Modeling Economic Life…

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# Modeling Economic Life… - PowerPoint PPT Presentation

Modeling Economic Life…. Total EAC = EAC of Ownership + EAC of Operation Objective is to determine the optimum number of years to own an asset that minimizes the Total EAC. This is called the Economic Service Life of the asset. Conceptually…. Over time…

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Presentation Transcript

Modeling Economic Life…

Total EAC = EAC of Ownership + EAC of Operation

Objective is to determine the optimum number of years to own an asset that minimizes the Total EAC.

This is called the Economic Service Life of the asset.

Conceptually…

Over time…

Cost of ownership typically decreases.

Cost of operation typically increases.

ESL is the lifetime with minimum EAC!

ESL Example

An existing process has a market value of \$15 000, but to provide the required physical service level for the next 5 years will require an immediate overhaul for \$4 000. The salvage value at the end of the current year is estimated to be \$1 100, and that will decrease by \$100/year thereafter.

Operating and Maintenance costs are currently \$4 000 yearly, but these costs will increase by \$2 100/year after this year.

Determine the economic service life and resultant total EAC of the asset if the MARR is 8%, compounded annually.

DIAGRAM:

\$1 100 – \$100(n-1)

1

2

3

n = 1, 2, … 5 yrs

0

\$4 000

\$4 000

\$2 100

\$15 000

(19 000)

(1.0800)

(0)

(1.0000)

=\$15 184

(.5608)

(.4808)

(.4808)

=\$13 087

(.3880)

(.9487)

(.3080)

=\$12 507

(.3019)

(1.4040)

(.2219)

=\$12 518

(.2505)

(1.8465)

(.1705)

ESL = 4 yrs, this is the EACDefender if used through the end ofthe 4th year!

Finding EAC for lifetimes of 1 through 5 years:

EAC1=(4000+15000)(A/P,8%,1)+4000+2100(A/G,8%,1) -1100(A/F,8%,1)

=\$23 420

EAC2=(19 000)(A/P,8%,2)+4000+2100(A/G,8%,2) -1000(A/F,8%,2)

EAC3=(19 000)(A/P,8%,3)+4000+2100(A/G,8%,3) -900(A/F,8%,3)

EAC4=(19 000)(A/P,8%,4)+4000+2100(A/G,8%,4) -800(A/F,8%,4)

EAC5=(19 000)(A/P,8%,5)+4000+2100(A/G,8%,5) -700(A/F,8%,5)

ESL Example w/ Challenger

The Best Challenger process to replace the Defender has a 1st cost of \$65 000, and an expected physical lifetime of 15 years. The same MARR of 8% (compounded annually) has been used to run an ESL study on the Challenger using the best data available today, and it looks like the ESL is 8 years.

For the 8 year Challenger ESL, the operating and maintenance costs are expected to be \$3 500 yearly; and the salvage value at the end of the 8 years is estimated to be \$5 000.

Determine how to replace the existing process.

Defender:

\$800

1

2

3

n = 4 yrs

0

\$4 000

\$4 000

\$2 100

\$15 000

Challenger:

\$5 000

1

2

3

n = 8 yrs

0

\$3 500

\$65 000

Comparing Best Challenger

Opportunity Cost View for “Outsider”:

Challenger:

\$5 000

1

2

3

n = 8 yrs

0

Minimum

EACDefender

\$3 500

\$65 000

Ownership Life – b/c less than EACChallenger

Finding EAC for Challenger w/ lifetime of 8 years:

EACChallenger 8=(65 000)(A/P,8%,8) + 3 500 - 5 000(A/F,8%,8)

=\$14 340

(.1740)

(.0940)

Choose to keep Defender today, plan to replace defender at start of year 6

(but recheck next year to see if anything has changed!)

Questions?
• For online Replacement HW problem:
• Assume all equipment may be salvaged at 10% of first cost, including existing system.