650
URBAN TRAN/CE SYSTEMS
ENGINEERING
Lecture 8
STOCHASTIC PROCESSES AND
QUEUING THEORY
Review of Probability Theory
Event a specified outcome of a random phenomenon
Simple event: single possible outcome
3 heads from 3 coins HHH
Compound event:
2 or more simple events
2 heads w/ 3 coins
HHT HTH THH
Sample point simple event
STOCHASTIC PROCESSES AND
QUEUING THEORY (contd)
Sample space collection of all possible sample points
3 coins S={HHH, HHT, HTH, THH, TTH, HTT, THT, TTT}
Venn diagram
1
2
3
4
5
6
S
E1
Complimentary event for every E there exists a complimentary event E which
consists of all sample points in S which are not in E .
Probability Axioms
NONNEGATIVITY: the probability of any event in S is greater than or equal to zero,
and less than or equal to one.
0P(Ei)1
ADDITIVITY:
EiS
If two events Ei and Ej are both in S but have no sample points in
common, the probability that at least one of the events occurs is the
sum of P(Ei) and P(Ej).
P(Ei or Ej) = P(EiEj) = P(Ei) + P(Ej)
That is, Ei and Ej are mutually exclusive EiEj = .
Ej
Ej
Probability Axioms (contd)
COMPLETENESS OF THE SAMPLE SPACE:
P(S) = 1
P ( ) = 0
P(AB) = P(A) + P(B) - P(AB)
If A = {E1, E2, ..., En} and EiEj =
P(A) =
P( E )
i
i
Marginal Probabilities
unconditional probabilities that do not depend on other events
e.g.,
P(x = 5 on a die) = 1/6
However, P(x = 5x 3) = 1/5.
CONDITIONAL PROBABILITIES:
P(AB) = probability of A given B.
P( A B) =
P( A B)
P( B)
P(B)>0
B
A
AB
P(AB) = probability of falling in ///// given you are in B.
Marginal Probabilities (contd)
Example:
Thesis on the success of OR/MS projects vs. formalization of OR/MS
procedures.
100 companies surveyed
70 report success
50 report formalization
A Company reports success
B Company reports failure
C Company has formalized OR/MS procedures
D Company has not formalized OR/MS procedures
From above: P(A) = 0.70
P(D) = 0.50
How many reported failure? P(B) = 0.30?
Marginal Probabilities (contd)
Fill in the table below:
Success A
Failure B
Marg. Prob.
Formalized
C
.40
.10
.50
Not Formalized
D
.30
.20
.50
Also given:
P(AC) = 0.40
P(AC) = P(AC)/P(C) = 0.40/0.50 = 0.80
P(AD) = P(AD)/P(D) = 0.30/0.50 = 0.60
Marg. Prob.
.70
.30
Marginal Probabilities (contd)
NOTE:
if two events are mutually exclusive
P(AB) = P() = 0
P(AB) = P(AB)/P(B) = P(BA).
ADDITIVE LAW:
P(AB) = P(A) + P(B) - P(AB)
B
A
AB
Marginal Probabilities (contd)
MULTIPLICATIVE LAW:
P(AB) = P(A)*P(BA)
or
= P(B)*P(AB)
EXAMPLE:
60% read Morning Bugle
50% read Afternoon Bugle
10% read both
P(AB) = P(A) + P(B) - P(AB)
= 60 + 50 - 10 = 100%
Independence
A and B are independent eve nts if
P(AB) = P(A) or P(BA) = P(B)
2 coins
P(AB) = P(A)*P(BA) = P(A)*P(B)
Note: Mutually exclusive events are not independent.
P(AB) = 0 P(A)*P(B)
P(AB) = 0 P(A) or P(B)
Example:
P( 5 heads) =
11111
1
=
2 2 2 2 2 32
Independence (contd)
Example:
A = first card is a club
B = second card is a club
P(AB) = P(A)*P(BA)
13 12 156
=
=
5.9%
52 51 2652
If replaced?
2
13
P(AB) = P(A)*P(B) = 6.3%
52
BAYES THEOREM
P( A i B) =
(
P( A i ) P B A i
)
P( A ) P( B A )
n
j=1
j
j
where A j are mutually exclusive events.
