1. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-1 Developed By: Dr. Don Smith, P.E. Department of Industrial Engineering Texas A&M University College Station, Texas Executive Summary Version Chapter 6 Annual Worth Analysis

2. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-2 LEARNING OBJECTIVES 1.One life cycle 2.CR and AW calculation 3.Alternative selection by AW 4.AW of permanent investment CR = Capital Recovery AW = Annual Worth

3. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-3 Sct 6.1 Advantages and Uses of Annual Worth  Popular analysis technique  Easily understood -- results are reported in $ per time period, usually $ per year  Eliminates the LCM problem associated with the present worth method  Only have to evaluate one life cycle

4. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-4 AW Calculation from PW or FW AW Calculation from PW or FW  Computation from PW or FW  AW = PW(A/P,i%,n) or  AW = FW(A/F,i%,n)  If AW determined for alternative comparison, equal service assumption requires that  n = LCM number of years  AW converts all cash flows to their end of period equivalent amounts in $ per year

5. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-5 AW Value from Cash Flows  AW values can be calculated directly from cash flows for only one life cycle  Not necessary to consider the LCM of lives as is in PW or FW analysis  For alternative comparison, select the alternative with the best AW value

6. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-6 AW and Repeatability Assumption  If two or more alternatives have unequal life estimates, only evaluate the AW for one life cycle of each alternative  The annual worth of one cycle is the same as the annual worth of all future cycles (from repeatability assumption)

7. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-7 Repeatability Assumption  Given alternatives with unequal lives, the assumptions are: 1.The services provided are needed forever 2.The first cycle of cash flows is repeated for all successive cycles in the same manner 3.All cash flows will have exactly the same estimated values in every life cycle. Note: The third assumption may be unrealistic in many problems encountered in industry

8. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-8 One or More Cycles One or More Cycles Cycle 1Cycle 2Cycle k Annualize any one of the cycles AW assumes repeatability of cash flows Find the annual worth of any given cycle ($/period) …

9. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-9 6 Year & 9 Year Alternatives (Ex 6.1)  For PW or FW analysis, need an 18 year study period  3 life cycles of the 6 year project  2 life cycles of the 9 year project  Means a lot of calculation effort! 6 year Project 9 year Project

10. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-10 Ex 6.1 -- Using AW Analysis  If the cash flow patterns are assumed to remain the same for the 6 and 9 year projects for future cycles, then for AW method Project A: 6 years Project B: 9 years Find the AW of any 6-year cycle Find the AW of any 9-year cycle Compare AW A value for 6 years with AW B for 9 years to select the better alternative

11. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-11 Advantages/Applications of AW  Applicable to a variety of engineering economy studies such as:  Asset replacement  Breakeven analysis  Make-or-Buy decisions  Studies dealing with manufacturing costs  Economic value added (EVA) analysis

12. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-12 Sct 6.2 Calculating Capital Recovery and AW  An economic alternative should have the following cash flow estimates made  Initial investment -- P  Estimated future salvage value -- S  Estimated life -- n  Interest rate -- i% (this is usually the MARR)  Estimated annual operating costs – AOC  Capital Recovery (CR) is the annualized equivalent of the initial investment P and the future salvage value S for n years at i%

13. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-13 CAPITAL RECOVERY COST It is important to know the equivalent annual cost of owning an asset This cost is called “Capital Recovery” or CR CR is determined using {P, S, i, and n}... P is initial purchase price at time 0 Estimated salvage value at time n is S 0 1 2 …………………………. n-1 n

14. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-14 CAPITAL RECOVERY COST Given: Convert to: 0 1 2 3 n-1 n An amount of A per year (CR) P 0 1 2 3 n-1 n ……….S

15. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-15 Comparing CAPITAL RECOVERY with AW  CR is a cost, so it carries a negative sign  CR is the annual equivalent (an A value) that represents the implied ‘cost’ of an asset for n years at i% with a first cost of P and a salvage value of S in year n  CR does NOT include annual operating costs, AOC  To obtain AW once CR is determined, calculate AW = - CR - AOC where AOC itself is an annual equivalent amount (same each year)

16. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-16 Capital Recovery Calculations COMPUTING CR FOR INVESTMENTS WITH A SALVAGE VALUE Method 1  Compute equivalent annual cost of the investment P and subtract the equivalent annual savings of the salvage value S. This is P(A/P,i,n) - S(A/F,i,n)  Determine CR as the negative (cost) of this relation CR = -[P(A/P,i,n) - S(A/F,i,n)]

17. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-17 More Commonly Used CR Calculation Method 2  Subtract salvage value S from original cost P and calculate the equivalent annual cost of (P-S)  Add to that the interest which the salvage value would return each year, S(i) CR = -[(P - S)(A/P,i,n) + S(i)]  Excel can be used to find CR: =PMT(i%,n,P,-S)

18. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-18 CR Amount: What It Means CR is the annual cost associated with owning a productive asset over n time periods at interest rate i% per period Equivalently, CR may be interpreted as the minimum amount of money an investment must earn each of n years to recover the initial cost at a return of i% Why? Remember, the purchase of assets to conduct business involves a commitment of the owner's funds. As such, an investment is a commitment of the owner’s funds over n time periods. Thus, the owners expect a return on that investment.

19. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-19 CR and AW Computation: Example 6.2  P = $12.46 millionS = $0.5 million  n = 8 yearsi = 12%  AOC = $0.9 million per year  CR by method 2: CR = -[(12.46-0.5)(A/P,12%,8)+ 0.5(0.12)] = $-2.47 million per year  AW = CR – AOC = - 2.47 - 0.9 = $ - 3.37 million per year

20. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-20 Sct 6.3 Evaluating Alternatives Using AW  For mutually exclusive alternatives, select one with lowest AW of costs(service) or highest AW of net incomes (revenue)  This means, select the numerically largest AW alternative  If AW < 0 at MARR, the (revenue) alternative is not economically justifiable, since initial investment P is not recovered over n years at the required rate of MARR = i% per year

21. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-21 AW Method: Example 6.4(a)  Alternative A has two components with n values of 8 and 12 years; Alternative B has n = 24 years  Selection using AW chooses B with the lower equivalent annual costs  Only one life cycle of each asset was considered  PW analysis would require using LCM of 24 years AlternativeLivesCRAW A8 & 12$-24,424$-36,724 B24$-27,146$-29,646

22. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-22 AW Method: Example 6.4(b)  Specified study period of only 6 years reduces time to recover investment, so AW values of costs go up  Now, select A since it has lower AW of costs  Bigger impact on CR for B since recovery time is reduced from 24 to only 6 years  Check out the computer solution for this example Alternative Study Period CR over 6 years AW over 6 years A6$-26,382$-38,682 B6$-43,386$-45,886

23. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-23 Special Cases for AW Analysis  If cash flow repeatability assumption can’t be made, specify a study period of n years and perform analysis with this n in all computations (Example 6.4(b) did this)  If projects are independent, select all with AW > 0 at i = MARR, provided no budget limit is defined. If budget limited, use techniques of chapter 12

24. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-24 Sct 6.4 AW of a Permanent Investment  If an investment has no finite cycle (or a very long estimated life) it is called a perpetual or permanent investment  If “P” is the present worth of the cost of the investment, then the AW value is P times i AW =A = P(i)  AW is actually the amount of interest P would earn each year, forever  See Examples 6.5 and 6.6 for illustrations Remember: P = A/i from the previous chapter

25. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-25 Chapter Summary  AW method is often preferred to the PW method  AW deals with only one life cycle of an alternative  AW offers an advantage for comparing different-life alternatives  Assumption for AW method: Cash flows in one cycle are assumed to replicate themselves in future cycles  For infinite life alternatives, simply multiply P by i to get AW value

26. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 6-26 Chapter 6 End of Slide Set