1.
Input Analysis
SUBJECT :- Modelling Simulation and Analysis :TE504
FACULTY GUIDE :- Prof. Dr. L.B. Zala
PREPARED BY:-
Mansi C. Rupani (17TS808)
Bhavik A. Shah (17TS809)
CIVIL ENGG. DEPARTMENT
BIRLA VISHVAKARMA MAHAVIDYALAYA ENGG. COLLEGE
VALLABH VIDYANAGAR-388120
M.TECH - TRANSPORTATION ENGINEERING
K. Salah

2.
2
Driving simulation models
Stochastic simulation models use random variables to represent inputs such as
inter-arrival times, service times, probabilities of storms, proportion of ATM
customers making a deposit.
We need to know the distribution family and parameters of each of these
random variables.
Some methods to do this:
•Collect real data, and feed this data to the simulation model. This is called trace-
driven simulation.
•Collect real data, build an empirical distribution of the data, and sample from
this distribution in the simulation model.
•Collect real data, fit a theoretical distribution to the data, and sample from that
distribution in the simulation model.
We will examine the last two of these methods.

3.
3
Why Is This an Issue?
Service Times
(in Days)
0
50
100
150
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
Days
Number
The above graph shows service times (LOS) for a hospital in Ontario. How would you
model a patient’s length of stay?
N = 546

4.
4
Option 1: Trace
Patient LOS
1 4
2 1
3 7
4 6
5 31
6 1
7 4
8 2
9 6
10 5
11 1
12 1
13 6
14 1
15 2
We could attempt a trace. In this scheme, we
would hold this list of numbers in a file.
When we generate the first patient, we would
assign him or her an LOS of 4; the 2nd
1; the 3rd
7;
…
Traces have the advantage of being simple and
reproducing the observed behaviour exactly.
Traces don’t allow us to generate values outside
of our sample.
Our sample is usually of a limited size, meaning
that we may not have observed the system in all
states.

5.
5
Option 2: Empirical Distribution
LOS Count Frequency Cum. Freq
0 0 0.000 0.000
1 139 0.255 0.255
2 120 0.220 0.474
3 69 0.126 0.601
4 48 0.088 0.689
5 30 0.055 0.744
6 25 0.046 0.789
7 21 0.038 0.828
8 12 0.022 0.850
9 11 0.020 0.870
10 8 0.015 0.885
11 9 0.016 0.901
12 10 0.018 0.919
13 7 0.013 0.932
14 0 0.000 0.932
15 1 0.002 0.934
16 2 0.004 0.938
17 4 0.007 0.945
18 2 0.004 0.949
19 3 0.005 0.954
20 12 0.022 0.976
21 5 0.009 0.985
22 0 0.000 0.985
23 2 0.004 0.989
24 0 0.000 0.989
25 2 0.004 0.993
26 0 0.000 0.993
27 1 0.002 0.995
28 2 0.004 0.998
29 1 0.002 1.000
30 0 0.000 1.000
546
A second idea might be to use an empirical
distribution to model LOS.
We will use the LOS and cumulative frequency as
an input to our model.
For instance, assume we pick a random number
(say .62) as F(x). This corresponds to x of
between 3 and 4 (~3.3 by interpolation).
The LOS will represent x, and the cumulative
frequency represents F(x).
Empirical distributions may however, be based on
a small sample size and may have irregularities.
The empirical distribution cannot generate an x
value outside of the range [lowest observed value,
highest observed value].

6.
6
Option 3: Fitted Distribution
31 intervals of w idth 1 betw een 1 and 31 1 - Logarithmic Series
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
Frequency-Comparison Plot
Interval Midpoint
Proportion
1.50 5.50 9.50 13.50 17.50 21.50 25.50 29.50

7.
7
Why Theoretical Distributions?
• Theoretical distributions “smooth out” the irregularities
that may be present in trace and empirical distributions.
• Gives the simulation the ability to generate wider range of
values.
– Test extreme conditions.
• There may be a compelling reason to use a theoretical
distribution.
• Theoretical distributions are a compact way to represent
very large datasets. Easy to change, very practical.

8.
8
Describing Distributions
A probability distribution is described by its family and its parameters. The
choice of family is usually made by examining the density function, because this
tends to have a unique shape for each distribution family.
Distribution parameters are of three types:
location parameter(s) γ
scale parameter(s) β
shape parameter(s) α
x
f(x)
γ1 γ3γ2
x
f(x)
β 1
β 2
β 3
x
α 1
α 2
α 3
f(x)

9.
9
Examples: continuous distributions
x
f(x)
ba
Uniform distribution (a is the location parameter; (b-a) is the shape parameter)
ab −
1




≤≤
−=
otherwise
bxa
abxf
0
1
)(
0
( )
0
x a
x a
F x a x b
b a
x b
<
 −
= ≤ ≤
−
 >
[ ]
( )
2
: ,
:
2
:
12
Range a b
a b
Mean
b a
Variance
+
−
Uses: 1st
model in which only a, b are known.
Essential for generation of other
distributions.

