Chapter 6 Annual Equivalent Worth Criterion Chapter

Chapter 6 Annual Equivalent Worth Criterion

Chapter 6 Annual Equivalence Analysis q q q Annual equivalent criterion Applying annual worth analysis Mutually exclusive projects

Annual Worth Analysis Principle: Measure an investment worth on annual basis Find the net present worth of the original series and multiply by the capital recovery factor Annual Equivalent (AE) = PW (A/P, i%, N)

Annual Worth Analysis Evaluating a single revenue project AE > 0 accept the investment AE = 0 indifferent AE < 0 reject the investment Comparing multiple alternatives Revenue project: alternative with highest AE is selected Service project: alternative with lowest annual equivalent cost is selected

Annual Worth Analysis Benefit: By knowing the annual equivalent worth, we can: • • • Seek consistency with the annual report format Determine the unit cost (or unit profit) Facilitate the unequal project life comparison

Computing Equivalent Annual Worth $9 PW(15%) = $6. 946 $5 0 $12 $10 $8 1 2 3 4 5 6 $3. 5 $6. 946 $15 A = $1. 835 0 0 1 2 3 4 5 AE(15%) = $6. 946(A/P, 15%, 6) = $1. 835 6

Annual Equivalent Worth - Repeating Cash Flow Cycles $800 $700 $500 $400 $1, 000 Repeating cycle

• First Cycle: PW(10%) = -$1, 000 + $500 (P/F, 10%, 1) +. . . + $400 (P/F, 10%, 5) = $1, 155. 68 AE(10%) = $1, 155. 68 (A/P, 10%, 5) = $304. 87 • Both Cycles: PW(10%) = $1, 155. 68 + $1, 155. 68 (P/F, 10%, 5) = $1, 873. 27 AE(10%) = $1, 873. 27 (A/P, 10%, 10) = $304. 87

n n n When only costs are involved, the AE method is called the annual equivalent cost. Revenues must cover two kinds of costs: Operating costs (incurred by operation of plants or equipment) and Capital costs (incurred by purchasing the assets). Capital costs are one-time costs, whereas operating costs are recurring. Annual Equivalent Costs Annual Equivalent Cost Capital costs + Operating costs

Capital (Ownership) Costs n n Def: Owning an equipment is associated with two transactions—(1) its initial cost (I) and (2) its salvage value (S). Capital recovery cost: the annual equivalent of a capital cost S 0 N I 0 1 2 3 CR(i) N

Example What is the equivalent annual cost of a machine that has an initial cost of $8, 000, a salvage value of $500 after 8 years and annual operating costs of $900 if MARR=20%?

Solution AE = 8000 (A/P, 20%, 8) + 900 – 500 (A/F, 20%, 8) = 8000 (0. 2606) + 900 – 500 (0. 0606) = $2954. 5 OR AE = (8000 - 500) (A/P, 20%, 8) + 500 (0. 2) + 900 = 7500 (0. 2606) + 100 + 900 = $2954. 5

Example

Example

Applying Annual Worth Analysis Some economic analysis problems can be solved more efficiently by annual worth analysis • Unit Cost (Unit Profit) Calculation • Make-or-Buy Decision

Unit Cost (Unit Profit) Calculation n n Determine the number of units to be produced each year Identify the cash flow series Calculate the annual worth Divide the annual worth by the number of units to be produced each year. If the number of units varies, then convert to equivalent annual units.

Example Equivalent Worth per Unit of Time $24, 400 $27, 340 $55, 760 0 1 $75, 000 2 3 Operating Hours per Year 2, 000 hrs. Compute the equivalent savings per machine hour with i=15%.

Example Equivalent Worth per Unit of Time PW (15%) = -75, 000+24, 400(P/F, 15%, 1) +27, 370(P/F, 15%, 2)+55, 760(P/F, 15%, 3) = $3, 553 AE (15%) = $3, 553 (A/P, 15%, 3) = $1, 556 Savings per Machine Hour = $1, 556/2, 000 = $0. 78/hr.

Example: Mutually Exclusive Alternatives with Equal Project Lives Standard Motor 25 HP Size $13, 000 Cost 20 Years Life $0 Salvage 89. 5% Efficiency $0. 07/k. Wh Energy Cost Operating Hours 3, 120 hrs/yr. Premium Efficient Motor 25 HP $15, 600 20 Years $0 93% $0. 07/k. Wh 3, 120 hrs/yr. (a) At i= 13%, determine the operating cost per k. Wh for each motor. (b) At what operating hours are they equivalent?

