Consider a segmented paging system with the segment table and the page table wit
ID: 644746 • Letter: C
Question
Consider a segmented paging system with the segment table and the page table within a segment stored in memory. First part of the address is segment number, second part is the page number within a segment and the third part is the offset within a page.
a) If a memory reference takes 200 nano seconds, how long does a segmented page memory reference take?
b) If we add Translation Lookaside Buffer (TLB), and 80% of all memory references are found in the TLB, what is the effective memory reference time in each case? (Assume that finding a segmented page-table entry in the TLBs takes zero time, if the entry is there. Otherwise memory tables must be searched to determine the physical address. )
Explanation / Answer
Answer:
For 400 nanoseconds: 200 ns to access the page table plus 200 ns to access the word in memory.
250 nanoseconds: 80% of the time it's 200 ns, and the other 25% of the time it's 400ns, so the equation is:
E.A. = (.80*200)+(.25*400)
If the time to access the associative registers is actually 2 ns. Then the Answe would be
e.a. = 2 + (.80*200)+(.25*400)