कम्प्युटर र सूचना प्रविधि (प्राविधिक)
कम्प्युटर र सूचना प्रविधि (प्राविधिक)
Computer & IT (Technical) · ५,०७६ प्रश्न
५,०७६ प्रश्न उपलब्ध। मोड छान्नुहोस्:
कम्प्युटर र IT: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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Demand paging means that:
सही उत्तर: घ) A page is brought into memory only when it is referenced
In demand paging a lazy swapper loads a page only when the process actually accesses it, causing a page fault the first time.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Memory access time is 200 ns, the average page-fault service time is 8 ms and the page-fault rate is 0.001. The effective access time is approximately:
सही उत्तर: ग) 8.2 microseconds
EAT = (1 - 0.001) × 200 + 0.001 × 8,000,000 ns = 199.8 + 8000 = 8199.8 ns ≈ 8.2 µs.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a paged system, the hardware register that points to the page table of the currently running process is the:
सही उत्तर: ख) Page-Table Base Register (PTBR)
The PTBR holds the starting address of the current process's page table; on a context switch only this register needs to be changed.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A page table that has one entry for each physical frame of memory, rather than one for each virtual page, is called a(n):
सही उत्तर: क) Inverted page table
An inverted page table keeps one entry per physical frame (storing process ID and page number), greatly reducing table size.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In segmentation, if the offset in a logical address is greater than or equal to the segment's limit:
सही उत्तर: घ) A trap (addressing error) is raised to the operating system
The hardware compares the offset with the limit; an out-of-range offset causes a trap so the process cannot access memory outside its segment.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a segmentation system, segment 2 has base 4300 and limit 400. What is the physical address of the logical address (segment 2, offset 53)?
सही उत्तर: ग) 4353
Since offset 53 is less than the limit 400, physical address = base + offset = 4300 + 53 = 4353.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A logical address space has 64 pages of 1024 words each and is mapped onto a physical memory of 32 frames. How many bits are there in the logical address?
सही उत्तर: ग) 16
64 pages need 6 bits and 1024 words need 10 bits of offset, so the logical address has 6 + 10 = 16 bits.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A system uses 32-bit logical addresses, a page size of 2 KB and page-table entries of 4 bytes. What is the size of a single-level page table for one process?
सही उत्तर: ग) 8 MB
Number of pages = 2^32 / 2^11 = 2^21; table size = 2^21 × 4 bytes = 2^23 bytes = 8 MB.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Free partitions are 100, 500, 200, 300 and 600 KB, and requests of 212, 417, 112 and 426 KB arrive in order. Which placement strategy succeeds in allocating all four requests?
सही उत्तर: ग) Best fit
Best fit places 212 in 300, 417 in 500, 112 in 200 and 426 in 600; first, next and worst fit all leave 426 KB unallocated.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Memory partitions of 100, 500, 200, 300 and 600 KB (in that order) are free. Processes of 212, 417, 112 and 426 KB arrive in that order. Using first fit, which process cannot be allocated?
सही उत्तर: घ) 426 KB
First fit: 212 goes to 500 (288 left), 417 to 600 (183 left), 112 to the 288 KB hole; no hole of 426 KB remains.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
यी अभ्यास प्रश्न हुन्, आधिकारिक प्रश्न होइनन्। पाठ्यक्रम, विज्ञापन र नतिजाका लागि सम्बन्धित निकायको आधिकारिक वेबसाइट हेर्नुहोस्। उत्तरमा गल्ती भेटे प्रश्नमुनिको "टिप्पणी / गल्ती रिपोर्ट" मा लेख्नुहोस्।
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