Annual Equivalent Worth Criterion Engineering Economy

Annual Equivalence Analysis Annual equivalent criterion Provides basis for measuring investment worth by determining equal payments on an annual basis We first find the NPV for the original series then multiply that by the capital recovery factor Applying annual worth analysis We use the following formula to evaluate the investment AE(i)= PW(i)(A/P, i, N) When evaluating a single project if AE(i)>0 accept if equal to zero indifferent if less than zero reject investment Mutually exclusive projects If we are comparing a service project we choose the one with the least annual equivalent cost

Example Computing Annual Equivalent Worth $9.0 $12 $10 $5 $8 PW(15%) = $6.946 1 2 3 4 5 6 $3.5 $6.946 $15 A = $1.835 1 2 3 4 5 6 AE= $6.946(A/P, 15%, 6) = $1.835

Example Economics of Installing A Feed-water Heater Install a 150MW unit: Initial cost = $1,650,000 Service life = 25 years Salvage value = 0 Efficiency 55% Expected improvement in fuel efficiency = 1% Fuel cost = $0.05kWh Load factor = 85% Determine the annual worth for installing the unit at i = 12%. If the fuel cost increases at the annual rate of 4%, what is AE(12%)?

Calculation of Annual Fuel Savings Required input power before adding the second unit: Required input power after adding the second unit: Reduction in energy consumption: 4,870kW Annual operating hours: Annual Fuel Savings : .

Annual Worth Calculations (a) with constant fuel price: (b) with escalating fuel price: Cash Flow Diagrams

Benefits of Annual Worth Analysis Why use Annual worth while we know that the project is accepted or rejected by looking at the PW? Principle: Measure investment worth on annual basis Benefit: By knowing annual equivalent worth, we can: Seek consistency of report format Determine unit cost (or unit profit) Facilitate unequal project life comparison

Annual Equivalent Worth - Repeating Cash Flow Cycles $800 $800 $700 $700 $500 $500 $400 $400 $400 $400 $1,000 $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) = $304.78 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) = $304.78

Annual Equivalent Cost When only costs are involved, the AE method is called the annual equivalent cost. Revenues must cover two kinds of costs: Operating costs and capital costs. Capital costs Annual Equivalent Costs + Operating costs

Capital (Ownership) Costs Def: Owning of an equipment is associated with two transactions—(1) its initial cost (I) and (2) its salvage value (S). Capital costs: Taking these items into consideration, we calculate the capital recovery costs as: S N I 0 1 2 3 N CR(i) Remember from Ch2 that: (A/F, i, N)= (A/P, i, N)-i

Value after three years Will Your Car Hold its Value? Vehicle Value after three years Monthly lease Acura MDX 4WD $21,375 588 Toyota 4Runner 4WD 4-door Limited (with moonroof) $18,750 646 Lexus RX300 2 WD 651 GMC Envoy 4WD 4door SLT (with Bose audio, CD changer and moonroof) $15,050 733 Mitsubishi Montero 4WD 4 door Limited $15,225 743 Ford Explorer 4WD 4-door Eddie Bauer edition (with moonroof, rear air and third seat) $14,425 754 Isuzu Trooper 4WD 4-door Limited $11,675 823 Source: Automotive Lease Guide

Example - Capital Cost Calculation for Mini Cooper Given: I = $19,800 N = 3 years S = $12,078 i = 6% Find: CR(6%) $12,078 3 $19,800

The Cost of Owning the Rest of a Vehicle SEGMENT BEST MODELS ASKING PRICE PRICE AFTER 3 YEARS CR (6%) Compact car Mini Cooper $19,800 $12,078 $3,614 Midsize car Volkswagen Passat $28,872 $15,013 $6,086 Sports car Porsche 911 $87,500 $48,125 $17,618 Near luxury car BMW 3 Series $39,257 $20,806 $8,151 Luxury car Mercedes CLK $51,275 $30,765 $9,519 Minivan Honda Odyssey $26,876 $15,051 $5,327 Subcompact SUV Honda CR-V $20,540 $10,681 $4,329 Compact SUV Acura MDX $37,500 $21,375 $7,315 Full size SUV Toyota Sequoia $37,842 $18,921 $8,214 Compact truck Toyota Tacoma $21,200 $10,812 $4,535 Full size truck Toyota Tundra $25,653 $13,083 $5,488

The Cost of Owning the Rest of a Vehicle

Example

Where to Apply the AE Analysis Unit cost (or profit) calculation Outsourcing (Make-Buy) Decision Pricing the Use of an Asset Contemporary Engineering Economics, 5th edition, © 2010

Unit Cost (Unit Profit) Calculations Determine the number of units to be produced each year Identify the cash flow series associated with production Calculate the present worth of the project’s cash flow series at a given (i) Determine the equivalent annual worth Divide the equivalent annual worth by the number of units to be produced or serviced during each year

