1. Engineering Economics
1/11/2019
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.
Copyright (c) ; D.H. Jensen & K.D. Douglas

2. Conceptually… Over time… Cost of ownership typically decreases.
Cost of operation typically increases.
ESL is the lifetime with minimum EAC!

3. 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 $ 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.

4. 1 2 3
n = 1, 2, … 5 yrs
$4 000
DIAGRAM:
$1 100 – $100(n-1)
$2 100
$15 000
Finding EAC for lifetimes of 1 through 5 years:
EAC1=( )(A/P,8%,1) (A/G,8%,1) -1100(A/F,8%,1)
=$23 420
(19 000)
(1.0800)
(0)
(1.0000)
EAC2=(19 000)(A/P,8%,2) (A/G,8%,2) -1000(A/F,8%,2)
=$15 184
(.5608)
(.4808)
EAC3=(19 000)(A/P,8%,3) (A/G,8%,3) -900(A/F,8%,3)
=$13 087
(.3880)
(.9487)
(.3080)
ESL = 4 yrs, this is the EACDefender if used through the end of the 4th year!
EAC4=(19 000)(A/P,8%,4) (A/G,8%,4) -800(A/F,8%,4)
=$12 507
(.3019)
(1.4040)
(.2219)
EAC5=(19 000)(A/P,8%,5) (A/G,8%,5) -700(A/F,8%,5)
=$12 518
(.2505)
(1.8465)
(.1705)

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.

6. Comparing Best Challenger
Opportunity Cost View for “Outsider”: 1 2 3
n = 4 yrs
$4 000
Defender:
$800
$2 100
$15 000 1 2 3
n = 8 yrs
$3 500
Challenger:
$5 000
$65 000

7. Finding EAC for Challenger w/ lifetime of 8 years:1 2 3
n = 8 yrs
$3 500
Challenger:
$5 000
$65 000
Finding EAC for Challenger w/ lifetime of 8 years:
EACChallenger 8=(65 000)(A/P,8%,8) (A/F,8%,8)
=$14 340
(.1740)
(.0940)
Lifetime
EACDefender 1
$23 420 2
$15 184 3
$13 087 4
$12 507 5
$12 518
Choose to keep Defender today, plan to replace defender at start of year 6
(but recheck next year to see if anything has changed!)
Minimum
EACDefender
Ownership Life – b/c less than EACChallenger

8. What if the Challenger maintenance costs were $1 825 lower?1 2 3
n = 8 yrs
$ $1 825 = $1 675
Challenger:
$5 000
$65 000
What if the Challenger maintenance costs were $1 825 lower?
Finding EAC for Challenger w/ lifetime of 8 years:
EACChallenger 8=(65 000)(A/P,8%,8) (A/F,8%,8)
=$12 515
(.1740)
(.0940)
Lifetime
EACDefender 1
$23 420 2
$15 184 3
$13 087 4
$12 507 5
$12 518
Choose to keep Defender today, plan to replace defender at start of year 5
Minimum
EACDefender
Ownership Life – b/c less than EACChallenger

9. What if the Challenger maintenance costs were $1 850 lower?1 2 3
n = 8 yrs
$ $1 850 = $1 550
Challenger:
$5 000
$65 000
What if the Challenger maintenance costs were $1 850 lower?
Finding EAC for Challenger w/ lifetime of 8 years:
EACChallenger 8=(65 000)(A/P,8%,8) (A/F,8%,8)
=$12 490
(.1740)
(.0940)
Lifetime
EACDefender 1
$23 420 2
$15 184 3
$13 087 4
$12 507 5
$12 518
Choose to replace Defender today, plan to replace Challenger at start of year 9
Minimum
EACDefender
EACDefender at ESL is greater than EACChallenger