Lecture 3. Contents. Conflict free access

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1 Lecture 3 Conflict free access Contents Fixed resource allocation FDMA DMA Generalized DMA CDMA Dynamic resource allocation Reservation Packet scheduling 7..8

2 M/G/ Poisson arrival process Single server General interarrival time distribution λ arrival rate of the packets General service time distribution Service time of the packets are independent and identically distributed random variables X i service time of customer i X () s Laplace transform of the service time distribution E{exp(-sX i )} X mean service time E{X i } X second moment of service time E{X i } M/G/ Define Wi Waiting (queuing) time of customer i Ri Residual service time when customer i arrives X i Service time of customer i Ni Number of customers in the system upon arrival of customer i. Residual Service time X X X X 3 i X i + X n Customer i arrives R i 7..8 i Wi = Ri + Xi 4 j= i N i

3 M/G/ Expected waiting (queuing) time i E{ Wi} = E{ Ri} + E Xi = R+ λ X W j = i Ni N E{ Ri} = limn E X = λ X N i= λ X W =, ρ = λ X ρ Pollaczek-Khinchin otal time spend in the system = waiting time + service time λ X D = W + X = X +, ρ = λ X ρ ρ = Utilization Fraction of time the server is busy M/G/ Number of customers in the system N We can model the system at departure time and extend the results to all points of time: in the case of Poisson arrival the distribution of N at departure times is the same as at arbitrary points of time (PASA) the N t follows a discrete time Markov process at departure times: N k : number of customers after the departure of a customer k V k : number of arrivals during the service time of customer k, V k Poisson(λ) N N + V N = k k+ k k + V k+ Nk =

4 M/G/ Expressing the steady state of the Markov-chain describing N, we get the z-transform of the distribution of N (Q(z)=E{z N }) ( ) ( ) ρ ( z) Qz ( ) = X λ λz Pollaczek-Khinchintransform X λ λ z z ( ) L-transform of the waiting time W pdf s ( ρ ) W () s = s λ λx () s L-transform of the time spend in the system D=W+X pdf s ( ρ ) D () s = W () s X () s = X () s s X λ+ λ () s Fixed resource allocation 4

5 Fixed resource allocation System parameters Fixed data rate R Fixed message size P Number of user M If the whole bandwidth would be allocated just to single user, the transmission time of the packet would be P/R Each user generates packets according to Poisson process with intensity λ FDMA FDMA system Bandwidth is divided equally among the M users he system can be modeled as M parallel M/D/ queueing systems Data rate per user R/M Service time =MP/R Laplace transform of the service time distribution: X(s)=E{exp(-s)}= exp(-s) λ λ λ Τ Τ Τ M

6 Expected packet delay FDMA λ X λ ρ MP ρ D= W + X = X + = + = + = + λ ρ R ρ λ X Conflict free system: hroughput S = Utilization ρ = λ Normalized delay = Expected delay / Deleay in single user system ˆ D S S M DFDMA = = M + = M = + P/R S S S ( ) 7..8 FDMA 4 3 M= M=5 M= M=5 M= FDMA Maximum load Normalized delay hroughput Delay is rather insensitive Delay increases quickly to the throughput and cannot be tolerated

7 FDMA Delay distribution s s( ρ) s( ρ) e s( ρ) D () s = X () s = = s X s λ λ () s s λ λ e λ+ s λ e Moments m m d E{ D } = ( ) D ( s) ds s= ( ) Calculation of the moments typically require (repeated) application of the Hôpital s rule lim f () s '() lim f s s = s g() s g'() s f() =, g() = s Variance of the packet delay Normalized on = 4 3 var{d} var(x) E{D} EX For high load he jitter is high - - For small load he jitter is small hroughput

8 Variance of the packet delay Symbolic tools such as Maple or Mathematica can be utilized to find the moments Here is example for calculating variance of the packet delay using Matlab s Symbolic oolbox (i.e. Maple) syms s lambda rho=lambda; D=s(-rho)/(lambda+(s-lambda)exp(s)); dd=diff(d); %Derivative dd=diff(dd); %Second derivative m=-limit(dd,s,); %First moment m=limit(dd,s,); %Second moment vard=simplify(m-m^) %Variance S=:.:.99; %hroughput vector %Numerical solution mean_delay=double(subs(subs(m,,),lambda,s)); var_delay=double(subs(subs(vard,,),lambda,s)); %Plot results semilogy(s,var_delay,s,mean_delay) legend('var(d)','e{d}',) xlabel('hroughput') Flexible spectrum use World Radio Conference 7 will identify new specrum for next generation communication systems (IM-A) entative spectrum requirements may be MHz/operator Very high peak data rate requirements ( Mbit/s vehicular, Gbit/s nomadic) he available spectrum may not allow full spectrum to all operators Benefits from availability of full spectrum for a user user impatience the hardware is there, why not let the user benefit from it Spectrum sharing between operators becomes a viable option to investigate