Example Unit Profit per Machine Hour When Annual Operating Hours remain Constant Step 1: Determine the annual volume. 3,000 hours per year Step 2: Obtain the equivalent annual worth. PW (12%) = $30,065 AE (12%) = $30,065 (A/P, 12%, 4) = $9,898 Step 3: Determine the unit profit (savings per machine hour). Savings per Machine Hour = $9,898/3,000 = $3.30/hour Project Cash Flows & Operating Hours

Example 6.6 Unit Profit per Machine Hour When Annual Operating Hours Fluctuate Step 1: Determine the annual volume. Year 1: 3,500 hours Year 2: 4,000 hours Year 3: 1,700 hours Year 4: 2,800 hours Step 2: Obtain the equivalent annual worth. AE (12%) = $30,065 (A/P, 12%, 4) = $9,898 C[(3,500)(P/F,12%,1) + (4,000)(P/F,12%,2) + (1,700)(P/F,12%,3) + (2,800)(P/F,12%,4)](A/P,12%,4) = 3,062.95C Step 3: Determine the unit profit (savings per machine hour). Savings per Machine Hour C = $9,898/3,062.95 = $3.23/hour Project Cash Flows & Operating Hours

Make-or-Buy Decision The unit cost comparison between the two options requires the use of annual worth analysis: Determine the time span for which the part will be needed Determine the annual quantity of the part Obtain the unit cost of purchasing the part Determine the cost of the equipment, manpower and all other resources needed for the make Estimate the net cash flows associated with the make Compute the annual equivalence cost of producing the part Compute the unit cost of making the part by dividing the annual equivalent cost by the required annual quantity Choose the option with the smallest unit cost

Issues to consider when buying The quality of the product The reliability of the supplier providing the needed quantities in time If the difference is not much and we are looking for good quality we can ignore the difference

Example Outsourcing the Manufacture of Cassettes and Tapes Make Option Buy Option

Solution: Make Option: Buy Option: AEC(14%) = $4,582,254 Unit cost: $4,582,254/3,831,120 = $1.20 Buy Option: AEC(14%) = $4,421,376 $4,421,376/3,831,120 = $1.15 Figure: 06-06

Pricing the Use of an Asset The cost per square foot for owning and operating a real property (example, rental fee) The cost of using a private car for business (cost per mile) The cost of flying a private jet (cost per seat) The cost of using a parking deck (cost per hour)

Example: Pricing an Apartment Rental Fee Investment Problem: Building a 50-unit Apartment Complex At Issue: How to price the monthly rental per unit? Land investment cost = $1,000,000 Building investment cost = $2,500,000 Annual upkeep cost = $150,000 Property taxes and insurance = 5% of total investment Occupancy rate = 85% Study period = 25 years Salvage value = Only land cost can be recovered in full Interest rate = 15% Contemporary Engineering Economics, 5th edition, © 2010

Solution: Ownership cost: Annual O&M Cost Total Equivalent Annual Cost CR(15%) = ($3,500,000 - $1,000,000)(A/P, 15%, 25) + ($1,000,000)(0.15) = $536,749 Ownership cost: Annual O&M Cost Total Equivalent Annual Cost Required Monthly Charge AEC(15%) = $536,749 + $325,000 =$861,749

Example Mutually Exclusive Alternatives with Equal Project Lives Standard Premium Motor Efficient Motor 25 HP 25 HP $13,000 $15,600 20 Years 20 Years $0 $0 89.5% 93% $0.07/kWh $0.07/kWh 3,120 hrs/yr. 3,120 hrs/yr. Size Cost Life Salvage Efficiency Energy Cost Operating Hours (a) At i= 13%, determine the operating cost per kWh for each motor. (b) At what operating hours are they equivalent?

Solution: Determine total input power Conventional motor: (a) Operating cost per kWh per unit Determine total input power Conventional motor: input power = 18.650 kW/ 0.895 = 20.838kW PE motor: input power = 18.650 kW/ 0.93 = 20.054kW

Determine total kWh per year with 3120 hours of operation Conventional motor: (3120 hrs/yr)* (20.838 kW) = 65,018 kWh/yr PE motor: ( 3120 hrs/yr) *(20.054 kW) = 62,568 kWh/yr Determine annual energy costs at $0.07/kwh: Conventional motor: $0.07/kwh  65,018 kwh/yr = $4,551/yr PE motor: $0.07/kwh  62,568 kwh/yr = $4,380/yr

Capital cost: Conventional motor: $13,000(A/P, 13%, 20) = $1,851 PE motor: $15,600(A/P, 13%, 20) = $2,221 Total annual equivalent cost: Notice that the total output power is 25HP X 0.746KW/HP X 3120 hours/year = 58,188 KWh/year AE(13%) = $4,551 + $1,851 = $6,402 Cost per kwh = $6,402/58,188 kwh = $0.11/kwh AE(13%) = $4,380 + $2,221 = $6,601 Cost per kwh = $6,601/58,188 kwh = $0.1134/kwh

