MadeEasy Test Series: Operating System - Process Schedule

Question

Consider four process all are arriving at time zero, with total execution time of 20, 10, 10 and 20 unit
respectively. Each process spends the first 20% of execution time doing CPU, the next 60% of doing I/O
computation and the last 20% of time doing CPU computation again. The operating system uses longest
time first scheduling algorithm and schedules a new process either when running process get blocked I/O
or when the running process finishes its CPU burst.
Assume that I/O operations can be overlapped as much as possible. The average TAT of the system
given by ______ unit.
[Note: When same burst occurs for multiple process high priority given to lowest process id] (upto one
decimal place)

Answer 1

if a process is excuting cpu bound no other process execute the cpu bound.

if a process is excuting cpu bound other process executing i/o bound .

if one process completes cpu ,i/o and cpu bound then only it completes its total excution

	AT	CPU	I/O	CPU	CT	TAT=(CT-AT)
P1	0	4	12	4	20	20
P2	0	2	6	2	22	22
P3	0	2	6	2	24	24
P4	0	4	12	4	26	26

calculate for remaining processes also.

Each process spends 20% during cpu bound.so cpu execution time=totalexecutiontime*20%

                                                                                             =20*20%=4

Each process spends 60% during cpu bound.so cpu execution time=totalexecutiontime*60%

                                                                                                    =20*60%=12

Each process spends 20% during cpu bound.so cpu execution time=totalexecutiontime*20%

                                                                                             =20*20%=4

so we get table using that table apply LJSF algorithm.GANTT CHART

p1

p4

p2

p3

////

p1

p2

p3

p4

o           4             8          10          12            16           20          22         24         26

SO AVERAGE TAT=20+22+24+26/4=23.

% OF CPU IDLE IS 4/26=15.38%.   //  "///" CPU IS IDLE.ONLY I/O OPERATIONS ARE EXECUTING

Comments

The operating system uses longest 
time first scheduling algorithm and schedules a new process either when running process get blocked I/O 
or when the running process finishes its CPU burst. (Mentioned in question)

This means when one process is scheduled, then another process is scheduled next only if the previous process has voluntarily released control of cpu i.e. its cpu burst is over or if it is moved to wait state for i/o. 

This follows non -preemptive longest job first.

In preemptive longest job first i.e. even if current process havent completed its cpu burst, it will be preempted by another process having longest next cpu burst.

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The question is copied wrong. It was LRTF in the made easy question. Can you show LRTF implementation?

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Please see the pic below: