ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendation New homework IRR Depreciation

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 2 Recommendation National Gallery, Sainsbury wing– at Trafalgar Square –Free –Great impressionists (although there is a lot else there such as medieval and renaissance paintings) –Slide show at break

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 3 Practice Problem You are considering four types of engineering designs. The project lasts 10 years with the following estimated cash flows. The interest rate (MARR) is 15%. Which of the four is more attractive? Project ABCD Initial cost $150$220$300$340 Revenues/ Year $115$125$160$185 Expenses/ Year $70$65$60$80 IRR (%)

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 4 ItemsCMS OptionFMS Option Annual O&M costs: Annual labor cost$1,169,600$707,200 Annual material cost832,320598,400 Annual overhead cost 3,150,0001,950,000 Annual tooling cost470,000300,000 Annual inventory cost141,00031,500 Annual income taxes1,650,0001,917,000 Total annual costs$7,412,920$5,504,100 Investment$4,500,000$12,500,000 Net salvage value$500,000$1,000,000 Example 7.13 Incremental Analysis for Cost- Only Projects

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 5 nCMS OptionFMS Option Incremental (FMS-CMS) 0-$4,500,000-$12,500,000-$8,000, ,412,920-5,504,1001,908, ,412,920-5,504,1001,908, ,412,920-5,504,1001,908, ,412,920-5,504,1001,908, ,412,920-5,504,1001,908, ,412,920-5,504,100 $2,408,820 Salvage+ $500,000+ $1,000,000 Incremental Cash Flow (FMS – CMS)

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 6 Solution:

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 7 Summary Rate of return (ROR) is the interest rate earned on unrecovered project balances such that an investment’s cash receipts make the terminal project balance equal to zero. Rate of return is an intuitively familiar and understandable measure of project profitability that many managers prefer to NPW or other equivalence measures. Mathematically we can determine the rate of return for a given project cash flow series by locating an interest rate that equates the net present worth of its cash flows to zero. This break-even interest rate is denoted by the symbol i*.

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 8 Internal rate of return (IRR) is another term for ROR that stresses the fact that we are concerned with the interest earned on the portion of the project that is internally invested, not those portions that are released by (borrowed from) the project. To apply rate of return analysis correctly, we need to classify an investment into either a simple or a nonsimple investment. A simple investment is defined as one in which the initial cash flows are negative and only one sign change occurs in the net cash flow, whereas a nonsimple investment is one for which more than one sign change occurs in the net cash flow series. Multiple i*s occur only in nonsimple investments. However, not all nonsimple investments will have multiple i*s either.

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 9 For a pure investment, the solving rate of return (i*) is the rate of return internal to the project; so the decision rule is: If IRR > MARR, accept the project. If IRR = MARR, remain indifferent. If IRR < MARR, reject the project. IRR analysis yields results consistent with NPW and other equivalence methods. For a mixed investment, we need to calculate the true IRR, or known as the “return on invested capital.” However, if your objective is simply to make an accept or reject decision, it is recommended that either the NPW or AE analysis be used to make an accept/reject decision. To compare mutually exclusive alternatives by the IRR analysis, the incremental analysis must be adopted.

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 10 Despite a strong academic preference for NPV, surveys indicate that executives prefer IRR over NPV. [6] Apparently, managers find it easier to compare investments of different sizes in terms of percentage rates of return than by dollars of NPV. However, NPV remains the "more accurate" reflection of value to the business. IRR, as a measure of investment efficiency may give better insights in capital constrained situations. However, when comparing mutually exclusive projects, NPV is the appropriate measure. [6]

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 11 Under Italian law, anyone who works at the same job for more than 36 months is considered an employee, entitled by law to four weeks of annual vacation, 11 paid holidays, maternity leave, state pension contributions and severance pay if dismissed.

