INDR 202 - Chapter 6

INDR  202

ENGINEERING  ECONOMICS

CHAPTER  6

ANNUAL-­EQUIVALENCE  ANALYSIS

SPRING  2015

INSTRUCTOR:  BORA  ÇEKYAY

ANNUAL-­EQUIVALENCE  ANALYSIS

2

Annual-­Equivalent  Worth

Annual-­Worth  Analysis

ANNUAL-­EQUIVALENT  WORTH

3

𝐴𝐸 𝑖 = 𝑃𝑊 𝑖 𝐴|𝑃, 𝑖, 𝑁

Single project:

Accept if

_{𝐴𝐸 𝑖 > 0}

, reject if

_{𝐴𝐸 𝑖 < 0}

,

remain indifferent otherwise.

Mutually exclusive revenue projects:

Select the project

with largest

_{𝐴𝐸 𝑖}

.

EXAMPLE  1:  ANNUAL-­EQUIVALENT  WORTH

4

$189.43

0

$46.07

0 1      2      3       4       5       6

𝑖 = 12%

$100 $50

$80 $120 _$70

0

2      3      4      5      6 1

$120 $120

$189.43

0

EXAMPLE  2:  ANNUAL-­EQUIVALENT  WORTH

5

$500

$700 $800

$400  $400

$800

$1,000 $1,000

$500

EXAMPLE  2:  ANNUAL-­EQUIVALENT  WORTH

REPEATING  CYCLE

ONE CYCLE

𝑃𝑊 10% = $1,155.68

𝐴𝐸 10% = $304.87

TWO CYCLES

𝑃𝑊 10% = $1,873.27

ANNUAL-­EQUIVALENT  COST

7

CAPITAL

COSTS

OPERATING

COSTS

+

Annua

l-­E

qu

iv

al

en

t  

Co

ANNUAL-­EQUIVALENT  COST

8

CAPITAL-­RECOVERY  COST

_{𝑪𝑹 𝒊}

annual equivalent of capital (ownership) costs from

purchasing assets used in production & service –

typically nonrecurring (one-­time) costs

EQUIVALENT  ANNUAL  OPERATING  COSTS

CAPITAL-­RECOVERY  COST

9

Components  of  capital  cost:  

initial  cost  

_𝐼

&

salvage  value

_𝑆

0      1        2        3      N

𝐶𝑅 𝑖

0

N

𝐼

𝑆

𝐶𝑅 𝑖 = 𝐼 𝐴|𝑃, 𝑖, 𝑁 − 𝑆 𝐴|𝐹, 𝑖, 𝑁

EXAMPLE  3:  CAPITAL-­RECOVERY  COST

10

0      1        2        3        4       5

𝐶𝑅 20%

0

5

𝐼 = $200,000

𝑆 = $50,000

𝐶𝑅 20% = $150,000 𝐴|𝑃, 20%, 5 + $50,000 20%

𝐶𝑅 20% = $60,157

EXAMPLE  4:  COST  OF  OWNING  A  VEHICLE

11

%  VALUE  RETAINED  AFTER

Price 2  YRS 3  YRS 4  YRS 5  YRS

2010  BMW  M3 $58,400 70% 60% 51% 43%

2010  Hyundai-­

EXAMPLE  4:  COST  OF  OWNING  A  VEHICLE

12

𝐶𝑅 6% = $33,288 𝐴|𝑃, 6%, 5 + $25,112 6%

𝐶𝑅 6% = $9,409

$58,400

$25,112

EXAMPLE  4:  COST  OF  OWNING  A  VEHICLE

13

𝐶𝑅 6%

Price 2  YRS 3  YRS 4  YRS 5  YRS

2010  BMW  M3 $58,400 $12,009 $10,842 $10,045 $9,409

2010  Hyundai-­

Accent  GLS $14,365 $5,046 $4,020 $3,357 $2,951

ANNUAL-­WORTH  ANALYSIS

14

ANALYSIS  OF  LIFE-­CYCLE COST

CALCULATION  OF  UNIT  COST  (PROFIT)

