MadeEasy Test Series: Operating System - Process Schedule

Question

Consider four processes 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 are 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)

my ans is 16.5 but ans is given 24.5

Answer 1

0---P1--4---P4---8----P2---10---P3---12----no process---16---P1---20---P4---24---P2---26---P3---28

At 16 P1 finishes it IO

At 20 P4 finishes it IO

At 16 P2 finishes it IO

At 18 P3 finishes it IO

Avg turnaround time=20+24+26+28/4=24.5

Comments

Turn around time is waiting + execution time thts why i used to add waiting time with Total execution time to calculate turn around time.

As u also said turn around time includes waiting time burst time n io.

Ohh yes i didnt add IO execution..

But yah completion time-arrival tim for TAT is easy one

Thank you

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if we are ask to solve using LRTF then what are the possible wording can be used in the question.

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@Shubhanshu 34.5 I think