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 ________.

To calculate the Effective (Average) Memory Access Time (EMAT):

Given:

• TLB lookup time, $t_{\text{TLB}} = 20\text{ ns}$

• Main memory access time, $t_m = 100\text{ ns}$

• TLB hit ratio, $h = 0.95$

• TLB miss ratio, $1 - h = 0.05$

• Page fault rate (on memory/page table access upon TLB miss), $p = 0.10$

• Disk transfer time per page = $5000\text{ ns}$

Page Fault Service Time ($S$):

• For $20\%$ of page faults, a dirty page must be written back and the new page read in (2 disk transfers = $2 \times 5000 = 10000\text{ ns}$).

• For $80\%$ of page faults, only the new page is read in (1 disk transfer = $5000\text{ ns}$).

Cases for Memory Access:

1. TLB Hit (probability = $0.95$):

   $$\text{Time} = t_{\text{TLB}} + t_m = 20 + 100 = 120\text{ ns}$$

2. TLB Miss (probability = $0.05$): First, TLB is looked up ($20\text{ ns}$), then page table in main memory is accessed ($100\text{ ns}$):

   • No Page Fault (probability = $1 - p = 0.90$):

     $$\text{Time} = 20 + 100 + 100 = 220\text{ ns}$$

   • Page Fault (probability = $p = 0.10$):

     $$\text{Time} = 20 + 100 + S + 100 = 20 + 100 + 6000 + 100 = 6220\text{ ns}$$

Overall EMAT: