Chapter 5

1. Chapter 5 Analysis of Engineering Business Decisions

2. Two kind of problems Cost • Find the least cost machine; find the most economical way of doing something. • Cash flows involve only cost, possibly with the exception of a salvage value at the end of an item’s useful life. Investment • Find the most profitable investment = The investment that has the highest present worth. • Cash flows involve both costs and revenues. Usually there are high initial costs (capital investment) followed later by revenues. This class will mostly concern repeatable cost problems.

3. Techniques • PW Present worth • AW Annual worth • Capitalized Worth • Present worth with an infinite time horizon and repeatability • Incremental Investment Analysis • IRR with delta method Used for cost-based or investment-based problems Used only for investment problems

4. Repeatability:Does the problem repeat? Cost problems (repeatability often useful) • Are the cash flows necessary to provide essential equipment that must be replaced when no longer operational? • If a machine must be purchased again and again at regular intervals, then the cash flows for purchase, O&M, and salvage repeat. Investment problems • Usually does not apply.

5. Repeatability: analysis • It will be easier to compare AW of costs than PW. • The AW converts all the cash flows into a steady annualized cost. • The AW can be used to take account of differences in life cycles and convert cash flows into common units. • If PW must be used: • The PW study period is critical and must be the same for each alternative studied. Otherwise the comparison is biased (example: you can not directly compare $5000 for 10 years service vs. $4000 for 7 years. You need common units first) • The simplest way is to align the PW study period in years, with the life cycles of the equipment so that a whole number of life cycles is considered. • If the PW study period does not match the life cycles of the alternatives, then end-of-study market values may be needed for the portions of analysis where the study period does not match the life cycle. • CW or Capitalized Worth is PW with an infinite study period.

6. Cost-based example Company needs to buy an Industrial Drill MARR = 10% Drill must be kept running (Repeatability) Mitsubishi Life 4 years Initial Cost $200 000 O&M $40,000/year Salvage value $50,000 • General Electric • Life 6 years • Initial Cost $250 000 • O&M $50,000/year • Salvage value $30,000

7. Compare different methods • Annual worth method • PW method with 12 year study period • PW method with 5 year study period • CW method

8. Annual worth method AW(Cost)= (Initial Cost)*(A/P,marr%,life) - (SalvageValue)*(A/F,marr%,life) +Annual O&M cost Mitsubishi AW = ($200,000)*(A/P,10%,4)-($50,000)*(A/F,10%,4)+$40,000 =($200,000)*(0.3155)-($50,000)*(0.2155)+$40,000 =$63,100-$10775+$40000=$92325 ($/year) GE AW = ($250,000)*(A/P,10%,6)-($30,000)*(A/F,10%,6)+$50,000 = ($250,000)*(0.2296)-($30,000)*(0.1296)+$50,000 =$57400-$3888+$50000=$103512 ($/year) The Mitsubishi unit has the lowest AW, and therefore the lowest cost.

9. PW method with 12 year study period In 12 years, the company will go through 3 life cycles of Mitsubishi drill vs. 2 life cycles of GE drill We’ll use an excel spreadsheet for this

10. PW method with 5 year study period Mitsubishi: Need to buy a 2nd drill at end of year 4, then sell it at end of year 5. GE drill: Need to buy only 1 drill, sell it at end of year 5 Do we know the market values of each drill when they are sold as used? If yes, use them. If no, estimate.

11. Estimation of Used Value • The most accurate used “market values” are used market prices, at which a good actually sold. (example: prices from completed trades or auctions) This data can be difficult to find. • Used “market values” are not: • Ask prices in a newspaper. Asking prices are usually a bit too high. • Historical prices. Used market prices change over time. • Prices estimated from a formula (unless that formula comes from statistical research of the used market… and then it will be a different formula for every product)

12. Textbook Method • Is only a rough estimate. • You need the new price, the salvage value S, the Life L, and an interest rate to use the formula. • Assume the item generates annual valueA over the life L. Find A from A = (A/P, i%, Life) * New Price - (A/F,i%,L)*SThis particular “A” value is often called the “Capital Recovery value” • Suppose that N years are left in the Life. • ThenUsed Value = A * (P/A,i%,N) + S*(P/F,i%,N)

13. CW method The company buy drills forever. Two ways to solve 1. Use AW for 1 life cycle;then CW=AW/i 2. Or use PW directly over an infinite study period Hint: Use (P/A,i%,)=1/i and the compounding-interval trick for adjusting i% (i*=(1+i)N-1)for non-yearly cash flows such as purchasing a new drill every 4 or 6 years.

14. Let’s switch problems and cover one last topic left over from last week.

15. Incremental Investment Analysis Procedure • Sort investments from low to high by capital requirements • Determine IRR, select investment with lowest capital requirement having IRR>MARR. Delete investments with IRR<MARR. • Call this investment the current alternative • Calculate deltas and IRR(deltas) of all remaining investments vs. current alternative. • Eliminate any investments where IRR(delta)<MARR. • If any of the remaining investments have IRR(delta)>MARR, then select the investment with lowest capital requirement as the “new” current alternative and go back to 3.

16. Summary • Repeatable cost-based economic problems are common to equipment selection and certain public projects (choosing a road surface or hill slip cover) • Common problem variables: • MARR (I%) • the lifetime for each alternative choice of equipment (L) • Purchase price P for each alternative • Salvage value F for each choice of equipment • Annual O&M Cost for each choice of equipment • Goal: find lowest cost alternative • Techniques • PW for N years – create an N year plan for each alternative. Compare the PW of each plan • AW – Consider a single cycle (buy, use, salvage) of each alternative of equipment and calculate the AW of each alternative. • CW – Consider plans involving an infinite stream of replacement purchases for each alternative. Calculate the PW of each alternative. • In theory, these can all produce the same answers. • In practice, concerns such as actual values for used equipment can produce differences.