Engineering Economic Analysis Chapter 6 Annual Cash Flow Analysis Donald G. Newnan San Jose State University Ted G. Eschenbach University of Alaska Anchorage Jerome P. Lavelle North Carolina State University Neal A. Lewis University of New Haven Copyright Oxford University Press 2017

Chapter Outline Annual Cash Flow Calculations Annual Cash Flow Analysis Analysis Period Analysis Period Equal to the Alternative Lives Analysis Period Different from the Alternative Lives Infinite Analysis Period Analyzing Loans Annuity Due Copyright Oxford University Press 2017

Learning Objectives Define equivalent uniform annual cost & benefit Express problem as annual cash flow equivalent Conduct equivalent uniform annual worth analysis Compare alternatives With lives that are equal, a common multiple, or infinite With a fixed study period Develop & use spreadsheets to analyze loans Use annuity due for beginning of period cash flows Copyright Oxford University Press 2017

Vignette: Energy Star Appliances US Department of Energy goal: reduce home’s energy use for appliances & electronics by 20%. New appliances meeting energy efficiency criteria earn ENERGY STAR label. Adoption slow: Steady electricity prices (only 1 – 3% increase/year) Many investment decisions are made by home builders, landlords, & property managers—not consumers Relative short period of residency (US average < 8 years) Copyright Oxford University Press 2017

Vignette: Energy Star Appliances Typical American household 400 loads of laundry per year 40 gallons of water per load ENERGY STAR  reduce water & energy consumption by 40% Copyright Oxford University Press 2017

Vignette: Energy Star Appliances Major appliances last 7 to 15 years; refrigerators longest For major appliances what should be considered in keep & move vs. leave or sell when moving? Which appliance replacement decision for a clothes washer or a refrigerator is more sensitive to electricity price changes? How would you explain this to someone not skilled in engineering economics? EnergyGuide labels are not required for TVs, ranges, ovens, & clothes dryers. Why do you think this is true? Copyright Oxford University Press 2017

Example 6-1 Annual Cash Flow Calculations Machine costs $30,000; O&M = $2000; saves $10,000 in labor costs. Salvage value at 5 years = $7000. What is EUAW at 7%? Copyright Oxford University Press 2017

Example 6-2 Annual Cash Flow Calculations Consider only capital costs from Example 6-1. Machine costs $30,000. Salvage value at 5 years = $7000. What is EUAC or capital recovery cost at 10%? Copyright Oxford University Press 2017

Annual Cash Flow Calculations P n S 4 1 2 3 A n-1 n  𝐸𝑈𝐴𝐶=𝑃( 𝐴 𝑃,𝑖,𝑛)−𝑆( 𝐴 𝐹,𝑖,𝑛) (6-1) 𝐸𝑈𝐴𝐶=(𝑃−𝑆)( 𝐴 𝐹,𝑖,𝑛)+𝑃(𝑖) (6-3) 𝐸𝑈𝐴𝐶=(𝑃−𝑆)( 𝐴 𝑃,𝑖,𝑛)+𝑆(𝑖) (6-4) Copyright Oxford University Press 2017

Example 6-3 Annual Cash Flow Calculations Year Maintenance & Repair Cost 1 545 2 590 3 635 4 680 5 725 i = 7% First find PW using the NPV function. Then convert to EUAC using the PMT function. Copyright Oxford University Press 2017

Example 6-4 Add an overhaul in year 3 Maintenance & Repair Cost 1 545 2 590 3 635+750 4 680 5 725 i = 7% Copyright Oxford University Press 2017

EAUC for a simple overhaul A machine requires a $100K overhaul in year 5 of 8 year life. EAUC of overhaul at i = 10%? $62.09K $133.10K $11.64K $18.74K I don’t know how to do this C Copyright Oxford University Press 2017

EAUC for a simple overhaul A machine requires a $100K overhaul in year 5 of 8 year life. EAUC of overhaul at i = 10%? $62.09K $8.74K $11.64K $18.74K I don’t know how to do this =PV(10%,5,0,-100) = $62.09K =PMT(10%,8,-62.09) = $11.64K C Copyright Oxford University Press 2017

Annual Cash Flow Analysis Situation Criterion Neither input nor output fixed Maximize EUAW (Equivalent Uniform Annual Worth) EUAW = EUAB − EUAC Fixed input Maximize EUAB (Equivalent Uniform Annual Benefits) Fixed output Minimize EUAC (Equivalent Uniform Annual Costs) Copyright Oxford University Press 2017

