Three processes arrive at time zero with CPU bursts of 16, 20 and 10 milliseconds. If the scheduler has prior knowledge about the length of the CPU bursts, the minimum achievable average waiting time for these three processes in a non-preemptive scheduler (rounded to nearest integer) is ___________ milliseconds.

To minimize the average waiting time for a set of processes arriving at the same time ($t = 0$), the optimal non-preemptive scheduling algorithm is Shortest Job First (SJF).

Let the processes be:

• $P_1$ with burst time $BT_1 = 10\text{ ms}$

• $P_2$ with burst time $BT_2 = 16\text{ ms}$

• $P_3$ with burst time $BT_3 = 20\text{ ms}$

Scheduling in non-decreasing order of burst times ($P_1 \to P_2 \to P_3$):

• $P_1$: Starts at $t = 0$, finishes at $t = 10$.

  $$\text{Waiting Time}(P_1) = 0 - 0 = 0\text{ ms}$$

• $P_2$: Starts at $t = 10$, finishes at $t = 10 + 16 = 26$.

  $$\text{Waiting Time}(P_2) = 10 - 0 = 10\text{ ms}$$

• $P_3$: Starts at $t = 26$, finishes at $t = 26 + 20 = 46$.

  $$\text{Waiting Time}(P_3) = 26 - 0 = 26\text{ ms}$$

Total waiting time:

Average waiting time:

Therefore, the minimum achievable average waiting time is 12 milliseconds.