Three processes A, B and C each execute a loop of 100 iterations. — GATE CS 2014 Set 2 Q32

Three processes $A$ , $B$ and $C$ each execute a loop of $100$ iterations. In each iteration of the loop, a process performs a single computation that requires $t_c$ CPU milliseconds and then initiates a single I/O operation that lasts for $t_{io}$ milliseconds. It is assumed that the computer where the processes execute has sufficient number of I/O devices and the OS of the computer assigns different I/O devices to each process. Also, the scheduling overhead of the OS is negligible. The processes have the following characteristics:

$\begin{array}{|c|c|c|} \hline \textbf{Process id} & \textbf{$t\_c$} & \textbf{$t_{io}$} \\ \hline \textbf{A} & \text{100 ms} & \text{500 ms} \\ \hline \textbf{B} & \text{350 ms} & \text{500 ms}\\ \hline \textbf{C} & \text{200 ms} & \text{500 ms} \\ \hline \end{array}$

The processes $A$ , $B$ , and $C$ are started at times $0$ , $5$ and $10$ milliseconds respectively, in a pure time sharing system (round robin scheduling) that uses a time slice of $50$ milliseconds. The time in milliseconds at which process C would complete its first I/O operation is ___________.

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Answer: 1000

To find the completion time of the first I/O operation of process $C$ , we trace the CPU scheduling using Round Robin with a time quantum of $50\text{ ms}$ .

The arrival times are: $T_A = 0\text{ ms}$ , $T_B = 5\text{ ms}$ , and $T_C = 10\text{ ms}$ .

The step-by-step execution trace is as follows:

$[0 - 50\text{ ms}]$ : $A$ executes. Remaining CPU time for $A = 50\text{ ms}$ . Ready queue at $t=50$ : $[B, C, A]$ .
$[50 - 100\text{ ms}]$ : $B$ executes. Remaining CPU time for $B = 300\text{ ms}$ . Ready queue at $t=100$ : $[C, A, B]$ .
$[100 - 150\text{ ms}]$ : $C$ executes. Remaining CPU time for $C = 150\text{ ms}$ . Ready queue at $t=150$ : $[A, B, C]$ .
$[150 - 200\text{ ms}]$ : $A$ executes and completes its first CPU burst of $100\text{ ms}$ . $A$ initiates its I/O at $t = 200\text{ ms}$ . Ready queue: $[B, C]$ .
$[200 - 250\text{ ms}]$ : $B$ executes. Remaining CPU time for $B = 250\text{ ms}$ . Ready queue at $t=250$ : $[C, B]$ .
$[250 - 300\text{ ms}]$ : $C$ executes. Remaining CPU time for $C = 100\text{ ms}$ . Ready queue at $t=300$ : $[B, C]$ .
$[300 - 350\text{ ms}]$ : $B$ executes. Remaining CPU time for $B = 200\text{ ms}$ . Ready queue at $t=350$ : $[C, B]$ .
$[350 - 400\text{ ms}]$ : $C$ executes. Remaining CPU time for $C = 50\text{ ms}$ . Ready queue at $t=400$ : $[B, C]$ .
$[400 - 450\text{ ms}]$ : $B$ executes. Remaining CPU time for $B = 150\text{ ms}$ . Ready queue at $t=450$ : $[C, B]$ .
$[450 - 500\text{ ms}]$ : $C$ executes and completes its first CPU burst of $200\text{ ms}$ .

At $t = 500\text{ ms}$ , process $C$ initiates its first I/O operation. Since the I/O operation takes $t_{io} = 500\text{ ms}$ and there is no I/O resource contention, the completion time of $C$ 's first I/O operation is:

$500\text{ ms} + 500\text{ ms} = 1000\text{ ms}$

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GATE CS 2014 Set 2 Q32 — CPU Scheduling

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