COMP 346 Lecture Notes - Lecture 10: Rtp1, Delaware Route 1

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Sen'it c T i tttt?
, T
(Pi)
: the ûi n olI!it oftim e a pr
ocess needs [o be h [llc mn n ing sjl nlc
:
nllo c alcd the CPU) before it is completed
J: tiit Til t Be. w (Pi) : the total amocl p t 1hal ili e proces s needeAl to wait i [ H1 n o nr t rnnír ] g
statc h om the n l om e nt it is r e ady u11 timl c it is coniplcted
R
:
v } m1 s e Time : R
(Pi) : tile totaB an l ount 0 lin i e ñon l the monle n r t he proces s ís
rca{]y w i [i l it is conside>
edßm h eßrsítìme by the CPU woca[ed [hc CPU ìor tile
1"li
)
T1tmt 1 rtJ11ntt Til11e : TÏR. Pi
)
: the amo u nt oflime between the pr o cess first enters tlie
æ a dy mte unti1 the m o ment t he proyes s exits the n m ning state br [1
1e ]astlim e
(tc r m in 1t cd )
Ainl a n HmH1a
Scll e dulil \ g I ul 7
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%tlfer W& ;
NottPrel: }Bïpfive Scll=dttlittg
. Secûn. c :
me
m
; $ Comc I
\ \r$ Scn'c[wcm Schc
:
hllin
An E1ampla Load/ lin v T
*IPJ nnü
0 350{NgLk
125Y r cdpm pcL I
2 475
3 250
4 75
-m
rers Schüdu11
350 475 950 12D0 1275
Po lm 1 í. 1 p) I
l csl üÐnsc T i m e :
R (Po) =0 !g?
y 3 5 0 R
(P2) = 4 7 5 R
(P3) = 9 50
: R(P.) = 12 0 0
Average Respo n se Time-0 + 350 + 475 + 950> 12 0 0 = 2975 / 5 = 595
Wait T i m
w (Po) · 0 1w
(Pr) = 350 w (P:) = 41 5
:
(Yi
) = 950 n120 0
Ave r age Wait Tinw-0 + 350 + 475 + 95 0 + \ 200 = 2975 / 5 = 595
Turnarou n u T i m
T
(Pn) = 350 T(Pl! = 475 T(P 1) = 950 1(P]) = 12 0 0 T[P,) = 1275
AveragerlrmnroL1ml T ime-350 + 475 + 950 + 1200 + 12 7 5 = 425 0 / 5 = 8 5 0
Co m p 3461546Operating Syste m
Ain\an 1\ ann a
Scheduling 2 of 7
I
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