Example Mutually Exclusive Alternatives with Unequal Project Lives Model A: 0 1 2 3 $12,500 $5,000 $5,500 $6,000 $2000 Required service Period = Indefinite Analysis period = LCM (3,4) = 12 years $1500 Model B: 0 1 2 3 4 $5,500 $4,000 $4,500 $5,000 $15,000

$2000 $12,500 $5,000 $5,500 $6,000 0 1 2 3 Model A: First Cycle: PW(15%) = -$12,500 - $5,000 (P/F, 15%, 1)- $5,500 (P/F, 15%, 2) - $6,000 (P/F, 15%, 3) + $2,000 (P/F, 15%, 3) = -$23,637 AE(15%) = -$23,637(A/P, 15%, 3) = -$10,352 With 4 replacement cycles: PW(15%) = -$23,637 [1 + (P/F, 15%, 3) + (P/F, 15%, 6) + (P/F, 15%, 9)] = -$64,531 AE(15%) = -$64,531(A/P, 15%, 12) = -$10,352

$15,000 $4,000 $4,500 0 1 2 3 4 $5,000 Model B: First Cycle: PW(15%) = - $15,000 - $4,000 (P/F, 15%, 1) - $4,500 (P/F, 15%, 2) - $5,000 (P/F, 15%, 3)- $4,000 (P/F, 15%, 4) = -$27,456 AE(15%) = -$27,456(A/P, 15%, 4) = -$12,024 With 3 replacement cycles: PW(15%) = -$27,456 [1 + (P/F, 15%, 4) + (P/F, 15%, 8)] = -$74,954 AE(15%) = -$74,954(A/P, 15%, 12) = -$12,024

We choose Model A since it has less cost although it has shorter life-spans 0 1 2 3 $4,000 $4,000 4 5 6 $5,000 $5,000 $5,500 $5,500 $12,500 $12,500 $4,000 $4,000 7 8 9 $5,000 $5,000 $5,500 $5,500 $12,500 $12,500 Model A $4,000 $4,000 10 11 12 $5,000 $5,000 $5,500 $5,500 $12,500 $12,500 $4,000 $4,000 $5,000 $5,000 $5,500 $5,500 0 1 2 3 4 $12,500 $12,500 $4,000 $4,000 5 6 7 8 $4,000 $4,000 $4,500 $4,500 $5,000 $5,000 $4,000 $4,000 $15,000 $15,000 $4,000 $4,000 9 10 11 12 $4,500 $4,500 $5,000 $5,000 $15,000 $15,000 $4,000 $4,000 Model B $4,000 $4,000 $4,500 $4,500 $5,000 $5,000 $15,000 $15,000

Minimum Cost Analysis Concept: Total cost is given in terms of a specific design parameter (provide required functional quality with the lowest cost) Goal: Used to determine optimal engineering design parameter that will minimize the total cost Typical Mathematical Equation: 𝑨𝑬 𝒊 =𝒂+𝒃𝒙+ 𝒄 𝒙 where x is common design parameter Analytical (optimal) Solution:

Typical Graphical Relationship Total Cost Capital Cost Cost ($) O & M Cost Design Parameter (x) Optimal Value (x*)

Example Optimal Cross-Sectional Area Decision Problem: A constant electric current of 5,000 amps is to be transmitted a distance of 1,000 feet from a power station to a substation. Find: the optimal size of a copper conductor Relevant Physical and Financial Data Copper price: $8.25/lb Resistance: 0.8145x10-5in2/ft Cost of energy: $0.05/kWh Density of copper: 555 lb/ft Useful life: 25 years Salvage value: $0.75/lb Interest rate: 9% Power Transmission Substation Power Plant 1,000 ft. 5,000 amps 24 hours 365 days Cross-sectional area

Operating Cost (Energy Loss) Energy loss in kilowatt-hour (L) I = current flow in Amps R = resistance in ohms T = number of operating hours A = cross-sectional area

Material Costs Material weight in pounds Material cost (required investment) Total material cost = 3,854A($8.25) = $31,797A Salvage value after 25 years: ($0.75)(3,854A)

Capital Recovery Cost 2,890.6 A 25 31,797 A Given: Initial cost = $31,797A Salvage value = $2,890.6A Project life = 25 years Interest rate = 9% Find: CR(9%) 2,890.6 A 25 31,797 A

Total Equivalent Annual Cost AEC = Capital cost + Operating cost = Material cost + Energy loss Find the minimum annual equivalent cost

Optimal Cross-Sectional Area Figure: 06-09

Summary 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. 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. The equation for CR(i) is CR(i)= (I – S)(A/P, i, N) + iS, where I = initial cost and S = salvage value.

Summary 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.