184 views
A computer system has a page size of 1024 bytes and maintains page table for each process in the main memory.The overhead required for doing a lookup in the page table is $500ns$.To reduce this overhead, the computer has a TLB that caches 32 virtual pages to physical page frame mappings.A TLB lookup requires $100ns$.What TLB hit-rate is required to ensure an average virtual address translation time of $200ns$?
| 184 views
0
.75 ???
0
0.75...?
0
0
yes it should be 80%
+3
$200=100\times h+\left ( 1-h \right )\times 600$

h=0.80

So, Average virtual address translation time(and not Average Memory Access time) is asked.

If TLB hit, it takes only 100ns

otherwise, it takes 100+500=600ns

$200=h(100)+(1-h)(600)$

$h=0.8$
0
Guys correct me if I am wrong.
0
Actually I was thinking TLB access require memory access

but isnot TLB lookup require memory access?
0

@ mam

I m also thinking the same..TLB contains just the frequently used page table entries,Finally we have to get the page from memory!

0

@srestha-No TLB lookup is performed by hardware.No Memory operation is required in it. Here they have just asked the effective time in which we can get the physical address from the logical address.

+1 vote
It should be 0.8

Note that here, we are asked to calculate average virtual address translation time. So the equation would be like this:

Let the hit rate be $x$.

$200 = x(100) + (1-x)(100+500)$.

Solving this should get you the value of $x$ as $0.8$.

Here, we don't add the time to access the page from the memory because we only care about the address translation time.
by
0
ditto!!
OR we can do like this

P>>Hit rate

Avg access Time= Access time for TLB+(1-P)*(500)

200=100+(1-P)*500

100=500-500*P

500*P=400

P=0.8