9 Flexible spectrum use Assume that available bandwidth is MHz allowing peak data rate of Mbit/s for vehicular users he number of competing operators is Option : Both operators get 5 MHz band and are not allowed to share the spectrum Option : Operators share the spectrum (e.g. by sharing the hardware) Licensed bands M operators Offered traffic of operator i: λ i Packet delay i X λi X λ Di = X + λi = M + λi ( M ) ( M ) Maximum offered traffic (S i = λ i M<) λi M λ λ λ Μ ΜΤ ΜΤ ΜΤ M

10 Shared spectrum Aggregate traffic λ=σ i λ i Packet delay λ X D= X + λ X i i λ S i i = + λi = + S i i i λ λ λ Μ Τ M Maximum offered traffic (S=S +S <) λ i < i Shared spectrum wo operators share the spectrum ideally based on traffic Packet size: octets 5 Mean packet delay (ms) S 7..8 S

11 Shared spectrum M= Capacity region in terms of offered traffic λ Stable capacity region for fixed spectrum allocation Dividing the resources beforehand 7..8 Causes truncking losses Stable capacity region for flexible spectrum use In heavy load, spectrum sharing does not help λ+ λ < λ If the loading of the cells is uneven spectrum sharing helps DMA In time division multiple access, the time axis is divided into time slots which are preassigned to the different users. he slot assignment follow predefined pattern that repeat itself in a cycle or frame Frame Frame ime 7..8

12 DMA Assume that the MAC layer packet size is selected to be equal to the slot size =P/R he frame size is c =M Each packet in the queue will add c to the waiting time Arrival can take place at any time during the frame Poisson arrivals: Arrivals are uniformly distributed over the time axis => Remaining time till the beginning of next frame is uniformly distributed between and c c Waiting time Service time ime st packet for user Additional delay U(, c ) arrives nd packet for user 7..8 arrives 3 c DMA Waiting time till the end of the frame is uniformly distributed between and c. Hence, the expected time is c c Wc = t dt = c = M c Waiting time in the queue W = ρ ρ q c M, ρ λc λm = = = ( ρ) ( ρ) Service (transmission time) is Expected packet delay st packet for user arrives ρ M D= Wc+ Wq + = M+ M+ = + ( ρ) ( ρ) Additional delay U(, c ) nd packet for user arrives c Waiting time Service time ime

13 DMA ˆ D MP D = = +, S =λ R Normalized packet delay DMA P/R ( S) 4 3 M= M=5 M= M=5 M= DMA Normalized delay hroughput DMA vs FDMA Comparison between FDMA and DMA ˆ M DFDMA = + S ˆ = + D DMA ( S ) ˆ ˆ P M D = D + Dˆ R FDMA DMA DMA he expected delay of the DMA is always less than that of FDMA for M>. he difference grows linearly as a function of users For high load the factor /(-S) dominates, and the relative difference becomes smaller

14 DMA vs FDMA 4 3 M= M=5 M= M=5 M= FDMA (continuous line), DMA (dotted line) Normalized delay hroughput DMA In general the messages are much bigger than MAC layer slot capacity. Hence transmission of a single message requires multiple MAC layer frames. Let L denote the length of the message in time slots. he generating function of L (z-trandform of its pdf) is thus l= { } l Lz ( ) = Pr L= lz d d L = L( z), L + L = L( z) dz z= dz z= Messages arrive according to Poisson process with intensity λ

15 Message delay D is defined as the time elapsing between the message arrival epoch until the transmission of the last packet of the message is completed. Consider an arbitrary tagged message arriving at c w seconds after the beginning of the (j+)st frame. Assume that the tagged message is the ( k +)st message arriving during that frame. he number of messages k arriving before the tagged message follows Poisson distribution with parameter λ w ( ) c DMA Let q j denote the number of packets waiting transmission at the beginning of the (j+)st frame Arrival of k messages agged message arrives q j c w Frame j+ Frame j+ w DMA Delay D M/G/ k D = w + max q, + L + L + { } + j c c i k c c i= q j = j+ L = w ( q j ) k L agged messagek k = + k L k = + = { c L c 3 D L k c ( L + ) k c L k + c c

16 DMA Uniformly distributed time till the beginning of the next frame Probability distibution function f (), w t = t c Characteristic function (L-transform) s c sw st e W () s = E{ e } = e dt = s c M/D/ queue with service time expressed in terms of frame lengths ρ=λ c D ( ) / / () s L s M G () s ρ = s L λ λ () s Characteristic function of the constant - c s( ) s( ) { } c c () s = E e = e c DMA Packet delay D = w + D M / G/+ c Characteristic function of the packet delay hat is, sd sw sd M / G/ s ( c ) { } { } { } { } D () s = E e = E e E e E e = W () s D () s () s e = M / G/ ( ρ ) s ( c ) sc L () s sc s λ λl s ρ e = sc s e () c s λ λl s e () s ( c ) L () s ρ e D () s e L () s sc s ( c ) = c s λ λl () s

