Annual Equivalence Analysis Chapter 6 Fundmentals of Engineering Economics Copyright © 2008

Brand ABrand B $55/Year$85/Year $799$699 Critical Issue - How to Determine the Operating Cost per Hour? Two different brands of refrigerator Purchase cost:  Brand A: $799  Brand B: $699 Annual operating cost:  Brand A: $55/year  Brand B: $85/year Service life: 25 years Salvage value: 0

Annual Worth Analysis Principle: Measure an investment worth on annual basis Benefit: By knowing the annual equivalent worth, we can: Seek consistency of report format Determine the unit cost (or unit profit) Facilitate the unequal project life comparison

Process of Calculating the AE Value PW(i) Project Cash Flow Equivalent Annual Cash Flow AE(i) = PW(i)(A/P, i, N)

Example 6.1 Cost of implementing the project is $159,000. boilers have remaining service life of 12 years. Any upgrade have No salvage value at end of 12 years Annual electricity use in the boiler house is expected to be reduced from 410,000 KW to 1800,000 KW as a result of the upgrade. This is equivalent to $14,000 per year. The savings is expected to increase at annual rate of 4% as the cost of electricity increase over time Coal use will be 2% lower due to the projected improvement in boiler efficiency. This corresponds to a cost of reduction of $40,950 per year. The savings is projected to increase as the coal prices increase at a n annual rate of 5% If the hospital uses 10% interest rate for any project justification, what would be the annual equivalent energy savings due to the improvement?

Example 6.1 Finding AE by Conversion from NPW

Benefits of AE Analysis AE is used when it is more useful to presents annual cost or benefit of an ongoing project rather than its overall cost or benefit. When life-cycle cost analysis is desired When there is a need to determine unit costs or profits When project lives are unequal

$500 $700 $800 $400 $500 $700 $800 $400 $1,000 Annual Equivalent Worth with Repeating Cash Flow Cycles First Cycle Second Cycle

First Cycle: PW(10%) = -$1,000 + $500 (P/F, 10%, 1) $400 (P/F, 10%, 5) = $1, AE(10%) = $1, (A/P, 10%, 5) = $ Both Cycles: PW(10%) = $1, $1, (P/F, 10%, 5) = $1, AE(10%) = $1, (A/P, 10%,10) = $ Calculation:

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 Operating costs + Annual Equivalent Cost

Capital (Ownership) Costs

SEGMENTBEST MODELSASKING PRICE PRICE AFTER 3 YEARS Compact carMini Cooper$19,800$12,078 Midsize carVolkswagen Passat $28,872$15,013 Sports carPorsche 911$87,500$48,125 Near luxury carBMW 3 Series$39,257$20,806 Luxury carMercedes CLK$51,275$30,765 MinivanHonda Odyssey $26,876$15,051 Subcompact SUVHonda CR-V$20,540$10,681 Compact SUVAcura MDX$37,500$21,375 Full size SUVToyota Sequoia$37,842$18,921 Compact truckToyota Tacoma$21,200$10,812 Full size truckToyota Tundra$25,653$13,083 Source: “Will your car hold its value? A New study does the math,” The Wall Street Journal, August 6, 2002 Will Your Car Hold Its Value?

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

Example 6.2 Ferguson company is considering an investment in computer aided design equipment. The equipment will cost $110,000 and will have a five year useful economic life It has $10,000 terminal disposal price Generate annual operating saving of $33,000 Should it be purchased at interest rate of 15%?

Example 6.2 Justifying a project with the AE Method

Practice Problem Compute the annual equivalent worth of the following project cash flows at an interest rate of 15%. $100 $140 $180 $220 $ $300

Practice Problem You purchased an industrial robot at $200,000 to automate a part of manufacturing process in your assembly line. If the robot will be used 10 years and the expected salvage value of the robot is 10% of the initial cost, what would be the capital cost of owning the robot at an interest rate of 15%?

