Subject Name: Probability and Queueing Theory
UNIT –I :
1.
Subject Code: MA2262 Question Bank 2012 - 2013
RANDOM VARIABLES
PART - A
2
, determine the variance of X.
2t
1
2
3
2
2
t

t t t
Mx  t  

 1    1          
t
2t

 2
2 2 2
2 1  
 2
1  t  1  t2  3  t3 
 1         
2  1!  2  2!  4  3! 
If the r.v has the mgf M x  t  
 tr 
 r  coefficient of   in M X (t),
 r! 
1 1 1
2
Var(X)   2   1     .
2 4 4
2.
1 
Find the mgf of the r.v whose moments are  r 
1
1
,  2 
2
2
(r  1)!
.
2r


 tr 
 t r  (r  1)! 1!  t 0  2!  t1  3!  t 2 
M X (t)     r    
 0   1   2   
2r
2  0!  2  1!  2  2! 
r  0  r! 
r  0  r! 
2
3
t t
t
 1 2    3    4   
2 2
2
3.
A continuous r.v X has the pdf f (x) 
t

 1  
 2
2

4
2  t
2
.
x 2 e x
, x  0 . Find the rth moment of X about the
2
origin.
 r  E  X r  
4.
Let M x  t  



2 x

1
r
r  x e
x
f
(x)
dx

x
dx

e  x x r  2 dx

0  2 

2 0
1
1
   r  3   r  2  !
2
2
1
such that t  1, be the mgf of r.v X. Find the mgf of
1 t
Y = 2X +1.
5.
6.
et
 1 
M Y  t   M 2X 1  t   e t M X  2t   e t 

.
1  t  t 2t 1  2t
Find the mgf of X whose pdf is given by f(x) = 1 - |x| , - 1 < X < 1.

0
1
et  e t  2
tx
tx
M x  t    e f (x) dx   e (1  x) dx   e tx (1  x) dx 
.
t2

1
0
The mean of a Binomial distribution is 20 and standard deviation is 4. Find the parameters of
the distribution.
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Subject Name: Probability and Queueing Theory
 np = 20 and
Subject Code: MA2262 Question Bank 2012 - 2013
npq  4  npq 16  (20)q  16  q 
4
5
1
4 1
 .  np  20  n 100
 100 and
are the parameters.
5
5 5
One percent of jobs arriving at a computer system need to wait until weekends for scheduling,
owing to core-size limitations. Find the probability that among a sample of 200 jobs there are
no jobs that have to wait until weekends.
p = 0.01, n = 200,  = np = 2,
X is the no. of jobs that have to wait
 x
2
x
e 
e (2)
e2 (2)0
P(X  x) 

 P(X  0) 
 e 2  0.1353.
x!
x!
0!
The number of monthly breakdown of a computer is a r.v having a Poisson distribution with
mean equal to 1.8. Find the probability that this computer will function for a month with
only one breakdown.
Mean =  = np = 1.8
e  x e1.8 (1.8) x
e1.8 (1.8)1
P(X  x) 

 P(X  1) 
 0.2975.
x!
x!
1!
Let one copy of a magazine out of 10 copies bears a special prize following Geometric
distribution, Determine its mean and variance.
Sol:
1
9
1
q
p
q
 Mean   10,
Variance  2  90
10
10
p
p
If the probability is 0.05 that a certain kind measuring device will show excessive drift, what
is the probability that the sixth of these measuring devices tested will be the first to show
excessive drift?
p = 0.05, q = 0.95, x =6,
P(X=x) = p qx-1 = (0.05) (0.95)5 = 0.0387.
4
If X is a Uniformly distributed r.v with mean 1 and variance , find P(X<0).
3
p 1  q  1 
7.
8
9.
10.
11.
ab
b  a 4
 1  a  b  2 and variance =
  b  a 4
Mean =
2
12
3
By solving the above eqns. We get a = -1 and b = 3
1
1
f ( x) 
in a  x  b  ,  3  x  3
ba
6
0
0
1
1 0
1
P(X  0)  f (x)dx   dx   x 1  .
4
4
4
1
1
If a r.v ‘X’ is uniformly distributed over (-3,3), then compute P ( | X – 2 | < 2).
1
1
f ( x) 
in a  x  b  ,  3  x  3
ba
6
P ( | X – 2 | < 2) = P ( -2 < X – 2 < 2) = P ( 0 < X < 4)
3
3
1
1 3 1
=  f (x)dx  dx  x 0  .
6
6
2
0
0
2
12.
13. The time required to repair a machine is exponentially distributed with parameter
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
1
 = . What is the conditional probability that a repair takes at 11h given that its direction
2
exceeds 8h?
x
  2x 

 2
e
1 e 
P( X  11 / X > 8) = P(X > 3) =  f ( x) dx  
dx  
 e 1.5  0.2231


2
2  1/ 2
3
3


14. Suppose the length of life of an appliance has an exponential distribution with mean 10 years.
What is the probability that the average life time of a random sample of the appliances is
atleast 10.5?
x
1
1 10
x
  , f (x)  e , x  0  f (x)  e , x  0
10
10
x


1 10
P(X  10.5)  f (x)dx 
e dx  e 1.05  0.3499
10
10.5
10.5
15. Write down the moment generating function of Gamma distribution.
Mx  t  
1
1  t 

, t 1
16. If a r.v X follows Gamma distribution with variance 3, find P( | X| < 1).
 = variance = Var(X) = 3
 e x x 1
 e x x 2
 e x x 2
x

0
x

0
x0



f (x)   ()
  (3)
 2
 0
 0
 0
otherwise
otherwise
otherwise



1
P ( |X| < 1) = P ( -1< X < 1) =   f (x)dx
0
1
1
0
1
1
  f (x)dx   f (x)dx  0 
 5e 1  2
e x x 2
dx

 0.0803.
0 2
2
17. Write the Physical conditions of Binomial distribution.
(a) There are two possible outcomes (b) Probability of success is constant for each and every
trial
(c ) No. of Trials are independent and finite.
18 Find the expected value of the number of times one must throw a die until the outcome one
has occurred 4 times.
r  4, p  1
6 are the parameters of Negative binomial distribution
r
4
E ( x)  Mean  
 24
1
p
6
 
19. If X is a random variable with cdf as F(x), show that the random variable Y = F(x) is
uniformly distributed in (0,1).
Y = F(x), F is a monotonic function. Ry = {0,1}
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
dy d

( F ( x)  f ( x) , as F(x) is cdf of X with the pdf of f(x).
dx dx
1 , 0  y  1
dx
1
1
 f ( x)
 f ( x)

dy
dy
f ( x) 0 , otherwise
dx
20. State Weibull Distribution.
The continuous random variable X has a Weibull distribution, if its density function is
given by

x  1 e x , x  0

f ( x)  
where   0 and   0

, elsewhere
0
21. What are the applications of Weibull distribution?
(i)
It can be applied to reliability problems like time to failure of a component.
(ii)
It provides a close approximation to the distribution of the lifetime of a
component.
(iii) The failure rate is not a constant but is proportional to the rate of change of a
time exponent.
PART B
1
The incidence of an occupational disease in an industry is such that the worker have 25%
change of suffering from it. What is the probability that out of 5 workers, at the most two
contract that disease.
X : number of workers contracting the diseases among 5 workers
Then, X is a binomial variate with parameter n=5 and
p = P[a worker contracts the disease] = 25/100 = 0.25
The probability mass function (p.m.f) is P(x) = 5Cx(0.25)x (0.75)5-x, x=0,1,2,….5
The probability that at the most two workers contract the disease is
P[ X  2]  p (0)  p (1)  p(2)
g ( y )  f ( x)
 5C 0 (0.25) 0 (0.75) 5  5C1 (0.25)1 (0.75) 4  5C 2 (0.25) 2 (0.75) 3
2
3
 0.2373  0.3955  0.2637
 0.8965
In a large consignment of electric lamps, 5% are defective. A random sample of 8 lamps is
taken for inspection. What is the probability that it has one or more defectives?
X: number of defective lamps
Then, X is B(n=8, p=5/100 = 0.05)
The p.m.f. is – p(x) = 8C x (0.05) x (0.95) 8 x , x  0,1,2,...8
P[sample has one or more defectives] = 1-P[no defectives] = 1-p(0)
= 1 - 8C0 (0.05)0(0.95)8 = 1 – 0.6634 = 0.3366
In a Binomial distribution the mean is 6 and the variance is 1.5. Then, find (i) P[X=2] and
(ii) P[X≤2].
Let n and p be the parameters. Then,
Mean = np = 6
Variance = npq = 1.5
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
Variance npq 1.5 1



Mean
np
6
4
Therefore, q = ¼ and p = ¾
Therefore, Mean = n*3/4 = 6
That is, n = 24/3 = 8
The p.m.f is ------- p(x) = 8Cx (3/4)x (1/4)8-x, x=0,1,2,…….8
(i)
P[X=2] = 8C2 (3/4)2 (1/4)6 =252/65536=0.003845
(ii)
P[X≤2] = p(0)+p(1)+p(2) = 8C0 (3/4)0 (1/4)8 + 8C1 (3/4)1 (1/4)7 + 8C2 (3/4)2 (1/4)6
= 277/65536 = 0.004227
4
The number of accidents occurring in a city in a day is a Poisson variate with mean 0.8. Find
the probability that on a randomly selected day. (i)There are no accidents (ii)There are
accidents
Let X: number of accidents per day.
Then, X ~ P(λ=0.8).
The p.m.f. is –
e 0.8 (0.8) x
p ( x) 
, x  0,1,2,3,....
x!
(i)
Probability that a particular day three are no accidents is P[no accidents] =
P[X=0]=p(0)
e 0.8 (0.8) 0
=
 e 0.8  0.449
0!
P[accidents occur] = 1-P[no accidents] = 1-p(0) = 1-0.449 = 0.551
5
The number of persons joining a cinema queue in a minute has Poisson distribution with
parameter 5.8. Find the probability that (i) no one joins the queue in a particular minute
(ii)2 or more persons join the queue in the minute.
Solution:
Let X : number of persons joining the queue in a minute Then, X ~ P(λ=5.8).
e 5.8 (5.8) x
, x  0,1,2,3,....
The p.m.f is p( x) 
x!
(i)
P[no one joints the queue] = P[X=0] = p(0)
e 5.8 (5.8) 0
=
0!
 5 .8
= e =0.003
P[two or more join] = P[ X  2]  1  P[ X  2]  1  { p (0)  p (1)}
 e 5.8 (5.8) 0 e 5.8 (5.8)1 
 5.8
 1 

  1  e 1  5.8
0!
1!


