ISRO CSE 2016 | Question: 22

Question

Let the page fault service time be $10$ ms in a computer with average memory access time being $20$ ns. If one page fault is generated for every $10^6$ memory accesses, what is the effective access time for the memory?

• A. 21.4 ns
• B. 29.9 ns
• C. 23.5 ns
• D. 35.1 ns

Comments

B)30*10^-9=29

Answer 1

Answer is B

given that one page fault for every 10^6 memory access, so the probability of page fault is  p-> 1/10⁶   

page fault service time is = 10ms-> 10 * 10⁶  ns ( note : given options are in Nano seconds better if you convert this to nano seconds at first).

memory access time is = 20ns

effective access time would be = p * page fault service time + (1-p) * memory access time

=> 1/10^6 *10*10^6+(1-1/10^6)20=>30-1/10^6=> closer answer is 29.9 ns 

Comments

what about this formula? -> memory access time + p * page fault service time.

This is same as ->  (1-p) * memory access time + p( memory access time + page fault service time )

Above formula is applied in case of hierarchical memory organization.

The formula which you have used is applied when all memory modules are each directly connected to CPU.

But in question, there is no mention of which memory organization it is ......So what made you apply that formula. What made you assume that it is the 2nd case and not the 1st one. Please help me .....its getting irritating now

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memory access time should be 2*(memory access time) . one access for “Page table” and second for “main memory” , since it is not mentioned that Page table is stored in a register it is by default stored in main memory.

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@rajsh3kar
why not.. ?

EMAT = pf(ps+ma)+(1-pf)(ma)

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https://pages.cs.wisc.edu/~solomon/cs537/html/paging.html
 

There are several different page table organizations use in actual computers. One approach is to put the page table entries in special registers. This was the approach used by the PDP-11 minicomputer introduced in the 1970's.

If Paage table is stored in registers. EMAT = hit *(Time to access ain mem (for content aaccess)) + fault (time to service page)  
We can ignore time taken to access page table as it is in cpu registeres which has very high speeds. Note: Page table is still accessed in both fault and hit case. 

Also on a sidenote, in my opinion since this is a non standard approach which sometimes comes in ISRO exams, i dont think it will ever come in gate like this. Unless they menton explicitly. 

Answer 2

hope it might help.............

Answer 3

I think correct answer is 30 ns. Part (B) is more closer to it so it could be selected.

Explanation -> 

I am assuming memory  has hierarchical organization. In all cases (Fault or not)  page will be accessed form main memory. 

Notice - Page fault service time does not include memory access time. Please correct me if i am wrong.

(Refer -http://stackoverflow.com/questions/40027359/what-is-page-fault-service-time  and 

https://gateoverflow.in/2122/gate2011-20-ugcnet-june2013-ii-48?show=126742#c126742 )

Case 1(No Page Fault): -  20 ns.

Case 2(Page Fault): - (20 + 10*10^6) ns.

Effective memory access  time  = (1- (1/10^6))*(Case 1 time)  + (1/10^6)*(Case 2 time)  = 20 + 10  = 30 ns. 

Comments

nice... this is the correct way!!

Answer 4

B)29.9 ns

https://gateoverflow.in/2122/gate2011_20

Answer 5

Ans (B)

Refer this, 

Answer 6

EMAT=Frequency of page fault  x Page fault service time + Memory access time =>p x Ps +m

EMAT=p x Ps + m

EMAT= 1/10^6 x 10x10^-3 + 20 ns=29ns