17 Expected message delay D= E w + D + { M / G/ c} Normalized delay DMA d λc L D () s c L ds s= = = + + ( λl c ) ˆ D M L ρ D= = M L + + L ρ Generalized DMA he allocation of single slot within the frame to each user is reasonable if the communication requirements are homogeneous. In data transmission, the required capacity of the users vary and it thus makes sense to allocate different number of slots to users based on their needs. Example: User : 3 slots, User : slots, User 3 and 4: slot Frame Frame Slot

18 Generalized DMA Let d(k) denote the time between two consecutive slots allocated to an user Frame 4 4 d() d() d(3) Hofri and Rosberg has shown that the message delay can be minimized by allocating the slots uniformly d(k)=d for all k (if possible) Frame Generalized DMA Slot length: = Number of slots in a frame: M=4 Number of slots allocated to an user: K=4 Message size is uniformly distributed between and 4 It can be seen from the figure that the impact of slot assignment is not very high Optimal allocation

19 Radio channel Fast fading cause packet loss In pedestrian channel, the transmission time interval (I) / slot length is shorter than the coherence time of the channel and thus the packet loss process is correlated. In data channels, the effect of fading is mitigated by using Automatic Repeat request (ARQ). his translates the packet loss to packet delay. Second generation cellular systems utilized traditional ARQ in which erroneous packet is discarded by the receiver and retransmitted by the transmitter Modern wireless networks utilizes hybrid ARQ, in which the receiver combines the original packet with the retransmitted copy of the packet (chase combining) or additional code bits provided by the transmitter (incremental redundancy) ARQ Assume for simplicity that the message is one MAC layer packet long and the MAC packet contains error detection coding such that the receiver is able to detect lost packets. In FDD systems, we can utilize time offset between uplink and downlink such that the ARQ feedback delay is less than the frame length Uplink Downlink NACK ACK

20 ARQ We could model the channel state as a Markov chain, in which each state corresponds to different block error rate BLER. (Gilbert-Elliot channel model) he simplest channel model consists of two sates: Good (no error) and Bad (packet lost) p P P p =-p Good Bad p =-p p Block error probability PBLER = P Average length of an error burst B = p In steady state pp = Pp P + P = p P = P, P = p p + p ARQ he service time distribution of the packet is geometric P l = Pr{ L= l} = l P( p) p l =,3,4... Channel was bad during the first X attempt stayed bad for l- slots and finally became good again

21 ARQ ransmission time in terms of frames l l ( ) ( ) l= l= L = P + P l p p = P p P + P l p p = + P pp P p l l L = P + P l ( p ) p = P p P + P l ( p ) p l= l= ( p ) = P pp+ P + p p Packet delay c λ L D = c L + + ( λl c ) ARQ Retransmissions contribute to the load ρ = λl c L 4 3 Packet delay 6 4 B..4 BLER B..4 BLER

22 CDMA Single rate CDMA system Spreading factor is S=W/R ransmit power control is utilized to requlate the receive power Q Noise power is ν Other-to-own cell interference ratio is i Required Signal-to-noise+interference ratio (SINR) at the receiver is Γ Q Γ= S ( + i) ( M ) Q+ υ ΓS υ Q = +ΓS ΓS ( + i) M +ΓS CDMA Maximum tolerable noise raise over thermal (Ro) (also known as noise raise) is η ( + i) ( M ) Q+ υ η η M υ η ΓS ( + i) + ΓS Hence, the Ro constraint defines the number of parallel channels. Now for given chip rate W, SINR-target Γ and Ro η, the system appears as M parallel channels with each having roughly portion /M of the overall bandwidth η W R = for large M η Γ M M M Hence, with fixed rate and number of channels, the packet delay performance of CDMA is similar to FDMA

23 CDMA vs FDMA In modern CDMA systems, the data rate of the user is controlled by packet scheduler and the data rate can be set based on the current interference level (i(t) and M(t)) his avoids the trunking losses caused by the fixed channel allocation used in FDMA Hence, CDMA is more flexible than FDMA CDMA vs DMA If dynamic conflict free packet scheduler is used also in DMA type of slotted system, then both CDMA (simultanous transmissions of several users) and DMA (allocating all the resources to signle user at the time) approaches has their merits. In brief, CDMA provides gains if the interference power is large and uncontrollable (uplink) while DMA is beneficial if the interference can be kept small (downlink). More detailed analysis is provided in the course S-7.36 Radio Resource Management Methods