9.8 Consider a logical-address space of eight pages of 1,024 words each, mapped
ID: 3773046 • Letter: 9
Question
9.8 Consider a logical-address space of eight pages of 1,024 words each, mapped onto a physical memory of 32 frames.
a.How many bits are in the logical-address?
b.How many bits are in the physical address?
9.10Consider a paging system with the page table stored in memory.
a.If a memory reference takes 200 nanoseconds, how long does a paged memory reference take?
b.If we add TLBs, and 75 percent of all page-table references are found in the TLBs, what is the effective memory reference time? (Assume that finding a page-table entry in the TLBs takes zero time, if the entry is there.)
Explanation / Answer
Answer for Question 9.8)
a. Logical address will have
3 bits to specify the page number (for 8 pages) .
10 bits to specify the offset into each page (210 =1024 words) = 13 bits.
b. For (2^5) 11 32 frames of 1024 words each (Page size = Frame size)
We have 5 + 10 = 15 bits.
Answer for Question 9.9)
a) Ans is 2 memory accesses: page lookup followed by actual access => 2*200ns = 400ns
b) Ans is 75% * TLB hit-time + 25% * TLB miss-time = 75% * 200ns + 25% * 400ns = 250ns
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