1. No category

INSTITUT FÜR
KOMMUNIKATIONSNETZE
UND RECHNERSYSTEME
Universität Stuttgart
Prof. Dr.-Ing. Dr. h. c. mult. P. J. Kühn
Sample Solution
Teletraffic Theory and Engineering
Date:
March 5, 2008
Problem 1
System Analysis by Phase-Type Distributions
Part 1
Analysis of the Queueing System H2/M/1–s with Finite Capacity
Question 1
State-Transition Diagram
28 Points
a)
The phase model of H2 is
q1
M
λ1
q2
M
λ2
b)
λ 1 , in case Phase 1 was selected; λ 2 , in case Phase 2 was selected.
c)
S = { ( X , I ) 0 ≤ X ≤ s + 1, I = 1, 2 }
X is the number of requests in the system; I the currently selected Phase
for the next arrival.
d)
State transition diagram
q1λ1
0,1
µ
q2λ1
q1λ1
q2λ1
q1λ2
0,2
µ
q2λ2
e)
Question 2
µ
1,1
q1λ1
µ
2,1
q2λ1
q1λ2
µ
1,2
q2λ2
q1λ1
3,1
µ
q2λ1
q1λ2
µ
2,2
q2λ2
q1λ2
3,2
q2λ1
µ
q1λ2
4,2
q2λ2
Blocking occurs in states ( 4, 1 ) and ( 4, 2 ) with transitions
( 4, i 1 ) → ( 4, i 2 ) , i 1 , i 2 = 1, 2
Characteristic Performance Values
11 Points
4
a)
q1λ1
4,1
Y = E[X ] =
∑ 1 ⋅ ( px1 + px2 )
x=1
= 1 – p 01 – p 02
q2λ2
4
b)
Ω =
∑ ( x – 1 ) ⋅ ( px1 + px2 )
x=1
c)
q1 q2
1
------ = E [ T A ] = ----- + ----λA
λ1 λ2
q1 q2
Ω
Little’s Law: w 1 = ------ =  ----- + -----
λA
λ1 λ2
d)
4
∑ ( x – 1 ) ⋅ ( px1 + px2 )
x=1
No Markov arrivals, therfore PASTA not applicable.
p 41 λ 1 dt + p 42 λ 2 dt
p 41 λ 1 + p 42 λ 2
B = ----------------------------------------------------------------=
------------------------------------------------------4
4
4
4
λ 1 dt
∑ px1 + λ2 dt ∑ px2
x=0
x=0
λ1
∑ px1 + λ2 ∑ px2
x=0
x=0
p 41 λ 1 dt + p 42 λ 2 dt
q 1 q2
Alternative: B = -------------------------------------------- =  ----- + ----- ⋅ ( p 41 λ 1 dt + p 42 λ 2 dt )
λA
λ1 λ 2
Question 3
Solution for the Probabilities of State
18 Points
a)
Let i = 2, 1 if i = 1, 2 .
λ i ⋅ p 0i = µ ⋅ p 1i
( λ i + µ ) ⋅ p xi = µ ⋅ p x + 1, i + q i λ i ⋅ p x – 1, i + q i λ i ⋅ p x – 1, i
( q i λ i + µ ) ⋅ p si = q i λ i ⋅ p s – 1, i + q i λ i ⋅ p s – 1, i + q i λ i ⋅ p si
b)
x = 0
x = 1, …, s – 1
x = s
Take p 01 and p 02 for granted. Express p x + 1, i for x = 0, …, s – 1 by
{ p 01, p 02, …, p x – 1, 1, p x – 1, 2 } and thus by { p 01, p 02 } . Express p s1 with
the equation for x = s by { p s – 1, 1, p s – 1, 2, p s2 } and thus by { p 01, p 02 } .
This yields a second different expression for p s1 .
s
Together with the normalization condition
2
∑ ∑ pxi
= 1 , which reduces
x = 0i = 1
to a linear combination of p 01 and p 02 that equals 1, solve the equations for
p 01 , p 02 , and p s1 .
Problem 1
Page 2
Part 2
Phase-Type Queueing Systems with Non-Renewal Arrival Processes
Question 4
