# Universität Stuttgart Sample Solution

```INSTITUT FÜR
KOMMUNIKATIONSNETZE
UND RECHNERSYSTEME
Universität Stuttgart
Prof. Dr.-Ing. Andreas Kirstädter
Sample Solution
"Teletraffic Theory and Engineering" (TTE)
"Performance Modelling & Simulation" (PMS)
Date:
March 10, 2010
Problem 1
Markovian Service Systems
Part 1
Single Server Queuing Systems
Question 1
18 Points
a)
State Transition Diagram for Model A
λ
0
b)
µ
λ
1
µ
λ
x
µ
µ
λ
x+1
µ
State Transition Diagram for Model B
λ
0
λ
(1–q)ε
λ
...
1
qε
c)
λ
...
qε
λ
x
qε
(1–q)ε
λ
x+1
qε
(1–q)ε
qε
(1–q)ε
Probability distribution for Model A
i
p A ( i ) = P { X = i } = ( 1 – ρ A )ρ A , i = 0, 1, ..., where ρ A = λ ⁄ µ
d)
e)
f)
Probability distribution for Model B
ρ
ρ i
p A ( i ) = P { X = i } =  1 – -----B-  -----B- , i = 0, 1, ..., where ρ B = λ ⁄ ( µε )

q  q 
Comparison
ρB
p A ( i ) = p B ( i ) ⇒ ρ A = ------ ⇒ µ = qε
q
Model A:
∞
E[ X] =
∑ x ⋅ pA ( x )
x=0
ρA
ρA
= … = Ω + A = ρ A ( 1 – ρ A ) ----------------------2- + ρ A = --------------1 – ρA
( 1 – ρA )
1
1
E [ T F ] = --------------- ⋅ --1 – ρA µ
Model B:
ρB ⁄ q
ρB
E[X]
1
1
E [ X ] = ---------------------- = --------------- ; E [ T F ] = ------------- = --------------- ⋅ --1 – ρB ⁄ q
q – ρB
λ
q – ρB ε
Question 2
13 Points
Delay Analysis of Models A and B
a)
Probability density function of waiting times WRT all requests for Model A
From eq. (3.1) of chapter 4.1 where W = ρA:
w A ( t ) = ( 1 – W )δ ( t ) + Wµ ( 1 – ρ A )e
– µ ( 1 – ρ A )t
= ( 1 – ρ A )δ ( t ) + ρ A ( 1 – ρ A )µe
b)
( 1 – ρ A )µ
LST φ A ( s ) = ( 1 – ρ A ) ⋅ 1 + ρ A --------------------------------s + ( 1 – ρ A )µ
c)
1st
moment:
(1)
wA
= wA
dφ ( s )
= – ------------ds
;2
nd
– µ ( 1 – ρ A )t
moment:
s=0
(2)
wA
2
d φ( s)
= --------------2
ds
s=0
– ( 1 – ρ A )µ ⋅ 1
ρ A ( 1 – ρ A )µ
ρA
dφ ( s )
1
------------- = ρ A ---------------------------------------2- = – ---------------------------------------2- ⇒ w A = --------------- ⋅ --ds
1 – ρA µ
[ s + ( 1 – ρ A )µ ]
[ s + ( 1 – ρ A )µ ]
2
2ρ A ( 1 – ρ A )µ
d φ(s )
--------------=
--------------------------------------2
3
ds
[ s + ( 1 – ρ A )µ ]
d)
Output Processes:
Model A:The output process is Markov, i.e. the inter-departure time CDF is
P{TD ≤ t} = 1 – e–λt, due to the Output Theorem by P. J. Burke.
Model B:The output process is more complicated due to correlated
departures.