10.
10
Examples: continuous distributions
x
f(x)
Exponential distribution (one scale parameter)
β
1
2
0
01)(
0
0
1
)(
σµβ
β
β
β
==





≥−=





≥=
−
−
otherwise
xexF
otherwise
xexf
x
x
[ ]
2
: 0,
:
:
Range
Mean
Variance
β
β
∞
Uses: Inter-arrival times.
Notes: Special case of Weibull and Gamma (α =
1 , β = β)
If X1, X2, …, Xm are independent expo(β)
then X1+ X2+ …+ Xm is distributed as an
m-Erlang or gamma(m, β)

11.
11
Examples: continuous distributions ...
Gamma distribution (one scale parameter,one shape parameter)
( )
( )
( )1
0
1
!
0( )
0
1 0
0
Note: F(x) has no closed form solution if is not a positive integer.
j
j
x
x
x
j
x e
xf x
otherwise
F x e x
otherwise
α
α α β
β
β
β
α
α
−
=
−
− −
−

 >=  Γ



∑
=  − >


x
α=1
α=2
α=3
f(x) β=1
[ ]
( ) ( )
2
1
0
: 0,
:
:
Γ α = α-1 ! if αisa positiveinteger
otherwiset
Range
Mean
Variance
t e dtα
αβ
αβ
∞
− −
∞
∫
Uses: Task completion time.
Notes: For positive integer (m) gamma (m, β) is an m-erlang.
If X1, X2, …, Xn are independent gamma(αi,β) then X1+
X2+ …+ Xn is distributed as an gamma (α1 + α2 +…+ αn,
β)

12.
12
Examples: continuous distributions ...
Weibull distribution (one scale parameter,one shape parameter)
( )
1
0( )
0
1 0
0
x
x
x e xf x
otherwise
e xF x
otherwise
α
α
βα α
β
αβ
 
− 
− −  
 
− 
 

 >= 


 − >= 

x
α=1
α=2
f(x) β=1
[ ]
22
: 0,
1
:
2 1 1
: 2
Range
Mean
Variance
β
α α
β
α α α α
∞
 
Γ 
 
      
Γ − Γ     
      
Uses: Task completion time; equipment failure time
Notes: The expo(β) and the Weibull(1, β) are the same
distribution.

13.
13
Examples: continuous distributions ...
Normal distribution (one scale parameter,one shape parameter)
( )
( )
2
2
2
1
( )
2
.
x
f x e
F x No closed form
µ
σ
πσ
−
−
= 

=
x
f(x)
[ ]
2 2
: ,
:
:
Range
Mean or
Variance a or
β µ
σ
−∞ ∞
Uses: Errors from a set point. Quantities that are the sum of a
large number of other quantities
Notes:

14.
14
Examples: discrete distribution
Bernoulli distribution
1 0
( ) 1
0
0 0
( ) 1 0 1
1 1
p if x
f x p if x
otherwise
if x
F x p if x
if x
− =

= =


<

= − ≤ <
 ≥
x
p(x)
p
1-p
0 1
{ }
( )
: 0,1
:
: 1
Range
Mean p
Variance p p−
Uses: Outcome of an experiment that either succeeds or fails.
Notes: If X1, X2, …, Xt are independent Bernoulli trials, then
X1+X2+ … + Xt is binomially distributed with
parameters (t, p).

15.
15
Examples: discrete distribution
Binomial Distribution
( ) { }
( )
0
1 if x 1,2,...,
( )
0 otherwise
0 0
( ) 1 0
1
t px
x
t ii
i
t
p p t
f x x
if x
t
F x p p if x t
x
if x t
−
−
=
 
− ∈ =  


<

 
= − ≤ ≤  
 
 >
∑
x
p(x)
p
0 1
{ }
( )
: 0,1,...,
:
: 1
Range t
Mean tp
Variance tp p−
Uses: Number of defectives in a batch of size t.
Notes: If X1, X2, …, Xm are independent and distributed
bin(ti,p), then X1+X2+ … + Xm is binomially distributed
with parameters (t1 + t2+ …+ tm, p).
The binomial distribution is symmetric only if p = 0.5.
2 3 4 5
t = 5
p = 0.5

16.
16
• Poisson
• Pareto

17.
17
Selecting an Input Distribution - Family
• The 1st
step in any input distribution fit procedure is to hypothesize a
family (i.e. exponential).
• Prior knowledge about the distribution and its use in the simulation
can provide useful clues.
– Normal shouldn’t be used for service times, since negative values can be
returned.
• Mostly, we will use heuristics to settle on a distribution family.

