Consider a paging system that uses 1-level pa...
Consider a paging system that uses 1-level page table residing in main memory and a TLB for address translation. Each main memory access takes 100 ns and TLB lookup takes 20 ns. Each page transfer to/from the disk takes 5000 ns. Assume that the TLB hit ratio is 95%, page fault rate is 10%. Assume that for 20% of the total page faults, a dirty page has to be written back to disk before the required page is read in from disk. TLB update time is negligible. The average memory access time in ns (round off to 1 decimal places) is ________.
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Consider a paging system that uses 1-level page table residing in main...
EMAT = 0.95 × (20 + 100) + 0.05 × (0.9 × (20 + 100 + 100) + 0.1 × [0.2 ×
(20 + 100 + 5000 + 5000) + 0.8 × (20 + 100 + 5000)] = 154.5 ns
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Consider a paging system that uses 1-level page table residing in main...
The average memory access time can be calculated by considering the time taken for different operations involved in the paging system. Let's break down the calculation step by step:

1. TLB Lookup Time:
- TLB lookup takes 20 ns.
- As the TLB hit ratio is given as 95%, the TLB hit time is 20 ns * 0.95 = 19 ns.

2. Page Fault Time:
- Page fault rate is given as 10%, which means 10% of the memory accesses result in page faults.
- For 20% of the page faults, a dirty page needs to be written back to disk before the required page is read in from disk.
- So, the effective page fault rate requiring disk I/O is 10% * 20% = 2%.
- Each page transfer to/from the disk takes 5000 ns.
- Therefore, the average time for page fault handling is 5000 ns * 2% = 100 ns.

3. Memory Access Time:
- Considering the TLB hit ratio and page fault rate, the average memory access time can be calculated as follows:
- TLB hit time: 19 ns
- Page fault time: 100 ns
- Memory access time without page fault: 100 ns (assuming it takes the same time as page fault time for simplicity)
- Memory access time with page fault: 5000 ns (page transfer time)
- The probability of a page fault occurring is the page fault rate, which is 10%.
- Therefore, the average memory access time is calculated as follows:
- Average memory access time = (TLB hit time + (page fault rate * memory access time with page fault) + ((1 - page fault rate) * memory access time without page fault))
- Average memory access time = (19 ns + (10% * 5000 ns) + (90% * 100 ns))
- Average memory access time = 19 ns + 500 ns + 90 ns
- Average memory access time = 609 ns (rounded off to 1 decimal place)

The correct answer is given as 154.5 ns, which seems to be incorrect based on the given information and calculations. The accurate average memory access time calculated above is 609 ns.
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Consider a paging system that uses 1-level page table residing in main memory and a TLB for address translation. Each main memory access takes 100 ns and TLB lookup takes 20 ns. Each page transfer to/from the disk takes 5000 ns. Assume that the TLB hit ratio is 95%, page fault rate is 10%. Assume that for 20% of the total page faults, a dirty page has to be written back to disk before the required page is read in from disk. TLB update time is negligible. The average memory access time in ns (round off to 1 decimal places) is ________.Correct answer is '154.5'. Can you explain this answer?
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