Example 6-5 i = 7% Copyright Oxford University Press 2017

Example 6-6, i = 8% Plan A Plan B Plan C Installed equipment cost $15,000 $25,000 $33,000 Material & labor savings/year $14,000 $9,000 Annual operating expenses $8,000 $6,000 Salvage value $1,500 $2,500 $3,300 Plan A Plan B Plan C Material & labor savings/year $14,000 $9,000 Salvage value * (A/F, 8%, 10) 104 172 228 EUAB = $14,104 $9,172 $14,228 Installed cost * (A/P, 8%, 10) $2,235 $3,725 $4,917 Annual operating expenses 8,000 6,000 EUAC = $10,235 $9,725 $10,917 EUAW = EUAB – EUAC = $3,869 -$553 $3,311 Copyright Oxford University Press 2017

Example 6-6, Spreadsheet Solution Copyright Oxford University Press 2017

Example 6-7 Pump A Pump B Initial cost $7000 $5000 Salvage value $1200 $1000 Useful life, in years 12 6 Annual cost for 6 years for Pump B is Copyright Oxford University Press 2017

Example 6-7 The same as the 6-year analysis The annual cost for 12 years for Pump B is The same as the 6-year analysis Copyright Oxford University Press 2017

Annual Cash Flow & Analysis Period Analysis period = alternative lives Use that analysis period Analysis period = common multiple of lives or continuing requirement or infinite Assume identical replacements & each annual equivalent based on that alternative’s life Some other analysis period Estimate terminal values for all alternatives at end of analysis period Copyright Oxford University Press 2017

Example 6-8 Analysis Period for a Continuing Requirement at 7% Pump A Pump B Initial cost $7,000 $5,000 Salvage value $1,500 $1,000 Useful life (years) 12 9 𝐸𝑈𝐴𝐶 𝐴 = 7000−1500 𝐴 𝑃 ,7%,12 +1500 7% =$797 𝐸𝑈𝐴𝐶 𝐵 = 5000−1000 𝐴 𝑃 ,7%,9 +1000 7% =$684 Copyright Oxford University Press 2017

Pumps: Brass, 3 years, $12K Stainless, 4 years, $15K EAC of brass vs stainless at 8% Brass = $3.62K Stainless = $5.82K Brass = $4.53K Stainless = $4.66K Brass = $4.66K Stainless = $4.53K None of the above I don’t know Copyright Oxford University Press 2017

Pumps: Brass, 3 years, $12K Stainless, 4 years, $15K EAC of brass vs stainless at 8% Brass = $3.62K Stainless = $5.82K Brass = $4.53K Stainless = $4.66K Brass = $4.66K Stainless = $4.53K EAC (brass) = PMT(8%,3,-12) = $4.66 EAC (stainless) = PMT(8%,4,-15) = $4.53 Copyright Oxford University Press 2017

Example 6-9 Infinite Analysis Period Tunnel Pipeline Initial cost $5.5 million $5 million Maintenance Useful life Permanent 50 years Salvage value 𝐸𝑈𝐴𝐶 𝑇𝑢𝑛𝑛𝑒𝑙 =𝑃 𝑖 =5.5𝑚𝑖𝑙𝑙𝑖𝑜𝑛(6%)=$330,000 𝐸𝑈𝐴𝐶 𝑃𝑖𝑝𝑒𝑙𝑖𝑛𝑒 =5.5𝑚𝑖𝑙𝑙𝑖𝑜𝑛( 𝐴 𝑃, 6%,50)=$317,000 Copyright Oxford University Press 2017

Example 6-10 Analysis Period Alternatives Alt. 1 Alt. 2 Initial Cost $50,000 $75,000 Estimated salvage value at end of useful life $10,000 $12,000 Useful Life 7 years 13 years Estimated market value, end of 10-year $20,000 $15,000 𝑁𝑃𝑊 1 =−50,000+ 10,000−50,000 𝑃 𝐹 ,8%,7 +20,000 𝑃 𝐹 ,8%,10 =−$64,076 𝐸𝑈𝐴𝐶 1 = 𝑁𝑃𝑊 1 𝐴 𝑃 ,8%,10 =64,076 0.1490 =$9,547 𝑁𝑃𝑊 2 =−75,000+15,000 𝑃 𝐹 ,8%,10 =−$69,442 𝐸𝑈𝐴𝐶 2 = 𝑁𝑃𝑊 2 𝐴 𝑃 ,8%,10 =69,442 0.1490 =$10,347 Copyright Oxford University Press 2017

Example 6-11 Analyzing Loan with Spreadsheet: Amortization Schedule Loan $2400, 6% nominal annual rate, 6 monthly payments Copyright Oxford University Press 2017

Example 6-12 Finding a Balance Due How much owed ½ way through a 48-month loan of $15,000 at 9% nominal? Monthly Payment Copyright Oxford University Press 2017

Example 6-14 Annuity Due for a Lease Find PW for year & equivalent monthly cost; lease payments = of $1200/month; interest = 1%/month Use Type = 1 EU monthly Cost = 1200(1+0.01) = $1212 Copyright Oxford University Press 2017