Generalization of the Hyperexponential Arrival Process
22 Points
a)
H2 as PH-RP
λ1
λ2
M1
M2
1
q1
q2
1
3
b)
H2 as PH-MRP
λ1
λ2
M1
M2
q1
q2
q2
q1
c)
Alternating PH-MRP
λ1
λ2
M1
M2
1
1
Question 5
Queueing Analysis with Non-Renewal Arrivals
14 Points
a)
S = { ( X , I ) 0 ≤ X ≤ s + 1, I = 1, 2 }
X is the number of requests in the system; I the currently selected Phase
for the next arrival.
b)
State Transition diagram
0,1
µ
λ1
1,1
λ1
λ2
0,2
Problem 1
µ
µ
2,1
λ1
λ2
1,2
µ
µ
λ2
2,2
µ
3,1
λ1
λ2
3,2
Page 3
Question 6
Clustered Arrival Processes
13 Points
a)
Define λ 1 = 1 ⁄ E [ T A1 ] and λ 2 = 1 ⁄ E [ T A2 ] .
λ2
λ2
λ2
λ1
M2
M3
M4
M1
1
1
1
1
b)
S = { ( X , I ) 0 ≤ X ≤ s + 1, I = 1, 2, 3, 4 }
X is the number of requests in the system; I the currently selected Phase
for the next arrival.
Problem 1
Page 4
Problem 2
Ethernet LAN Design
Part 1
Switch Design
Question 1
Individual arrivals at ports are Poisson processes. Therefore the combined arrival process at the switch is due to the superposition theorem again Poisson.
4 Points
N
The rate λ of this process is λ =
Question 2
16 Points
∑i = 1 λ i .
Queueing Model
a)
Each output port needs depending on the link model (not given) no queue or
its own queue. Incoming frames must be queued in front of the switching
fabric. Due to the FIFO strategy (no scheduling required on incoming frames) one queue is sufficient here. Thus, 1 or N + 1 queues are needed.
…
b)
…
…
M
The output process of the M/M/1 switching fabric is Poisson again according to Burke’s theorem. It is split probabilisticly towards the output queues
according to the p i . Due to the splitting theorem, the arrival process is there
is Poisson again, with rates p i λ .
c)
d)
Question 3
5 Points
The system is stable if all queues are stable. As no link model is given, only
λh < 1 must hold.
h
Flow time theorem: t S = t W = --------------1 – λh
Queueing Model
a)
Similar to 3a), but this time no queue is needed in front of switching fabric,
as it can process all incoming requests immediately. Thus, 0 or N queues
are needed.
D
D
…
…
…
…
b)
Problem 2
Output queues must be stable only. As no link model is given the system
can be assumed stable.
Page 5
Question 4
Switching Fabric
7 Points
a)
Little’s Law: E [ X ] = λd
b)
Probability distribution is identical to state distribution of M/M/n loss
system with n → ∞ .
( λd ) x
-------------( λd ) x –λd
x!
p x = lim ---------------------=
-------------- e , thus Poisson distributed.
x!
n→∞ n
( λd ) i
∑ -----------i!
i=0
Part 2
Topology Analysis
Question 5
Transmission Time T T
14 Points
a)
 1
 - e – x / b , for x ≥ 0
g(x) =  -b

