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Use the slotted ALOHA model with one outstanding packet per active station and a collision channel: a slot with exactly one attempt succeeds, whereas an idle slot or a slot with two or more attempts delivers nothing. Let $Y_t$ be independent Poisson random variables of mean $\nu$, independent also of the transmission choices. Arrivals join after the current slot's attempts. Given $(S_t,N_t)=(S,N)$, the attempt count has a binomial distribution with parameters $N$ and $1/S$. Consequently, for $N\ge 1$,For $N=0$, set $p_0=1$ and $p_1=0$. The equation $N_{t+1}=N_t+Y_t-1_{\{Z_t=1\}}$ adds arrivals and removes precisely one packet following a successful slot. The estimate $S_t\ge 1$ controls the common transmission probability and updates using the idle, success, or collision feedback. The conditional distribution of the next state depends only on $(S_t,N_t)$, so this pair is a Markov chain.

Away from the reflecting floor in $S$, its one-step conditional expectations areWhen $N$ and $S$ are large with $N/S\approx \kappa$, the Poisson limit theorem gives $p_0\approx e^{-\kappa }$ and $p_1\approx \kappa e^{-\kappa }$. Treating these approximate drifts as derivatives motivates the fluid approximation of slotted ALOHA:The floor at $S=1$ is negligible in this interior scaling. This calculation motivates the ordinary differential equations; it is not a proof of a stochastic fluid limit.

For $(a,b,c)=(2-e,0,1)$, putThen $g^{\prime }(\kappa )=(e-2+\kappa )e^{-\kappa }>0$ and $g(1)=0$. On the domain $s>0$, introduce time $\tau$ by $d\tau /dt=1/s$. Differentiating $\kappa =n/s$ giveswhere $r(\kappa )=\kappa [1-(e-2+\kappa )e^{-\kappa }]$.

The function $r$ is strictly increasing from zero to infinity, and $r(1)=e^{-1}$. To check the monotonicity explicitly,Indeed, with $d=3-e\in (0,1)$,because its discriminant is $d^2-4d<0$; and $e^{\kappa }\ge 1+\kappa$. Thus the scalar ordinary differential equation for $\kappa$ points toward the unique value ${\kappa }_{\ast }$ satisfying $r({\kappa }_{\ast })=\nu$. Every trajectory with $s(0)>0$, $n(0)\ge 0$ has bounded $\kappa$ and $\kappa (\tau )\to {\kappa }_{\ast }$. Positivity of $s$ follows from $s(\tau )=s(0)\exp ({\int }_0^{\tau }g(\kappa (u))du)$.

If $0\le \nu <e^{-1}$, then ${\kappa }_{\ast }<1$ and $g({\kappa }_{\ast })<0$. Thus $s(\tau )$ decays exponentially for large $\tau$, andThis is finite-time draining of a fluid model. The unmodified equations have $n/s$ undefined at the origin, so the statement that trajectories converge to the origin needs this endpoint interpretation; continuing with an absorbing zero trajectory is an additional fluid-model convention, not a solution of the displayed equations at the origin.

If $\nu >e^{-1}$, then ${\kappa }_{\ast }>1$ and $g({\kappa }_{\ast })>0$. Now $s(\tau )$ grows exponentially, physical time tends to infinity, andIn particular the trajectory escapes to infinity. Even without the ratio analysis, $\dot{n}\ge \nu -e^{-1}>0$, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$.

For the actual Markov chain, it is essential not to replace the finite-$N$ success probability by $e^{-1}$: a single backlogged packet can succeed with probability one. Instead, for $N\ge 2$, maximizing over $S\ge 1$ gives the ALOHA throughput boundIf $\nu >e^{-1}$, choose $K_0$ and $r_0<\nu$ so that $p_1(N,S)\le r_0$ for all $N>K_0$, uniformly in $S$. For sufficiently small $\theta >0$,because its derivative at zero is $r_0-\nu <0$. Independence of arrivals and attempts giveswhenever $N>K_0$. For any integer $K\ge K_0$, stop at the stopping time ${\sigma }_K=\inf \{t:N_t\le K\}$. The stopped process $e^{-\theta N_{t\wedge {\sigma }_K}}$ is a nonnegative supermartingale. The optional stopping theorem, first at bounded times and then by a limit, yields the exponential-supermartingale escape boundFrom every state with $N\le K$, a Poisson distribution arrival burst has a uniformly positive probability of sending the next backlog to at least $K+m$, for any fixed $m>0$. From there the probability of never revisiting $N\le K$ is at least $1-e^{-\theta m}$. By the Markov property at successive visits, the probability of infinitely many visits to this strip is zero: each visit has a uniform positive chance of being the last. This holds for every integer $K\ge K_0$, soTherefore the Markov chain is transient for every finite choice of $a,b,c$. The proof is uniform in $S$ and does not infer stochastic transience merely from the approximate ordinary differential equations.