can you please answer #5 Consider the following set of processes, with the lengt
ID: 3849333 • Letter: C
Question
can you please answer #5
Consider the following set of processes, with the length of the CPU burst time given in milliseconds. Processes may begin executing 1 ms after they arrive (i.e., a process arriving at time 5 could start executing at time 6) Any process arriving at time 0 is in the ready queue when the scheduler makes a decision about the first process to run. Determine the turnaround time for each process using each of the following scheduling algorithms: (i) first-come, first-serve (FCFS). (ii) shortest job first (SJF), (iii) shortest time to completion first (STCF), (iv) round-robin (RR) with time quantum -1 ms, and (v) a non- preemptive priority scheme in which lower numbers indicate higher priority. To break ties between processes (same burst time/priority), use first-come, first-serve ordering. a. Say the currently running process has 16 active pages, P0-P15, all of which have their reference bits set to 1. If the operating system uses the clock algorithm for page replacement, the pages are ordered numerically around the "clock (P0 is first, P1 is second, etc.), and the "clock hand" currently points to P6, which page will be replaced if a frame is needed to bring in a new page? Explain your answer for full credit. b. A portion of the currently running process's page table is shown below:Explanation / Answer
i) FCFS
0--1--------------------11-------------------14-------------------21--------------------22----------------27
Turnaround time
P1 = 11-0 = 11 ms
P2 = 14-0 = 14 ms
P3 = 21-2 = 19 ms
P4 = 22-4 = 18 ms
P5 = 27-6 = 21 ms
ii) SJF
0--1--------------------4-------------------11-------------------12--------------------17----------------27
Turnaround time
P1 = 27-0 = 27 ms
P2 = 4-0 = 4 ms
P3 = 11-2 = 9 ms
P4 = 12-4 = 8 ms
P5 = 17-6 = 11 ms
iii) Shortest Time to Completion First(STCF)
0--1-----------------4-------------------5--------------6-----------------11----------------17---------------27
Turnaround time
P1 = 27-0 = 27 ms
P2 = 4-0 = 4 ms
P3 = 17-2 = 15 ms
P4 = 6-4 = 2 ms
P5 = 11-6 = 5 ms
iv) Round Robin(quantum = 1)
Turnaround time
P1 = 27-0 = 27 ms
P2 = 11-0 = 11 ms
P3 = 24-2 = 22 ms
P4 = 6-4 = 2 ms
P5 = 22-6 = 16 ms
v) non-preemptive priority
0--1--------------------11-------------------12-------------------19--------------------24----------------27
Turnaround time
P1 = 11-0 = 11 ms
P2 = 27-0 = 27 ms
P3 = 19-2 = 17 ms
P4 = 12-4 = 8 ms
P5 = 24-6 = 18 ms
P1 P2 P3 P4 P5Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.