MERCER EGR 312 - Replacement Study

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Replacement StudySlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Replacement StudyFundamental IssueOver time, company assets need to be replaced due to:•Reduced performance•Altered requirements•ObsolescenceA replacement study answers the question of when, not if, to replace.A replacement study is an application of the AW method of comparing unequal-life alternatives.Replacement StudyDefinitionsDefender – currently installed assetChallenger – the “best” alternative to replace the defenderAW – primary economic measure of comparison between defender and challenger.Economic Service Life (ESL) – the number of years at which the lowest AW of cost for an alternative occurs Defender First Cost – the current market value of the defenderChallenger First Cost – the cost of replacing the defender with the challengerReplacement StudyEconomic Service Life (ESL)– the number of years at which the lowest AW of cost for an alternative occurs •The AW of the annual operating costs tends to increase as the asset ages•The capital recovery cost tends to decrease as the capital cost is spread over more yearsAnnual Worth of an Alternative0501001500 10 20YearsAW ($)Capital RecoveryAW of AnnualOperating CostTotal AWReplacement StudyEconomic Service Life (ESL)Total AW = -capital recovery – AW of annual operating costsTotal AWkWhere P = initial investment or current market valueSk = salvage value or market value after k yearsAOCj = annual operating cost for year jAnnual Worth of an Alternative0501001500 10 20YearsAW ($)Capital RecoveryAW of AnnualOperating CostTotal AW),,/(),,/(),,/(),,/(1kiPAjiFPAOCkiFASkiPAPkjjkReplacement StudyEconomic Service Life (ESL) - ExampleA manufacturing process is being considered for replacement. The current market value is $13,000. What is the economic service life of this defender if the interest rate is 10% per year?Year jMVjAOCjCapital RecoveryAW of AOCTotal AWk1 9000 -2500 -5300 -2500 -78002 8000 -2700 -3681 -2595 -62763 6000 -3000 -3415 -2717 -6132 ESL = 34 2000 -3500 -3670 -2886 -65565 0 -4500 -3429 -3150 -6579CR(1) = -$13000(A/P,10%,1)+$9000(A/F,10%,1) = -$5300CR(2) = -$13000(A/P,10%,2)+$8000(A/F,10%,2) = -$3681AOC(1) = -$2500AOC(2) = [-$2500(P/F,10%,1)-$2700(P/F,10%,2)](A/P,10%,2) = -$2595AOC(3) = [-$2500(P/F,10%,1)-$2700(P/F,10%,2)-$3000(P/F,10%,3)](A/P,10%,3) = -$2717Replacement StudyPerforming the StudyTwo types of study: •No study period specified•Study period specifiedNo study period specified:1. Perform ESL analysis on defender and on challenger2. If AWC < AWD, then replace defender with challenger and end study.Else, go to step 3.3. Keep defender for a year, then obtain new market values and estimated AOCs for defender and best challenger. Go to step 1.Replacement StudyPerforming the Study - Examplei = 10% ChallengerChallenger Year kMarket ValueAOCTotal AW if owned k years0 50,000$ -1 40,000$ (5,000)$ (20,000)$ 2 32,000$ (7,000)$ (19,524)$ 3 25,600$ (9,000)$ (19,245)$ 4 20,480$ (11,000)$ (19,123)$ 5 16,384$ (13,000)$ (19,126)$ DefenderDefender Year kMarket ValueAOCTotal AW if owned k years0 15,000$ -1 12,000$ (20,000)$ (24,500)$ 2 9,600$ (8,000)$ (18,357)$ 3 7,680$ (12,000)$ (17,307)$ Because AWD < AWC: -$17,307 < -$19,123, keep defender for another year.Replacement StudyPerforming the Study – Example (one year later)However, one year later AWD > AWC: -$20,819 > -$19,123, therefore replace defender with challenger. i = 10% ChallengerChallenger Year kMarket ValueAOCTotal AW if owned k years0 50,000$ -1 40,000$ (5,000)$ (20,000)$ 2 32,000$ (7,000)$ (19,524)$ 3 25,600$ (9,000)$ (19,245)$ 4 20,480$ (11,000)$ (19,123)$ 5 16,384$ (13,000)$ (19,126)$ Defender - 1 Year LaterDefender Year kMarket ValueAOCTotal AW if owned k years0 12,000$ -1 2,000$ (12,000)$ (23,200)$ 2 -$ (16,000)$


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