Lecture No 27

 

Held on: Wednesday, November 8, 2000

 

Notes Prepared by:Saket Jain

 

 

 

 

 

 

                                       Queuing theory

 

Queuing Jargon 

 

Example:   Time to failure of a machine has distribution:    f(t) = 0.1e^-0.1t   for t ³ 0.

(a) Find the probability that it fails in 5 minutes.

(b) Given that the machine has lasted for three hours, find the probability that it will last for another 6 hours.

                                                     

Answer:

 (a)        Mean life  =   ò   t * f (t) dt  integrated over the entire possible range of  time , i.e. ,from 0 to  ¥ .

 

For an exponential distribution, mean life  T =  1/ l =  10 hrs.

        P(T£ 5)  =          le^{-l t} dt   

 

(b)     Similarly , we can find the probability of not failure of machine before 6 hrs.

 

        P(T³ 6)  =   ò le^{-l t} dt 

 

 

       Pure Birth process

 

Poisson process is a pure birth process having following characteristics: -

1.                      Only arrivals

2.                      Interarrival time having exponentially distribution

 

These have the following probability distribution of finding  n  number of units at any time  t . Due to the memorylessnes of exponential distributions these probabilities are independent of previous events.

P_n (t)  =      (l t)ⁿ e^{- l t}

                             (N-n)!

 

Pure Death process

 

Poisson arrivals can also lead to a pure death process at some storage facility.

Example:

Consider an inventory store where some amount of inventory is stored in the beginning and as customers took away the units the stock size went down. There is no replenishment of the stock until it is finished.

 

Let initially N no of units were there in the stocks.

Customer demand rate is Poisson distribution with mean rate of demand = m

 

Find the remaining stock after time t.

 

We will find the probability of having n no. of units after time t , i.e.,

 

P_n (t) =    (m t)^N-n _*  e^-mt     =  probability that n stock is left.

                (N-n)! 

 

 

 

Parameters to express a queue

 

Any queue can be represented by its characteristics as shown below:

The   

(a/b/c) : (d/e/f)      for example:    (M/D/10) : (FCFS/N/¥)   where

 

1.      a: interarrival time distribution         

           M : Memoryless

           D : Deterministic

           G: General

           GI: General independent

           E_k: Earlang k distribution

 

2.    b: Service time distribution

 

3.    c: No. of servers in the queue

              

4.    d: Service discipline                  FCFS/LCLS/SIRO/PS

 

5.    f: Size of calling source            Limited or infinite

 

 

 

Birth and Death processes

 

Arrival rates are exponentially distributed with mean arrival rate = l_n. Also the service rate is exponentially distributed with mean service rate = m_n .

 

We will consider n people in the system where there are different states possible as shown below:    

 

 

 

                               l₀                                  l₁                               l₂                           l₃                

 

 

 

                                                                                                            ……

                                                                                                                                                   

 

 

                      m₁                          m₂                                        m₃                         m₄

 

 

Future is still independent of whatever happened in the past but the mean rates of arrival and service which characterises any distribution are different in different state and they are not independent of the state.

 

l_n  and  m_n  are the total rates in any system

C =  No. of servers

m_n  = i*m   if i £ C

 m_n  = C*m   if  i ³ C

 

In a stable system

N_i  =  no. of times we came to state n from either n-1 or n+1 state

N_o  =  no. of times we left state n to go to either n-1 or n+1 state

N   =  total no. of states visited during the stabilisation process

Then,

 

ï N_i - N₀ ï  £  1  then,

 

ï N_i - N₀ ï    £    1      also,     1             0   as N is very large because

      N                   N                  N             the system has stabilised.

 

 

Now to find the fraction of time the process is there in state  n  is P_n .

To find this fraction again consider three consecutive states when the system is stable:

                           l_n-1                                                 l_n 

                                                                                                                                        

                                                                            

 

                               m_n                                                          m_n+1

 

Let T  is the total time for which we have observed the system.

T_n-1 is the time we have been in state n-1

T_n+1 is the time we have been in state n+1

Then  P_n =  T_n ¸ T

 

Now the number of visits to state n from state n-1 = l_n-1T_n-1

Similarly number of visits to n from n+1 = m_n+1T_n+1

Þ  Total no. of visits to state  n  =   l_n-1T_n-1 + m_n+1T_n+1

Similarly the total no. of visits from the state n to either n-1 or n+1 states   =  l_nT_n + m_nT_n

 

Þ      l_n-1T_n-1 + m_n+1T_n+1     =        l_nT_n + m_nT_n

                      T                                      T           

Þ     l_n-1P_n-1 + m_n+1P_n+1  =  ( l_n + m_n ) P_n

Þ     l₀P₀  =  m₁P₁

Now , similarly we can prove that   P₂   =   l₀l₁

                                                                     m₁m₂              

Þ     From mathematical induction  we can prove that

    P_n  =      l₀l₁l₂l₃l₄l₅l_{6………………..} l_n-1   * P₀          "n

                  m₁m₂m₃m₄m₅m₆m_{7…………………}m_n

 

Now,    å P_i  = 1

 

Þ      P₀  =                              1                

                          l₀l₁l₂l₃l₄l_5……….. l_i-1       + 1

                                       m₁m₂m₃m₄m₅m_{6…………}m_i