A scheduling algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (the lowest priority). The scheduler re-evaluates the process priorities every $T$ time units and decides the next process to schedule. Which one of the following is TRUE if the processes have no I/O operations and all arrive at time zero?

Consider the execution flow:

1. All processes arrive at time $t = 0$ with an initial priority of $0$.

2. The scheduler selects a process to run for a time slice of $T$ units.

3. During these $T$ units, the running process accumulates no additional waiting time, whereas all non-running (waiting) processes have their waiting time increased by $T$, thus increasing their priority.

4. At $t = T$, the scheduler picks the process with the highest priority (the one that has waited the longest).

5. This process then runs for $T$ units, while the others accumulate more wait time. After every process has executed once for $T$ units, the cycle repeats in the same order.

This behavior is identical to the Round-Robin scheduling algorithm with a time quantum of $T$.

Hence, option (B) is correct.