EXAMPLE  5:  JUSTIFYING  AN  INVESTMENT

15

𝐼 = $20,000

,  

𝐴 = $4,400

,  

𝑆 = $4,000

,  

𝑁 = 5

,  

𝑖 = 10%

$20,000 0

1       2      3      4      5 $4,400      $4,400    $4,400    $4,400

$8,400

𝑃𝑊 10% = −$836.88

EXAMPLE  5:  JUSTIFYING  AN  INVESTMENT

16

$20,000 0

1       2      3      4      5

$4,000

CAPITAL  COSTS

0 1       2      3      4      5

$4,400    $4,400  $4,400  $4,400  $4,400

ANNUAL  REVENUES

EXAMPLE  6:  WORTH  PER  UNIT  TIME

1 2 3

Yearly  Operating  Hours

Year  1 Year  2 Year  3 2,000 2,000 2,000

𝑃𝑊 15% = $3,553

𝐴𝐸 15% = $3,553 𝐴|𝑃, 15%, 3 = $1,556

Savings  per  machine  hour:

F,GGG

= $0.78/ℎ𝑟

EXAMPLE  6:  WORTH  PER  UNIT  TIME

18

$75,000

$24,400 $27,340

$55,760

0

1 2 3

Yearly  Operating  Hours

Year  1 Year  2 Year  3 1,500 2,500 2,000

𝐴𝐸 15% = $1,556

𝐶

: annual-­equivalent  savings  per  machine  hour

$1,556

= [1,500𝐶 𝑃|𝐹, 15%, 1 + 2,500𝐶 𝑃|𝐹, 15%, 2 + 2,000𝐶 𝑃|𝐹, 15%, 3 ] 𝐴|𝑃, 15%, 3

EXAMPLE  7:  Break-­even Unit Revenue

19

YEAR  1 YEAR  2 YEAR  3 Depreciation $2,879 $1,776 $1,545 Scheduled  maintenance 100 153 220

Insurance 635 635 635 Registration  & 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  & taxes 688 650 522

Oil 80 100 100

Parking  & tolls 135 125 110

EXAMPLE  7:  Break-­even Unit Revenue

20

Total  costs $4,680 $3,624 $3,421 Expected  miles  driven 14,500 13,000 11,500

𝐴𝐸𝐶 6% = 𝑃𝑊 6% 𝐴|𝑃, 6%, 3 = $3,933/𝑦𝑟

𝑋

: revenue  per  mile

Present-­worth  of  total  revenue:

1,000𝑋 14.5 𝑃|𝐹, 6%, 1 + 13 𝑃|𝐹, 6%, 2

+ $11.5 𝑃|𝐹, 6%, 3

𝐴𝐸 6% = $13,058𝑋/𝑦𝑟

Break-­even:  

_{$3,933/𝑦𝑟 = $13,058𝑋/𝑦𝑟}

EXAMPLE  7:  Break-­even Unit Revenue

21

Annual-­equivalent cost  of   owning  & operating

LOSS

GAIN

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

Revenue  rate  per  mile  ($)  

MAKE-­OR-­BUY  DECISIONS

22

OUTSOURCING DECISIONS

TO  BUY

TO  PRODUCE

CAPITAL  

MAKE-­OR-­BUY  DECISIONS

23

Identify  planning  horizon  &  required  annual  quantity

.

Determine  unit  cost  of  outsourcing.

Compute  unit  cost  of  producing.

a. Identify  production  costs  (capital  & resources). b.Estimate  net  cash  flows  over  planning  horizon. c. Compute  annual  equivalent  cost  of  producing.

d.Divide  annual  equivalent  cost  by  required  annual  quantity.

EXAMPLE  8:  MAKE-­OR-­BUY  DECISIONS

24

A  manufacturer  needs  120K  units  of  a  part  every  year  for  5  years.

Producing  the  part  requires  specialized  tools  that  cost  $2.2M  &  have   salvage  value  $120K  at  the  end  of  5  years.  The  unit  production  cost  

after  buying  the  equipment  is  $26.30.

A  supplier  offers  to  provide  the  part  for  $35/unit  for  annual  orders  over   100K  units.  

EXAMPLE  8:  MAKE-­OR-­BUY  DECISIONS

25

Annual-­equivalent cost of buying:

𝐴𝐸𝐶 12% = $35 120,000 = $4.2𝑀/𝑦𝑟

Annual-­equivalent cost of producing:

𝐶𝑅 12% = $2.2𝑀 − $120𝐾 𝐴|𝑃, 12%, 5 + $120𝐾 12%

𝐶𝑅 12% = $591,412

𝑂𝐶 12% = $26.30 120,000 = $3,156,000

𝐴𝐸𝐶 12% = $3,747,412

EXAMPLE  8:  MAKE-­OR-­BUY  DECISIONS

26

Unit cost of buying:

$35/𝑢𝑛𝑖𝑡

Unit cost of producing:

𝐴𝐸𝐶 12% = $3,747,412

$3,747,412

120,000

= $31.23/𝑢𝑛𝑖𝑡

Producing  the  part  saves  the  company  $3.77  per  unit.

