Question

For the system in problem 4, suppose a main memory access requires 30ns, the page fault rate is .01%, it costs 12ms to access a page not in memory (this time includes the time necessary to transfer the page into memory, update the page table, and access the data). Also suppose a TLB hit requires 7ns, the cache miss rate is 3%, the TLB hit rate is 95%, a cache hit requires 15 ns. On a TLB or cache miss, the time required for access includes a TLB and/or cache update, but the access is not restarted. On a page fault, the page is fetched from disk, all updates are performed but the access is restarted . All references are sequential (no overlap, nothing done in parallel)
a.) Calculate the time for a TLB hit and a cache hit.

b.) Calculate the EAT (effective access time) for a TLB hit.

Answer 1

Answer:

a. 7.75

b. 24.4

Explanation:

The Operating system uses virtual memory and page tables maps these virtual address to physical address. TLB works as a cache for such mapping.

program >>> TLB >>> cache >>> Ram

A program search for a page in TLB, if it doesn't find that page it's a TLB miss and then further looks for the page in cache.

If the page is not in cache then it's a cache miss and further looks for the page in RAM.

If the page is not in RAM, then it's a page fault and program look for the data in secondary storage.

So, typical flow would be

Page Requested >> TLB miss >> cache miss >>main memory>> page fault >> looks in secondary memory.

Here,

Main memory access time= 30 ns

Page fault rate=.01%

page fault service time= 12ns

TLB access time=7 ns

TLB hit rate= .95%

TLB miss rate =1-.95=.05%

cache access time = 15 ns

cache miss rate= .3%

cache hit rate = 1-.3=.97%

So,

a) TLB hit time= TLB access time = 7 ns

cache hit time = TLB hit rate * TLB access time + TLB miss rate * ( TLB access time + cache hit time)

= .95 * 7 + .05 * (7+15)

= 7.75 ns

b) EAT for TLB hit= 7ns

Total EAT = TLB hit rate *( TLB access time + Cache hit rate * cache access time + cache miss rate * (cache + main memory access time))+ TLB miss rate ( TLB access time + main memory access time + cache hit rate * cache access time + cache miss rate ( cache + main memory access time))

= .95 *( 7 + (.97*15) + .03(15+30))+ .05*(7+30+(.97*15) + .03 ( 15 + 30))=24.4 ns