18.
Summary Statistics
Function Summary Statistic Dist
Type
Comments
Minimum, Maximum X(1), X(n) C, D Estimates range
Mean µ
( )X n C, D Central tendency
Median x0.5
( )
1
2
1
2 2
ˆ
1
2
n
n n
X if n is odd
x n
X X if n is even
+ 
 
 
   
+   
   



=  
 + 
   
C, D Central tendency
Variance σ2
( )2
s n C, D Measure of variability
Coeffient of Variation
(cv) =σ/µ ( )
( )
2
s n
X n
C Measure of variability
cv = 1 implies exponential
If cv > 1, then lognormal is a
better model than gamma or
weibull.
Lexus ratio
τ= σ2
/µ
( )
( )
2
s n
X n
D Measure of variability
τ= 1 implies Poisson
τ< 1 implies Binomial
τ> 1 implies Negative Binomial
Skewness
v =
( )
( )
3
2
3
2
E X µ
σ
−
( )
( )( )
( )( )
3
2
3
1
2
1
ˆ
n
i
i
X X n
n
v n
S n
=
−
=
∑
C, D Measure of symmetry
v = 0 for normal distributions
v > 0 for right skewed distribution
v < - for left skewed distributions
17

19.
19
Summary Example
Consider the following sample
CV suggests NOT
exponential
Conclusion: Gamma? Weibull? Beta?
Sample Value Sample Value
1 0.0065 16 0.0826
2 0.0118 17 0.0964
3 0.0135 18 0.1010
4 0.0241 19 0.1083 Min: 0.0065
5 0.0281 20 0.1258 Max: 0.2714
6 0.0285 21 0.1319 Mean: 0.1035
7 0.0347 22 0.1458 Median: 0.0739
8 0.0372 23 0.1884 Variance: 0.0066
9 0.0380 24 0.1930 CV: 0.7875
10 0.0428 25 0.1958 Lexus: 0.0642
11 0.0496 26 0.1985 Skewness: 0.7190
12 0.0521 27 0.2017
13 0.0544 28 0.2517
14 0.0650 29 0.2616
15 0.0652 30 0.2714
Skew suggests a left skewed family.
Continuous data

20.
20
Draw a Histogram
Histogram
0
2
4
6
8
10
12
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60
Value
Count
• Law & Kelton suggest
drawing a histogram.
• Use the plot to “eyeball”
the family.
• Law and Kelton suggest
trying several plots with
varying bar width.
• Pick a bar width such that
the plot isn’t too boxy, nor
too scraggly.

21.
21
Histogram Example – I
Histogram (Bin Width = .025)
0
1
2
3
4
5
6
7
8
0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.450
0.500
0.550
0.600
Value
Count

22.
22
Histogram Example – II
Histogram (Bin Width = .05)
0
2
4
6
8
10
12
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60
Value
Count

23.
23
Histogram Example – III
Histogram (Bin Width = .1)
0
2
4
6
8
10
12
14
16
18
0.00 0.10 0.20 0.30 0.40 0.50 0.60
Value
Count

24.
24
Sturge’s Rule
Histogram (Bin Width = .06)
0
2
4
6
8
10
12
14
0.060 0.120 0.180 0.240 0.300
Value
Count
Select k (# bins): Int(1 + 3.322log10n), where n = number of samples.
We will guess at a gamma distribution.

25.
25
Parameter Estimation
• To estimate parameters for our distribution, we use the
Maximum Likelihood Estimators (MLE’s) for the selected
distribution.
• For a gamma we’ll use the following approximation.
Calculate T:
( )( ) ( )
1
1 1
2.64
1 ln(.1035) ( 2.6534)
ln ln
n
i
i
T
X n X
n =
= = =
− −
− ∑
Use table 6.20 to obtain α. α = 1.464
Calculate B: ( ) .1035ˆ .069
ˆ 1.464
X n
β
α
= = =

26.
26
A Note on MLEs
• Maximum likelihood
estimator (MLE)
• Method for determining
parameters of a
hypothesized distribution.
• We assume that we have
collected n IID
observations.
( ) ( ) ( ) ( )1 2 ... nL p X p X p Xθ θ θθ =
We define the likelihood function:
The MLE is simply the value of theta
that maximizes L(θ) over all values of
θ.
In simple terms, we want to pick an
MLE that will give us a good estimate
of the underlying parameter of
interest.

