Transcript No Slide Title

Question 11 – 3
λ
0.4
a) P0  1   1 
 0.3333
μ
0.6
λ2
(0.4)2
b) L q 

 1.3333
μ(μ  λ)
0.6(0.6 0.4)
λ
0.4
c) L  L q   1.3333
2
μ
0.6
Lq
1.3333
d) Wq 

 3.3333min
λ
0.4
1
1
e) W  Wq   3.3333
 5min
μ
0.6
f) Pw
λ
0.4


 0.6667
μ
0.6
Question 11 – 9
λ
2.2
a) P0  1   1 
 0.56
μ
5

2.2
b) P1   P0 
(0.56)  0.2464
5

2
2
3
3

 2.2 
c) P2    P0  
 (0.56)  0.1084
 5 


 2.2 
d) P3    P0  
 (0.56)  0.0477
 5 

Question 11 – 9 cont.
e) P (Moret han2 wait ing)  P (Moret han3 are in syst em)
 1  (P0  P1  P2  P3 )  1  0.9625
 0.0375
λ2
(2.2)2
f) L q 

 0.3457
μ(μ  λ) 5(5  2.2)
Wq 
Lq
λ
 0.157hours (9.43minut es)
Question 11 – 11 a.
λ  2.5 μ  60 10  6 cust omersper hour
λ2
(2.5)2
Lq 

 0.2976
μ(μ  λ)
6(6  2.5)
λ
L  L q   0.7143
μ
Wq 
Lq
λ
 0.1190hours (7.14minut es)
1
W  Wq   0.2857hours
μ
λ
2.5
Pw 

 0.4167
μ
6
Question 11 – 11 b, c.
b) No; Wq  7.14minut es.Firm should increase
t hemean servicerat e() for t heconsult antor
hire a secondconsult ant.
c) μ  60 8  7.5 cust omersper hour
λ2
(2.5)2
Lq 

 0.1667
μ(μ  λ) 7.5(7.5 2.5)
Wq 
Lq
 0.0667hours (4 minut es)
λ
T heservicegoal is being met .
Queueing Theory: Part II
Elementary Queueing Process
Served customers
Queue
Customers
Queueing system
C
C
CCCCCCC C
C
Served customers
S
S
S
S
Service
facility
L,W , Lq , and Wq .
n is a constant  for all n.
Relationships between
Assume that
In a steady-state queueing process,
L  W .
Lq  Wq .
Assume that the mean service time is a constant,
1  for all n  1. It follows that,
W  Wq 
1

.
The Birth-and-Death Process
Most elementary queueing models assume that the
inputs and outputs of the queueing system occur
according to the birth-and-death process.
In the context of queueing theory, the term birth
refers to the arrival of a new customer into the
queueing system, and death refers to the departure
of a served customer.
The birth-and-death process is a special type of
continuous time Markov chain.
0 1 2
State: 0
1
2
3
1  2 3
n
and
n
n2 n1 n
 n-2 n-1
n
n+1 
n1 n n1
are mean rates.
Rate In = Rate Out Principle.
For any state of the system n (n = 0,1,2,…),
average entering rate = average leaving rate.
The equation expressing this principle is called the
balance equation for state n.
State
Rate In = Rate Out
0
1P1  0 P0
1
0 P0  2 P2  (1  1 ) P1
2
1P1  3 P3  (2  2 ) P2


n–1
n

n2 Pn2  n Pn  (n1  n1 ) Pn1
n1Pn1  n1Pn1  (n  n ) Pn

State:
0:
1:
2:
0
P1 
P0
1
1
1
10
1
P2 
P1 
( 1 P1  0 P0 ) 
P1 
P0
2
2
2
 2 1
210
2
2
1
P2 
P0
P3 
P2  (  2 P2  1 P1 ) 
3
3  2 1
3
3
To simplify notation, let
n 1n 2  0
Cn 
,
 n  n 1  1
for
n = 1,2,…
and then define Cn  1 for n = 0.
Thus, the steady-state probabilities are
Pn  Cn P0 , for n = 0,1,2,…
The requirement that

P
n 0
n
1
implies that


  Cn  P0  1,
 n 0 

so that
1


P0    Cn  .
 n 0 

The definitions of L and Lq specify that


n 0
ns
L   nPn , Lq   (n  s) Pn .
W
L
Wq 

Lq

,
 is the average arrival rate. n is the mean
arrival rate while the system is in state n. Pn is the
proportion of time for state n,

   n Pn .
n 0
The Finite Queue Variation of the M/M/s Model]
(Called the M/M/s/K Model)
Queueing systems sometimes have a finite queue;
i.e., the number of customers in the system is not
permitted to exceed some specified number. Any
customer that arrives while the queue is “full” is
refused entry into the system and so leaves forever.
From the viewpoint of the birth-and-death process,
the mean input rate into the system becomes zero
at these times.
The one modification is needed