0 , for x < 0

b)

C
X

– -----L-t

P { T T ≤ t } = P  ------ ≤ t  = P { X ≤ C L t } =  1 – e b , for t ≥ 0
 CL 

0
, for t < 0


1
 1--- e – --h- t , for t ≥ 0
b
h = ------ ; f T(t) =  h
CL

 0 , for t < 0
∞
φ T(s) =
∫
0
c)
Question 6
∞
e –st
⋅ f T(t) dt =
…
∫
0
∞
1

1
1 – --- t
1 – s + --h- t
1⁄h
e –st ⋅ --- e h dt = --- ∫ e
dt = -----------------h
h
s+1⁄h
…
0
M/M/1–∞ pure delay system
Transmission Delay (points A i to B i )
25 Points
x
a)
TF
x
= TT +
∑ TT is Erlang- x + 1 distributed
i=1
b)
Geometric distribution with P { X = x } = ( 1 – ρ )ρ x .
∞
G(z) =
∑ P { X = x }z x
x=0
Problem 2
∞
= (1 – ρ)
∑ ( ρz ) x
1–ρ
= --------------1 – ρz
x=0
Page 6
X
c)
T F = TT +
∑ TT
i=1
d)
e)
1⁄h
1–ρ
1 ⁄ h ⋅ (1 – ρ)
φ(s) = φ T(s) ⋅ G(φ T(s)) = ------------------ ⋅ ------------------------------- = ----------------------------------------------s+1⁄h
1⁄h
s + (1 ⁄ h – ρ ⋅ 1 ⁄ h)
1 – ρ -----------------s+1⁄h
1 ⁄ h ⋅ (1 – ρ)
= ---------------------------------------s + 1 ⁄ h ⋅ (1 – ρ)
h
The PDF is negative exponentially distributed with mean ------------ .
1–ρ
fF(t)
1/h(1–ρ)
t
0
Question 7
3 Points
Propagation Delay (points B i to C i )
The PDF only consists of a dirac impulse at d. f P(t) = δ(t – d)
fP(t)
1
0
Question 8
d
t
Complete Delay (points A i to C i )
17 Points
– 1 ⁄ h ( 1 – ρ ) ⋅ ( t – d ) , for t ≥ d

P { TD ≤ t } = P { TF ≤ t – d } =  1 – e
0
, for t < d

– 1 ⁄ h ( 1 – ρ ) ⋅ ( t – d ) , for t ≥ d

f(t) = f F(t – d) =  1 ⁄ h ( 1 – ρ ) ⋅ e
0
, for t < d

f(t)
1/h(1–ρ)
0
a)
Problem 2
d
t
The output process at point B i is Poisson with rate λ , due to Burke’s theorem. The added constant propagation delay does not change the interarrival
time of the frames, so the output process at point C i is still Poisson with the
same rate.
Page 7
b)
Question 9
1 ⁄ h ⋅ (1 – ρ)
T D = T F + d , thus ψ(s) = φ(s) ⋅ e – sd = ---------------------------------------- ⋅ e –sd
s + 1 ⁄ h ⋅ (1 – ρ)
c)
Between points C i and A i + 1 is another infinite server with negative exponentially distributed processing delay. According to Burke’s theorem for
n → ∞ the output process is Poisson with rate λ .
a)
Occupancy is defined by transmission of frames. Therefore the M/M/1
delay system defines the occupancy of the link.
b)
h ⋅ λ 12 = h ⋅ λ 01 , therfore ρ 12 = λ 01 h
10 Points
h ⋅ λ 23 = h ⋅ ( λ 02 + λ 12 ) ( 1 – q 20 ) , therefore
ρ 23 = ( 1 – q 20 ) ( λ 02 h + ρ 12 ) = h ( 1 – q 20 ) ( λ 02 + λ 01 )
h ⋅ λ 34 = h ⋅ ( λ 03 + λ 23 ) ( 1 – q 30 ) , therefore
ρ 34 = ( 1 – q 30 ) ( λ 03 h + ρ 23 ) = h ( 1 – q 30 ) ( λ 03 + ( 1 – q 20 ) ( λ 02 + λ 01 ) )
h ⋅ λ 45 = h ⋅ ( λ 04 + λ 34 ) ( 1 – q 40 ) , therefore
ρ 45 = ( 1 – q 40 ) ( λ 04 h + ρ 34 )
= h ( 1 – q 40 ) ( λ 04 + ( 1 – q 30 ) ( λ 03 + ( 1 – q 20 ) ( λ 02 + λ 01 ) ) )
+
–
+
–
4
3
Question 10 ψ 15(s) = p 1 ⋅ ψ(s) + p 1 ⋅ ( ψ(s) ) = ψ(s) ( p 1 + p 1 ⋅ ( ψ(s) ) )
4 Points
Problem 2
1 ⁄ h ⋅ (1 – ρ)
1 ⁄ h ⋅ (1 – ρ) 3
+
–
= ----------------------------------------  p 1 + p 1 ⋅  ---------------------------------------- 
 s + 1 ⁄ h ⋅ ( 1 – ρ ) 
s + 1 ⁄ h ⋅ (1 – ρ) 
Page 8