1. Annual Equivalent Worth Criterion
Engineering Economy

2. 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

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

4. 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%)?

5. 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 : .

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

7. 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

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

9. First Cycle:
PW(10%) = -$1,000 + $500 (P/F, 10%, 1)
$400 (P/F, 10%, 5)
= $1,155.68
AE(10%) = $1, (A/P, 10%, 5) = $304.78
Both Cycles:
PW(10%) = $1, $1, (P/F, 10%, 5)
= $1,873.27
AE(10%) = $1, (A/P, 10%,10) = $304.78

10. 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

11. 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 N
CR(i)
Remember from Ch2 that:
(A/F, i, N)= (A/P, i, N)-i

12. 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

13. 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

14. 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

15. The Cost of Owning the Rest of a Vehicle

16. Example

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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

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

24. 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

25. 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)

26. 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

27. 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

28. 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?

29. Solution: Determine total input power Conventional motor:
(a) Operating cost per kWh per unit
Determine total input power
Conventional motor:
input power = kW/ = kW
PE motor:
input power = kW/ 0.93 = kW

30. Determine total kWh per year with 3120 hours of operation
Conventional motor:
(3120 hrs/yr)* ( kW) = 65,018 kWh/yr
PE motor:
( 3120 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

31. 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

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

33. $2000
$12,500
$5,000
$5,500
$6,000
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

34. $15,000
$4,000
$4,500
$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

35. We choose Model A since it has less cost although it has shorter life-spans
$4,000
$4,000
$5,000
$5,000
$5,500
$5,500
$12,500
$12,500
$4,000
$4,000
$5,000
$5,000
$5,500
$5,500
$12,500
$12,500
Model A
$4,000
$4,000
$5,000
$5,000
$5,500
$5,500
$12,500
$12,500
$4,000
$4,000
$5,000
$5,000
$5,500
$5,500
$12,500
$12,500
$4,000
$4,000
$4,000
$4,000
$4,500
$4,500
$5,000
$5,000
$4,000
$4,000
$15,000
$15,000
$4,000
$4,000
$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

36. 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:

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

38. 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: x10-5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

39. 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

40. 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)

41. 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

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

44. Optimal Cross-Sectional Area
Figure: 06-09

45. 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.

46. 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.