Question:hard
Consider a hard disk with a rotational speed of 15000 rpm. The time to move the read/write head from a track to its adjacent track is 1 millisecond. Initially, the head is on track 0. The number of sectors per track is 400. The sector size is 1024 bytes. It is necessary to transfer data from 10 randomly located sectors in each of the following tracks in the order: 5, 12 and 7. The total time for the data transfer (in milliseconds) from the hard disk is _________. (rounded off to one decimal place)
Consider a hard disk with a rotational speed of 15000 rpm. The time to move the read/write head from a track to its adjacent track is 1 millisecond. Initially, the head is on track 0. The number of sectors per track is 400. The sector size is 1024 bytes. It is necessary to transfer data from 10 randomly located sectors in each of the following tracks in the order: 5, 12 and 7. The total time for the data transfer (in milliseconds) from the hard disk is _________. (rounded off to one decimal place)
Show Hint
Add seek time between tracks 0-5-12-7, then for each of the 10 randomly placed sectors per track add an average half-revolution wait (since requests are served in given order, not reordered), plus the small per-sector transfer time.
Updated On: Jul 22, 2026
Show Solution
Correct Answer: 77.3
Solution and Explanation
Step 1: Set up the problem as a sum over 30 individual sector accesses.
Instead of grouping by track, treat the whole job as 30 separate sector fetches: 10 on track 5, 10 on track 12, 10 on track 7, fetched strictly in that order.
Step 2: Head-movement time.
The head starts at track 0 and steps to track 5 (5 moves), then track 5 to track 12 (7 moves), then track 12 to track 7 (5 moves), each move costing 1 ms. Total head-movement time = (5 + 7 + 5) x 1 ms = 17 ms.
Step 3: Model the rotational wait as an expectation.
One full revolution takes 60000/15000 = 4 ms, giving 4/400 = 0.01 ms per sector-width. For any single requested sector whose angular position relative to the head is unknown (random) at the moment it is requested, model the wait time W as a continuous random variable uniformly distributed on [0, 4) ms, with density f(w) = 1/4 for 0 <= w < 4. Its expectation is E[W] = integral from 0 to 4 of w x (1/4) dw = (1/4) x [w^2/2] from 0 to 4 = (1/4) x 8 = 2 ms.
Step 4: Sum the expected rotational wait over all 30 accesses.
Expected total rotational wait = 30 accesses x 2 ms = 60 ms.
Step 5: Sum the transfer time over all 30 accesses.
Transfer time = 30 x 0.01 ms = 0.3 ms.
Step 6: Combine all three components.
Total expected time = head-movement (17 ms) + expected rotational wait (60 ms) + transfer time (0.3 ms) = 77.3 ms. \[ \boxed{77.3 \text{ ms}} \]
Instead of grouping by track, treat the whole job as 30 separate sector fetches: 10 on track 5, 10 on track 12, 10 on track 7, fetched strictly in that order.
Step 2: Head-movement time.
The head starts at track 0 and steps to track 5 (5 moves), then track 5 to track 12 (7 moves), then track 12 to track 7 (5 moves), each move costing 1 ms. Total head-movement time = (5 + 7 + 5) x 1 ms = 17 ms.
Step 3: Model the rotational wait as an expectation.
One full revolution takes 60000/15000 = 4 ms, giving 4/400 = 0.01 ms per sector-width. For any single requested sector whose angular position relative to the head is unknown (random) at the moment it is requested, model the wait time W as a continuous random variable uniformly distributed on [0, 4) ms, with density f(w) = 1/4 for 0 <= w < 4. Its expectation is E[W] = integral from 0 to 4 of w x (1/4) dw = (1/4) x [w^2/2] from 0 to 4 = (1/4) x 8 = 2 ms.
Step 4: Sum the expected rotational wait over all 30 accesses.
Expected total rotational wait = 30 accesses x 2 ms = 60 ms.
Step 5: Sum the transfer time over all 30 accesses.
Transfer time = 30 x 0.01 ms = 0.3 ms.
Step 6: Combine all three components.
Total expected time = head-movement (17 ms) + expected rotational wait (60 ms) + transfer time (0.3 ms) = 77.3 ms. \[ \boxed{77.3 \text{ ms}} \]
Was this answer helpful?
0
Questions Asked in GATE CS exam
A database has 25,000 fixed-length records of 100 bytes (primary key 15 bytes). The data file is block-aligned so each record is fully contained within a block. The file is indexed by a primary index file, also block-aligned and ordered. Block size is 1024 bytes and a block pointer is 5 bytes. Each index entry stores the block’s anchor key (15 bytes) and a pointer (5 bytes). Binary search on an index file of \(b\) blocks needs \(\lceil \log_2 b \rceil\) block accesses in the worst case. Given a key, the number of block accesses required to identify the data file block that may contain the record, in the worst case, is ____________.
- GATE CS - 2023
- Database Management Systems