The queue that shrank

Good news at the ticket counter this morning: a queue of eight people, which had been there since opening, has disappeared. The counter finishes with one person every two minutes.

Good news at the ticket counter: the queue of $8$ people has gone. The counter finishes with $1$ person every $2$ min.

Queue $L$; leaving $s=\frac{1}{2}$, arriving $a=\frac{1}{4}$ (people per min).

A newcomer joins the queue only every four minutes. So in four minutes two people leave and one arrives: the queue gets one person shorter every four minutes.

A newcomer joins only every $4$ min. In $4$ min, $2$ leave and $1$ arrives: the change in the queue is $-1$ per $4$ min.

$\Delta t=4\Rightarrow \Delta L=(a-s)\Delta t=-1$.

The queue started at eight, fell to seven after four minutes, to six after eight, and so on. After thirty-two minutes nobody was left. As a formula: on average the length is eight minus a quarter of the time in minutes.

Length $L$ at time $t$ (minutes): $L(0)=8,L(4)=7,L(8)=6$. So $L(t)=8-\frac{t}{4}$ and $L(32)=0$.

$L(t)=8-\frac{t}{4},0\le t\le 32,L(32)=0$.

Things would be different if a newcomer arrived every minute. Then the queue would grow by half a person each minute: eight plus half the time, and thirty-eight people after one hour. That queue would grow for ever.

With a newcomer every $1$ min, each minute adds $1-\frac{1}{2}=\frac{1}{2}$ of a person. So $L(t)=8+\frac{t}{2}$ and $L(60)=38$: it grows for ever.

$a=1\Rightarrow L(t)=8+\frac{t}{2},L(60)=38$.