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 12

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 13 DEPRECIATION FIXED ASSETS LOOSE THEIR VALUE - BEST EXEMPLIFIED BY PERSONAL AUTOMOBILES THIS LOSS IS CALLED DEPRECIATION ECONOMIC DEPRECIATION - GRADUAL LOSS OF UTILITY WITH TIME = PURCHASE PRICE - MARKET VALUE ACCOUNTING (OR ASSET) DEPRECIATION -- ALLOCATING THE COST OF THE ASSET OVER ITS DEPRECIABLE LIFE. TRIES TO MATCH ECONOMIC DEPRECIATION

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 14 WHAT IS DEPRECIABLE? ITEMS THAT ARE: 1. USED IN BUSINESS AND HELD FOR THE PRODUCTION OF INCOME 2. HAVE A DEFINITE SERVICE LIFE LONGER THAN A YEAR 3.MUST WEAR OUT, DECAY, BE USED UP, BECOME OBSOLETE 4.MUST COST MORE THAN $1000

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 15 WHO CAN DEPRECIATE ASSETS? ANYONE WHO USES THE ASSET FOR BUSINESS PURPOSES AND MEETS THE RULES ON THE PREVIOUS SLIDE

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 16 COST BASIS COST BASIS IS: THE TOTAL COST OF THE ASSET OVER ITS LIFE –INITIAL INVESTMENT –FREIGHT –SITE PREPARATION –INSTALLATION –TRADE-IN ALLOWANCE OF THE PREVIOUS EQUIPMENT IT DOES NOT INCLUDE OPERATING EXPENSES

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 17 OTHER DEFINITIONS THE BOOK VALUE EQUALS THE COST BASIS LESS THE ACCUMULATED TOTAL DEPRECIATION SALVAGE VALUE - ESTIMATED VALUE AT THE END OF AN ASSET’S LIFE DEPRECIABLE LIFE - NUMBER OF YEARS THAT THE ASSET CAN BE USED ASSET DEPRECIATION RANGE (ADR) - PRESCRIBED GUIDELINES BY THE IRS FOR DEPRECIABLE LIFE IRS - INTERNAL REVENUE SERVICE -- ENFORCES TAX RULES CREATED BY CONGRESS AND COLLECTS THE TAXES

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 18 DEPRECIATION METHODS DEPRECIATION IS CALCULATED TWO WAYS, EACH WITH ITS OWN PURPOSES BOOK DEPRECIATION –FINANCIAL REPORTS –INCOME STATEMENTS –REFLECTS ACTUAL LOSS IN VALUE –STATE TAXES AND FEDERAL TAXES BEFORE 1981 TAX DEPRECIATION –TO CALCULATE TAXES FOR THE IRS –TAKES ADVANTAGE OF TAX RULES –USUALLY LEADS TO BETTER CASH POSITION IN EARLIER YEARS

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 19 BOOK DEPRECIATION METHODS STRAIGHT LINE (SL) ACCELERATED –DECLINING BALANCE (DB) –DECLINING BALANCE WITH CONVERSION TO SL UNIT OF PRODUCTION DEPLETION

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 20 DECLINING BALANCE METHOD USES A FIXED FRACTION (alpha) OF THE BEGINNING BOOK BALANCE EACH YEAR: MULTIPLIERS: –1.5 OR 150% DB –2.0 OR 200% OR DOUBLE DECLINING METHOD (DDB)

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 21 DECLINING BALANCE WITH CONVERSION TO STRAIGHT LINE WHAT DO WE DO IF THE DB METHOD DOES NOT RESULT IN THE ESTIMATED SALVAGE VALUE? 1. IF B N > S –WE HAVE NOT DEPRECIATED THE ASSET FULLY –SWITCH TO STRAIGHT LINE THE FIRST YEAR THAT THE STRAIGHT LINE DEPRECIATION IS GREATER THAN THE DB METHOD 2. B N < S –STOP DEPRECIATING WHEN YOU GET TO S EXAMPLES 9.3 AND 9.4

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 22 Example 9.3 Cost basis of the asset I = $10,000 Useful life N = 5 years Estimated salvage value S = $2,000

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 23 Example 9.4 –Declining Balance Cost basis of the asset I = $10,000 Useful life N = 5 years Estimated salvage value, S = $778 D 1 =

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 24 Changing to straight line WE DID NOT DEPRECIATE THE ASSET FULLY AND MUST SWITCH TO STRAIGHT LINE

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 25 UNITS - OF - PRODUCTION METHOD ALLOCATES THE DEPRECIATION IN PROPORTION TO THE UNITS PRODUCED AS A FRACTION OF THE TOTAL UNITS EXAMPLE 9.7

