Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 35 3 Solution Created 2026-10-03 Updated 2026-10-14
A random access network shares one communication channel among an unbounded collection of possible transmitters. Each packet occupies one slot; exactly one attempt succeeds, an idle slot carries nothing, and a collision delivers none of its attempted packets. Assume perfect slot synchronization, immediate idle/success/collision feedback, no channel errors, and independent random choices by packets. The important quantities are throughput, the size of the retransmission backlog, and delay. A good one-slot success probability alone does not establish stability of the backlog.
For a precise infinite-population slotted ALOHA model, let be independent Poisson random variables of mean , representing new packets which attempt immediately in slot . Let be the backlog before those arrivals. Each old packet independently retries with a constant probability , so conditional on , its retry count is , independent of . If records a successful slot, thenThis includes a new packet in the backlog precisely when it has not succeeded; an old packet is removed precisely when it succeeds. The total backlog is a Markov chain in this constant-probability model. Its idle and success probabilities arewith the second term in interpreted as zero. A success may be either one new packet and no old attempt, or one old attempt and no new packet.
The fluid approximation of slotted ALOHA replaces the binomial retry count by a Poisson count when there are many packets with small retry probability. If is total attempt intensity, the approximate success probability is . Differentiation givesThis is the limiting ALOHA throughput bound, not an exact upper bound for every finite packet population: one isolated packet which certainly transmits succeeds with probability one. In a fluid model, when the equation has two roots. The root below one is locally stable for constant , while the root above one is unstable, because the drift derivative is . These approximations explain a low-backlog operating region and a congestion threshold; they do not imply a stationary law for the infinite-population chain.
Indeed, for fixed , the exact success probability tends to zero as , so the exact drift tends to . One can prove instability, rather than relying only on this positive drift. Fix and use . Since is zero or one,The first term is strictly less than one and the second tends to zero. Hence is a nonnegative supermartingale outside a sufficiently large finite set, where is the first entrance into that set, say . The optional sampling theorem for a supermartingale, first at bounded stopping times, givesThis is an exponential-supermartingale escape bound. The chain is irreducible: a no-arrival slot with exactly one old retry can reduce any positive state by one, and Poisson arrivals can reach arbitrarily high states. Therefore the escape probability proves transience and excludes a positive recurrent Markov chain, for every and fixed . A locally stable fluid operating point can consequently coexist with eventual stochastic congestion. At , once at least two packets are backlogged, they collide forever, so constant unit retry probability also fails.
A state-dependent retry rule removes this fixed- congestion mechanism. As an ideal benchmark with exact backlog information and known arrival intensity , useThis is backlog-aware ALOHA with arrival-adjusted attempt probabilities. Its total attempt intensity approaches one: the mean old-attempt count is and the new-attempt count has mean . Substitution in the exact success formula yieldsThus the drift tends to . For the Lyapunov function , the drift is bounded above by a negative constant outside a finite set, and the expected one-step backlog is finite at every state. Foster theorem therefore gives positive recurrence. This proves a genuine stochastic stability result for this specified, informed controller. Real transmitters generally observe feedback rather than the exact backlog; estimating the backlog and using a probability of order its reciprocal motivates adaptive ALOHA, but an estimate's dynamics must be included in the state and analyzed before claiming the same stability result.
A different nonconstant rule uses a packet's own collision history. In geometric-attempt exponential backoff, a packet at stage attempts with probability , where . A collision advances every attempted packet by one stage, a success removes its packet, and new arrivals enter stage zero. For definiteness, in this version arrivals join at the end of a slot and first become eligible in the next slot. The full Markov chain state is the sequence of stage populations; the total backlog alone is insufficient. Conditional on that state,These formulas follow from independent attempts and selecting which one packet succeeds. They reduce approximately to for many small probabilities with . The retry probabilities decrease after failure, preventing repeated aggressive attempts by the same packets. Binary exponential backoff instead chooses uniform random counters from expanding finite windows; that counter model requires residual counters as well as collision stages in its state and is not identical to geometric waiting.
Backoff also exposes a distinction between eventual delivery and acceptable delay. In the simplifying model where each attempt independently fails with a fixed probability , the probability of reaching stage is and its mean waiting time is . The mean delay under independent exponential-backoff failures is thereforeAlthough a packet succeeds almost surely whenever , its mean delay can be infinite. In an actual random access network, failures are correlated with the other stage populations, so this fixed- calculation is a diagnostic approximation, not a proof of network stability. These models show why random-access analysis must track the retry rule, feedback information and full state, and assess throughput, stability and delay separately.