In real-time operating systems, which of the following is the most suitable scheduling scheme?
| A. | round-robin |
| B. | first-come-first-served |
| C. | preemptive |
| D. | random scheduling |
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Option: C Explanation : Click on Discuss to view users comments. |
Dirty bit is used to show the
| A. | page with corrupted data |
| B. | the wrong page in the memory |
| C. | page that is modified after being loaded into cache memory |
| D. | page that is less frequently accessed |
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Option: C Explanation : Click on Discuss to view users comments. |
If there are 32 segments, each of size 1 K byte, then the logical address should have
| A. | 13 bits |
| B. | 14 bits |
| C. | 15 bits |
| D. | 16 bits |
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Option: C Explanation : Th specify a particular segment, 5 bits are required (since 25 = 32). . Having selected a page, to select a particular byte one needs 10 bits (since 210 = 1 k byte). So, totally 5 + 10 =15 bits are needed. Click on Discuss to view users comments. |
The first-fit, best-fit and the worst-fit algorithm can be used for
| A. | contiguous allocation of memory |
| B. | linked allocation of memory |
| C. | indexed allocation of memory |
| D. | all of these |
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Option: A Explanation : Click on Discuss to view users comments. |
Consider a computer with 8 Mbytes of main memory and a 128 K cache. The cache block size is 4 K. It uses a direct mapping scheme for cache management. How many different main memory blocks can map onto a given physical cache block ?
| A. | 2048 |
| B. | 256 |
| C. | 64 |
| D. | None of these |
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Option: C Explanation : Click on Discuss to view users comments. Amit Kumar said: (1:51pm on Monday 5th October 2015)
No of blocks in cache= 128k/4k = 32 and 1 main memory=8 Mb/128k=8*2^20/128*2^10 = 64 cache memories So no of blocks in main memory=32*64 So Number of main memory blocks map onto 32 blocks of cache=64*32 and number of main memory blocks map onto any given physical cache block=64*32/32=64
Debendra said: (7:49pm on Sunday 19th June 2016)
No of Frames in Main memory= 8*2^20 Bytes/4*2^10Bytes=2^11No of Cache Blocks:128*2^10/4*2^10=32=2^5So Ans: 2^11/2^5=2^6=64
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