The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
0 votes
119 views
Suppose: TLB lookup time = 20 ns

TLB hit ratio = 80%

Memory access time = 75 ns

Swap page time = 500,000 ns

50% of pages are dirty

OS uses a single level page table

What is the effective access time (EAT) if we assume the page fault rate is 10%? Assume the cost toupdate the TLB, the page table, and the frame table (if needed) is negligible.
in Operating System by (333 points) | 119 views
0

1 Answer

0 votes

page fault = 0.1

nopagefault = 0.9

dirtypage = 0.5

nodirtypage = 0.5

tlb hit ratio = 0.8

tlb miss ratio = 0.2

tlb access time = 20ns

memory access time = 75ns

swap page time = swap page from main memory to disk for update = swap update page from disk to main memory = 500,000 ns

EMAT=(TLB hit ratio) * (TLB access time + page access time) + (TLB miss ratio) * [nopagefault*(TLB access time + main memory access time) + pagefault * {(nodirtypage*swap page time) + (dirty page*(swap page from main memory to disk for update)* (swap update page from disk to main memory) )

 

=0.8(20+75)+0.2[0.9(20+75+75)+0.1(0.5×500000+0.5(2×500000))]0.8(20+75)+0.2[0.9(20+75+75)+0.1(0.5×500000+0.5(2×500000))]

=0.8×95+0.2[153+0.1(250000+500000))]

=15106.6ns

by (27 points)

Related questions

+3 votes
1 answer
2
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,339 questions
55,763 answers
192,337 comments
90,771 users