 1  0.003  6.8  1  0.0204  0.9796
6.
2 percent of the fuses manufactured by a firm are expected to be defective, Find the
probability that a box containing 200 fuses contains (i) defective fuses (ii) 3 or more
defective fuse
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Subject Name: Probability and Queueing Theory
7.
8.
Subject Code: MA2262 Question Bank 2012 - 2013
Solutions:
2 percent of the fuses are defective. Therefore, probability that a fuses is defective is
p = 2/100 = 0.02
Let X denote the number of defective fuses in the box of 200 fuses. Then, X is
B(n = 200, p = 0.02)
Here, p is very small and n is very large. Therefore, X can be treated as Poisson variate with
parameter λ=np = 200 * 0.02 = 4.
e 4 4 x
The p.m.f. is ---- p( x) 
, x  0,1,2,3,...
x!
P[box has defective fuses] = 1-P[no defective fuses]
= 1 – p(0) = 1 – e-4[(40)/0! = 1 – 0.0183 = 0.9817
P[3 or more defective fuses] = 1-P[less than 3 defective fuses]
= 1 – [p(0)+p(1) +p(2)] = 1 – e-4[1+4+8] = 1 – 0.0183*13
= 1-0.2379 = 0.7621
The probability that a razor blade manufactured by a firm is defective is 1/500. Blades are
supplied in packets of 5 each. In a lot of 10,000 packets, how many packets would
(i)
be free defective blades?
(ii) contains exactly one defective blade?
(e-0.01=0.99)
Let X be the number of defective blades in a packet of 5 blades. Then, X is B (n = 5, p =
1/500)
Since p is very small and n is sufficiently large, X is treated as Poisson variate with
e 0.01 (0.01) x
parameter λ=np = 5*(1/500) = 0.01, p( x) 
, x  0,1,2,3,...
x!
(i) P[ no defective blades] = p(0) = e-0.01(0.01)0/0! = 0.99
The number of packets which will be free of defective blades is
N * P[no defective blades] = 10000*0.99 = 9900
(ii) P[one defective blade] = p(1) = e-0.01(0.01)1/1! = 0.0099
The number pf packets which will have one defectives blade is
N * P[one defective blade] = 10000*0.0099 = 99
A random variable X has a uniform distribution over (-3, 3). Compute
1
(1) P( X  2)
(2) P X  2
(3) Find k for which P( X  k ) 
3
X is uniformly distributed over (-3,3)
1
 ,
f ( x)   6
0
3 x  3
otherwise
2
2
2
1
5
1 
(1) P ( X  2)   f ( x)dx   dx   x  
6
 6  3 6
3
3
2)
P X  2  P(2  x  2) 
2
1
2
dx 
6
3
2

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3) Given
Subject Name: Probability and Queueing Theory
3
1
Subject Code: MA2262 Question Bank 2012 - 2013
1
 6dx  3  3  k  2  k  1
k
9.
The daily consumption of milk in excess of 20000 litres in a town is approximately
exponentialy distribution with parameter 1/3000. The town has a daily stock of 35,000 litres.
What is the probability that of 2 days selected at random the stock is insufficient for both
days?
If Y denotes daily consumption of milk then X = Y – 2000 follows an exponential
distribution with parameter 1/3000. Then we have f ( x) 
1  x / 3000
e
; x  0.
3000
P(stock insufficient for one day) = P(Y>35000) = P(X+20000 > 350000)
1   x / 3000
e
dx  e 5
= P ( X  15000 ) 

3000 15000
 
5 2
10
Therefore, P(stock insufficient for two days ) = e
e
10. The length of the shower in a tropical island in a rainy season has an exponential distribution
with parameter 2, time being measured in minutes. What is the probability that it will last for
atleast one more minute?
2 x
Let X be the length of the shower, then f ( x)  2 e ; x  0
P(Shower will last for at least one more minute provided shower has lasted for 2 minutes) =

P( X  3 / X  2)  P( X  1) [ by memoryless property].  2 e  2 x dx  e  2  0.1353
1
11. In a factory the daily consumption of electric power in millions of kilowatt hours can be
treated as a random variable having Erlang distribution with parameters  = 1 / 3 and k = 4.
If the power supplied to the factory is 12 million kilowatt hours, what is the probability that
this power supply will be inadequate on any given day?
Let X be the power consumption in millions of kilowatt hours. Then pdf of X is given by
f(x)
4
1
 
x
3  3 3

 f ( x) 
x e ; x0
3!
4
1  1  3 
P(the power supply is inadequate) = P ( X  12)     x e 3 dx = 0.4335
3!12 3 
x
12. In a certain locality the daily consumption of water in millions of litres, can be treated as a
random variable having a gamma distribution with an average of 3 million litres. If the
pumping station of the locality has a daily supply capacity of a millions litres. What is the
probability that this water supply will be inadequate on any given day?
It is given that X be the daily water consumption of the locality in million litres and
X follows gamma distribution with average of 3 million litres.
Therefore Mean =  = 3 million litres.
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
e  x x  1
ex x2
; x  0  f ( x) 
The density function of X is f ( x) 
(  1)!
(3  1)!
P(Water supply of 4 million litres inadequate on a given day)
= P ( X  4)  P (4  X  ) 

ex x2
4
 2!  13 e  0.2376
4
  
,  , find the probability density function of
 2 2
13. If X is uniformly distributed in 
Y = tanX
  
, 
 2 2
Given X is uniformly distributed in 
A random variable X is said to have a continuous uniform distribution over an interval (a,
b) if its probability density function is


 1
,  x 

 1
1 

, a  x  b 


2
2  ,  x 
f ( x)   b  a
  
 
2
2
2
2
0
0
, otherwise 
, otherwise
0
, otherwise
Given y = tanx, tan-1y = x and 
x


2
x

2
, if

 
 
 y  tan      , x   y  tan    
2
2
 2
 2
The pdf of Y is
f(y)  f(x)
dx  1  1
 1  1 
 
 i.e f(y)   
, x
dy    1  y 2
   1  y 2 
1  1 
,    x  
 
i.e. f(y)     1  y 2 

0 , otherwise

14. The life time of a component measured in hours in weibull distributed with parameters
  0.2,   0.5 . Find (a) the mean life time of the component (b) the probability that such
a component will be more than 200 hours.
(a) E ( x)  (0.2) 2 (3)  50 hours

(b) P( X  200) 
 (0.1) x
0.5
e ( 0.2) x dx  0.059
0.5
200
15. In a company 5% defective components are produced. What is the probability that at least 5
components are to be examined in order to get 3 defectives?
Here p = 0.05, q = 0.95
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013

= Required probability = P(X = 5) + P(X = 6) + ……. =
 ( x  1) C2 (0.05) 3 (0.95) x3
x 5
= 1
4
 ( x  1) C2 (0.05) 3 (0.95) x3 = 0.9995
x 3
1.
UNIT –II TWO DIMENSTIONAL RANDOM VARIABLES
PART - A
The joint probability mass function of (X,Y) is P(x,y) = k ( 2 x+3y), X = 0,1,2 and Y = 1,2,3. Find k.
3
We know that
3
 P(x
j 1 j 1
Y/X
0
1
2
i,
yj) 1
1
3k
5k
7k
15k
72k = 1  k 
2.
2
6k
8k
10k
24k
3
9k
11k
13k
33k
18k
24k
30k
72k
1
72
If X and Y are independent continuous random variables, Show that E(Y/X) = E (Y).
y
E(Y/X) 
f y ( y / x) dy
(conditiona l p.d . f )
x
y
  y f y ( y ) dy
(m arg inal probabilit y )
y
 E ( y)
3.
Let X and Y have joint density function f  x, y   2, 0  x  y  1 .
Find the marginal density
function. Find the conditional density function Y given X  x .
Marginal density function of X is given by
fX  x  f  x 

 f  x, y  dy

1
1
  f  x, y  dy  2dy  2  y  x
1
x
x
 2 1  x  ,0  x  1.
Marginal density function of Y is given by
fY  y   f  y  

 f  x, y  dx

y
  2dx  2 y, 0  y  1 .
0
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
Conditional distribution function of Y given X  x is f
4.
5.
 y x   f f x ,xy   2 12 x   1 1 x
.
If Y = -2x + 3, Find the Covariance of(X,Y)
Cov(X,Y) = E(XY) – E(X) E(Y)
= E(X(-2X+3)) – E(X){E(-2X+3)}
= E (-2X2 + 3X) - E(X) {-2XE(X)-3}
= -2E(X2) +3 E( X) + 2 (E(X))2 - 3E(X)
= 2(E(X))2 – 2 E(X2) = -2 var(X)
Write the statement of central limit theorem.
If X is the mean of the random sample of size n taken from a population having the mean  and
the finite variance  2 , then Z 
6.
X 

x
is a random variable whose distribution function
approaches that of the standard normal distribution as n --> ∞
Find the value of k if f(x,y) =k (1-x) (1-y) for 0<X,Y<1 is to be joint density function.
By the property of joint density function, we have
1 1
 f ( x, y)dxdy  1
 k   (1  x)(1  y )dxdy  1
R
0 0
1 1
 k   (1  x  y  xy )dxdy  1
0 0
y
 1
 k  1   y  dy  1
2
2
0 
1
1
 y y2 y2 
 k     1
4 0
2 2
7.
1

x2
x2 y 
 k   x   xy 
 dy  1
2
2 0
0 
1
y
1
 k    y  dy  1
2
2
0
1
1 1 1
 k(   )  1
2 2 4
 k4
If the joint P.d.f of (X,Y) is given by f(x,y) = 24y(1-x), 0 ≤ X,Y ≤ 1 Find E(XY).
1 1
1 1
0 0
0 0
1
1
E ( XY )    xy 4(1  x)dxdy  4   ( xy  x 2 y )dxdy
x 2 y x3 y 1
y y
 4 (

)0 dy  4  (  ) dy
2
3
2 3
0
0
1
 y2 y2 
 4  
6 0
 4
8.
 E ( xy ) 
1
3
Find the variance of X – Y from the following data.  x  5,  y  4,  xy 
1
.
8
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
 2 x  y   2 x   2 y  2Cov( x  y)
5
Cov( x  y)   xy  x  y 
2
5
 2 x  y  25  16  2  36
2
 VAR ( X  Y )  36
9.
Find the covariance of X and Y if the joint probability density function is
f ( x, y )  1, 0  x  1 & 0  y  1
Cov(X,Y) = E(XY) - E(X)E(Y)
1
1
 x2 
1
1
E ( XY )    xydxdy     ydy   ydy 
2 0
20
4
0 0
0 
1 1
1
1
1
1
1
E ( X )   xdx 
, E (Y )   xdx 
2
2
0
0
 Cov( X , Y ) 
10.
1 1
 0
4 4
If the tangent of the angle between the lines of regression y on x and x on y is 0.6 and  x 
1
y .
2
Find the correlation coefficient.
 x y 1  r 2 
tan   2
 x   2 y  r 
 1.5r  1  r 2
11.
 0.6 
 r  2, 0.5
1
 x y
2
1 2
 x  2y
4
2
1
1  r 2  2 1  r 2 

 5

r


 r 
4
1, r  0.5
If joint pdf of the random variable (x,y) is given by f(x,y) = kxy e  ( x
2
y2 )
, x  0, y  0 . Find the
value of k.

