Birth and Death Processes
Example: A Barber Shop A small town barbershop has two barbers and an additional chair
for 1 waiting customer. If a customer arrives when there are 3 customers in the shop, the
customer leaves.
• The customers arrive according to a homogeneous Poisson process. The average time
between arrivals is 30 minutes.
• The two barbers are Andy and Bob. Andy completes serving a customer at a rate of
2 per hour. Bob completes a serving a customer at a rate of 1 per hour.
• Because Andy is faster than Bob, if only one customer is in the store, Andy handles
the customer.
Assume that inter-arrival times and service times are independent exponential random variables.
(a) Draw a state diagram with possible states and corresponding birth/death rates.
Solution:
2
2
2
1
0
2
2
3
3
3
E[B] = 30min = 1/2 hrs. implying rate λ = 2 per hour.
Let TA = time until Andy finishes and TB = time until Bob finishes.
For k = 1, D = TA ; since µA = 2 per hour, D ∼ Exp(2)
For k = 2, 3, D = min{TA , TB } giving D ∼ Exp(1 + 2 = 3); since µA = 2 and µB = 1
(b) What is the (large t) probability that the shop is empty?
1 2·2 2·2·2
+
+
2 2·3 3·3·2
4
8
= 1+1+ +
6 18
28
=
9
S = 1+
implying that p0 =
1
S
=
9
28
(c) Let X be number of customers in the store.
pk = P [X = k] =
k
pk = P (X = k)
λ0 λ1 . . . λk−1
p0
µ1 µ2 . . . µk
0
1
9
28
9 2
28 2
1
What is the pmf of X?
9
28
2
·
2
3
9
28
3
·
4
9
(d) What is the (large t) probability that an arriving customer is turned away?
An arriving customer is lost when the shop is full i.e., X = 3
p3 = P (X = 3)
= 4/28 = 1/7
(e) What is the distribution of the number of customers arriving in the first
3 hours? (Name the distribution and it parameter value(s))
Let Y (t) = # of arrivals in t hours; note that the number of arrivals is not the same
as the number that enter the shop.
Y (t) is a homogeneous Poisson process with rate λ = 2 per hour.
Thus Y (3) ∼ P oisson(λt = 2 × 3)
(f) What is the probability that no customer arrives in a 3 hour time interval?
What is the expected number of arrivals in 3 hours?
Since Y (3) ∼ P oisson(6),
P (Y (3) = 0) =
60 −6
e
0!
E[Y (3)] = 6
(g) What is the distribution of the time until the 5th customer arrives? (Name
the distribution and it parameter value(s); As in part (f), an arrival does
not necessarily enter the barbershop)
Let O5 = the time until the 5th customer arrives.
O5 ∼ Erlang(5, 2) since k = 5 and λ = 2
.
(h) What is the probability that the time until the 5th customer arrives is less
than 2.5 hours?
From above, P (O5 ≤ 2.5) = FX (2.5) where X ∼ Erlang(5, 2)
FX (2.5) can be computed using the Poisson cdf, since FX (2.5) = 1 − FY (5 − 1)
where Y ∼ P oi(2 × 2.5) i.e., we need to look up the Poisson table for FY (4) where
Y ∼ P oi(5).
From the tables, this is equal to .44 so the required probability is 1 − .44 = .56.
2