Analysis of Engineering Business Decisions Chapter 5 Analysis of Engineering Business Decisions

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.

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

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.

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.

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

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

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.

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

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.

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)

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 value A over the life L. Find A from A = (A/P, i%, Life) * New Price - (A/F,i%,L)*S This particular “A” value is often called the “Capital Recovery value” Suppose that N years are left in the Life. Then Used Value = A * (P/A,i%,N) + S*(P/F,i%,N)

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.

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

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.

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.