Chapter 2
Machine Interference Model
Long Run Analysis
Deterministic Model
Markov Model
Problem Description
•
•
•
•
Group of m automatic machines
Operator must change tools or perform minor repairs
How many machines should be assigned to one operator?
Performance measures
– Operator utilization:  = fraction of time the operator is busy
– Production rate: TH = # finished items per unit time
– Machine availability:  = TH/G, where G is the gross production
rate, or the production rate that would be achieved if each machine
were always available
• Note: In this queuing system, the machines are the
customers!
IE 512
Chapter 2
2
Long Run Analysis
Each machine has gross production rate h
Pn is the proportion of time that exactly n machines are down:

m
n 0
Pn  1
Then, given Pn,
TH   n 0  m  n  hPn
m
m
nP
TH  n 0  m  n  hPn


 1  n
mh
mh
n 0 m
  1  P0
m
IE 512
Chapter 2
3
Eliminate some unknowns
Suppose the mean time to repair a machine is 1/, and the
mean time between failures for a single machine is 1/.
D  t  = avg. # of repairs in (0,t] = t
m
A  t  = avg. # of failures in (0,t] =  n0  m  n  Pnt  mt
In the long run, assuming the system is stable,
D  t   A  t  , so that  t   mt , and


m
h
TH   mh  

IE 512
  1  P0 is still unknown...
Chapter 2
4
Queuing Measures of Performance
•
L = average # of machines waiting for service
   n  1 P   nP  1  P 
m
m
n 1
n 1
n
 m 1     
•

m
n 1
nPn  m 1   
T = average downtime duration of a machine
m
Total machine-hrs down in (0,t]  n1 nPnt m 1   t 1 




A t 
•
0
N = average number of machines down

•
n
 mt
 mt

W = average duration of waiting time for repair
T 
IE 512
1


1 


1

Chapter 2
5
Little’s Formula
Observe from the previous equations:
N  T   m
 1  1 
L   m 
    m W
   
where  m
is the total average number of failures per unit time
= the arrival rate of customers to the queuing system
Little’s formula relates mean # of customers in system to
mean time a customer spends in the system.
IE 512
Chapter 2
6
A Deterministic Model
Suppose each machine spends exactly 1/ time units working
followed by exactly 1/ time units in repair. Then if
 m  1
1


1

and we could stagger the failure times, we would have no
more than one machine unavailable at any time, so that
m  1 
m


 h m h
 1 1 
, 

, TH  

 
m   
  
  
(Otherwise,
IE 512
  1,  

h
, TH 
m

Chapter 2

7
A Markov Model
 j

exp    be the time between the (n-1)st repair
Let n
and the nth failure of machine j, and
Sn exp    be the time duration of the nth repair (indep.)
The time until the first failure is min 11 ,1 2 ,...,1 m exp  m 
N(t) = # of machines down at time t follows a CTMC with
S = {0, 1, …, m} and

 m
 


Q
 0


 0
IE 512
m
   m  1    
0
 m  1 
0
0

0
0
   m  n     
m  n 
0

0
Chapter 2
0 
0 


0 


  
8
Steady-State Probabilities
pn  lim Pr  N  t   n 
t 
satisfy the balance equations
or level-crossing equations
m p0   p1
m p0   p1
  m  1     p
1
 m  1  p1   p2
 m p0   p2
  m  n      p   m  n  1  p
n
n 1
  pn 1
 pm 1   pm
 pm   pm1
IE 512
 m  n   pn   pn1
Chapter 2
9
Solution
n
m!   
pn 
  p0 and
 m  n !   
Say ˆ 

m!   

 n0 pn  1  p0   
 
m

n
!



 n 0
m
m
n




1
1

k
k!
and let G  ˆ , k    n0
ˆ n , so that p0   G  ˆ , m  

 k  n !
Then G  ˆ , k   1  k ˆ G  ˆ , k  1 , k  0,1,...
 G  ˆ , m  1


1 
1



1 
 
m m ˆ m ˆ  G  ˆ , m  
G  ˆ , m 
IE 512
Chapter 2
10
Erlang Distribution
If failure and/or repair times are not exponential, can fit an
Erlang distribution by matching moments:
Solve X  k  , S 2  k  2 simultaneously for k and 
Big advantage: Can still model as a CTMC.
Consider time to machine failure (each machine) as Erlangk.
Can think in terms of k phases in the time to failure, where
the time the m/c spends in each phase is exponential (k):
1
Mean time spent in each phase =
k
1
1

k 
Mean total time to failure =
k
IE 512
Chapter 2
11
Expanded State Definition
Mi(t) = # of machines operating in phase i at time t
(so N  t   m   i 1 M i  t  )
k
and M 1  t  , M 2  t  ,..., M k  t  is a CTMC with


S   l1 , l2 ,..., lk  , 0  li  m, i  1,..., k and 0   i 1 li  m
k
For example, if k = 2, then a single machine without
interference follows the CTMC (1 = failed state):
0;1
2

IE 512
0;2
Chapter 2
2
1
12
Transitions among States (k=2)
Steady state probabilities:
p  l1 , l2   Pr li machines are operating in phase i, i  1, 2
Say l  l1  l2: Rate out of state  l1 , l2  is   2  l1  l2     2l
Rate into state
l1 1, l2 

2(l1+1)
 l1 , l2 
2(l2+1)
IE 512
 l1  1, l2 1
Chapter 2
l1, l2  1
13
Balance Equations
l 0
1  l  m 1
lm
 p  0, 0   2 p  0,1
   2l  p  l1 , l2    p  l1  1, l2   2  l1  1 p  l1  1, l2  1
2  l2  1 p  l1 , l2  1
2 mp  l1 , l2    p  l1  1, l2   2  l1  1 p  l1  1, l2  1
This system of equations (for any k) has the solution:
p  l1 , l2 ,..., lk  
IE 512
 
 l
k
l1 !l2 ! lk !
Chapter 2
p  0, 0,...0 
14
SS Number of Machines Working
From the previous equation and p  l   
l1
l2
 p l , l ,..., l 
1
2
k
lk
get p  l   Pm l (probability that l machines are working)


=
p 0
l!
 l

(Note: typo in (2.47), p.33 text)
Find probabilities by normalizing to 1.
This distribution is independent of k or any other
characteristics of the failure time distribution. It can be
shown that the same state distribution holds for any failure
time distribution!
IE 512
Chapter 2
15