Economic Analysis in Transportation Systems

Tapan K. Datta, Ph.D., P.E.

CE 7640: Fall 2002

Chapter 8- Depreciation Concepts

Decreasing of the value of the machine, property, etc.

Replacement is higher than the amount of depreciation of the real estate.

Income generating property

Depreciation

Depreciation: Decrease in value. Also called prepaid cost allocated to

current operational expense.

Methods of Allocating Depreciation Expense

A = Present Worth at i % B = Straight Line MethodC = Multiple Straight Line MethodD = Sum of the Year Digit MethodE = Declining Balance Method

AB

C

DE

X2

X1

Terminal Value

Init

ial C

ost

B = Dep. base including terminal value

Bd = Depreciable base = B - T

Bx = Unallocated portion of Dep. Base at

age X

T = Terminal value

B - Bx = Total depreciation at age X

D = Annual depreciation allocation

Methods of Allocating Depreciation Expense

Dx = Accumulated Dep. Allocation at age X

X= Age of the property

n = Probable service life

f = Depreciation rate per year

Methods of Allocating Depreciation Expense

Present Worth MethodVariant of the sinking fund method In terms of depreciation symbols and

say rate of return ‘r’ is analogous to ‘i’ then the following can be written:

B = Bd + T

(1+r)n - 1[B = A

r(1+r)n+] T

(1+r)n

Present Worth MethodSolving for A gives:

r(1+r)n

[A = Bd

(1+r)n - 1

+] rT

The present value at any age x, i.e. the depreciated value, would be:

Bx = A T(1+r)n-x

(1+r)n-x - 1 r(1+r)n-x[ ] +

Substituting the value of A,

(1+r)n-x - 1 r(1+r)n-x[ ] r(1+r)n

Bx = Bd rT[ ]+ (1+r)n - 1

T (1+r)n-x

+

Present Worth Method

Capital Recovery Factor =

(1+r)n - (1 + r)x

Bx = BdT[ ] +

(1+r)n - 1

(1+r)n - 1

r(1+r)n

Straight Line MethodConstant rate applied to constant base

Annual Dep. Allocation = D = (B -T)/n = Bd/n

Total allocation at age X is Dx = Bd (X/n)The unallocated base at age X is

Bx = B - Dx = B - Bd (X/n)

= (Bd + T) - Bd (X/n)

= Bd (1 - (X/n)) + T

Bx = Bd + T(n - X) n

Declining Balance Method

Depreciation Method is applied to the remaining balance.

For example: 12% per year allocation on $100 is:

Year Dep. Amount Rem. Balance

1 $ 12 $ 88.00

2 88*.12 = $ 10.56 $ 77..44

3 77.44 *.12 = 9.29 $ 68.15

4 68.15 * .12 = 8.18 $59.97

:

Declining Balance Method (Continued)

Unallocated portion of the base is

Bx = B (1 - f)x

T = B (1 - f)n

f = 1 - T/Bn

Sum of the Years Digits Method

Total Dep. = Sum of yearly digits over the service life

Suppose, Service life = 10 yearsSum of the digits (SOD) = 1 +2 + 3 + …+10 = 55

Or, SOD = N(N+1) 2

1st year depr. = 10/55= 0.1818182nd year depr. = 9/55 = 0.1636363rd year depr. = 8/55 = 0.1454544th year depr. = 7/55 = 0.1272735th year depr. = 6/55 = 0.1090916th year depr. = 5/55 = 0.0909097th year depr. = 4/55 = 0.0727278th year depr. = 3/55 = 0.0545459th year depr. = 2/55 = 0.03636410th year depr. = 1/55 = 0.018181

Sum of the Years Digits Method

A $ 10,000 property will be depreciated as follows:

Year Amount ($) 1 1818.18 2 1636.36 3 1454.54 4 1272.73 5 1090.91 6 909.09 7 727.27 8 545.45 9 363.6310 181.81

Sum of the Years Digits Method

Sinking Fund Method

Annual year end deposit to a fund to accumulate to “F” in ‘n’ years at ‘i’ interest rate is:

A = F i

(1+i)n - 1

A = (B - T)

i (1+i)n - 1

F = A (1+i)n - 1 i

Sinking Fund Depreciation Method

In terms of depreciation symbols

Dx = A

(1+i)x - 1

i

Dx = (B - T)

i (1+i)n - 1 [ ]

(1+i)x - 1

i[ ]

Dx = (B - T) (1+i)n - 1 [ ]

(1+i)x - 1

Dx = Bd

(1+i)n - 1 [ ]

(1+i)x - 1

Sinking Fund Depreciation Method

Bx = (Bd + T) - Bd

Bx = Bd

(1+i)n - (1+i)x (1+i)n - 1

(1+i)n - 1 [ ]

(1+i)x - 1

+ T[ ]

Service Life: of a physical property is that period of time extending from date of installation to the date of retirement from service. It is the actual measured usage whether profitable or not

Physical Life: that period of time the property exists, not necessary in a usable condition.

Service Life of Physical Property

Economic Life: of a physical property is that period of time extending from date of installation into service to that date when the property is no longer profitable to use.

