ME test os

Question

I am getting answer  11 sec but it is given none ..pls check

Comments

i can't see ..plz upload better pic or take a snapshot on zoom it

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11 is correct @Arnabi

Answer 1

CPU Scheduling will be:

${P_{1}}^{4}{P_{2}}^{5}{P_{3}}^{6}{P_{1}}^{10}{P_{2}}^{11}{P_{4}}^{15}{P_{2}}^{16}{P_{4}}^{18}{P_{2}}^{19}$

TAT for P1 = 10 - 0.0 = 10
TAT for P2 = 19 - 0.4 = 18.6
TAT for P3 = 6 - 1.0 = 5
TAT for P4 = 18 - 7.6 = 10.4

$\therefore$ avg. TAT $= \frac{10+18.6+5+10.4}{4} = \frac{44}{4} = 11$ sec

Comments

yeah thank u for checking lokesh..:)