CPU Scheduling: SJF & SRTF | L 7 | Operating System | GATE 2022

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[@RandomMan271]

Lecture starts at 7:00

[@HarshSharma-tq7gg]

for SJF : avg. WT=1.6 ; avg. TAT=4.6 ; L = 15 ; throughput = 1/3 ;

[@akshaykushwaha7003]

Start 7:00

[@PEAKY_BLINDERS_-nt4iv]

Akshay bhai OP !

[@detector_aryan]

TQ

[@facta2z692]

gate clear hpgya?

[@RohanKumar-ig6wo]

Mind blowing! Hats off for ur efforts ❤️

[@Ghkfdvbj]

How does OS know burst time before it execute on CPU.

[@replyingtomemeansyourstupi8996]

Experience

[@akhil2721]

May be it would depend on the size of processes. Larger the size of processes more time it would take.

[@sayantanmanna1360]

Head towards the next lecture to unveil this mystery 😅

[@thanatos221]

Pcb record it

[@ankitkhewle873]

It is the problem with SJF and SRTF as OS don't know burst time, practical implementation of SJF and SRTF is not possible..

[@jiwnaniakshay8774]

Hw answers: 1. For SJF: avg TAT= 4.6 avg WT= 1.6 2. No of context switches= 2 3. For SRTF: Avg WT= 5.5

[@facta2z692]

bhai gate clear hogya kya

[@digbijayagudu7454]

SRTF average waiting time is 22/7=3.25 your answer is wrong

[@milindjain544]

May be a silly doubt, but ready queue is a queue means first in first out according to it order should be p1,p2,p3 . 12:41

[@suyashrahatekar4964]

I dont think the queue data structure and the actual data structure used to implement ready queue in OS are the same. Your answer would be right only in case of FCFS.

[@educationhacks5320]

1Q.Avg turn around time is = (12+13+14)/3 2Q.22/4 3Q.2 Context switches

[@digbijayagudu7454]

Correct brother same answer

[@__singhayush__]

For that LRTF question, consider the process with the highest burst time will get scheduled first and if two processes have same burst times then the process with the lower pid will get scheduled first. This way you will get AVG TAT = 13

[@crackthemind9821]

Thanks brother... But sir ne LRTF to nhi padhaya hai i think

[@__singhayush__]

Padhaya nhi but ek question homework me diya hai iss video me...uska hi answer ye tha

[@crackthemind9821]

Thanks... Bro... Atleast you answered... Kya kar rahe ho abhi??

[@venom-kl3un]

@@crackthemind9821 abhi to three sum kr rha hu bro

[@neha_coding5563]

HW ans : STRF no.of context switches=2 LRTF avg.TAT= 13 STRF avg.wt=22/4=5.5milliseconds

[@Ghkfdvbj]

How come LRTF answer is 13, i am getting like (14+12+8)/3

[@ramkumarramkumar-du5il]

@@Ghkfdvbj i'm also getting the same bro

[@suyashrahatekar4964]

@@Ghkfdvbj check your gannt chart, u didnt understand the algorithm carefully. Once P1 and P2 are at same remaining time, P1 will get priority for 1ms and P2 will wait, once all three are at same remaining time, P0 will get priority over other two for 1ms. The gantt chart will have many such 1ms sections.

[@shakuntalakamade]

same!!!

[@vaibhavmetange6875]

Q1)No of context switch =2 Q2) Avg TAT =13 Q3) Avg WT = 5.5

[@vibhurathore63]

how 2nd answer came mine is coming 11.34

[@nithinm0836]

@@vibhurathore63 It is longest 'remaining' time first algo. So it would be p2 p1p2 p1p2 p0 p1p2 p0 p1p2 0 4 5 6 7 8 9 1011 121314 after p2 completes 4, p2 and p1 both will be having 4. Its a tie-breaker, so go with process id p1. p1 executes for 1 and becomes 3. Now p2 already with 4 is longest, so again p2 is executed. And this thing continues, till all the processes finishes.

[@agnidebmukherjee9756]

@@nithinm0836 But all 3 arrive at the same time So 2 will execute completely first then 1 and then 0 .

[@joseph2073]

@@agnidebmukherjee9756 yes bro .. so avg tat , the answer is 11.3 right?

[@VishuDarwaja]

14:26 Thats what she said

[@honey.hemant_]

Thank you Sir ☺️

[@soumyadeepbarik5711]

GATE-2006 QUESTION , the SCHEDULING ALGORITHM , behaves like NON-PREEPTIVE , evenagter being, S.R.T.F , how can it be possible?

[@coder1163]

❤love you sir

[@muskansharma4991]

Thanks 😊

[@Pragyanam833]

For sjf: avg tat= 4.6 Avg wt=1.6 SL=15-0=15 Throughput= 5/15= 1/3 2nd hw: avg tat= 13 unit Q.3. no. Of contex switch : 2 Q.4 avg wt = 5.5

[@Patrick-qt9ow]

Thanks You Sir

[@anushasrivastava3848]

For SJF: Avg. TAT= 4.6 Avg. WT = 1.6 L= 15 Throughput= 1/3

[@atiyawarsi111]

He teaches the best

[@mdghufranalam7369]

❤️🙏

[@yeaahhhpriya]

start at 6:56

[@aniketbasu3865]

dimag khul gaya pura

[@networking236]

From 3-5,process p4 does context switching for 2 times using itself...so total switching might be 8... doesn't it so

[@55swatantrasingh77]

If a process P1 terminated after completion and P2 comes in CPU then context switching wil happen or not???

[@networking236]

Context saving doesn't occur so no switching hppn

[@durgfestivalvibes]

Thanks sir

[@haryanvimukeshbhodiwaal689]

❤️❤️❤️❤️🔥

[@aritrasaha3938]

For the H/W LRTF question: The gantt chart will look like- p3 - 0 to 4 p2 - 4 to 5 p3 - 5 to 6 p2 - 6 to 7 p3 - 7 to 8 p1 - 8 to 9 p2 - 9 to 10 p3 - 10 to 11 p1 - 11 to 12 p2 - 12 to 13 p3 - 13 to 14 TAT = (12+13+14)/3 = 13 units

[@TarunKumar-cn6in]

Last problem was interesting

[@avinashgupta2308]

Notes ?

[@Piya_Journal_Unfolded]

Homework ka answer h....

[@UnacademyComputerScience]

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[@RohitRaj-t4c5v]

10:25 sir ye aap kya chatwaa rahe wo???🤣🤣🤣🤣🤣🤣😂😂😂😂

[@funtoos3627]

Sir sometime your voice gets high and after few seconds it gets low , it's too much irritating. Please try to speak with same frequency 🙏 Otherwise you are amazing and your way of teaching is excellent 👌

[@HarshKumarVermaVlog]

😂😂😂😂

[@soumyadeepbarik5711]

BATTER k jaayga BAT-WOMAN v bol shakte the

[@suyashrahatekar4964]

seriously?? 😑 kuch accha comment kr dete isse accha , HW answer comment krte , doubt puchte , lekin nhi hame toh bakchodi par focus krna h bc.

[@Patrick-qt9ow]

Thanks You Sir