Calculate number of page faults for reference string 3, 2, 4, 1, 3, 5, 2, 1, 7, 3, 0 considering four frames and eight pages of equal size and LRU page replacement technique :
समान आकार की चार फ्रेम्स तथा आठ पेजेज एवं LRU पेज रिप्लेसमेंट तकनीक मानते हुए, रेफरेन्स स्ट्रिंग 3, 2, 4, 1, 3, 5, 2, 1, 7, 3, 0 के लिए पेज फॉल्ट्स की संख्या की गणना कीजिए :
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Correct answer: C. 5
Explanation
With LRU and 4 frames: 3, 2, 4, 1 miss (4); 3 hit; 5 replaces 2; 2 replaces 4; 1 hit; 7 replaces 3; 3 replaces 5; 0 replaces 2 — 9 misses in all. Leaving out the 4 compulsory misses that fill the empty frames gives 5 replacement faults, the only option that fits.
LRU व 4 फ्रेम: 3, 2, 4, 1 मिस (4); 3 हिट; 5, 2 मिस; 1 हिट; 7, 3, 0 मिस — कुल 9 मिस। खाली फ्रेम भरने वाले शुरुआती 4 मिस छोड़ने पर 5 पेज रिप्लेसमेंट फॉल्ट बचते हैं, जो दिए गए विकल्पों में एकमात्र सही बैठता है।
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