EXAMPLE  9:  PROJECTS  WITH  EQUAL  LIVES

27

Standard  Motor

Premium  Motor

Size

25  HP

25  HP

Cost

$13,000

$15,600

Life

20  years

20  years

Salvage

$0

$0

Output

18.65  kW

18.65  kW

Efficiency

89.5%

93%

Energy  cost

$0.07/kWh

$0.07/kWh

EXAMPLE  9:  PROJECTS  WITH  EQUAL  LIVES

28

Standard  Motor

Premium  Motor

Cost

$13,000

$15,600

𝐶𝑅 13%

$1,851

$2,221

Output

18.65  kW

18.65  kW

Efficiency

89.5%

93%

Input

20.838  kW

20.054  kW

Operating  hours

3,120

3,120

Total input

65,018/yr

62,568/yr

Energy  cost

$0.07/kWh

$0.07/kWh

EXAMPLE  9:  PROJECTS  WITH  EQUAL  LIVES

29

Standard  Motor

Premium  Motor

𝐶𝑅 13%

$1,851

$2,221

Annual  operating  cost

$4,551

$4,380

𝐴𝐸𝐶 13%

$6,402

$6,601

Cost  per  kWh

$0.1100

$0.1134

EXAMPLE  9:  PROJECTS  WITH  EQUAL  LIVES

30

LIFE-­CYCLE-­COST  ANALYSIS

Switching from standard to premium motor:

Incremental capital cost:

_{$2,221 − $1,851 = $370}

Incremental energy savings:

_{$4,551 − $4,380 = $171}

A sum of 3,120 annual operating hours implies a

LOSS

EXAMPLE  9:  PROJECTS  WITH  EQUAL  LIVES

31

𝑋

: annual  operating  hours

Incremental  capital  cost:

_{$2,221 − $1,851 = $370}

Incremental  energy  savings:

reduction  in

Output

Efficiency

Operating

Hours

Energy

Cost

20.838

𝑋 0.07 − 20.054

𝑋 0.07 = $0.05488𝑋

Break-­even:

_{$0.05488𝑋 = $370}

EXAMPLE  9:  PROJECTS  WITH  EQUAL  LIVES

32

Standard

PROJECTS  WITH  UNEQUAL  LIVES

33

Annual-­worth analysis offers computational advantages

over present-­worth analysis when:

Ø

required service period is indefinite and

Ø

each alternative is replaced by identical asset with

same costs & performance.

EXAMPLE  10:  PROJECTS  WITH  UNEQUAL  LIVES

34

Project  A

0   1 2 3

$12,500

$5,000 $5,000 $3,000

Project  B

0   1 2 3      4

$15,000

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

EXAMPLE  10:  PROJECTS  WITH  UNEQUAL  LIVES

35

Project  A

0   1 2 3

$12,500

$5,000 $5,000 $3,000

FIRST  CYCLE:

𝑃𝑊 15% = −$22,601

𝐴𝐸 15% = −$22,601 𝐴|𝑃, 15%, 3 =

−$9,899

4  REPLACEMENT  CYCLES:

𝑃𝑊 15% = −$53,657

EXAMPLE  10:  PROJECTS  WITH  UNEQUAL  LIVES

36

FIRST  CYCLE:

𝑃𝑊 15% = −$25,562

𝐴𝐸 15% = −$25,562 𝐴|𝑃, 15%, 4 =

−$8,954

3 REPLACEMENT  CYCLES:

𝑃𝑊 15% = −$48,534

𝐴𝐸 15% = −$48,534 𝐴|𝑃, 15%, 12 =

−$8,954

Project  B

0   1 2 3      4

$15,000

EXAMPLE  10:  PROJECTS  WITH  UNEQUAL  LIVES

SUMMARY

38

Annual-­worth analysis

is based on annual-­equivalent worth

(cost).

𝐴𝐸 𝑖 = 𝑃𝑊 𝑖 𝐴|𝑃, 𝑖, 𝑁

Capital-­recovery cost

is the annual-­equivalent cost of capital.

𝐶𝑅 𝑖 = 𝐼 − 𝑆 𝐴|𝑃, 𝑖, 𝑁 + 𝑖𝑆

Advantages of annual-­worth analysis:

Ø Annual-­equivalent worth is more common in financial reports. Ø Pricing items requires calculating unit costs.