Annual Worth Analysis Chapter 6 Fundamentals of Engineering Economics Copyright © 2008

 Unit Cost (Unit Profit) Calculation  Make or Buy Decision Applying Annual Worth Analysis

Example 6.3 Unit Profit per Machine Hour A firm is looking to invest in a robot to avoid workers accidents The investment will cost $1 Million and the salvege value is $100,000 after 5 years service life The robot will reduce labor costs, insurance, accidents,..etc The savings are $800,000 a year Additional operating and maintenance costs amount to $300,000 Suppose the robot will operate for 2000 hours per year Compute the equivalent savings per machine hour at i=15%

Example 6.3 Unit Profit per Machine Hour

Example 6.5 Unit Cost: Make or Buy  Financial Facts: Annual volume = 120,000 units  Buy Option:  Unit price = $35  Make Option:  Required Investment in specialized tools: $2.2M, service life = 5 years, salvage value = $0.12M  Variable production cost: $26.30 per unit

Solution: Select the Make Option Buy Option:  AEC(12%) = ($35/unit)×(120,000 units/year) = $4,200,000/year  Unit cost = $35/unit Make Option:  Capital recovery cost: CR(12%) = ($2,200,000 - $120,000)(A/P,12%,5) ($120,000) = $591,412  Production cost: OC(12%) = ($26.30/unit)×(120,000 units/year) = $3,156,000/year  Total equivalent annual cost: AEC(12%) = $591,412 + $3,156,000 = $3,747,412 Unit cost = $3,747,412/120,000 = $31.23/unit

Life-Cycle Cost Analysis  At Issue: To compare different design alternatives, each of which would produce the same number of units (constant revenues), but would require different amounts of investment and operating costs.  What to do: Compute the annual equivalent cost of the design alternatives over the entire service life and select the alternative with the least cost.

StandardPremium Motor Efficient Motor 25 HP (18.65kW)25 HP $13,000 $15,60020 Years$0 89.5%93%$0.07/kWh3,120 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 the two types of motors equally economical? Example 6.6 Life-Cycle Cost Analysis

Solution: Step 1 – Calculate the Required Input Power (a): Operating cost per kWh Determine the required input power  Conventional motor: input power = kW/ = kW  PE motor: input power = kW/ 0.93 = kW

 Determine the total kWh per year with 3120 hours of operation  Conventional motor: 3,120 hrs/yr ( kW) = 65,018 kWh/yr  PE motor: 3,120 hrs/yr ( 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 Step 2: Determine the Annual Energy Cost

Step 3: Determine the Total Equivalent Annual Cost  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:  Conventional motor: AE(13%) = $4,551 + $1,851 = $6,402 Cost per kWh = $6,402/58,188 kWh = $0.11/kWh  PE motor: AE(13%) = $4,380 + $2,221 = $6,601 Cost per kWh = $6,601/58,188 kWh = $0.1134/kWh Note: The total output power is 58,188kWh.

(b) Break-Even Analysis The break- even number of operating hours for the PE motors: 6,742 hours per year

 Model A: $150,000 $55,000  Model B: $230,000 $30,000 Example 6.7 AEC Comparison – Unequal Project Lives Required service Period = Indefinite Analysis period = LCM (3,5) = 15 years $15,000 $35,000 5 $30,000

Model A: $150,000 $55,000 $15,000

Model B: $230,000 $30,000 $35,000 5

Least Common Multiple Service Period of 15 Years LCM = 15 years AEC(15%) A = $116,377 AEC(15%) B = $93,422

Practice Problem The City of Auburn has decided to build a softball complex and the city council has already voted to fund the project at the level of $800,000 (initial capital investment). The city engineer has collected the following financial information for the complex project. Annual upkeep costs: $120,000 Annual utility costs: $13,000 Renovation costs: $50,000 for every 5 years Annual team user fees (revenues): $32,000 Useful life: Infinite Interest rate: 8% If the city can expect 40,000 visitors to the complex each year, what should be the minimum ticket price per person so that the city can breakeven? In answering the question, identify the range that contains the solution. (a) Price: greater than $2.50 and less or equal $3.00 (b) Price: greater than $3.00 and less or equal $3.50 (c) Price: greater than $3.50 and less or equal $4.00 (d) Price: greater than $4.00 and less or equal $4.50

