Final Exam
1. Consider a single server queue with no waiting room. Two independent Poisson arrival streams visit this
queue. The first stream customers arrive at a rate of 1/hour and require exponentially distributed service with
mean time equaling 30 minutes. The second stream customers arrive at a rate of 2 per hour and they require
exponential amount of service with rate 1/3 per hour. No waiting is allowed in this system so the customer
forced to wait leaves the system. The first stream customers have preemptive priority over second stream
customers, i.e., if an arriving first stream customer sees the second stream customer at service it immediately
displaces it and starts service while the displaced customer leaves the system.
a) What are the states for this CTMC? (2 points)
States: 0,1,2. Where 0 is when there is no one in the system, 1 when there is a type 1 customer and 2 when
there is a type 2 customer.
b) Draw the rate diagram for this CTMC (3 points)
The following table shows the rates at which the system goes from states in the first column to states in the
first row.
0
1
2
0

1
2
1
2


2
1/3
1

c) Find the steady state probabilities for this CTMC (3 points)
Balance equations yield:
2P(1) = P(0) + P(2)
4P(2)/3 = 2P(0)
P(0) + P(1) + P(2) =1
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So that P(2) = 2/5, P(1) = 1/3 and P(0) = 4/15
d)
What is the rate at which customers of stream 2 complete service (i.e., on average how many customers
of stream 2 complete service in 1 hour)? What proportion of customers of stream 2 get full service from the
server? (4 points)
Type 2 customers enter the system only when the system is in state 0. Hence, they arrive at a rate 2P(0)=8/15
per hour.
A type 2 customer is fully served if no type 1 customer arrive before the end of time of service. Hence, only a
fraction (1/3) / (1+1/3) = ¼ of type 2 customers that start service are allowed to get full service. Therefore,
type 2 customers complete service at a rate 2/15 per hour.
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 Winter '11
 Ramesh
 Probability theory, #, 1 Hour, 6 minutes, jobshop

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