Chapter 6

Annual Equivalence Analysis

„ Annual equivalent
criterion
„ Applying annual
worth analysis
„ Mutually exclusive
projects

1
 Annual Worth Analysis
Principle: Measure investment worth on annual
basis

Benefit:
• Annual reports, yearly budgets
• Seek consistency of report format
• Determine unit cost (or unit profit)
• Facilitate unequal project life
comparison
2
Annual Equivalent Worth
AE(i)=PW(i)*(A|P,i,N)

If AE(i) > 0 , accept

AE(i) < 0 , reject
AE(i) = 0 , remain indifferent

Since (A|P,i,N) > 0 for –1<i<∞

AE(i) > 0 PW(i) > 0
3
 Computing Equivalent Annual Worth

$120
$80 $70
0 1
2 3 4 5 6
$50
$189.43 $100
A = $46.07

0 1 2 3 4 5 6
0

PW(12%) = $189.43 AE(12%) = $189.43(A/P, 12%, 6)

= $46.07

4
Annual Equivalent Worth - Repeating
Cash Flow Cycles
$800 $800
$700 $700
$500
$400 $400 $500 $400 $400

$1,000 $1,000

Repeating cycle

5
• 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

6
Annual Equivalent Cost
„ When only costs are

Annual Equivalent Costs

involved, the AE Capital
method is called the costs
annual equivalent +
cost.
Operating
„ Revenues must costs
cover two kinds of
costs: Operating
costs and capital
costs.
7
 Capital and Operating Costs
„ Capital costs are incurred by purchasing assets to
be used in production and service. Normally, they are
nonrecurring (one-time) costs.
„ Operating costs are incurred by the operation of
physical plant or equipment needed to provide
service (e.g. labor and raw materials). Normally, they
recur for as long as an asset is owned.
„ Operating costs are on annual basis anyway. Annual
equivalent of a capital cost is called
capital recovery cost, CR(i).

Remember: (A|P, i, N) is called the capital recovery factor.

8
 Capital (Ownership) Costs
Def: The cost of owning an S
equipment is associated with 0
two transactions—(1) its initial N
cost (I) and (2) its salvage value
(S).
I
„ Capital costs: Taking into these
sums, we calculate the capital
0 1 2 3 N
costs as:
CR(i) = I( A / P, i, N) − S( A / F, i, N)
= (I − S)( A / P, i, N) + iS CR(i)

since (A|F,i,N)=(A|P,i,N)-i
9
Example - Capital Cost
Calculation
$50,000
„ Given:
I = $200,000 0
N = 5 years 5
S = $50,000
i = 20%
„ Find: CR(20%) $200,000

CR (i ) = ( I - S ) ( A / P, i, N ) + iS
CR ( 20%) = ($200,000 - $50,000 ) ( A / P, 20%, 5)
+ (0.20)$50,000
= $60,157
Annual cost of owning the asset at 20% 10
 Justifying an investment based on AE
Method

Given: I = $20,000,
S = $4,000, N = 5
years, i = 10%
Find: see if an annual
revenue of $4,400 is
enough to cover the
capital costs.
Solution:
CR(10%) = $4,620.76
Conclusion: Need an
additional annual
revenue in the amount
of $220.76. 11
Example - Capital Cost
Calculation for Mini Cooper
$12,078
„ Given:
I = $19,800 0
N = 3 years 3
S = $12,078
i = 6%
„ Find: CR(6%) $19,800
CR (i ) = (I -S ) (A/P , i , N ) + iS
CR (6%) = ($19,800 - $12,078) (A /P , 6%, 3)
+ (0.06)$12,078
= $3,613.55

12
Applying Annual Worth
Analysis

•Unit Cost (Profit) Calculation

• Unequal Service Life Comparison

• Minimum Cost Analysis

13
Equivalent Worth per Unit of Time
$55,760
$24,400 $27,340
0

1 2 3
$75,000
Operating Hours per Year
2,000 hrs. 2,000 hrs. 2,000 hrs.
• PW (15%) = $3553
• AE (15%) = $3,553 (A/P, 15%, 3)
= $1,556
Note: 3553/6000=0.59/hour:
• Savings per Machine Hour instant savings in present
= $1,556/2,000 worth for each hourly use;
does not consider the time
= $0.78/hr.
over which the savings occur 14
 Equivalent Worth per Unit of Time
(cont’d)
$55,760
$24,400 $27,340
0

1 2 3
$75,000
Operating Hours per Year
1,500 hrs. 2,500 hrs. 2,000 hrs.

