Page 111 - DCAP601_SIMULATION_AND_MODELING

Unit 7: Simulation of Queuing System (I)

that is Notes

dPn(t) 1  1 1  1
 P n 1 (t)     P (t)  P n 1 ...(9)
n
dt      
Equation (9) holds for all n > 0. When n = 0, the contributions made by the P terms would be
n-1
zero. Therefore
dP (t)  1 P (t)  1 P (t) ...(10)
0
dt  1  0
P (t) can be obtained by solving the two differential Equations (9) and (10). We are most interested
n
in a steady state solution. If  < 1, after the passage of a sufficiently long time the queue would
reach an equilibrium, and P (t) would converge to a constant. Thus we can set
n
dP
n
 0
dt
Applying this to Eq. (10) and (9) gives, respectively


P  P (t)   P 0 ...(11)
1
0

and P n 1  (1  ).P   .P n 1 ...(12)

n

Repeated substituion of Eq. (11) into (12) gives
n
Pn =  P ,
0
for all n > 0 and  < 1. ...(13)
Clearly

 P  1
n

n 0
Therefore


n
P 0   1

n 0
1

or P  1, since < 1
0
1  
Therefore,
P = (1 – ),
0
the same result as in Eq. (8).
The average number of customers in the system is given by

 
 n.P  P 0 n.  ...(14)
n
n
n 0 1  

The average number of customers in the system but not being served (i.e., the average queue
length) is the difference of the average number of customers in the system and the probability
that at least one customer is in the system (which is nothing but ). Thus

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