ESD: Recitation #5
The barbershop revisited
Photo removed due to copyright restrictions.
Infinite number of waiting seats
• One barber, infinite number of chairs for
waiting customers.
• Prospective customers arrive in a Poisson
manner at the rate of λ per hour.
• It takes the barber 1/µ on average to serve a
customer (λ = 0.9 x µ).
• No prospective customer is ever lost.
• What is the average number of customers?
Model
λ
0
λ
1
µ
λ
2
µ
λ
3
µ
4
µ
…
Solving (1)
• What is the probability that N customers
are in the barbershop?
!
!
P1 = P0 ;P2 = P1
µ
µ
" ! %N
PN = $ ' P0
#µ&
)
* Pi = 1 ( P0 =
i= 0
1
1
1
=
=
= 0.1
k
)
)
10
"!%
k
0.9
*$# µ '& *
k= 0
k= 0
Solving (2)
• Average number of customers:
# " &N
PN = 0.1! % ( = 0.1! 0.9 N
$µ'
)
)
k= 0
k= 0
E[Nb _ customers] = * k.Pk =0.1! * k ! 0.9 k = 9
Different service completion rate
• One barber, two chairs for waiting customers.
• Prospective customers arrive in a Poisson
manner at the rate of λ per hour.
• It takes the barber 1/µ on average to serve a
customer. The service completion rate is
described by a second order Erlang pdf.
Assume λ = µ.
• Prospective customers finding the barbershop
full are lost forever.
• What is the average number of customers?
Model
0
2.µ
λ
λ
λ
λ
λ
1
2
3
4
5
2.µ
2.µ
2.µ
2.µ
Arrival rate: fA(x) = λ.e- λ.x, x ≥ 0
Service rate: fS(x) = 4.µ2.x.e-2.µ.x, x ≥ 0
6
2.µ
Solving (1)
• Steady-state probabilities:
!.P0 = 2.µ.P1
(2.µ + !).P1 = 2.µ.P2
(2.µ + !).P1 = 2.µ.P2
(2.µ + !).P2 = ! .P0 + 2.µ.P3
(2.µ + !).P3 = ! .P1 + 2.µ.P4
(2.µ + !).P4 = ! .P2 + 2.µ.P5
2.µ.P5 = ! .P3 + 2.µ.P6
2.µ.P6 = ! .P4
Solving (2)
• Calculate P0:
6
1
3
5
11
21
11
65
" Pk = 1 ! P0 + 2 P0 + 4 P0 + 8 P0 + 16 P0 + 32 P0 + 32 P0 = 16 P0
k= 0
6
16
" Pk = 1 ! P0 = 65
k= 0
Solving (3)
• Average number of customers:
" 8 12 %
16
E[Nb _ customers] = 0 ! + 1! $ + '
# 65 65 &
65
" 2 11 %
" 21 11 %
+2 ! $ + ' + 3 ! $
+
'
# 13 65 &
# 130 130 &
22
E[Nb _ customers] =
( 1.692
13
Additional barber
• Two barbers:
– Adam (takes 1/µ1 on average to serve a customer)
– Ben (takes 1/µ2 on average to serve a customer)
• One chair for waiting customers.
• Prospective customers arrive in a Poisson
manner at the rate of λ per hour.
• Prospective customers finding the barbershop
full are lost forever
Modeling the system
λ
0,0
µ1
µ2
λ
1,0
µ1
µ2
λ
0,1
µ2
λ
λ
1,1
µ1
2,0
2,1
µ1
State of the system: Si,j
i: number of people being serviced by or waiting for Adam
j: number of people being serviced by Ben
The question
• Suppose λ = µ1 = µ2.
• What is the probability that Ben is busy
at a random time?
Solving (1)
• Steady-state probabilities:
!.P0,0 = µ1 .P1,0 + µ 2 .P0,1
(µ1 + ! ).P1,0 = ! .P0,0 + µ1 .P2,0 + µ 2 .P1,1
(µ1 + ! ).P2,0 = ! .P1,0 + µ 2 .P2,1
(µ1 + µ 2 ).P2,1 = ! .P2,0 + !.P1,1
(µ1 + µ 2 + ! ).P1,1 = ! .P0,1 + µ1 .P2,1
(µ 2 + ! ).P0,1 = µ1 .P1,1
Solving (2)
• P{Ben busy} = P0,1 + P1,1 + P2,1
• Using:
P0,0 + P1,0 + P2,0 + P2,1 + P1,1 + P0,1 = 1
• We find:
P{Ben _ busy} =
8
4
22
42
+
+
=
! 0.326
129 43 129 129