•Let C denote the equivalent annual savings per machine hour

• $1,556=[(C)(1500)(P|F,15%,1)
+(C)(2500)(P|F,15%,2)
+(C)(2000) (P|F,15%,3)] (A|P,15%,3)
C=$0.79/hr

15
 Breakeven Analysis
Problem: Year Miles Total costs
(n) Driven (Ownership
& Operating)
At i = 6%, what
1 14,500 $4,680
should be the
2 13,000 $3,624
reimbursement
3 11,500 $3,421
rate per mile so
that Sam can
break even? Total 39,000 $11,725

16
 Breakeven Analysis
First Year Second Year Third Year
Depreciation $2,879 $1,776 $1,545
Scheduled maintenance 100 153 220
Insurance 635 635 635
Registration and taxes 78 57 50
Total ownership cost $3,693 $2,621 $2,450
Nonscheduled repairs 35 85 200
Replacement tires 35 30 27
Accessories 15 13 12
Gasoline and taxes 688 650 522
Oil 80 100 100
Parking and tolls 135 125 110
Total operating cost $988 $1,003 $971
Total of all costs $4,680 $3,624 $3,421
Expected miles driven 14,500 miles 13,000 miles 11,500 miles 17
• Equivalent annual cost of owning and operating the car
[$4,680 (P/F, 6%, 1) + $3,624 (P/F, 6%, 2) +
$3,421 (P/F, 6%, 3)] (A/P, 6%,3)
= $3,933 per year

• Equivalent annual reimbursement

Let X = reimbursement rate per mile
[14,500X(P/F, 6%, 1) + 13,000X(P/F, 6%, 2)
+ 11,500 X (P/F, 6%, 3)] (A/P, 6%,3)
= 13.058X
• Break-even value
13.058X = 3,933
18
X = 30.12 cents per mile
 Annual equivalent reimbursement as a
function of cost per mile
Annual equivalent
cost of owning and
operating ($3,933)
Gain
Annual equivalent ($)

4000

3000 Loss Minimum

reimbursement
8X
13,05 requirement ($0.3012)
2000
Annual reimbursement
1000 amount

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

Reimbursement rate ($) per mile (X)

19
 Mutually Exclusive Alternatives
with Equal Project Lives
Standard Premium
Motor Efficient Motor
Size 25 HP 25 HP
Cost $13,000 $15,600
Life 20 Years 20 Years
Salvage $0 $0
Output 18.65 kW per motor 18.65 kW per motor
Efficiency 89.5% 93%
Energy Cost $0.07/kWh $0.07/kWh
Operating Hours 3,120 hrs/yr. 3,120 hrs/yr.
(a) At i= 13%, determine the operating cost per kWh for each motor.
(b) At what operating hours are they equivalent? 20
 Solution:
(a) Operating cost per kWh per unit

output power
Input power =
% efficiency

Determine total input power

Conventional motor:

input power = 18.65 kW/ 0.895 = 20.838kW

PE motor:

input power = 18.65 kW/ 0.930 = 20.054kW

21
 Solution:
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 22
 Solution:
„ 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 = $0.1100/kwh
(where 58188=18.65*3120 is the total output power per year)
PE motor:
AE(13%) = $4,380 + $2,221 = $6,601
Cost per kwh = $6,601/58,188 kwh = $0.1134/kwh 23
 Solution:

„ Savings from switching from conventional to PE

motors?
„ Incremental capital cost required=2221-1851=370
„ Incremental energy savings=4551-4380=171
„ Hence, at 3120 annual operating hours, a loss of 199
for each motor.

24
Solution:

(b) Break-even
operating
hours = 6,742

25
Mutually Exclusive Alternatives with
Unequal Project Lives

1. Required service period = Indefinite

2. Each alternative replaced by identical asset

with same costs and performance

Then,

We may solve for AE of each project based on

its initial life span rather than on LCM of
project lives.

26
 Mutually Exclusive Alternatives with
Unequal Project Lives
Model A: Required service
0 1 2 3
Period = Indefinite
$3,000
$5,000 $5,000 Analysis period =
$12,500 LCM (3,4) = 12 years
Model B: (Least common
multiple)
0 1 2 3 4

$2,500
$4,000 $4,000 $4,000
$15,000

27
 Model A:
0 1 2 3

$3,000
$5,000 $5,000
$12,500
• 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 28
 Model B:
0 1 2 3 4

$2,500

$4,000 $4,000 $4,000

$15,000
• 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
29
IF:
1. The service of the selected alternative required on a continuous basis
2. Each alternative replaced by identical asset with same costs and
performance

30
 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.

