introduction to wireless networks - chinese university of ...cslui/csc6480/wireless.pdf ·...

Introduction to Wireless Networks

John C.S. Lui

Department of Computer Science & EngineeringThe Chinese University of Hong Kong

www.cse.cuhk.edu.hk/∼cslui

John C.S. Lui (CUHK) Advanced Topics in Network Analysis 1 / 36

Outline

Outline

1 Aloha and Slotted Aloha

2 Slotted ALOHA with Delay

3 Stability in ALOHA


Aloha and Slotted Aloha

GoalsFor "pure" ALOHA, users transmit any time they desire.For "slotted ALOHA, the channel is to "slot" time into segmentswhose duration is exactly equal to transmission time of a singlepacket.The aim is to derive the

throughputdelaythroughput-delay tradeoffstability issue



AnalysisLet say transmission of a packet takes P secs. The vulnerableperiod is 2P, as shown:

P

2P

vulnerableperiod for

slotted ALOHAP

vulnerable period forpure ALOHA

time

Define:S denote the throughput of the channel (e.g., average number ofsuccessful transmission per transmission period P).G denote the average channel traffic (e.g., number of packettransmission attempted per P sec).



Analysis: Pure ALOHAAssume the total traffic (G) is Poisson and entering the channel isan independent process generated by an infinite population, then

S = G Prob[no additional packet in vulnerable period] = Ge−2G.(1)

This was obtained by Abramson for ALOHA.What is the maximum S? Taking ∂S

∂G and equate to 0, maximumoccurs at G = 1/2 and we have:

S∗ =12

e−1 ≈ 0.184. (2)

Comment on the physical meaning and the efficiency of pureALOHA.



Analysis: Slotted AlohaTime is "slotted" and all users are synchronized to these slotintervals.When a packet arrives within a slot, user delays the transmissionuntil next time slot. Therefore, the vulnerable period is P only.Using similar assumption that the total traffic is a Poisson process:

S = Ge−G. (3)

To obtain the maximum value of S, use similar technique, themaximum occurs at G = 1, we have

S∗ = e−1 = 0.368. (4)

It is twice that of pure ALOHA.



Throughput for pure and slotted ALOHA

Consider the throughput vs. offered traffic



Analysis: continueConsider a finite population model with M independent users in aslotted ALOHA.User’s packet transmission (old or new packet) as a sequence ofindependent Bernoulli trials, or Gm be the probability the mth usertransmits a packet in any give slot, where m = 1, . . . , M.Gm can be viewed as the average traffic (per slot) for the mth user.So the average offered load is G =

∑Mi=1 Gm.

Let Sm be the probability that the mth user’s packet is successfullytransmitted. Similarly, S =

∑mi=1 Sm is the system throughput (per

slot). We have:

Sm = Gm∏i 6=m

(1−Gi) m = 1, 2, . . . , M. (5)

The set of M equations has a solution set, {Sm}, which defines theallowable mixtures of source rates which this channel can support.



Example 1Assume all users are statistically identical. Then Sm = S/M andGm = G/M, we have

S = G(

1− GM

)M−1

.

Also, as M →∞, S → Ge−G. So as M increases, throughput offinite user model approaches to the Poisson throughput of infiniteusers.The above express, maximum S∗ occurs when M = 1 and S = G.Define g =

∏Mi=1(1−Gi), we have

Sm =Gm

1−Gmg. m = 1, 2, . . . , M.

This set of M equations define the set of achievable throughput{Sm} in terms of offer load {Gm}.



Condition to achieve maximum SPrevious M equations define a region in the M−dimensionalspace whose coordinates are G1, . . . , GM .The boundary to this region defines the maximum throughput forSm when all other Si(i 6= m) are fixed at allowable values. Thiscan be found by setting the Jacobian (determinant) J to zero with:

∂Sj

∂Gi=

{ g1−Gi

i = j−gGi

(1−Gj )(1−Gi )i 6= j

After some algebra, we find

J = gM−2

∣∣∣∣∣∣∣∣∣(1−G1) −G1 −G1 · · · −G1−G2 (1−G2) −G2 · · · −G2

......

......

...−GM −GM −GM · · · (1−GM)

∣∣∣∣∣∣∣∣∣= gM−2(1−G1 −G2 − · · · −GM).



Condition to achieve maximum SApplying J = 0, we have the following general condition on the setof offered loads that achieves the set of maximum throughputvalues:

G =M∑

m=1

Gi = 1.

The above equation gives us our throughput performance contour.

Example 1Let M = 1, we then have S = G = 1.This is obvious the best we can do that a user never collides ordestroys his own packet.