Solution: (a) Operating cost per k. Wh per unit Determine total input power 1 HP = 0. 7457 k. W 25 HP=18. 65 k. W Ø Standard motor: input power = 18. 650 k. W/ 0. 895 = 20. 838 k. W Ø PE motor: input power = 18. 650 k. W/ 0. 93 = 20. 054 k. W

Determine total k. Wh per year with 3120 hours of operation Ø Standard motor: 3120 hrs/yr (20. 838 k. W) = 65, 018 k. Wh/yr Ø PE motor: 3120 hrs/yr (20. 054 k. W) = 62, 568 k. Wh/yr Determine annual energy costs at $0. 07/kwh: Ø Standard motor: $0. 07/kwh 65, 018 kwh/yr = $4, 551/yr Ø PE motor: $0. 07/kwh 62, 568 kwh/yr = $4, 380/yr

Capital cost: Ø Standard motor: $13, 000(A/P, 13%, 20) = $1, 851 Ø PE motor: $15, 600(A/P, 13%, 20) = $2, 221 Total annual equivalent cost = Capital Cost + Energy Cost Ø Standard motor: AE(13%) = $4, 551 + $1, 851 = $6, 402 Cost per output kwh = $6, 402/58, 188* kwh = $0. 11/kwh Ø PE motor: AE(13%) = $4, 380 + $2, 221 = $6, 601 Cost per output kwh = $6, 601/58, 188 kwh = $0. 1134/kwh (* 58, 188 kwh/year = 3120 hours/year*25 HP*0. 7457 kwh/HP)

(b) break-even Operating Hours = 6, 742

Example: Mutually Exclusive Alternatives with Unequal Project Lives Model A: 0 1 $5, 000 $12, 500 2 $5, 000 Required service Period = Indefinite 3 $3, 000 Analysis period = LCM (3, 4) = 12 years Model B: 0 1 2 3 4 $2, 500 $4, 000 $15, 000 $4, 000

Model A: 0 1 2 3 $3, 000 $12, 500 $5, 000 • First Cycle: PW(15%) = -$12, 500 - $5, 000 (P/A, 15%, 2) - $3, 000 (P/F, 15%, 3) = -$22, 601 AE(15%) = -$22, 601(A/P, 15%, 3) = -$9, 899 • With 4 replacement cycles: PW(15%) = -$22, 601 [1 + (P/F, 15%, 3) + (P/F, 15%, 6) + (P/F, 15%, 9)] = -$53, 657 AE(15%) = -$53, 657(A/P, 15%, 12) = -$9, 899

Model B: 0 1 2 3 4 $2, 500 $15, 000 $4, 000 • First Cycle: PW(15%) = - $15, 000 - $4, 000 (P/A, 15%, 3) - $2, 500 (P/F, 15%, 4) = -$25, 562 AE(15%) = -$25, 562(A/P, 15%, 4) = -$8, 954 • With 3 replacement cycles: PW(15%) = -$25, 562 [1 + (P/F, 15%, 4) + (P/F, 15%, 8)] = -$48, 534 AE(15%) = -$48, 534(A/P, 15%, 12) = -$8, 954

Summary n Annual equivalent worth analysis, or AE, is—along with present worth analysis—one of two main analysis techniques based on the concept of equivalence. The equation for AE is AE(i) = PW(i)(A/P, i, N). AE analysis yields the same decision result as PW analysis.

n The capital recovery cost factor, or CR(i), is one of the most important applications of AE analysis in that it allows managers to calculate an annual equivalent cost of capital for ease of itemization with annual operating costs. n The equation for CR(i) is CR(i)= (I – S)(A/P, i, N) + i. S, where I = initial cost and S = salvage value.

n AE analysis is recommended over NPW analysis in many key real -world situations for the following reasons: 1. In many financial reports, an annual equivalent value is preferred to a present worth value. 2. Calculation of unit costs is often required to determine reasonable pricing for sale items. 3. Calculation of cost per unit of use is required to reimburse employees for business use of personal cars. 4. Make-or-buy decisions usually require the development of unit costs for the various alternatives.