WKT

  f (x, y)dxdy  1
   kxye  (x
0 0
 k  xe
0
12.
 y2 )
dxdy  1
0 0


2
 x2

dx  ye
 y2

dy  1
0
k
1
4

k
k
  e  t dt  e  u du  1
20
20
 k4
If X and Y are random Variables, Prove that Cov  X , Y   E  XY   E  X  E Y 
cov  X , Y   E  X  E  X   Y  E Y   
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 11 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
 E  XY  XY  YX  XY 
 E  XY   XE Y   YE  X   XY
 E  XY   XY  XY  XY
13.
14.
Write any two properties of regression coefficients.
1. Correlation coefficient is the geometric mean of regression coefficients
2. If one of the regression coefficients is greater than unity then the other should be less than
unity.
Find the mean of x and y, given two regression lines are x + 6y = 4 and 2x + 3y = -1
 
Since the regression lines pass through x, y , we have
x  6y  4
and
2x  3y  1
Solving we get x  2 and y  1
15.
Let the joint probability density function of random variable X and Y be given by
8xy, 0  y  x  1
f ( x, y )  
. Find the Marginal probability densty of X.
 0 , otherwise .

x
 y2 
 = 4x3
fx(x) = f(x) =  f ( x, y )dy =  8 xydy = 8x 
2
 0

0
16.
17.
18.
x
Find the value of k from the f(x , y) = kxy for x = 1, 2, 3 and y = 1, 2, 3.
y/x
1
1
k
2
2k
3
3k
2
2k
4k
6k
3
3k
6k
9k
36k = 1, k = 1/36
If the following equations are regression lines. Identify the regression equation on y on x :
3x + 4y = 5 and 2x + y =3
Let y on x: 4y = 5-3x, y = 5/4-3/4 x
byx = -3/4
x on y : x = ½(3-y) = 3/2 – ½ y
bxy = -1/2
2
r = bxy byx = -3/4 * -1/2 = 3/8 <1
4y = 5-3x is regression equation y on x
Find the marginal density function of y if the joint probability density function of a
two dimensional random variable (X , Y) is f(x,y) = 2 ; 0<x<1, 0<y<1
1
fy(y) =
 2dx = 2
0
19.
The regression line between two random variable X and Y is given by 3x + y =10,
3x + 4y = 12. Find the correlation coefficient between X and Y
x on y : x = 10/3 – 1/3 y
bxy = -1/3
y on x : y = 12/4 –3/4 x
byx = -3/4
r2 = bxy byx = (–1/3 ) (–3/4) = 1/4 = 0.25.
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Page 12 of 48
Subject Name: Probability and Queueing Theory
20.
Since bxy and byx are both negative r = -0.5
If the joint pdf of (X, Y) is f(x,y) = ¼ , 0<x, y<2, Find p(X + Y  1 )
1 1 y
P(X + Y  1 ) =
1
1
1
1
dxdy = 1/4  ( x)10 y dy = 1/4  (1  y)dy = 1/4(y – y2/2) 0
4
0
0

0 0
P(X + Y ) = 1/8
= 1/4 (1-1/2) = 1/8
01.
Subject Code: MA2262 Question Bank 2012 - 2013
PART - B
The joint probability mass function of (X,Y) is given by
p ( x , y )  k ( 2 x  3 y ), x  0 ,1 ,2 ; y  1, 2, 3. Find all the marginal distributions and conditional
probability distributions. Also find the probability distribution of X + Y and Min(X,Y).
X
0
1
2
1
3k
5k
7k
2
6k
8k
10k
3
9k
11k
13k
Y
To find the value of k:
We know that SUM OF ALL PROBABILITIES = 1
So, 3k + 5k + 7k + 6k + 8k + 10k + 9k + 11k + 13k = 1
 72k = 1
 k
1
72
Marginal distribution of X:
X
0
P(X = x)
18
1
24
72
2
30
72
72
Marginal distribution of Y:
Y
1
P(Y = y)
15
2
24
72
3
33
72
72
Conditional distribution of X given Y = 1
X
0
1
2
P(X = x/Y = 1)
3
15
5
15
7
15
Conditional distribution of X given Y = 2
X
0
1
2
P(X = x/Y = 2)
6
24
8
24
10
24
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
Conditional distribution of X given Y = 3
X
0
1
2
9
P(X = x/Y = 2)
11
33
13
33
33
Conditional distribution of Y given X = 0
Y
1
2
3
3
18
P(Y = y/X = 0)
6
9
18
18
Conditional distribution of Y given X = 1
Y
1
2
3
5
P(Y = y/X = 1)
8
24
11
24
24
Conditional distribution of Y given X = 2
Y
1
2
3
7
P(Y = y/X = 2)
10
30
30
13
30
Probability distribution of Z = X + Y
Overall Outcomes: {(0,1), (0,2), (0,3), (1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}
Values of Z = X + Y
Outcomes
P[Z = X + Y]
1
(0,1)
3/72
2
(0,2), (1,1)
11/72
3
(0,3), (1,2), (2,1)
24/72
4
(1,3), (2,2)
21/72
5
(2,3)
13/72
02.
Probability distribution of Z = Min(X,Y)
Overall Outcomes: {(0,1), (0,2), (0,3), (1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}
Values of Z = Min (X,Y)
Outcomes
P[Z = X + Y]
0
(0,1), (0,2), (0,3)
18/72
1
(1,1), (1,2), (2,1), (1,3)
31/72
2
(2,2), (2,3)
23/72
Given table below shows the joint distribution of (X,Y). Obtain (i) the mean of X & Y
(ii) Standard deviations of X & Y (iii) P(X ≤ 1/Y = 2) (iv) P(X +Y ≤ 2) and (v) E(XY).
X
Y
0
1
2
0
0.02 0.08
1
0.05 0.20 0.25
2
0.03 0.12
0.10
0.15
Marginal Distribution of X:
X:
0
1
2
P(X = x): 0.10 0.40 0.50
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
Marginal Distribution of Y:
Y:
0
1
2
P(Y = y): 0.20 0.50 0.30
Mean of X and Y:
E(X) = (0)(0.1) + (1)(0.4) + (2)(0.5) = 1.4
E(Y) = (0)(0.2) + (1)(0.5) + (2)(0.3) = 1.1
Standard Deviations of X and Y:
E(X2) = (0)2(0.1) + (1)2(0.4) + (2)2(0.5) = 2.4
E(Y2) = (0)2(0.2) + (1)2(0.5) + (2)2(0.3) = 1.7
Var(X) = E(X2) – {E(X)}2 = 2.4 – (1.4)2 = 0.44
Var(Y) = E(Y2) – {E(Y)}2 = 1.7 – (1.1)2 = 0.49
Standard Deviation of X =
0.44
Standard Deviation of Y = 0.49
PX  1 / Y  2
P X  1 / Y  2  
P( X  1, Y  2) P0, 2  P1, 2 0.15


 0.5
P( Y  2)
P( Y  2)
0.30
P(X + Y ≤ 2)
Overall Outcomes: {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2)}
Favourable Outcomes: {(0,0), (0,1), (0,2), (1,0), (1,1), (2,0)}
P(X + Y ≤ 2) = 0.48
3.
E(XY) = (0)(0)(0.02) + (0)(1)(0.05) + (0)(2)(0.03) + (1)(0)(0.08) + (1)(1)(0.20) + (1)(2)(0.12) +
(2)(0)(0.10) + (2)(1)(0.25) + (2)(2)(0.15)
= 1.54
If X and Y are two random variables having joint probability density function
1
  6  x  y  , 0  x  2, 2  y  4
f  x, y    8
 0
, otherwise
(iii) P  X  1/ Y  3 .
P  X  1  Y  3 
y 3
Find (i) P  X  1  Y  3 (ii) P  X  Y  3
x 1
  f  x, y  dxdy
y  x 
y 3 x 1

1
  8  6  x  y  dxdy
y  2 x 0
3 1
1
    6  x  y  dxdy
820
1
3

1 
x2
   6 x   xy  dy
82
2
0
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Page 15 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
3
3
1 11 y y 2 
1 11 
 
    y  dy  
8 2
2 2
822

3
P  X  1  Y  3 
8
1 3 x
1
(ii). P  X  Y  3     6  x  y  dydx
8
0 2
3 x
1
1 
y2 
  6 y  xy   dx
80
2 2
2
1

3  x

1 
  6  3  x   x  3  x  
 12  2 x  2 dx
8 0 
2

1

9  x2  6 x 

1 
2
  18  6 x  3x  x 
 10  2 x   dx
80
2


1

1 
9 x2 6 x
  18  9 x  x 2     10  2 x dx
80
2 2 2


1
1 7
x2 

4
x


 dx
8 0  2
2
1
1  7 x 4 x 2 x3 
1 7
1
  
    2 
8 2
2
6 0 8  2
6
1  21  12  1  1  10  5
.
 
  
8
6
 8  6  24
P  x  1  y  3

(iii). P X  1
Y 3
P  y  3


The Marginal density function of Y is fY  y  
2
 f  x, y dx
0
2
1
 6  x  y dx
8
0

2

1
x2
 6 x   yx 
8
2
0
1
12  2  2 y 
8
5 y

, 2  y  4.
4

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Page 16 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
x 1 y 3

P X 1
Y 3

1
  8  6  x  y  dxdy
x 0 y  2
y 3

fY  y  dy
y 2
4.
If
the
3
3
8
8
3

3
5

y


1
y2 
dy
5y  
2  4 
4 
2 2
3 8 3
   .
8 5 5
X
distribution
functions
of
joint
and
x
y

 1  e  1  e  , x  0, y  0
F  x, y   
, otherwise

0
Find the marginal density of X and Y . Are X and Y independent?
Find P (1 < X < 3, 1 < Y < 2).


Given F  x, y   1  e  x 1  e  y
Y

 1  e x  e y  e x y 
The joint probability density function is given by
f  x, y  
 2 F  x, y 
xy
 
1  e x  e y  e x  y  

xy
  y

e  e  x  y  

x
 x  y 

, x  0, y  0
e
 f  x, y   
, otherwise

0

2
(ii) The marginal probability function of X is given by
f  x  f X  x




0

 f  x, y  dy   e
 x y 
dy

 e  x  y  


 1  0

 x y
  e   

0
x
e , x0
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is
given
by
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
The marginal probability function of Y is
f  y   fY  y 


 f  x, y  dx


  e x  y  dx  e x  y  
0

0
 e y , y  0
 f  x  f  y   e  x e  y  e    f  x, y 
 X and Y are independent.
(iii) P 1  X  3,1  Y  2  P 1  X  3  P 1  Y  2 [Since X and Y are independent]
 x y
3
2
  f  x dx   f  y  dy.
1
1
3
2
  e  x dx   e  y dy
1
1
3
2
 e   e y 

 

 1 1  1 1
x


 e3  e 1 e 2  e 1

 e5  e4  e3  e2
5.
6 x  y
,0  x  2, 2  y  4 for a bivariate (X , Y ) , find the correlation coefficient
8
E  XY   E  X  E Y 
Correlation coefficient  XY 
If f  x , y  
 XY
Marginal density function of X is
fX  x 

4

2
6  2x
 6 x y 
 f  x, y  dy    8  dy  8
Marginal density function of Y is
fY  y  



10  2 y
 6 x y 
f  x, y  dx   
 dx 
8
8

0
2
2
2
 6  2x 
 dx
 8 
Then E  X   xf X  x  dx  x 


0
0
2
1  6 x 2 2 x3 
 


8 2
3 0
1
16  1 20 5
 12    

8
13  8 3 6
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
4
1 10 y 2 2 y 3 
17
 10  2 y 
E Y    y 

 dy  
 
8 2
3 2
6
 8 
2
4
2
1  6 x3 2 x 4 
 6  2x 

E  X    x f x  x  dx   x 
 dx  
 1
8 3
4 0
 8 
0
0
2
2
2
2
 
E Y
2
2
4
1 10 y 3 2 y 4 
25
 10  2 y 
y 

 dy  
 
8 3
4 2 3
 8 
2
4
2
 
Var  X     E X
2
X
 
Var Y     E Y
2
Y
2
2
2
2
11
5
  E  X    1    
36
6
2
25  17 
11
  E Y   
  
3  6
36
2
 6 x y 
E  XY     xy 
 dxdy
8


2 0
4 2
2
4
1  6 x 2 y x3 y x 2 y 2 
 


 dy
82 2
3
2 0
4
6.
4
1 12 y 2 8 y 2 2 y 3 
1 
8
2


  12 y  y  2 y dy  

8 2
3 2
3 2
8 2
3

1
64 128
16 16  1  56 
 96 

 24      
8
3
3
3 3  8 3 
7
E  XY  
3
7  5  17 


E  XY   E  X  E Y  3  6 
 6 
 XY 

 XY
11 11
6 6
1
 XY   .
11
Using the given information given below compute x , y and r .
Also compute  y when
 x  2, 2 x  3 y  8 and 4 x  y  10 .
When the regression equations are known the arithmetic means are computed by solving the
equation.
2 x  3 y  8 ------------ (1)
4 x  y  10 ------------ (2)
(1)  2  4 x  6 y  16 ------- (3)
 2  3  5 y  6
y
6
5
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
6
5
18
 2x  8 
5
11
x
5
11
6
i.e. x 
&y
5
5
To find r , Let 2 x  3 y  8 be the regression equation of X on Y .
3
2x  8  3y  x  4  y
2
3
 bxy  Coefficient of Y in the equation of X on Y  
2
Let 4 x  y  10 be the regression equation of Y on X
 y  10  4 x
 byx  coefficient of X in the equation of Y on X  4 .
Equation 1  2 x  3    8
r   bxy byx
 3
      4 
 2
 2.45

bxy & byx are negative

Since r is not in the range of  1  r  1 the assumption is wrong.
Now let equation 1 be the equation of Y on X
8 2x
y 
3 3
 byx  Coefficient of X in the equation of Y on X
byx  
2
3
from equation (2) be the equation of X on Y
bxy  
1
4
r   bxy byx
2
1
     0.4081
3
4
To compute  y from equation  4  byx  
But we know that byx  r
y
x

2
   0.4081 y
3
2
2
3
  y  3.26
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Subject Name: Probability and Queueing Theory
7.
If
Subject Code: MA2262 Question Bank 2012 - 2013
and Y are independent exponential variates with parameters 1, find the pdf of
U  X Y .
Given that X and Y are exponential variates with parameters 1
X
f X  x   e x , x  0, fY  y   e y , y  0
Also f XY  x, y   f X  x  f y  y  since X and Y are independent
 e  xe  y
 x y
 e   ; x  0, y  0
Consider the transformations u  x  y and v  y
 x uv, y  v
v
x x
  x, y  u v 1  1
J


1
  u, v  y y 0 1
u v
fUV  u, v   f XY  x, y  J  e x e y  eu v e v
 e  u  2 v  , u  v  0, v  0
In Region I when u  0
f u  
RI
R II
v= - u


u
u
 f  u, v  dv   e
u
.e 2 v dv
u

 e 2 v 
e 

 2   u
eu
eu
0  e2u  

2
2
u
In Region II when u  0

f  u    f (u , v)dv
0

 e
0
 u  2 v 
dv 
eu
2
 eu
 2 , u  0
 f  u    u
e , u  0
 2
8.
A random sample of size 100 is taken from a population whose mean is 60 and variance is 400. Using
central limit theorem, with what probability can we assert that the mean of the sample will not differ
from   60 by more than 4?
Given that n  100 ,   60 ,  2  400
Since the probability statement is with respect to mean, we use the Linderberg-levy form of central
limit Theorem.
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Subject Name: Probability and Queueing Theory
X
Subject Code: MA2262 Question Bank 2012 - 2013
 2 
2
.
N   ,  i.e. X follows normal distribution with mean '  ' and variance
n
 n 
400 

i.e. X N  60,

100 

X N  60, 4 
 X will not differ from 
 mean of the sample will not 
P
 P


 differ from 60 by more than 4 
   60 by more than 4 


 P X  4
 P  4  X    4
 P  4  X  60  4
64  60 
 56  60
 P 56  X  64   P 
z
2 
 2
 P  2  Z  2
 2P 0  Z  2  2  0.4773  0.9446
1
UNIT – III MARKOV PROCESS AND MARKOV CHAINS
PART - A
Consider the random process X (t )  cost    where  is a random variable with density
1 

   . Check whether or not the process is wide sense stationary.
function f    ,
 2
2
1 

 
Given X (t )  cost    and f    ,
 2
2

EX (t ) 
 X (t ) f ( )d , 


2
1


2
2

sin(t    2
, 
2
1

1

2
 cos(t   )d

2
cos t  ( cos t )

1

2 cos t  constant
So, X(t) is not WSS.
Consider a random Process X (t )  coswt    , where w is a real constant and  is a uniform

variable in 0 , 

. Show that X(t) is not wide sense stationary.
2
Given X (t )  coswt    .

Since  is uniformly distributed in 0 , 
2 we have f ( )  2 , 0    2
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Subject Name: Probability and Queueing Theory

EX( t ) 
2
 X(t )f ()d 
0

2

Subject Code: MA2262 Question Bank 2012 - 2013

2
2
 cos( wt  )d   sin( wt   0 2
0
1
cos wt  sin wt )  cons tan t

So, X(t) is not WSS.
Define a Random Process.
A random process or Stochastic process X(s,t) is a function that maps each element of a
sample space into a time function called sample function.
Define First order stationary and Second order stationary process.
A random process is said to be stationary to order one if the first order density functions
defined for all the random variables of the process are same.
A random process is said to be stationary to order two if for all t1 , t 2 and  its second order

3
4
5
6
7
density functions satisfy the condition f x x1 , x2 ; t1 , t 2   f x ( x1 , x2 ; t1   , t 2   ).
Define Markov Process.
A random process or Stochastic process X(t) is said to be a Markov process if given the
value of X(t), the value of X(v) for v > t does not depend on values of X(u) for u < t. In
other words, the future behavior of the process depends only on the present value and not
on the past value.
Define Markov Chain.
A Markov process is called Markov chain if the states X i  is discrete no matter whether ‘t’
is discrete or continuous.
Define Chapman-Kolmogrov Equation
The Chapman-Kolmogrov equation provides a method to compute the n-step transition
probabilities. The equation can be represented as Pijnm  k 0 pikn pkjm  n, m  0 .

8
9
10
11
12
13
When do you say that a Markov chain is irreducible?
The Markov chain is irreducible if all states communicate with each other at some time.
When do you say the Markov chain is regular?
A regular Markov chain is defined as a chain having a transition matrix P such that for
some power of P, it has only non-zero positive probability values.
When do you say that state ‘i’ is periodic and aperiodic?
Let A be the set of all positive integers such that pii( n )  0 and ‘d’ be the Greatest Common
Divisor(G.C.D.) of the set A. We say state ‘i’ is periodic if d>1 and aperiodic if d = 1.
Define Poisson process.
Poisson process is a counting point process representing the number of occurrences of
certain event in a finite collection of non overlapping statistically independent time arrivals.
What are the properties of Poisson process?
(a)The poisson process is not a stationary process. It is vivid from the expressions of
moments of poisson process that they are time dependent.
(b)The poisson process is a Markov process.
Determine whether the given matrix is irreducible or not.
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Subject Name: Probability and Queueing Theory
0.3 0.7 0 
P  0.1 0.4 0.5
 0 0.2 0.8
0.3 0.7 0 
P  0.1 0.4 0.5
 0 0.2 0.8
Here
14
Pij( n)
Subject Code: MA2262 Question Bank 2012 - 2013
0.16 0.49 0.35
P  0.07 0.33 0.60
0.02 0.24 0.74
2
and
 0 ,  i, j . So, P is irreducible.
When do you say the Markov chain is homogeneous?
If the one-step transition probability does not depend on the step i.e.
pij (n  1, n)  pij (m  1, m)
15
16
17
then the Markov chain is called a homogeneous Markov chain.
What are the different types of Random process.
Continuous Random process, Discrete Random process, Continuous Random sequence and
Discrete Random sequence.
Define Birth and Death process.
If X(t) represents the number of individuals present at time t in a population in which two
types of events occur – one representing birth which contributes to its increase and the
other representing death which contributes to its decrease, then the discrete random process
{X(t)} is called the birth and death process.
A housewife buys 3 kinds of cereals A, B and C. She never buy the same cereal in successive
weeks. If she buys cereal A, the next week she buys B. However if she buys B or C the next
week she is 3 times as likely to buy A as the other cereal. Construct the Transition Probability
Matrix.
A B C
A 0

P  B 34

C 3
 4
18
1
0
1
4
0

1 
4
0 
1 
 0
Let A  
 be a Stochastic matrix. Check whether it is regular.
1 / 2 1 / 2
1 
 0
1 / 2 1 / 2 
A 
and A 2  

.
1 / 2 1 / 2
1 / 4 3 / 4
Since all entries in A 2 are positive, A is regular.
19
The number of particles emitted by a radioactive source is Poisson distributed. The source
emits particles at the rate of 6 per minute. Each emitted particle has a probability of 0.7 of
being counted. Find the probability that 11 particles are counted in 4 minutes.
The number of particles N(t) emitted is poisson with parameter p = 6(0.7) = 4.2
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Page 24 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
e 4.2t 4.2t 
m!
11
4.2 ( 4 )

e
4.2(4) 
P( N (4)  11) 
 0.038.
11!
Prove that the sum of two independent Poisson process is a Poisson process.
Let X 1 (t ) and X 2 (t ) be the Poisson process with mean 1 and  2 respectively.
m
P( N (t )  m) 
20
  X 1 ( t ) ( w)  e  1t (1e
iw
X
and
)
2 (t )
( w)  e  2t (1e
iw
)
Since X 1 and X 2 are independent,  X 1 ( t )  X 2 ( t ) ( w)   X 1 ( t ) ( w) .  X 2 ( t ) ( w)
 ( 1  2 ) t (1eiw )
1.
=e
Therefore X 1 + X 2 follows Poisson Distribution with mean 1   2 .
PART B
Find the mean , autocorrelation and auto covariance of the Poisson process
The probability law of the Poisson process {X(t)} is the same as that of a Poisson
e  t ( t ) n
distribution with parameter  t.  Pn (t )  P( X (t )  n) 
, n  0,1, 2...
n!
The mean of the Poisson Process is given by


x 0
x 0
E ( X 2 (t ))   x 2 p( x)   [ x( x  1)  x] p( x)


  [ x( x  1)] p( x)   xp( x)
x 0
x 0

  [ x( x  1)].
e
 t
x 0
( t ) x
 E ( x)
x!
( t ) x  2
 t
x  2 ( x  2)!

 e   t ( t ) 2 
 e  t (t ) 2 .et  (t )  (t ) 2  (t )


x 0
x 0
E ( X 2 (t ))   x 2 p ( x)   [ x( x  1)  x] p ( x)


  [ x( x  1)] p ( x)   xp ( x)
x 0
x 0

  [ x( x  1)].
x 0
e
 t
( t ) x
 E ( x)
x!
( t ) x  2
 t
x  2 ( x  2)!

 e   t ( t ) 2 
 et (t )2 .et  (t )  (t )2  (t ) q
Var(X) = E ( X 2 )  [ E ( X )]2  (t ) 2  t  (t ) 2  t
The auto correlation of the Poisson process is given by
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
RXX (t1 , t2 )  E[ X (t1 ). X (t2 )]
 E[ X (t1 )[ X (t2 )  X (t1 )  X (t1 )]
 E[ X (t1 )( X (t2 )  X (t1 ))]  E[( X 2 (t1 )]
 E[ X (t1 ) E[ X (t2  t1 )]  E[ X 2 (t1 )]
X(t) is a process of independent increments.
RXX (t1 , t2 )  t1. (t2  t1 )   2t12  t1 ,if t2  t1
  2t1t2   2t12   2t12  t1
  2t1t2  t1
  2t1t2   min{t1 , t2 )
The auto covariance of the poisson process is given by
C XX (t1 , t2 )  RXX (t1 , t2 )  E[ X (t1 )]E[ X (t2 )]
  2t1t2  t1  (t1.t2 )
 t1 ,
if t2  t1
  min{t1 , t2 }
2.
The transition probability matrix of a Markov chain {Xn}, n = 1,2,3  having three states 1,
2 and 3 is
 0.1 0.5 0.4
p  0.6 0.2 0.2 and the initial distribution is P 0   0.7 0.2 0.1 .


