qwerty123
Branch Unspecified
26 Apr 2012

# anybody solved the question with detailed answer

Suppose the cache and main memory access times are 100 ns and 1200 ns respectively. The cache is used for paging and the hit ratio of finding the page table entry in the cache is 98%, what is the effective paged memory access time?

(A) 240 ns (B) 220 ns(C) 250 ns (D) 1200 ns

Anoop Kumar

Branch Unspecified
26 Apr 2012
I am getting 124nm by fomula: tav= h * tc + (1 - h)*(tc + tm)
H is hit ratio: 98%=0.98
Tc=100ns, Tm=1200nm
if its 90% then we get 220ns , one of above answer.
Reason: If data is found only in cache then only tc will count otherwise we need to access main memory and hence (Tc+Tm).

nareshkumar6539

Branch Unspecified
26 Apr 2012
I think that is not correct formula for effective paged memory access time

Teff= h(c+m)+(1-h)(c+2m)
c+m means to check frame number avilable in cache it will take c sec if avilable to access that frame number it will take m sec.total it will take c+m
c+2m means to check page number available in cache it will take c sec if it is not there it will check in page table for this it will take m sec and to access that frame number it will take m sec
total it will take c+2m
according to this formula i am getting 1324ns

Anoop Kumar

Branch Unspecified
26 Apr 2012
hmm... its right.I misunderstood the question I think.
your fomula is for accessing actual memory acces time:
pointer go to page memory and then using page pointer it will access the actual memory.

by my formula :
It only getting time taken to find page pointer time (or actual memory access without paging).I thought you are asking only for page pointer access.😳

qwerty123

Branch Unspecified
09 May 2012
Suppose the cache and main memory access times are 100 ns and 1200 ns respectively. The cache is used for paging and the hit ratio of finding the page table entry in the cache is 98%, what is the effective paged memory access time?

(A) 240 ns (B) 220 ns(C) 250 ns (D) 1200 ns

Anoop Kumar

Branch Unspecified
10 May 2012