3/16/2016rd1 Engineering Economic Analysis Chapter 13  Replacement Analysis

3/16/2016rd2 … from the Deacon/s Masterpiece by Oliver Wendell Holmes … Now in building of chaises, I tell you what,- There is always somewhere a weakest spot,—- In hub, tire, felloe, in spring or thill,- In panel, or crossbar, or floor, or sill,- In screw, bolt, thoroughbrace,—lurking still,- Find it somewhere, you must and will,—- Above or below, or within or without,—- And that's the reason, beyond a doubt,- A chaise breaks down, but doesn't wear out. You see, of course, if you 're not a dunce, How it went to pieces all at once,-- All at once, and nothing first,-- Just as bubbles do when they burst. -

3/16/2016rd3 Reasons for Replacement Analysis Physical Impairment (Deterioration) Altered Requirements (Obsolescence) Technology (Obsolescence Financing (taxes, lease versus buy) Replacement study is designed to decide to retain or to replace now. If to replace, then done. If not, then revisit annually.

3/16/2016rd4 Types of Lives Economic life – resulting in minimum equivalent uniform annual cost (EUAC) Ownership life – date of acquisition and date of abandonment or replacement Physical life – original acquisition to disposal Useful life – time an asset is kept in productive services

3/16/2016rd5 Replacement Factors Recognition and acceptance of past errors unable to see future, bad estimates Sunk costs (forget the past, and the book value) Use existing asset value and outsider viewpoint Economic life of best challenge Remaining economic life of defender Income tax considerations (sunk cost relevant here)

Replacement Assumptions Infinite Horizon (repeatability) No technological change Do Nothing option is not feasible (or Keep defender is the do nothing option) Economic life of asset is the age at which costs are minimized. 3/16/2016rd6

3/16/2016rd7 Identify Participants of Comparison De fender Marginal Cost Increasing Defender Marginal Cost Data Find lowest EUAC for Defender no Find EUAC over Given Life yes AT# 1 Compare next year marginal cost with C-EUAC AT# 2 Compare lowest D-EUAC with C-EUAC at min cost life Lacking defender data, compare D-EUAC over life with C-EUAC at minimum cost life Available Not available Defender Best Challenger

Annualize the Costs Find the annual cost for the cash flow below. n cf-15M-500K-825K 7.68M & -1.36M 3/16/2016rd8 $15M $500K 825K 1.36M $7.68M

Economic Life You bought a car for $3K. Determine economic life at 12%. Year Annual Expenses Market Value Loss in Market Value Loss in interest Total Marginal costs Find EUAC of Total Marginal Costs. (EUAC '( ) 12) 3/16/2016rd9

Problem First cost = $1050K; Salvage value = $225K at any time; O&M = $235K with $75K gradient. MARR = 10%. Find economic service life. nCRO&MCR+O&MEUAC *** /16/2016rd10

3/16/2016rd11 Example Defender has AOC of $60K per year, a life of 5 years with zero salvage. Present market value is $30K. Challenger costs $120K with AOC of $30K and a salvage after 5 years of $50K. Use before tax MARR of 20%. AT#3 PW D (20%) = -30K – 60K(P/A, 20%, 5) = -$209, PW C (20%) = -120K - 30K(P/A, 20%, 5) + 50K(P/F, 20%, 5) = -$189, Choose the challenger.

Example Defender bought 3 years ago for $120K with expected life of 10 years and $25K salvage value with AOC of $30K. Current book value is $80K and current life is 3 years. Challenger available for $100K with trade-in of $70K for defender. Useful life is 10 years, salvage value is $20K and ASOC is $20K. Market value for defender is $70K. DefenderChallenger First cost$70K$100K AOC 30K 20K Salvage 10K 20K Life (years) /16/2016rd12

3/16/2016rd13 Example First cost (sunk 5 years)$12,500 DepreciationStraight Line (salvage = $2,500) Market Value$8,000 Useful Life10 years Annual Operating Cost$3,000 Annual Benefit$4,500 Market Value after life$1,500 Combined tax rate34% n BTCFDep TITax Rate (34%)ATCF 0 8K $7830 (sell) 0 -8K $7830 (keep) Cost Basis = 12, * (12,500 –2500) / 10 = $7500 $8000 – 7500 = $500 (depreciation recapture)

3/16/2016rd14 Example 13-1 $7500(A/P) $900(A/G) $500+$400(A/G, 8%, n) n CRWR Maint Operating Total EUAC

