Lecture no46 replacement fundamentals

Replacement Analysis Fundamentals Lecture No 46 Chapter 14 Contemporary Engineering Economics Copyright © 2016 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Chapter Opening Story A university medical center is considering replacing an old steam-driven chiller at $7.7 million  At issue: o What basis they make the replacement decisions? o How much savings in energy cost would justify the purchase of the new absorption chiller? th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Replacement Terminology • • • Defender: an old machine • future decisions Challenger: a new machine Current market value: selling price of the defender in the market place Sunk cost: any past cost unaffected by any • Trade-in allowance: value offered by the vendor to reduce the price of a new equipment th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 14.1: Sunk Cost associated with an Asset’s Disposal  Given: o o o Original investment = $20,000 Current market value = $10,000 Repair cost made in the past = $5,000  Find: (a) Sunk cost, (b) Relevant cost for replacement analysis th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution  Relevant Cost for Replacement Analysis: o o o o Lost investment value, $10,000 Repair cost made, $5,000 Total sunk cost = $15,000 Current market value = $10,000 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Opportunity Cost Approach • Basic Principle: Treat the proceeds from sale of the old machine as the investment required to keep the old machine • Compute the AEC for each alternative and select the one with the minimum AEC th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 14.2: Opportunity Cost Approach Given: o o Defender • • • • Market price: $10,000 Remaining useful life: years Salvage value: $2,500 O&M cost: $8,000 Challenger • • • • Cost: $15,000 Useful life: years Salvage value: $5,500 O&M cost: $6,000 Find: Replace the defender now? th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution: Replace the Defender th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Economic Service Life • • Definition: Economic service life is the remaining useful life of an asset that results in the minimum annual equivalent cost Annual Equivalent Cost (AEC) AEC = Capital Cost + Operating Cost th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Mathematical Relationship • Capital Cost • Operating Cost • Total Cost • Objective: Find n* that minimizes AEC(i) n* th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 14.3: Economic Service Life for a Lift Truck  Given: o o o o I = $18,000 i = 12% Salvage value = −20% over the previous year O&M = $3,000 during the first year, and 15% increase over the previous year thereafter  Find: Economic Service Life th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Cash flows if you kept n = or n = n = 1: $11,400 n = 2: $3,000 $18,000 $11,520 $3,000 $3,450 $18,000 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved AEC Calculation If You Kept the Truck for Years  Ownership Cost CR(12%) = ($18,000 − $11,520)( A / P ,12%,2) +(0.12)($11,520) = $5,760  Operating Cost OR(12%) = [ $3,000(P / F ,12%,1) + $3,450(P / F ,12%,2)] ×(A / P ,12%,2) = $3,212  Annual Equivalent Cost AEC(12%)n=2 = $5,760 + $3,212 = $8,429 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Conversion of an Infinite Number of Replacement Cycles to Infinite AEC Streams th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Economic Service Life Calculation Using Excel  Economic Service Life = Years with AEC(12%) = $7,977  What It Really Means  You purchase a brand new lift truck for every years, assuming that the future replacement cost as well as operating costs remain constant Then the equivalent annual cost of owning and operating the truck is $7,977 th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Sensitivity of Economic Service Life o o For an asset with non-increasing operating cost, keep the asset as long as it lasts If everything remains the same, a higher interest rate will tend to extend the economic service life (or defer the replacement decision) th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved ... Relevant cost for replacement analysis th Contemporary Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution  Relevant Cost for Replacement Analysis:... Engineering Economics, edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Replacement Terminology • • • Defender: an old machine • future decisions Challenger: a new machine... considering replacing an old steam-driven chiller at $7.7 million  At issue: o What basis they make the replacement decisions? o How much savings in energy cost would justify the purchase of the new absorption