0.3 0.4 0.3
Find PX 3  2, X 2  3, X 1  3, X 0  2 and PX 2  3.
 0.1 0.5 0.4
0.43 0.31 0.26


2 
p  0.6 0.2 0.2 , p  0.24 0.42 0.34
0.3 0.4 0.3
0.36 0.35 0.29
0 
PX 2  3, X 1  3, X 0  2 =
PX 2  3=
3.
0 
2 
1
13
p  p

1
1
23
33
p  p p = 0.2 X 0.2 X 0.3 = 0.012
   
   
p  p  p  p
2
0
2
0
3
2
23
3
33
= 0.7 x 0.26 + 0.2 x 0.34 + 0.1 x 0.29
= 0.279
The process {X(t)} whose probability distribution under certain condition is given
 at n1
, n  1,2,3

n 1
(
1

at
)
PX (t )  n 
. Show that the process is not stationary.
at

, n 0

1  at
E [ X(t)] =  nPn
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Subject Name: Probability and Queueing Theory
=
1
1  at 2
Subject Code: MA2262 Question Bank 2012 - 2013
2


at
 at 
 3
1 2
  ...
1  at   1  at 


= 1 = a constant
 n2Pn
=  n(n  1) - n Pn
2
E [ X (t)] =
4.
= 2 ( 1+ at) – 1 = 1 + 2at
V[X(t) = 1 + 2at - 1 = 2at
X{t} is not stationary.
A man either drives a car or catches a train to go to office each day. He never goes two days
in a row by train but if he drives one day, then the next day he is just as likely to drive again
as he is to travel by train. Now suppose that on the first day of the week, the man tossed a
fair die and drove to work if and only if ‘6’ appeared. Find the probability that he takes a
train on the third day. Also find the probability that he drives to work in the long run.
Train –T and Car – C
T C
States of Xn-1
T  0
1
C
2
1
1

2
1
1
, P(CC) =
2
2
Initial state probability distribution is obtained by throwing a die.
1
Probability of going by car = P(6) =
6
1 5
 Probability of going by train =1 - =
6 6
5 1
The first day state distribution is P(1) = 
 6 6 
P(TT) = 0, P(TC) = 1, P(CT) =
5
Second day state probability P(2) = P(1) P = 
6
1  0
1
6  
2
1   1 11 
1 = 

 12 12 
2
 1 11   0
P(2) = P(1) P = 
1
12 12  
2
1
 11 13 
1 = 
 24 24 

2
11
P(he takes a train on the third day) =
24
Let    1  2  be the limiting form or long run probability distribution. Then  p  
 1
2
0
1

2
1
1   1  2 

2
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Subject Name: Probability and Queueing Theory
1
3
1  ,  2 
Subject Code: MA2262 Question Bank 2012 - 2013
2
3
2
.
3
E-mail passing through a server computer is classified into two types: “short” messages
(mostly plain text), and “long” messages (usually including graphics or audio data). The
message arrivals are modelled as two independent Poisson processes, with rates 20 per hour
(for the short messages) and 15 per hour (for the long messages). Find the probabilities that:
(a) less than 1 minute elapses between two messages.
(b) there are more than 4 long messages in any given 10-minute period.
(c) the next message to arrive is long.
(d) exactly 2 of the next 5 messages are short.
(a) The message arrivals (ignoring whether they are short or long) are a Poisson process
of rate 35 per hour, or 35/60 per minute. So the inter-arrival time T is distributed as
exponential(35/60).
We want P (T ≤1) which is 1 − exp(−35/60) = 0.4420.
(b) The number N of long messages in the 10-minute period is distributed Poisson(15/6). So
P(driving in the long run) =
5.
k
6.
 15 

15 
4

6  6 
P (N > 4) = 1 −  e
= 0.1088
k!
K 0
(c) This is 15/(15 + 20) = 3/7 = 0.4286.
(d) This is P (N = 2) where N ~Bin(5, 4/7),
2
2
 5 4   3 
4320
      
 0.2570 .
16807
 2  7   7 
The Notorious News Corporation monitors breaking news stories around the world. It has
found that most types of major news stories occur as Poisson processes. This is true of
political stories (which have a rate of  P = 4/day), business stories (  B = 6/day), and humaninterest stories (  H = 5/day); moreover, these three types of stories occur independently of
each other. Find
(a) The mean number of stories (of any of these types) in each day.
(b) The probability that no new story breaks in a given hour.
(c) The probability that the next story to break will be political.
(d) The probability that the next 4 stories to break will all be of the “human interest” type.
(e) Cov(X, Y ), where X is the number of political stories that occur today, and Y is the total
number of stories that occur today.
a) The combined process of all stories is also a Poisson process, with rate
 =  P +  B +  H = 15/day. The actual number of stories breaking in any given
day is thus distributed Poisson(15), and so the expected number is 15.
(b) The number N of stories in 1 hour is distributed Poisson(15/24). So
0
 15 

15 

24 
24 
 0.5353
P(N = 0) = e
0!
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
p
 4 / 15

4
 H 
(d) This is 
  1 / 81
  
(c) This is
7.
(e) Let Z be the number of non-political stories that occur today. Then Y = X + Z, so
Cov(X, Y ) = Cov(X,X + Z)
= Cov(X,X) + Cov(X,Z)
= Var(X) + 0
=4
since X and Z are independent and X ~Poisson(4).
Consider a random process X(t) = Bcos(50t + ) where B and ø are independent random
variables. B is a random variable with mean 0 and variance 1. Ø is uniformly
distributed in the interval [-  ,  ]. Find mean and auto correlation of the process.
Given that B is a R.V with mean 0 and 1
(i.e) E(B) = 0 and 𝜎 2 (B) = 1
1
, −𝜋 ≤ ∅ ≤ 𝜋
Since is uniformly distributed in [-𝜋, 𝜋] is pdf is given by f   
2
Consider E(X(t)] = E[Bcos(50t+ )]
= E(B).E(cos(50t+ )],  B AND are independent
 E(B)=0
=0
Hence the mean, E[X(t)] = 0
The autocorrelation is defined as,
R (t1,t2) = E(X(t1),X(t2)]
= E(Bcos(50t1+ ). Bcos(50t2+ )]
= E[B2 cos(50t+ ) cos(50t+ )]
=
𝐸(𝐵2 )
1
2
E[cos(50(t1+t2)+2 ) + cos(50(t1-t2)]
= 2{E[cos(50(t1+t2)+2 ) + cos(50(t1-t2)]}
 𝜎 2 (B) = 1
 E(B2)-E(B)2 = 1
 E(B) = 0
 E.(B2) = 1
𝜋
1
1
= 2 {(∫−𝜋 COS(50(t1 + t 2 ) + 2φ) 2𝜋 𝑑𝜑 + COS(50(𝑡1 − 𝑡2 ))
sin(50(𝑡1 +𝑡2 )+2∅ 𝜋 1
)−𝜋 2𝜋 + cos(50(𝑡1 − 𝑡2 )}
2
1 1
= 2 [4𝜋 (sin(50(𝑡1 + 𝑡2 )) − sin(50(𝑡1 + 𝑡2 )) + cos(50(𝑡1 − 𝑡2 ))]
1
= 2 cos(50(𝑡1 − 𝑡2 ))
1
∴ R (t1,t2) = 2 cos(50(𝑡1 − 𝑡2 )) which is a function of 𝑡1 − 𝑡2
1
= 2 {(
8.
Find the auto correlation function of the periodic time function {X(t)} = Asin𝝎𝒕
Given X(t)= A sinwt
X(t) X(t+𝜏) = Asinwt Asinw(t+ 𝜏)
= A2 sinwt sinw(t+ 𝜏)
= A2 sin(wt+w 𝜏) sinwt
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
1
= 𝐴2 2 [cos(𝑤𝑡 + 𝑤𝜏 − 𝑤𝜏) − cos(𝑤𝑡 + 𝑤𝜏 + 𝑤𝜏)]
1
= 𝐴2 2 [cos(𝑤𝜏) − cos(2𝑤𝑡 + 𝑤𝜏)]
The time autocorrelation function
1 𝑇
Rxx(t,t+𝜏) = 2𝑇 ∫−𝑇 𝑋(𝑡)𝑋(𝑡 + 𝜏)𝑑𝑡
𝑇 𝐴2
1
= 2𝑇 ∫−𝑇
1 𝐴2
= 2𝑇
2
2
[𝑐𝑜𝑠𝑤𝜏 − cos(2𝑤𝑡 + 𝑤𝜏)]𝑑𝑡
𝐴2
𝑇
𝑇
∫−𝑇 𝑐𝑜𝑠𝑤𝜏𝑑𝑡 + 4𝑇 ∫−𝑇 cos(2𝑤𝜏 + 𝑤𝜏) 𝑑𝑡
𝐴2
𝐴2 𝑠𝑖𝑛(2𝑤𝑇+𝑤𝜏 𝑇
]−𝑇
2𝑤
=(2𝑇)(2𝑇) 𝑐𝑜𝑠𝑤𝜏[𝑡]𝑇−𝑇 − 4𝑇 [
𝐴2
= (2𝑇)(2𝑇) (2𝑇)𝑐𝑜𝑠𝑤𝜏=
=
=
=
=
=
=
1
2
3
4
5
𝐴2
2
𝐴2
2
𝐴2
2
𝐴2
2
𝐴2
2
𝐴2
2
𝐴2
2
𝐴2
8𝑤𝑇
[sin(2𝑤𝑡 + 𝑤𝜏)]𝑇−𝑇
𝐴2
𝑐𝑜𝑠𝑤𝜏-8𝑤𝑇 [sin(2𝑤𝑇 + 𝑤𝜏) − sin(−2𝑤𝑇 + 𝑤𝜏)
𝐴2
𝑐𝑜𝑠𝑤𝜏-8𝑤𝑇 2 cos [
2𝑤𝑇+𝑤𝜏−2𝑤𝑇+𝑤𝜏
2
] 𝑠𝑖𝑛 [
2𝑤𝑇+𝑤𝜏+2𝑤𝑇−𝑤𝜏
2
]
𝐴2
𝑐𝑜𝑠𝑤𝜏-4𝑤𝑇 𝑐𝑜𝑠𝑤𝜏𝑠𝑖𝑛2𝑤𝑇
𝐴2
𝑐𝑜𝑠𝑤𝜏-4𝑤𝑇 𝑐𝑜𝑠𝑤𝜏sin[2
2𝜋
𝑇
𝑇]
2 

 w  2f  T 
𝐴2
𝑐𝑜𝑠𝑤𝜏-4𝑤𝑇 𝑐𝑜𝑠𝑤𝜏sin 4𝜋
𝑐𝑜𝑠𝑤𝜏 − 0
𝑐𝑜𝑠𝑤𝜏
UNIT – IV: QUEUEING THEORY
PART - A
What are the basic characteristics of queueing system?
The basic characteristics of queueing system are Arrival time pattern, Service time pattern,
Number of services channels, Capacity of the system, Service discipline.
What do you mean by transient state and steady state queueing system?
Transient state: System depends on time t, Steady state: System independent of time t.
What do you mean by traffic intensity?
Arrival rate
The traffic intensity  
Service rate
Write down Little’s formula.
L

L s   Ws , L q   w q , Ws  Wq  , Wq  q .


