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 |
Answer : C Explanation : 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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Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