Intuition:
P(AB) = P(B)*P(AB) = P(A)*P(BA)
P( A ) P( B A )
P( A B)
P(AB) =
=
P( B)
P( B)
Since A and A are mutually exclusive events
P(B) P(AB) + P( A B)
= P(BA)*P(A) + P(B A )*P( A )
P(AB) =
P( A ) P( B A )
P( B A ) P( A ) + P( B A ) P( A )
BAYES THEOREM (contd)
Example:
100 bins with 10 chips each
T ype I:
T ype II:
5B+5W
8B+2W
(70%)
(30%)
Bin picked at random and asked whether Type I or II
1 chip drawn and is black
Define events:
A Type I
C Type II
B black chip
W white chip
BAYES THEOREM (contd)
P(AB) =
P( A ) P( B A )
P( A ) P( B A ) + P( C) P( B C)
P(BA) = 50%
P(BC) = 80%
P(A) = 70%
P(C) = 30%
P(AB) =
5070
= 59.3%
5070 + 3080
P(CB) =
3080
= 1- P(AB) = 40.7%
5070 + 3080
PROBABILITY DISTRIBUTIONS
Random Variable a function whose numeral value depends on the outcome of
some experiment.
PROBABILITY DISTRIBUTION:
relates a probability to the values that a random variable can take
on.
DISCRETE DISTRIBUTIONS
- random variables are discrete.
4 coins
X= # of heads
{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH,
TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT,
TTTH, TTTT}
PROBABILITY DISTRIBUTIONS
(contd)
24=16
X
0
1
2
3
4
5
Probability
1/16
4/16
6/16
4/16
1/16
0
CDF = Cumulative Probability Distribution
CDF = P(Xx) =
P( x )
xi x
i
CDF
1/16
5/16
11/16
15/16
16/16 = 1
1
BINOMIAL DISTRIBUTION
Bernoulli trial:
a random phenomenon involving two mutually exclusive and
exhaustive events
P probability of success
(1-P) probability of failure
Binomial:
[
n independent Bernoulli trials
]
P( x ) = n C x p x q n x
x number of successes
n number of trials
p probability of success
q (1 p) probability of failure
n C x number of combinations of n t aken x at a time
n!/ [ x !( n x ) !]
m! = m ( m ( 1) m 2) 1
0! = 1
BINOMIAL DISTRIBUTION (contd)
Example:
20 Light bulb shipment, 10% historically defective
x = # of defects
p = 0.1
p( x ) =
0
1
2
3
0.1216
0.2702
0.2852
0.1901
q = 0.9
n = 20
20!
0.1x 0.9 ( 20 x )
x !( 20 x ) !
4
5
6
7
0.0898
0.0319
0.0089
0.0020
8
9
10
11
0.0004
0.0001
0.0000
0.0000
POISSON DISTRIBUTION
Simeon Poisson (1837)
1. During some time interval t , the probability of an occurrence of an event is
constant, regardless of when t starts.
2. The occurrence of an event is independent of any other occurrence of the event.
3. If t is chosen such that the probability of an event within t is small, then the
probability that the event will occur is approximately equal to the width of the
interval.
P( event in t ) = 0.005
P( event in 2 t ) = 0.01
PROBABILITY MASS FUNCTION (PMF)
e k
P( X = k ) =
> 0, k = 0,1,2, . . .
k!
= mean number of events in a time interval
k = # of events
POISSON DISTRIBUTION (contd)
Example: =3 cars/minute @ toll gate
X= arrivals/minute
0
1
2
3
4
5
P(x)
0.0498
0.1494
0.2240
0.1680
0.1008
0.1008
X
6
7
8
9
10
11
P(x)
0.0504
0.0216
0.0081
0.0027
0.0008
0.0002
CONTINUOUS PROBABILITY
DISTRIBUTIONS
- continuous variables
Probability Density Functions (PDF) vs. PMF
f(x)
UNIFORM DISTRIBUTION
x
f(x)
1
( b a) = 1
Area =
ba
1/(b-a)
a
b
x
f ( x) =
1
ba
0
a xb
otherwise
CONTINUOUS PROBABILITY
DISTRIBUTIONS (contd)
CDF (Cumulative Distribution Function)
F ( x ) = P( X x ) = 0
xa
=
ba
=1
x<a
a xb
xb
Expected Value =
xf ( x ) dx =
x
Variance
2
2
= x 2 f ( x ) dx x = x
Uniform:
1
( b + a)
2
13
b a3
2
x = 3
( b a)
x =
(
)
Exponential Distribution
-popular for interarrival and service times
-related to Poisson distribution
e x
f ( x) =
0
x0
x<0
f(x)
x
x = 1
F ( x ) = 1 e x
2
x = 1 2
x0
NORMAL DISTRBUTION
Gaussian distribution
can be used to approximate both the binomial and Poisson distribution.
many random phenomena behave normally
due to the Central Limit Theorem, if x is a variable with x and variance x2 , then the
x2
mean of random samples of size n is N ( x ,
), irrespective of the distribution of the
n
random variable.