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 26 HISTORY OF DEPRECIATION METHODS UNTIL STRAIGHT LINE ONLY ADDED ACCELERATED METHODS: –DECLINING BALANCE (DB) –DOUBLE DECLINING METHOD (DDB) –SUM OF THE YEARS’ DIGITS (SOYD) (WE DID NOT COVER) 1981 –REPLACED BY ACCELERATED COST RECOVERY SYSTEM (ACRS) –1986 MODIFIED ACRS = MACRS

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 27 MODIFIED ACCELERATED RECOVERY SYSTEM (MACRS) SALVAGE VALUE ALWAYS ZERO RECOVERY PERIOD -- ARBITRARY LIFE FOR THE INVESTMENT, NOT NECESSARILY EQUAL TO ACTUAL LIFE –EIGHT CATEGORIES: 3, 5, 7, 10, 15, 20, 27.5, 39 YEARS 3, 5, 7 AND 10 YEARS - USE 200%DDB, SWITCH TO SL 15 AND 20 YEARS -- USE 150%DDB AND SWITCH TO SL 27.5 YEARS RESIDENTIAL RENTAL -- SL 39 YEARS COMMERCIAL BUILDINGS -- SL USE TABLE 9.3 ON PAGE 448

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 28 MACRS TABLE

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 29 HALF YEAR CONVENTION FOR EQUIPMENT PLACED IN SERVICE USING MACRS: ALL ASSETS ARE ASSUMED TO BE PLACED IN SERVICE IN THE MIDDLE OF THE YEAR ALL SALVAGE VALUE IS ZERO ONLY A HALF YEARS’ DEPRECIATION IS ALLOWED THE FIRST YEAR THE REMAINING HALF YEAR IS ALLOWED FOLLOWING THE END OF THE RECOVERY PERIOD FOR REAL PROPERTY, USE MID - MONTH INSTEAD OF HALF YEAR

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 30 EXAMPLE 9.8 ASSET = $10,000 MACRS CLASS 5 YEARS

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 31 DEPLETION APPLIES TO NATURAL RESOURCES OBJECTIVE IS THE SAME AS DEPRECIATION COST DEPLETION WORKS JUST LIKE UNITS OF PRODUCTION - EQUATION 9.9: COST DEPLETION =(ADJUSTED BASIS OF MINERAL PROPERTY)x(NUMBER OF UNITS SOLD) / (TOTAL NUMBER OF RECOVERABLE UNITS) EXAMPLE 9.10 PERCENTAGE DEPLETION - PORTION OF GROSS INCOME, LIMITED BY 50% OF THE TAXABLE INCOME WITHOUT THE DEPLETION ALLOWABLE PERCENTAGES ARE GIVEN IN TABLE 9.5. THESE RANGE FROM 5% TO 22% EXAMPLE 9.11

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 32 Cost depletion vs. percentage depletion

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 33 REPAIRS OR IMPROVEMENTS For tax depreciation purposes, repairs or improvements made to any property are treated as separate property items. The recovery period for a repair or improvement to the initial property normally begins on the date the repaired or improved property is placed in service. The recovery class of the repair or improvement is the recovery class that would apply to the property if it were placed in service at the same time as the repair or improvement. Example 9.12 (modified)

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 34 EXAMPLE 9.12 In January 2004, Kendall Manufacturing Company purchased a new numerical control machine at a cost of $60,000. The machine had an expected life of 10 years at the time of purchase and a zero expected salvage value at the end of the 10 years. For book depreciation purposes, no major overhauls had been planned over the 10-year period, and the machine was being depreciated toward a zero salvage value, or $6,000 per year, with the straight-line method. For tax purposes, the machine was classified as a 7-year MACRS property.

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 35 EXAMPLE 9.12 CONTINUED

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 36 EXAMPLE 9.12 CONTINUED In December 2006, however, the machine was thoroughly overhauled and rebuilt at a cost of $15,000. It was estimated that the overhaul would extend the machine’s useful life by 5 years.

ENGINEERING ECONOMICS ISE460 SESSION 16 June 23, 2015 Geza P. Bottlik Page 37 EXAMPLE 9.12 CONTINUED