GATE IT 2008 | Question: 16

Question

A paging scheme uses a Translation Look-aside Buffer (TLB). A TLB-access takes $10$ ns and the main memory access takes $50$ ns. What is the effective access time(in ns) if the TLB hit ratio is $\text{90\%}$ and there is no page-fault?

• A. $54$
• B. $60$
• C. $65$
• D. $75$

Comments

Good question to keep in short notes.

---

We always need to have 1 TLB and 1 MM access (to get frame from MM), when there is miss we need to have 1 extra MM access (to get page table).

So EMT = 10 + 50 + 0.1(50) = 55ns

---

no @smsubham its not like that . if its a hit then you need to access TLB and main memory once but if its a miss then you have already accessed TLB which didnt turn out to be worthy so u need to access main memory twice (once for the page table and second time for the data).

---

Option C)

---

if there is no page fault then 100% chance of main memory access. That's the main crux of the question otherwise formula putting things.

---

TLB stores address translations only, not data.

So even on a hit, you must still go to main memory to fetch data.

---

So for this question pehle aap agr hit hua too tlb pr jayenga then content ko access ke liye hm jayenge main memory pe or maan lo ki miss ho gaya too fir hm jayenge MM pe but hmko pta kaise lga ki miss hai too for that hm pehle jb tlb pr gaye tb fir mm pe or fir ek baar or mm pe to access the content ye hai scinerio bhaiya or didi logo

 

Answer 1

Effective access time $=$ hit ratio $\times$ time during hit $+$ miss ratio $\times$ time during miss

In both cases TLB is accessed and assuming page table is accessed from memory only when TLB misses.

$= 0.9 \times (10+50) + 0.1 \times (10 + 50 + 50)$

$= 54 + 11=65$

Correct Answer: $C$

Comments

to access page table and content

---

@Rackson

$1^{st}\  50ns$ to access page table
$2^{nd} 50ns$ to access actual page present in frame

---

because one for page table access and another is for main memory access

Answer 2

Answer is (C) Part.

Just one more approach -->

$Effective Access Time = Address Translation Time + Main Memory Access Time$

= $10 + 0.1(50) + Main Memory Access Time = 15 + 50 = 65ns$

Comments

Nice font:)

Answer 3

$EAT\Rightarrow 0.9\times (10+50)+0.1(10+50+50)=54+11=65$

Answer 4

answer :- (c) 65

reason-

Given : TLB access time =  10 ns

main memory access time = 50 ns

TLB hit ratio= 90%  i.e (0.9)

now we first need to find miss ratio that is [1- TLB hit ratio]

miss ratio= 1- 0.9= 0.1

now calculating the effective access time by formula,

hit ratio X (TLB access time + memory access time) + miss ratio X (TLB access ratio + 2 X memory access ratio)

= 0.9X(10 ns + 50 ns)+ 0.1 X (10 ns + (2 X 50 ns))

=0.9 X 60 ns + 0.1 X 110ns

=54 ns + 11 ns

=65 ns

Answer 5

"there is no page-fault"  means we should just put

10ns + 50ns =60 ns

Comments

thank you sir It clears alot.

---

By default we've to assume 1 level paging?

---

Yes, if nothing is talking about the number of paging level than, we can take 1 level paging.