Part 2
Markovian Queuing Networks
Question 3
State Analysis
16 Points
2ρ A
1
(2)
⇒ w A = ---------------------- ⋅ -----2
2
( 1 – ρA ) µ
a)
State X = (X1, X2, X3), where Xi = number of jobs in station i, i = 1, 2, 3
b)
Conservation of flow equations:
Station 1: λ1 = λ0 + λ2 + λ3q31
Station 2: λ2 = λ1q12
Station 3: λ3 = λ1(1 – q12)
λ0
⇒ λ 1 = ---------------------------------------------------- ; λ2 = q12λ1 ; λ3 = (1 – q12)λ1
1 – q 12 – q 31 ( 1 – q 12 )
c)
Utililzation factors: ρ1 = λ1/µ1, ρ2 = λ2/µ2, ρ3 = λ3/µ3
Stability condition: ρi < 1, i = 1, 2, 3
⇒
q 12 λ 0 ⁄ µ 2
( 1 – q 12 )λ 0 ⁄ µ 3
λ0 ⁄ µ1
---------------------------------------------------- < 1 ; ---------------------------------------------------- < 1 ; ---------------------------------------------------- < 1
1 – q 12 – q 31 ( 1 – q 12 )
1 – q 12 – q 31 ( 1 – q 12 )
1 – q 12 – q 31 ( 1 – q 12 )
µ
µ3


⇒ λ 0 < min  µ 1 D, ------2-D, ---------------D  , where D = 1 – q 12 – q 31 ( 1 – q 12 )
q 12 1 – q 12 

Problem 1
Page 2
d)
Individual state probability distributions:
x
x
x
p 1 ( x 1 ) = ( 1 – ρ 1 )ρ 1 1 ; p 2 ( x 2 ) = ( 1 – ρ 2 )ρ 22 ; p 3 ( x 3 ) = ( 1 – ρ 3 )ρ 33
e)
Overall state distribution:
3
p ( x 1, x 2, x 3 ) =
∏ pi ( xi )
x
x
x
= ( 1 – ρ 1 ) ( 1 – ρ 2 ) ( 1 – ρ 3 )ρ 1 1 ρ 22 ρ 33
i=1
f)
Average number of jobs
ρ1
ρ2
ρ3
per station: E [ X 1 ] = -------------- , E [ X 2 ] = -------------- , E [ X 3 ] = -------------1 – ρ1
1 – ρ2
1 – ρ3
ρ1
ρ2
ρ3
in system: E [ X ] = E [ X1 ] + E [ X 2 ] + E [ X3 ] = -------------- + -------------- + -------------1 – ρ 1 1 – ρ2 1 – ρ 3
Question 4
Delay Analysis
8 Points
a)
b)
ρ1
ρ2
ρ3
E[X]
1
E [ T F ] = ------------- = … = -----  -------------- + -------------- + --------------
λ0
λ 0  1 – ρ 1 1 – ρ 2 1 – ρ 3
Consideration of one specific job:
First i disk I/Os (station 2), then j I/Os (station 3)
b1) pi,j = q12i (1 – q12)j q31j–1 (1 – q31)
b2) The job is serviced i+j times in station 1.
1
1
1
1
1
1
b3) E [ T F ] i, j = ( i + j ) ⋅ -------------- ⋅ ----- + i ⋅ -------------- ⋅ ----- + j ⋅ -------------- ⋅ ----1 – ρ1 µ1
1 – ρ2 µ2
1 – ρ3 µ3
Problem 1
Page 3
Problem 2
Random Processes in Networks
Part 1
Network Transfer Delays
Question 1
Modeling Transfer Delays
16 Points
a)
Phase model for TF (with TV1):
k phases
D
...
M
M
1/ε
1/ε
M
1/ε
(Erlang-k)
b)
Phase model for TF (with TV2):
M
1/ε1
M
1/ε2
q
D
1–q
(H2)
c)
Probability density functions and Laplace-Stieltjes Transforms:
φC(s) = 1·e–sd
TC: fC(t) = δ(t – d);
k–1
ε k
φ V1 ( s ) =  -----------
 s + ε
( εt )
– εt
TV1: f V1 ( t ) = s ( t ) ⋅ ε ------------------e ;
( k – 1 )!
