SRTF schedule

Question

2)4 processes arriving at time 0,0,2,5 with total execution time 10,20,30,40 units respectively. Each process spends the first 20% of execution time doing I/O, 70% time doing computation and last 10% time doing I/O again.The percentage of CPU idle time for the schedule length using shortest remaining time first is _________________

Answer 1

In Process1 first 2 unit for IO, next 7 unit for computation, next 1 unit for IO

In Process2 first 4 unit for IO, next 14 unit for computation, next 2 unit for IO

In Process3 first 6 unit for IO, next 21 unit for computation, next 3 unit for IO

In Process4 first 8 unit for IO, next 28 unit for computation, next 4 unit for IO

IO time will overlap with if some other process want to compute then

Gantt chart

0______2(I/O time)|_____________P₁_______________9(Process1)

__________P₂______________23(Process2)________P₃_________44(Process3)___________________

P_{4_____________}72(Process4)_________4(I/O)_________76

So, among 76 unit time

6unit used for IO

72 unit require for process to complete

So, in IO time in percent $\frac{6}{76}\times 100=7.8$

Comments

the answer had 6/72 * 100

so got confused

Answer 2

Process	AT	I/O	CPU	I/O
P1	0	2	7	1
P2	0	4	14	2
P3	2	6	21	3
P4	5	8	28	4

 USING SRTF:

IDLE	P1	P2	P3	P4	IDLE
0-2	2-9	9-23	23-44	44-72	72-76

%of time cpu idle = idle time / total cpu time

                         = 6/76*100 =7.89%

Comments

According to RBR, last idle time need not to be consider to compute idle time % because it can be overlapped with some other process, can someone clarify it.

in that case ans should be 2/72*100