Consider an (M/M/1) queueing system. If   6 &   8 , find the probability of more than
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 30 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
10 customers in the system.
10 1
6

11
6
P(n  10)         0.75
8

In a given (M / M / 1) : (  / FCFS) queue   0.6 , what is the probability that the queue
contains 5 or more customers?
11
5
7
8
9

P(n  5)      5  (0.6) 5

What is the probability that a customer has to wait more than 15 min. to get his service
Completed in ( M / M / 1) : ( / FIFO) queue system if   6 / min . &   10 / hr ?
1
P( w s  15 min)  P( w s  hr ) , w s is exp onential with parameter   
4


 (  )
 (106)
 ( )
(  ) 

4
4
  (  )e
d   e

e

e
 e 1

1


1
4
4
A duplicating machine maintained for office use is operated by an office assistant. If jobs
arrive at a rate of 5 per hour and the time to complete each job varies according to an
exponential distribution with mean 6 min., find the percentage of idle time of the machine in
a day.
1
( M / M / 1) : ( / FIFO) model.   5 / hr. ,   min. 10 / hr.
6

5
P0  1   1   0.5 , 50% of the time machine is idle.

10
Derive the average no. of customers in the system for ( M / M / 1) : ( / FIFO)

L   n pn  0.p0  1.p1  2.p2  3.p3  .....
s n0
n
2
3





  n   p0  p 0   2    3    ....
n 1



 


10
11
2

 
  


1
1

 p0 1  2    3    ....  p 0
.
2  Ls  1 
2 

 
  

 
    

1

1

 
 




Write down the Kendall’s notations for queueing model.
(a / b / c) : (d / e) is the Kendall’s notation
Where (a) is Arrival Pattern ,(b) Service pattern, (c) No. of servers, (d) Capacity of the
system,
(e) Service discipline.
A one person barber shop has six chairs to accommodate people waiting for a haircut. Assume
customers who when all six chairs are full leave without entering the barber shop, customer
arrive at the average rate 3/hr and customer spend an average of 15 minutes in the barber
chair, what is the effective arrival rate?
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 31 of 48
Subject Name: Probability and Queueing Theory
  3 / hr &   4 / hr , K
Subject Code: MA2262 Question Bank 2012 - 2013


 6  1  7 , P0 
 0.2778
k 1

1  

1
The Effective arrival rate =    1  P0   4 1  0.2778  2.888
12
13
In a railway marshaling yard, goods trains arrive at a rate of 30 trains per day: Assume that
the inter arrival time follows an exponential distribution and the service time distribution is
also exponential with an average of 36 minutes .Calculate the probability that the yard is
empty.
1 
  0.75 , P0 
 0.2649
1   K 1
If   4 / hr &   12 / hr in an M / M / 1 : 4 / FIFO model, Find the probability that there is no
customer in the system. If    ,what is the value of this probability?


 1 




p0  
k 1 
 

1    

  

, if   
i.e. =
1   412 
1   412 
5
 0.669
1
1

k 1 5
What is the effective arrival rate for (M / M / 1) : (4 / FCFS) queueing model?
The effective arrival rate is     (1  p0 )
If   , p 0 
14
1

 4 1



where p0   1 


41
 
1    
  
15
if   
if   
For (M / M / S) : ( N / FIFO) Model, write down the formula for Ls , L q .
s
L s  Lq 
16
L

Ws  s ,Where




k s
Lq  p0  
 (k  s)(1  )k s 
2 1  

  s!1   
s 1



   s   (s  n)p n  and  
s
 n 0

Find the probability that an arriving customer is forced to join the queue of M/M/C model.
Arriving customers have to join the queue if
No. of customers in the system  no. of servers = C
P(n  c)  1  p0  p1  p2  ....  pc1 
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 32 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
2
3
c 1


1
1 
1 
1 
1   p0 
p0    p0    p0  ..... 
p

0
 
1! 
2!   
3!   
(c  1)!   


17
c 1
 1  1    2 1   3
1  
 1  p0 1 
       ..... 
  
(c  1)!    
 1!  2!    3!   
What is the probability that an arrival to an infinite capacity 3 server poisson queue, with

2
1
 and p 0  enters the service without waiting.
c 3
9
Arriving customer shall enter the service without waiting if
No. of customer in the system  No. of server ( =3 )
n
2
18
1 
1
1
 Pn    p 0
 P(n  3)  p0  p1  p2  p0 
p0    p0
n!  
1
2 
1
1 1 1 5
 2
 2

Given
 
   2 and c  3  P(n  3)   2.  4 
9
9 2 9 9
c 3
3 3

The local one person barber shop can accommodate a maximum of five people at a time (4
waiting and 1 getting haircut).Customers arrive according to a poisson distribution with
mean 5/ hr. The barber cuts hair at an average rate of 4/hr. (Exponential service time).What
fraction of the potential customers are turned away?
1  
  5 ,   4 , K  5 P0 
K 1  0.0888,
1   
19
20
1
5

P  n  5    P0  0.2711

A two person barber shop has 5 chairs to accommodate waiting customers, Potential
Customers, who arrive when all 5 chairs are full, leave without entering the shop. Customers
arrive at an average rate of 4 / hr and spend an average of 12 min. in the barber chair. Find 
(M/M/2) with infinitecapacity : Accommodation for 5 customers to wait +2 chair = 7
1
 4

4
  4 / hr ,   / min.  5 / hr , s  2, k  7,   0.8 ,  

 0.4
12
 5
s  2  5
There are three typists in an office. Each typist can type an average of 6 letters per hour. If
letters arrive for being typed at the rate of 15 letters per hour, find the traffic intensity.
 ( M / M / 3) : ( / FIFO) model.

 2.5
  15 / hr. &   6 / hr , s  3   2.5 &

 0.833

s
3
PART - B
Arrivals at a telephone booth are considered to be Poisson with an average time of 12 min.
between one arrival and the next. The length of a phone call is assumed to be distributed
exponentially with mean 4 min.
(a) Find the average number of persons waiting in the system.
(b) What is the probability that a person arriving at the booth will have to wait in the queue?
(c) What is the probability that it will take him more than 10 min. altogether to wait for the
phone and complete his call?
(d) Estimate the fraction of the day the phone will be in use.
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 33 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
This is an infinite queueing model with single server
Arrival rate  
(a) Ls 

1
1
/ min. , Service rate  = / min.
12
4
 0.5 customer
1 1

4 12
(b) P(n > 0) =1 - P (n = 0) = 1 - P (no customer in the system) = 1- p 0
1
    12 1
= 1  1    

   1 3
4
 

1
12
(c) P(Ws > 10) = e
(  )10
=e
5
3  0.1889
(d) P(phone will be idle) = P(n = 0) = Po = 1 
 2

 3
2 1

3 3
Customers arrive at a one man barber shop according to a poisson process with a
mean inter arrival time of 20 min. Customers spend an average of 15 min. in the barber chair,
then
(a) What is the probability that a customer need not wait for a hair cut ?
(b) What is the expected no. of customers in the barber shop and in the queue?
(c) How much time can a customer expect to spend in the barber shop?
(d) Find the average time that the customers spend in the queue.
(e) What is the probability that there will be 6 or more customers waiting for service?
.· P (Phone will be in use) = 1 
2
This is an infinite queueing model with single server  
1
1
/ min. and   / min.
15
20
1

1
(a) p(n  0)  1   1  20  ,
1

4
15
1


3 9
20
(b) Ls 

 3, Lq  Ls   3    2.2
   115  120

4 4
1
1
(c) Ws 

 60min.
 1  1
15 20
1

20

 45 min .
(d) Wq 
 (   ) 1  1 1 
  
15  15 20 
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 34 of 48
Subject Name: Probability and Queueing Theory
6
1
    20 
(e) p(n  6)     
  0.1779
    115 
There are three typists in an office. Each typist can type an average of 6 letters per hour. If
letters arrive for being typed at the rate of 15 letters per hour,
(a) What fraction of the time all the typists will be busy?
(b) What is the average number of letters waiting to be typed?
(c) What is the average time a letter has to spend for waiting and for being typed .
This is an (M / M / 3) : ( / FIFO) model.
6
3
Subject Code: MA2262 Question Bank 2012 - 2013
  15 / hr. &   6 / hr, s  3, 

 2.5
 2.5 &

 0.833

s
3
(a) All the typists will be busy if there are at least 3 customers (letters) in the system
p(n  3)  p(3)  p(4)  p(5)  .....  1  p 0  p1  p 2 

 s1
1 n
p0   
 n 0 n !  n




s

1

 s
 s !
 
1   
 s  
1




3
 1
1  1   

1


.1 


2 
3


0
!
1
!

2
!


3
!






 1  s  


 
2

2.52   2.53
1

 1  2.5 



2   6 (1  0.833) 

2
1

2
p0  2.5 p0 ,
p2 
p0  2.5 p0
2
1! 
2
2! 
(b) Waiting to be typed (queue)
p1 
Lq 
(c) Ws 
4
s  1
 s  s! 
s 1
p0
1


1  s 


2

1
 2.54 p0  3.5078
3 6
1
 1
Lq    3.5078 2.5  0.4005 hr.


  15
A two person barber shop has 5 chairs to accommodate waiting customers. Potential
customers, who arrive when all 5 chairs are full, leave without entering shop . Customers
arrive at average rate of 4 / hr and spend average of 12 min. in the barber chair. Compute
p 0 , p1 , p 7 , and Lq
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 35 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
This is an (M / M / 2) : (7 / FIFO)
1
 4

4
  4 / hr,  
/ min.  5 / hr, s  2, k  7,   0.8,  

 0.4
12
 5
s 2  5
ns
n2
1  s 1 1 n
 s k      1 1 n
2 7    
(a)
 


     
  
p0  n0 n !  n  s s ! n s  s    n0 n !  n  2 2! n2  s  

 



1


 1  (0.8)  (0.8) 2 (0.4)0  (0.4)1  (0.4) 2  (0.4)3  (0.4) 4  (0.4)5   2.311488
2


1
 p0 
 0.4289
2.311488
1

(b) p1    p0  0.3431

1
1!
(c) p7 
1
s ns
7

1
  p0  5 0.87 0.4289  0.0014
s!  
2 2!
s


(d) Lq  p0  
1  k s  (k  s)(1  )k s   0.1462
2 
   s ! 1   
5
Derive p 0 , Ls , Lq ,Ws ,Wq
for ( M / M / 1) : ( / FIFO) queueing model.
By birth and death process pn 
n  
 n
&
0 1 2 ......n1
  .......n times
p0 
p0
  ........n times
 0 1  2 .... n1
n  
 n
n


n
p0    p0
n


p0  p1  p 2  .........  1
2
3



 p0  p 0    p 0    p 0  ......  1



       2   3

 p0 1           .........   1
   



 
 p0 1  
 
1
 1  p0  1 


To find L s :
2
3



Ls  E(N)   n p n  0.p 0  1.p1  2.p 2  ........  1. p 0  2.   p 0  3   p 0  .......



2
2
2


   


 

 p 0 . 1  2.  3.    .......   p 0 . 1    1   . 1   




 


   

St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 36 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
To find Lq , Ws , W q
Using Littlle’s formula



2
,

 
  
(   )
L q  Ls 
6
for ( M / M / s ) : ( / FIFO) queueing model
Derive p0 , Ls , Lq ,Ws ,Wq
By birth and death process pn 
n  
 n
L
1
Ws  s 


0 1 2 ......n1
p0
1  2  3 .... n
n
n  
s 
&

1

Pn  

1
 
2 
 
2 
, ns
, ns
.......ntimes
p0
3 ........ntimes
.......n times
p0
3  ........s times
 n
p

n 0
n ! 
Pn  
n

 s ! s n s  n p0

, ns
,n  s
, ns
,n  s
p0  p1  p2  .....  ps  ps 1  ps 2  ....  1
p0 

1 !
p0 
n
2
s
 s1
p

.....

p

p0  .....  1
0
0
2!  2
s ! s
s .s !  s 1
s 1

1  
p0     p0.  
s!   
n 0   
s
2

   
1 
    ....   1
 s  s 





 s 1   n 1    s  1
p0        
s!     1  
n0   


s

To find Lq :



 s 1   n 1    s  1

   1  p0         
s!     1  
n0   




s









1

 n 1
n  2
Lq   (n  s) p n  0.ps  1.ps 1  2.ps  2  ...... 
p0 
p0  .....
s ! s  n 1
s ! s2 n  2
n s
1




 n 1

 n 1
 n 1

p
1

2.

......

p
1


p
0
0


0

s
s ! s  n 1 
s ! s  n 1  s 
s ! s n 1

Using Little’s formula we can find Lq , Ws , Wq
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s 2 2
(s  ) 2
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
L
L


q
L  L  ,W  s , W 
, W W 
s
q 
s
q
s
q 


7
A car service station has 2 bays offering service simultaneously. Because of space constraints,
only 4 cars are accepted for servicing. The arrival pattern with 12cars per day. The service
time in both the bays is exponentially distributed with μ=8cars per day per bay .Find the
average number of cars in the service station, the average number of cars waiting for service
and the average time a car spends in the system.
This is a multiple server model with finite capacity.
Arrivalrate λ=12/day and Service rate μ=8/day ,S=2,K=4
1


n
s
ns 
1
 s 1
1
1 k  
2
2 
 1.5 1



p0          
 1 
 1.5  1   0.75    0.75  


 n  0 n!    s !    n  s    
1 2





s



1   k  s   k  s 1     k  s 
P0  0.1960 ,  
, Lq  P0  
2 
s
   s !1   
 0.75 
2
2
2
 1- 0.75 -2  0.25 0.75  =0.4134car.
Lq =0.1960 1.5 
2  

 2  0.25 


s 1
1
Ls =Lq  s   s  n  pn  0.4134  2  2  n  pn  2.4134  2  2 p0  p1   1.73cars.
n 0
n 0


Ls
where     s    s  n  pn   8  2   2 p0  p1    10.512

 n 0

s 1
Ws =
1.73
 0.1646 day
10.512
Average time that a car has to spend in the system = 0.16
Patients arrive at a clinic according to Poisson distribution at the rate of 30 patients per hour.
The waiting room does not accommodate more than 14 patients. Examination time per
patient is exponential with a mean rate of 20 per hour.
(a) Find the effective arrival rate at the clinic.
(b)What is the probability that an arriving patient will not wait?
(c)What is the expected waiting time until a patient is discharged from the clinic?
Arrival rate   30 / hr., Service rate  20 / hr. , M / M /1: K / FIFO model.
1- λ
1- 3
μ
2 =0.00076
   , p0 
=
16
k+1
1- 3
2
1- λ
μ
Ws 
8
 
 
(a) The effective arrival rate is    1  p0   20 1  0.00076  19.98hr.
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
(b) P  a patient will not wait   p  n  0  p0  0.00076
K 1
16

3

 K  1  
16  


2

(c) Ls 

 (3)    16  13patients
K 1
 

3
1  
1  
2

L
13
Ws  s 
 0.65hr or 39min.
  19.98
1
2
3
4
5
6
7
8
UNIT– V: NON- MARKOVIAN QUEUES AND QUEUE NETWORKS
PART - A
Explain M/G/1 model.
It is a non Markovian queueing model. Where the arrival pattern M is Poisson, the Service
time distribution G follows any general distribution and the number of servers is one.
What are the differences between Markovian and non Markovian queueing model?
Markovian queueing model: Service time distribution follows Poisson distribution.
Non Markovian queueing model: Service time distribution follows any general distribution.
Write down Pollaczek-khintchin formula.
 2  Var(T)  E(T) 2 
Ls   E(T) 
2 1  E(T)
Write down Little’s formula relation in M/G/1 model.
Lq

L s   Ws , L q   w q , Ws  Wq  , Wq 
.


Write down Pollaczek-khintchin formula for the case when service time distribution is
Erlang distribution with k phases.

2 1  k 
1
1
E (T )  , Var (T ) 
,
L


s

k 2
 2k (    )
A one man barber shop takes exactly 25 minutes to complete one hair cut. If customers arrive
at the barber shop in a Poisson at an average rate of one in every 40 minutes, how long on the average a
customer spends in the shop?
25
625
55
1


E(T) = 25, Var (T) = 0,   / min , Ls 
.
40 (1600)2(1  25/ 40) 48
40
Write short notes on queue networks.
Network of queues can be described as a group of nodes, where each of the nodes represents a
service facility of the same type. The customers may enter the system at some node, can traverse from
node to node in the system and finally can leave the system from any node. Also customers can
return
to the nodes already visited.
Define series queues (Tandem queues) with an example.
In series queues , there are a series of service stations through which each calling unit must
progress prior to leaving the system. Ex. A physical examination for a patient where the patient
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Subject Name: Probability and Queueing Theory
9
under
goes a series of stages, lab tests, ECG , X ray etc.
What is the joint probability that there are m customers at station 1 and n customers at station
2 for a 2 stage series queue?
m
10
11
12
13
14
Subject Code: MA2262 Question Bank 2012 - 2013
n
  
     
 

Pmn    1     1 
 1   1    2    2 
Consider a service facility with two sequential stations with respective service rates of 3/min
and 4/min. The arrival rate is 2/min. What is the average waiting time of the system if the
system could be approximated by a two series Tandem queue?
1  3/ min. , 2  4/ min.,   2/ min. .
Average waiting time of customers in the system =Ws(station 1) + Ws(station 2)
1
1
1
1



 1.5min.
=
1    2   3  2 4  2
A TVS company in Chennai containing a repair station shared by a large number of
machines has 2 sequential stations with respective service rates of 3 per hour and 4/hr. the
failure rate of all the machines is 1 per hour. Assuming that the system behavior can be
approximated by a 2 stage tandem queue find the probability that the service stations are
idle.
m
n
 
   
 
1  3/ hr. & 2  4/ hr. ,   1/ hr. , Pmn    1     1  
 1   1    2    2 
0
0
1  1 1  1
P00    1     1    0.5
 3  3  4  4
What do you mean by bottle neck of a network?
The service station for which the utilization factor is maximum among all the other service
stations of the network is called the bottle neck of a network.
A company's repair section has 2 sequential stations with respective service rates of 4/hour
and 5/ hour. The cumulative failure rate of all the machines is 1/hour. Assuming the system
behavior can be approximated by the two stage tandem queue find the bottle neck of the
repair facility.
 1  1


 ,  ,since

, the service station 1 is the bottle neck of the repair facility.
1 4  2 5
1  2
Write the steady state equations for series queues with blocking.
Let  be the arrival rate at station 1 which follows Poisson distribution and service time is
exponential with parameters 1 and  2 respectively. Pn1n2 represent the probability that
there are n1 customers in station 1 and n2 customers in station 2.
 P0,0   2 P0,1  0,  1 P1,0   2 P1,1  P0,0  0,  (   2 ) P0,1  1 P1,0   2 Pb,1  0
 ( 1   2 ) P1,1  P0,1  0,   2 Pb,1  1 P1,1  0
15
Define Jackson networks.
Networks preserving the following characteristics are called Jackson networks.
1. Arrivals from outside through node i follow a Poisson process with mean arrival rate ri.
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
2. Service times at node i are independent and exponentially distributed with parameter
i
16
17
18
3.The probability that a customer who has completed service at node i will go to next
node j is
Pij , i = 1,2,...k , j= 0,1,2,...k.
Define open Jackson networks.
In open Jackson networks, the arrivals from outside to the node i is allowed and once a
customer gets the service completed at node i , joins the queue at node j with probability Pij
or leaves the system with probability Pi0.
Define closed Jackson networks.
In a closed Jackson networks, no customer may enter the system from outside and r i0 = 0
for all i , i.e. No customer may leave the system.
Write down the flow balance equation of open Jackson network.
k
 j  rj   i Pij , j  1,2,...k , where rj is the arrival rate to node j ,  j is the total arrival rate
i1
19
of customers to node j and Pij is the probability that a departure from server i joins the
queue at server j.
Write down the flow balance equation of closed Jackson network.
k
 j   i Pij , j  1,2,...k ,where i is the total arrival rate of customers to node i and Pij is
i 1
the probability that a departure from server i joins the queue at server j.
20. Write down the applications of network queues.
Telecommunication field, Bioinformatics, biomedical Industry.
PART - B
1
In a heavy machine shop, the overhead crane is 75% utilised. Time study observations gave
the average slinging time as 10.5 minutes with a standard deviation of 8.8 minutes. What is
the average calling rate for the services of the crane and what is the average delay in getting
service? If the average service time is cut to 8.0 minutes with a standard deviation of 6.0
minutes, how much reduction will occur on average in the delay of getting served?
The service time is not following exponential distribution. Hence it is a  M / G /1 queuing
model
1
1
  0.75, E T   10.5,  

,     0.0714min. or 4.286 hr.
E T  10.5
Average delayin gettingservice
2

   0.75 
  
60 

Wq 

 0.449 hrs. or 26.8min.
2 1   
2  4.29 1  0.75 
If the service time is cut to 8 minutes with
60
4.29
  6min., then μ= =7.5/hr. and  
 0.571
8
7.5
2
2
2
 4.29 
2  8.8 
2
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013


2 2
2
 2 2   2  2  2 2   2  1     


2 1   
2  1   
2  1   
Now average delay in getting service is
 
2
2

0.571 1   0.75  6
60 


 0.1386 hrs. or 8.32min. ,Reduction 26.8  8.32  18.5min .
2  7.5 1  0.751
2
A one-man barber shop takes exactly 25 minutes to complete one hair-cut. If customers
arrive at the barber shop in a Poisson fashion at an average rate of one every 40 minutes, how
long on the average a customer in the spends in the shop. Also, find the average time a
customer must wait for service?
The service time is constant. Hence it is a  M / D /1 queuing model. E  T  =25 , Var  T  =0 ,
Arrival rate λ=
1
Ls   E  T  

/min.
40
 2  E 2  t   Var T  
2 1   E  T  

2
3
 1   2 
   25  0  55
1
Ls 
 customers.
 25    40  1
40
48


2 1   25  
 40

 55 
Ls  48 
1
1
Ws 

 45.8min.,Wq  Ws   Ws 
 45.8  25  20.8 min.
1
  1 

 
E T 
 40 
A patient goes to a single doctor clinic for a general check up has to go through 4 phases. The
doctor takes on the average 4 minutes for each phase of the check up and the time taken for
each phase is exponentially distributed. If the arrivals of the patients at the clinic are
approximately Poisson at the average rate of 3 per hour, what is the average time spent by a
patient (i) in the examination (ii) waiting in the clinic?
The clinic has 4 phases ( each having different service nature) in series as follows:
Considering all the phases (each with exponential service time) together as a “one server”
we shall take it as a server with Erlang service time. So this is a  M / Ek /1 model with
E T  
k


4
k
4
 16 min . and Var(T) 
=
= 64 min.
2
1
1
θ
 
 
4
 16 
1
3
3
60
the arrival rate   / hr. 

