How to calculate turnaround time? (Process Scheduling - Operating System)

Question

A and B are the only two processes in the system. Both are in the ready state. A arrives before B but the difference in arrival time is negligible. A is a CPU bound process while B is a I/O bound process. The pseudocode for A and B are as follows:

A:

repeat 2 times:

compute (takes 4 unit time)

write result to disk (takes 1 unit time)

exit (takes negligible time)

 

B:

repeat 2 times:

compute (takes 2 unit time)

write result to disk (takes 4 unit time)

exit (takes negligible time)

 

Assume that

(i) I/O can be done concurrently, i.e., the time taken to write to disk will not be prolonged if both processes write to disk at the same time;

(ii) the running time of A and B can be estimated accurately;

(iii) time quantum for preemptive scheduler is 1 unit time.

 

For each of the following algorithms:

(a) Round Robin;

(b) Shortest Remaining Time First;

 

Compute the turnaround time for A and B.

 

Link: http://www.comp.nus.edu.sg/~ooiwt/cs2106/1112s1/tut05.pdf

Answer 1

For RR

$_{0}A_{1}B_{2}A_{3}B_{4}A_{6}//_{7}A_{8}B_{9}A_{10}B_{11}A_{13}$

at t=4 B goes for I/O which will be completed at t=8

at t=6 A goes for I/O which will be completed at t=7

at t=11 B goes for I/O which will be complete at t=15

at t=13 A goes form I/O which will be complete at t=14

TAT of A = Completion - Arrival = 14-0=14

TAT of B = Completion - Arrival = 15-0=15

SRTF

$_{0}A_{1}B_{3}A_{6}//_{7}A_{8}B_{10}A_{13}$

at t=10 B goes for I/O complete at 14

at t=13 A goes for I/O complete at 14

TAT A=B=14

or

$_{0}A_{1}B_{3}A_{6}//_{7}B_{9}A_{13}$

at t=9 B goes for I/O complete at 13

at t=13 A goes for I/O complete at 14

TAT A=14 B=13

Comments

Thanks a lot!

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@Tesla!
for SJF why did u considered that A has come first
since it was given that DIFFERENCE IN THEIR ARRIVAL IS NEGLIGIBLE;
and even if A gets CPU first, how are u so sure that it will run for 1 ms only and then B will run??
i mean A can run for less than 1ms too!

according to me
Turnaround time for A=13; B=12