31
 Summary
„ 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.

32
 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. 33
Example 1:
Susan wants to buy a car that she will keep for
the next four years. She can buy a Honda Civic
at 15000 and then sell it for 8000 after four
years. If she bought this car, what would be her
annual ownership cost (capital recovery cost)?
Assume that her interest rate is 6%.

34
CR(6%) = 15000( A | P, 6%, 4) − 8000( A | F , 6%, 4) = 2500.2
or
(15000− 8000) ( A | P, 6%, 4) + (0.06)(8000) = 2500.2

35
Example 2:
The following cash flows represent the potential annual
savings associated with two different types of production
processes, each of which requires an investment of
12000. Assuming an interest rate of 15%, complete the
following tasks: (a) Determine the equivalent annual
savings for each process. (b) Determine the hourly
savings for each process assuming 2000 hours of
operation per year. (c) Determine which process should
be selected.

n A B
0 -12000 -12000
1 9120 6350
2 6840 6350
3 4560 6350
4 2280 6350
36
(a)
⎛ 9120 6840 4560 2280 ⎞
AE A (15%) = ⎜ − 12000 + + + + 4 ⎟
( A | P,15%, 4) = 1893.14
⎝ 1.15 1.15 1.15 1.15 ⎠
2 3

AEB (15%) = ( − 12000 + 6350( P | A,15%, 4) ) ( A | P,15%, 4) = 2147.08

(b)
1893.14
Process A : = 0.95 / hour
2000
2147.08
Process B : = 1.07 / hour
2000

(c) Select B.

37
Example 3:
Norton Auto Parts, Inc. is considering two different forklift
trucks for use in its assembly plant.
Truck A costs 15000 and requires 3000 annually in
operating expenses. It will have a 5000 salvage
value at the end of its 3-year service life.
Truck B costs 20000 and requires 2000 annually in
operating expenses. It will have a 8000 salvage
value at the end of its 4-year service life.
The firm’s MARR is 12%. Assuming that the trucks are
needed for 12 years and no significant changes are
expected in the future price and functional capacity of
both trucks, select the most economical truck based on
AE analysis.

38
 Capital Recovery Cost

AE A (12%) = (15000 − 5000) ( A | P,12%, 3) + (0.12)(5000) + 3000 = 7763

AEB (12%) = (20000 − 8000) ( A | P,12%, 4) + (0.12)(8000) + 2000 = 6910.4

Truck B is more economical.

39
Example 4:
Your company needs a small front loader for
handling bulk materials at the plant. It can be
leased from the dealer for three years for $4050
per year including all maintenance. It can also
be purchased for $14,000. You expect the
loader to last for six years and to have a salvage
value of $3000. You predict that maintenance
will cost $400 the first year and increase by $200
per year in each year after the first. Your MARR
is 15%. Use annual equivalence analysis to
determine whether to lease or buy the loader.
What is the shortest project life for which the
annual worths you have calculated are exactly
correct?
40
 Present worth

AE (15%) = (14000 + 400 ( P | A,15%, 6) + 200 ( P | G,15%, 6) − 3000 ( P | F ,15%, 6) ) ( A | P,15%, 6)

= 4175.46

LEASE.

Exactly correct for 6 years (or an integer multiple of 6)

41
Example 5:
Consider the following cash flows for two types of models. Both
models will have no salvage value upon their disposal (at the end of
their respective service lives). The firm’s MARR is known to be 15%.
(a) Notice that models have different service lives. However, model A
will be available in the future with the same cash flows. Model B is
available now only. If you select model B now, you will have to
replace it with model A at the end of year 2. If your firm uses the
present worth as a decision criterion, which model should be selected
assuming that your firm will need either model for an indefinite
period? (b) Suppose that your firm will need either model for only two
years. Determine the salvage value of model A at the end of year two
that makes both models indifferent (equally likely).

Project cash flow

n A B
0 -6000 -15000
1 3500 10000
2 3500 10000
3 3500 42
Option 1 : Model A now and Model A forever
Option 2 : Model B now and Model A forever

For one cycle :

PWA (15%) = −6000 + 3500 ( P | A,15%, 3) = 1991.2
PWB (15%) = −15000 + 10000 ( P | A,15%, 2) = 1257
AE A (15%) = 1991.2 ( A | P,15%, 3) = 872.15
AEB (15%) = 1257 ( A | P,15%, 2) = 773.18

(a)
872.15
PWOpt1 (15%) = = 5814.33
0.15
872.15
PWOpt 2 (15%) = 1257 + ( P | F ,15%, 2) = 5653.47
0.15

(b)
− 6000 + 3500 ( P | A,15%, 2) + S ( P | F ,15%, 2) = 1257
S = 2072.54 43