Example 2The M users form two groups such that m1 of them have a rate S1and m2 = M−m1 of them have a rate S2.S = m1S1 + m2S2 and G = m1G1 + m2G2. The M equationsreduce

S1 = G1(1−G1)m1−1(1−G2)

m2

S2 = G2(1−G2)m2−1(1−G1)

m1 .

This defines the set of permitted rates {S1, S2} as a function ofoffer loads {G1, G2}.Imposing the optimization condition G = 1, we may solve for S1and S2 for any given value of G1 ∈ [0, 1/m1] andG2 = (1−m1G1)/m2.



Throughput Contour:


Slotted ALOHA with Delay

Slotted ALOHA with DelayAssume an infinite number of users who collectively form a sourcethat generates packets independent of the channel.This source generates V packets per slot from the distributionvk = P[V = k ] with mean of S packets per slot.Packet is of constant length requiring 1 slot time to transmit.Let R be the number of slots it takes to check whether atransmitted packet is successful or not (e.g., satellite).If the packet was destroyed, user retransmits and randomlychoose one of the next K slots to retransmit (so as to avoidcolliding again with other collided packets). Thus, theretransmission will take place either R + 1 to R + K slots after theinitial transmission.Total packets (new and retransmit pkt) per slot is L, wherepk = P[L = k ] with mean traffic of G packets per slot.What is the maximum throughput S, the average packet delay T?



AnalysisGoal: derive pk = P[exactly k transmitted packets per slot] or itsZ−transform. Consider the following figure:

currentslottagged slot

R slotsK slots

.... ...

The traffic of "current slot" is due to (a) retransmission from any ofthe "K-slots" and (b) new packets generate in the current slot.Let qm = Probability that exactly m packets which are transmittedin the current slot due to retransmission from the tagged slot, form = 0, 1, 2, . . ..Define Z-transforms:

P(z) =∑∞

k=0 pk zk ; q(z) =∑∞

m=0 qmzm; V (z) =∑∞

k=0 vk zk .



Since transmission in slots are independent, let us first derive q0:

q0 = p0 + p1 +∞∑

k=2

pk

(1− 1

K

)k

.

For q1, we have:

q1 =∞∑

k=2

pkk(

1K

) (1− 1

K

)k−1

.

For the rest qm, where m ≥ 2, we have:

qm =∞∑

k=m

pk

(km

) (1K

)m (1− 1

K

)k−m

m ≥ 2.



The Z−transform of qm, or q(Z ) is

q(Z ) = q0 + q1Z +∞∑

m=2

[ ∞∑k=m

pk

(km

) (1K

)m (1− 1

K

)k−m]

Z m

Adding and subtracting m = 0 and m = 1 terms, we have

q(Z ) = q0 + q1Z −∞∑

k=0

pk

(1− 1

K

)k

− z∞∑

k=1

pkk(

1K

)m (1− 1

K

)k−1

+∞∑

m=0

[ ∞∑k=m

pk

(km

) (1K

)m (1− 1

K

)k−m]

Z m



Substituting q0 and q1 and interchanging the order of summation in thedouble sum:

q(Z )=p0+p1−[p0+p1

(1− 1

K

)]−z

p1

K+∞∑

k=0

pk

k∑m=0

(km

) ( zK

)m(

1− 1K

)k−m

.

Reorganizing the binomial expansion, we get

q(Z ) =p1

K(1− Z ) +

∞∑k=0

pk

[1− 1

K+

zK

]k

Taking the definition of P(Z ), we have:

q(Z ) =p1

K(1− Z ) + P

(1− 1− Z

K

).



Now, the number of packets to be transmitted in the current slot is thedue to (1) the K slots in the tagged slot region; (2) number of newlygenerated packets in the current slot, expressing the relationship ofthese random variables:

p = q + · · · q + v .

Expressing in Z− transform, we have:

P(Z ) = [q(Z )]K V (Z ).

Or

P(Z ) =

[p1

K(1− Z ) + P

(1− 1− Z

K

)]K

V (Z ).

Since P(Z ) is a moment generating function, we can obtain variousmoments and “crank” out probabilities.



Analysis with Delay in Retransmission

Since S is the throughput of the channel (e.g., average number ofsuccessful transmission per transmission period P), and G is theaverage offered traffic (e.g., number of packet transmission attemptedper P sec), we have:

S/G : probability of successful packet transmission,G/S : average number of transmission until success.

Let E be the average number of extra transmissions a packet incurs(i.e., collision), we have:

1 + E = G/S.

