MadeEasy Test Series: Operating System - Page Replacement

Question

Consider a demand paged memory system, page table is held in registers. It takes 800 nsec to service a page fault if empty page is available or replaced page is not modified and 950 nsec if the replaced page is modified, main memory access time is 120 nsec. If page to be replaced is modified 85% of time and page faultrate is 20% then average memory access time is ________. (Upto 1 decimal place)

Comments

https://gateoverflow.in/669/gate2000-2-22

see sachin mittal sir comment on this ques

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obviously wrong solution correction required

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I think actual answer should be 441.5 ns.

i.e., 0.80( 120 ns) + 0.20 ( 800ns + 0.85 (950ns) + 0.15 ( 800ns))

Answer 1

Emat = P*( Page fault service time + Memory access time ) + (1-P)( Memory access time )

          = P*(Page fault service time) + Memory access time

Page fault service time  = pm(Time taken when page modified) + (1-pm)(Time taken when page not modified)

                                         = 0.85*950 + 0.15*800 = 927.5

Emat = 0.2*927.5+120 = 185.5 + 120 = 305.5