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Question 11 – 3
0.66670.6
0.4
μ
λP f)
5min0.6
13.3333
μ
1W We)
3.3333min0.4
1.3333
λ
L Wd)
20.6
0.41.3333
μ
λLL c)
1.33330.4)0.6(0.6
(0.4)
λ)μ(μ
λL b)
0.33330.6
0.41
μ
λ1P a)
w
q
q
22
q
0
Question 11 – 9
0477.0)56.0(P
1084.0)56.0(P
2464.0)56.0(P
3
0
3
2
0
2
0
5
2.2P d)
5
2.2P c)
5
2.2P b)
0.565
2.21
μ
λ1P a)
3
2
1
0
Question 11 – 9 cont.
minutes) (9.43 hours 0.157λ
L W
0.34572.2)5(5
(2.2)
λ)μ(μ
λL f)
0.0375
0.96251)PPP(P1
system)in are 3 than P(More waiting)2 than P(More e)
22
q
3210
Question 11 – 11 a.
0.41676
2.5
μ
λP
hours 0.2857μ
1WW
minutes) (7.14 hours 0.1190λ
LW
0.7143μ
λLL
0.29762.5)6(6
(2.5)
λ)μ(μ
λL
hourper customers 61060μ 2.5λ
w
q
q
22
q
Question 11 – 11 b, c.
met. being is goal service The
minutes) (4 hours 0.0667λ
L W
0.16672.5)7.5(7.5
(2.5)
λ)μ(μ
λL
hourper customers 7.5860μ c)
.consultant second a hire
or consultant for the )( rate servicemean the
increase should Firm minutes. 7.14 WNo; b)
22
q
q
Queueing Theory: Part II
Elementary Queueing Process
C C C C C C C
CCCC
SS ServiceS facilityS
Customers
Queueing system
Queue
Served customers
Served customers
Relationships between and,,, qLWL .qW
.WL
Assume that is a constant for all n.
In a steady-state queueing process,
n
.qq WL
Assume that the mean service time is a constant, for all It follows that,
.1
qWW
.1n1
The Birth-and-Death Process
Most elementary queueing models assume that the inputs and outputs of the queueing system occur according to the birth-and-death process.
In the context of queueing theory, the term birth refers to the arrival of a new customer into the queueing system, and death refers to the departure of a served customer.
The birth-and-death process is a special type of continuous time Markov chain.
State: 0 1 2 3 n-2 n-1 n n+1
2n 1n n210
1 2 3 1n n 1n
n n and are mean rates.
Rate In = Rate Out Principle.
For any state of the system n (n = 0,1,2,…),
average entering rate = average leaving rate.
The equation expressing this principle is called the balance equation for state n.
State
0
1
2
n – 1
n
0011 PP
Rate In = Rate Out
1112200 )( PPP
2223311 )( PPP
11122 )( nnnnnnn PPP
nnnnnnn PPP )(1111
)(1
)(1
11223
23
23
00112
12
12
01
01
PPPP
PPPP
PP
0123
0122
3
2
012
011
2
1
PP
PP
State:
0:
1:
2:
To simplify notation, let
,11
021
nn
nnnC for n = 1,2,
…
and then define for n = 0.
Thus, the steady-state probabilities are
1nC
,0PCP nn for n = 0,1,2,…
The requirement that
10
nnP
implies that
,100
PCn
n
so that
.1
00
nnCP
The definitions of L and specify thatqL
.)(,0
sn
nqn
n PsnLnPL
,
LW
LW
.0
n
nnP
is the average arrival rate. is the mean arrival rate while the system is in state n. is the proportion of time for state n,
nnP
The Finite Queue Variation of the M/M/s Model]
(Called the M/M/s/K Model)
Queueing systems sometimes have a finite queue; i.e., the number of customers in the system is not permitted to exceed some specified number. Any customer that arrives while the queue is “full” is refused entry into the system and so leaves forever.
