M7. Assume that we have a paged memory system with associativeregisters to hold
ID: 3613252 • Letter: M
Question
M7.
Assume that we have a paged memory system with associativeregisters to hold the most active page-table entries and all ofpage-table entries in main memory. (The associative registersare also called the translation lookasidebuffer.) Access time is 8 nanoseconds for theassociative registers and 80 nanoseconds for the main memory. When a page is referenced, a search is first made of theassociative registers and if that fails, main memory is thenaccessed.
(a) What is the effective access time for a page-tableentry if 90% of all memory references find their entriesin the associative registers? Note: consider the time required to access the page table, but not thetime required for the subsequent access to the page itself.
(b) Repeat part (a) for 50%.
(c) What do the results from parts (a) and (b) suggest about theusefulness of associate registers for storing page-tableentries?
Explanation / Answer
To get the page number from associative memory = 8 nanoseconds
To read the desired word from memory = 80 nano seconds
Time = 8+80= 88 nano seconds.
Now the time when not in associative memory is
Time = 8+80+80= 168 nano seconds
[One memory access extra is required to read the page table frommemory.]
Effective access time = Page number in associative memory + Pagenumber not in associated memory.
Page number in associative memory = 0.9*88
Page number not in associated memory = 0.1 * 168.
Effective access time = 0.9 * 88 + 0.1 * 168 = 96 nanoseconds
To get the page number from associative memory = 8 nanoseconds
To read the desired word from memory = 80 nano seconds
Time = 8+80= 88 nano seconds.
Now the time when not in associative memory is
Time = 8+80+80= 168 nano seconds
[One memory access extra is required to read the page table frommemory.]
Effective access time = Page number in associative memory + Pagenumber not in associated memory.
Page number in associative memory = 0.5*88
Page number not in associated memory = 0.5 * 168.
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