1
20
/ min.
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
 2  E 2  t   Var  T  


Ls   E  T  
2 1   E  T  
2
 1   2
  16  64  14
1

 customers.
16    20  1
20
5


2 1  (16) 
 20

 14 
 
L
1
1
5
Ws  s     56 min,
Wq  Ws   Ws 
 56  16  40 min.
1
  1 

 
E T 
 20 
4
A car wash facility operates with only one bay. Cars arrive according to a Poisson fashion
with a mean of 4 cars per hour and may wait in the facility’s parking lot if the bay is busy.
The parking lot is large enough to accommodate any number of cars. Find the average time a
car spends in the facility, if the time for washing and cleaning a car is constant of 10 minutes.
In this model the service time is not varying and constant. Hence it is  M / D /1
queuing model.
E T   10 min. and Var  T  = 0
The arrival rate   4 / hr. 
1
15
/ min.
2
Ls   E  T  
5
1  2 
  10  0 
  1 10   15  
 1.33cars.
 
15
 1

2 1  (10) 
 15

 2  E 2  t   Var  T  

2 1   E  T  
By Little’s formula
4
 
L
1
1
3
Ws  s     20 min.,Wq  Ws   Ws 
 20  10  10 min.
1
 1

 
E T 
 15 
A car wash facility operates with only one bay. Cars arrive according to a Poisson fashion
with a mean of 4 cars per hour and may wait in the facility’s parking lot if the bay is busy.
The parking lot is large enough to accommodate any number of cars. Find the average
number of cars waiting in the parking lot, if the time for washing and cleaning a car follows a
discrete distribution with values equal to 4,8,15 minutes and corresponding probabilities 0.2,
0.6 and 0.2.
This is  M / G /1 queuing model as the service time is a discrete distribution as given below:
T: 4
8
15
P (T): 0.2
0.6
0.2
So the mean and variance are calculated as
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
E (T )  4  0.2   8  0.6   15  0.2   8.6 min
 
E T 2  42  0.2   82  0.6   152  0.2   86.6 min
 
2
Var (T )  E T 2  E (T ) 2  86.6  8.6  12.64
The arrival rate   4 / hr. 
1
15
/ min.
2
6
1  2
  8.6  12.64 
 2  E 2  t   Var  T   1
15 



Ls   E T  
  8.6  
 1.024customers.
15
2 1   E  T  
 1

2 1   8.6  
 15

1
 
 

15
Lq  Ls 
 1.024     0.451customer.

 1 
 
 8.6 
Derive the Pollaczek-Khinchine formula.
Let N and N  be the number of customers in the system at times t and t+T, when two
consecutive customers have just left the system after getting service.
Thus T is the random service time, which is a continuous random variable. Let
f  t  , E T  ,V T  be the probability density function, mean and variance of the service
time T. Let M be the number of customers arriving in the system during service time T.

M,
 N 1  M ,
N  
if
if
N  0
N  0
where M is a discrete random variable taking the values 0,1,2,…
The same can be written as
 1,
N   N  1  M   .... (1), where   
 0,
if
if
N 0
N  0
Note that by the definition of , 2 =  and N = 0
Squaring both sides of 1, we have,
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Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
2
 N 2   M  1   2  2 N  M  1  2 N   2  M  1 
N 2
2
 N 2   M  1    2 N  M  1  2  M  1 
 N 2  M 2  2 M  1  2 N  M  1   2 M  1 
2 N 1  M   N 2  N 2  M 2  2 M  1   2 M  1  .......  2 
Taking Expectations on the both sides of (1) we get,
E  N    E  N   1  E  M   E  
In steady state the probability that the number of customers in the system
E  N   E  N 
and
is constant
   
E N 2  E N 2 ........(3)
and substituting in previous equation we have E    1  E  M  .....(4)
Taking Expectations on the both sides of (2) we get
     
2E  N 1  M   E N 2  E N 2  E M 2  2E  M   1   2E  M   1 E   ..
using (3) we have
 
2E  N 1  M   E M 2  2E  M   1   2E  M   1 E   ..
using (4) we have
 
2
 E  M 2   2E  M   1  2E  M   2E  M   1  E  M 
2
 E  M 2   2E  M   E  M 
2 E  N 1  M   E M 2  2 E  M   1   2 E  M   1 1  E  M  
2 E  N 1  M 
Since the number of arrivals (M) to a system is independent of the number of customers
already in the system (N) we have
 
2 E  N  1  E  M   E  M 2   2 E 2  M   E  M 
E  M 2   2E 2  M   E  M 
EN 
......(5)
2
2 E  N  E 1  M   E M 2  2 E  M   E  M 
2 1  E  M 
Since the arrivals in service time T follows Poisson process with parameter  .
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Subject Name: Probability and Queueing Theory

Subject Code: MA2262 Question Bank 2012 - 2013

E  M T   T ,E M 2 T   2T 2  T , 2M T  T
E  M   E  T    E T 
  

 
E M 2  E  2T 2  T   2 E T 2   E T    2  var T   E 2 T    E T 


E 2  M    2 E 2 T 
substituting in (5) we have,
EN
 2  var  T   E 2 T     E T   2 2 E 2 T    E T 



2 1   E  T  
 2  var  T   E 2 T    2 E T   2 2 E 2 T 



2 1   E  T  
 2  var  T   E 2 T    2 E T  1  E T  



2 1   E  T  
 2  var  T   E 2 T   2 E  T  1  E  T  



2 1   E  T  
2 1   E  T  
 2  var  T   E 2 T  

  E T
EN 
 
2 1   E  T  
7
A repair facility by a large number of machines has two sequential stations with respective
rates one per hour and two per hour. The cumulative failure rate of all the machines is 0.5
per hour. Assuming that the system behavior may be approximated by the two-stage tandem
queue, determine the average repair time.
  0.5, 1  1, 2  2, 1  0.5,  2  0.25
Note that each station is a (M/M/1) queue model
The average length of the queue at station 1
1
0.5
LS1 
1  1

1  0.5
1
The average length of the queue at station 2
2
0.25
1
LS 2 


1   2 1  0.25 3
Ws1 
LS

1

1
0.5
2
Ws2 
LS 2

1

3 2
0.5 3
The total repair time of the network is  Ws1  Ws 2  2 
2
3

8
3
hours
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Subject Name: Probability and Queueing Theory
8
Subject Code: MA2262 Question Bank 2012 - 2013
Find the average response time for a two-stage tandem open network with feedback.
By Jackson’s result the two-queues at each node will behave like independent M/M/1
queues and hence the probability that there are k0 customers at node 0 and k1 customers at
node 1 is given by


p k 0 , k1  1   0  k0 1  1  k1 where 0  0 ,
1  1
1
0
0
1

 



0 , 1 be the arrival rates at nodes 0 and 1 respectively.
In steady state the arrival rate and departure rate at the respective nodes coincide.
The node 0 has arrival from outside with a rate  and the arrival rate from the feedback (
that is departure rate from node 1 which is nothing but 1 ) and so
0    1 when a job is completed at node 0, it will reach node 1 therefore the number of
jobs
at node 1 is p1 and so the average arrival rate to node 1 is given by
p



and 1  1 which implies that
1  0 p1 thus: 0 
1  p1
p0
0
p1
0 p0
p0 1
B0 denote the total service time at node 0 for a job then


p
1
E  B0   0 
Similarly E  B1   1  1

p0 0

p0 1
The average response time is computed as
Ls  Ls 2 1  
 
L
E  R  s  1
  0  1 


 1  0 1  1 
0 

p
, 1 


9
1
p0 0  

 
 
1
p0 1

p1
 
 
E B0
E B1

1   E B1
1   E B0
There are 2 salesmen in a super market. Out of the 2 salesman, one is incharge of billing and
receiving payment while other salesman is incharge of weighing and delivering the items. Due
to lack of space, Only one customer is allowed to enter the shop, only if the billing clerk is
free. The customer who has finished is billing job has to wait until the delivery section
becomes free. If customer arrive according to Poisson process at a 1/hour and the service
times of 2 clerks are independent and have exponential rates of 3/hour. Find
(i) The proportion of customers who enter the supermarket.
(ii)The average number of customer in the system.
(iii)The average amount of time a customer spends in the shop.
St. Joseph’s College of Engineering & St. Joseph’s Institute of Technology
Page 47 of 48
Subject Name: Probability and Queueing Theory
Subject Code: MA2262 Question Bank 2012 - 2013
This problem is sequential queuing model with blocking Arrival rate   1/ hr.
Service time of 1 salesman 1  3/ hr.,Service time of 2 salesman 2  3/ hr.
µ1
(0, 1)
(0, 0)
µ2
(b, 1)
µ1
µ2
λ
(1, 0)
(1,1)
µ1
µ2
From the steady state equation with   1, 1  3, 2  3
we have p0,0  2 p0,1 .....(1)
3 p1,0  p0,0  2 p1,1
.......(2)
3 p0,1  3 p1,0  2 pb,1 .......(3)
5 p1,1  p0,1
2 pb,1  3 p1,1
.......(4)
.. .....(5)
From the boundary condition
p0,0  p1,0  p0,1  p1,1  pb,1  1.....(6)
20
8
10
2
3
, p1,0  , p0,1  , p1,1  , pb,1 
43
43
43
43
43
20 10 30


(i)Proportion of customers entering the shop is equal to p0,0  p0,1 
43 43 43
28
(ii) Ls =   m+n  pm, n  1 p1,0  p0,1   2  p1,1  pb,1  
43
m n
L
30
(iii) Ws  s , A  Average rate entering of super market A  1 ,
A
43
Solve the above equations we get p0,0 
Ws 
28
43  14 hrs.=56min.
30
15
43
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