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Assignment 11

Assignment 11. Q2) A memory system has a memory access time of 250 nanoseconds and a page fault service time of 50 milliseconds.

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Assignment 11

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  1. Assignment 11

  2. Q2) A memory system has a memory access time of 250 nanoseconds and a page fault service time of 50 milliseconds. b) If there are no page faults, the effective access time is equal to the memory access time. Suppose that even with page faults, we do not want the effective access time to degrade by more than 10 %. What should the page fault rate be to achieve this? We want EAT < (memory access time + 10% of memory access time) (1-p) * memory access time + p*page fault service time <( 250 ns + 25 ns ) (1-p)*250+p*50000000 <275 250 – 250 p + 50000000 p <275 250 +p(– 250 + 50000000) <275 250+p * 49999750 <275 p * 49999750 < 275 – 250 p * 49999750 <25 p < 25 / 49999750  p < 5 * 10-7

  3. Q3 EAT = TLB hit ratio * (TLB lookup time + memory access time) + (1- TLB hit ratio – page fault rate) * (TLB lookup time + PT lookup time + memory access time) + page fault rate *(TLB lookup time + PT lookup time + swap page out time + swap page in time + memory access time)

  4. Assignment 12

  5. Q2 Block number = address/block size Byte offset = address – address at the beginning of the block= address – (block number*block size)

  6. K  M  G  T Q3 210 B = KB 220 B = MB 230 B = GB 240 B = TB Maximum number of bytes addressed by n direct pointersis = Number of direct pointers * BlockSize Maximum number of bytes addressed by a single indirect pointer is = NumberOfEntriesInOneBlock * BlockSize = (BlockSize / PointerSize) * BlockSize Maximum number of bytes addressed by a double indirect pointer is = NumberOfEntriesInOneBlock2* BlockSize = (BlockSize / PointerSize)2* BlockSize Maximum number of bytes addressed by a triple indirect pointer is = NumberOfEntriesInOneBlock3* BlockSize = (BlockSize / PointerSize)3 * BlockSize

  7. 8KB/4B= 2K entry(pointer) = 2000 entry Block = 8K 16MB i-node 8KB/4B= 2K entry(pointer) =2000 entry 4 bytes direct 32 GB 104KB indirect 64 TB 8KB/4B= 2K entry(pointer)= 2000 entry double triple

  8. Block = 8K i-node 4 bytes direct 104KB 13 * 8KB = 104 KB indirect double triple

  9. 8KB/4B= 2K entry(pointer) = 2000 entry Block = 8K 16MB i-node 4 bytes direct 2000 * 8 KB = 16000 KB  covert to MB = 16 MB indirect double triple

  10. Block = 8K i-node 8KB/4B= 2K entry(pointer) =2000 entry 4 bytes direct 32 GB indirect 2000  2000*2000 * 8Kentry entry 4,000,000 * 8K 32,000,000 KB convert to GB by /1,000,000 32 GB double triple

  11. Block = 8K 2000  2000*2000entry entry = 4,000,0000 4,000,000*2000 * 8K entry entry 8,000,000,000 * 8K = 64,000,000,000 KB convert to TB by /1000,000,000 = 64 TB i-node 4 bytes direct indirect 64 TB 8KB/4B= 2K entry(pointer)= 2000 entry double triple

  12. K  M  G  T 210 B = KB 220 B = MB 230 B = GB 240 B = TB Maximum number of bytes addressed by n direct pointersis = Number of direct pointers * BlockSize = 13* 8 KB = 104 KB Maximum number of bytes addressed by a single indirect pointer is = NumberOfEntriesInOneBlock * BlockSize = (BlockSize / PointerSize) * BlockSize = (8 KB / 4 B )* 8 KB = 2 K * 8 KB = 2000 * 8 KB = 16000 KB = 16 MB

  13. K  M  G  T 210 B = KB 220 B = MB 230 B = GB 240 B = TB Maximum number of bytes addressed by a double indirect pointer is = NumberOfEntriesInOneBlock2* BlockSize = (BlockSize / PointerSize)2* BlockSize = ( 8 KB / 4 B)2 * 8 KB = ( 23 . 210/ 22)2 * (23 . 210 B) = ( 211 )2 * 213 B = 222 * 213 B = 22 * 220 * 23*210B = 25 * 230 B = 32 GB Maximum number of bytes addressed by a triple indirect pointer is = NumberOfEntriesInOneBlock3* BlockSize = (BlockSize / PointerSize)3 * BlockSize = ( 8 KB / 4 B)3 * 8 KB = ( 23 . 210/ 22)3* (23 . 210 B) = ( 211 )3* 213 B = 233 * 213 B = 23 * 230 * 23*210B = 26 * 240 B = 64 TB

  14. Maximum file size is 64 TB + 32 GB + 16 MB + 104 KB

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