Practice Question (Gate 1993)
Duration: 3 min
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This educational video provides a detailed walkthrough of a GATE 1993 computer organization problem involving moving arm disk storage. The instructor systematically breaks down the given specifications to calculate two key performance metrics: average latency (P) and data transfer rate (Q). The lecture focuses on converting units and applying standard formulas for disk access time and throughput.
Chapters
0:00 – 2:00 00:00-02:00
The session begins with the instructor reading the problem statement displayed on the screen. He lists the specifications: 'Number of tracks/recording surface = 200', 'Disk rotation speed = 2400 rpm', and 'Track storage capacity = 62,500 bits'. He underlines the target variables, 'average latency' and 'data transfer rate'. He starts the calculation by converting the rotation speed from rpm to rotations per second, writing '2400 / 60' on the board. This yields 40 rotations per second. Consequently, he calculates the time for one full rotation as 1/40 seconds. To make this more readable, he converts the time into milliseconds by multiplying by 1000, resulting in '25 ms' for one full rotation.
2:00 – 2:40 02:00-02:40
The instructor proceeds to find the average latency, explaining that it is half the time of one full rotation. He writes '12.5 ms' on the board, identifying this as the value for P. Next, he tackles the data transfer rate (Q). He establishes a relationship where 25 ms corresponds to the transfer of 62,500 bits. To find the rate per second, he sets up a proportion: 1 second corresponds to (62,500 / 25) * 1000 bits. He performs the division 62,500 / 25 to get 2,500, and then multiplies by 1000 to get 2,500,000. He writes the final answer as '= 2.5 x 10^6 b/s', completing the calculation for Q.
The video effectively demonstrates how to derive disk performance parameters from basic physical specifications. By first determining the rotational period, the instructor establishes the foundation for calculating both latency and transfer rates. The step-by-step board work clarifies unit conversions between seconds and milliseconds, ensuring students understand the magnitude of the values involved. This methodical approach is crucial for solving similar problems in computer architecture exams.