TV2: f V2 ( t ) = s ( t ) ⋅ ( qε 1 e
–ε1 t
+ ( 1 – q )ε 2 e
– ε2 t
);
ε2
ε1
φ V2 ( s ) = q ------------- + ( 1 – q ) ------------s + ε1
s + ε2
d)
Sketches of PDFs
fF(t)
log(fF(t))
D + Ek
D + H2
qε 1 + ( 1 – q )ε2
qε 1
( 1 – q )ε2
0
Problem 2
d
k
--- + d
ε
t
0
t
d
Page 4
Question 2
18 Points
Characteristic Values
a)
b)
E[TF] = E[TC] + E[TV1] = d + k · 1/ε
VAR[TF] = VAR[TC] + VAR[TV1] = 0 + k · VAR[1 phase] = k · 1/ε2
–ε 1
–ε2
dφ ( s )
φ' V2 ( s ) = ------------- = q --------------------2 + ( 1 – q ) -------------------2
ds
( s + ε1 )
( s + ε2 )
2
2ε 1
2ε 2
d φ( s)
φ'' V2 ( s ) = --------------= q --------------------3- + ( 1 – q ) --------------------32
ds
( s + ε1 )
( s + ε2 )
E [ T V2 ] = – φ' V2 ( 0 ) = q ⁄ ε 1 + ( 1 – q ) ⁄ ε 2
2
2
2
E [ T V2 ] = φ'' V2 ( 0 ) = 2q ⁄ ε 1 + 2 ( 1 – q ) ⁄ ε 2
2
q ( 2 – q ) 1 – q 2q ( 1 – q )
2
2
VAR [ T V2 ] = E [ T V2 ] – ( E [ T V2 ] ) = ------------------- + ------------- – ----------------------2
2
ε1 ε2
ε1
ε2
c)
E[TF] = E[TC] + E[TV2] = d + q / ε1 + (1 – q) / ε2
2
q ( 2 – q ) 1 – q 2q ( 1 – q )
VAR [ T F ] = VAR [ T C ] + VAR [ T V2 ] = 0 + ------------------- + ------------- – ----------------------2
2
ε1 ε2
ε2
ε1
∞
d)
BD =
∞
∫ f F ( t ) dt
=
t0
= qe
∞
∫
t0 – d
–ε1 ( t0 – d )
∫
f V2 ( τ ) dτ =
[ qε 1 e
–ε1 τ
+ ( 1 – q )ε 2 e
– ε2 τ
] dτ
t0 – d
+ ( 1 – q )e
–ε2 ( t0 – d )
–ε2 ( t0 – d )
Remark: B D ≈ ( 1 – q )e
Illustratration (not required):
for ε1 >> ε2, q in reasonable range
f(t)
qε 1 + ( 1 – q )ε2
prob. mass BD
0
d
t0
0
t0 – d
Part 2
Sensor Networks
Question 3
State Analysis of a Limited Buffer Sensor Node
12 Points
τ
τ=t–d
i
a)
Problem 2
t
( λt ) –λt
(Poisson distribution)
p i ( t ) = -----------e
i!
Page 5
∞
b)
∑ i ⋅ pi( c )
E[N ] =
= … = λc (see chapter 2.2)
i=0
∞
2
E[N ] =
∑i
2
2
⋅ p i ( c ) = … = ( λc ) + λc
i=0
c)
d)
 pj ( c )

qj ( c ) = 
∞
s–1
pi ( c ) = 1 – ∑
pi ( c )
 ∑

i=s
i=0
for j = 0, 1, …, s – 1
for j = s
Mean number of data units transferred per polling period (i.e. queued at
s
time c): E [ X ] =
∑j = 0 j ⋅ q j ( c )
Mean number of arriving data units per polling period: E[N] = λc
The probability of loss corresponds to the fraction of untransmitted data
units:
s
E[N ] – E[X ]
1
B = -------------------------------- = 1 – ------ ⋅ ∑
j ⋅ qj ( c )
E[N ]
λc
j=0
Note: B > 0 since
∞
E[N ] =
∑i = 0
i ⋅ pi ( c ) > ∑
s
i=0
i ⋅ pi( c ) + ∑
∞
i = s+1
s ⋅ pi ( c ) = E [ X ] ,
i
( λc ) –λc
where p i ( c ) = ------------e > 0 for i = 0, 1, …, ∞
i!
Question 4
10 Points
State Analysis of an Unlimited Buffer Sensor Node
a)
Moment Method:
mean queue length: Ω
mean waiting time: w
balance equation: w = c / 2 + Ω · c,
where c / 2 is mean of residual time to next polling instant
Little’s law: Ω = λ · w
b)
c
⇒ w = c / 2 + λw · c ⇒ w (1 – λc) = c / 2 ⇒ w = ----------------------2 ( 1 – λc )
Comparison with M/D/1
ρ
λc
w M ⁄ D ⁄ 1 = -------------------- ⋅ h = ----------------------- ⋅ c
2(1 – ρ )
2 ( 1 – λc )
2
wM ⁄ C ⁄ 1 – wM ⁄ D ⁄ 1
c
λc
c ( 1 – λc )
c
= ----------------------- – ----------------------- = ----------------------- = --2 ( 1 – λc ) 2 ( 1 – λc )
2 ( 1 – λc )
2
⇒ wM/C/1 = c/2 + wM/D/1
Reason: In wM/D/1 service starts immediately when the server is available.