From the viewpoint of the birth-and-death process, the mean input rate into the system becomes zero at these times.
The one modification is needed
0
n
for n = 0, 1, 2,…, K-1
for n K.
Because for some values of n, a queueing system that fits this model always will eventually reach a steady-state condition, even when
0n
.1k
Question 1Consider a birth-and-death process with just three attainable states (0,1, and 2), for which the steady-state probabilities are P0, P1, and P2, respectively. The birth-and-death rates are summarized in the following table:
Birth Rate Death RateState
012
110
_22
(a)Construct the rate diagram for this birth-and-death process.
(b)Develop the balance equations.(c)Solve these equations to find P0 ,P1 , and P2.(d)Use the general formulas for the birth-and-death
process to calculate P0 ,P1 , and P2. Also calculate L, Lq, W, and Wq.
Question 1 - SOLUTINONSingle Serve & Finite Queue(a) Birth-and-death process
0 1 2
10
21
11
22
(b) In Out
1
2
321
2
210
21
120
01
PPP
PP
PPP
PP (1)Balance
Equation(2)
(3)
(4)
(b)
02
02
020
020
4
12
12
2
32
)2
1(32
PP
PP
PPP
PPP
)2()1(2
1 (1) From 01
PP
7
41
4
124
14
1
2
1
00
000
PP
PPP
From (4)
so
)7
4(
4
1
7
1)
7
4(
2
1
7
2
7
4210 PPP
6
1
76
71
3
2
6
4
6
7
7
4
7
6
7
24
7
21
7
41
7
1)1(
7
10
)1()0(7
4
7
2
7
2
)2(7
1)1(
7
20
)2()1()0(
1100
21
210
q
LW
LW
PP
PPL
PPPL
Question 2Consider the birth-and-death process with the following
mean rates. The birth rates are =2, =3, =2, =1, and =0 for n>3. The death rates are =3 =4 =1
=2 for n>4.
(a)Construct the rate diagram for this birth-and-death process.
(b)Develop the balance equations.(c)Solve these equations to find the steady-state probability
distribution P0 ,P1, …..(d)Use the general formulas for the birth-and-death process
to calculate P0 ,P1, ….. Also calculate L ,Lq, W, and Wq.
0 1 2 3n
n1 2 3
Question 2 - SOLUTION
(a)
0 1 2
2
3
3
4
3 4
2
1
1
2
(b)
43
342
231
120
10
21
222
613
642
32
PP
PPP
PPP
PPP
PP
143210 PPPPP
(1)
(2)
(3)
(4)
(5)
(6)
(c)
02
02
020
2
1
24
442
PP
PP
PPP
)3
2(642)2(
)1(3
2
020
01
PPP
PP
231 63)3( PPP
03
030
030
32
)2
(6)3
2(3
PP
PPP
PPP
342 222)4( PPP
040
040
340
22
)(22)2
1(2
22)2
1(2
PPP
PPP
PPP
04 2
1PP So,
11
3
22
6
16
226
36346
)2
11
2
1
3
21(
2
1
2
1
3
2
)6(
0
0
0
0
0000043210
P
P
P
P
PPPPPPPPPP
So,
22
3)
11
3(
2
1,
11
322
3)
11
3(
2
1,
11
2)
11
3(
3
2,
11
3
403
210
PPP
PPP
(d)
11
20
22
40
22
12186422
12
11
9
22
6
11
2
)22
3(4)
11
3(3)
22
3(2)
11
2(1
P4P3P2P1P0L 43210
11
18
22
36
22
66121211
3
22
6
11
6
11
6
)11
3(1)
22
3(2)
11
3(3)
11
3(2
11
12
22
24
22
9123
22
9
11
6
22
3
)22
3(3)
11
3(2)
11
3(1
3210
33221100
4321
PPPP
PPPPLq
3
2
18
12
11181112
9
10
18
20
11181120
LW
LW
