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Replacement Analysis
ENGM 661
Chapter 11 - Replacement
Given a cash flow profile for current and replacement equipment, be able to use either the cash flow or the outsider viewpoint to determine the better alternative.
Given a cash flow profile, be able to determine the optimal replacement interval for a piece of equipment.
Learning Objectives
1. Current asset, which we call the defender, has developed several deficiencies,
2. Potential replacement assets, which we call the challengers, are available which have a number of advantages over the defender
3. Changing external environment, including a) customer preferences and expectations b) requirementsc) new, alternative ways of obtaining the
functionality provided by the defenderd) increased demand that cannot be met
with the current equipment
Replacement decisions occur for a variety of reasons, including:
1. the cash flow approach or insider approach
2. opportunity cost approach or outsider viewpoint approach.
If performed correctly, the two approaches will yield the same recommendation. (The essential difference in the two relates to how the salvage value of the defender is treated.)
Two approaches for replacement analyses:
As equipment ages, O&M costs increase and CR cost decreases.
Replacement / Challenge
Example Car grows older and needs repairsat engine overhaul time should we fix or replace?
Replacement / Challenge
Example Car grows older and needs repairsat engine overhaul time should we fix or replace?
Note: sunk costs are unrecoverable
Example Just put $800 in car, engine needs overhaul, should we repair or replace?The $800 just invested has no bearing number is not part of analysis.
Example: Replacement
Chemical Plant owns filter press purchased 3 yearsago. Operating expense started at $4,000 per year2 years ago and has increased by $1,000 per year.The press could last 5 more years with an estimated salvage of $2,000 at that time. Current market value of the press is $9,000. A new press can be purchased for $36,000 with an estimated life of 10 years. Annual operating costs are 0 in year 1 growing by $1,000 per year.
Cash Flow ApproachEnd of Operating and End of Operating and SalvageYear, t Maintenance Costs Year, t Maintenance Costs Value, St
-2 -4,000 -2-1 -5,000 -10 -6,000 0 36,0001 -7,000 1 0 30,0002 -8,000 2 -1,000 24,6003 -9,000 3 -2,000 19,8004 -10,000 4 -3,000 15,6005 -11,000 5 -4,000 12,000
6 -5,000 9,0007 -6,000 6,6008 -7,000 4,8009 -8,000 3,60010 -9,000 3,000
Replacement (Cash Flow)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
Replacement (Cash Flow)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
0 1 2 3 4 5
4,000
12,000
Replace
36,000
9,000
1,000
Replacement (Cash Flow)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
NPW = -7,000 (P/A, 15,5) - 1,000 (P/G, 15,
5) + 2,000 (P/F, 15,
5)
= ($28,246)
Replacement (Cash Flow)
0 1 2 3 4 5
4,000
12,000
Replace
36,000
9,000
1,000
NPW = 9,000 - 36,000 -1,000 (P/G,
15,5) + 12,000 (P/F,
15, 5)
= ($26,809)
Replacement (Cash Flow)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
0 1 2 3 4 5
4,000
12,000
Replace
36,000
9,000
1,000
NPWK = ($28,246) NPWR = ($26,809)
Replacement (Cash Flow)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
0 1 2 3 4 5
4,000
12,000
Replace
36,000
9,000
1,000
NPWK = ($28,246) NPWR = ($26,809)
Choose Replace
Replacement (Cash Flow)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
0 1 2 3 4 5
4,000
12,000
Replace
36,000
9,000
1,000
NPWK = ($28,246) NPWR = ($26,809)
Note: NPWR - NPWK = $ 1,437
Replacement (Outsider View)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
9,000
0 1 2 3 4 5
4,000
12,000
Replace
36,000
1,000
Replacement (Outsider View)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
9,000
NPW = - 9,000 -7,000 (P/A, 15,5) - 1,000 (P/G, 15, 5) + 2,000 (P/F, 15,
5)
= ($37,246)
Replacement (Outsider View)
0 1 2 3 4 5
4,000
12,000
Replace
36,000
1,000
NPW = - 36,000 -1,000 (P/G,
15,5) + 12,000 (P/F,
15, 5)
= ($35,809)
Replacement (Outsider View)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
9,000
0 1 2 3 4 5
4,000
12,000
Replace
36,000
1,000
NPWK = ($37,246) NPWR = ($35,809)
Replacement (Outsider View)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
9,000
0 1 2 3 4 5
4,000
12,000
Replace
36,000
1,000
NPWK = ($37,246) NPWR = ($35,809)
Choose Replace
Replacement (Outsider View)
0 1 2 3 4 5
7,000
11,000
2,000
Keep
9,000
0 1 2 3 4 5
4,000
12,000
Replace
36,000
1,000
NPWK = ($37,246) NPWR = ($35,809)
Note: NPWR - NPWK = $ 1,437
Optimal Replacement
Suppose we have a compressor which costs $2,000 and has annual maintenance costs of $500 increasing by $100 per year. MARR=20%.