n  
0
for n = 0, 1, 2,…, K-1
for n  K.
Because n  0 for some values of n, a
queueing system that fits this model always will
eventually reach a steady-state condition, even
when    k  1.
Question 1
Consider a birth-and-death process with just three attainable
states (0,1, and 2), for which the steady-state probabilities
are P0, P1, and P2, respectively. The birth-and-death rates are
summarized in the following table:
State
0
1
2
Birth Rate
1
1
0
Death Rate
_
2
2
(a)Construct the rate diagram for this birth-and-death
process.
(b)Develop the balance equations.
(c)Solve these equations to find P0 ,P1 , and P2.
(d)Use the general formulas for the birth-and-death
process to calculate P0 ,P1 , and P2. Also calculate L, Lq,
W, and Wq.
Question 1 - SOLUTINON
Single Serve & Finite Queue
(a) Birth-and-death process
0  1
0
1  1
1
1  2
(b)
In
2
2  2
2 P1  P0
Out
(1)
1P0  2 P2  3P1
(2)
P1  2 P2
(3)
P0  P1  P2  1
(4)
Balance
Equation
1
(b) From (1)  P1  P0
2
(1)  (2)
P0  2 P2
P0  2 P2
2 P2
P2
1
 3( P0 )
2
3
 P0
2
1
 P0
2
1
 P0
4
From (4)
1
1
P0  P0  P0  1
2
4
4  2 1
4
P0  1  P0 
4
7
so
4
P0 
7
2 1 4
1 1 4
P1   ( ) P2   ( )
7 2 7
7 4 7
L  P0 (0)  P1 (1)  P2 (2)
2
1
 0  (1)  (2)
7
7
2 2 4
  
7 7 7
Lq  P1 (0)  P2 (1)
1
1
 0  (1) 
7
7
4 2 42 6
  0 P0  1 P1  1   1  

7
7
7 7
L 47 4 2
W     
 76 6 3
Lq 1  6  1
Wq 

 
 7 7 6
Question 2
Consider the birth-and-death process with the following
mean rates. The birth rates are 0 =2, 1 =3, 2=2, 3 =1,
and n =0 for n>3. The death rates are 1=3  2=4 3 =1
n =2 for n>4.
(a)Construct the rate diagram for this birth-and-death
process.
(b)Develop the balance equations.
(c)Solve these equations to find the steady-state probability
distribution P0 ,P1, …..
(d)Use the general formulas for the birth-and-death process
to calculate P0 ,P1, ….. Also calculate L ,Lq, W, and Wq.
Question 2 - SOLUTION
3
2
(a)
0
1
3
(b)
2
3
2
4
1
2 P0  3P1
(1)
2 P0  4 P2  6 P1
(2)
3P1  1P3  6 P2
(3)
2 P2  2 P4  2 P3
(4)
1P3  2 P4
(5)
P0  P1  P2  P3  P4  1
1
(6)
4
2
(c)
2
P1  P0  (1)
3
2
(2)  2 P0  4 P2  6( P0 )
3
2 P0  4 P2  4 P0
4 P2  2 P0
1
P2  P0
2
(3)  3P1  P3  6P2
P0
2
3( P0 )  P3  6( )
3
2
2 P0  P3  3P0
P3  P0
(4)  2P2  2P4  2P3
1
2( P0 )  2 P4  2 P3
2
1
2( P0 )  2 P4  2( P0 )
2
P0  2 P4  2 P0
1
P4  P0
So,
2
(6) 
2
1
1
P0  P1  P2  P3  P4  P0  P0  P0  P0  P0
3
2
2
2 1
1
 (1    1  ) P0
3 2
2
6 43 63

P0
6
22

P0
6
1
6
3
So, P0 

22 11
3
2 3
2
1 3
3
P0  , P1  ( )  , P2  ( ) 
11
3 11 11
2 11 22
3
1 3
3
P3  P0  , P4  ( ) 
11
2 11 22
(d)
L  0P0  1P1  2P2  3P3  4P4
2
3
3
3
 1( )  2( )  3( )  4( )
11
22
11
22
2 6 9 12
   
11 22 11 22
4  6  18  12 40 20



22
22 11
Lq  0 P1  1P2  2 P3  3P4
3
3
3
 1( )  2( )  3( )
11
11
22
3 6 9 3  12  9 24 12

 



22 11 22
22
22 11
  0 P0  1 P1  2 P2  3 P3
3
3
3
3
 2( )  3( )  2( )  1( )
11
11
22
11
6 6 6 3
  

11 11 22 11
12  12  6  6 36 18



22
22 11
20
L 11 20 10
W 


 18 18 9
11
12
Lq 11 12 2
Wq 

 
 18 18 3
11