Problem 1
Denote by Pn , the long-term probability of having n people in the system
Pn = lim P (X(t) = n).
t→∞
Sometimes it is referred to as the steady state probabilities. For example, P6 = 0.5 implies
that the probability of finding 6 people in the system is 0.5. Let’s try to derive equations
for Pn under the assumption of Poisson arrival and departures with constant rate λ and µ
(as you did in the project). Consider P0 . The rate at which system leaves the 0 state is
λP0 (exits per minute). On the other hand, state n = 0 can only be entered by a departure
from state n = 1. The rate at which system goes from state n = 1 to n = 0 is µ, and the
proportion of time the system spends in state n = 1 is P1 , so the rate at which the system
arrives in state n = 0 is µP1 . Since the rate at which the system enters and leaves a certain
state has to be equal, we have
λP0 = µP1 .
Now consider state n = 1. The rate of leaving this state is (λ + µ)P1 (note the system
leaves the system either by arrival or departure). The process can enter this state either by
departing P0 (with rate λP0 ) or departing P2 (with the rate µP2 ). Equating rates we find
(λ + µ)P1 = λP0 + µP2 .
1) Argue similarly and derive the equations for each state Pn .
2) Assume λ < µ and derive analytical expressions for all Pi , i = 0, 1, 2, ..... Note
your answers will only depend on λ and µ. Plot i vs. Pi for certain large range of i (say
i = 1 : 3000).
3) What will you get if you apply your results from part 2 for the case λ > µ?
4) To treat the case λ ≥ µ we will assume that the system can not have more than N
customers, i.e., if the system has N customers then there is no inflow.
a) Write down the equations for such system.
b) Find analytical expressions for Pi , i = 1, 2, ...N .
c) Plot Pi for N = 500.
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