TLB ,VGATE

Question

@Arjun sir, I solved it by using the same concept of gate 2003 78,79 ..but techtud marked it as wrong..this qs has only one confusion which is how to use page table walk and tlb update...I used it in the part of L3 ache miss of Tlb miss.. and used this formula ... 

Tavg = Tlb hit( Tlb time +L1 hit (cache time) + L1 miss(L2 hit(cache time) + L2 miss( L3 hit (cache time) + L3  miss(cache time+memory time ))))

+ Tlb miss(Tlb time + Memory time + L1 hit (cache time) + L1 miss(L2 hit(cache time) + L2 miss( L3 hit (cache time) + L3  miss(cache time+page table walk and Tlb update ))))

Sir,pls check this

Comments

whats the answer given?

Answer 1

Problem of using formula- we must know what's the use of a page table. It is to get actual physical address. TLB is a quick look-up for page table. 

Regarding question- it's a really well framed question like in GATE. Only issue is for me to read the small font. 

So, memory access works like this - First look in cache and then go to RAM. But before looking in cache we need physical address- because cache uses physical address as given in question. (it can use virtual addressing or virtually index and physical tag also but this question is clear- physical index, physical tag). 

Now what happens when a TLB miss happens? We look in page table which is in main memory. Can this page table be cached? Yes, quite often. So, determining the time becomes complex. But see the question- it directly gives the time during a TLB miss- so we are saved. Page walk just means looking up in page table. 

If you haven't understood so far I suggest to better skip this portion for GATE. There is no use other than getting negative if you read below without understanding above. 

Average memory access time = Avg. address translation time + Avg. data access time

$= \left(0.95 \times 1 + 0.05 \times (1+200)\right) +  0.95 \times 1 + 0.05 \times 0.80 \times (1 + 8 )+ 0.05 \times 0.2 \times 0.5 \times (1+8+50) + 0.05 \times 0.2 \times 0.5 \times (1+8+50+100) 
\\= 11 + 0.95 + 0.36 + 0.295+0.795
\\=13.40 ns$

Comments

yes sir,but now only 10 days left,so i am leaving it..It is time consuming also ,,but a very good problem

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sir in case of  TLB Miss time shouldn't be 201 ??
i.e. (0.95×1+0.05×(200+1)) in place of (0.95×1+0.05×200).

TLB access time + TLB Miss rate * TLB update time = (1+0.05×200)
is itn't ?
 

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yes, I was careless. Corrected now :)

Answer 2

Time = TLB access time + Miss rate TLB * TLB update time + L1 access time + Miss Rate L1 * L2 Access Time + Miss Rate of L1 * Miss Rate L2 * L3 Access Time +  Miss rate L3 * Memory Access time

= 1 + 0.05*200 + 1 + 0.05*8 + 0.05*0.20*50 + 0.50*0.20*0.50*100
= 1 + 10 + 1 + 0.40 + 0.50 + 0.50
= 13.40 ms

Comments

previously i didnt see that carefully,today i saw and understood.thanks yaar. :)

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Without the use of brackets formula written is wrong. Values are placed correctly, though.