3/16/2016rd15 Example 13-1

3/16/2016rd16 Marginal Costs Challenger First Cost = $25K; Annual O&M = $2K with -$500 gradient; Annual Risk $5K for 3 yrs; then Gradient Market value ($ )K from 1 to 7 years; MARR = 15% n Lost MarketLost InterestO&MRiskTotal MC EUAC 1$7K 0.15*25K=$3750 $2.0K$5.0K$17,750 17, K K 5.0K 15,200 16, K K 5.0K 13,950 15, K K 6.5K 14,350 15, K K 8.0K 14,900 15,427 *** 6 1K K 9.5K 15,600 15, K K 11.0K 16,950 15,582 (AGP (List-pgf '( ) 15) 15 7)  $15, (EUAC cash-flow 15)

3/16/2016rd17 Example 13-3 Defender's Market value= $15K; Annual O&M = $10K with a $1500 gradient; Market value ($ )K from 1 to 5 years; MARR = 15% n Lost MarketLost Interest O&MTotal Marginal Cost 1$1K 0.15*15K=$2250 $10.0K$13, K K 14, K K 15,950 > $15,427 ** 4 1K K 17, K K 18,650 Note that the marginal costs are increasing each year => AT #1

Problem 13-4 Maintenance costs are expected to be higher. Fully depreciated and other factors irrelevant. Cost this year is expected to be $800. AT#1 one year, 3/16/2016rd18

3/16/2016rd19 Problem First cost = $1050K; Salvage value = $225K at any time O&M = $235K with $75K gradient. MARR = 10%. Find economic service life. nCRO&MEO&MEUAC 1$930K235K235K1165K K310K270.71K K K385K305.24K659K K460K338.59K K K535K370.76K K *** K610K401.77K K CR = (P – S)(A/P, i%, n) + Si 3/16/2016rd19

3/16/2016rd20 nOperating Cost MaintMVLost MV Lost int Total Marginal Cost NPWEUAC 0120K 115K 9K85K35K24K83K K10K65K20K17K62K K12K50K15K13K57K K18K40K10K 58K K20K35K 5K 8K58K K25K30K 5K 7K67K K30K25K 5K 6K76K Big-J Construction Problem MARR = 20%

Problem MARR = 10% YearD-MC D-EUAC C-EUAC 1$2500 $ $ *** *** a)What is the lowest EUAC of the defender? $ b)What is minimum cost life of challenger? $2600 c)When should defender be replaced with challenger? Never 3/16/2016rd21

Problem n MCD-DEUAC-C MARR = 10% 1 $3000$ (EUAC '( ) 10)  Shouldn't replace Defender at all, but had year 1 for D > 3300 etc., replace then/now. 3/16/2016rd22

3/16/2016rd23 Problem Old forklift MARR = 8% AT#1 n Maint Cost /year YearBTCF Depc TI Tax 40% ATCF $160-$240 Challenger YearBTCF Depc TI Tax 40% ATCF $ $ EUAC = 6500(A/P, 8%, 10) – 230 = $ – $230 = $ => keep old forklift for another year.

Example The economic service life given current market value of asset is $15K and expected cash flows shown below and MARR at 10% is YearSalvage ValueO&M Cost 1 $10K$50K 2 8K 53K 3 5K 60K 4 0K 68K a) i year b) 2 years c) 3 years d) 4 years EUAC 1 = (15 – 10)K(A/P, 10%, 1) + 10K* K = $56.50K EUAC 2 = (15 – 8)K (A/P, 10%, 2) + 8K* K = 56.36K *** EUAC 3 = (15 – 5)K (A/P, 10%, 3) + 5K* K = 58.54K 3/16/2016rd24

Example When should the defender be replaced? a) now b) 1 year from now c) 2 years from nowd) 3 years from now 3/16/2016rd25 YearAW-D CostAW-C Cost 1$24K$31K 2 25K 28K 3 26K 25K 4 27K 25.9K 5 28K 27.5K

Economic Life Asset bought at $8K at 8% cost of money. n Aexpense 3K3K3.5K4K4.5K5.52K MV Loss in MV Loss in Int Total Margin (EUAC '( ) 8) $ years 3/16/2016rd26

PW of ATCF Present Worth of After Tax data, find economic life of each and when to replace defender. YearDefenderChallenger (mapcar #' AGP '( ) (list-of 5 12) (upto 5))  ( ) (mapcar #' AGP '( ) (list-of 3 12)(upto 3)) /16/2016rd27