SOLVED PROBLEMS 3; 4; 7; 16; 24; 28; 29; 31; 35;

6.3) Consider the following cash flows and compute the equivalent annual worth at i = 12% n A n Investment A n Revenue 0-$25,000 1-$10,000$14,000 2$13, $8,000 6$5,500 SOLUTION PROBLEM

SOLUTION

PROBLEM SOLUTION PW (9%) = -$1,000 + $800 (P/F, 9%, 1) + $1,100 (P/F, 9%, 2) + $1,600 (P/F, 9%, 3) + $800 (P/F, 9%, 4) + $800 (P/F, 9%, 5) PW (9%) = -$1,000 + $734 + $926 + $1,236 + $567 + $520 = $2,983 AE (9%) = $2,983 (A/P, 9%, 5) = $766.92

PROBLEM SOLUTION

PROBLEM SOLUTION

PROBLEM

SOLUTION PROBLEM

6.29 Taka Company, a farm-equipment manufacturer, currently produces 30,000 units of gas filters for use in its lawnmower production annually. The following costs are reported, according to the previous year’s production: It is anticipated that gas-filter production will last five years. If the company continues to produce the product in-house, annual direct-material costs will increase at a rate of 5%. (For example, the annual direct-material costs during the first production year will be $63,000.) In addition, direct- labor costs will increase at a rate of 6% per year, and variable-overhead costs will increase at a rate of 3%, while fixed-overhead costs will remain at the current level over the next five years. Masami Company has offered to sell Taka Company 30,000 units of gas filters for $30 per unit. If Taka Company accepts the offer, some of the facilities currently used to manufacture the gas filters could be rented to a third party at an annual rate of $30,000. In addition, $3.50 per unit of the fixed- overhead costs applied to gas-filter production would be eliminated. The firm’s interest rate is known to be 15%. What is the unit cost of buying the gas filters from the outside source? Should Taka accept Masami’s offer? Why or why not? ITEMEXPENSE Direct Materials$60,000 Direct Labor$180,000 Variable Overhead (power & water)$135,000 Fixed Overhead (light & heat)$105,000 Total Cost$480,000

6.29 SOLUTION: Option 1: Purchase units from Masami Unit cost = $30 + ($105,000 – $30,000)/30,000 – $3.5 = $29 Option 2: Make units in-house PW DM (15%) = $63,000 (P/A 1, 5%, 15%, 5) = $230,202 PW DL (15%) = $190,800 (P/A 1, 6%, 15%, 5) = $709,433 PW VO (15%) = $139,050 (P/A 1, 3%, 15%, 5) = $490,847 AEC TC (15%) = [$230, $709, $490,847] (A/P, 15%, 5) + $105,000 = $1,430,482 (0.2983) + $105,000 = $426,713 + $105,000 = $531,713 Unit cost = $531,713/30,000 = $17.72 (rounded off to 2 decimal points) Option 2 is a better choice.

SOLUTION

6.35) A large state university, currently facing a severe parking shortage on its campus, is considering constructing parking decks off campus. A shuttle service composed of minibuses could pick up students at the off-campus parking deck and quickly transport them to various locations on campus. The university would charge a small fee for each shuttle ride, and the students could be quickly and economically transported to their classes. The funds raised by the shuttle would be used to pay for minibuses, which cost about $150,000 each. Each minibus has a 12-year service life, with an estimated salvage value of $3,000. To operate each minibus, the following additional expenses must be considered: If student pay 10 cents for each ride, determine the annual ridership (i.e., the number of shuttle rides per year) required to justify the shuttle project, assuming an interest rate of 6%. ITEM ANNUAL EXPENSES Driver$40,000 Maintenance$7,000 Insurance$2,000 SOLUTION

HOMEWORK PROBLEMS DUE DATE IS Saturday, May 7 th, 2011 class time 10, 21, 27, 30,43