We want to derive E . To do this, let us define:q : P[newly generated packet is successfully transmitted]qt : P[previously collided packet is successfully transmitted]



Let xi be the probability that a packet collides exactly i times beforesuccessful transmission, we have:

xi = (1− q)(1− qt)i−1qt i = 1, 2, . . . ,

So

E =∞∑

i=1

i xi =1− q

qt.

We can now express S, the channel throughput:

S =G

1 + E= G

(qt

qt + 1− q

). (6)

Remark: If we can derive q and qt , then we have S.



Derive q: Based on previous analysis and definition of qm, we have

q = qK0 e−S

where q0 can be expressed as:

q0 = e−G + Ge−G +∞∑

i=2

Gi

i!e−G

(1− 1

K

)i

= e−G + Ge−G + e−G[eG(1−1/K ) − 1−G

(1− 1

K

)]= e−G/K +

GK

e−G.

Substituting this into the expression of q, we have:

q =

(e−G/K +

GK

e−G)K

e−S. (7)



Derive qt : the probability that previously collided packet is successfullytransmitted. It has three factors:

no packet which collided with it in the tagged slot again collideswith it in the current slot.no packets from one of the K−1 other possible tagged slotscollides with this packet in the current slot.no new packet is generated in the current slot.

Let qc be the probability of the first event, we have

qc =∞∑

k=1

(Gk

k !

) (e−G

1− e−G

) (K − 1

K

)k

=e−G/K − e−G

1− e−G .

For second event, the probability is qK−10 .

For third event, the probability is e−S.We have

qt = qcqK−10 e−S =

[e−G/K − e−G

1− e−G

] [e−G/K +

GK

e−G]K−1

e−S. (8)



Note that Equations (6), (7) and (8) need to be solved numerically soas to derive S, q and qt .

To derive T , the average number of slots to successfully deliver apacket, we know that:

the time needed for one successful transmission (R + 1) slots,the extra transmission due to collision.

Therefore, T is:

T = (R+1)+

[K +1

2+(R+1)

]E = (R+1)+

1−qqt

[R+1+

K−12

]. (9)

Remark: We now have the performance measures of S and T as afunction of K , R and G.



To illustrate (a) S vs. G; (b) effect of K ; (c) effect on T .



Two groups of users: a group of "small" users (m1 = ∞) and one largeuser (m2 = 1), with S∞ = limm1→∞m1S1 and their G∞ = m1G1. Thelarge user throughput is S2 and offered load is G2. In herT = (S∞T∞ + S2T2)/S.


Stability in ALOHA

ModelPrevious analysis assumes equilibrium conditions. Now consider

Slotted ALOHA with large (but finite) number (M) of activeterminals.Each terminal has one buffer space and only when the buffer isempty a new packet can be generated with probability σ per slot.Let N(t) be the r.v. representing the total number of nonemptyterminals (or channel backlog) and S(t) be the combined inputrates of packets into all terminals at time t . E.g., N(t) = n, then[N(t), S(t)] = [n, (M − n)σ]. This is a linear feedback model.Packets that collide are retransmitted (after a round-trippropagation delay of R slots) in one of the next K slots, eachbeing chosen at random with probability 1/K .The retransmission takes place on the average R + (K + 1)/2slots after the previous transmission.


Stability in ALOHA

AnalysisThe model is hard to analyze. We approximate the backlogged packetindependently retransmits with probability p where

p =1

R + (K + 1)/2.

Define Sout be the throughput rate of the channel, which is theprobability of exact one successful packet transmission in a slot. If[N(t), S(t)] = [n, (M−n)σ], then

Sout = (1− p)n(M − n)σ(1− σ)M−n−1 + np(1− p)n−1(1− σ)M−n (10)

In the limit when M →∞ and σ → 0 such that Mσ = S < ∞, we havethe infinite population model in which packets are generated over thechannel at the Poisson rate S. We rewrite the equation to

Sout = (1− p)nSe−S + np(1− p)n−1e−S. (11)

This expression is very accurate even for finite M if σ << 1 and if wereplace S = Mσ by S = (M − n)σ.


Stability in ALOHA

For a fixed K , we show the behavior of Sout as function of the channelload [n, S]

There is an equilibrium contour in the (n, S) "phase plane" onwhich the input rate S is equal to the system throughout Sout .In the shaded region, Sout > S; elsewhere Sout < S (the systemcapacity is exceeded!).


Stability in ALOHA

The area of the shaded region (where Sout > S) may be increased byincreasing K


Stability in ALOHA

We define the channel load line in the (n, S) plane as the linen = M − (S/σ). A channel is said to be "stable" when its load lineintersects the equilibrium contour in exactly one place.


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