economic analysis in transportation systems tapan k. datta, ph.d., p.e. ce 7640: fall 2002
Post on 20-Dec-2015
220 views
TRANSCRIPT
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