( x x ) 2
2
1
e 2 x
- x
2
x x
Z=
N ( 0,1)
x
standard normal (tables)
f ( x) =
NORMAL DISTRBUTION (contd)
Example:
Scores N (450, (50)2)
Z = (x-450)/50
a) Score > 550
Z=(550-450)/50=2
P(x>550) = P(Z>2) = 1-P(Z2) = 1-0.9772 = 2.3%
b) 400 Score 500
Z400= -1
Z500= +1
P(400x500) = P(-1Z1)
= P(Z1) - [1 - P(Z1)]
= 2P(Z1) - 1 = 2*(0.8413) - 1
= 0.6828
DISCRETE RANDOM VARIABLES
A random variable is DISCRETE if it can assume only discrete values, x1 , x2 , ... .
A discrete random variable X is characterized by the fact that we know the probability that X=xi (written as
P(X=xi)).
P(X=xi) probability mass function (pmf) for the random variable X.
The Cumulative Distribution Function F(x) for any random variable X is defined by
F(X)=P(X = xi).
F( x) =
CDF for a discrete random variable X,
P( X = x )
k
all x having
xk x
Example: X the number of dots that show when a die is tossed.
For I =1,2,3,4,5,6,
1
P(X=xi) = 1/6
F(x)
CDF
4/5
3/5
2/5
1/5
X
6 4/5
6 1/5
5 3/5
5
4 2/5
3 4/5
3 1/5
2 3/5
2
1 2/5
4/5
1/5
0
CONTINUOUS RANDOM VARIABLES
if, for some interval, the random variable X can assume all values of the interval, then X
is a CONTINOUS random variable.
Probability density function (pdf)
P(xXx+) f(x)
Area 1 = P(axa+) f(a)
Area 2 = P(bxb+) f(b)
F(x)
1
2
a
b
P( a x b) = f ( x ) dx
a
Cumulative Distribution Function:
F ( a ) = P( x a ) =
a
f ( x) dx
b
x
CONTINUOUS RANDOM VARIABLES
(contd)
Example:
if 0 x 1
find F(x)
otherwise
F(x)
2 x
Given f ( x ) =
0
for a0, F(a) = 0.
a
for 0a1, F (a ) = 2 xdx = a 2
0
for a1, F(a) = 1.
Also,
1
P x
4
[x ]
2 3/ 4
1/ 4
=
3
4
3
= 2 xdx
4 1
4
9
11
=
16 16 2
a2
a
1
x
MEAN AND VARIANCE OF A
RANDOM VARIABLE
MEAN
DISCRETE RANDOM VARIABLE
For a discrete random variable x,
E ( x ) = x k P( X = x k )
all k
CONTINOUS RANDOM VARIABLE
E ( x ) = xf ( x ) dx
VARIANCE
DISCRETE RANDOM VARIABLE
Var X = [ x k
all k
E ( x)
] P( x x )
2
k
CONTINUOUS RANDOM VARIABLE
2
Var X = [ x E ( x) ] f ( x ) dx
Standard Deviation
x = Var X
MEAN AND VARIANCE OF A
RANDOM VARIABLE (contd)
Example:
Consider the discrete random variable x having
P( x = i ) =
1
for i = 1, 2, 3, 4, 5, 6. Find E ( x ) and Var ( x ) .
6
21 7
1
E ( x ) = ( 1 + 2 + 3 + 4 + 5 + 6) =
= = 3.5
6
62
35
1
2
2
2
Var ( x ) = ( 1 3.5) + ( 2 3.5) ++( 6 3.5) =
6
12
[
]
MEAN AND VARIANCE OF A
RANDOM VARIABLE (contd)
Example:
Find the mean and variance for the continuous random variable x having the following
density function:
2 x if 0 x 1
f ( x) =
0
otherwise
2x3 2
E ( x ) = x ( 2 x ) dx =
=
3 3
0
1
1
2
1
2
4
4
Var ( x ) = x 2 xdx = x 2 x + 2 xdx
3
3
9
0
0
1
2 x 4 8x 3 8x 2
1
=
+
= 18
9
18 0
4