In wM/C/1 service starts only at the next clock instant; the mean
time to the next clock instant is c/2.
Problem 2
Page 6
Problem 3
Discrete Event Simulation Method
Remark
This problem is based on a lecture part on simulation which differs from the contents taught since winter term 2010/2011. Consequently, its questions may not
be covered by current lectures.
Part 1
Generation of Stochastic Variables
Question 1
Inversion Method
9 Points
a)
Derivation of T from RN in CDF time diagram:
Remarks:
Known for F(t):
F(t)
P{t1 < T ≤ t2}
= F(t1) – F(t2)
↔ F(t1) < RN ≤ F(t2)
1
RN
RN
F(t1)
F(t2)
t
0
t1
t2
Mapping Process:
RN = F(T)
⇒ T = F–1(RN)
(inversion method)
T
b)
Inversion method:
b1) F(t) = 1 – e–λt ⇒ RN = 1 – e–λT ⇒ T = –1/λ · ln(1 – RN)
Note: As (1 – RN) is also uniformly distributed in (0, 1]:
T = –1/λ · ln( RN) (This avoids a substraction!)
1
b2) F ( t ) = ----------- ( t – a ) for a ≤ t ≤ b
b–a
T–a
F(T) = RN ⇒ ------------ = RN ⇒ T = a + RN · (b – a)
b–a
Illustration:
F(t)
1
RN
0
a
b
t
T
c)
t
F ( t ) = 1 – exp  ---
β
α
⇒ RN = 1 – e
T
---
 β
α
⇒ T = β ⋅ ( ln ( 1 – R N ) )
Note: We can again substitute 1 – RN by RN: T = β ⋅ ( ln ( RN ) )
Problem 3
1
--α
1
--α
Page 7
Question 2
8 Points
Generator Method
a)
Phase model for T:
M
1/ε1
M
1/ε2
q1
1–q1
b)
CDF: F ( t ) = 1 – q 1 e
–ε1 t
Remark: RN = 1 – q 1 e
– ( 1 – q 1 )e
–ε1 T
– ε2 t
– ( 1 – q 1 )e
–ε2 T
cannot be inverted analytically
Generator method: T is ( 1 ⁄ ε 1 ) ln ( RN 1 ) with probability q1
( 1 ⁄ ε 2 ) ln ( RN 2 ) with probability 1 – q1
c)
Flow Chart:
Start
RN := RNG
true
RN ≤ q1
RN := RNG
false
RN := RNG
1
ε1
1
ε2
T := ----- ln(RN)
T := ----- ln(RN)
End
Part 2
Measurement of Variables in an Event-by-Event System Simulation
Question 3
Measurement of TW
10 Points
Problem 3
a)
- Waiting time samples are taken when a customer is taken out of the queue.
- The arrival time (i.e. the instant when the customer is placed in the queue).
- Sample value TW := departure time (dequeue) – arrival time (enqueue)
b)
- Variables:
STW = sum of all waiting times during the batch run
NW = number of customers having to wait during batch run
- Updating for each new sample value TW:
STW := STW + TW
NW := NW + 1
- Mean waiting time at the end of the batch run: E[TW] := STW / NW
Page 8
c)
Question 4
9 Points
- Additional variable STW2 = sum of squares of waiting times
- Updating: STW2 := STW2 + TW2
- Variance computation (at end of batch):
E[TW2] := STW2 / NW
VAR[TW] := E[TW2] – (E[TW])2
Measurement of State Distribution and Average Number of Customers in the
System
a)
- Samples are taken on each state change, i.e. on arrival and departure events
- Sample value TS = the duration of the last state
- TS = time ω of current state change – time α of previous state change
- The time α of the last state change must be registered (auxiliary variable)
Illustration:
X
...
0
...
α
ω
t
b)
- Variable S[X] = array of sums of the periods where the system is in state x
- Updating: S[X] := S[X] + TS (where X is the state between α and ω)
c)
P[X] := S[X] / Tbatch, where Tbatch = duration of the batch
d)
E[X ] =
∑
X ⋅ P[ X]
all X
Problem 3
Page 9
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