Solution

The correct answer is Option 1) 1.024 sec

Key Points

• Process size (to swap): 15 MB
• Disk transfer rate: 30 MB/s
• Average rotational latency: 12 ms = 0.012 s
• No head seek ⇒ ignore seek time; only latency + transfer time matter.
• Swapping a process in/out involves two I/O operations: write to disk (swap out) and read back (swap in).

Step 1: Transfer time for 15 MB

Transfer time per operation = Size / Rate = 15 / 30 = 0.5 s.

Step 2: Add rotational latency per operation

Each I/O pays one latency: 0.012 s.

So, time per operation (either swap out or swap in): 0.5 + 0.012 = 0.512 s.

Step 3: Total swap time (out + in)

Total = 2 × 0.512 = 1.024 s.

Important Points

• Transfer time scales with size and bandwidth; latency is a fixed overhead per I/O.
• If seek time were present, it would be added once per I/O as well.
• Always convert milliseconds to seconds when summing with transfer time.