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Replacement Analysis

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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.

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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) requirements c) new, alternative ways of obtaining the functionality provided by the defender d) increased demand that cannot be met with the current equipment

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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.)

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As equipment ages, O&M costs increase and CR cost decreases.

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Example Car grows older and needs repairs at engine overhaul time should we fix or replace?

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Example Car grows older and needs repairs at 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.

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Chemical Plant owns filter press purchased 3 years ago. Operating expense started at $4,000 per year 2 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.

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,000 11,000 2,000 Keep

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,000 11,000 2,000 Keep ,000 12,000 Replace 36,000 9,000 1,000

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,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)

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,000 12,000 Replace 36,000 9,000 1,000 NPW = 9, ,000 -1,000 (P/G, 15,5) + 12,000 (P/F, 15, 5) = ($26,809)

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,000 11,000 2,000 Keep ,000 12,000 Replace 36,000 9,000 1,000 NPW K = ($28,246) NPW R = ($26,809)

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,000 11,000 2,000 Keep ,000 12,000 Replace 36,000 9,000 1,000 NPW K = ($28,246) NPW R = ($26,809) Choose Replace

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,000 11,000 2,000 Keep ,000 12,000 Replace 36,000 9,000 1,000 NPW K = ($28,246) NPW R = ($26,809) Note: NPW R - NPW K = $ 1,437

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,000 11,000 2,000 Keep 9, ,000 12,000 Replace 36,000 1,000

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,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)

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,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)

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,000 11,000 2,000 Keep 9, ,000 12,000 Replace 36,000 1,000 NPW K = ($37,246) NPW R = ($35,809)

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,000 11,000 2,000 Keep 9, ,000 12,000 Replace 36,000 1,000 NPW K = ($37,246) NPW R = ($35,809) Choose Replace

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,000 11,000 2,000 Keep 9, ,000 12,000 Replace 36,000 1,000 NPW K = ($37,246) NPW R = ($35,809) Note: NPW R - NPW K = $ 1,437

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Suppose we have a compressor which costs $2,000 and has annual maintenance costs of $500 increasing by $100 per year. MARR=20%. Then:

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26 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.

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

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28 ESL Example w/ Challenger The Best Challenger process to replace the Defender has a 1 st 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 $ Determine how to replace the existing process.

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

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30 EAC Challenger 8 =(65 000)(A/P,8%,8) (A/F,8%,8)=$ (.1740)(.0940) Finding EAC for Challenger w/ lifetime of 8 years: n = 8 yrs $3 500 Challenger:$5 000 $ LifetimeEAC Defender 1$ $ $ $ $ Minimum EAC Defender 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 EAC Challenger

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