Solutions
Queueing Theory
1
17.2-2) Newell and Jeff are the two barbers in a barber shop they own and operate. They
provide two chairs for customers who are waiting to begin a haircut, so the
number of customers in the shop varies between 0 and 4. You are given the
following steady state probabilities for state n=number of customers.
n
Pn
0
1
2
3
4
.0625 .2500 .3750 .2500 .0625
Soln:
n
0
1
2
3
4
Sum =
pn
0.063
0.250
0.375
0.250
0.063
1.000
a. L   npn  2.0
npn
0.00
0.25
0.75
0.75
0.25
2.000
(n-2)pn
0.00
0.00
0.00
0.25
0.13
0.375
# Served Pn
0.00
0.25
0.75
0.50
0.13
1.625
b. Lq  (no.inqueue) pn  0.375
c. Avg. no. served  (no. served ) pn 1625
.
d. Since this is a finite queue, we can not use Little’s formula directly.
    n pn  4 p0  4 p1  4 p2  4 p3
 4(1  p4 )
 4(1.0625)  3.75
Solutions
Queueing Theory
17.2-2) Soln:
d. Since this is a finite queue, we can not use Little’s formula directly.
    n pn  4 p0  4 p1  4 p2  4 p3
 4(1  p4 )
 4(1.0625)  3.75
From Little’s
W
Wq 
L


Lq

2
 0.53
3.75

0.375
 010
.
3.75
e. E [ service time] 
W  Wq 
1

1

1

 W  Wq
 0.53  010
.  0.63
1
Solutions
Queueing Theory
1
17.5-5) A service station has one gasoline pump. Cars wanting gasoline arrive according to
a Poisson process at a mean rate of 15 per hour. However, if the pump already is
being used, these potential customers may balk. In particular, if there are n cars
already at the service station, the probability that an arriving potential customer
will balk is n/3 for n=1, 2, 3. The time required to service a car has exponential
distribution with mean 4 minutes (=15).
a. Construct the rate diagram
b. develop the balance equations
c. find the steady state probabilities
d. find the expected waiting time for cars that stay.
Soln:
15
0
a.
10
1
15
5
2
15
3
15
b.
State
0
1
2
3
Balance
15P1 = 15P0
(10+15)P1 = 15P0 + 15P2
(5+15)P2 = 10P0 + 15P3
15 P3 = 5P2
P1 = P0
P2 = 2/3P1
P3 = 70/45P2
Solutions
17.5-5)
Queueing Theory
15
10
0
State
0
1
2
3
c.
5
1
15
2
15
3
15
Balance
15P1 = 15P0
(10+15)P1 = 15P0 + 15P2
(5+15)P2 = 10P0 + 15P3
15 P3 = 5P2
n
0
1
2
3
Sum =
1
P1 = P0
P2 = 2/3P1
P3 = 70/45P2
p'n
pn
np n
(n-1)p n
1.000
1.000
0.667
1.555
4.222
0.237
0.237
0.158
0.368
1.000
0.00
0.24
0.32
1.10
1.658
0.00
0.00
0.16
0.74
0.893
Solutions
Queueing Theory
1
17.5-5)
c.
d.
n
0
1
2
3
Sum =
L  1655
.
p'n
pn
np n
(n-1)p n
1.000
1.000
0.667
1.555
4.222
0.237
0.237
0.158
0.368
1.000
0.00
0.24
0.32
1.10
1.658
0.00
0.00
0.16
0.74
0.893
Lq  0.893
  15(.237)  10(.237)  5(.157)
 6.71
W
Wq 
1

L 1655
.

 0.247
 6.71
Lq


0.893
 0133
.
6.71
 W  Wq  0.247  0133
.  0114
.
Solutions
Queueing Theory
1
17.5-6) A maintenance person has the job of keeping two machines in working order. The
amount of time that a machine works before breaking down has an exponential
distribution with a mean of 10 hours. The time then spent by the maintenance
person to repair the machine has an exponential distribution with a mean of 8
hours.
A.
b.
c.
d.
e.
Show that this process fits the birth death process and give rates
calculate pn
calculate L, Lq, W, Wq
determine the portion of time the maintenance person is busy
determine the proportion of time a machine is working
Soln:
2/10
0
1
1/8
State
0
1
2
1/10
2
1/8
Balance
2/10P1 = 1/8P0
(1/8+1/10)P1 = 2/10P0 +1/8P2
1/10)P1 = 1/8P2
P1 = 1.6P0
P2 = 1.28P0