final - Operating Systems - Lecture Notes, Study notes of Operating Systems

The main point in the lecture notes of the Operating Systems are:Final, Shared, Unshared, Page of Data, Compiler, Editor, Thrashing, Editor Text, Shell and Editor, Upper Bound

Typology: Study notes

2012/2013

Uploaded on 05/08/2013

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10 pages of text: shared
1 page of data: unshared x 100 = 100
pages.
Shell: 10 pages of text, 1 page of data.
20 pages of shared text.
2 pages of data x 100 = 200 pages.
Editor
10 pages of shared text
4 pages of data x 100 = 400
Compiler
= 300 pages
2 pages of unshared data
Assignment
Problem 2
No thrashing means that we have the pages
for the problem at hand available.
shell text, editor text, and one of compiler
or assignment text: 40 pages of read-only
text.
data for shell and editor: 300 pages
data for one of compiler or run: 400 pages
Need:
This is an upper bound; it is likely that people
will be doing different things and that whole
sections of the shell, editor, and/or compiler
will be swapped out.
COMP111 review session
Sunday, December 16, 2012
4:25 PM
Review Page 1
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10 pages of text: shared 1 page of data: unshared x 100 = 100 pages. Shell: 10 pages of text, 1 page of data. 20 pages of shared text. 2 pages of data x 100 = 200 pages. Editor 10 pages of shared text 4 pages of data x 100 = 400 Compiler 1 page of unshared text. = 300 pages 2 pages of unshared data Assignment Problem 2 No thrashing means that we have the pages for the problem at hand available. shell text, editor text, and one of compiler or assignment text: 40 pages of read-only text. data for shell and editor: 300 pages data for one of compiler or run: 400 pages Need: This is an upper bound; it is likely that people will be doing different things and that whole sections of the shell, editor, and/or compiler will be swapped out. COMP111 review session Sunday, December 16, 2012 4:25 PM

will be swapped out. This is the minimum to insure no thrashing, but one can have less and still not thrash if user patterns are sufficiently different. Problem 2b: New λ' = λ*100 = 200 c/s. Response time = 1/4 second Queue reaches steady state only if p=λ/μ< Prob(queue length is k) = p k (1-p) where p=λ/μ The mean jobs in system are n=p/(1-p) (including jobs in queue and jobs being processed). (L in Little's law) The mean waiting-line length (jobs not yet being served) is w=p 2 /(1-p) (including jobs in process of being queued and dequeued) For M/M/1 queue, Mean time in system = n/λ (W in Little's law) Mean wait time before service = w/λ Little's laws for M/M/1 queue: Mean time in system = 1/4 second = n/λ = (p)/((1-p)λ). p = λ/μ. Solve for μ.