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Answer (Detailed Solution Below)

Option 1 : 8.00 ms / 8.00 मिलीसेकंड

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Detailed Solution

सही उत्तर 8.00 ms / 8.00 मिलीसेकंड है।

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PrepSnapprepsnap.in • Part -A

Question:

Four processes P1 (arrival 0, burst 7), P2 (arrival 2, burst 4), P3 (arrival 4, burst 1), P4 (arrival 5, burst 4) are scheduled with non-preemptive Shortest Job First. What is the average turnaround time in milliseconds? / चार प्रोसेस P1 (0,7), P2 (2,4), P3 (4,1), P4 (5,4) को नॉन-प्रीएम्प्टिव शॉर्टेस्ट जॉब फर्स्ट से शेड्यूल किया गया। औसत टर्नअराउंड समय (मिलीसेकंड में) क्या है?

1.
8.00 ms / 8.00 मिलीसेकंड
2.
7.00 ms / 7.00 मिलीसेकंड
3.
8.75 ms / 8.75 मिलीसेकंड
4.
6.50 ms / 6.50 मिलीसेकंड

Correct Answer: 8.00 ms / 8.00 मिलीसेकंड

Non-preemptive SJF: P1 runs 0-7 (only arrival at time 0). At 7, arrived jobs are P2 (burst 4), P3 (burst 1), P4 (burst 4); shortest is P3, runs 7-8. P2 and P4 tie at burst 4, so earlier arrival P2 runs 8-12, then P4 runs 12-16. Completion times: P1 = 7, P3 = 8, P2 = 12, P4 = 16. Turnaround times: 7, 4, 10, 11. Average = 32/4 = 8.00 ms. 7.00 ms is the preemptive SRTF answer and 8.75 ms is the FCFS answer. / नॉन-प्रीएम्प्टिव SJF: P1 (0-7); समय 7 पर P3 सबसे छोटा (7-8); फिर P2 (8-12), फिर P4 (12-16)। पूर्णता समय 7, 8, 12, 16; टर्नअराउंड 7, 4, 10, 11; औसत = 32/4 = 8.00 मिलीसेकंड।

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