27.
27
MLE for an Exponential Distribution
( ) ( ) ( ) ( )1 2 ... nL p X p X p Xθ θ θθ =
1 2
1
1
1 1 1...
n
n
i
i
X X X
X
n
e e e
e
L β β β
β
β β β
β
β
                    
     
=
− − −
−
−
 
 
 
− − −
∑
=
=
We want to estimate ß:
So, the problem is to maximize
the rhs of the above equation. To
make things simpler we’ll
maximize ln(L(ß))
1
1ln
n
i
i
XMax n β
β
=
 
 
 
−− ∑
Take the 1st
derivative and set =
0. Solve for ß
( )
2
2
1
1
1
. .
1
1
1
0
n
i
i
n
i
i
n
i
i
i e X
X
X
X
n
n
n
β
β
β
β
β
=
=
=
+
=
=
= −
∑
∑
∑
In general the MLEs are difficult to
calculate for most distributions. See
Chapter 6 of Law and Kelton.

28.
28
Determining the “Goodness” of the Model
• As you might imagine, determine whether our hypothesize model is
“good” is fraught with difficulties.
• Law and Kelton suggest both heuristic and analytical tests to
determine “goodness”.
• Heuristic tests:
– Density/Histogram over plots.
– Frequency comparisons.
– Distribution function difference plots.
– Probability-Probability Plots (P-P plots).
• Analytical tests:
– Chi-Squared tests.
– Kolmogorov-Smirnov tests

29.
29
Frequency Comparisons
Frequency Comparison
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.06 0.12 0.18 0.24 0.3
x
hj,rj(%)
Data Dist
F(bj-1) - F(bj)
x Data Dist
0 0 0
0.060 0.433 0.385
0.120 0.633 0.688
0.180 0.767 0.850
0.240 0.900 0.931
0.300 1.000 0.968
Cum. Distn Functions

30.
30
Distribution Function Difference Plot
Plot the CDF of the hypothesized distribution – the CDF as observed in the data.
If the fit is perfect, this plot should be a straight line at 0.
L&K suggest the difference should be less than .10 for all points.
Distribution Function Difference Plot
-0.2
-0.1
0
0.1
0.2
0 0.1 0.2 0.3 0.4 0.5
x
F'(x)-Fn(x)

31.
31
P-P Plot
P-P Plot
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
Fn(x) - Data
F(X)-Distribution
Note: We are plotting F(x) vs. F

32.
32
Q-Q Plot
L&K note that Q-Q plots tend to emphasize errors in the tail.
Our fitted distribution doesn’t look appropriate in the upper tail.
Q Q plot
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500
X(i)s
F-1((i-.5)/n)

33.
33
Analytic Test – Chi Square Test
1. Divide the range into k adjacent intervals.
2. Nj = # of Xi’s in jth interval.
3. Determine the proportion of Xi’s that should fit into the jth
interval. For continuous data equal-probability approach is
recommended. Pj’s are set to be equal values
For continuous R.V.;
( )
( )
1
1
ˆ
ˆ ˆ
j
j j
a
a
j j
n np
n f x dx
n F a F a
−
−
  
  
  
=
=
= −
∫
For discrete R.V.; Pj; Total probability that the random var.
will take values in jth interval = P(X=x| x>=aj-1, x<=aj)

34.
34
Analytic Test – Chi Squared Test
1. The critical test statistic is:
( )
2
2
0 i
1
i
where O is the observed num of samples in theith interval
E is theobserved num of samples in theith interval
j
i i
i i
O E
E=
−
Χ =∑
2. A number of authors suggest that the
intervals be grouped such that Ei is >= 5
in all cases.
3. H0: X conforms to the assumed
distribution.
H1: X does not conform to the assumed
distribution.
Reject if X0
2
> X2
k-1,1-α

35.
35
Example
k=5 Pj=1/k=.2
Cumulative a(j) s Observed Frequency Exp. Frequency (O-E)^2/E
0.2 0.033599 6 6 0
0.4 0.063509 7 6 0.166667
0.6 0.101155 5 6 0.166667
0.8 0.160816 4 6 0.666667
0.999999 infinity 8 6 0.666667
SUM 1.666667
X critical (5-1,.95) 9.48 accepted

