Question 42

Given processes:

• $A$: Arrival 0, Burst 6
• $B$: Arrival 3, Burst 2
• $C$: Arrival 5, Burst 4
• $D$: Arrival 7, Burst 6
• $E$: Arrival 10, Burst 3

Execution trace:

• $[0, 3]$: Process $A$ runs. At $t = 3$, remaining burst of $A = 6 - 3 = 3$.
• At $t = 3$: $B$ arrives with burst 2. Since $2 < 3$, $B$ preempts $A$.
• $[3, 5]$: Process $B$ runs. At $t = 5$, $B$ completes ($CT_B = 5$).
• At $t = 5$: $C$ arrives with burst 4. Remaining: $A (3), C (4)$. $A$ runs because $3 < 4$.
• $[5, 7]$: Process $A$ runs. At $t = 7$, remaining burst of $A = 3 - 2 = 1$.
• At $t = 7$: $D$ arrives with burst 6. Remaining: $A (1), C (4), D (6)$. $A$ has shortest remaining time.
• $[7, 8]$: Process $A$ runs and completes at $t = 8$ ($CT_A = 8$).
• At $t = 8$: Ready queue has $C (4), D (6)$. $C$ runs.
• $[8, 10]$: Process $C$ runs. At $t = 10$, remaining burst of $C = 4 - 2 = 2$.
• At $t = 10$: $E$ arrives with burst 3. Remaining: $C (2), E (3), D (6)$. $C$ has shortest remaining time.
• $[10, 12]$: Process $C$ runs and completes at $t = 12$ ($CT_C = 12$).
• At $t = 12$: Ready queue has $E (3), D (6)$. $E$ runs.
• $[12, 15]$: Process $E$ runs and completes at $t = 15$ ($CT_E = 15$).
• $[15, 21]$: Process $D$ runs and completes at $t = 21$ ($CT_D = 21$).

Turnaround time ($TAT = CT - AT$):

• $TAT_A = 8 - 0 = 8$
• $TAT_B = 5 - 3 = 2$
• $TAT_C = 12 - 5 = 7$
• $TAT_D = 21 - 7 = 14$
• $TAT_E = 15 - 10 = 5$

Sum of Turnaround Times = $8 + 2 + 7 + 14 + 5 = 36\text{ msec}$.