> If the processor is always busy, that means there’s never even an instant when a newly arriving task can be assigned immediately to the processor. That means

> If the processor is always busy, that means there’s never even an instant when a newly arriving task can be assigned immediately to the processor. That means that, whenever a task finishes processing, there must always be a task waiting in the queue; otherwise the processor would have an idle moment. And by induction, we can show that a system that’s forever at 100% utilization will exceed any finite queue size you care to pick: This argument is incorrect because it proves too much. If tasks arrive predictably, say at the rate of one per minute, and the processor completes tasks at the same rate, you can have maximum throughput and a finite queue -- indeed, the queue never has to exceed two waiting tasks. The unbounded queue growth problem only arises when tasks arrive unpredictably. Since this argument proves the same conclusion without regard to predictability, it's incorrect. Here's how I think it actually works: if tasks randomly come in and out at the same rate, the size of the queue over time can be modeled as an unbiased random walk with a floor at zero. Elementary techniques prove that such a random walk will hit zero in finite time, and will therefore hit zero infinitely many times. These correspond to the idle times of the processor. The only way to avoid hitting zero over and over is if tasks come in faster than they come out, which leads to tasks piling up over time.