TLB hit ration and memory lookup time

Question

Consider a system where TLB lookup time is $25$ ns and memory access time is $200$ ns, respectively. Assuming a virtual address space of $2$ KB, page size of $32$ bytes, and a PTE size of $2$ bytes, what is the minimum TLB hit ratio that results in an average v2p (virtual to physical) translation latency of $185$ ns?

Comments

I think it should be $0.6$.

VA = 11 bit.

Page size = Frame size = $32$ Byte which require 5 bit so for indexing pages we have 6 bit. PTE = 2 Byte.

Doing Paging for the first time -> Page table size = $2^7 Byte $ which is greater than $2^5 Byte$ which means we have to apply paging one more time.

Now, Page table size = $2^3 Byte \le 2^5Byte$ which is Ok. Hence we have two level-paging here.

In the question, it is asking Translation time not EMAT. So, we have to calculate for v2p translation time only. Thus

$185 = x(25) + (1-x)(25+200*2)\\185 = 25x + 425- 425x \\ 185-425 = 25x-425x\\x = \frac{240}{400}  = \frac{3}{5} = 0.6$.

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Thanks I understood it all wrong . Very useful information.

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Shubhanshu

why you have taken x(25). It should be x(25 + 200)

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TLB is present inside the Cache so you don't need to access Main Memory for it whereas on a Miss, you need to access Two Levels of Page Table and hence 2 * Main Memory Access.

Answer 1

hit ratio is 1/5.