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Birth and Death Processes

Example: A Barber Shop  A small town barbershop has two barbers and an additional chairfor 1 waiting customer. If a customer arrives when there are 3 customers in the shop, thecustomer leaves.

•   The customers arrive according to a homogeneous Poisson process. The average timebetween 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 handlesthe customer.

Assume that inter-arrival times and service times are independent exponential random vari-ables.

(a)   Draw a state diagram with possible states and corresponding birth/death rates.

Solution:

22

0

2

 

32 3

E [B] = 30min = 1/2 hrs. implying rate  λ  = 2 per hour.Let T A = time until Andy finishes and  T B  = time until Bob finishes.

For  k  = 1,  D  =  T A; since  µA = 2 per hour, D  ∼   Exp(2)

For  k  = 2, 3, D = min{T A, T B}  giving  D  ∼   Exp(1 + 2 = 3); since  µA = 2 and  µB  = 1

(b)   What is the (large   t) probability that the shop is empty?

S    = 1 + 1

2 +

 2  · 2

2 · 3 +

 2  · 2 · 2

3 · 3 · 2

= 1 + 1 + 4

6 +

  8

18

=   289

implying that  p0 =   1

S   =   9

28

(c)   Let   X   be number of customers in the store. What is the pmf of   X ?

 pk = P [X  = k] = λ0λ1 . . . λk−1µ1µ2 . . . µk

 p0

k   0 1 2 3 pk = P (X  = k)   9

28

9

28

2

2

9

28 ·   2

3

9

28 ·   4

9

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(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 sameas the number that enter the shop.

Y  (t) is a homogeneous Poisson process with rate  λ  = 2 per hour.

Thus Y  (3)   ∼   Poisson(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

0!e−6 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) = F X (2.5) where  X  ∼   Erlang(5, 2)

F X (2.5) can be computed using the Poisson cdf, since   F X (2.5) = 1  −  F Y   (5 −  1)where Y    ∼   Poi(2 × 2.5) i.e., we need to look up the Poisson table for  F Y   (4) whereY    ∼   P oi(5).

From the tables, this is equal to .44 so the required probability is 1  − .44 = .56.

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