Service Life of Physical Property

Service Life

0 10 20 30 40

Useable Life

Economic Life

Retirement Date

35

Demolish

Time

Factors Influencing Retirement of a Property

1. Physical wear and tear

2. Traffic Accidents – can cause premature retirement

3. Natural Catastrophes – Flood, Earthquakes, Hurricane, etc.

(Look for the history)

General Obsolescence – Unfit for current use, or technology advancement.

Factors Influencing Retirement of a Property

(Continued)

Retirement

Retirement decisions are a management decision

For example, when a highway no longer renders useful and satisfactory service may be abandoned, re-used elsewhere,

completely removed or rebuilt

No universal rules

Concrete pavement collect data from different resources to estimate the service life

Traffic Signal Collect data – come up with complete life cycle then find the

average life Traffic Signs

The face can deteriorate because of sun, weather or accidents Collect data on when signs where placed and when they were

removed, and calculate the average service life

Retirement

Determining average service life Compares over a period of years accumulated

installations or units of service with accumulated retirements

The average service life (turnover period) is given by that period of time it takes to accumulate future retirements which total the accumulated number of units in service at the beginning date.

Turnover Method

The average service life is the horizontal distance between the two curves starting at any chosen time on either curve

When the installations of units or retirements are not known from the beginning, the average service life may be determined by plotting the accumulated retirements curve backward to where it intersects the curve of units in service

Method does not result in a survivor curve

Turnover Method

Fig 9-1. Accumulated placement and retirement curves plotted for determining average life by the turnover method.

Turnover Method does not result in a survivor curve because ages of the retirement are not used

Turnover Method

Turnover Method

Accumulated # Installed

# In

stal

led

or

Ret

ired

1965 1970Time

Accumulated # Retired

Average Service Life

1966 1967 19691968

1906 1966

Average Life

Accumulated # Constructed

# in Service

Accumulated # Retired

4334 59

# C

onst

ruct

ed o

r R

etir

ed

Turnover MethodWhen number of units of retirements are not known: Calculate backwards

Steps Involved

From individual year’s No. of Installations or Retirements, calculate No. of Cumulative Installed and Cumulative Retired

Draw a graph with X-axis having Years and Y-axis having Numbers Installed or Numbers Retired

From the graph, find the Average Life at any point of time

Example

No. No. Cumulative

Year Installed Retired Installed Retired

1 500 0 500 0

2 500 0 500+500=1000 0

3 1500 1000 1000+1500=2500 0+1000=1000

4 1500 1500 2500+1500=4000 1000+1500=2500

5 1000 1500 4000+1000=5000 2500+1500=4000

Fig 9-1. Accumulated placement and retirement curves plotted for determining average life by the turnover method.

Installation and Retirement Curves

0

1000

2000

3000

4000

5000

6000

1 2 3 4 5

Year

Cum

. Num

ber I

nsta

lled

or R

etire

d

Year

Cum Installed

Cum RetiredAverage Life = 1 yr

Actuarial Method This method evolves around yearly

additions and placements, the property units in service at specific dates and retirements.

Objectives1. To determine average service life of

property2. To determine the dispersion of ages at

retirement

Fig 9-2. A typical survivor curve and its derived curves

Example

YEAR

NUMBER IN

SERVICENUMBER ADDED

NUMBER RETIRED

PERCENT IN SERVICE

1950 0 10 - 1001955 25 15 0 1001960 30 10 5 30/35 = 861965 39 10 1 39/45 = 871970 45 10 4 45/55 = 821975 55 20 10 55/75 = 731980 60 20 15 60/95 = 631985 50 10 20 50/105 = 471990 30 0 20 30/105 = 291995 10 0 20 10/105 = 102000 1 0 9 1/105 = 1

0102030405060708090

100

0 10 20 30 40 50

Age in Years

% S

urv

ived

Survivor Curve

Survivor Curve

Shows the number of units of property that survive in service at given ages

Area under this curve is a direct measure of service life of the property units

The probable life of the serving units at any age can also be calculated from the remaining area by dividing the remaining area by the amount surviving at that age.

Individual Units Methods

Data frequently shows only the number of units retired during a given year or series of years together with the age of each unit at its retirement

If these retirements are arranged in order of their ages and then summed from the oldest to the youngest, a survivor curve can be plotted.

Survivor Curve Calculated by the Individual Units Method

Individual Units MethodNo. of units retired during a year or years.

With age of each unit.

No. of years of life ‘N’

Frequency‘f’

Cumulative

N*fNumbers Percent

%

25

10152025303540

26

152520151052

28

234868839398100

989077523217720

430

15037540037530017580

Total 100 1889

Mean = M = 1889/100 = 18.89 years

0102030405060708090

100

0 5 10 15 20 25 30 35 40

Age in Years

% S

urv

ived

Mean = 18.89 = Avg. Service LifeSurvivor Curve

Frequency Curve

98%

52%

32%

90%

77%

7%2%

0%

17%

Per

cen

t

Age (Years)

Probable Life Curve

Survivor Curve

Curve shows percent of units continued in service to any given age

Average life = Total service in unit yearsTotal units retired

Original–Group Method

When the number of units placed in service at a given date is known, together with the number of those units remaining in service at successive later observation dates, a survivor curve can be constructed covering the experience of this original group over the years for which the data are compiled.

Survivor curve calculated by the original-group method

Perc

ent S

urvi

ving

Age, Years