Question 51

The processes, arrival times, and bursts are:

• $P_1$: Arrival = $0$, Burst = $3$
• $P_2$: Arrival = $1$, Burst = $1$
• $P_3$: Arrival = $3$, Burst = $3$
• $P_4$: Arrival = $4$, Burst = $Z$

Let us trace SRTF scheduling step by step:

• At $t = 0$: Only $P_1$ has arrived (remaining burst = $3$). $P_1$ starts execution.
• At $t = 1$: $P_2$ arrives with burst $1$. $P_1$'s remaining burst is $3 - 1 = 2$. Since $P_2$'s remaining burst ($1$) is less than $P_1$'s ($2$), $P_1$ is preempted and $P_2$ runs.
• At $t = 2$: $P_2$ completes its burst of $1$.
  • Completion time of $P_2 = 2$.
  • Turnaround time of $P_2 = 2 - 1 = 1$.
  • Waiting time of $P_2 = 1 - 1 = 0$.
    $P_1$ resumes with remaining burst $2$.
• At $t = 3$: $P_3$ arrives with burst $3$. $P_1$'s remaining burst is $2 - 1 = 1$. Since $1 < 3$, $P_1$ continues.
• At $t = 4$: $P_1$ completes its burst.
  • Completion time of $P_1 = 4$.
  • Turnaround time of $P_1 = 4 - 0 = 4$.
  • Waiting time of $P_1 = 4 - 3 = 1$.

At $t = 4$, $P_4$ arrives with burst $Z$. Ready queue holds $P_3$ (burst $3$) and $P_4$ (burst $Z$).
We are given that the average waiting time is $1\text{ ms}$:

If $Z < 3$, $P_4$ is dispatched before $P_3$:

• $P_4$ runs from $t = 4$ to $4 + Z$. Completion time = $4 + Z$.
  $$WT(P_4) = (4 + Z) - 4 - Z = 0$$
• $P_3$ runs from $t = 4 + Z$ to $7 + Z$. Completion time = $7 + Z$.
  $$WT(P_3) = (7 + Z) - 3 - 3 = 1 + Z$$

Sum of waiting times:

Checking consistency: $Z = 2 < 3$, so $P_4$ is indeed scheduled before $P_3$. Hence, $Z = 2$.