Question 31
If there are 32 segments, each size 1 k bytes, then the logical address should have
Step-by-Step Solution
Detailed Explanation:
Segment bits = 5 ($2^5$ = 32), offset bits = 10 ($2^10$ = 1024). Total = 15 bits.
If there are 32 segments, each size 1 k bytes, then the logical address should have
Detailed Explanation:
Segment bits = 5 ($2^5$ = 32), offset bits = 10 ($2^10$ = 1024). Total = 15 bits.
In a lottery scheduler with 40 tickets, ticket distribution among 4 processes P1, P2, P3, P4 for 10%, 5%, 60%, 25% is:
Detailed Explanation:
P1 = 40 * 0.10 = 4, P2 = 40 * 0.05 = 2, P3 = 40 * 0.60 = 24, P4 = 40 * 0.25 = 10.
Which data structure is best suited for ready queue in Round-Robin CPU scheduling?
Detailed Explanation:
Circular queue naturally supports cyclic round-robin process scheduling.
A hard disk system has the following parameters :
Number of tracks =500
Number of sectors/track =100
Number of bytes /sector =500
Time taken by the head to move from one track to adjacent track =1 ms
Rotation speed =600 rpm.
What is the average time taken for transferring 250 bytes from the disk ?
Detailed Explanation:
Average Seek Time = 250 tracks * 1 ms = 250 ms.
Average Rotational Latency = 1/2 of 100 ms = 50 ms.
Transfer time for 250 $B = (250 / 50,000) \cdot 100 ms = 0.5 ms$.
Total time = 250 + 50 + 0.5 = 300.5 ms.
Counting semaphore initial value = 7. After 20 P operations and 15 V operations, resulting value is:
Detailed Explanation:
7 - 20 + 15 = 2.