Q h 7thECE2006

A logical address space of 12 pages of 1024 bytes each mapped to a physical memory of 64 frames. What would be the minimum number of bits in logical address?

Q i 7thECE2006

What would be the minimum number of bits in physical address?

Q 2 7thECE2006

A demand paged virtual system has the following parameters

- Time to serve page faults:
- When a new frame is allocated: 12 mili seconds
- When a modified page is replaced: 20 mili seconds

- Memory access time: 200 nano seconds
- While serving a page fault:
- Probability of allocating a new frame: 30%
- Probability of allocating a modified frame: 70%

Find the maximum page fault rate such that degradation in memory access time is not more than 20%.

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Q h 7thECE2006 and Q i 7thECE2006

Think of the Page Table.

If 12pages are to be represented how many bits required? Ans:4bits

If 64 frames are to be represented how many bits required? Ans:6bits

Now add it with the offset. 1024 require 10bits

So for Logical 14 and for Physical 16bits minimum.

Q 2 7thECE2006

Just multiply the related probability with the time to service.

so 30×12=360 mili sec and 70×20= 1400 mili sec

to service 100 page faults: 360+1400=1760 mili sec

for 1 page fault 17.6 mili sec or 17600 nano sec.

Now put EAT formula(p->pagefault rate; ma->memory access time):

EAT=(1-p)x ma + p x pagefault time

EAT=(200 – 200p) + 17600p

EAT= 200 + 17400p

Now degradation in performance is permitted upto 20% so memory access of 200 nano sec may go up another 40 nano sec so maximum up to 240 nano sec.

We will put the value at the place of EAT.

240>=200 + 17400p

p<= 40/17400

p<= 1/435

(page fault should be less than one pf in 435 page accesses) got it?