapter 14
Contemporary Engineering Economics
Copyright © 2016

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Contemporary Engineering Economics, 6 edition
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Chapter Opening Story
A university medical center is considering replacing
an old steam-driven chiller at $7.7 million.

 At issue:
o What basis do they make the replacement
decisions?
o How much savings in energy cost would justify
the purchase of the new absorption chiller?

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Contemporary Engineering Economics, 6 edition
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Replacement Terminology
•
•
•

Defender: an old machine

•

future decisions

Challenger: a new machine
Current market value: selling price of the
defender in the market place

Sunk cost: any past cost unaffected by any

•

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

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Contemporary Engineering Economics, 6 edition
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Example 14.1: Sunk Cost associated with an Asset’s Disposal

 Given:
o
o
o

Original investment = $20,000
Current market value = $10,000
Repair cost made in the past = $5,000

 Find: (a) Sunk cost, (b) Relevant cost for replacement analysis

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Contemporary Engineering Economics, 6 edition
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Solution



Relevant Cost for Replacement Analysis:

o
o
o
o

Lost investment value, $10,000
Repair cost made, $5,000
Total sunk cost = $15,000
Current market value = $10,000

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Contemporary Engineering Economics, 6 edition
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Opportunity Cost Approach

•

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

•

Compute the AEC for each alternative and select the one with the minimum
AEC.

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Contemporary Engineering Economics, 6 edition
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Example 14.2: Opportunity Cost Approach

Given:
o

o

Defender

•
•
•
•

Market price: $10,000
Remaining useful life: 3 years
Salvage value: $2,500
O&M cost: $8,000

Challenger

•

•
•
•

Cost: $15,000
Useful life: 3 years
Salvage value: $5,500
O&M cost: $6,000

Find: Replace the defender now?
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Contemporary Engineering Economics, 6 edition
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Solution: Replace the Defender

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Contemporary Engineering Economics, 6 edition
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Economic Service Life

•
•

Definition: Economic service life is the remaining useful life of an asset that
results in the minimum annual equivalent cost.
Annual Equivalent Cost (AEC)
AEC = Capital Cost + Operating Cost

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Mathematical Relationship

•

Capital Cost

•

Operating Cost

•

Total Cost

•

Objective: Find n* that minimizes AEC(i)

n*

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Example 14.3: Economic Service Life for a Lift Truck

 Given:
o
o
o
o

I = $18,000
i = 12%
Salvage value = −20% over the previous year
O&M = $3,000 during the first year, and 15% increase over the previous year
thereafter

 Find: Economic Service Life

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Cash flows if you kept n = 1 or n = 2
n = 1:
$11,400

0
1

n = 2:
$3,000

$18,000

$11,520

0

1

2

$3,000
$3,450

$18,000

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AEC Calculation If You Kept the Truck for 2 Years


Ownership Cost

CR(12%) = ($18,000 − $11,520)( A / P ,12%,2)
+(0.12)($11,520)
= $5,760



Operating Cost

OR(12%) = [ $3,000(P / F ,12%,1) + $3,450(P / F ,12%,2)]
×(A / P ,12%,2)
= $3,212



Annual Equivalent Cost

AEC(12%)n=2 = $5,760 + $3,212
= $8,429

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Conversion of an Infinite Number of Replacement Cycles to Infinite AEC Streams

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Economic Service Life Calculation Using Excel


Economic Service Life = 6 Years with AEC(12%) = $7,977



What It Really Means



You purchase a brand new lift truck for every 6 years, assuming
that the future replacement cost as well as operating costs
remain constant. Then the equivalent annual cost of owning
and operating the truck is $7,977.

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Sensitivity of Economic Service Life

o
o

For an asset with non-increasing operating cost, keep the asset as long as it lasts.
If everything remains the same, a higher interest rate will tend to extend the economic
service life (or defer the replacement decision).

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Contemporary Engineering Economics, 6 edition
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