Universität Stuttgart Sample Solution
Transcription
Universität Stuttgart Sample Solution
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