36.
36
Example 6.15
T.2Tablefrom204.27.1for
.
042.)20/21ln(399.0
020.)20/11ln(399.0
19,..,2,1)20/1ln(399.0
issatisfy thmuste-1/)(ˆ
on)distributiunboundedisl(Expoentia,0
ok)arewe5(95.10219*05.0
interval.eachfor
05.0/1sointervals20wantWe20Set
0.399withlexponentiaisondistributiFitted
6.7)Tableinis(Data219
01.01,120
2
1
399.0
-
200
==
=−−=
=−−=
=−−=
==
∞==
>===
===
=
=
−−XueCritcalval
a
a
jja
akjaF
aa
np
kpk
n
j
j
a
j
j
j
j
α
β

37.
37
Example 6.15
aj’s
Observed Expected
k
# of
intervals
Less than the
critical value

38.
38
Example 6.16
T.2Tablefrom605.4
0.1forvalueCritical
6.13in tablegivenintervals3withupendWe
valuean thisgreater thbemusts'
dist.)fittedtheof(Mode0.3460)P(x
intervalsintodatathegroupingbyequalroughlys'themaketry toWe
intervalsyprobabilitequalexactlymakenotcanWe
eom(0.346)isondistributiFitted
size)Demand6.9Tableinis(Data156
01.01,13 =
=
==
=
−−X
p
p
g
n
j
j
α

39.
39
Example 6.16
Determine the intervals
so that Pj’s are more or less
equal
Less than the
critical value

40.
40
P value
Alpha
Critical valueTest statistic
calculated
Chi-square density with k-1 d.f.
P value; Cumulative Prb. to the right of
the test stat.
if P value > alpha Don’t reject Ho

41.
41
The Kolomogorov-Smirnov (K-S) Test
The K-S test compares an empirical distribution
function(F^(x)) with a hypothesized distribution function
(Fn(x)).
The K-S test is somewhat stronger than the Chi-Square test.
This test doesn’t require that we aggregate any of our samples
into a minimum batch size.
We define a test statistic Dn:
( ) ( )
^
supn n
x
D F x F x
 
= − 
 
The largest value over all x

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K-S Example
Let’s say we had collected 10 samples (0.09, 0.23, 0.24, 0.26,
0.36, 0.38, 0.55, 0.62, 0.65, and 0.76) and have developed an
emprical cdf.
We want to test the hypothesis that our sample is drawn from a
U(0,1) distribution
f(x) = 1 0 <= x <= 1
F(x) = x 0 <= x <= 1
Where
H0 : Fn(x) = F^
(x)
H1 : Fn(x) <> F^
(x)

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K-S Example
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2
F^
(x)
Fn(x)

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K-S Example
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2
Dn is simply the largest of the gaps
between F^
(x) and Fn(x). Remember – we
need to find the Dn
-
and Dn
+
at every
discontinuity

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K-S Example
Sample x F^(x)
(Fit
Distribution)
Fn(x)
(Empirical)
Dn
-
Dn
+
0 0 0 0
1 0.09 0.09 0.1 0.09 0.01
2 0.23 0.23 0.2 0.13 0.03
3 0.24 0.24 0.3 0.04 0.06
4 0.26 0.26 0.4 0.04 0.14
5 0.36 0.36 0.5 0.04 0.14
6 0.38 0.38 0.6 0.12 0.22
7 0.55 0.55 0.7 0.05 0.15
8 0.62 0.62 0.8 0.08 0.18
9 0.65 0.65 0.9 0.15 0.25
10 0.76 0.76 1.0 0.14 0.24
Our Dn is the largest of the Dn
-
, Dn
+
columns. In this case Dn = 0.25.

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K-S Test
• The value of Dnwhen appropriately adjusted
(see next slide), is found to be < the critical
point (1.224) for α = 0.10.
• Thus we cannot reject H0.

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Adjusted Critical Values
• Law and Kelton present
critical values for the K-S
tables that are non-
standard.
• They use a compressed
table based on work from
Stevens (1962).
Case Mulitply
Dn by
All parameters in
F^(x) known
Normal Distn
Expo Distn
0.11
0.12 nn D
n
 
+ + ÷
 
0.85
0.01 nn D
n
 
− + ÷
 
0.2 0.5
0.26nD n
n n
  − + + ÷ ÷
  
0.11
0.12
0.11
10 0.12 .25 (3.32)(.25) 0.83
10
nn D
n
 
+ + 
 
 
+ + = = 
 
For tables of critical values see
pgs 365-367 of L&K

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THANK YOU
FOr BeAriNg.
Mansi C. Rupani (17TS808)
Bhavik A. Shah (17TS809)