Then:
Optimal ReplacementOptimal Replacement
t 4 Yr 5 Yr 6 Yr 7 Yr 8 Yr 9 Yr 10 Yr0 2,000 2,000 2,000 2,000 2,000 2,000 2,0001 500 500 500 500 500 500 5002 600 600 600 600 600 600 6003 700 700 700 700 700 700 7004 800 800 800 800 800 800 8005 900 900 900 900 900 9006 1,000 1,000 1,000 1,000 1,0007 1,100 1,100 1,100 1,1008 1,200 1,200 1,2009 1,300 1,30010 1,400
NPV = 3,624 3,986 4,321 4,628 4,907 5,159 5,385
Optimal ReplacementOptimal Replacement
t 4 Yr 5 Yr 6 Yr 7 Yr 8 Yr 9 Yr 10 Yr0 2,000 2,000 2,000 2,000 2,000 2,000 2,0001 500 500 500 500 500 500 5002 600 600 600 600 600 600 6003 700 700 700 700 700 700 7004 800 800 800 800 800 800 8005 900 900 900 900 900 9006 1,000 1,000 1,000 1,000 1,0007 1,100 1,100 1,100 1,1008 1,200 1,200 1,2009 1,300 1,30010 1,400
NPV = 3,624 3,986 4,321 4,628 4,907 5,159 5,385EUAC = ($1,400) ($1,333) ($1,299) ($1,284) ($1,279) ($1,280) ($1,336)
=NPV(.2,C5:C13)+C4 = PMT(.2,4,C14)
26
ESL ExampleESL ExampleAn 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.
27
01 2 3
n = 1, 2, … 5 yrs
$4 000
DIAGRAM: $1 100 – $100(n-1)
$2 100$15
000
$4 000
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)
(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 of the 4th year!
Finding EAC for lifetimes of 1 through 5 years:
28
ESL Example w/ Challenger
ESL Example w/ ChallengerThe 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.
Comparing Best ChallengerComparing Best Challenger
Opportunity Cost View for “Outsider”:
01 2 3
n = 4 yrs
$4 000
Defender: $800
$2 100$15
000
$4 000
01 2 3
n = 8 yrs
$3 500
Challenger: $5 000
$65 000
30
EACChallenger 8=(65 000)(A/P,8%,8) + 3 500 - 5 000(A/F,8%,8) =$14 340(.1740) (.0940)
Finding EAC for Challenger w/ lifetime of 8 years:
01 2 3
n = 8 yrs
$3 500
Challenger: $5 000
$65 000
Lifetime EACDefender
1 $23 420
2 $15 184
3 $13 087
4 $12 507
5 $12 518
Minimum
EACDefender
Choose to keep Defender today, plan to replace defender at start of year 6
(but recheck next year to see if anything has changed!)
Ownership Life – b/c less than EACChallenger
References:
1. Principles of Engineering Economics 6th Edition (White, Case, Pratt)
2. D. H Jensen & K.D. Douglas3. Dr. Stuart Kellogg
